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Title: Video claims shooter dressed as police
Source: [None]
URL Source: https://duckduckgo.com/?q=LAS+VEGAS ... a=videos&iax=1&iai=qMxn7hpXmk4
Published: Oct 8, 2017
Author: Planet X Investigations
Post Date: 2017-10-08 15:41:01 by A K A Stone
Keywords: None
Views: 11623
Comments: 186

Video claims shooter dressed as police


Poster Comment:

Video claims shooter dressed as police

Post Comment   Private Reply   Ignore Thread  


TopPage UpFull ThreadPage DownBottom/Latest

#1. To: A K A Stone (#0) (Edited)

>>PLANET X INVESTIGATIONS

SMH.  LOL.

VxH  posted on  2017-10-08   15:43:32 ET  (1 image) Reply   Trace   Private Reply  


#2. To: VxH (#1)

Why the silly meaningless image?

A K A Stone  posted on  2017-10-08   15:44:56 ET  Reply   Trace   Private Reply  


#3. To: VxH (#1)

People just refuse to admit that this was all a Klingon plot.

At least you see the bigger picture, unlike all these sheep.

Tooconservative  posted on  2017-10-08   15:45:02 ET  Reply   Trace   Private Reply  


#4. To: VxH (#1)

You didn't have time to watch the video yet.

Debunk it please.

A K A Stone  posted on  2017-10-08   15:46:17 ET  Reply   Trace   Private Reply  


#5. To: A K A Stone (#2)

Your Vegas video was produced by the Niburu/Planet X kooks.

Tooconservative  posted on  2017-10-08   15:47:13 ET  Reply   Trace   Private Reply  


#6. To: Tooconservative, VXH, nolu chan, buckeroo (#3)

There is some interesting footage at the 9 minute mark.

A K A Stone  posted on  2017-10-08   15:48:24 ET  Reply   Trace   Private Reply  


#7. To: Tooconservative (#5)

Niburu/Planet X kooks.

Never heard of either of them. Just clicking on youtube videos and seeing what is out there.

Just because I post this doesn't mean I believe it.

Just checking out both sides.

A K A Stone  posted on  2017-10-08   15:49:19 ET  Reply   Trace   Private Reply  


#8. To: A K A Stone (#4)

You didn't have time to watch the video yet.

I've watched the video it's recycled from enough.

Again. The sound of the "weapons fire" he's attributing to the "shooter" is the local crack of the bullets as they impact and/or pass by.

There are two sets of audio events to listen for:

* the firecracker-like sound of the bullets. * the lower pitched muzzle report.

Count the number of events and observe the patterns.

VxH  posted on  2017-10-08   15:51:38 ET  Reply   Trace   Private Reply  


#9. To: VxH (#8)

10:54 is interesting.

A K A Stone  posted on  2017-10-08   15:53:04 ET  Reply   Trace   Private Reply  


#10. To: A K A Stone (#2) (Edited)

Why the silly meaningless image?

"Planet X / Nibburu"

https://www.google.com/search? source=hp&q=Planet+X+%2F+Nibburu

Not a credible source.

VxH  posted on  2017-10-08   15:54:18 ET  Reply   Trace   Private Reply  


#11. To: VxH (#10)

Not a credible source

Perhaps you are correct. I've never heard of them.

There is a video they have and it has some interesting shots.

Care to comment or you just going to call me conspiracy theorist?

No matter either way.

I think one person did so far.

A K A Stone  posted on  2017-10-08   15:56:02 ET  Reply   Trace   Private Reply  


#12. To: A K A Stone (#9) (Edited)

10:54 is interesting.

These Niburu/Planet X people are kooks and frauds.

You can't assume they are honest or that they haven't manipulated the audio or video. They're complete liars about everything.

And they've taken time out from their usual kooking about Niburu and Planet X and used their expertise to produce just one Vegas shooting video, the one you posted.

They're kooks. It's a kookarama. They live to kook. They are trying to branch out from kooking about Niburu/Planet X into kooking about the Vegas massacre. But it is still just kookery. Because they are raging kooks.

This video doesn't really rate its own thread, though I do like VxH's Klingon bird of prey.

Tooconservative  posted on  2017-10-08   15:57:13 ET  Reply   Trace   Private Reply  


#13. To: A K A Stone (#9)

What he's claiming doesn't match the audio.

He's focusing on the sound of the bullets - claiming the shots are local.

Completely ignores the associated pattern of report that follows the bullet pattern.

The report is sub-sonic. The bullets are traveling faster than the speed of sound. That's why there are TWO distinct and separate groups of sound events associated with the same shots.

VxH  posted on  2017-10-08   15:58:41 ET  Reply   Trace   Private Reply  


#14. To: Tooconservative (#12)

You gotta admit even if it is kooky. It kind of looks like they are shooting.

Or did you watch it?

A K A Stone  posted on  2017-10-08   15:58:59 ET  Reply   Trace   Private Reply  


#15. To: VxH (#13)

What he's claiming doesn't match the audio.

He's focusing on the sound of the bullets - claiming the shots are local.

Nothing about sound. Video and flashes in the crowd.

A K A Stone  posted on  2017-10-08   15:59:43 ET  Reply   Trace   Private Reply  


#16. To: A K A Stone (#14)

You gotta admit even if it is kooky. It kind of looks like they are shooting.

They're kooks. I don't pay attention to kooks. Not my job.

Tooconservative  posted on  2017-10-08   16:02:00 ET  Reply   Trace   Private Reply  


#17. To: A K A Stone (#11) (Edited)

call me conspiracy theorist?

Nah. It's easy for someone to think they hear something (and be misled) when they don't understand the acoustics.

Especially when the Niburu cultists / grifters start injecting their click- bait input.

VxH  posted on  2017-10-08   16:03:08 ET  Reply   Trace   Private Reply  


#18. To: A K A Stone (#15)

Nothing about sound.

Uhuh.

And understanding the the audio is critical to understanding what's being presented.

VxH  posted on  2017-10-08   16:05:18 ET  Reply   Trace   Private Reply  


#19. To: VxH, tooconservative (#17)

Nah. It's easy for someone to think they hear something (and be misled) when they don't understand the acoustics.

"There is a principle which is a bar against all information, which is proof against all arguments and which can not fail to keep a man in everlasting ignorance-that principle is contempt prior to investigation." --HERBERT SPENCER

A K A Stone  posted on  2017-10-08   16:09:20 ET  Reply   Trace   Private Reply  


#20. To: A K A Stone (#19)

"There is a principle which is a bar against all information, which is proof against all arguments and which can not fail to keep a man in everlasting ignorance-that principle is contempt prior to investigation listening to a pack of goddamned kooks." --Tooconservative

Tooconservative  posted on  2017-10-08   16:13:00 ET  Reply   Trace   Private Reply  


#21. To: A K A Stone (#19) (Edited)

Like I said, I've seen the video these Niburuspiracists recycled - when it was first published.

I understood the audio then, and it hasn't changed.

The guy wants you to believe the "shooter" made sounds that

A: were not local gunfire.
B: are associated with the pattern of reports that is coming from the hotel.

Eventually somebody will compile a timeline with links to every burst of gunfire. I believe the evidence will all point back to those two windows. Haven't seen any evidence to the contrary yet.

VxH  posted on  2017-10-08   16:18:54 ET  Reply   Trace   Private Reply  


#22. To: VxH (#21)

Ok tell me about your audio.

Where were was the taxi driver specifically at the first five burst fired.

Show me where you checked the information. How did you find out where the taxi driver was?

Then also since you are so smart. Do your echo test from where this video was shot.

You are to lazy to dispute anything in this video. Even if you are right.

So you are probably to busy or lazy to do what I asked. I seriously doubt you did your own research.

A K A Stone  posted on  2017-10-08   16:58:27 ET  Reply   Trace   Private Reply  


#23. To: Tooconservative (#20)

There is a principle which is a bar against all information, which is proof against all arguments and which can not fail to keep a man in everlasting ignorance-that principle is contempt prior to investigation listening to a pack of goddamned kooks." --Tooconservative

plagiarized.

Also for someone who talks about everything posted. It is funny you can't debunk an image in a video. That is unlike you TC.

A K A Stone  posted on  2017-10-08   16:59:42 ET  Reply   Trace   Private Reply  


#24. To: A K A Stone (#22)

How did you find out where the taxi driver was?

I watched the Cab-driver video. Then I used google earth to view the 3d image of the hotel and surrounding area.

VxH  posted on  2017-10-08   17:36:48 ET  Reply   Trace   Private Reply  


#25. To: A K A Stone (#22) (Edited)

You are to lazy to dispute anything in this video.

I've already disputed it.

You're too lazy to go do the research to learn and understand what you're hearing.

VxH  posted on  2017-10-08   17:49:02 ET  Reply   Trace   Private Reply  


#26. To: VxH (#24) (Edited)

Then I used google earth to view the 3d image of the hotel and surrounding area.

Google Earth and Google Maps Street View were both helpful.

Tooconservative  posted on  2017-10-08   17:53:50 ET  Reply   Trace   Private Reply  


#27. To: VxH, tooconservative (#25)

I've already disputed it.

Then you should both email me your results privately. I want to see if they are the same.

Oh you can't.

Ok post what you found and describe it please.

Why can't you do that?

A K A Stone  posted on  2017-10-08   18:12:36 ET  Reply   Trace   Private Reply  


#28. To: VxH, A K A Stone (#10)

[VxH #10] "Planet X / Nibburu"

https://www.google.com/search? source=hp&q=Planet+X+%2F+Nibburu

Not a credible source.

Says the person with the image from the credible source, pbase and Klingons.

https://a4.pbase.com/o10/92/33392/1/166356364.c4kuLP7o.Klingons.jpg

Why is pbase and Klingons more credible than Youtube?

https://www.youtube.com/watch?v=FNPrR9pP51E

Las Vegas Cover Up: Video Shows 3 Extra Missing Windows On West Side of Mandalay Bay Hotel

Qronos16
Published on Oct 8, 2017

Antifa Literature Video footage of the Mandalay Bay’s exterior captured the morning after the massacre proves that three windows on the west face of the building were knocked out in addition to the two windows the media reported to be broken in the shooter’s suite and the adjacent room.

- - - - - - - - - -

https://www.youtube.com/watch?v=QO_wq9dv8Gc

Mandalay Bay Casino: New Images and Questions *GRAPHIC*

Headlines With A Voice
Published on Oct 3, 2017

- - - - - - - - - -

nolu chan  posted on  2017-10-08   19:06:11 ET  (1 image) Reply   Trace   Private Reply  


#29. To: nolu chan (#28)

Says the person with the image from the credible source, pbase and Klingons.

I demand that you seriously consider that this was a Klingon massacre. I read somewhere that they may have conspired with some Lutherans.

Tooconservative  posted on  2017-10-08   19:16:38 ET  Reply   Trace   Private Reply  


#30. To: nolu chan, VxH, A K A Stone, misterwhite (#28)

That three-missing-windows video is total crap. It could easily be faked. And it is distinctly unreal looking. The actual building has windows that are deep gold in color, yet we see a fully illuminated blue sky above.

You could fake that video and the windows without a lot of software in a half hour or so.

Take a look at the Qronos16 YouTube channel. It's stuffed full of every kind of kookery that you can name. Menacing asteroids, insidious solar storm, sinister earthquakery, false flaggotry, End Times babbling, Fukushima coverage. Oh, yeah, and highly professional coverage of the Vegas massacre.

Do you seriously expect anyone to credit anything to a YouTube channel of such odious and hackneyed kookery?

YouTube is not a repository of knowledge. It's the Wikipedia of video. And, no, adding 4 billion nutjobs to the internet does not necessarily create more knowledge. A lot of it is just wrong and stupid. Or lures to try to get clicks out of gullible people. People can make significant money doing this.

This is bottom-feeding on YouTube, all in the wake of a human tragedy. So, yes, the loons and kooks are out in full force.

Tooconservative  posted on  2017-10-08   19:34:48 ET  Reply   Trace   Private Reply  


#31. To: A K A Stone (#27)

It's disputed if you know what you're hearing.

Why don't you know that?

VxH  posted on  2017-10-08   19:35:19 ET  Reply   Trace   Private Reply  


#32. To: A K A Stone (#22)

Another copy of the same video - claiming to be

[Original footage of las vegas shooting 50 filled 200 injured]

https://www.youtube.com/embed/kcjYefWRsKU? version=1&showinfo=0

Doubful that it's "original" - but anyhow observe the sustained burst in question, between:

https://youtu.be/kcjYefWRsKU?t=8 and
https://youtu.be/kcjYefWRsKU?t=22

There isn't any discernible echo.   Probably because this is the aiming point and not the origination point.

Here's what I see between the last bullet sound and the last report:

 
T1T2Elapsed TimeTotal Distance
17.6918.761.071208.80ft
 

Where T1 is the last bullet sound and T2 is the last report sound.

1.07 seconds between T1 and T2 = 1208 ft. 

1208 ft from the Mandalay Bay, per Google earth, puts us right about where the video is being taken.

 

It's difficult to count the number of bullet events because they are in close synchronization with the report events - but I count somewhere between 80 - 95; making this probably one of the longer events I counted in the cab- driver videos - 94 rounds, which I suspect is from a 100 round drum magazine.

Do we see a belt fed weapon or a drum in the "PLANET X INVESTIGATIONS" video?  Nope. 

More evidence the Niburutards are FOS, as usual. 

==============

Distribution:

A K A Stone,Tooconservative,nolu chan, buckeroo,KlingonAmbassador

VxH  posted on  2017-10-08   19:38:13 ET  Reply   Trace   Private Reply  


#33. To: Tooconservative, VxH, A K A Stone, misterwhite (#30)

That three-missing-windows video is total crap. It could easily be faked.

Youtube is not exactly a reliable source. It may be useful for discussion and consideration.

nolu chan  posted on  2017-10-08   19:46:52 ET  Reply   Trace   Private Reply  


#34. To: VxH, A K A Stone (#32)

Doubful that it's "original" - but anyhow observe the sustained burst in question, between:

It it is on Youtube, it is NOT the original, nor a bit for bit copy of the original.

nolu chan  posted on  2017-10-08   19:48:42 ET  Reply   Trace   Private Reply  


#35. To: A K A Stone, VxH (#22)

Where were was the taxi driver specifically at the first five burst fired.

Show me where you checked the information. How did you find out where the taxi driver was?

This video I posted to the other thread gives a very good idea of where the taxi was and when. He identifies specific Mandalay Bay landmarks in the taxi driver video.

https://libertysflame.com/cgi-bin/readart.cgi?ArtNum=53000&Disp=42#C42

nolu chan  posted on  2017-10-08   20:15:37 ET  Reply   Trace   Private Reply  


#36. To: nolu chan (#34)

nor a bit for bit copy of the original.

Yeah I see/acknowledge what you're talking about regarding generation loss now -- Compression artifacts introduced by rendering from a compressed source.

VxH  posted on  2017-10-08   20:17:01 ET  Reply   Trace   Private Reply  


#37. To: VxH, Klingon Ambassador, Kahless the Unforgettable (#32)

Going out to all Klingons.

A K A Stone  posted on  2017-10-08   20:22:42 ET  Reply   Trace   Private Reply  


#38. To: nolu chan (#35)

Yep. Same idea. I just looked for the covered carport with the curved road.

VxH  posted on  2017-10-08   20:30:08 ET  Reply   Trace   Private Reply  


#39. To: nolu chan (#33)

It may be useful for discussion and consideration.

It's useful to discuss videos that may be fakes amid a stream of other contradictory videos?

I don't think so.

Tooconservative  posted on  2017-10-08   20:31:07 ET  Reply   Trace   Private Reply  


#40. To: VxH, A K A Stone, Tooconservative (#32)

There isn't any discernible echo.   Probably because this is the aiming point and not the origination point.

Here's what I see between the last bullet sound and the last report:

 
T1T2Elapsed TimeTotal Distance
17.6918.761.071208.80ft
 

Where T1 is the last bullet sound and T2 is the last report sound.

1.07 seconds between T1 and T2 = 1208 ft. 

1208 ft from the Mandalay Bay, per Google earth, puts us right about where the video is being taken.

What do you mean by the last bullet? It that the time of the sound as recorded on that video, or what is it?

Is T1 a gunshot and T2 an echo.

What do you mean by report, if not an echo?

And just how did you calculate such a remarkable result for distance?

Don't tell me. 1,125 feet per second (the speed of sound) x 1.07 = 1,203.75 and you rounded it off to 1,208 feet. Forget the rounding. Why does that measure the distance from the video taker to the shooter?

https://en.wikipedia.org/wiki/Speed_of_sound

In dry air at 0 °C (32 °F), the speed of sound is 331.2 metres per second (1,087 ft/s; 1,192 km/h; 741 mph; 644 kn). At 20 °C (68 °F), the speed of sound is 343 metres per second (1,125 ft/s; 1,235 km/h; 767 mph; 667 kn), or a kilometre in 2.91 s or a mile in 4.69 s.

nolu chan  posted on  2017-10-08   20:52:32 ET  Reply   Trace   Private Reply  


#41. To: Tooconservative (#39)

It may be useful for discussion and consideration.

It's useful to discuss videos that may be fakes amid a stream of other contradictory videos?

It [Youtube] may be useful for discussion and consideration to determine whether some content about something or other is worth considering and researching.

In the same way, reddit and 4chan cannot be taken at face value.

In a thread full of unreliable videos, it may be useful to point out that contrary videos can be produced from Youtube. However, the one video I posted appears useful to get the geography and locate where the taxi driver was relative to the 32nd floor windows and in trying to figure out what area of the hotel is depicted in certain portions of her video. Whether one chooses to believe the flashes are muzzle shots or something else, they are not coming from the 32nd floor. There is a certain usefulness to that video.

Youtube may be useful or entertaining. As proof of things, it is unreliable.

nolu chan  posted on  2017-10-08   21:10:06 ET  Reply   Trace   Private Reply  


#42. To: nolu chan, A K A Stone, Tooconservative, Klingon Ambassador (#40) (Edited)

>>What do you mean by report, if not an echo?

As any Boy Scout with one of these...

...should know:

https://duckduckgo.com/? q=An+explosive+noise%3A+the+report+of+a+rifle

>>And just how did you calculate such a remarkable result for distance? { blah blah blah wikispew}

I used the same speed I used in my other analysis (appended below) - based upon an air temperature of 72 degrees

Then I calculated the difference in time between the last bullet sound (T1) and the corresponding last report sound (T2).

T2-T1 = time the  report traveled = 1.07

1.07 * FPS of 1130.8 = 1208.8

==========================


An analysis of two sequential burts of gunfire between: 
["Taxi Driver Video" the Zapruder Film of the Las Vegas shooting UNCUT / UNEDITED]
https://www.youtube.com/watch ?v=mBbOFwWquAw&feature=youtu.be&t=1m7s
and
https://www.youtube.com/watc h?v=mBbOFwWquAw&feature=youtu.be&t=1m24s
===============  
T1: Time from start of video (1minute N seconds) at the time of the last shot in the burst.
T2: Time from the start of the video (1minute N seconds) at the time of the echoed sound event corresponding to T1. 
TempF: the air temperature (72 degrees F) 
FPS:   1130 ft per second -- The speed of sound at 72 degrees F
Elapsed Time:  T2 minus T1, the number of seconds between the last shot, and the echo of the last shot in each burst. 
Total Distance:  Elapsed Time * FPS = the total distance traveled between T1 and T2.
Echo Distance  = The distance the echo traveled from the aiming point back to the point of origin.
===============

Conclusion:  Burst B is NOT two weapons being fired simultaneously.  It is one weapon being fired at a more distant target.  The longer distance, observable in the period between Burst B's T1 and T2, manifests a corresponding longer period of reverb.  It is the reverb that is being incorrectly interpreted as a second weapon (and second shooter) firing at the same time.

Research resources:
https://www.google.co m/search?biw=1544&bih=856&q=Forensic+Acoustics+gunfire
http://www.physic sclassroom.com/mmedia/waves/er.cfm
https://www.timeanddate.com/weather/usa/las- vegas/historic
http://www.csgnetwork.c om/soundspeedcalc.html

VxH  posted on  2017-10-08   21:29:32 ET  (3 images) Reply   Trace   Private Reply  


#43. To: All (#42) (Edited)

 
T1T2Elapsed TimeTotal Distance
17.68965518.7586211.0689661208.80

Here's a snapshot with increased decimal precision and less rounded confusion :-}

VxH  posted on  2017-10-08   21:51:22 ET  Reply   Trace   Private Reply  


#44. To: nolu chan (#41)

However, the one video I posted appears useful to get the geography and locate where the taxi driver was relative to the 32nd floor windows and in trying to figure out what area of the hotel is depicted in certain portions of her video. Whether one chooses to believe the flashes are muzzle shots or something else, they are not coming from the 32nd floor. There is a certain usefulness to that video.

Unless it is fake.

How do you know that that is the original video? I haven't found an original source of it.

Tooconservative  posted on  2017-10-08   22:03:57 ET  Reply   Trace   Private Reply  


#45. To: VxH (#42)

Conclusion: Burst B is NOT two weapons being fired simultaneously. It is one weapon being fired at a more distant target. The longer distance, observable in the period between Burst B's T1 and T2, manifests a corresponding longer period of reverb. It is the reverb that is being incorrectly interpreted as a second weapon (and second shooter) firing at the same time.

At some point fairly early in the attack, he must have shot at the aviation fuel tanks. I haven't heard any reporting on that other than that a couple of bullets hit one tank and one bullet did penetrate the tank but caused no explosion (which happens mostly in movies).

That is a nice set of links you've curated. I'm not sure if anyone will examine them. So much easier to go click up another kook video off YouTube than to do any serious science reading.

Tooconservative  posted on  2017-10-08   22:12:48 ET  Reply   Trace   Private Reply  


#46. To: All (#43) (Edited)

17.689655 18.758621 1.068966 1208.80

Hmmm.

Actually that's not quite the correct calculation required here.

I'm making an error by treating the T1 as the firing time.

Need to update that to incorporate the muzzle velocity....

VxH  posted on  2017-10-08   22:18:14 ET  Reply   Trace   Private Reply  


#47. To: Tooconservative (#45) (Edited)

At some point fairly early in the attack, he must have shot at the aviation fuel tanks.

There is a video (somewhere in the haystack now) that has audio of what sounds like several single hp rifle cracks at the start of the shooting - presumably at the fuel tanks.

VxH  posted on  2017-10-08   22:22:32 ET  Reply   Trace   Private Reply  


#48. To: Tooconservative (#44)

Unless it is fake.

lol.

A K A Stone  posted on  2017-10-09   7:09:29 ET  Reply   Trace   Private Reply  


#49. To: A K A Stone (#48)

Laugh all you want.

If we have tons of contradictory YouTube videos based on this taxi driver video (and none of them prepared from the original video), then most of them must be wrong or scammers trying to draw clicks.

The best you can say is that the ones you are posting are somehow better than the others. And you can't offer any evidence that they were not faked or that they were prepared using authentic undoctored footage.

You're reaching into the YouTube jar and trying to draw out a winner almost at random, mostly based on how clickbaitish their titles are. Maybe there are no winners in the YouTube raffle; maybe all it has are the consolation prizes for dummies.

Tooconservative  posted on  2017-10-09   7:19:16 ET  Reply   Trace   Private Reply  


#50. To: Tooconservative (#49)

Laugh all you want.

If we have tons of contradictory YouTube videos

The taxi video was there right from the start.

You think she was in on it?

A K A Stone  posted on  2017-10-09   7:23:04 ET  Reply   Trace   Private Reply  


#51. To: Tooconservative (#49)

And you can't offer any evidence that they were not faked

You want me to prove a negative?

You can't prove they are fake. Someone on the internet will if they are fake.

The taxi video has been there right from the start. The video chan posted lines up with it.

They are not fake and you know it.

A K A Stone  posted on  2017-10-09   7:24:32 ET  Reply   Trace   Private Reply  


#52. To: A K A Stone (#50)

The taxi video was there right from the start.

Right where? Where are you saying she posted the original video? Where is it? Either you can point me to it or you're just making crap up because you want to believe it or you just don't want to admit to being taken in like a rube with inauthentic video footage.

Among the earliest postings of it seems to be video distributed by Las Vegas Journal-Review. I think the taxi driver sold it to them but I can't confirm that. However, LVJR did not initially distribute the entire video but offered cut-down versions that were edited.

We have no idea whether we have ever seen the original video as she recorded it. We don't even know what brand and model her phone was or what resolution the video was.

Tooconservative  posted on  2017-10-09   7:33:25 ET  Reply   Trace   Private Reply  


#53. To: A K A Stone (#51) (Edited)

They are not fake and you know it.

I know no such thing. And neither do you.

And you're just blustering now because you finally realize you've been had by these YouTube scammers like the Niburu/Planet X cultists.

Tooconservative  posted on  2017-10-09   7:34:27 ET  Reply   Trace   Private Reply  


#54. To: Tooconservative (#52)

We have no idea whether we have ever seen the original video as she recorded it.

This was the earliest version I saw:

https://www.youtube.com/watch?v=WNiRr763gJA&t=3s

Unfortunately no longer available.

VxH  posted on  2017-10-09   8:06:58 ET  Reply   Trace   Private Reply  


#55. To: nolu chan (#41)

Youtube may be useful or entertaining. As proof of things, it is unreliable.

Youtube is just a tool. 

How reliable or useful it is for discerning the truth is relative to the integrity of the individuals and organizations using it.

Certainly not any less reliable than the arrogant, self- serving, propaganda-parrots perched atop their "Newseum"

https://en.wikipedia.org/wiki/N ewseum

VxH  posted on  2017-10-09   8:16:21 ET  (1 image) Reply   Trace   Private Reply  


#56. To: VxH (#54) (Edited)

That video was posted by BlazingPress.com, a generic rightwing/Christian site. They posted it originally on Tuesday, 10/3/17. You can see a better version of the page (not the video) at Archive.org. That is the same page archived on 10/4.

Very very unlikely that is the original video.

LVNR posted their copy on 10/4 or 10/3, can't recall which.

It could be that these all came from a local network news affiliate who bought it off the taxi driver and then everyone just captured it and published it as their own content. Certainly, that is what Blazing Press did.

This all underlines my previous point: we don't know which of these videos are the original or if there is an original reliable copy of the taxi driver's video. I'm starting to doubt that there is.

Maybe the Klingons and Lutherans have silenced our intrepid taxi driver, threatening to take away her spicy tacos.

Tooconservative  posted on  2017-10-09   8:20:32 ET  Reply   Trace   Private Reply  


#57. To: Tooconservative (#56)

Very very unlikely that is the original video.

Yep.

Maybe the Klingons and Lutherans have silenced our intrepid taxi driver, threatening to take away her spicy tacos.

I'm not convinced it was tacos. Burritos are an equally plausible source of flatulent reverberation in Las Vegas.

VxH  posted on  2017-10-09   8:40:56 ET  Reply   Trace   Private Reply  


#58. To: Tooconservative (#56) (Edited)

 

Cori Langdon October 3 at 8:58pm

Here is my video. It has been edited per the request of the passengers to bleep out their names and end the video before their faces are seen. No other editing has been done.

https://www.facebook.com/cori.l angdon

Looks as clear as the vids I upload to FB from my phone.

VxH  posted on  2017-10-09   9:36:39 ET  Reply   Trace   Private Reply  


#59. To: VxH (#58)

Cori Langdon October 3 at 8:58pm

Here is my video. It has been edited per the request of the passengers to bleep out their names and end the video before their faces are seen. No other editing has been done.

That doesn't mean that Fakebook didn't recompress it. She didn't offer the original footage at all.

When you upload to Fakebook or YouBoob, they create their own versions in various resolutions as I detailed in another post. When YouBoob first started to offer HD videos, they did offer the actual original footage exactly as uploaded. They discontinued that some time back and now re-encode everything. I think Facebook does the same thing.

If the media and all the YouBoobers just grabbed her Facebook video, then it is possible, likely even, that no one has had an actual first-gen copy of her video.

Tooconservative  posted on  2017-10-09   9:43:29 ET  Reply   Trace   Private Reply  


#60. To: Tooconservative (#59)

That doesn't mean that Fakebook didn't recompress it.

Right - but it's probably as close to the "original" source as we're going to get.

If the media and all the YouBoobers just grabbed her Facebook video, then it is possible, likely even, that no one has had an actual first-gen copy of her video

Yep.

VxH  posted on  2017-10-09   9:49:38 ET  Reply   Trace   Private Reply  


#61. To: nolu chan (#40)

I was hiking in the mountains and I yelled, "Hello". Right after that I heard four other people yell 'hello', one after the other. And here I thought I was alone.

misterwhite  posted on  2017-10-09   10:32:39 ET  Reply   Trace   Private Reply  


#62. To: Tooconservative (#30)

That three-missing-windows video is total crap. It could easily be faked.

Notice the windows are completely missing. The shooter's windows were jagged.

misterwhite  posted on  2017-10-09   10:36:01 ET  Reply   Trace   Private Reply  


#63. To: Tooconservative (#53)

And you're just blustering now because you finally realize you've been had by these YouTube scammers like the Niburu/Planet X cultists.

Enter Screen

How was I had when I changed their title which was 100 percent proof to "video says"?

You still were unable to disprove the images and you dodged it like a little girly man.

I think they were police officers who turned around and pointed their guns.

But what was the flash? None of you two pussies dared to refute it. Girlymen.

A K A Stone  posted on  2017-10-09   13:44:39 ET  Reply   Trace   Private Reply  


#64. To: A K A Stone (#63)

Shouldn't you be hiding in your bunker so Planet X doesn't hit you in the head?

Tooconservative  posted on  2017-10-09   13:48:50 ET  Reply   Trace   Private Reply  


#65. To: Tooconservative (#64)

I'm always in my bunker here on Q'onoS. Chancellor Gowron wouldn't have it any other way.

A K A Stone  posted on  2017-10-09   14:07:17 ET  Reply   Trace   Private Reply  


#66. To: Tooconservative (#64)

Regardless of your lame humor. You still haven't told me why those two guys in police uniforms look like they turn around and fire on the crowd.

I'm not saying that is what they are doing and never said that. I just wondered if anyone smarter then me (not you TC, don't make me laugh, i'm thinking of others) knew what was going on in that section of the video.

It is above your pay grade so don't worry about it. Maybe someone else knows since you don't.

A K A Stone  posted on  2017-10-09   14:10:28 ET  Reply   Trace   Private Reply  


#67. To: A K A Stone (#66)

I'm not saying that is what they are doing and never said that. I just wondered if anyone smarter then me (not you TC, don't make me laugh, i'm thinking of others) knew what was going on in that section of the video.

I can't help you. I don't watch kook videos.

Tooconservative  posted on  2017-10-09   14:13:21 ET  Reply   Trace   Private Reply  


#68. To: Tooconservative (#67) (Edited)

I can't help you. I don't watch kook videos.

I know you can't help me. I'm not looking for someone who likes to remain ignorant and boast about their ignorance. The video was uploaded by a person who was really there, unlike you who weren't.

A K A Stone  posted on  2017-10-09   14:20:13 ET  Reply   Trace   Private Reply  


#69. To: A K A Stone (#68)

The video was uploaded by a person who was really there, unlike you who weren't.

No, it was uploaded by known kooks who probably manipulated it. And you.

Tooconservative  posted on  2017-10-09   14:25:19 ET  Reply   Trace   Private Reply  


#70. To: Tooconservative (#69)

No, it was uploaded by known kooks who probably manipulated it. And you.

So they were there? You are misinformed. Which usually isn't the case.

A K A Stone  posted on  2017-10-09   14:27:52 ET  Reply   Trace   Private Reply  


#71. To: VxH (#43)

I used the same speed I used in my other analysis (appended below) - based upon an air temperature of 72 degrees

Then I calculated the difference in time between the last bullet sound (T1) and the corresponding last report sound (T2).

Your arithmetic is fine, your logic is a FAIL.

You failed to answer the question: "Why does that measure the distance from the video taker to the shooter?"

You have still not identified what sounds you refer to.

What is the first sound of a bullet? The initial sound wave as heard at the recording location?

What is the last sound of the bullet? An echo? Another bullet?

In this case, the video taker was 400 yards away from the shooter.

The sound wave originated at position a and traveled 400 yards to get to the videotaker at position b. That took ~1.07 seconds for the sound to travel.

If you measure the first sound of the bullet as when it traveled faster than sound, and then struck something or whizzed past making a sound, your ~1.07 second measurement is impossible as the elapsed time difference would be the time it took the sound wave to travel (~1.07s) minus the time it took the bullet to travel. Your calculation as the bullet traveling at aproximately the speed of light.

The bullet and the sound both travel the same path at the same time, at different velocities. As the sound takes ~1.07 seconds, the difference in their arrival times cannot be ~1.07s.

Two cars travel an 80 mile strip. One travels 80 mph and crosses the finish line in 1 hour. The other travels 40 miles an hour and crosses the finish line in 2 hours. If the slow car went half the speed of the faster car, and the time difference was 1 hour, the distance can be calculated as the velocity of the faster vehicle (80 mph) divided by the velocity of the slower vehicle (40 mph) time the time difference (1 hour).

Your calculation is good arithmetic but gibberish logic. ~1.07 seconds is simply the time for sound to travel from Mandalay Bay to the target recording location.

An echo of a sound originating at position a results from the sound wave traveling the distance originating point a to recording position b, proceeding an unidentified distance to the reflective surface c, and returning to position b. The sound must arrive/leave recording position b at ~1.07 seconds, travel to a reflective surface, and return. The elapsed time must exceed 1.07s.

In measuring the distance of a lightning bolt, you can use the sighting of the lightning bolt, the light traveling at the speed of light, as the originating time of the initiating event. You can count until the slower moving sound wave arrives, and calculate distance to the lightning. This is because the velocity of light is so great, its travel time over relatively short distances is negligible.

Unfortunately, even with a Boy Scout merit badge, an audio recording from 400 yards away does not identify distance of the originating event. Bullets do not travel at the speed of light, or anywhere near it.

The way you know the distance from the Mandalay Bay to the event location is by measuring it.

nolu chan  posted on  2017-10-09   18:52:32 ET  Reply   Trace   Private Reply  


#72. To: Tooconservative (#44)

Unless it is fake.

How do you know that that is the original video? I haven't found an original source of it.

The Mandalay Bay is a public location that is clearly documented by, among others, Google. For the geography of the location, it does not appear necessary to present a half-dozen sources to verify that the building is situated as it is, where it is. There is nothing controversial about the layout of the land.

I have yet to see any depiction of Mandalay Bay without an entrance and a taxi stand, but if you know of one, please present it.

nolu chan  posted on  2017-10-09   19:00:38 ET  Reply   Trace   Private Reply  


#73. To: Tooconservative, nolu chan, aka stone (#46) (Edited)

I'm making an error by treating the T1 as the firing time.

Need to update that to incorporate the muzzle velocity....

Ok, so I was on the right track.  I found this research paper:

[Using Sound of Target Impact for Acoustic Reconstructions of Shooting Events]

http://www.btgresearch.org /impactsound.pdf

"For example, if the microphone is adjacent to the victim (such as a 911 recording might be), the equation for determining the distance becomes: t=tb - ts= d/Vb-d/Vs If the muzzle blast duration obscures the sound of the bullet hitting the target, simple inspection of the sound waveform is insufficient. "

I took their formula, built a spreadsheet,  and plugged in 223 balistic data generated via shooterscalculator.com:

Important to note:

* Presently we don't have information regarding specificaly which weapons and amunition were used.  So the ballistic data was generated with a guestimate 223 configuration.

* My DAW (Sonar) doesn't appear to have the capability of capturing a sound spectrogram like the ones the authors of the study produced; but after reading their commentary on the blast noise obscuring impact noise, I filtered the crowd noise, and filtered/looked alternately for the report and then the high energy impact sounds - and I revised T1 and T2 accordingly.

More accurate results could possibly be obtained if the corresponding burst sequence on the Taxi-Driver video is identified and aligned, as the taxi-driver's audio contains only the muzzle blast and echo.  It doesn't have the crowd and impact noise to obscure the muzzle events.

VxH  posted on  2017-10-09   19:50:48 ET  (1 image) Reply   Trace   Private Reply  


#74. To: nolu chan (#71) (Edited)

{ blah blah blah }

You didn't read this...

https://libertysflame.com/cgi-bin/readart.cgi? ArtNum=53025&Disp=46#C46

...before that screed did you.

:-p

VxH  posted on  2017-10-09   20:04:13 ET  Reply   Trace   Private Reply  


#75. To: nolu chan, VxH (#71)

Unfortunately, even with a Boy Scout merit badge, an audio recording from 400 yards away does not identify distance of the originating event. Bullets do not travel at the speed of light, or anywhere near it.

You're wrong. He is adjusting for the speed of sound on the night in LV. The speed of sound is 1100 feet/minute as a decent estimate anywhere (1,087 ft/s more accurately) but he is correcting for altitude and weather conditions. This has nothing to do with speed of light since the muzzle flashes weren't visible from the concert area in any video I've seen.

If the GPS position is being continuously recorded or if the people recording at the concert (hundreds or thousands) were recording video, you can readily determine their position and distance from the shooter pretty easily, just based on the time at which the sound of gunshots arrive, knowing the distance of their source. They can only be described as a portion of an arc, bounded by the concert area rectangle, that constitutes a circle around the shooter's location in the hotel. That's a pretty sound way to determine a fixed location.

This is true of all these A/V recordings of the event. You can determine very accurately their location from the video and the audio timing, even more so if a bullet flies past their camera.

VxH does have a sound method for analysis but only if he can gather enough A/V recordings from the concert and anywhere else, like the taxi driver. I am curious that we have seen no reporting that police requested the concert goers to upload their videos from that night. Maybe the sheriff and the FBI have really dropped the ball on this. Usually, the FBI is pretty sharp on audio lab stuff.

Tooconservative  posted on  2017-10-09   20:08:37 ET  Reply   Trace   Private Reply  


#76. To: Tooconservative (#75)

we have seen no reporting that police requested the concert goers to upload their videos from that night

Grossly uninformed again.

A K A Stone  posted on  2017-10-09   21:14:06 ET  Reply   Trace   Private Reply  


#77. To: VxH, nolu chan, buckeroo (#1)

Does your image represent your Klingon teleportion theory?

A K A Stone  posted on  2017-10-11   0:36:35 ET  Reply   Trace   Private Reply  


#78. To: VxH (#42)

A K A Stone  posted on  2017-10-11   0:36:57 ET  Reply   Trace   Private Reply  


#79. To: A K A Stone (#77)

Where is Mr. Alien Conspiracy.. err I mean Mr. HealthGuy's DATA?

Also, is he misinterpreting the supper-sonic cracks of bullets passing by as "pavement hits"? - probably.

VxH  posted on  2017-10-11   9:15:35 ET  Reply   Trace   Private Reply  


#80. To: A K A Stone (#77)

I don't believe this crackpot's lamebrain hypothesis. He is claiming verifiable acoustic signature analysis based on somebody's cell phone audio capture that is not controlled in anyway.

BULLSHIT.

buckeroo  posted on  2017-10-11   18:54:02 ET  Reply   Trace   Private Reply  


#81. To: Tooconservative, VxH (#75)

You're wrong. He is adjusting for the speed of sound on the night in LV. The speed of sound is 1100 feet/minute as a decent estimate anywhere (1,087 ft/s more accurately) but he is correcting for altitude and weather conditions. This has nothing to do with speed of light since the muzzle flashes weren't visible from the concert area in any video I've seen.

He is absolutely wrong and just bullshitting again. Your observations are irrelevant to his claim being accurate. His claim is 100% impossible.

At #42, Vxh compares two times from the Taxi Driver video:

An analysis of two sequential burts of gunfire between:

["Taxi Driver Video" the Zapruder Film of the Las Vegas shooting UNCUT / UNEDITED]

https://www.youtube.com/watch ?v=mBbOFwWquAw&feature=youtu.be&t=1m7s

and

https://www.youtube.com/watc h?v=mBbOFwWquAw&feature=youtu.be&t=1m24s

At #43, citing #42, he claims a total distance of 1208.80 feet.

He is measuring distance of sound, not light/bullet speed or travel distance to a target.

If, as claimed, the total distance for the sound is 1,208.80 feet, that includes the echo time according to his own graphic in the same post.

His graphic indicates that the total distance is twice the echo distance, or the echo distance (presumably unobstructed) is half the total distance.

He adjusted the distance by 100%.

This purports to be from the taxi video. The sound of the gunshot was recorded almost immediately. The recorded returning sound is from an unknown reflecting distance.

If the reflecting surface was plucked from his butt to be the bandstand, the sound would have traveled 1208 feet to the bandstand, and 1208 feet back to the taxi, if the path back at ground level were unobstructed.

You cannot adjust to a 1,208 foot total distance, unless the sound turned around and came back unobstructed at 604 feet. Unless the sound somehow returned at the speed of light, instead of the speed of sound, the total distance must be double the 1,280 feet that he claimed. He is again incapable of describing the data on his own graphic.

Of course, the sound could have bounced off something 640 feet away and come back to the taxi. The Reflecting surface is unknown. What sound cannot do is zing into something 1,280 feet away, and come back from that point, and travel a total distance of 1,280 feet.

nolu chan  posted on  2017-10-12   16:34:10 ET  (1 image) Reply   Trace   Private Reply  


#82. To: nolu chan (#81) (Edited)

You're putting your words in his mouth.

Nice straw man you've got there.

Tooconservative  posted on  2017-10-12   16:45:39 ET  Reply   Trace   Private Reply  


#83. To: nolu chan (#81) (Edited)

What sound cannot do is zing into something 1,280 feet away,

SMH.

Attention to detail isn't your forte.

Neither 1280 nor 1208 are on the graphic

What does this mean?

VxH  posted on  2017-10-12   17:02:17 ET  (1 image) Reply   Trace   Private Reply  


#84. To: buckeroo (#80) (Edited)

He is claiming verifiable acoustic signature analysis based on somebody's cell phone audio capture that is not controlled in anyway.

Even if the audio capture hasn't been altered there are still parameters required to make the ballistic calculations to "prove" his conclusion of a second shooter.

http://www.jbmballistics.com/cgi-bin/jbmtraj-5.1.cgi

Nobody knows what weapon or ammunition was used - so any attempt to fill in those parameters is, at best, only a guess.

VxH  posted on  2017-10-12   17:06:19 ET  Reply   Trace   Private Reply  


#85. To: VxH (#74)

You didn't read this...

As I can do simple calculation, unlike you, I need no additional source to discover the bullshit.

Sound cannot travel 1280 feet to a target and come back to the taxi, and travel a total distance of 1280 feet. Not even a super genius can make sound do that. But that is what you claimed, that the sound traveled a total distance of 1280 feet, based solely on the taxi driver recording.

nolu chan  posted on  2017-10-12   17:23:24 ET  Reply   Trace   Private Reply  


#86. To: VxH (#83)

What does this mean?

It means that total distance the time traveled is two times the echo travel distance. T1 on the taxi recording is almost instantaneous after the shot, a very short travel distance. Then the sound traveled 1208 feet to the target area, and at least 1208 feet back (unobstructed). Do arithmetic much?

The total distance will be double the distance if unobstructed and reflected by something at the target area straight back to the taxi. If the sound travels another mile and gets reflected back from a distant building, it would take even longer.

nolu chan  posted on  2017-10-12   17:31:46 ET  Reply   Trace   Private Reply  


#87. To: nolu chan (#86) (Edited)

No, it means the number "1208" (and 1280) isn't referenced by that graphic because the audio data on that graphic pertains to a completely different discussion of a completely different audio data set.

VxH  posted on  2017-10-12   17:44:22 ET  Reply   Trace   Private Reply  


#88. To: nolu chan (#86) (Edited)

total distance will be double the distance

What is 2612.57 / 2?
What is 7369.77 / 2?

This is an open book quiz. The answers are ECHO'd on the graphic.

VxH  posted on  2017-10-12   17:54:49 ET  Reply   Trace   Private Reply  


#89. To: Tooconservative (#82)

You're putting your words in his mouth.

Nice straw man you've got there.

No strawman or putting words in his mouth.

Here are his exact words.

At #42 he stated, all in reference to the Taxi Driver video:

T1: Time from start of video (1minute N seconds) at the time of the last shot in the burst.
T2: Time from the start of the video (1minute N seconds) at the time of the echoed sound event corresponding to T1.
TempF: the air temperature (72 degrees F)
FPS: 1130 ft per second -- The speed of sound at 72 degrees F
Elapsed Time: T2 minus T1, the number of seconds between the last shot, and the echo of the last shot in each burst.
Total Distance: Elapsed Time * FPS = the total distance traveled between T1 and T2.
Echo Distance = The distance the echo traveled from the aiming point back to the point of origin.

T1 is at the taxi. T2 is out and back and also recorded at the taxi.

Echo distance is NOT the distance traveled from the aiming point to the point of origin. It is the time between T1 as recorded at the taxi, and T2 as recorded at the taxi. T1 and T2 are derived solely from the recording at the taxi, and do not tell either the point of origin or the point of aim.

T2 could have reached the taxi by being reflected off anything the sound wave hit.

If the distance between the taxi and the target were 400 yards, then for the sound to have come from the target area back to the taxi would require a total travel of 800 yards.

At #43, he claims, in refrence to T1 and T2 on the taxi recording: "Total distance:
1208.80"

T2 - T1 is the difference in the time of arrival at the taxi, and the time to travel to a reflective surface and to make the return trip to the taxi. If that indicates the total distance sound traveled of 1280 feet, the sound would have to have been reflected from not more than 640 feet away.

#42. To: nolu chan, A K A Stone, Tooconservative, Klingon Ambassador (#40) (Edited)

>>What do you mean by report, if not an echo?

As any Boy Scout with one of these...

...should know:

https://duckduckgo.com/? q=An+explosive+noise%3A+the+report+of+a+rifle

>>And just how did you calculate such a remarkable result for distance? { blah blah blah wikispew}

I used the same speed I used in my other analysis (appended below) - based upon an air temperature of 72 degrees

Then I calculated the difference in time between the last bullet sound (T1) and the corresponding last report sound (T2).

T2-T1 = time the  report traveled = 1.07

1.07 * FPS of 1130.8 = 1208.8

==========================


An analysis of two sequential burts of gunfire between: 
["Taxi Driver Video" the Zapruder Film of the Las Vegas shooting UNCUT / UNEDITED]
https://www.youtube.com/watch ?v=mBbOFwWquAw&feature=youtu.be&t=1m7s
and
https://www.youtube.com/watc h?v=mBbOFwWquAw&feature=youtu.be&t=1m24s
===============  
T1: Time from start of video (1minute N seconds) at the time of the last shot in the burst.
T2: Time from the start of the video (1minute N seconds) at the time of the echoed sound event corresponding to T1. 
TempF: the air temperature (72 degrees F) 
FPS:   1130 ft per second -- The speed of sound at 72 degrees F
Elapsed Time:  T2 minus T1, the number of seconds between the last shot, and the echo of the last shot in each burst. 
Total Distance:  Elapsed Time * FPS = the total distance traveled between T1 and T2.
Echo Distance  = The distance the echo traveled from the aiming point back to the point of origin.
===============

Conclusion:  Burst B is NOT two weapons being fired simultaneously.  It is one weapon being fired at a more distant target.  The longer distance, observable in the period between Burst B's T1 and T2, manifests a corresponding longer period of reverb.  It is the reverb that is being incorrectly interpreted as a second weapon (and second shooter) firing at the same time.

Research resources:
https://www.google.co m/search?biw=1544&bih=856&q=Forensic+Acoustics+gunfire
http://www.physic sclassroom.com/mmedia/waves/er.cfm
https://www.timeanddate.com/weather/usa/las- vegas/historic
http://www.csgnetwork.c om/soundspeedcalc.html

VxH  posted on  2017-10-08   21:29:32 ET  (3 images) Reply   Trace   Private Reply  


#43. To: All (#42) (Edited)

 
T1T2Elapsed TimeTotal Distance
17.68965518.7586211.0689661208.80

Here's a snapshot with increased decimal precision and less rounded confusion :-}

VxH  posted on  2017-10-08   21:51:22 ET  Reply   Trace   Private Reply

nolu chan  posted on  2017-10-12   18:00:42 ET  (3 images) Reply   Trace   Private Reply  


#90. To: VxH (#88)

This is an open book quiz. The answers are ECHO'd on the graphic.

You are just yukon being yukon. This is not a quiz. Consider it your final exam. You FAILED.

If the distance from the taxi to the point the sound was reflected back is 400 yards, the sound had to travel a total distance of 800 yards, or about 2400 feet.

Keep digging. With yourself for an expert source, it is no wonder that your conclusion is impossible bullshit.

If T2 - T1 at the taxi were to indicate a total distance of sound travel of 400 yards, the sound went 200 yards to get to a reflective surface and 200 yards to come back to the taxi.

nolu chan  posted on  2017-10-12   18:08:29 ET  Reply   Trace   Private Reply  


#91. To: Tooconservative, VxH, A K A Stone (#75)

I am curious that we have seen no reporting that police requested the concert goers to upload their videos from that night. Maybe the sheriff and the FBI have really dropped the ball on this.

Who is we?

https://www.cnbc.com/2017/10/02/shooter-las-vegas-strip-police.html

Deadliest shooting in modern US history leaves at least 59 dead, 500+ injured in Las Vegas

Arjun Kharpal and| Michael Sheetz
CNBC
Published 2:11 AM ET Mon, 2 Oct 2017
Updated 1:53 PM ET Tue, 3 Oct 2017

[Excerpt]

The FBI requested that anyone with video or photos regarding the shooting call 1-800-225-5324.

nolu chan  posted on  2017-10-12   18:11:37 ET  Reply   Trace   Private Reply  


#92. To: VxH, Tooconservative, A K A Stone (#73)

* My DAW (Sonar) doesn't appear to have the capability of capturing a sound spectrogram like the ones the authors of the study produced....

Regarding your Digital Audio Workstation (DAW) for music, you might indicate what the abbreviation stands for, and with Sonar, you might indicate that it is a company brand name, and does not relate to SONAR.

Sonar DAW available starting at $49.95.

nolu chan  posted on  2017-10-12   18:14:27 ET  Reply   Trace   Private Reply  


#93. To: VxH, Tooconservative (#47)

There is also a noticeable lack of muzzle flashes from the 32nd floor.

There is a video (somewhere in the haystack now) that has audio of what sounds like several single hp rifle cracks at the start of the shooting - presumably at the fuel tanks.

Initial shots at 3:12.

Las Vegas Shooting First Actual Shots - Rare

Tall Ace of Spades
Published on Oct 7, 2017

Most footage doesn't capture the real, initial, non automatic shots shown here that the shooter probably used as test shots before using the bump stock and continuous fire when the next song starts. People are clearly trying to leave the scene shortly after the initial shots which start at 3:12.

nolu chan  posted on  2017-10-12   18:23:10 ET  Reply   Trace   Private Reply  


#94. To: A K A Stone, TooConservative (#0)

Regarding the video referenced by this thread: "Video claims shooter dressed as police"

This research paper - http:// www.btgresearch.org/AcousticReconstruction02042012.pdf describes, as part of their methodology, using an audio spectrogram to differentiate between impact sound and muzzle report sound.

The spectrogram capabilities of my DAW are very limited so I looked around for a free utility and, not finding anything suitable, decided to build my own utility for creating a spectrogram. 

Here's what I'm seeing in the relevant section of audio referenced in the  "Video [that] claims shooter dressed as police".

* This audio was filtered to remove as much of the crowd noise as possible.
* Lower frequency sounds (the muzzle reports) are graphed at the top - higher energy sounds towards at the bottom.
* More work to be done adding time and frequency indexes, zooming, anotating etc but it looks promising.

This is just a progress report. I will document the complete methodology as time allows.

VxH  posted on  2017-10-12   18:35:33 ET  (1 image) Reply   Trace   Private Reply  


#95. To: nolu chan (#89) (Edited)

Yep. You've already demonstrated you can cut and paste.

Two completely different sets of sound events and data being discussed and confused there.

Now again - per the data in my graphic:

What is 2612.57 / 2?
What is 7369.77 / 2?

This is an open book quiz. The answers are ECHO'd on the graphic.

VxH  posted on  2017-10-12   18:38:36 ET  Reply   Trace   Private Reply  


#96. To: VxH (#95)

Yep. You've already demonstrated you can cut and paste.

Two completely different sets of sound events and data being discussed and confused there.

yukon, you have derived T1 and T2 from the Taxi recording. The difference in the times does not reveal point of origin or point of aim. The difference in the two times can be used to derive the length of time for sound to travel to a reflective surface and return. A total distance of 1,280 feet would be for a round trip. Half of that would put the reflective point at 640 feet, and your bullshit would show a shooter midway between the taxi and the taxi and the music venue. FAIL.

nolu chan  posted on  2017-10-12   19:03:53 ET  Reply   Trace   Private Reply  


#97. To: nolu chan (#96)

Where are you getting 1,280 from?

VxH  posted on  2017-10-12   19:05:07 ET  Reply   Trace   Private Reply  


#98. To: VxH (#97)

Where are you getting 1,280 from?

1208. Feel better?

It is still bullshit.

T1 And T2 were explicitly derived from the same Taxi tape.

If the total distance the sound traveled is 1208 feet, and the sound went from the taxi and came back, the furthest point the sound got to was about 604 feet away.

Prove that wrong.

nolu chan  posted on  2017-10-12   19:26:05 ET  Reply   Trace   Private Reply  


#99. To: nolu chan (#98) (Edited)

1208. Feel better?

1208 isn't not on my graphic/meme with the "Taxi Driver Video" URL referenced at the bottom though, is it.

Why is that?

VxH  posted on  2017-10-12   19:39:42 ET  Reply   Trace   Private Reply  


#100. To: nolu chan (#98) (Edited)

T1 And T2 were explicitly derived from the same Taxi tape.

Nope. T1 and T2 are just generic labels for Time 1 and Time 2.

Look at my meme again - note the rows Labeled "Burst A" and "Burst B"

There's a column for T1 and a column for T2

"Burst A" has a T1 and T2, and "Burst B" has a T1 and T2

See the ECHO distance and Total distance columns?

VxH  posted on  2017-10-12   19:45:07 ET  (1 image) Reply   Trace   Private Reply  


#101. To: VxH, A K A Stone (#100)

T1 And T2 were explicitly derived from the same Taxi tape.

Nope. T1 and T2 are just generic labels for Time 1 and Time 2.

In your #42, you provided two Youtube links which were broken with extra spaces in the middle. I have them below in a usable form. Anyone may observe clearly that the first is a link to 1m7s mark of the video, and the second is a link to the 1m24s mark of the same video, i.e. 14 seconds apart. Points 14 seconds apart were chosen because REASONS.

Anyone going to either link will find themselves in the Taxi Lady video.

https://www.youtube.com/watch?v=mBbOFwWquAw&feature=youtu.be&t=1m7s

https://www.youtube.com/watch?v=mBbOFwWquAw&feature=youtu.be&t=1m24s

https://libertysflame.com/cgi-bin/readart.cgi?ArtNum=53025&Disp=42#C42

And here are your exact words at #42 (emphasis added)

I used the same speed I used in my other analysis (appended below) - based upon an air temperature of 72 degrees

Then I calculated the difference in time between the last bullet sound (T1) and the corresponding last report sound (T2).

[graphic omitted]

T2-T1 = time the report traveled = 1.07

1.07 * FPS of 1130.8 = 1208.8

==========================

[graphic omitted]

An analysis of two sequential burts of gunfire between:
["Taxi Driver Video" the Zapruder Film of the Las Vegas shooting
UNCUT / UNEDITED]
https://www.youtube.com/watch ?v=mBbOFwWquAw&feature=youtu.be&t=1m7s
and
https://www.youtube.com/watc h?v=mBbOFwWquAw&feature=youtu.be&t=1m24s
===============
T1: Time from start of video (1minute N seconds) at the time of the last shot in the burst.
T2: Time from the start of the video (1minute N seconds) at the time of the echoed sound event corresponding to T1.
TempF: the air temperature (72 degrees F)
FPS: 1130 ft per second -- The speed of sound at 72 degrees F
Elapsed Time: T2 minus T1, the number of seconds between the last shot, and the echo of the last shot in each burst.
Total Distance: Elapsed Time * FPS = the total distance traveled between T1 and T2.
Echo Distance = The distance the echo traveled from the aiming point back to the point of origin.
===============

NOTE: "1.07 * FPS of 1130.8 = 1208.8" is incorrect. 1.07 * 1130.8 = 1209.956.

NOTE: In the next mention FPS is 1130 ft per second at 72 degrees.

NOTE: If you are measuring time between two different gunshots, rather than a shot and its own echo, you cannot derive distance. You would be measuring the time between two shots, saying nothing of distance about either one.

The sounds are all from the Taxi Lady recording.

These sounds do not give the aiming point, or point of origin of the shots.

These sounds give the time sounds were recorded at the taxi.

A sound that traveled 1.07 seconds at the speed of sound went 1,209.1 feet. (1,130 * 1.07)

You cannot measure the Echo Distance from the aiming point back to the point of origin as there is no recording at the point of origin, the point of origin being the 32nd floor (supposedly). Both links go to the taxi video. The Echo Distance is from the point the sound reflected back, to the taxi location where it was recorded, following the path of the sound at ground level back to the taxi.

What is recorded on the Taxi Lady video is the sound that traveled from the 32nd floor to the taxi, and whatever may have come from elsewhere as a sound reflected back. For a sound and its echo to show up on the taxi video at a 1.07 second interval, it had to travel to a reflective surface and back in 1.07 seconds, going a total distance of 1,209.1 feet. The event venue was about 400 yards away.

For any recorded echo, the sound of the shot had to travel to the taxi, and the sound also had to travel from the 32nd floor to a reflecting surface and come back to the taxi. If the echo came from the venue area, 400 yards away, the echoed sound had to travel 400 yards to a reflective surface, then turn around and travel at least 400 yards, if the path were unobstructed at ground level, to the taxi location.

If 1,209.1 feet were one way, the round trip out and back to the 32nd floor would be about 2,418.2 feet. The path back to the taxi, if unobstructed would be somewhat shorter, as 1,209.1 would be the hypotenuse of a triangle, with the distance back to the taxi being the long side of the right triangle, if unobstructed. Any obstructions at ground level would cause the sound to take an indirect path back to the taxi.

You can make pretty graphics, and wonderful word salads, and throw around terms like relativity, but you cannot do simple calculation.

Your analysis is pretty, but it is complete bullshit. No echo traveled to the venue and returned in 1.07 seconds. For a sound recorded in the taxi, there can be no corresponding echo of that sound recorded 1.07 seconds later in the taxi, if the echo came from the venue area.

As relativity goes, you're relatively incompetent.

nolu chan  posted on  2017-10-13   23:15:05 ET  Reply   Trace   Private Reply  


#102. To: nolu chan (#101) (Edited)

>>NOTE: "1.07 * FPS of 1130.8 = 1208.8" is incorrect. 1.07 * 1130.8 = 1209.956.

Psst -

T1 T2 Elapsed Time Total Distance
17.689655 18.758621 1.068966 1208.80
https://libertysflame.com/cgi-bin/readart.cgi? ArtNum=53025&Disp=43#C43

I've already acknowledged that 1208 and the associated 1.07 seconds was the wrong calculation.

https://libertysflame.com/cgi- bin/readart.cgi?ArtNum=53025&Disp=46#C46

I've move on with the correct calculation:

https://libertysflame.com/cgi-bin/readart.cgi? ArtNum=53025&Disp=73#C73

But hey - You go ahead and obsess on 1208.

VxH  posted on  2017-10-14   1:37:01 ET  Reply   Trace   Private Reply  


#103. To: nolu chan (#101) (Edited)

You are confusing TWO separate analysis -

The one on the left pertains to audio from video recorded at the target - which is the subject of this thread "Video claims shooter dressed as police"

The one on the right pertains to audio recorded by the Taxi Driver.

Please explain to the class what the 1.0612 value under Ts=d/Vs is, and why T= Tb-Ts is the CORRECT calculation, incorporating muzzle velocity and ballistic deceleration, required to determine the distance to the shooter.

VxH  posted on  2017-10-14   1:51:21 ET  (2 images) Reply   Trace   Private Reply  


#104. To: VxH (#103)

Please explain to the class what the 1.0612 value under Ts=d/Vs is, and why T= Tb-Ts is the CORRECT calculation, incorporating muzzle velocity and ballistic deceleration, required to determine the distance to the shooter.

It means you can put numbers into somebody's chart and present columns in fancy colors and lines with shadows. And then you can ask questions about what it all means.

The link up top looks interesting but does not work.

nolu chan  posted on  2017-10-14   4:22:29 ET  Reply   Trace   Private Reply  


#105. To: nolu chan, A K A Stone, TooConservative (#104) (Edited)

It means you can put numbers into somebody's chart

Nope.

1. That's not what 1.0612  is.

2. It's not "somebody's chart" - It's a spreadsheet I created per the research methodology described on the "link" which works just fine:

http://www.btgresearch.org/impactsound.pdf

http://www.btgresearch.org/AcousticReconstruction02042012.pdf

VxH  posted on  2017-10-14   9:13:07 ET  Reply   Trace   Private Reply  


#106. To: misterwhite (#62)

I was hiking in the mountains and I yelled, "Hello". Right after that I heard four other people yell 'hello', one after the other. And here I thought I was alone.

Not only did four people answer, but they answered from four different distances.

nolu chan  posted on  2017-10-15   1:10:40 ET  Reply   Trace   Private Reply  


#107. To: All (#105)

{ crickets crickets crickets }

VxH  posted on  2017-10-15   9:45:18 ET  Reply   Trace   Private Reply  


#108. To: VxH, A K A Stone, Tooconservative, sneakypete (#107)

https://libertysflame.com/cgi-bin/readart.cgi?ArtNum=53025&Disp=103#C103

The first thing wrong with your spreadsheet chartoon is that column 3 is defined as [Vel x + y in ft/s].

You do not define x or y.

You did not provide the exact rifle and barrel length.

As for ammunition, your chartoon specified .223.

In your spreadsheet chartoon, you specified 62 grain, but fail to say what or why. A popular 223 is Remington Metal Case, 55 grain. It provides 3239 fps.

Heavier 62 grain ammunition tends to go slower.

Do you think .223 American Eagle (Federal) FMJ 62 grain ammunition was used. It provides 3,020 initial velocity.

Do you think Remington Core-Lokt 62 grain ammunition was used. It provides 3100 fps initial velocity.

Perhaps an Ultramax FMJ 62 grain. That provides 2925 fps.

Perhaps a UMC Remington Flat Base Closed Tip 62 grain. That provides 3100 fps.

Perhaps USA (Winchester) FMJ 62 grain. That provides 3100 fps.

http://gundata.org/ballistic-calculator/

That was all the .223, 62 grain ammo I saw. What 62 grain ammunition did you calculate?

You claim an initial velocity of 3,240 fps.

https://www.federalpremium.com/ammunition/rifle/family/american-eagle/american-eagle-rifle/ae223

223 Rem. Full Metal Jacket Boat-Tail ammunition provides 3240 muzzle velocity, however, it is 55 grain, not 62 grain.

You did not provide calculations on the spreadsheet chartoon you created.

Your showing of two different times on the taxi driver video is pure bullshit.

You have no idea of the delay of the muzzle blast to the taxi, or the route the sound took. You do not account the time that sound is traveling at least 341 feet while the bullet is traveling.

You have no idea of where the sound turned around to come back to the taxi, or what path the sound took to return to the taxi.

Your echo chartoon describes itself as "Test for ECHO."

The formula you are making believe you are using specifies things that have nothing to do with taxi driver distant echo recordings.

"A microphone was placed near the muzzle to record both the muzzle blast and the sound of the bullet hitting the deer."

The recording for this method is of the nearby muzzle blast and the sound of the bullet hitting the target. In the Las Vegas taxi, you have a shooter 338 feet up, and the taxi is down, to the side, and out from the hotel. The gunman is around the corner from the taxi, and you have an echo of the muzzle blast coming back, from an unknown reflecting surface, via an unknown indirect route to the taxi which is at ground level and not line of sight.

http://gundata.org/blog/post/223-ballistics-chart/

.223 Remington, Remington Metal Case, 55gr

400 [Range]
-31.7981 [Drop inches]
1588 [Velocity]
308 [Energy]
532 [Time, Milliseconds)

You claim to rely on a specific research paper.

http://www.btgresearch.org/impactsound.pdf

Using Sound of Target Impact for Acoustic Reconstructions of Shooting Events

By: Michael Courtney, PhD, and Amy Courtney, PhD, Ballistics Testing Group, Western Carolina University, Cullowhee, NC

Key Words: acoustic, reconstruction, shooting

Abstract

The sound of a bullet hitting a target is sometimes discernable in an audio recording of a shooting event and can be used to determine the distance from shooter to target. This paper provides an example where the microphone is adjacent to the shooter and presents the simple math needed in cases where the microphone is adjacent to the target.

[...]

Introduction

With surveillance systems becoming more ubiquitous in society, the number of shooting events being captured on audio is rapidly increasing. The sound of a bullet hitting a living target is very loud, almost as loud as the muzzle blast. This allows determination of the distance between the shooter and target if the time of the muzzle blast and target strike can be determined from an audio recording. If the location of the target is known, this greatly narrows possible locations of the shooter.

Method

Deer were shot with a muzzleloader shooting saboted .40 caliber pistol bullets impacting at velocities typical of the .40 S&W cartridge (1350 fps for a 135 grain bullet). A microphone was placed near the muzzle to record both the muzzle blast and the sound of the bullet hitting the deer. The time recorded between the muzzle blast and bullet striking the target represents the sum of the bullet time of flight (tb) and the time for the sound to return to the microphone from the target (ts),

t = ts + tb = d/Vs + d/Vb

where d is the target distance, Vs is the velocity of sound, and Vb is the average bullet velocity over the distance.

Vb depends on the distance, because the bullet is slowing in flight due to air resistance. Consequently, this equation must be solved using a ballistic calculator [1] and an iterative technique where one guesses different distances and computes the resulting ts and tb until there is agreement with the observed total time. The ballistic calculator requires knowing the muzzle velocity, ballistic coefficient of the bullet, air temperature, relative humidity, barometric pressure, and altitude. Converging on a distance is not hard since the total time is a monotonic and nearly linear function of the distance.

[...]

Note: the formula cited above is NOT the one cited by you. It relies on a microphone close to the muzzle also being close enough to the target to pick up the sound of the bullets striking, not an echo of the muzzle blast returning from some unknown point.

= = = = = = = = = =

http://www.btgresearch.org/AcousticReconstruction02042012.pdf

At page 5:

Discussion

These results show that it is possible to use an audio recording of a shooting event to accurately determine the distance between the target and shooter. In cases where the location of the microphone is different, the mathematical details are different, but the ideas are the same. For example, if the microphone is adjacent to the victim (such as a 911 recording might be), the equation for determining the distance becomes:

t = tbts = d/Vbd/Vs

Directly above is the precise formula you have cited on your spreadsheet chartoon:

Did you have a victim in the taxi? This formula only applies where the microphone is close to the victim (the target).

The discussion continues,

If the muzzle blast duration obscures the sound of the bullet hitting the target, simple inspection of the sound waveform is insufficient. Filtering techniques or spectrogram generation might recover the time of the target hit [1], or determination of the target hit might not be possible. However, in cases where the microphone is adjacent to the target and the bullet is supersonic, the sound of the bullet hitting the target occurs first, so it cannot be obscured by the muzzle blast.

The last part is what The Health Ranger used, and what I expanded upon.

At page 6,

A significant weakness in the study is the placement of the microphone near the muzzle of the gun, an unlikely location in most forensic cases, except for possible reconstructions of self-defense claims where the event is captured on a recording of the emergency call, officer involved shootings where the event is captured on a duty radio or other nearby microphone, and cases where the distance to the target is important to determining whether a soldier followed the applicable rules of engagement.

The only audio recordings you linked to are that of the Taxi Driver. The Taxi Driver recordings are irrelevant to the formula you cite.

T2 - T1 is not the time some report traveled. It is time difference between the time of the bullet arrival report and the sound arrival report. If your dippy calculation were correct, and the bullet were subsonic, T1 would be larger than T2, the the time some mythical report traveled would be a negative number. T2 - T1 is larger or smaller depending on the difference between the speed of the bullet and the speed of sound. If the difference of the flight time of the bullet and sound at 1200 feet were .2 seconds, you would calculate (T2 - T1) the muzzle report traveled 400 yards in .2 seconds. The speed of sound would still take 1.06 seconds to travel 1200 feet. Regardless of what T2 - T1 indicates, the muzzle report will travel 1200 feet in 1.06 seconds. The difference between the two report times indicates the difference in the velocity of time and the velocity of the bullet. If the sound took 1.06 seconds, the distance was 1200 feet. With the known distance, the velocity of the bullet can be calculated. If the bullet took 0.2 seconds less, the bullet made the trip in 0.86 seconds. Your T2 - T1 calculation is complete nonsense.

Picking back up with your spreadsheet chartoon, you seem to have some fascination with the number 1.062.

This number is not directly relevant to the Las Vegas shootings, but is the time for sound, adjusted for specific conditions, to travel 400 yards or 1200 feet at 1130 feet per second. Nothing in particular is known to have happened where the bullet traveled 400 yards or 1200 feet at ground level.

1200 feet is not the distance from the window in Paddock's room to anything in particular identified with the case.

https://www.nytimes.com/interactive/2017/10/02/us/mandalay-bay-vegas-shooting.html

Las Vegas Shooting: Chaos at a Concert
and a Frantic Search at Mandalay Bay

New York Times
UPDATED 10:30 PM ET, OCT. 2, 2017

Here is where 400 yards or 1200 feet come from. It is an estimated distance between Mandalay Bay and the Harvest Festival stage, at ground level, based on a Google image.

nolu chan  posted on  2017-10-23   19:43:06 ET  (3 images) Reply   Trace   Private Reply  


#109. To: nolu chan (#108)

DOOD!

Target shooting was invented to keep people like you distracted so you don't get bored and hurt yourselves.

If you don't already do it,start tomorrow! You owe it to your obsessive self.

In the entire history of the world,the only nations that had to build walls to keep their own citizens from leaving were those with leftist governments.

sneakypete  posted on  2017-10-23   19:46:47 ET  Reply   Trace   Private Reply  


#110. To: nolu chan (#108) (Edited)

Did you have a victim in the taxi?

No - as I've explained multiple times before.

The two images are of two SEPARATE analysis.

The one on the left references Audio taken by the taxi driver, the one on the right references audio recorded on the field.

 

 

>>The only audio recordings you linked to are that of the Taxi Driver. 

Bzzt.  FAIL again.

[Original footage of las vegas shooting 50 filled 200 injured]

https://youtu.be/kcjYefWRsKU?t=8  and 
https://youtu.be/kcjYefWRsKU? t=22

That's the audio referenced by the 2nd meme.


VxH  posted on  2017-10-23   20:30:15 ET  (2 images) Reply   Trace   Private Reply  


#111. To: nolu chan (#108) (Edited)

 

So - how'bout you go download the free tools required to create an amplitude graph of this section of audio:

https://youtu.be/kcjYefWRsKU?t=8   and  
https://youtu.be/kcjYefWRsKU? t=22

and tell us what the time you measure between the last bullet sound and the last report sound is? 

Let's see if it's between 0.6126 and 0.7378

Then you can tell us all what that means.

VxH  posted on  2017-10-23   21:12:38 ET  (1 image) Reply   Trace   Private Reply  


#112. To: VxH (#110)

https://libertysflame.com/cgi-bin/readart.cgi?ArtNum=53025&Disp=32#C32

https://libertysflame.com/cgi-bin/readart.cgi?ArtNum=53025&Disp=32#C32

Where T1 is the last bullet sound and T2 is the last report sound.

1.07 seconds between T1 and T2 = 1208 ft.

1208 ft from the Mandalay Bay, per Google earth, puts us right about where the video is being taken.

No, nitwit. 1.07 seconds would be the time differential between the flight time of the bullet and the time of sound to reach the recording point.

Sound travels 1208 feet in 1.07 seconds.

If the sound went 1208 feet and arrived in 1.07 seconds, and the bullet arrived 1.07 seconds before the sound, the bullet took 0.00 seconds in flight time.

Try again.

nolu chan  posted on  2017-10-24   0:43:58 ET  Reply   Trace   Private Reply  


#113. To: nolu chan (#112) (Edited)

LOL.  You're like a senile backseat driver who obsessively complains about (keeps pasting from) a wrong turn taken by a taxi driver 1.07 years ago while they were delivering you, on time (relatively speaking), to a compulsory (coincidental to the phase of the Moon) monthly appointment with your neurologist/psychotherapist.

I made a wrong turn, corrected it - and moved on.

https://libertysflame.com/cgi-bin/readart.cgi? ArtNum=53025&Disp=46#C46

YOU Try again.

Go ahead - download those tools and show us how its done!

YOU - create an amplitude graph of this section of audio:

https://youtu.be/kcjYefWRsKU?t=8    and   
https://youtu.be/kcjYefWRsKU? t=22

and tell us what the time you measure between the last bullet sound and the last report sound is?  

Let's see if it's between 0.6126 and 0.7378

 

VxH  posted on  2017-10-24   9:50:44 ET  (2 images) Reply   Trace   Private Reply  


#114. To: VxH (#113) (Edited)

I made a wrong turn, corrected it - and moved on.

You drove into a ditch. You are still in it, spinning your wheels.

nolu chan  posted on  2017-10-25   2:09:38 ET  Reply   Trace   Private Reply  


#115. To: VxH (#110)

The one on the left references Audio taken by the taxi driver

Yeah, you make believe that the sounds recorded in the taxi can yield an accurate measurement of distance.

The problem, of course, is related in your reference study.

http://www.btgresearch.org/AcousticReconstruction02042012.pdf

Using Sound of Target Impact for Acoustic Reconstructions of Shooting Events

At page 2:

A microphone was placed a few centimeters from the muzzle to record both the muzzle blast and the sound of the bullet hitting the deer. The time recorded between the muzzle blast and bullet striking the target represents the sum of the bullet time of flight (tb) and the time for the sound to return to the microphone from the target (ts),

t = ts + tb = d/Vs + d/Vb

where d is the target distance, Vs is the velocity of sound, and Vb is the average bullet velocity over the distance.

At page 5:

These results show that it is possible to use an audio recording of a shooting event to accurately determine the distance between the target and the shooter. In cases where the location of the microphone is different, the mathematical details are different, but the ideas are the same.

At page 6:

A significant weakness in the study is the placement of the microphone near the muzzle of the gun, an unlikely location in most forensic cases....

Your recording is at a taxi nearly 400 feet away from the muzzle. You can do all the calculations you desire and the microphone will be no closer to the muzzle. The muzzle was likely around the corner, about 340 feet up, and some angular distance away from the microphone in the taxi. There was no direct path for the sound to reach the taxi.

The taxi did not pick up the sounds of the bullets striking people on the ground over 1200 feet away. The muzzle blast echoed back, but you do not know where from, or what path it took to the taxi at ground level.

A taxi recording indicates the muzzle blast with a delay by the time the sound took to reach the taxi, about .35 seconds at 400 feet. During that delay, the muzzle blast is on its way to some reflective surface which redirects the sound by some route to the taxi at ground level.

The elapsed time at the 400 foot distant taxi is not the elapsed time of the muzzle burst soundwave out and back. You ignored the ~0.35 second initial delay to reach the taxi, and you have no idea what reflective surface(s) redirected the sound before the echo arrived at the taxi.

You do have a nice picture with circles on it though.

nolu chan  posted on  2017-10-25   2:26:19 ET  Reply   Trace   Private Reply  


#116. To: nolu chan (#115) (Edited)

The problem, of course, is related in your reference study.

Nope. I only referenced that study for the 2nd analysis - of audio/video recorded on the field. That analysis was done to refute the claim that a 2nd shooter, dressed as a policeman, was firing on the field. The observable difference between the bullet and report sound events succinctly refutes the asserted "proof" of a 2nd shooter on the field.

Regarding my initial meme "TEST FOR ECHO" -- Balistic data is NOT required to determine the total distance sound traveling from, and echoing back to, the ORIGIN point which is essentially where the Taxi driver was.

VxH  posted on  2017-10-25   9:01:10 ET  Reply   Trace   Private Reply  


#117. To: nolu chan (#115)

 

>>The taxi did not pick up the sounds of the bullets striking people on the ground over 1200 feet away

I never said it did.

The ECHO observed and discussed in the "TEST FOR ECHO" meme is not the sound of bullet impact - it is the reflected report.

You do understand what REPORT is now, right?

At least that's progress!

VxH  posted on  2017-10-25   9:13:45 ET  Reply   Trace   Private Reply  


#118. To: nolu chan (#115) (Edited)

>>Your recording is at a taxi nearly 400 feet away from the muzzle

"our reporting suggests that Paddock was positioned directly above the camera at this point"

https://www.nytimes.com/video/us/100000005473328/las-vegas- shooting-timeline-12-bursts.html

1. That's not 400 feet.

2. Imagine a firehose at an elevated position on the shore of a lake pointing the stream of water at a point on the surface 1200 feet away.  

Now, for the water, substitute the stream of shockwaves coming from the muzzle - and for the surface of the lake - the ground. 

3. What's your explanation for the difference between the echo times observed for Burst A and Burst B?

The shooter hasn't moved.  The taxi hasn't moved.  What's different?  The angle the hose is pointing?

VxH  posted on  2017-10-25   9:44:17 ET  Reply   Trace   Private Reply  


#119. To: A K A Stone (#4)

Debunk it please.

 


What would the Elapsed time between the Last Report sound event and the Last Bullet sound event be if, as the video alleged, the guy "dressed as police" was shooting?

VxH  posted on  2017-10-25   11:38:17 ET  (1 image) Reply   Trace   Private Reply  


#120. To: VxH (#116)

Regarding my initial meme "TEST FOR ECHO" -- Balistic data is NOT required to determine the total distance sound traveling from, and echoing back to, the ORIGIN point which is essentially where the Taxi driver was.

The taxi was hardly at or near the origin point. It was 338+ feet away. The study placed a microphone "a few centimers" from the muzzle, not 10,302+ centimeters away.

The taxi was 338 feet down, some distance out from the building. Even if directly opposite the taxi, the muzzle had to travel 338+ feet to get to the taxi. That still take 0.299 seconds for the muzzle blast to reach the taxi from 338 feet.

Adjusting for your cited claim (at #118) that NYT reporting "suggests that Paddock was positioned directly above the camera at this point," with the taxi directly below the window, your blather has not materially changed the problem with your chartoon. The taxi microphone was not a few centimeters from the muzzle, it was over 338 feet away.

Yeah, you make believe that the sounds recorded in the taxi can yield an accurate measurement of distance.

The problem, of course, is related in your reference study.

http://www.btgresearch.org/AcousticReconstruction02042012.pdf

Using Sound of Target Impact for Acoustic Reconstructions of Shooting Events

At page 2:

A microphone was placed a few centimeters from the muzzle to record both the muzzle blast and the sound of the bullet hitting the deer. The time recorded between the muzzle blast and bullet striking the target represents the sum of the bullet time of flight (tb) and the time for the sound to return to the microphone from the target (ts),

t = ts + tb = d/Vs + d/Vb

where d is the target distance, Vs is the velocity of sound, and Vb is the average bullet velocity over the distance.

At page 5:

These results show that it is possible to use an audio recording of a shooting event to accurately determine the distance between the target and the shooter. In cases where the location of the microphone is different, the mathematical details are different, but the ideas are the same.

At page 6:

A significant weakness in the study is the placement of the microphone near the muzzle of the gun, an unlikely location in most forensic cases....

Your recording is at a taxi over 338 feet away from the muzzle. You can do all the calculations you desire and the microphone will be no closer to the muzzle. The muzzle was likely around the corner, about 338 feet up, and some angular distance away from the microphone in the taxi.

The taxi did not pick up the sounds of the bullets striking people on the ground over 1200 feet away. The muzzle blast echoed back, but you do not know where from, or what path it took to the taxi at ground level.

A taxi recording indicates the muzzle blast with a delay by the time the sound took to reach the taxi, about .299 seconds at 338 feet. During that delay, the muzzle blast is on its way to some reflective surface which redirects the sound by some route to the taxi at ground level.

The elapsed time at the 338+ foot distant taxi is not the elapsed time of the muzzle burst soundwave out and back. You ignored the ~0.299 second initial delay to reach the taxi, and you have no idea what reflective surface(s) redirected the sound before the echo arrived at the taxi.

You do have a nice picture with circles on it though.

Also, as there was no firearm seen protruding from any window, if Paddock was the shooter, he and the firearm were inside the room. The sound of muzzle blast had to travel out through the hole in the window in a directional manner. No straight path to the taxi was available.

nolu chan  posted on  2017-10-25   16:15:25 ET  Reply   Trace   Private Reply  


#121. To: VxH (#117)

The ECHO observed and discussed in the "TEST FOR ECHO" meme is not the sound of bullet impact - it is the reflected report.

The report of what, reflected from what surface?

nolu chan  posted on  2017-10-25   16:18:02 ET  Reply   Trace   Private Reply  


#122. To: nolu chan (#120) (Edited)

It was 338+ feet away.

Which is irrelevant for the purpose of measuring the elapsed time between the initial Report sound event and the corresponding Echo event at the same location, at the bottom of the wall directly beneath the shooter.

VxH  posted on  2017-10-25   16:21:03 ET  Reply   Trace   Private Reply  


#123. To: nolu chan (#121) (Edited)

he report of what, reflected from what surface?

Explain the difference between the Echos observed in Burst A and Burst B.


 

The shooter is in the same position. The taxi is in the same position.  The "reflective surfaces" haven't moved.  And the surface of the ground, along which the report shockwave radiated back from SOMEWHERE (probably closely relative to the aiming point),  hasn't  moved either.

VxH  posted on  2017-10-25   16:25:17 ET  (1 image) Reply   Trace   Private Reply  


#124. To: VxH (#123)

[VxH #117] The ECHO observed and discussed in the "TEST FOR ECHO" meme is not the sound of bullet impact - it is the reflected report.

[nolu chan #121] The report of what, reflected from what surface?

[VxH #123] Explain the difference between the Echos observed in Burst A and Burst B.

The question was,

The report of what, reflected from what surface?

I am not interested in your evasive non-answering invitations to a snipe hunt.

If you do not know what the report was, or you do not know what it reflected from, say so. In such case, your times are meaningless for your chosen calculations.

nolu chan  posted on  2017-10-25   19:05:49 ET  Reply   Trace   Private Reply  


#125. To: VxH (#122)

Which is irrelevant for the purpose of measuring the elapsed time between the initial Report sound event and the corresponding Echo event at the same location

Which is irrelevant unless you had a microphone at the location of the muzzle blast, not 338+ feet away.

At ~.299s the sound reached the taxi. Also at .299s the sound had traveled 338 feet toward whatever reflective surface it found in the distance.

As the taxi is not at the location of the muzzle blast initiation, and your nonsense does not meet the conditions of the study which stipulated a microphone a few centimeters from the muzzle. The taxi was over 10,000 centimeters away.

nolu chan  posted on  2017-10-25   19:14:22 ET  Reply   Trace   Private Reply  


#126. To: nolu chan (#124) (Edited)

The report of what, reflected from what surface?

The Report observable at T1 in Burst A and Burst B - which correspond to the observable ECHOED events at T2 in Burst A and Burst B.

VxH  posted on  2017-10-25   19:17:27 ET  Reply   Trace   Private Reply  


#127. To: nolu chan (#125) (Edited)

the conditions of the study which

LOL. Have your donkey look again.

The formula you pulled out of the study is to determine the distance to the SHOOTER from the microphone.

The study does not address the calculation of the distance to the target from the microphone.

Also the ECHO'd events are not Bullet impacts - they are the echo'd muzzle blast shock waves.

VxH  posted on  2017-10-25   19:19:55 ET  Reply   Trace   Private Reply  


#128. To: VxH, A K A Stone (#119)

What would the Elapsed time between the Last Report sound event and the Last Bullet sound event be....

It would be a positive number expression of time.

On your spreadsheet chartoon, notice that you calculate T = Tb - Ts.

You calculate elapsed time as the time it took the bullet to travel, minus the time it took the sound to travel.

The correct formula should be T = Ts – Tb = d/Vs – d/Vb.

As the bullet is supersonic, and sound is a constant, the sound would travel 400 yards in 1.06s and the bullet would travel in less than 1.06s. Subtracting 1.06 from a smaller number will always yield a negative number.

At 1200 feet, you calculate Tb as 0.448578s, and Ts as 1.062s and calculate the T as -0.6126, negative 0.6126 seconds. The average donkey could recognize that something is wrong when the result is negative time.

Just what do you think happens in negative 0.6126 seconds?

You could at least recognize that if you get a negative number, you have stated the required formula backwards, and you proceeded to perform the calculation backwards, and present the bass ackwards result of your misunderstanding of the study you looked at.

Moreover, while you state backwards that T = Tb - Ts, your spreadsheet never defines what T is supposed to represent. Negative 0.6126 is the time of what? What is the significance of this negative 0.6126 seconds (other than to demonstrate you did not understand the reference study)?

nolu chan  posted on  2017-10-25   19:23:56 ET  Reply   Trace   Private Reply  


#129. To: nolu chan (#128) (Edited)

ust what do you think happens in negative 0.6126 seconds?

LOL.

The difference in time, 0.6126, is an ABSOLUTE value. Sign is irrelevant.

VxH  posted on  2017-10-25   19:27:54 ET  Reply   Trace   Private Reply  


#130. To: nolu chan (#128) (Edited)

your spreadsheet never defines what T is supposed to represent.

T is the absolute value of the difference between Tb and Ts -- which corresponds to the elapsed time between T1 and T2 that is OBSERVED in the audio recording's amplitude graph.

It is that correspondence that is then used to find the range - which was generated from the ballistic data for intervals of 75 feet.

VxH  posted on  2017-10-25   19:31:38 ET  Reply   Trace   Private Reply  


#131. To: nolu chan (#128) (Edited)

>>The correct formula should be T = Ts – Tb = d/Vs – d/Vb.

Nope.

You're not even reading from the relevant part of the paper - where the microphone adjacent to the victim scenario is discussed.

http://ww w.btgresearch.org/AcousticReconstruction02042012.pdf

That's the same formula I have in my illustration:

VxH  posted on  2017-10-25   19:59:45 ET  (2 images) Reply   Trace   Private Reply  


#132. To: A K A Stone, tooconservative (#131)

If the guy "dressed like police" was a shooter as alleged - the Elapsed Time between Bullet Impact sound events and Muzzle Report sound events would be much nearer to zero than what the audio shows.

So, Is that "DEBUNKED" enough for ya?

VxH  posted on  2017-10-25   20:40:32 ET  Reply   Trace   Private Reply  


#133. To: VxH (#132)

I think everyone wants to grab the mantle of science for themselves and deride each other as kooks. Which may be mostly true.

OTOH, the official investigation is so miserably bad that you can't blame people for making up their own explanations when the FBI has botched it this badly.

Tooconservative  posted on  2017-10-25   21:02:09 ET  Reply   Trace   Private Reply  


#134. To: VxH (#129)

The difference in time, 0.6126, is an ABSOLUTE value. Sign is irrelevant.

It is relevant when calculated and displayed on a spreadsheet. You explicitly calculated for and displayed the negative value. Had you calculated for an absolute value, a negative value would not appear. If your formula in the cell does not say it is an absolute value, you do not produce an absolute value.

nolu chan  posted on  2017-10-25   21:46:29 ET  Reply   Trace   Private Reply  


#135. To: VxH (#131)

You're not even reading from the relevant part of the paper - where the microphone adjacent to the victim scenario is discussed.

No, I read the correct part.

The formula is correct for a supersonic bullet only if the value is explicitly expressed as an absolute. Otherwise, the correct value is derived by changing the formula. Either will work. You did neither and derived negative times and published them that way.

nolu chan  posted on  2017-10-25   21:49:46 ET  Reply   Trace   Private Reply  


#136. To: nolu chan (#135) (Edited)

>>No, I read the correct part.

LOL

Here's what you quoted:

====================

A microphone was placed a few centimeters from the muzzle to record both the muzzle blast and the sound of the bullet hitting the deer. The time recorded between the muzzle blast and bullet striking the target represents the sum of the bullet time of flight (tb) and the time for the sound to return to the microphone from the target (ts),
t = ts + tb = d/Vs + d/Vb

nolu chan     posted on  2017-10-25   16:15:25 ET

https://libertysflame.com/cgi- bin/readart.cgi?ArtNum=53025&Disp=120#C120

====================

And this is the correct section of the paper that deals with a microphone AT THE TARGET, the scenario where the "shooter dressed as police" video in question was taken.

http://ww w.btgresearch.org/AcousticReconstruction02042012.pdf


 

>>The formula is correct for a supersonic bullet only if the value is explicitly expressed as an absolute

The formula is fine just the way the authors of the paper wrote it.  The negative time is perfectly acceptable IF you actually understand what the value and chart are saying:

The sound of the Report (Ts) event was recorded 0.6126 seconds after Tb and that time differential corresponds to a distance of 1200 ft from the shooter.


VxH  posted on  2017-10-25   23:33:41 ET  (1 image) Reply   Trace   Private Reply  


#137. To: Tooconservative (#133)

I think everyone wants to grab the mantle of science for themselves and deride each other as kooks. Which may be mostly true.

This is a good example of why Jefferson wanted WE common people educated.  We're not supposed to be waiting for government "experts" to give us our opinions.


"... finally, that truth is great and will prevail if left to herself, that she is the proper and sufficient antagonist to error, and has nothing to fear from the conflict, unless by human interposition disarmed of her natural weapons, free argument and debate, errors ceasing to be dangerous when it is permitted freely to contradict them. "
 
"I HAVE SWORN UPON THE ALTAR OF GOD ETERNAL HOSTILITY TO EVERY FORM OF TYRANNY OVER THE MIND OF MAN"
--Thomas Jefferson, 1786
 

VxH  posted on  2017-10-25   23:39:46 ET  (1 image) Reply   Trace   Private Reply  


#138. To: VxH (#137)

I've always thought it likely that Jefferson's first objection to modern American government would be to the monument they built in his memory. He just wasn't that kind of guy.

I think that famous personalities in history could only speak to their own times. We like to imagine or pretend that they were some wise sages, imparting timeless wisdom for the ages, blah-blah-blah, insert three fingers and think deep thoughts, etc.

The truth is that the great men of history were creatures of their own times and only capable of speaking to the great issues of their times. And placing them on pedestals and trying to pretend that they were speaking to the issues of our time is just laziness or political hackery.

You might just as well start asking WWJT (What Would Jesus Tweet?). Well, obiously, Jesus wouldn't tweet anything and he wouldn't even imagine it. And if he could imagine it, he wouldn't bother to tweet.

We are all inescapably products of our own times.

Tooconservative  posted on  2017-10-25   23:50:43 ET  Reply   Trace   Private Reply  


#139. To: Tooconservative (#138)

I've always thought it likely that Jefferson's first objection to modern American government would be to the monument they built in his memory.

Have you ever been there?

It's not the man/statue that's being memorialized so much as the foundational American ideals for which there are trail-heads on each of the four walls.

TRUTH IS GREAT AND WILL PREVAIL.

People who don't understand Science will be bamboozled by those who do, or worse.

VxH  posted on  2017-10-25   23:59:32 ET  Reply   Trace   Private Reply  


#140. To: Tooconservative, A K A Stone (#138) (Edited)

The difference in time between the Bullet sound event and the Muzzle Report sound event corresponds to a shooter distance of approximately 1350 feet away.

Is the guy dressed like a policeman 1350 feet away?

NO.

So the claim of the video subject of this thread, that "shooter dressed as police" - is DEBUNKED.

VxH  posted on  2017-10-26   0:06:25 ET  Reply   Trace   Private Reply  


#141. To: VxH (#140)

So the claim of the video subject of this thread, that "shooter dressed as police" - is DEBUNKED.

I never grasped why they thought there was any factual basis for this claim to begin with.

It just seemed like Kookbait to me, beginning to end. It was intended to appeal to the kooks and make some money for the con-men who created it when the kooks kept clicking on and reposting their kookery around the internet but especially on Fakebook and Twit-ter.

Tooconservative  posted on  2017-10-26   0:15:25 ET  Reply   Trace   Private Reply  


#142. To: Tooconservative (#141)

I never grasped why they thought there was any factual basis for this claim to begin with.

I give them the benefit of the doubt.

They mistook the bullet events as gunfire coming from the flashlight bearing police officer.

Then their imagination took over.

VxH  posted on  2017-10-26   0:26:16 ET  Reply   Trace   Private Reply  


#143. To: VxH (#142)

They have a need for their kookery. It is integral to their worldview, to their opinion of themselves and their denigrating opinions of others.

They start jonesing for a kookery within a few hours of any major tragedy. It happens over and over. 9/11, Newtown, Boston marathon, Vegas massacre, you name it. It is a very consistent pattern. It's a noxious kind of neurotic behavior.


Bring me . . . a Kookery!

And the wailing and gnashing of teeth if someone points out they are actual kooks ... oy vey.

Tooconservative  posted on  2017-10-26   0:37:36 ET  (1 image) Reply   Trace   Private Reply  


#144. To: Tooconservative (#143)

They start jonesing for a kookery within a few hours of any major tragedy

They are a product of the culture that created them.

Bezmenov described the process.

Marcuse and Co. implemented it.

Now we have multiple generations of individuals raised by individuals who are disconnected from reality.

The Soviets were planning on that not ending well for us.

VxH  posted on  2017-10-26   0:41:46 ET  Reply   Trace   Private Reply  


#145. To: VxH (#136)

LOL

Here's what you quoted:

====================

A microphone was placed a few centimeters from the muzzle to record both the muzzle blast and the sound of the bullet hitting the deer. The time recorded between the muzzle blast and bullet striking the target represents the sum of the bullet time of flight (tb) and the time for the sound to return to the microphone from the target (ts), t = ts + tb = d/Vs + d/Vb

nolu chan posted on 2017-10-25 16:15:25 ET

https://libertysflame.com/cgi-bin/readart.cgi?ArtNum=53025&Disp=120#C120

You are an ass. How many times did #120 explicitly say it was about your lame attempt to use the taxi video?

#120. To: VxH (#116)

Regarding my initial meme "TEST FOR ECHO" -- Balistic data is NOT required to determine the total distance sound traveling from, and echoing back to, the ORIGIN point which is essentially where the Taxi driver was.

The taxi was hardly at or near the origin point. It was 338+ feet away. The study placed a microphone "a few centimers" from the muzzle, not 10,302+ centimeters away.

The taxi was 338 feet down, some distance out from the building. Even if directly opposite the taxi, the muzzle had to travel 338+ feet to get to the taxi. That still take 0.299 seconds for the muzzle blast to reach the taxi from 338 feet.

Adjusting for your cited claim (at #118) that NYT reporting "suggests that Paddock was positioned directly above the camera at this point," with the taxi directly below the window, your blather has not materially changed the problem with your chartoon. The taxi microphone was not a few centimeters from the muzzle, it was over 338 feet away.

Yeah, you make believe that the sounds recorded in the taxi can yield an accurate measurement of distance.

The problem, of course, is related in your reference study.

http://www.btgresearch.org/AcousticReconstruction02042012.pdf

Using Sound of Target Impact for Acoustic Reconstructions of Shooting Events

At page 2:

A microphone was placed a few centimeters from the muzzle to record both the muzzle blast and the sound of the bullet hitting the deer. The time recorded between the muzzle blast and bullet striking the target represents the sum of the bullet time of flight (tb) and the time for the sound to return to the microphone from the target (ts),

t = ts + tb = d/Vs + d/Vb

where d is the target distance, Vs is the velocity of sound, and Vb is the average bullet velocity over the distance.

At page 5:

These results show that it is possible to use an audio recording of a shooting event to accurately determine the distance between the target and the shooter. In cases where the location of the microphone is different, the mathematical details are different, but the ideas are the same.

At page 6:

A significant weakness in the study is the placement of the microphone near the muzzle of the gun, an unlikely location in most forensic cases....

Your recording is at a taxi over 338 feet away from the muzzle. You can do all the calculations you desire and the microphone will be no closer to the muzzle. The muzzle was likely around the corner, about 338 feet up, and some angular distance away from the microphone in the taxi.

The taxi did not pick up the sounds of the bullets striking people on the ground over 1200 feet away. The muzzle blast echoed back, but you do not know where from, or what path it took to the taxi at ground level.

A taxi recording indicates the muzzle blast with a delay by the time the sound took to reach the taxi, about .299 seconds at 338 feet. During that delay, the muzzle blast is on its way to some reflective surface which redirects the sound by some route to the taxi at ground level.

The elapsed time at the 338+ foot distant taxi is not the elapsed time of the muzzle burst soundwave out and back. You ignored the ~0.299 second initial delay to reach the taxi, and you have no idea what reflective surface(s) redirected the sound before the echo arrived at the taxi.

You do have a nice picture with circles on it though.

Also, as there was no firearm seen protruding from any window, if Paddock was the shooter, he and the firearm were inside the room. The sound of muzzle blast had to travel out through the hole in the window in a directional manner. No straight path to the taxi was available.

nolu chan  posted on  2017-10-25   16:15:25 ET  Reply   Trace   Private Reply  

nolu chan  posted on  2017-10-26   2:34:30 ET  Reply   Trace   Private Reply  


#146. To: nolu chan (#145) (Edited)

>>How many times did #120 explicitly say { blah blah blah }

You might want to rethink the value of quoting yourself to "prove" what someone else said.  Doesn't seem to be working very well for you.

The Taxi Video applies to The Test for ECHO meme - not to the video/audio being discussed in this thread which asserts that "shooter dressed as police".

That "shooter dressed as police" ASSertion is CLEARLY refuted by the audio data.  

Audio data that I've analysed using the correct forumula - which works just fine without your tweakage.

http://ww w.btgresearch.org/AcousticReconstruction02042012.pdf

And that IS the same formula I have in my illustration:


VxH  posted on  2017-10-26   11:16:27 ET  (2 images) Reply   Trace   Private Reply  


#147. To: VxH (#146)

Why do you keep posting this chartoon when all your data is not only wrong, but farcical? The only things you proved is that you do not know how to calculate the average velocity of an imaginary bullet and you are hopeless at spreadsheets. Your entertainment value as a useful idiot is over for now, and you will never figure it out without more help. Help is on the way, grasshopper.

Columns 1, 2, and 3 are direct entry of data generated by entering imaginary data into a generator at http://www.shooterscalculator.com/. I replicated the data taken from the calculator with “My BB's.” If I input initial velocity as 3240 fps, and other data, and call it “My BB's,” I can show a chart for magical bb’s.

http://www.shooterscalculator.com/ballistic-trajectory-chart.php?t=34fa8220

The Shooter’s Calculator only provides a result based on user input. It does not present a spreadsheet with the formulas to generate the data. The data from the Calculator can be cut and pasted into a spreadsheet, or entered by direct entry; this produces data in the cells, but no spreadsheet formulas in the cells. The chart states the speed of sound as 1130 feet per second (fps).

The remaining 4 columns, (4, 5, 6, 7) were generated by VxH.

Column 6 uses 1130.8 fps to calculate the time for sound to travel the distance stated in Column 1.

Column 4 is labeled as (Avg V) Vb. This column purports to present the average velocity of the bullet to cover the distance for the row it is in. All of the data in this column is epically wrong as the methodology of calculation is absurdly wrong.

To calculate the average velocity of the bullet, divide distance by time.

Instead of this, a personal misbegotten formula was used. Probably a pocket calculator for each cell in Column 4 was used to perform the calculations, and the data was directly entered into the cells by hand.

For the first two data rows, sum 3240 and 3163 and divide by 2. 6403/2 yields the 3201 in Column 4.

For the first three data rows, sum 3240+3163+3088 for 9491. 9491 / 3 yields the 3163.6667 in Column 4.

And so on, and so forth. All the calculated Column 4 data (average Vb), is garbage.

The chosen methodology was to sum the velocity given for each distance, and divide by the number of elements summed. This produces nonsensical data.

Example: You drive a car 100 miles at 80 mph. You drive another 100 miles at 20 mph. With this bogus methodology, 80 + 20 = 100, divide by 2, and your average velocity was 50 mph. Not.

In the real world, you drove 100/80 or 1.25 hours at 80 mph. You drove 100/20 or 5 hours at 20 mph. And you drove 200 miles in 6.25 hours. Your average speed was 200/6.25, or 32 mph.

Column 4, in addition to using an absurd methodology for its calculations, also incorporates two summing errors for the velocities taken from Column 3, at 900 feet and 1275 ft. In each case, the actual sum was 1 less than that calculated.

Spreadsheet formulas are not prone to fat finger syndrome, and do not make such errors, but someone with a pocket calculator or pen and paper does. The data was typed in after external calculation.

Where you calculate 2367.5926 average Vb at 1950 feet, 1950/1.211933 (the velocity of the bullet in Column 5), it yields 1608.9998 fps, remarkably close to the 1609 in Column 3. But then, the elapsed time in Column 2 is 0.86, not 1.21933. It is a conundrum how the bullet traveled for 1.21933 seconds in an elapsed time of 0.86 seconds.

Of course, when you use Column 1 1950 ft and Column 3 1609 fps to derive the time of flight, the formula is d/Vb, and Vb is the Average Velocity.

The bullet will travel 1905 feet distance (Col 1) in 0.86 sec time (Col 2) in 1905/0.86 or 2267.4418 average Vb. Stated in your headnote is Tb is d/Vb.

It is noteworthy that you used Column 3 as the "average" velocity of the bullet in order to derive the other average velocity of the bullet in Column 4.

Column 5 (Tb) incorporates the garbage data from Column 4 into its calculations, and all the resulting calculated data is wrong. GIGO.

Column 7 (T = Tb – Ts) incorporates the garbage data from Column 5 and all the calculated data is wrong. GIGO.

The chart is multicolor and pretty, but the data for the imaginary bullet is demonstrably wrong in every column you created, except for column 6 where you succeeded in dividing the distance by 1130.8.

nolu chan  posted on  2017-10-28   15:12:38 ET  (1 image) Reply   Trace   Private Reply  


#148. To: nolu chan, A K A Stone (#147)

BTW = The Elapsed time between T1 and T2 0.689655 is quite quite sufficient for debunking the title of the video "shooter dressed as police".

Even without the ballistic data (which is calculated correctly for the parameters entered) - the difference between the bullet sound event and the report sound event puts the distance of the shooter at least 784 feet.

Is the guy "dreesed as police" 784 feet away? NOPE.

Video status = DEBUNKED.

VxH  posted on  2017-10-28   16:27:41 ET  Reply   Trace   Private Reply  


#149. To: nolu chan (#147) (Edited)

To calculate the average velocity of the bullet, divide distance by time.

Psst.

Please tell the class why, using your calculation, at 75 ft, the bullet has ACCELERATED to 3750fps, decerates to 3000fps at 150 seconds... and then accelerates to 3214fps at 225 feet... etc accelerating and decelerating and accelerating.

Is it a magic bullet?

The time in the chart rendered by the ballistic calculator only has 2 decimals of precision.

Calculating the average per the reported velocity is thus more accurate.

 

 

VxH  posted on  2017-10-28   16:47:57 ET  (1 image) Reply   Trace   Private Reply  


#150. To: VxH (#148)

Even without the ballistic data (which is calculated correctly for the parameters entered)

Which is only as valid as the improbable or impossible data you entered.

As I demonstratred, the same data entered for My BB's produces a chart with the same data for BB's.

http://www.shooterscalculator.com/ballistic-trajectory-chart.php?t=34fa8220

Come on. Question my analysis of how you made a botch of the Average Bullet Velocity. Give us your methodology and formula.

Why were all your calculations wrong except for distance divided by time?

nolu chan  posted on  2017-10-28   17:00:50 ET  Reply   Trace   Private Reply  


#151. To: nolu chan (#150) (Edited)

>>Question my analysis of how you made a botch of the Average Bullet Velocity.  

LOL.  OK - please tell the class why the bullet accelerates / decelerates / accelerates repeatedly when your "analysis" is applied?

The time in the chart rendered by the ballistic calculator only has 2 decimals of precision.

Calculating the average per the reported velocity is thus more accurate.

VxH  posted on  2017-10-28   17:04:31 ET  (1 image) Reply   Trace   Private Reply  


#152. To: nolu chan (#147) (Edited)

>> and the data was directly entered into the cells by hand.

Bzzzt.  Fail again.

http://www.shooterscalculator.com/ballistic-trajectory-chart.php

Drag Function: G1
Ballistic Coefficient: 0.300
Bullet Weight: 62 gr
Initial Velocity: 3240 fps
Sight Height : 1.5 in
Shooting Angle: -33°
Wind Speed: 0 mph
Wind Angle: 0°
Zero Range: 100 yd
Chart Range: 750 yd
Maximum Range: 50002 yd
Step Size: 25 yd
Corrected For Atmosphere
Adjusted BC: 0.33
Altitude: 2000 ft
Barometric Pressure: 29.92 Hg
Temperature: 72° F
Relative Humidity: 21%
Speed of Sound: 1130 fps

RangeTimeVel[x+y]
(ft) (s) (ft/s)
00.003240
750.023163
1500.053088
2250.073014
3000.102941
3750.122870
4500.152799
5250.182730
6000.202662
6750.232595
7500.262529
8250.292465
9000.322401
9750.362337
10500.392275
11250.422214
12000.462154
12750.492095
13500.532036
14250.561979
15000.601923
15750.641867
16500.681813
17250.731760
18000.771708
18750.811658
19500.861609
20250.911561
21000.961515
21751.011470
22501.061426

VxH  posted on  2017-10-28   18:22:37 ET  (2 images) Reply   Trace   Private Reply  


#153. To: VxH (#151)

[Vxh #148] Even without the ballistic data (which is calculated correctly for the parameters entered)

- - - - - - - - - -

[VxH #149] Please tell the class why, using your calculation, at 75 ft, the bullet has ACCELERATED to 3750fps, decerates to 3000fps at 150 seconds... and then accelerates to 3214fps at 225 feet... etc accelerating and decelerating and accelerating. Is it a magic bullet?

[VxH #151] Calculating the average per the reported velocity is thus more accurate.

More accurate is to divide the distance by the velocity and get the time to more decimal places and eliminate the rounding error. Your bullshit methodology of summing velocities and dividing does not work. It is bullshit.

The stupid... it hurts!

The chart results are based on the data you entered.

As I demonstratred, the same data entered for My BB's produces a chart with the same data for BB's.

If the chart correctly calculated the ballistic data for the parameters you entered,

http://www.shooterscalculator.com/ballistic-trajectory-chart.php?t=34fa8220

Tell everybody how you derived average velocity.

Come on. Question my analysis of how you made a botch of the Average Bullet Velocity. Give us your methodology and formula.

Why were all your calculations wrong except for distance divided by time?

The data which you input did not come from any real life ammunition, you just entered stuff, as I did for My BB's. I just entered the same stuff you did, proving my bb's have an initial Vel[x+y] of 3240 fps. My BB's perform precisely as do your imaginary cartridge. Are you saying the ballistics chart you used produced invalid results?

If the chart results are valid, please tell the class why the chart indicates the bullet traveled 75 ft. in 0.02 seconds and that indicates average velocity d/time of 750/.02 = 3750 fps.

It's your data. If the ballistics chart calculated correctly, you should understand the chart you presented, and be able to explain the results given.

Do you think you are entitled to just use a nonsense formula which produces nosense results because you do not understand the chart data that you selected and presented?

The note at the bottom of the chart indicates:

Keep in mind this is an approximation....

Of course, the time of 0.02 could represent a figure rounded to two decimal places for presentation, and actually represent anything from 0.0150 to 0.0249.

75 feet divided by Vel[x] 3239 75/3239 feet, taken to six decimal places gives 0.0231552 seconds bullet travel time. Hot damn, it's within the rounding error.

At Vel[x+y] 3240 feet per second, and 75 feet distance, the time to six decimal places would be 0.0231481 seconds bullet travel time and hot damn, that's within the rounding error too.

Thank you, Lord.

At my #108 I asked,

As for column 3, "Vel[x+y] (ft/s)", you seem to have forgotten to give any definition of x or y on your chart.

That question met with resounding crickets.

A mystery, wrapped in an enigma, hidden by a conundrum, is why, at 75 feet, the chart indicates Vel(x) = 3239, Vel (y) = 5.70, and Vel[x+y] = 3240. Whatever can that strange arithmetic be?

You could have chosen to display Vel[x] or Vel[y], or Vel[x+y]. Why did you choose to display Vel[x+y] rather than say, Vel[x]? What is Vel[x], Vel[y], and Vel[x+y]?

nolu chan  posted on  2017-10-28   18:47:52 ET  Reply   Trace   Private Reply  


#154. To: nolu chan (#153) (Edited)

 

divide the distance by the velocity and get the time to more decimal places and eliminate the rounding error.

LOL.

So you're going to drive 99 miles at 2mph, then drive and 1 mile at 100mph.

You're going to divide 100 miles by what 100mph?

Here are the values of Nolu- Time calculated with your d/v brainstorm:

Ooops!

Congratulations! You "fixed" the rounding of 0.86 by transforming it into 1.2119328776 are you sure that works?

"Vel[x+y] (ft/s)", you seem to have forgotten to give any definition of x or y on your chart.

LOL. I know what they mean on the Ballistic calculator. Don't you?

Hint: They're Vectors.

And speaking of Vectors: If we treated each 75 ft segment as a vector and then calculated time as a function of the relationship between 75ft and the difference between  {Vn..Vn+1}... that might work a little better than your simple d/v idea.

 

VxH  posted on  2017-10-28   19:05:58 ET  (1 image) Reply   Trace   Private Reply  


#155. To: VxH (#154)

So you're going to drive 99 miles at 2mph, then drive and 1 mile at 100mph.

You're going to divide 100 miles by what 100mph?

No.

You measuring bullets at 75 foot increments. It is the DISTANCE that in the segment and remains the same. That is why your bullshit does not work and your question is bullshit. You stipulate a DISTANCE ratio of 99:1. Nice try.

It is not one distance and 99 times that distance. The distances on the chart are in precisely equal steps.

By your misbegotten Rube Goldberg "formula," the bullet traveled
0.023427s after 75 feet
0.047413s after 150 feet.

The correct times, carried to seven decimal places, are,
0.0237116s - 075 ft [d/Vb]
0.0485751s - 150 ft [d/Vb]

The bullets traveled the precise same distance in different times and velocities.

The distance segments are precisely the same; the elapsed time and velocity changes.

Over equal distances, your formula is still bullshit.

If you drive 99 miles at 99 mph, and 99 miles at 1 mph,

99/99 = 1 hour

and 99 miles at 1 mph.

99/1 = 99 hours

Your average speed is not 100/2 50 mph.

Your average speed is 198m/100h = 1.98 mph.

Duhhhh.

LOL. I know what they mean on the Ballistic calculator. Don't you?

Hint: They're Vectors.

I know. You just found out. You still did not explain why you used Vel[x+y] rather than Vel[x].

And speaking of Vectors: If we treated each 75 ft segment as a vector and then calculated time as a function of the relationship between 75ft and the difference between {Vn..Vn+1}... that might work a little better than your simple d/v idea.

Simple d/v is not my idea. You stated the Column 5 formula as Tb = d/Vb.

Your bullshit of summing the travel over one 75 foot segment with the travel time over the next 75 foot segment, and making believe that this is the formula to find the bullet average time of travel is still bullshit.

You did a nice job of chopping the formulas off the graphic posted above. Chopping them off does not make them go away. They are stated as:

d is the target distance.(range)
Vs is the velocity of sound, and
Vb is the average bullet velocity over the distance
Tb is d / Vb (time to cover distance @ Vb)
Ts is d/Vs (time to cover distance @Vs)

Where is the bullshit formula that has you summing the elapsed times of segments and dividing by the number of segments to get the average velocity?

You link to btgresearch and then put your own misbegotten formulas on the page, with your bullshit data, for an imaginary cartridge with imaginary properties tweaked to get the initial velocity you wanted at the time.

nolu chan  posted on  2017-10-28   23:28:01 ET  Reply   Trace   Private Reply  


#156. To: nolu chan (#155)

You still did not explain why you used Vel[x+y] rather than Vel[x].

Anybody who actually understands the ballistic calculator doesn't need an explanation.

VxH  posted on  2017-10-29   0:31:20 ET  Reply   Trace   Private Reply  


#157. To: nolu chan (#155) (Edited)

[duplicate]

VxH  posted on  2017-10-29   0:33:41 ET  Reply   Trace   Private Reply  


#158. To: nolu chan (#155) (Edited)

>>Simple d/v is not my idea.

Liar.

"more accurate is to divide the distance by the velocity and get the time to more decimal places

nolu chan   posted on  2017-10-28   18:47:52 ET 

https://libertysflame.com/cgi-bin/readart.cgi? ArtNum=53025&Disp=153#C153

Nolu-Time as d/v = FAIL

Which is column has values that closer to the values under Time?

Column I:  Nolu- Time
Column J:  d/Avg V

Column J does.

Why is that?

VxH  posted on  2017-10-29   0:43:51 ET  (1 image) Reply   Trace   Private Reply  


#159. To: All (#154)

>>speaking of Vectors...


VxH  posted on  2017-10-29   2:53:36 ET  (1 image) Reply   Trace   Private Reply  


#160. To: VxH, A K A Stone (#159)

speaking of Vectors...

The problem is that you are clueless and do not know what your are doing and do not know what a vector is. A vector is described by a line, not a point.

Here is a correct spreadsheet:


BALLISTICS DATA SPREADSHEET

A1 AVERAGE VELOCITY AND TIME DIFF OVER TOTAL DISTANCE AVERAGE VELOCITY FOR EACH 75 FEET SEGMENT
2
3 B C D E F G H I J K L
4 d Time (avg) Vel[x+y] Ts=d/Vs T=Tb - Ts Tb 75 ft Avg Velocity for Segment Segment Segment
5 (ft) d/Vel[x+y] (ft/s) d/Vs ABS(Tb-Ts) +C7-C6 75 foot segment distance begin end
6 0 0.0000 3240 +75/H4 +B7-B6 +K7+75 A7
7 75 0.0237 3163 0.0664 0.0427 0.0237 3163.0000 75 0 75
8 150 0.0486 3088 0.1327 0.0842 0.0249 3016.4744 75 75 150
9 225 0.0747 3014 0.1991 0.1245 0.0261 2876.1533 75 150 225
10 300 0.1020 2941 0.2655 0.1635 0.0274 2741.7798 75 225 300
11 375 0.1307 2870 0.3319 0.2012 0.0287 2617.2620 75 300 375
12 450 0.1608 2799 0.3982 0.2375 0.0301 2490.8930 75 375 450
13 525 0.1923 2730 0.4646 0.2723 0.0315 2378.2353 75 450 525
14 600 0.2254 2662 0.5310 0.3056 0.0331 2266.7686 75 525 600
15 675 0.2601 2595 0.5973 0.3372 0.0347 2160.0657 75 600 675
16 750 0.2966 2529 0.6637 0.3672 0.0364 2057.9351 75 675 750
17 825 0.3347 2465 0.7301 0.3954 0.0381 1967.1773 75 750 825
18 900 0.3750 2400 0.7965 0.4215 0.0403 1860.3774 75 825 900
19 975 0.4172 2337 0.8628 0.4456 0.0422 1777.1863 75 900 975
20 1050 0.4615 2275 0.9292 0.4677 0.0443 1691.5924 75 975 1050
21 1125 0.5081 2214 0.9956 0.4874 0.0466 1609.7315 75 1050 1125
22 1200 0.5571 2154 1.0619 0.5048 0.0490 1531.4566 75 1125 1200
23 1275 0.6089 2094 1.1283 0.5194 0.0518 1448.4509 75 1200 1275
24 1350 0.6631 2036 1.1947 0.5316 0.0542 1384.2156 75 1275 1350
25 1425 0.7201 1979 1.2611 0.5410 0.0570 1315.8863 75 1350 1425
26 1500 0.7800 1923 1.3274 0.5474 0.0600 1250.6135 75 1425 1500
27 1575 0.8436 1867 1.3938 0.5502 0.0636 1179.8360 75 1500 1575
28 1650 0.9101 1813 1.4602 0.5501 0.0665 1127.9144 75 1575 1650
29 1725 0.9801 1760 1.5265 0.5464 0.0700 1071.1245 75 1650 1725
30 1800 1.0539 1708 1.5929 0.5391 0.0738 1016.9418 75 1725 1800
31 1875 1.1309 1658 1.6593 0.5284 0.0770 973.8184 75 1800 1875
32 1950 1.2119 1609 1.7257 0.5137 0.0811 925.3285 75 1875 1950
33 2025 1.2972 1561 1.7920 0.4948 0.0853 879.1211 75 1950 2025
34 2100 1.3861 1515 1.8584 0.4723 0.0889 843.7085 75 2025 2100
35 2175 1.4796 1470 1.9248 0.4452 0.0935 802.5405 75 2100 2175
36 2250 1.5778 1426 1.9912 0.4133 0.0982 763.3722 75 2175 2250
37
38 Total
39 1.5780
40 SUM G7:G36



As a vector is described by a line and not a point, the Column D velocity at 75 feet describes the average bullet velocity for the segment from 0 to 75 feet, and the velocity at 150 feet describes the average bullet velocity from 0 to 150 feet, and so on.

The time for 75 feet indicates the elapsed time for 0 to 75 feet. The time for 150 feet indicates the elapsed time for 0 to 150 feet.

Column C, the time, is derived by dividing Column B (distance) by Column D. In your chart it is was rounded off to two decimal places. I took it to four decimal places.

Your added Rube Goldberg nonsense was not only wrong but unecessary. Average velocity at the stated distances was staring you in the face.

In Columns H thru L, I have provided the data for each 75-foot segment.

At 1575 feet, the bullet opens its largest gap on sound at 0.05502 seconds.

From 1575 to 1650 feet, the bullet travels at an average velocity of 1127.9144 fps, dipping below the speed of sound. After that, sound is traveling faster than the bullet and the gap diminishes.

nolu chan  posted on  2017-10-30   19:58:41 ET  Reply   Trace   Private Reply  


#161. To: nolu chan, A K A Stone (#160)

 

4dTime(avg) Vel[x+y]

5(ft)d/Vel[x+y](ft/s)

3219501.21191609

At 1950 ft your "spreadsheet" has an elapsed time of 1.2119 with a corresponding Vel[x+y] of 1609.

Meanwhile observe the corresponding elapsed time and Vel[x+y] generated by:

ShooterCalculator.com

19500.861609

Drag Function: G1
Ballistic Coefficient: 0.300
Bullet Weight: 62 gr
Initial Velocity: 3240 fps
Sight Height : 1.5 in
Shooting Angle: -33°
Wind Speed: 0 mph
Wind Angle: 0°
Zero Range: 100 yd
Chart Range: 750 yd
Maximum Range: 50002 yd
Step Size: 25 yd
Corrected For Atmosphere
Adjusted BC: 0.33
Altitude: 2000 ft
Barometric Pressure: 29.92 Hg
Temperature: 72° F
Relative Humidity: 21%

RangeTimeVel[x+y]
(ft)(s)(ft/s)
00.003240
750.023163
1500.053088
2250.073014
3000.102941
3750.122870
4500.152799
5250.182730
6000.202662
6750.232595
7500.262529
8250.292465
9000.322401
9750.362337
10500.392275
11250.422214
12000.462154
12750.492095
13500.532036
14250.561979
15000.601923
15750.641867
16500.681813
17250.731760
18000.771708
18750.811658
19500.861609
20250.911561
21000.961515
21751.011470
22501.061426

OOPS!

How'd ya manage to do that, Professor DonkyChan? 


 

VxH  posted on  2017-10-30   22:20:23 ET  Reply   Trace   Private Reply  


#162. To: VxH (#161)

I have one question. You keep changing your numbers. But they still always fit with your pet theory. Why is that

A K A Stone  posted on  2017-10-31   10:06:27 ET  Reply   Trace   Private Reply  


#163. To: A K A Stone, NoluChan (#162) (Edited)

Ballistic deceleration is a non-linear system.  Calculations we're using to model the system are thus only approximations.   Different functions and methods can be used to model the system with varying degrees of approximation

In this case, the data produced by ballistic chart generators like ShooterCalculator.com  can be used as a reference for approximate comparison. For example - the modeled values for Time should approximately correspond to the chart's reference values for Time.

Now ask Professor DonkeyChan why his calculation for time (1.2119 seconds at a distance of 1950ft and a Velocity of 1609fps) is so significantly different from the value for Time (0.86 seconds) in the ballistic chart at the same Distance (1950ft) and Velocity (1609fps)?

4dTime(avg) Vel[x+y]

5(ft)d/Vel[x+y](ft/s)

3219501.21191609

1609fps @ 1950ft would appear to be the INSTANTANIOUS velocity, BTW - not the average Velocity, as Professor DonkeyChan asserts.

VxH  posted on  2017-10-31   12:10:24 ET  Reply   Trace   Private Reply  


#164. To: VxH, A K A Stone (#161)

How'd ya manage to do that, Professor DonkyChan?

I used the precise data you provided and applied the correct formula, d/t = average velocity.

In the case that Column C contains instantaneous velocities, the data is unusable for calculation of average velocity, or to derive the time, to greater accuracy, or any result at all.

If the data in Column C is instantaneous velocity, the only way to calculate the average velocity is d/t, and the result for 75 yards would be 3750. While your chart fails to indicate any rounding has been performed, it is apparent that .02s appears rounded to two digits. Allowing for the rounding error, the result could be anything from 3000 fps to 4999.99 fps.

Note where the Khan Academy stated "your average velocity would be five meters per second, which doesn't necessarily equal the instantaneous velocities at particular points on your trip."

Why are you using instantaneous velocities at particular points to misstate average velocity?

nolu chan  posted on  2017-11-01   2:22:37 ET  Reply   Trace   Private Reply  


#165. To: A K A Stone, VxH (#162)

I have one question. You keep changing your numbers. But they still always fit with your pet theory. Why is that

What pet theory?

In all of his word salad, did he tell you his methodology to calculate the results for Average Velocity appearing in his Column D, or anyone else's methodology to calculate the Average Velocity appearing in his Column D?

[VxH] Ballistic deceleration is a non-linear system. Calculations we're using to model the system are thus only approximations. Different functions and methods can be used to model the system with varying degrees of approximation.

There is nothing quite like an approximation taken to four decimal places.

There is reference to a model system. Have you seen a model system to calculate the results for the Average Velocity in Column D?

Where did this model system come from, and what does this model system say?

Is there a link to a model system to calculate the Average Velocity as performed in Column D? Is there a link to any such system, model, or function?

While "different functions and methods can be used to model the system," have you seen any functions and methods that have been presented?

What you see is VxH or yukon's implementation of his model system to babble and obfuscate.

Assume the Velocities in Column C represent only the bullet velocity at the discrete time the bullet arrives at the Column A distance.

His spreadsheet calculation results indicate that he made up his own formula, which is sum the velocity at 0 feet and the velocity at 75 feet and divide by 2.

Sum the velocity given for one point, and that for another point, and divide by two to get the average velocity between the two.

If that works for 0 and 75 feet, it should work even better for 0 and 1950 feet.

Take velocity at 0 feet (3240 fps) and the velocity at 1950 feet (1609 fps), sum them (4849), divide by 2, and the result is 2424.5 fps average velocity for the range 0 to 1950 feet.

Why is his Column D average velocity (2367.5926) result for 1950 feet 56 fps different?

While "different functions and methods can be used to model the system with varying degrees of approximation," the link goes to a youtube video at the Khan Academy about instantaneous velocities — velocities at a single specific point. The video does not answer the question of how to calculate average velocity. It does sort of tell one how not to calculate average velocity. You do not have to watch it, there is a transcript.

Note where it says, "your average velocity would be five meters per second, which doesn't necessarily equal the instantaneous velocities at particular points on your trip."

If you want instantaneous velocities, and that is what is in Column C, we are not looking for a formula to calculate them. But we cannot use instantaneous velocities to derive average velocity.

The Khan Academy does not say that you can sum two instantaneous velocities and divide by two, and get an average velocity between the two points.

https://www.khanacademy.org/science/physics/one-dimensional-motion/displacement-velocity-time/v/instantaneous-speed-and-velocity

Video transcript

- [Instructor] Pretend you are a physics student. You are just getting out of class. You were walking home when you remembered that there was a Galaxy Wars marathon on tonight, so you'd do what every physics student would do: run. You're pretty motivated to get home, so say you start running at six meters per second. Maybe it's been a while since the last time you ran, so you have to slow down a little bit to two meters per second. When you get a little closer to home, you say: "No, Captain Antares wouldn't give up "and I'm not giving up either", and you start running at eight meters per second and you make it home just in time for the opening music. These numbers are values of the instantaneous speed. The instantaneous speed is the speed of an object at a particular moment in time.

And if you include the direction with that speed, you get the instantaneous velocity. In other words, eight meters per second to the right was the instantaneously velocity of this person at that particular moment in time.

Note that this is different from the average velocity. If your home was 1,000 meters away from school and it took you a total of 200 seconds to get there, your average velocity would be five meters per second, which doesn't necessarily equal the instantaneous velocities at particular points on your trip.

In other words, let's say you jogged 60 meters in a time of 15 seconds. During this time you were speeding up and slowing down and changing your speed at every moment. Regardless of the speeding up or slowing down that took place during this path, your average velocity's still just gonna be four meters per second to the right; or, if you like, positive four meters per second. Say you wanted to know the instantaneous velocity at a particular point in time during this trip. In that case, you'd wanna find a smaller displacement over a shorter time interval that's centered at that point where you're trying to find the instantaneous velocity. This would give you a better value for the instantaneous velocity but it still wouldn't be perfect. In order to better zero-in on the instantaneous velocity, we could choose an even smaller displacement over that even shorter time interval. But we're gonna run into a problem here because if you wanna find a perfect value for the instantaneous velocity, you'd have to take an infinitesimally-small displacement divided by an infinitesimally-small time interval. But that's basically zero divided by zero, and for a long time no one could make any sense of this. In fact, since defining motion at a particular point in time seemed impossible, it made some ancient Greeks question whether motion had any meaning at all. They wondered weather motion was just an illusion. Eventually, Sir Isaac Newton developed a whole new way to do math that lets you figure out answers to these types of questions. Today we call the math that Newton invented calculus. So if you were to ask a physicist:

"What's the formula for the instantaneous velocity?", he or she would probably give you a formula that involves calculus. But, in case some of you haven't taken calculus yet, I'm gonna show you a few ways to find the instantaneous velocity that don't require the use of calculus.

The first way is so simple that it's kind of obvious. If you're lucky enough to have a case where the velocity of an object doesn't change, then the formula for average velocity is just gonna give you the same number as the instantaneous velocity at any point in time.

If your velocity is changing, one way you can find the instantaneous velocity is by looking at the motion on an x-versus-t graph. The slope at any particular point on this position-versus-time graph is gonna equal the instantaneous velocity at that point in time because the slope is gonna give the instantaneous rate at which x is changing with respect to time.

A third way to find the instantaneous velocity is for another special case where the acceleration is constant. If the acceleration is constant, you can use the Kinematic Formulas to find the instantaneous velocity, v, at any time, t. (electronic music)

Video on YouTube

Creative Commons Attribution/Non-Commercial/Share-Alike

They talk about how to find instantaneous velocity. The first example does not apply because velocities are changing. The third example does not apply because the acceleration (or deceleration) is not constant, and the cited Kinematic formulas are used to find instantaneous velocity. For the third example, you need an x-versus-t graph. Seen one lately?

nolu chan  posted on  2017-11-01   2:30:50 ET  Reply   Trace   Private Reply  


#166. To: nolu chan, A K A Stone (#164) (Edited)

>>I used the precise data you provided and applied the correct formula, d/t = average velocity.

That'd be the CORRECT, non-linear, data produced by the Ballistic calculator - which you obviously applied the WRONG, linear, formula to "calculate" time (1.2119 seconds at a distance of 1950ft and a Velocity of 1609fps) which is significantly different from the value for Time  (0.86 seconds) in the ballistic chart at the same Distance (1950ft) and Velocity (1609fps).

ShooterCalculator.com Says:
Drag Function: G1
Ballistic Coefficient: 0.300
Bullet Weight: 62 gr
Initial Velocity: 3240 fps
Sight Height : 1.5 in
Shooting Angle: -33°
Wind Speed: 0 mph
Wind Angle: 0°
Zero Range: 100 yd
Chart Range: 750 yd
Maximum Range: 50002 yd
Step Size: 25 yd
Corrected For Atmosphere
Adjusted BC: 0.33
Altitude: 2000 ft
Barometric Pressure: 29.92 Hg
Temperature: 72° F
Relative Humidity: 21%

19500.861609

========================

Professor DonkeyChan says:

4dTime(avg) Vel[x+y]

5(ft)d/Vel[x+y](ft/s)

3219501.21191609

1.2119 seconds

nolu chan    posted on  2017-10-30   19:58:41 ET

https://libertysflame.com/cgi-bin/readart.cgi? ArtNum=53025&Disp=160#C160

========================

Meanwhile...

http://www.answers.com/Q /What_is_instantaneous_slope

As per what I said in https://libertysflame.com/cgi-bin/readart.cgi? ArtNum=53046&Disp=105#C105 the next step in the quest is to explore methods of deriving Time relative to the slope of the DIFFERENCE between Vmin and Vmax for a given vector segment.

VxH  posted on  2017-11-01   12:00:27 ET  (1 image) Reply   Trace   Private Reply  


#167. To: VxH (#166)

Meanwhile...

http://www.answers.com/Q /What_is_instantaneous_slope

As per what I said in https://libertysflame.com/cgi-bin/readart.cgi? ArtNum=53046&Disp=105#C105 the next step in the quest is to explore methods of deriving Time relative to the slope of the DIFFERENCE between Vmin and Vmax for a given vector segment.

To find what is at your source, we go to the link, which, like the link to the Khan Academy, only shows that you are bullshitting.

You seem to have a special affinity in providing cut and paste bullshit as as some sort of profound knowledge.

http://www.answers.com/Q/What_is_instantaneous_slope

Answer byBlue
Confidence votes 38.3K

The instantaneous slope of a curve is the slope of that curve at a single point. In calculus, this is called the derivative. It also might be called the line tangent to the curve at a point.

If you imagine an arbitrary curve (just any curve) with two points on it (point P and point Q), the slope between P and Q is the slope of the line connecting those two points. This is called a secant line. If you keep P where it is and slide Q closer and closer to P along the curve, the secant line will change slope as it gets smaller and smaller. When Q gets extremely close to P (so that there is an infinitesimal space between P and Q), then the slope of the secant line approximates the slope at P. When we take the limit of that tiny distance as it approaches zero (meaning we make the space disappear) we get the slope of the curve at P. This is the instantaneous slope or the derivative of the curve at P.

Mathematically, we say that the slope at P = limh—>0 [f(x+h) - f(x)]÷h = df/dx, where h is the distance between P and Q, f(x) is the position of P, f(x+h) is the position of Q, and df/dx is the derivative of the curve with respect to x.

The formula above is a specific case where the derivative is in terms of x and we're dealing with two dimensions. In physics, the instantaneous slope (derivative) of a position function is velocity, the derivative of velocity is acceleration, and the derivative of acceleration is jerk.

Of course, the calculus formula P = limh—>0 [f(x+h) - f(x)]÷h = df/dx, where h is the distance between P and Q, f(x) is the position of P, f(x+h) is the position of Q, and df/dx is the derivative of the curve with respect to x was not used anywhere in your spreadsheet, so you are just bullshitting.

Also,

When Q gets extremely close to P (so that there is an infinitesimal space between P and Q), then the slope of the secant line approximates the slope at P. When we take the limit of that tiny distance as it approaches zero (meaning we make the space disappear) we get the slope of the curve at P. This is the instantaneous slope or the derivative of the curve at P.

However, the slope of a bullet in flight is constantly changing, the deceleration is not constant, and the slope contains an infinite number of points.

Moreover, you have merely bullshitted and have not described any formula to obtain the average velocity of the bullet over a given range, using instantaneous velocities.

While you claim calculus formulas in your spreadsheet, you have yet to show a formula to sum changing parts of a spreadsheet column, i.e., sum row 1 and 2, sum row 1 thru 3, then row 1 thru 4, and so forth. I used such a formula and it showed that your column contained arithmetical errors not created by a spreadsheet formula. When you can program adding sums, I'll consider you doing calculus. As it is, you have not demostrated the ability to consistently add two numbers together, which is what you did to to sum that column. You added rows 1 and 2 to get the row 2 total; then you added row 3 to get the row 3 total, and so on, making two errors in 26 rows. You are fortunate it was now a thousand rows on a spreadsheet in a finance office.

The formula for calculating average velocity (d/t) is given by the Khan Academy in the video you referenced. In their example, they divide a distance fo 1000m by 200s and get an average velocity of 5 m/s, and then they explicitly state, that siad result "doesn't necessarily equal the instantaneous velocities at particular points."

The Khan Academy does not say that you can sum two instantaneous velocities and divide by two, and get an average velocity between the two points. See what you referenced. The first sentence is important — pretend you are a physics student.

https://www.khanacademy.org/science/physics/one-dimensional-motion/displacement-velocity-time/v/instantaneous-speed-and-velocity

Video transcript

- [Instructor] Pretend you are a physics student. You are just getting out of class. You were walking home when you remembered that there was a Galaxy Wars marathon on tonight, so you'd do what every physics student would do: run. You're pretty motivated to get home, so say you start running at six meters per second. Maybe it's been a while since the last time you ran, so you have to slow down a little bit to two meters per second. When you get a little closer to home, you say: "No, Captain Antares wouldn't give up "and I'm not giving up either", and you start running at eight meters per second and you make it home just in time for the opening music. These numbers are values of the instantaneous speed. The instantaneous speed is the speed of an object at a particular moment in time.

And if you include the direction with that speed, you get the instantaneous velocity. In other words, eight meters per second to the right was the instantaneously velocity of this person at that particular moment in time.

Note that this is different from the average velocity. If your home was 1,000 meters away from school and it took you a total of 200 seconds to get there, your average velocity would be five meters per second, which doesn't necessarily equal the instantaneous velocities at particular points on your trip.

In other words, let's say you jogged 60 meters in a time of 15 seconds. During this time you were speeding up and slowing down and changing your speed at every moment. Regardless of the speeding up or slowing down that took place during this path, your average velocity's still just gonna be four meters per second to the right; or, if you like, positive four meters per second.

Just as your instantaneous velocity at two discrete and infinitesimal points can not be summed and divided by two to obtain average velocity, the instantaneous slope at two discrete and infinitesimal points will be different and cannot be used to calculate the slope of a traveling bullet whose velocity is contantly changing.

While this bullshit about instantaneous slopes has diverted from your other bullshit about instantaneous velocities, you are still left searching to explain

(1) your calculation used to derive average velocity over the specified distances,

(2) your calculation used to change the formula for calculating average velocity over distance.

Your chart stipulated distance and time.

For 75 feet, you stipulated 0.02 seconds. This is your data, not mine.

Using the formula, d/t=V(avg), that is 3,750 feet per second average velocity.

If we assume that you meant the time to be anything between 0.015 and 0.025 seconds, that is 3000 - 5000 feet per second average velocity.

For 1950 feet, you stipulated 0.86 seconds, and an average velocity of 2367.5926 feet per second, obtained by a formula you can neither present nor explain, nor can you provide any citation to any authority for your bullshit calculation.

V(avg) = d/t = 1950/0.86 = 2267.4418 feet per second average velocity.

If we assume that you meant the time to be anything between 0.855 and 0.865, then,

V(avg) may equal 1950/0.0855 = 2280.7017

V(avg) may equal 1950/0.0865 = 2254.3353

Meanwhile, your bullshit 2367.5926 average velocity allows one to derive the time required to travel 1950 feet. 1950/2367.5926 = 0.823621429 seconds.

Indeed, your second time for Tb, the time of the bullet, in your column E, reflects a bullet flight time of 0.823621 seconds, giving three less decimal points than I did, but rounding the the same precise thing at your chosen four decimal places, indicating how you derived that bullshit Tb from the bullshit average velocity.

To check whether this bullshit time is not impossible with the stipulated data, one need only check if it is within the rounding possibilities of the stipulated data, i.e., from 0.855 to 0.865 seconds. Oh noes, your bullshit average velocity (0.823621) is not possible to reconcile with the stiplulated time, even allowing for the maximum rounding error. Your misbegotten time would round to 0.82 instead of 0.86.

You have yet to explain how you can stipulate a bullet time of 0.86 seconds, and through the magic of VxH formulas, transform that time into 0.823621 seconds, and then use that visibly bullshit time to perform further bullshit calculations.

If the bullet flew 1950 feet in 0.823621, why sure enough it went at an average velocity of 2367.5938 and covered 1950 feet.

However, at the stipulated time of 0.86 seconds, at the bullshit average velocity of 2367.5938 feet per second, the bullet would have flown 2036.1307 feet. The stipulated distance is 1950 feet.

At the maximum rounding down error to 0.855 seconds, at your bullshit average velocity of 2367.5926, the bullet would have flown 2024.292699 feet (0.855 x 2367.5926). The stipulated distance is 1950 feet.

With your stipulated data, you may not have more or less than 1950 feet. You may not have less than 0.855 seconds flight time, nor more than 0.865 seconds flight time. You cannot change the distance the bullet flew, nor do more than consider a rounding error on the time. Your calculated numbers fail miserably.

Your bullshit calculated numbers fall outside the maximum possible error attributable to a rounding error.

Your bullshit calculations result in a new time, not within any rounding error, replacing 0.86 with 0.823621.

Your bullshit average velocity over 1950 feet (2367.5926), at the maximum rounding error for stipulated time (0.86 rounded down to 0.855), requires the bullet to fly a minimum of 2024.292699 feet.

WHY IS YOUR CALCULATED DATA OUTSIDE THE POSSIBLE LIMITS OF A TIME ROUNDING ERROR???

nolu chan  posted on  2017-11-02   20:13:05 ET  Reply   Trace   Private Reply  


#168. To: nolu chan (#167) (Edited)

LOL. Idiot.

 


 

ShooterCalculator.com Says:
Drag Function: G1
Ballistic Coefficient: 0.300
Bullet Weight: 62 gr
Initial Velocity: 3240 fps
Sight Height : 1.5 in
Shooting Angle: -33°
Wind Speed: 0 mph
Wind Angle: 0°
Zero Range: 100 yd
Chart Range: 750 yd
Maximum Range: 50002 yd
Step Size: 25 yd
Corrected For Atmosphere
Adjusted BC: 0.33
Altitude: 2000 ft
Barometric Pressure: 29.92 Hg
Temperature: 72° F
Relative Humidity: 21%

19500.861609

========================

Professor DonkeyChan says:

4dTime(avg) Vel[x+y]

5(ft)d/Vel[x+y](ft/s)

3219501.21191609

1.2119 seconds

nolu chan     posted on  2017-10-30   19:58:41 ET

https://libertysflame.com/cgi-bin/readart.cgi? ArtNum=53025&Disp=160#C160



VxH  posted on  2017-11-03   13:45:22 ET  (1 image) Reply   Trace   Private Reply  


#169. To: nolu chan, A K A Stone, TooConservative (#167) (Edited)

Here is the ballistic data table generated for 1 yard intervals from 1875 to 1950 ft.

http://www.shooterscalculator.com/ballistic-trajectory-chart.php? pl=223+RemingtonLVA&presets=223+RemingtonLVA~%5BChart+Label%5D~G1~0.300~62~3240~ 100~1.5~-33~0~0~true~2000~72~29.92~21~true~1000~25&df=G1&bc=0.300&bw=62& amp; amp;vi=3240&zr=100&sh=1.5&sa=-33&ws=0&wa=0&cfa=on&alt=0& amp; amp;tmp=72&bar=29.92&hum=21&ssb=on&cr=1000&ss=1&chartColumns =Range~ft%3BTime~s%3BVel%5Bx%2By%5D~ft%2Fs&lbl=Test&submitst=+Create+Chart+

Drag Function: G1
Ballistic Coefficient: 0.300
Bullet Weight: 62 gr
Initial Velocity: 3240 fps
Sight Height : 1.5 in
Shooting Angle: -33°

Wind Speed: 0 mph
Wind Angle: 0°
Zero Range: 100 yd
Chart Range: 1000 yd
Maximum Range: 50002 yd
Step Size: 1 yd

Corrected For Atmosphere
Adjusted BC: 0.33
Altitude: 2000 ft
Barometric Pressure: 29.92 Hg
Temperature: 72° F
Relative Humidity: 21% 
Speed of Sound: 1130 fps

 
RangeTimeVel[x+y]
(ft)(s) (ft/s)
18750.811658
18780.811656
18810.821654
18840.821652
18870.821650
18900.821648
18930.821646
18960.831644
18990.831642
19020.831640
19050.831638
19080.831636
19110.831634
19140.841632
19170.841630
19200.841628
19230.841626
19260.841624
19290.851622
19320.851621
19350.851619
19380.851617
19410.851615
19440.861613
19470.861611
19500.861609

Now tell us, Professor DonkeyChan - from the data provided, what is the average Velocity for the 75 ft segment ending at 1950 ft (1950ft, being the point at which, BTW, the instantaneous velocity is 1609fps)?

VxH  posted on  2017-11-03   19:58:08 ET  Reply   Trace   Private Reply  


#170. To: VxH (#169)

Meaningless waste of time that proves nothing.

A K A Stone  posted on  2017-11-03   20:45:10 ET  Reply   Trace   Private Reply  


#171. To: A K A Stone (#170) (Edited)

2+2=4, Winstone.

VxH  posted on  2017-11-04   9:33:48 ET  Reply   Trace   Private Reply  


#172. To: VxH, A K A Stone (#169)

Unresponsive obfuscatory yukonesque bullshit

WHY IS YOUR CALCULATED DATA OUTSIDE THE POSSIBLE LIMITS OF A TIME ROUNDING ERROR???


Distance and time specified Time rounded to plus or minus maximum Avg velocity Avg velocity Avg velocity time for sound ABS Tb - Ts ABS Tb - Ts ABS Tb - Ts
time as given max possible min possible to travel dist time as given max possible min possible
B C D E F G H I J K LN O P
d Time Time -.005 Time +.005 Avg Vel unadj Avg Vel max Avg Vel min t for sound ABS Tdiff unadj ABS Tdiff MAX ABS Tdiff MIN VxH Avg Vel VxH Instant VxH Tdiff
(ft) (seconds) (seconds) b/c b/d b/e b/1130 ABS(c8-i8) ABS(d8-i8) ABS(E8-I8) Velocity Tb-Ts
7 0 0.00 3240
8 75 0.02 0.015 0.025 3750.00 5000.00 3000.0000 0.0664 0.0464 0.0514 0.0414 3201.5000 3163 0.0429
9 150 0.05 0.045 0.055 3000.00 3333.33 2727.2727 0.1327 0.0827 0.0877 0.0777 3163.6667 3088 0.0852
10 225 0.07 0.065 0.075 3214.29 3461.54 3000.0000 0.1991 0.1291 0.1341 0.1241 3126.2500 3014 0.1270
11 300 0.10 0.095 0.105 3000.00 3157.89 2857.1429 0.2655 0.1655 0.1705 0.1605 3089.2000 2941 0.1682
12 375 0.12 0.115 0.125 3125.00 3260.87 3000.0000 0.3319 0.2119 0.2169 0.2069 3052.6667 2870 0.2088
13 450 0.15 0.145 0.155 3000.00 3103.45 2903.2258 0.3982 0.2482 0.2532 0.2432 3016.4286 2799 0.2488
14 525 0.18 0.175 0.185 2916.67 3000.00 2837.8378 0.4646 0.2846 0.2896 0.2796 2980.6250 2730 0.2881
15 600 0.20 0.195 0.205 3000.00 3076.92 2926.8293 0.5310 0.3310 0.3360 0.3260 2945.2222 2662 0.3269
16 675 0.23 0.225 0.235 2934.78 3000.00 2872.3404 0.5973 0.3673 0.3723 0.3623 2910.2000 2595 0.3650
17 750 0.26 0.255 0.265 2884.62 2941.18 2830.1887 0.6637 0.4037 0.4087 0.3987 2875.5455 2529 0.4024
18 825 0.29 0.285 0.295 2844.83 2894.74 2796.6102 0.7301 0.4401 0.4451 0.4351 2841.3333 2465 0.4392
19 900 0.32 0.315 0.325 2812.50 2857.14 2769.2308 0.7965 0.4765 0.4815 0.4715 2807.4615 2401 0.4753
20 975 0.36 0.355 0.365 2708.33 2746.48 2671.2329 0.8628 0.5028 0.5078 0.4978 2773.8571 2337 0.5107
21 1050 0.39 0.385 0.395 2692.31 2727.27 2658.2278 0.9292 0.5392 0.5442 0.5342 2740.6000 2275 0.5454
22 1125 0.42 0.415 0.425 2678.57 2710.84 2647.0588 0.9956 0.5756 0.5806 0.5706 2707.6875 2214 0.5794
23 1200 0.46 0.455 0.465 2608.70 2637.36 2580.6452 1.0619 0.6019 0.6069 0.5969 2675.1176 2154 0.6260
24 1275 0.49 0.485 0.495 2602.04 2628.87 2575.7576 1.1283 0.6383 0.6433 0.6333 2642.8889 2095 0.6451
25 1350 0.53 0.525 0.535 2547.17 2571.43 2523.3645 1.1947 0.6647 0.6697 0.6597 2610.9474 2036 0.6768
26 1425 0.56 0.555 0.565 2544.64 2567.57 2522.1239 1.2611 0.7011 0.7061 0.6961 2579.3500 1979 0.7077
27 1500 0.60 0.595 0.605 2500.00 2521.01 2479.3388 1.3274 0.7274 0.7324 0.7224 2548.0952 1923 0.7378
28 1575 0.64 0.635 0.645 2460.94 2480.31 2441.8605 1.3938 0.7538 0.7588 0.7488 2517.1364 1867 0.7671
29 1650 0.68 0.675 0.685 2426.47 2444.44 2408.7591 1.4602 0.7802 0.7852 0.7752 2486.5217 1813 0.7956
30 1725 0.73 0.725 0.735 2363.01 2379.31 2346.9388 1.5265 0.7965 0.8015 0.7915 2456.2500 1760 0.8232
31 1800 0.77 0.765 0.775 2337.66 2352.94 2322.5806 1.5929 0.8229 0.8279 0.8179 2426.3200 1708 0.8499
32 1875 0.81 0.805 0.815 2314.81 2329.19 2300.6135 1.6593 0.8493 0.8543 0.8443 2395.7692 1658 0.8758
33 1950 0.86 0.855 0.865 2267.44 2280.70 2254.3353 1.7257 0.8657 0.8707 0.8607 2367.5926 1609 0.9008
34 2025 0.91 0.905 0.915 2225.27 2237.57 2213.1148 1.7920 0.8820 0.8870 0.8770
35 2100 0.96 0.955 0.965 2187.50 2198.95 2176.1658 1.8584 0.8984 0.9034 0.8934
36 2175 1.01 1.005 1.015 2153.47 2164.18 2142.8571 1.9248 0.9148 0.9198 0.9098
37 2250 1.06 1.055 1.065 2122.64 2132.70 2112.6761 1.9912 0.9312 0.9362 0.9262




Column B of above spreadsheet shows the specified distance and the specified time for that distance.

Column C shows the specified time for the distance traveled.

Column D shows the time rounded down to the minimum time possibly explained by rounding.

Column E shows the time rounded up to the maximum time possibly explained by rounding.

Column F shows the Average Velocity (d/t) calculated with the unadjusted time from Column C.

Column G shows the Average Velocity (d/t) calculated with the minimum time possible from Column D. This minimum time of flight shows the maximum possible average velocity of the bullet.

Column H shows the Average Velocity (d/t) calculsted with the maximum time possible from Column E. This maximum time of flight show the minimum average velocity of the bullet.

Column I shows the time for sound to travel the distance at 1130 fps.

Column J shows the time difference between the bullet and the sound using unadjusted time from Column C.

Column K shows the maximum possible time difference between the bullet and the sound using the time rounded down in Column D.

Column L shows the minimum possible time difference between the bullet and the sound using the time rounded up in Column E.

Column N states VXH Average Velocity using undisclosed math, presumably of Klingon origin.

Column O states the instantaneous velocities at the distances specified in Column B. These velocities reflect a specific and infinitesimal point it time only, and do not describe velocity at any other point in time.

Comparing Columns H and N, Column H calculates the maximum possible average velocity with the time round down as far is is possible. Column N is the average velocity claimed by VxH, using his secret Klingon mathematics.

Notice his secret method obtain an average velocity well below the maximum possible for 75 feet, but comes nearer to the maximum possible with every calculation, and at 1200 feet his calculations leave the realm of the possible.

At 1200 feet, at the specified time of 0.46 seconds, the average velocity would be 2608.70 feet per second (1200/0.46). Anyone can do the arithmetic. At 1200 feet, at 0.46 seconds rounded down as far as possible to 0.455 seconds (Column D), the maximum average veocity of 2637.36 feet per second (Column G) is achieved (2637.36/0.455). Anyone can still do the arithmetic. At this point, the VxH calculations exceed the possibilities of reality and achieve 2675.1176 feet per second.

After this point, every VxH calculation widens the error.

At 1950 feet, the Column G max average velocity is 2280.70 (1950/0.0855). After more calculations, the VxH error expands the difference to 2367.5926 feet per second.

If carried on to further distances, the error will simply keep increasing. He started with 64% of the maximum possible average velocity, and surpassed 100% of the maximum on his 16th calculation, and continued to surpass the maximum possible average velocity by a greater and greater amount.

Moreover, the distance and the time were a given.

....d ........ t ... Vxh

...ft........ sec .... sec per VxH

75 0.02 0.023427 150 0.05 0.047413 225 0.07 0.071971 300 0.10 0.097113 375 0.12 0.122843 450 0.15 0.149183 525 0.18 0.176138 600 0.20 0.203720 675 0.23 0.231943 750 0.26 0.260820 825 0.29 0.290357 900 0.32 0.320574 975 0.36 0.351496 1050 0.39 0.383128 1125 0.42 0.415484 1200 0.46 0.448578 1275 0.49 0.482427 1350 0.53 0.517054 1425 0.56 0.552465 1500 0.60 0.588675 1575 0.64 0.625711 1650 0.68 0.663578 1725 0.73 0.702290 1800 0.77 0.741864 1875 0.81 0.782303 1950 0.86 0.082621

Notice how VxH, in his calculations, at and after 1200 feet, reduces the time of flight of the bullet by more than any possible amount of rounding from two decimal places.

By 1950 feet, VxH has "rounded off" 0.86 and amazingly reduced the stated flight time to his own preferred 0.82621.

nolu chan  posted on  2017-11-05   1:40:06 ET  Reply   Trace   Private Reply  


#173. To: nolu chan (#172) (Edited)

>>By 1950 feet, VxH has "rounded off" 0.86

Nope. Not a rounding error Super Genius.  It's an artifact manufacted from the AVERAGING curve.

>>preferred 0.82621.

Nope. Try to keep up - - I've moved on with a revised curve that reconstructs time from Velocity and Distance,

As per what I said in https://libertysflame.com/cgi-bin/readart.cgi? ArtNum=53046&Disp=105#C105  the next step in the quest is to explore methods of deriving Time relative to  the slope of the DIFFERENCE between Vmin and Vmax for a given vector segment.

So, Professor DonkeyChan -- Here is the ballistic data table generated for 1 yard intervals from 1875 to 1950 ft.

http://www.shooterscalculator.com/ballistic- trajectory-chart.php? pl=223+RemingtonLVA&presets=223+RemingtonLVA~%5BChart+Label%5D~G1~0.300~62~3240~ 100~1.5~-33~0~0~true~2000~72~29.92~21~true~1000~25&df=G1&bc=0.300&bw=62& amp; amp; amp; amp;vi=3240&zr=100&sh=1.5&sa=-33&ws=0&wa=0&cfa=on&alt=0& amp; amp; amp; amp;tmp=72&bar=29.92&hum=21&ssb=on&cr=1000&ss=1&chartColumns =Range~ft%3BTime~s%3BVel%5Bx%2By%5D~ft%2Fs&lbl=Test&submitst=+Create+Chart+

Drag Function: G1
Ballistic Coefficient: 0.300
Bullet Weight: 62 gr
Initial Velocity: 3240 fps
Sight Height : 1.5 in
Shooting Angle: -33°

Wind Speed: 0 mph
Wind Angle: 0°
Zero Range: 100 yd
Chart Range: 1000 yd
Maximum Range: 50002 yd
Step Size: 1 yd

Corrected For Atmosphere
Adjusted BC: 0.33
Altitude: 2000 ft
Barometric Pressure: 29.92 Hg
Temperature: 72° F
Relative Humidity: 21% 
Speed of Sound: 1130 fps

 
RangeTimeVel[x+y]
(ft)(s) (ft/s)
18750.811658
18780.811656
18810.821654
18840.821652
18870.821650
18900.821648
18930.821646
18960.831644
18990.831642
19020.831640
19050.831638
19080.831636
19110.831634
19140.841632
19170.841630
19200.841628
19230.841626
19260.841624
19290.851622
19320.851621
19350.851619
19380.851617
19410.851615
19440.861613
19470.861611
19500.861609

Now tell us, Professor DonkeyChan - from the data provided, what is the average Velocity for the 75 ft segment ending at 1950 ft (1950ft, being the point at which, BTW, the instantaneous velocity is 1609fps)?

VxH  posted on  2017-11-05   6:36:04 ET  (1 image) Reply   Trace   Private Reply  


#174. To: VxH, A K A Stone (#173)

[VxH # 173] Nope. Try to keep up - - I've moved on with a revised curve that reconstructs time from Velocity and Distance,

You have explored methods which require require rewriting the bullet flight times far beyond any possible rounding error. Reconstructing the given travel time is VxH BULLSHIT.

YOUR CALCULATED DATA IS ALL BULLSHIT, AS ARE YOU

Time and distance are given. Rounding the given time up or down does not help your bullshit work. Your bullshit methodology changes the given 0.86 seconds elapsed time to 0.82 seconds.

There is no valid formula in the world for that.

Time and distance are given data. Average velocity equals distance divided by time. Distance in feet, divideded by time in seconds, yields velocity in feet per second.

WHY IS YOUR CALCULATED DATA OUTSIDE THE POSSIBLE LIMITS OF A TIME ROUNDING ERROR???


Distance and time specified Time rounded to plus or minus maximum Avg velocity Avg velocity Avg velocity time for sound ABS Tb - Ts ABS Tb - Ts ABS Tb - Ts
time as given max possible min possible to travel dist time as given max possible min possible
B C D E F G H I J K LN O P
d Time Time -.005 Time +.005 Avg Vel unadj Avg Vel max Avg Vel min t for sound ABS Tdiff unadj ABS Tdiff MAX ABS Tdiff MIN VxH Avg Vel VxH Instant VxH Tdiff
(ft) (seconds) (seconds) b/c b/d b/e b/1130 ABS(c8-i8) ABS(d8-i8) ABS(E8-I8) Velocity Tb-Ts
7 0 0.00 3240
8 75 0.02 0.015 0.025 3750.00 5000.00 3000.0000 0.0664 0.0464 0.0514 0.0414 3201.5000 3163 0.0429
9 150 0.05 0.045 0.055 3000.00 3333.33 2727.2727 0.1327 0.0827 0.0877 0.0777 3163.6667 3088 0.0852
10 225 0.07 0.065 0.075 3214.29 3461.54 3000.0000 0.1991 0.1291 0.1341 0.1241 3126.2500 3014 0.1270
11 300 0.10 0.095 0.105 3000.00 3157.89 2857.1429 0.2655 0.1655 0.1705 0.1605 3089.2000 2941 0.1682
12 375 0.12 0.115 0.125 3125.00 3260.87 3000.0000 0.3319 0.2119 0.2169 0.2069 3052.6667 2870 0.2088
13 450 0.15 0.145 0.155 3000.00 3103.45 2903.2258 0.3982 0.2482 0.2532 0.2432 3016.4286 2799 0.2488
14 525 0.18 0.175 0.185 2916.67 3000.00 2837.8378 0.4646 0.2846 0.2896 0.2796 2980.6250 2730 0.2881
15 600 0.20 0.195 0.205 3000.00 3076.92 2926.8293 0.5310 0.3310 0.3360 0.3260 2945.2222 2662 0.3269
16 675 0.23 0.225 0.235 2934.78 3000.00 2872.3404 0.5973 0.3673 0.3723 0.3623 2910.2000 2595 0.3650
17 750 0.26 0.255 0.265 2884.62 2941.18 2830.1887 0.6637 0.4037 0.4087 0.3987 2875.5455 2529 0.4024
18 825 0.29 0.285 0.295 2844.83 2894.74 2796.6102 0.7301 0.4401 0.4451 0.4351 2841.3333 2465 0.4392
19 900 0.32 0.315 0.325 2812.50 2857.14 2769.2308 0.7965 0.4765 0.4815 0.4715 2807.4615 2401 0.4753
20 975 0.36 0.355 0.365 2708.33 2746.48 2671.2329 0.8628 0.5028 0.5078 0.4978 2773.8571 2337 0.5107
21 1050 0.39 0.385 0.395 2692.31 2727.27 2658.2278 0.9292 0.5392 0.5442 0.5342 2740.6000 2275 0.5454
22 1125 0.42 0.415 0.425 2678.57 2710.84 2647.0588 0.9956 0.5756 0.5806 0.5706 2707.6875 2214 0.5794
23 1200 0.46 0.455 0.465 2608.70 2637.36 2580.6452 1.0619 0.6019 0.6069 0.5969 2675.1176 2154 0.6260
24 1275 0.49 0.485 0.495 2602.04 2628.87 2575.7576 1.1283 0.6383 0.6433 0.6333 2642.8889 2095 0.6451
25 1350 0.53 0.525 0.535 2547.17 2571.43 2523.3645 1.1947 0.6647 0.6697 0.6597 2610.9474 2036 0.6768
26 1425 0.56 0.555 0.565 2544.64 2567.57 2522.1239 1.2611 0.7011 0.7061 0.6961 2579.3500 1979 0.7077
27 1500 0.60 0.595 0.605 2500.00 2521.01 2479.3388 1.3274 0.7274 0.7324 0.7224 2548.0952 1923 0.7378
28 1575 0.64 0.635 0.645 2460.94 2480.31 2441.8605 1.3938 0.7538 0.7588 0.7488 2517.1364 1867 0.7671
29 1650 0.68 0.675 0.685 2426.47 2444.44 2408.7591 1.4602 0.7802 0.7852 0.7752 2486.5217 1813 0.7956
30 1725 0.73 0.725 0.735 2363.01 2379.31 2346.9388 1.5265 0.7965 0.8015 0.7915 2456.2500 1760 0.8232
31 1800 0.77 0.765 0.775 2337.66 2352.94 2322.5806 1.5929 0.8229 0.8279 0.8179 2426.3200 1708 0.8499
32 1875 0.81 0.805 0.815 2314.81 2329.19 2300.6135 1.6593 0.8493 0.8543 0.8443 2395.7692 1658 0.8758
33 1950 0.86 0.855 0.865 2267.44 2280.70 2254.3353 1.7257 0.8657 0.8707 0.8607 2367.5926 1609 0.9008
34 2025 0.91 0.905 0.915 2225.27 2237.57 2213.1148 1.7920 0.8820 0.8870 0.8770
35 2100 0.96 0.955 0.965 2187.50 2198.95 2176.1658 1.8584 0.8984 0.9034 0.8934
36 2175 1.01 1.005 1.015 2153.47 2164.18 2142.8571 1.9248 0.9148 0.9198 0.9098
37 2250 1.06 1.055 1.065 2122.64 2132.70 2112.6761 1.9912 0.9312 0.9362 0.9262




Column B of above spreadsheet shows the specified distance and the specified time for that distance.

Column C shows the specified time for the distance traveled.

Column D shows the time rounded down to the minimum time possibly explained by rounding.

Column E shows the time rounded up to the maximum time possibly explained by rounding.

Column F shows the Average Velocity (d/t) calculated with the unadjusted time from Column C.

Column G shows the Average Velocity (d/t) calculated with the minimum time possible from Column D. This minimum time of flight shows the maximum possible average velocity of the bullet.

Column H shows the Average Velocity (d/t) calculsted with the maximum time possible from Column E. This maximum time of flight show the minimum average velocity of the bullet.

Column I shows the time for sound to travel the distance at 1130 fps.

Column J shows the time difference between the bullet and the sound using unadjusted time from Column C.

Column K shows the maximum possible time difference between the bullet and the sound using the time rounded down in Column D.

Column L shows the minimum possible time difference between the bullet and the sound using the time rounded up in Column E.

Column N states VXH Average Velocity using undisclosed math, presumably of Klingon origin.

Column O states the instantaneous velocities at the distances specified in Column B. There velocities reflect a specific and infinitesimal point it time only, and do not describe velocity at any other point in time.

Comparing Columns H and N, Column H calculates the maximum possible average velocity with the time round down as far is is possible. Column N is the average velocity claimed by VxH, using his secret Klingon mathematics.

Notice his secret method obtain an average velocity well below the maximum possible for 75 feet, but comes nearer to the maximum possible with every calculation, and at 1200 feet his calculations leave the realm of the possible.

At 1200 feet, at the specified time of 0.46 seconds, the average velocity would be 2608.70 feet per second (1200/0.46). Anyone can do the arithmetic. At 1200 feet, at 0.46 seconds rounded down as far as possible to 0.455 seconds (Column D), the maximum average veocity of 2637.36 feet per second (Column G) is achieved (2637.36/0.455). Anyone can still do the arithmetic. At this point, the VxH calculations exceed the possibilities of reality and achieve 2675.1176 feet per second.

After this point, every VxH calculation widens the error.

At 1950 feet, the Column G max average velocity is 2280.70 (1950/0.0855). After more calculations, the VxH error expands the difference to 2367.5926 feet per second.

If carried on to further distances, the error will simply keep increasing. He started with 64% of the maximum possible average velocity, and surpassed 100% of the maximum on his 16th calculation, and continued to surpass the maximum possible average velocity by a greater and greater amount.

Moreover, the distance and the time were a given.

d ........ given..........Vxh bullshit

ft.......... sec..... sec

75 0.02 0.023427 150 0.05 0.047413 225 0.07 0.071971 300 0.10 0.097113 375 0.12 0.122843 450 0.15 0.149183 525 0.18 0.176138 600 0.20 0.203720 675 0.23 0.231943 750 0.26 0.260820 825 0.29 0.290357 900 0.32 0.320574 975 0.36 0.351496 1050 0.39 0.383128 1125 0.42 0.415484 1200 0.46 0.448578 1275 0.49 0.482427 1350 0.53 0.517054 1425 0.56 0.552465 1500 0.60 0.588675 1575 0.64 0.625711 1650 0.68 0.663578 1725 0.73 0.702290 1800 0.77 0.741864 1875 0.81 0.782303 1950 0.86 0.082621

Notice how VxH, in his calculations, at and after 1200 feet, reduces the time of flight of the bullet by more than any possible amount of rounding from two decimal places.

By 1950 feet, VxH has "rounded off" 0.86 and amazingly calculated, by secret methodology, the stated flight time to his own preferred 0.82621.

That is VxH bullshit. Not only wrong but impossible on its face.

[VxH]

    1950	0.86	1609

Now tell us, Professor DonkeyChan - from the data provided, what is the average Velocity for the 75 ft segment ending at 1950 ft (1950ft, being the point at which, BTW, the instantaneous velocity is 1609fps)?

The average velocity of any object covering 1950 feet in 0.86 seconds is 1950/0.86 = 2267.4419 feet per second. It could be a flying refrigerator. If it goes 1950 feet in 0.86 seconds, the average velocity is 2267.4419 seconds.

The object could have sped up and slowed down between 0 and 1950 feet in any manner and it makes no difference. If the object covers the 1950 feet in 0.86 seconds, the average velocity for the 1950 foot distance is 2267.4419 seconds.

Recall the Khan Academy video you previously referenced:

https://www.khanacademy.org/science/physics/one-dimensional-motion/displacement-velocity-time/v/instantaneous-speed-and-velocity

Video transcript

- [Instructor] Pretend you are a physics student. You are just getting out of class. You were walking home when you remembered that there was a Galaxy Wars marathon on tonight, so you'd do what every physics student would do: run. You're pretty motivated to get home, so say you start running at six meters per second. Maybe it's been a while since the last time you ran, so you have to slow down a little bit to two meters per second. When you get a little closer to home, you say: "No, Captain Antares wouldn't give up "and I'm not giving up either", and you start running at eight meters per second and you make it home just in time for the opening music. These numbers are values of the instantaneous speed. The instantaneous speed is the speed of an object at a particular moment in time.

And if you include the direction with that speed, you get the instantaneous velocity. In other words, eight meters per second to the right was the instantaneously velocity of this person at that particular moment in time.

Note that this is different from the average velocity. If your home was 1,000 meters away from school and it took you a total of 200 seconds to get there, your average velocity would be five meters per second, which doesn't necessarily equal the instantaneous velocities at particular points on your trip.

In other words, let's say you jogged 60 meters in a time of 15 seconds. During this time you were speeding up and slowing down and changing your speed at every moment. Regardless of the speeding up or slowing down that took place during this path, your average velocity's still just gonna be four meters per second to the right; or, if you like, positive four meters per second.

[snip]

The instantaneous velocity at 1950 feet is irrelevant to the calculation of the average velocity over the range 0 to 1950 feet.

nolu chan  posted on  2017-11-06   16:37:54 ET  Reply   Trace   Private Reply  


#175. To: nolu chan (#174)

beyond any possible rounding error

LOL. It's not a rounding error Professor DonkeyChan.

How you coming along with that Average Velocity for the 75 foot segment ending at 1950 ft?

VxH  posted on  2017-11-06   17:39:24 ET  Reply   Trace   Private Reply  


#176. To: VxH (#175)

How you coming along with that Average Velocity for the 75 foot segment ending at 1950 ft?

It came along quite well. Anything that travels 1950 feet in 1.06 seconds travels an average velocity of 2122.64 feet per second. The formula is distance divided by time.

How are you coming along with your bullet going splat at ~520 feet ground distance from Mandalay Bay?

How did you work out that negative 33º angle?

Side a represents the vertical height of Paddock's vantage point. At the 32nd floor, and at 10.9 feet per floor, (32-1) x 10.9 = 338 feet.

The VxH specified shooting angle was -33°. This should probably be expressed as a positive angle of declination. Not all ballistic calculators will even accept a negative angle value, but specify 0 to 90 degrees.

For another calculator, see:

http://gundata.org/blog/post/223-ballistics-chart/

It appears that VxH drew an imaginary horizontal line d at a vertical height of 338 feet from the ground, and an imaginary 338 foot line e down to the ground, bringing into view a rectangle with a mirror image triangle to that above.

VxH guessed 33º as the acute angle formed at the junction of sides c and imaginary side d at point B. VxH guessed very wrongly.

With a specified shooting angle of 33º at the junction of lines c and d, the angle made by sides c and b would also be 33º, and angle ß, made by sides a and c would be 57º. (The right angle at point A is 90º. The other two angles must add up to 90º.)

With side a being 338 feet, side b would be 520.4743578 feet, and side c would be 620.5944 feet.

As may be seen, disregarding gravity, if the bullet flew downward at the specified 33º from a height of 338 feet, it would fly a straight line of sight path into the ground at ~520 feet from the Mandalay Bay at ground level.

Calculating the bullet velocity after that point may be difficult, even with secret Klingon math.

nolu chan  posted on  2017-11-07   16:14:00 ET  (1 image) Reply   Trace   Private Reply  


#177. To: nolu chan (#176)

>>The formula is distance divided by time.

And when you don't have data for time with enough precision - what then, professor Donkeychan?

http://www.shooterscalculator.com/ballistic- trajectory-chart.php? pl=223+RemingtonLVA&presets=223+RemingtonLVA~%5BChart+Label%5D~G1~0.300~ 62~3240~ 100~1.5~-33~0~0~true~2000~72~29.92~21~true~1000~25&df=G1&bc=0.300&am p;bw=62& amp; amp; amp; amp;vi=3240&zr=100&sh=1.5&sa=-33&ws=0&wa=0&cfa=on&am p;alt=0& amp; amp; amp; amp;tmp=72&bar=29.92&hum=21&ssb=on&cr=1000&ss=1&char tColumns =Range~ft%3BTime~s%3BVel%5Bx%2By%5D~ft%2Fs&lbl=Test&submitst=+Create +Chart+

Drag Function: G1
Ballistic Coefficient: 0.300
Bullet Weight: 62 gr
Initial Velocity: 3240 fps
Sight Height : 1.5 in
Shooting Angle: -33°

Wind Speed: 0 mph
Wind Angle: 0°
Zero Range: 100 yd
Chart Range: 1000 yd
Maximum Range: 50002 yd
Step Size: 1 yd

Corrected For Atmosphere
Adjusted BC: 0.33
Altitude: 2000 ft
Barometric Pressure: 29.92 Hg
Temperature: 72° F
Relative Humidity: 21%  
Speed of Sound: 1130 fps

 
RangeTimeVel[x+y]
(ft)(s)(ft/s)
18750.811658
18780.811656
18810.821654
18840.821652
18870.821650
18900.821648
18930.821646
18960.831644
18990.831642
19020.831640
19050.831638
19080.831636
19110.831634
19140.841632
19170.841630
19200.841628
19230.841626
19260.841624
19290.851622
19320.851621
19350.851619
19380.851617
19410.851615
19440.861613
19470.861611
19500.861609

VxH  posted on  2017-11-07   16:38:52 ET  Reply   Trace   Private Reply  


#178. To: nolu chan (#176)

This should probably be expressed as a positive angle of declination.

BZZZT! Probably not, since using 0.33 instead of -0.33 produces a slower velocity:

(0.33) 1950 0.89 1489

(-0.33) 1950 0.86 1609

VxH  posted on  2017-11-07   17:11:24 ET  Reply   Trace   Private Reply  


#179. To: nolu chan (#176) (Edited)

It appears that VxH drew an imaginary horizontal line d at a vertical height of 338 feet from the ground, and an imaginary 338 foot line e down to the ground, bringing into view a rectangle with a mirror image triangle to that above.

VxH guessed 33º as the acute angle formed at the junction of sides c and imaginary side d at point B. VxH guessed very wrongly.

Yep. You're almost right - I've been meaning to take another look at the angle with a model in Google Sketchup... which says At 1290' from a height of 338' is -14.7

so...

But the reconstructed time could still more more accurate. Any luck figuring that CURVE out?

VxH  posted on  2017-11-07   18:56:55 ET  (2 images) Reply   Trace   Private Reply  


#180. To: VxH (#178)

This should probably be expressed as a positive angle of declination.

BZZZT! Probably not, since using 0.33 instead of -0.33 produces a slower velocity:

There is no 33 degree angle involved. Using your trajectory, Paddock would have come closer to shooting off his big toe than hitting anywhere in the festival venue.

Learn to read:

Declination is downward. Negative 33 degrees inclination is the same as 33 degrees declination. It is the difference between your preferred -33 degrees upward and 33 degrees downward.

33 degrees downward points down. Negative 33 degrees downward points up.

It is like walking negative 33 feet east is walking 33 feet west.

The VxH specified shooting angle was -33°. This should probably be expressed as a positive angle of declination. Not all ballistic calculators will even accept a negative angle value, but specify 0 to 90 degrees.

For another calculator, see:

http://gundata.org/blog/post/223-ballistics-chart/

The gundata entry specifies "Shooting angle (0..89)." Stick a negative value in there and it will reject.

As there is no such thing as a triangle with negative angles, the values for every angle of a triangle are positive values.

In a triangle with sides -3 and -4, a2 + b2 = c2 would yield a hypotenuse of positive 5.

BZZZT! Probably not, since using 0.33 instead of -0.33 produces a slower velocity:

(0.33) 1950 0.89 1489

(-0.33) 1950 0.86 1609

The data available at the links shows the only data change between my two charts is one is -33 and the other is 33. Not so for the chart you just created with yet another time calculation. You changed the input data.

http://www.shooterscalculator.com/ballistic-trajectory-chart.php?t=75dcb734

-33 degrees

http://www.shooterscalculator.com/images/trajectory/ballistic_trajectory_chart_75dcb734.png

This chart made with the original -33 data, shows the bullet rise above the original altitude, and remain above that altitude, for over 300 yards. This is an amazing feat for a bullet shot on a steep downward angle.

- - - - - - - - - -

http://www.shooterscalculator.com/ballistic-trajectory-chart.php?t=fd15e1b9

33 degrees

http://www.shooterscalculator.com/images/trajectory/ballistic_trajectory_chart_fd15e1b9.png

Amazingly, when the angle is changed from -33 to 33, the flight path does not change, according to your calculator.

- - - - - - - - - -

When your calculator permits you to enter ridiculous data, it provides you with ridiculous results.

As one may observe, it provides the precise same flight trajectory, whether at 33 or -33. When fired at 33 degrees downward, the bullet goes upward and remains above the original location for over 300 yards.

The data available at the links shows the only data change to be -33 to 33.

The time for 650 yards/1950 feet is 0.86 in either instance. The average velocity for any object that travels 1950 feet in 0.86 seconds is 1950/0.86 = 2267.4418 ft/sec.

- - - - - - - - - -

YOUR DATA WITH A NEW TIME SHOWS YOU ALTERED THE INPUT DATA.

And your flight trajectory chart for -33 degrees featuring a new time

http://www.shooterscalculator.com/ballistic-trajectory-chart.php?t=81e7edd6

http://www.shooterscalculator.com/images/trajectory/ballistic_trajectory_chart_81e7edd6.png

Notice that with the projectile purported fired at a 33 degree downward angle, the projectile maintains its original altitude for about 150 yards.

Of course, in your derivation of data, you changed the properties of the chart:

Drag Function: G1
Ballistic Coefficient: 0.300
Bullet Weight: 62 gr
Initial Velocity: 3240 fps
Sight Height : 1.5 in
Shooting Angle: -33°
Wind Speed: 0 mph
Wind Angle: 0°
Zero Range: 100 yd [Was 25 yd/75 feet]
Chart Range: 1000 yd[Was 750 yd]
Maximum Range: 50002 yd [Was 750 yd/2250 ft]
Step Size: 1 yd [Was 25 yd/75 ft]

Corrected For Atmosphere
Adjusted BC: 0.307 [was 0.300]
Altitude: 0 ft [was 2000 ft]
Barometric Pressure: 29.92 Hg
Temperature: 72° F
Relative Humidity: 21%
Speed of Sound: 1130 fps

- - - - - - - - - - - - - - - - - - - -

To see the prior input data, just look:

As for your magic ability to fire rounds down at 33º, and have them either rise or maintain altitude for hundreds of yards,

nolu chan  posted on  2017-11-08   23:50:49 ET  (4 images) Reply   Trace   Private Reply  


#181. To: VxH, A K A Stone (#179)

VxH guessed 33º as the acute angle formed at the junction of sides c and imaginary side d at point B. VxH guessed very wrongly.

Yep. You're almost right - I've been meaning to take another look at the angle with a model in Google Sketchup... which says At 1290' from a height of 338' is -14.7

Damn, you are more screwed up than I thought.

Now you are presenting a triangle with side a being 338 feet, side b being 1290 feet 7 inches (1290.5833 ft), and an hypotenuse of 1009 ft 4 in. I positively can't remember the last triangle I saw where the hypotenuse was shorter than one of the sides. Your Klingon math is magic, and Google Sketchup is a miracle worker.

If side a is 338 ft, and the angle of elevation of 14.7º rises to that height of 338 ft, side b will be 1264.283557 feet. So, the hypotenuse is impossible, and side b is whack by 26 feet.

If the angle of elevation is 14.7 degrees, and the range is 1290' 7", then the height would be 345.0311072 feet and each floor would be over 11 feet.

Put down Google sketch and pick up a scientific calculator and do some trigonometry. You can actually get correct results with trig.

You are supposed to be dazzling me with your math skills, not some shit like Google Sketchup.

Google Sketchup? Really?

nolu chan  posted on  2017-11-08   23:52:54 ET  (1 image) Reply   Trace   Private Reply  


#182. To: VxH, A K A Stone (#177)

The formula is distance divided by time.

And when you don't have data for time with enough precision - what then, professor Donkeychan?

By the way, if the given time and distance data was imprecise, what was used to calculate the instantaneous velocities?

[VxH at #175] LOL. It's not a rounding error Professor DonkeyChan.

If the given data is not subject to a rounding error, then it is totally accurate and precise, distance/time works fine, and your bogus calculated data is bullshit.

If the given data is subject to a rounding error, then it is accurate ± 0.005, in which case using distance/time works to define the possible range of average velocity within the rounding factor, and this also proves your bogus calculated data is bullshit.

nolu chan  posted on  2017-11-08   23:54:13 ET  Reply   Trace   Private Reply  


#183. To: nolu chan (#182)

[VxH at #175] LOL. It's not a rounding error Professor DonkeyChan.

If the given data is not subject to a rounding error, then it is totally accurate and precise, distance/time works fine, and your bogus calculated data is bullshit.

It's still not a rounding error, Super Genius. It's a result of the AVERAGING calculation that was used in the context of Time data that was truncated to an insufficiently precise (for the purpose of determining average velocity) 2 decimal points.

=SUM(C9:C10)/L9
=SUM(C$9:C11)/L10
=SUM(C$9:C12)/L11
etc. 
Where column L contains 1 @ row 8 and =+ L8+1,  =+L9+1 etc for rows 9..34

VxH  posted on  2017-11-09   12:34:42 ET  Reply   Trace   Private Reply  


#184. To: nolu chan (#181) (Edited)

and an hypotenuse of 1009 ft 4 in.

Pssst. Professor DonkeyChan.

1009 ft 4 in is the distance from the radius of that circle.

LOL. FAIL again vociferously pretentious NoluBUFFOON!

VxH  posted on  2017-11-09   12:37:05 ET  Reply   Trace   Private Reply  


#185. To: nolu chan, A K A Stone (#181) (Edited)

LOL. Do the Math, Professor DonkeyDung.

I didn't rescale the circle for the purpose of determining the angle. So - No bigee with the radius 324' 6" - the dimensions still add up for the hypotenuse.

VxH  posted on  2017-11-09   13:26:55 ET  (1 image) Reply   Trace   Private Reply  


#186. To: nolu chan (#181) (Edited)

Google Sketchup? Really?

Really!

VxH  posted on  2017-11-10   12:20:27 ET  Reply   Trace   Private Reply  


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