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Title: Once Again, President Trump Is Magnificently Right—This Time About Russia
Source: PJ Media
URL Source: https://pjmedia.com/spengler/once-a ... -right-this-time-about-russia/
Published: Jul 16, 2018
Author: David P. Goldman
Post Date: 2018-07-16 19:51:31 by nolu chan
Keywords: None
Views: 8537
Comments: 69

Once Again, President Trump Is Magnificently Right—This Time About Russia

By David P. Goldman
PJ Media
July 16, 2018

President Trump offended the entire political spectrum with a tweet this morning blaming the U.S. for poor relations with Russia. “Our relationship with Russia has NEVER been worse thanks to many years of U.S. foolishness and stupidity,” the president said, and he is entirely correct. By this I do not mean to say that Russia is a beneficent actor in world affairs or that President Putin is an admirable world leader. Nonetheless, the president displayed both perspicacity and political courage when he pointed the finger at the United States for mismanaging the relationship with Russia.

Full disclosure: I was a card-carrying member of the neoconservative cabal that planned to bring Western-style democracy and free markets to Russia after the fall of Communism. As chief economist for the supply-side consulting firm Polyconomics, I got an appointment as an adviser to Boris Yeltsin’s finance ministry and made several trips to Moscow. Of course, the finance ministry really was a family office for Yeltsin’s oligarch friends, who were too busy stealing Russia’s economy to listen to advice. The experience cured me of the neoconservative delusion that democracy and free markets are the natural order of things.

Unfortunately, the delusion that the United States would remake Russia in its own image persisted through the Bush and Obama administrations. I have no reason to doubt the allegations that a dozen Russian intelligence officers meddled in the U.S. elections of 2016, but this was equivalent of a fraternity prank compared to America’s longstanding efforts to intervene in Russian politics.

The United States supported the 2014 Maidan uprising in Ukraine and the overthrow of the Yanukovych government in the hope of repeating the exercise in Moscow sometime later. Then-Assistant Secretary of State Victoria Nuland pulled whatever strings America had to replace the feckless and corrupt Victor Yanukovych with a government hostile to the Kremlin. She didn’t say it in so many words, but she hoped the Ukraine coup would lead to the overthrow of Vladimir Putin. Evidently Nuland and her boss, Hillary Clinton, thought that the Ukraine coup would deprive Russia of its Black Sea naval base in Crimea, and did not anticipate that Russia simply would annex an old Russian province that belonged to Ukraine by historical accident.

[...]

The Maidan coup was the second American attempt to install a Ukrainian government hostile to Moscow; the first occurred in 2004, when Condoleezza Rice was secretary of State rather than Hillary Clinton. As I wrote in Asia Times a decade ago, “On the night of November 22, 2004, then-Russian president - now premier - Vladimir Putin watched the television news in his dacha near Moscow. People who were with Putin that night report his anger and disbelief at the unfolding 'Orange' revolution in Ukraine. ‘They lied to me,’ Putin said bitterly of the United States. ‘I'll never trust them again.’ The Russians still can't fathom why the West threw over a potential strategic alliance for Ukraine. They underestimate the stupidity of the West."

[snip]

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Comments (1-28) not displayed.
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#29. To: VxH (#28)

nolu chan  posted on  2018-07-17   15:20:45 ET  Reply   Trace   Private Reply  


#30. To: nolu chan (#29)
(Edited)

What time the herd legalize sodomy, again, in New Bolostan?

1963.

VxH  posted on  2018-07-17   15:33:59 ET  (1 image) Reply   Trace   Private Reply  


#31. To: VxH (#15)

I see you are still incapable of using the Google to find Ivan Petrovich Sidorov and still have no clue who it is. Pathetic.

nolu chan  posted on  2018-07-17   23:46:47 ET  Reply   Trace   Private Reply  


#32. To: nolu chan (#31)

I see you are still incapable of using the Google to find Ivan Petrovich Sidorov

Well wire up that calculator and show us how it's done, Igor!

VxH  posted on  2018-07-18   12:58:34 ET  Reply   Trace   Private Reply  


#33. To: VxH (#32)

Well wire up that calculator and show us how it's done, Igor!

Your apparent use of a calculator to search the internet may explain your continuing inability to identify Ivan Petrovich Sidorov.

Here's an Ivan Petrovich Sidorov that your calculator failed to find:

https://www.facebook.com/people/Ivan-Petrovich-Sidorov/100004557936311

nolu chan  posted on  2018-07-18   16:19:47 ET  Reply   Trace   Private Reply  


#34. To: nolu chan (#33) (Edited)

Yawn. A FB link with no content. Is he as Current as Vlad Putin?


"I thought about something just now:

The decision to nationalize this library was made by the first Soviet government, whose composition was 80-85 percent Jewish,"

Putin said June 13 during a visit to Moscow’s Jewish Museum and Tolerance Center.

https://www.youtube.com/watch?v=-pDtgWUtdUM

VxH  posted on  2018-07-19   13:17:28 ET  (1 image) Reply   Trace   Private Reply  


#35. To: VxH (#34)

Yawn. A FB link with no content. Is he as Current as Vlad Putin?

What a stupid comment. Ivan Petrovich Sidorov is always current. He may be one of the most cited people in Russian litigation. You may be one of the least competent users of Google ever.

nolu chan  posted on  2018-07-19   15:00:52 ET  Reply   Trace   Private Reply  


#36. To: VxH (#34)

nolu chan  posted on  2018-07-19   15:01:57 ET  Reply   Trace   Private Reply  


#37. To: nolu chan (#35) (Edited)

He may be one of the most cited people in Russian litigation.

And yet with regards to his/your alleged "debunking" of Golitsyn...

LOL. FAIL.

VxH  posted on  2018-07-19   16:20:27 ET  (1 image) Reply   Trace   Private Reply  


#38. To: VxH (#37)

And yet with regards to his/your alleged "debunking" of Golitsyn...

Pearce, Spymaster (2017) at p. 283:

Even when Golitsyn's claims that the Sino-Soviet splid had been a cunning ruse to foil the West were comprehensively debunked — something that should have sent alarm bells ringing — Angleton and his followers agreed that everything else must still be true. They'd invested their whole reputations in their pet defector.

- - - - - - - - - -

LOL. FAIL.

)))

Still too inept to effectively use the Google.

You appear to make pretty, if absurd, graphics with Google sketchup, even featuring a triangle with a side longer than the hypotenuse. You are a genius.

nolu chan  posted on  2018-07-19   23:40:54 ET  Reply   Trace   Private Reply  


#39. To: nolu chan (#38)

You appear to make pretty, if absurd, graphics with Google sketchup, even featuring a triangle with a side longer than the hypotenuse. You are a genius.

.

Do the math, Donkey Breath.

 

https://libertysflame.com/cgi- bin/readart.cgi?ArtNum=53025&Disp=186#C186

VxH  posted on  2018-07-20   9:13:33 ET  (1 image) Reply   Trace   Private Reply  


#40. To: VxH (#39)

)))

Still too inept to effectively use the Google.

You appear to make pretty, if absurd, graphics with Google sketchup, even featuring a triangle with a side longer than the hypotenuse. You are a genius.

https://libertysflame.com/cgi-bin/readart.cgi?ArtNum=53025&Disp=176#C176


#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

- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

https://libertysflame.com/cgi-bin/readart.cgi?ArtNum=53025&Disp=181#C181

#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  

nolu chan  posted on  2018-07-20   15:28:24 ET  (2 images) Reply   Trace   Private Reply  


#41. To: nolu chan (#29)

Hilarious.

Now that's clever.

Liberator  posted on  2018-07-20   15:33:28 ET  Reply   Trace   Private Reply  


#42. To: nolu chan (#40)

Do the math, Donkey Breath.

 

https://libertysflame.com/cgi- bin/readart.cgi? ArtNum=53025&Disp=186#C186

VxH  posted on  2018-07-20   18:04:06 ET  (1 image) Reply   Trace   Private Reply  


#43. To: VxH (#42)

Any right triangle must have sides and angles where the Sine, Cosine, Tangent, C-Tangent, Secand, and Co-secant correctly compute.

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.

Please do demonstrate the trigonometry behind your childish Google Sketchup nonsense.

Do the math. I will get you started.

Sine
SOH - opposite/hypotenuse

Cosine
CAH - adjacent/hypotenuse

Tangent
TOA - opposite/adjacent

Co-tangent
CAO - adjacent/opposite

Secant
SHA - hypotenuse/adjacent

Co-Secant
SHO - hypotenuse/opposite

You appear to make pretty, if absurd, graphics with Google sketchup, even featuring a triangle with a side longer than the hypotenuse. You are a genius.

https://libertysflame.com/cgi-bin/readart.cgi?ArtNum=53025&Disp=176#C176


#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

- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

https://libertysflame.com/cgi-bin/readart.cgi?ArtNum=53025&Disp=181#C181

#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  

nolu chan  posted on  2018-07-20   18:21:34 ET  (2 images) Reply   Trace   Private Reply  


#44. To: nolu chan (#43)

Do the math, Donkey Breath.

 

https://libertysflame.com/cgi- bin/readart.cgi? ArtNum=53025&Disp=186#C186

VxH  posted on  2018-07-20   19:25:22 ET  (1 image) Reply   Trace   Private Reply  


#45. To: VxH (#44)

You do the math.

- - - - - - - - - -

- - - - - - - - - -

- - - - - - - - - -

As soon as you publish your proof for your mathematical formula c < b.

Even a complete jackass, such as yourself, can see from your original play picture, from your #181 to which I linked, that you showed a hypotenuse of 1009' 4".

Your original play picture shows part of a circle with a radius of 338'.

Your second play picture displays the same radius of the circle as 338' on an up/down line. On a somewhat downward slanting horizontal line, you added a new measurement of the radius of that same circle at 324' 6". How the radius shrinks on one side by 11' 6" is not explained. It's a magic circle, or magic math.

1009' 4" + 338' is 1347' 4".

1009' 4" + 324' 6" is 1333' 10".

Keep changing that side radius distance, make believe it was part of the original hypotenuse, until you get your math to work.

What the magic circle with the changing radius is doing there is a mystery. The difference between level travel to end point and travel from elevation to end point is c - b, not the vertical distance of the elevation.

The 338' up/down radius of the circle indicates the elevation of the shooter, and the circle indicates you had an irresistable impulse to draw a circle and make believe the radius was something other than 338' at another angle. 338' was a stated distance of elevation. 324' 6" appears to be a figure plucked from your ass.

Given angles A (14.70°) and C (90.00°) and side a (338') —
side B and angles b and c may be readily calculated —
just not to the values you give them.

A = 14.70°
B = ?º
C = 90.00°

a = 338.00 ft
b = ? ft
c = ? ft

A + B + C = 180º
A + B = 90º
B = 90-Aº
B = 75.30º

b = sin(B) * a / sin(A)

c = sin(C) * a

nolu chan  posted on  2018-07-20   22:03:16 ET  (3 images) Reply   Trace   Private Reply  


#46. To: nolu chan (#45)

libertysflame.com/cgi-bin...? ArtNum=56433&Disp=46#C46

VxH  posted on  2018-07-20   23:14:41 ET  Reply   Trace   Private Reply  


#47. To: VxH (#46)

Hear ya go, Donkey Breath:

 
A338114244
B1290.58333333331665605.34027778
A^2+B^2
1779849.34027778


1334.1099430998



324.5+1009.33=1333.83
 


"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."

WRONG.

Your claim of a radius of 324' 6" is just bullshit pulled out of your ass.

Your incorrect hypotenuse is still 1009' 4" + the actual radius length 338' = 1347' 4". Your newly minted hypotenuse of 1333.83 feet, using data pulled from your ass, is still WRONG.

If you change the circle to a 324' 6" radius, the 1009' 4" section of the hypotenuse would need to expand 13' 6" to 1022' 10" to reach the radius circle.

In any case, I gave you the formulas and your "calculations" leave out any calculations and all your results are wrong.

You still have both 338 and 324' 6" in your number puzzle.

On the first line you list 338 and 114244. Nobody asked you what the square of 338 is.

On the second line you list 1290.5833333333 and 1665605.34027778. Nobody asked you what the square of 1290' 7" is.

You were asked to solve for side b, not pull the figure 1290' 7" out of your ass and square it.

Use the trig function and discover that the figure 1290' 7" is mathematically impossible. An angle of elevation of 14.7° does not reach an elevation of 338' at a distance of 1290' 7". I gave you the formula. I can't help it if you are too dumb, stupid, ignorant, and incompetent to use the formula. Pulling 1290 feet out of your ass is not a mathematical solution to side b.

On your third line you list the sum of 338 squared and 1290' 7" squared.

This would be the hypotenuse if you had properly solved for side b using the trig formula b = sin(B) * a / sin(A). That does not give 1290' 7".

The hypotenuse, solved with the formula c = sin(C) * a is not 1333.83.

Indeed, your wacky diagram at the bottom shows the distance of 1333' 11" and at the top of the rectangle shows 1009' 4" and 324' 6" which adds to 1333' 10". Of course, this relied on you fudging the 324' 6" distance by 13' 6". Unfortunately, c = sin(C) * a does not solve to 1333 anything using the known, given figures of A=14.70°, C=90°, and a=338'. The three given figures dictate what the remaining sides and angles MUST be.

Side a can be 338 or 324' 6", but it cannot be both. It is 338' as the given height of the window.

324' 6" is just a bullshit number, pulled out of your ass. It does not belong to anything but your imagination.

You are a mathematically incompetent nincompoop.

Trig, TRig, TRIG!!!

Do the math!!!!!

Or just admit that you do not know how to work with basic trig functions.

nolu chan  posted on  2018-07-21   1:03:21 ET  Reply   Trace   Private Reply  


#48. To: nolu chan (#47) (Edited)

GFYS Douchebag, The Sides add up.

libertysflame.com/cgi-bin...? ArtNum=56433&Disp=49#C49

VxH  posted on  2018-07-21   1:18:52 ET  Reply   Trace   Private Reply  


#49. To: VxH (#48)

GFYS Douchebag, The Sides add up.

The sides no longer add up to a right triangle, shithead. Now how, no way. What an arrogant but stupid prick who can't do simle math.

338 feet was calculated by a given floor/ceiling measurement of 10.9 feet per floor, multiplied by 31 = 337.9 feet, which would put Paddock standing on the 32nd floor.

After you shorten side a to 324.60 feet, and retain hypotenuse c at 1333.83, and side b at 1290.58 feet, it is impossible to retain a right traingle as it is a mathematical certainty that angle C will be greater than 90°. You will also have changed angle a to 14.08°, angle b to 75.36°. Angle c will be 90.56°. And Paddock will fall to the 30th floor.

As you changed the triangle so it is no longer a right triangle, the formula for right triangles a2 + b2 = c2 no longer works.

Congratulations, your spider infested mind just gave birth to a misshapen mess which I shall christen Gollum's Triangle.

As can readily be visualized, if you shorten side a by 13½ feet, and keep the dimensions of b and c, side a must leave its vertical position and fall away from point A as that is the only way the two lines remain connected. The 13½ foot shortening drops Paddock to the 30th floor, and the departure from the vertical drops him some more. Line B-C is supposed to be representing the elevation from the ground to Paddock's window. You can arbitrarily just change a figure on your cartoon, but Paddock's window did not actually move.

If you shorten side a to 324.60, and you retain the vertical side to a right triangle, and retain the angle of elevation at 14.70°, then you must get side b at 1237.30 feet and side c at 1279.17 feet. The hypotenuse, side c or line B-A has been shortened 54.66 feet.

))) I love how you think changing one measurement of a triangle does not change anything else.

There are only 3 sides and 3 angles to a triangle. All of the miscellaneous lines you drew beyond that are just surplus bullshit.

You have presented Gollum's Triangle, a misshapen pile of shit. Just because you draw it in the shape of a right triangle does not mean the associated data makes a right triangle possible.

Where's your MATH?

A = 14.70°
B = ?º
C = 90.00°

a = 324.60 ft
b = ? ft
c = ? ft

A + B + C = 180º
A + B = 90º
B = 90-Aº
B = 75.30º

b = sin(B) * a / sin(A)

c = sin(C) * a

nolu chan  posted on  2018-07-21   11:14:50 ET  (1 image) Reply   Trace   Private Reply  


#50. To: nolu chan (#49)

libertysflame.com/cgi-bin...? ArtNum=56433&Disp=53#C53

VxH  posted on  2018-07-21   21:43:15 ET  Reply   Trace   Private Reply  


#51. To: VxH (#50)

WRONG.

Get the fuck out of here.

If A=14.70°, side a = 338' and side b = 1290.58'

then angle B = 75.68° and angle C = 89.62°

and hypotenuse c = 1331.95'

Of course, with angle C being 89.62°, you have another misshapen Gollum Triangle, not a right triangle. With your given data, it is a mathematical impossibility to have a right triangle. It is impossible to have angle C be 90°. Trig does not lie. You do. Your stipulated angle of 14.70 and sides of 338 and 1290.58 cannot make a right triangle.

As you cannot possibly have a right triangle, you application of a formula applicable only to right triangles yields bullshit results.

You get two different figures for hypotenuse c, 1334.11 and 1333.83, both of them wrong.

With sides of 338, 1334.11 and 1290.58, you cannot get a right triangle with an angles of 14.70 and 75.30.

Provide 3 sides and 3 angles that are not mathematically impossible to work with each other to form a right triangle.

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You created two radii, one being 338' and the other 324.5'. All radii of the same circle are the same length. You get your choice of one.

If the radius is 338', then the hypotenuse is 1009.33 + 338 = 1347.33 as shown in your cartoon #1.

You can't fudge your figures by 13.5' by claiming two radii of different lengths.

Your figures are still a mess.

With angle A = 14.70°, angle C = 90° and side a = 338',

Angle B = 75.30°, side b = 1288.38' and hypotenuse c = 1331.98', as sure as

b = sin(B) * a / sin(A)

c = sin(C) * a

Try it with trig when you learn how. The figures actually work.

Side b is not 1290.5833 if the angle of elevation is 14.70° or you do not have a right triangle. You have yet to describe how you determined the length of side b is 1290.5833 feet.

Hypotenuse c is not 1334.1099.

Nor is hypotenuse c is 1009.33 + 338 = 1347.33' as per your first cartoon.

Nor can you even make believe hypotenuse c is 1009.33 + 324.5 = 1033.8 as per your revised cartoon with two different radii.

With a 324.5' radius you lose your right triangle.

With a 338' radius your bullshit cartoon yields a hypotenuse of 1347.33.

nolu chan  posted on  2018-07-22   0:29:38 ET  (2 images) Reply   Trace   Private Reply  


#52. To: nolu chan (#51) (Edited)

libertysflame.com/cgi-bin...? ArtNum=56433&Disp=55#C55

VxH  posted on  2018-07-22   16:31:23 ET  Reply   Trace   Private Reply  


#53. To: VxH (#52)

VxH #55 and #56

Title: RED SMOKE, COMMIE MIRRORS

With a 338' radius your bullshit cartoon yields a hypotenuse of 1347.33.

BZZZT!

324.5+1009.33= 1333.83

324.5+1009.33= 1333.83

324.5 is the imaginary radius of a circle in your cartoon when the radius is not 338 on the other side of the cartoon circle. It is useful to obtain imaginary results.

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VxH #56

Do the Math Comrade Donkey Breath.

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Very well shithead. Here is the math which confirms you are a mathematics illiterate shithead. The mission was to find a right triangle with Angle a being 14.70 degrees and side a being 338 feet long. You look like you are monkey, standing in a crowded stadium at centerfield, fucking a football. You FAILED.

The problem is given an elevation of 338 feet and an angle of elevation of 14.70 degrees, find the right triangle which accurately depicts the given two pieces of data, with accurate angles and accurate lengths of the remaining side and hypotenuse.

There are only six data points. They are the three sides of a right triangle representing length, and the three angles of the triangle. All else on your cartoon was surplus bullshit.

For the given data of 338 feet and 14.70°, and 90° for a right triangle, the other three data points are a calculated mathematical certainty. The certainty is that your stated results are incompatible with the given data points.

To make this simple for those not mathematically inclined, one may use CoSinCalc.com and enter data and let them present a triangle with all sides and angles presented. CoSinCalc.com gives results to two decimal places, plus the mathematical formulas used to derive the unknown data.

At the link is their depiction of the triangle that results from entering angle A 14.70° and angle B 90°, and side a 338 feet.

Angle B = 75.30°. Side b [AC] = 1288.38 feet. Hypotenuse c [AB] = 1331.98 feet.

http://cossincalc.com/#angle_a=14.7&side_a=338&angle_b=&side_b=&angle_c=90&side_c=&angle_unit=degree

Just hit the "Calculator" button to reset for data entry of whatever you choose, such as the VxH imaginary data. It does not work for a reason — it is bullshit.

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For the more mathematically inclined,

The sine, cosine, tangent, cosecant, secant and cotangent values of an angle are constants, not variables.

To find sin(14.70) on a scientific calculator

  • press the button marked sin.

  • the calculator displays sin(

  • enter 14.70)

  • the calculator shows sin(14.70)

  • press the button marked =

  • The calculator displays the solution 0.253757945

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sin(a) = opp/hyp

With a right triangle, the length of the side opposite angle a, divided by the length of the hypotenuse, equals 0.253757945.

cos(a) = adj/hyp

tan(a) = opp/adj

csc(a) = 1/sin(a) = hyp/opp

sec(a) = 1/cos(a) = hyp/adj

cot(a) = 1/tan(a) = adj/opp

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sin(14.70) is always 0.253757945

cos(14.70) is always 0.967267753

tan(14.70) is always 0.262345089

csc(14.70) is always 3.940763319

sec(14.70) is always 1.033839903

cot(14.70) is always 3.8117733280

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sin(75.30) = 0.9672677528

sin(90.00) = 1

In a right triangle, one with a 90° angle, given the length of one side, and either acute angle, one may accurately calculate the remaining angle and the lengths of the other two sides.

With a specific acute angle, such as 14.70°, the sides are always in a specific relational proportion. Knowing the length on any one side absolutely dictates the lengths of the other two sides, to be compatible with the given angle.

Point A of our triangle is at ground level. Point B is represents the stated elevation of 338 feet. Point C is at represents level directly below Point B.

A line drawn from point A rising at 14.70° will eventually rise to a height of 338 feet. Walking along the horizontal line AC eventually leads to the point where a vertical line upward would transect the rising line AB precisely where the elevation of 338 feet is reached.

The line BC ends at the given elevation of 338 feet.

The results from CoSinCalc.com were:

Angle A = 14.70
Angle B = 75.30°
Angle C = 90.00°

Side a (B to C) = 338 feet
Side b (A to C) = 1288.38
Hypotenuse c (A to B) = 1331.98

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Using the scientific calculator to more decimal places, the same calculations are:

Angle B = 180° - 14.70° - 90.00° = 75.30°

side b = sin(B) * a/sin(A)
side b = sin(75.30°) * 338 / sin(14.70°)
side b = 0.9672677528 * 338 / 0.253757945
side b = 326.9672677528 / 0.253757945
side b = 1288.500613 feet

side c = sin(C) * a/sin(A)
side c = sin(90.00°) * 338 / sin(14.70°)
side c = 338/0.253757945
side c = 1331.978 feet

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Now let us examne the VxH nonsense. His original angle of elevation was 33°, but I pointed out that would mean Paddock's shots would have been closer to his own big toe than the fairgrounds. Then the angle of elevation was changed by proclamation to 14.70°

VxH has never explained how his side b distance of 1290 feet 7 inches or 1290.5833333333 feet was derived. Although asked for, no calculation or explanation has been provided.

Using side 1290.583333333 and 338, and an angle of elevation of 14.70 degrees, the trig functions reveal that, at said angle of elevation, the elevation itself is not reached at a distance of 1290.583333333 feet, but at 1288.500613 feet. The side b length is whack by 2 feet.

https://libertysflame.com/cgi-bin/readart.cgi?ArtNum=56433&Disp=53#C53

His hypotenuse is calculated as a2 + b2 = c2 or 1334.1099430998. This is calculated using the mysteriously appearing side b data, which is whack by 2 feet.

Then it is alternately stated as 324.5+1009.33 = 1333.83

What 324.5 or 1009.33 represent is left a mystery. Added together, they equal 1333.83, representing nothing in particular.

The cartoon has a circle with radius 1 and radius 2; and a triangle with hypotenuse 1 and hypotenuse 2; and a side just dropped in with no explanation of how it was calculated, or if it was calculated or just proclaimed. In any case. the claimed side b length is whack by 2 feet.

When drawing a right triangle with side a of 338 feet, angle A of 14.70 degrees, and side b of 1290.5833333333, CosSinCalc.com provides the resulting Gollum Triangle:

http://cossincalc.com/#angle_a=14.7&side_a=338&angle_b=&side_b=1290.583333333&angle_c=&side_c=&angle_unit=degree

As one may see, with Angle a specified as 14.70°, and sides a and b specified as 338 feet and 1290.583333333 feet,

Angle B is 75.68° and
Angle C is 89.62°
and the triangle contains no angle of 90°
Oh shit, that's not good.

Hypotenuse c becomes 1331.95 feet, as opposed to his stated result of 1334.1099430998 feet. This is because his sides, combined with the specified angle of 14.70 degrees, requires that the 90° angle b change to 89.62°, and what was a vertical line to transect the hypotenuse is now tilted toward point A, shortening the length of the hypotenuse.

As I previously pointed out, a2 + b2 = c2 only works for right triangles. His length of hypotenuse c is calculated with an inapplicable formula as his side b, and when combined with the stipulated angle a of 14.70°, is incompatible with a right triangle.

We can try to fix this using angles 14.70° and 90° and side 1290.583333333 feet.

http://cossincalc.com/#angle_a=14.7&side_a=&angle_b=90&side_b=1290.583333333&angle_c=&side_c=&angle_unit=degree

Oh dear.

As one may see, with Angle a specified as 14.70°, and Angle c specified as 90°, and side b as 1290.583333333 feet, the right triangle is forced by data entry of the 90° angle, but side a, the elevation of 338 feet is now impossible and must be raised 7 inches.

When solving for an angle of elevation of 14.70° and an elevation of 338.00 feet, results incompatible with 14.70° or 338.00 feet are shit.

Well, hell, let's just try the three VxH sides:

http://cossincalc.com/#angle_a=&side_a=338&angle_b=&side_b=1290.583333333&angle_c=&side_c=1334.1099430998&angle_unit=degree

Entering side a=338 feet and side b=1290.583333333 feet and hypotenuse c=1334.1099430998 feet we get,

Angle A = 14.68°
Angle B = 75.32°
Angle C = 90.00°

Oh shit again! The resulting triangle is incompatible with the angle of elevation of 14.70°.

When solving for an angle of elevation of 14.70° and an elevation of 338.00 feet, results incompatible with 14.70° or 338.00 feet are shit.

The VxH Gollum Triangle is incompatible with the specified angle of elevation, the elevation itself, or the necessity of a 90 degree angle to make a vertical line and a right triangle.

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

nolu chan  posted on  2018-07-23   22:51:12 ET  Reply   Trace   Private Reply  


#54. To: VxH (#52)

I can observe the truth is that wormholes are all around us, only they're too small to see. Wormholes are very tiny. They occur in nooks and crannies in space and time. You might find it a tough concept, but stay with me.

Nothing is flat or solid. If you look closely enough at anything you'll find holes and wrinkles in it. It's a basic physical principle, and it even applies to time. Even something as smooth as a pool ball has tiny crevices, wrinkles and voids. Now it's easy to show that this is true in the first three dimensions. But trust me, it's also true of the fourth dimension. There are tiny crevices, wrinkles and voids in time. Down at the smallest of scales, smaller even than molecules, smaller than atoms, we get to a place called the quantum foam. This is where wormholes exist. Tiny tunnels or shortcuts through space and time constantly form, disappear, and reform within this quantum world. And they actually link two separate places and two different times.

But this is only introductory to the theory that "time is a derivative function of state-change which progresses relative to E within the inertial frame(s) it is observed in."

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https://libertysflame.com/cgi-bin/readart.cgi?ArtNum=55999&Disp=139#C139

#139. To: A K A Stone (#38)

You can believe God is a liar.

Or I can observe that Time is a derivative function of state-change which progresses relative to E within the inertial frame(s) it is observed in.

And that God's inertial frame isn't yours! :-/

VxH posted on 2018-06-14 20:12:43 ET Reply Trace Private Reply

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https://www.libertysflame.com/cgi-bin/readart.cgi?ArtNum=39740&Disp=62#C62

#62. To: Don (#58) (Edited)

Who is your God?

Who/whatever it is that says water freezes consistently at 32 degrees F, that objects fall and accelerate at 15 ft per second per second, that time is a derivative function of state change that progresses relative to E within the (billions of years old) inertial frame(s) where it is observed, that V = I x R, and that Nature selected HETEROsexual reproduction for humans - and not the mutual masturbation selected and worshiped by reprobate perverts of Nature.

The Creator of the self-evident laws of the universe who doesn't require interpretation by, or the intercession of, some parrot adorned in vestigial plumage left over from the Roman/Egyptian/Babylonian empires -- perched atop a self-serving, man-made, state-established, fallible and uninspiring religious hierarchic pyramid of Ba-al shyte.

The ONE God.

VxH posted on 2015-05-21 4:47:55 ET Reply Trace Private Reply

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http://www.freerepublic.com/focus/news/3007479/posts?page=7#7

To: BCW

Bullshyte.

Time is a derivative function of state change which progresses relative to E within the inertial frame(s) that it is observed within.

NO SALE

7 posted on 4/13/2013, 10:53:56 AM by TArcher ("TO SECURE THESE RIGHTS, governments are instituted among men" -- Does that still work?)
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http://freerepublic.com/focus/news/3561255/posts?page=75#75

To: mad_as_he$$

Try the English version:

Time is a derivative function of state change that progresses relative to E within the inertial frame(s) in which it is observed.

So simple a 2nd grader could understand it without being relatively special.

75 posted on 6/15/2017, 4:05:50 PM by HLPhat (It takes a Republic TO SECURE THESE RIGHTS - not a populist Tyranny of the Majority)
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http://www.freerepublic.com/focus/f-backroom/3548338/posts?page=92#92

To: bobby.223

That’s what “time travel” is super genius.

Fictional, government grant funded, Bullshyte.

90 posted on 4/29/2017, 7:13:24 PM by HLPhat (It takes a Republic TO SECURE THESE RIGHTS - not a populist Tyranny of the Majority)
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To: HLPhat

Hopefully some interesting “math” will fall out of this guy’s work!

String Theory may be a dead end in cosmology but the “math” behind it is finding uses elsewhere, e.g physics of surfaces and other “interfaces”.

91 posted on 4/29/2017, 7:21:49 PM by Reily
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To: Reily

Time is a derivative function of state change that progresses relative to E within the inertial frame(s) in which it is observed.

State change always goes forward.

Grant-funded mathemaconartists can’t change that self-evident fact.

92 posted on 4/29/2017, 7:38:32 PM by HLPhat (It takes a Republic TO SECURE THESE RIGHTS - not a populist Tyranny of the Majority)
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To: HLPhat

I’m sorry but the way you stated that makes no sense.

And new mathematical concepts may have use whether they are physically realizable or not.

I am amazed you can judge his work without having read a single paper produced.

Jealous perhaps?

93 posted on 4/29/2017, 8:01:39 PM by Reily
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To: Reily

It make sense just fine to anyone who understands the difference between special relativity and fictional Bullshyte.

94 posted on 4/29/2017, 8:04:48 PM by HLPhat (It takes a Republic TO SECURE THESE RIGHTS - not a populist Tyranny of the Majority)
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To: HLPhat

I understand both special & general relativity but am willing to listen to someone tell me something new. I will judge its validity only after that.

95 posted on 4/29/2017, 8:12:28 PM by Reily
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To: Reily

Then you shouldn’t have any trouble understanding that Time a derivative function of state-change that progresses relative to E within the inertial frames in which it is observed.

And that state change always moves forward.

96 posted on 4/29/2017, 8:22:40 PM by HLPhat (It takes a Republic TO SECURE THESE RIGHTS - not a populist Tyranny of the Majority)
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To: HLPhat

Not the way you state it!

Do you mean this d(Some State)/dt vs E - energy inside the inertial frames is greater then zero when t- time is greater then zero?

97 posted on 4/29/2017, 8:33:57 PM by Reily
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To: Reily

Nope I mean what I said. That T is a function of state-change that progresses relative to E.

Meanwhile, in the space-time of the emperor’s new lavender underwear..

“Most of the published paper is about picking holes in it. so, if anything, the paper says time machines are less possible than ever.”

What does this mean?

98 posted on 4/29/2017, 8:39:47 PM by HLPhat (It takes a Republic TO SECURE THESE RIGHTS - not a populist Tyranny of the Majority)
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To: HLPhat

I’m done talking to you.
You’re purposely being an pass!

99 posted on 4/29/2017, 8:43:23 PM by Reily
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https://www.facebook.com/ArrivalMovie/

FACEBOOK page of Arrival, the Movie.

See COMMENT by William Burke.

William Burke SMH. This movie exemplifies why kids these days can't even apply science well enough to understand what sex they are.

Time is a derivative function of state change that progresses relative to E within the inertial frame(s) in which it is observed.

State-change always moves forward.

Despite what FICTION psychotically imagines, there is no time travel and you can't tell the future because (surprise), it hasn't happened yet.

Reality, deal with it.

FAIL.
June 5 at 2:58am

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

Replies:

June 5 at 2:58am
David Bouley

David Bouley All based on what we think we know. But if it makes you more comfortable to live a closed-minded existence, by all means...

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July 14 at 3:18pm
William Burke

William Burke It's science FICTION.

Time is a derivative function of state change that progresses relative to E within the inertial frame(s) in which it is observed.

State-change always moves forward.

July 14 at 3:41pm

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nolu chan  posted on  2018-07-23   22:53:27 ET  Reply   Trace   Private Reply  


#55. To: nolu chan (#54)

libertysflame.com/cgi-bin...? ArtNum=56433&Disp=59#C59

VxH  posted on  2018-07-24   10:57:57 ET  Reply   Trace   Private Reply  


#56. To: VxH (#55)

LOL

Do the Math OComrade Donkey Breath.

A Squared + B Squared = C Squared.

http://mathforum. org/library/drmath/view/57232.html

EPIC FAIL. Your Dr. Math shows you how to do a2 + b2 = c2. It does not show you how to create a right triangle with acute angles of 14.70° and 85.30°. Your grammar school math does not work for that.

You still provide no means by which you derived side b of 1290.583333333 feet. Given only one side, 338 feet, it is not possible to derive either of the other two sides using a2 + b2 = c2. So you just made up 1290.583333333 which looks impressive with seven decimal places, but what did you use to "calculate" it. Don't be bashful. Do tell. Show us the "math" that you used.

Your right triangle, with those sides, is incompatible with the given angle of elevation of 14.70°. So, still, all you have is a pile of incorrect shit. At a 14.70° angle of elevation, and at a distance of 1290.583333333 feet, the altitude reached in not 368 feet.

Keep fucking that football while the stadium looks on. Keep trying with the grade school math, super genius. And remember you have specified angles of 14.70° and 90° and opposite the 14.70° angle you have a specified side of 338 feet. Those are given data, not variables.

As you know all about "Time is a derivative function of state-change which progresses relative to E within the inertial frame(s) it is observed in," surely you have the math skills to find the sides and angles of a right triangle.

Let's see your three rib-tickling sides again:

http://cossincalc.com/#angle_a=&side_a=338&angle_b=&side_b=1290.583333333&angle_c=&side_c=1334.1099430998&angle_unit=degree

Entering side a=338 feet and side b=1290.583333333 feet and hypotenuse c=1334.1099430998 feet we get,

Angle A = 14.68°
Angle B = 75.32°
Angle C = 90.00°

Still fucked! The resulting triangle is incompatible with the angle of elevation of 14.70°. Your triangle is the wrong size and shape.

When solving for an angle of elevation of 14.70° and an elevation of 338.00 feet, results incompatible with 14.70° or 338.00 feet are shit. You've got shit.

The VxH Gollum Triangle is incompatible with the specified angle of elevation, the elevation itself, or the necessity of a 90 degree angle to make a vertical line and a right triangle.

nolu chan  posted on  2018-07-24   11:57:10 ET  Reply   Trace   Private Reply  


#57. To: All (#56)

https://libertysflame.com/cgi-bin/readart.cgi?ArtNum=56433&Disp=61#C61

Yawn, as has been explained multiple times - the model was a SKETCH with left over elements that were not re-scaled when I updated the model with your 338 elevation.

Given angle A [14.70°] and side a [338 feet], you state side b at 1290.583333333 feet. Provide the calculations for that one without trig, Mr. Mathematical Super Genius.

VxH - mathematics super genius

The only mathematics super genius who

  • is fluent regarding Einstein's General Theory of Relativity

  • is fluent regarding Einstein's Special Theory of Relativity

  • understands what he means when he says, "Time is a derivative function of state-change which progresses relative to E within the inertial frame(s) it is observed in."

  • understands what he means when he says, "I've moved on with a revised curve that reconstructs time from Velocity and Distance, ..."

  • understands what he means when he says, "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."

And with all the mathematical super genius horsepower that implies, he has not a mathematical clue how to derive the sides and angles of a right triangle, given angle a and side a. It summons a vision of Einstein faced with the same problem, sitting there utterly stumped by a problem requiring nothing beyond high school math.

With angle a stated as 14.70 degrees, and side a stated as 338 feet, VxH simply summons side b at 1290.583333333 feet, an impossible value with the given, stated values and a right triangle. VxH did not use trig, and offers no explanation of how side b could be mathematically derived without trig. Indeed, he offers no explanation or computation regarding how side b was derived.

And jackass continues to make believe he has not been proven to be absolutely full of shit.

https://libertysflame.com/cgi-bin/readart.cgi?ArtNum=55999&Disp=139#C139

#139. To: A K A Stone (#38)

You can believe God is a liar.

Or I can observe that Time is a derivative function of state-change which progresses relative to E within the inertial frame(s) it is observed in.

And that God's inertial frame isn't yours! :-/

VxH posted on 2018-06-14 20:12:43 ET Reply Trace Private Reply

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https://libertysflame.com/cgi-bin/readart.cgi?ArtNum=53046&Disp=133#C133

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.

[...]

VxH posted on 2017-11-05 6:36:35 ET

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http://www.freerepublic.com/focus/f-backroom/3548338/posts?page=90#90

That’s what “time travel” is super genius.

Fictional, government grant funded, Bullshyte.

90 posted on 4/29/2017, 7:13:24 PM by HLPhat

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http://www.freerepublic.com/focus/f-backroom/3548338/posts?page=92#92

Time is a derivative function of state change that progresses relative to E within the inertial frame(s) in which it is observed.

State change always goes forward.

Grant-funded mathemaconartists can’t change that self-evident fact.

92 posted on 4/29/2017, 7:38:32 PM by HLPhat

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http://www.freerepublic.com/focus/f-backroom/3548338/posts?page=94#94

To: Reily

It make sense just fine to anyone who understands the difference between special relativity and fictional Bullshyte.

94 posted on 4/29/2017, 8:04:48 PM by HLPhat (It takes a Republic TO SECURE THESE RIGHTS - not a populist Tyranny of the Majority)
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http://www.freerepublic.com/focus/f-backroom/3548338/posts?page=133#133

What happens to M and T as E approaches [infinity], and what is the associative effect upon relative inertia?

Public school 2nd graders back circa 1967 were able to understand and articulate the answer.

But not you.

Gag on that, Jethro.

133 posted on 5/1/2017, 12:02:59 PM by HLPhat

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https://libertysflame.com/cgi-bin/readart.cgi?ArtNum=53046&Disp=143#C143

[...]

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.

[...]

nolu chan posted on 2017-11-08 23:55:22 ET

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https://libertysflame.com/cgi-bin/readart.cgi?ArtNum=53046&Disp=145#C145

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 [feet per second].

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 [feet per second].

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]

nolu chan posted on 2017-11-09 0:11:22 ET

- - - - - - - - - -

nolu chan  posted on  2018-07-25   16:15:19 ET  Reply   Trace   Private Reply  


#58. To: nolu chan (#57)


"With angle  a stated as 14.70 degrees, and side  a stated as 338 feet, VxH simply summons side b at 1290.583333333 feet, an impossible value with the given, stated values and a right triangle"

--- nolu chan   posted on  2018-07-25   16:13:22 ET

 

"An impossible value" says Professor Donkey Chan.

LOL.

Meanwhile, in reality land...

https://libertysflame.com/cgi-bin/readart.cgi? ArtNum=56433&Disp=66#C66

 



 

VxH  posted on  2018-07-26   11:37:20 ET  Reply   Trace   Private Reply  


#59. To: VxH (#55)

VxH simply summons side b at 1290.583333333 feet

[VxH #65] I didn't summon it, I sketched it.

Given angle A [14.70°] and side a [338 feet], you state side b at 1290.583333333 feet. Provide the calculations for that one without trig, Mr. Mathematical Super Genius.

HOW DID YOU DERIVE THE LENGTH OF 1290.583333333 FEET, Super Genius????

The resulting triangle is incompatible with the angle of elevation of 14.70°. Your triangle is the wrong size and shape.

Try CosSinCalc.com Why do your shit numbers NOT WORK?

When solving for an angle of elevation of 14.70° and an elevation of 338.00 feet, results incompatible with 14.70° or 338.00 feet are shit. You've still got shit.

Show us your math, shithead. Dazzle us with your brilliance using Google Sketchup. Show how you used that renowned mathematical tool.

nolu chan  posted on  2018-07-26   11:50:16 ET  (1 image) Reply   Trace   Private Reply  


#60. To: nolu chan (#59)

HOW DID YOU DERIVE THE LENGTH OF 1290.583333333 FEET, Super Genius????

 

"With angle   a stated as 14.70 degrees, and side  a stated as 338 feet, VxH simply summons side b  at 1290.583333333 feet, an impossible value with the given, stated values and a right triangle"

---  nolu chan    posted on  2018-07-25   16:13:22 ET


The "Impossible" 1290.583333333 feet?

LOL.

Do the math OCDonkey Clump.

VxH  posted on  2018-07-27   6:39:50 ET  (1 image) Reply   Trace   Private Reply  


#61. To: VxH (#60)

Your bullshit example here says,

REQUIRED DATA ENTRY

sin A = opp/hyp = a/c

sin B = adj/hyp = b/c

Where did you get the value of B which was data entry on this bullshit?????

You can draw whatever you want with your etch-a-sketch, but that does not mean it reflects reality.

Your right triangle calculator states that you must enter the values for TWO sides and one of the two angles, a or b. Your chosen calculator automatically enters 90 degrees for angle c, and is not subject to user modification.

Your problem is that you are given angle A of 14.70 degrees and side a of 338 feet, and you must solve for sides b and c. You cannot just pull 1290.583333333 out of your ass to nine decimal places, and just enter it and make believe you just solved for it.

With number of decimal places set to 1, the value of side b of 1290.583333333 is OBVIOUS DIRECT DATA INPUT.

With side a of 338 feet and angle A given as 14.70, the 90° angle C is impossible, as is the 75.30° angle B. As you have here used a calculator which does not calculate angles based on trig functions, it does not recognize angle incompatibility. On the bright side, your new calculator does give you the trig tables they say are needed for side and angle functions of a right triangle, as when you only know one side value. Try using them. If you can figure out how to use them, you will not look like such a fucking yukonesque idiot.

As you have chosen to use a CALCULATOR OF RIGHT TRIANGLES, the value of angle C is forced to 90°. Your chosen calculator makes it 90°. Your calculator assumes the jackass operator is actually inputting data for a right triangle.

http://www.csgnetwork.com/righttricalc.html

TAKE THAT SHIT AND GET IT OUT OF HERE. Try harder to present phony bullshit.

- - - - - - - - - -

sin(14.70°) is 0.253757945

The value of sin(14.70°) does not change.

You have sides:
a = 338
b = 1290.3333333
c = 1334.1

sin A = opp/hyp = a/c
338 / 1290583333333 = 0.261897075

sin(14.70) is 0.253757945
sin(14.70) is not 0.261897075

Congratulations, you established a new value for sin(14.70). Not.

Where sin A = opp/hyp = 0.253757945, the actual value of sin(14.70)
sin A = 338/hyp = 0.253757945
hyp = 338/0.253757945 = 1331.978

When the correct value of sin(14.70) is used, your opp/hyp vales are shown to be incompatible is angle A of 14.70°.

- - - - - - - - - -

sin(75.30) = 0.9966373868

The value of sin(75.30) does not change.

You have sides:
a = 338
b = 1290.583333333
c = 1334.1

sin B = adj/hyp = b/c

sin B = 1290.583333333 / 1334.1 = 0.9673812588

sin(75.30) is 0.9672677528
sin(75.30) is not 0.9673812588

Congratulations, you have established a new value for sin(75.30). Not.

Where sin(B) = adj/hyp = 0.9672677528, the actual value of sin(75.30)
sin(B) = 1290.58333333/hyp = 0.9672677528
hyp=1290.583333/0.9672677528 = 1334.256549

When the correct value for sin(75.30) is used, your adj/hyp values are shown to be incompatible with angle B of 75.30°.

- - - - - - - - - -

Just for good luck, let's try tan(14.70) = 0.2623450899

The value of tan(14.70) does not change.

You have sides:
a = 338
b = 1290.583333333
c = 1334.1

tan(a) = opp/adj = a/b

tan(14.70) = 338/1290.583333333 = 0.261897075

tan(14.70) is 0.2623450899
tan(14.70) is not 0.261897075

Congratulations, you have established a new value for tan(14.70). Not.

When the correct value for tan(14.70) is used, your opp/adj values are shown to be incompatible with angle A of 14.70°.

- - - - - - - - - -

None of the calculations, based on the side values, yields the correct sine or tangent value for the given angle.

Congratulations, you have proven that you cannot use a Right Triangle calculator to derive the values for a triangle not known to be a right triangle.

If you force a data entry of an angle of 90 degrees where no such angle exists, you get bullshit. Garbage in, garbage out. Your calculator states that angle C is ALWAYS 90 degrees and is not user modifiable.

What you excised from your calculator.

http://www.csgnetwork.com/righttricalc.html

Right Triangle Angle And Side Calculator

This calculator requires the use of Javascript enabled and capable browsers.

This calculator is designed to give the two unknown factors in a right triangle, assuming two factors are known. This calculator is for a right triangle only! The factors are the lengths of the sides and one of the two angles, other than the right angle. All values should be in positive values but decimals are allowed and valid. Fill in two (only two) values then click on Calculate. The other two other modifiable values will be filled in, along with the angle 3 field. In a triangle, all interior angles total to 180 degrees. No two angles can total to 180 degrees or more. Angle C is always 90 degrees; angle 3 is either angle B or angle A, whichever is NOT entered. Angle 3 and Angle C fields are NOT user modifiable.

Again, this right triangle calculator works when you fill in 2 fields in the triangle angles, or the triangle sides. Angle C and angle 3 cannot be entered.

In case you need them, here are the Trig Triangle Formula Tables, the Triangle Angle Calculator is also available for angle only calculations.

The problem presents sides b and c as unknown factors and cannot be solved by using your Right Triangle Angle and Side Calculator.

http://www.csgnetwork.com/trigtriformulatables.html

Trig Triangle Formula Tables

These tables are the formulae needed for side and angle functions of a right triangle. In case you need it, here is the Triangle Angle Calculator, and the Right Triangle Angle And Side Calculator.

[snip]

- - - - - - - - - - - - - - - - - - - -

You used a calculator which requires that you direct enter TWO known sides to a RIGHT triangle. You only had one known side of 338 feet.

Only one conclusion is possible. You fucked up again.

nolu chan  posted on  2018-07-27   22:22:52 ET  Reply   Trace   Private Reply  


#62. To: nolu chan (#61) (Edited)

You used a calculator which requires that you direct enter TWO known sides to a RIGHT triangle. You only had one known side of 338 feet.

LOL

Poor Nolu Numbnuts.

Two Sides were known:

338 elevation 1290.583333333 approximate distance to target on field.

Your donkey sure has a hard time with details and decimal precision settings.

Maybe you should shoot the donkey and get a cat?

VxH  posted on  2018-07-28   0:43:36 ET  Reply   Trace   Private Reply  


#63. To: VxH (#62)

Two Sides were known:

338 elevation 1290.583333333 approximate distance to target on field.

5 tenths
8 hundredths
3 thousandths
3 ten thousandths
3 hundred thousandths
3 millionths
3 ten millionths
3 hundred millionths
3 billionths

So you scraped your known approximate distance off an outhouse wall, where the approximate distance was given to nine decimal places, or measured to the billionth of a foot. Get the fuck out of here.

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

As for 1290.583333333 feet being a known approximate distance for the shooter, what source, or sources, did that figure come from, and why do you consider it so authoritative that you assume it as a given measure????

Did you scrape it off an outhouse wall at MIT, Rensselaer, or Cal Poly?

A Google search of "1290.583333333" hits nothing but the three current threads on Liberty's Flame.

Without quotation marks is the same.

1290.58 Las Vegas Shooter returns zero hits including 1290.58.

The only apparent source of your magic number, inaccurate to 9 decimal places, is yourself doing direct data entry of the bullshit magic number you invented.

When asked for your math, you sink to your usual yukon bullshit.

- - - - - - - - - -

Wikipedia, retrieved 28 July 2018.

https://en.wikipedia.org/wiki/2017_Las_Vegas_shooting

After Paddock used a hammer to break two of the windows in both of his suites,[4] he began shooting through them at 10:05 p.m.[26] He ultimately fired more than 1,100 rifle rounds[27] approximately 490 yards (450 m) into the festival audience.[28][29][30][c]

Approximately 490 yards is approximately 1470 feet.

- - - - - - - - - -

https://www.thespectrum.com/story/news/2017/10/05/las-vegas-shooting-four-common-questions-being-asked/736322001/

Las Vegas shooting: Answering 4 common questions

Lucas M Thomas, lthomas@dvtnv.com
Published 1:48 p.m. MT Oct. 5, 2017 | Updated 1:58 p.m. MT Oct. 5, 2017

[excerpt]

How far can rifle bullets travel, and how far away from the Mandalay Bay was the concert?

Clark County Sheriff Joseph Lombardo said the weapons found in Paddock's hotel room ranged in caliber from .223 to .308. The concert venue for the Route 91 Harvest Festival was approximately 1,100 feet away from the 32nd floor of the Mandalay Bay.

- - - - - - - - - -

https://www.latimes.com/nation/la-las-vegas-shooting-live-updates-the-trigonometry-of-terror-why-the-las-1507085772-htmlstory.html

The trigonometry of terror: Why the Las Vegas shooting was so deadly

By Geoffrey Mohan
Los Angeles Times
Oct 04, 2017 | 9:54 AM

Arthur B. Alphin is well acquainted with the trigonometry of terror.

The retired Army lieutenant colonel and West Point graduate, who has a mechanical engineering degree and specialized in ballistics, has testified in many multiple-shooting cases.

What he sees so far about Las Vegas shooter Stephen Paddock is a patient, well-trained gunner who did not pick and choose his targets, but held to a steady kill zone centered in the middle of thousands of concertgoers.

Once the trigger was pulled, simple laws of physics and trigonometry sealed the fate of more than 500 people who would fall wounded in the ensuing fracas — 59 of them fatally.

"He had a huge area of three, four or five football fields with people standing shoulder to shoulder," Alphin said. "He was not aiming at any individual person. He was just throwing bullets in a huge 'beaten zone.'"

Beaten zone is an infantry term dating to World War I. Shaped like the area a searchlight casts across a flat surface, it represents the area where bullets can strike, and moves substantially with tiny changes in the tilt of the gun.

If the shooter shifted by about 1 degree, or the width of two fingers held at arm's length, Alphin said, the beaten zone would fall outside the crowd.

"That's all the distance you have to move and you aren't hitting anybody," Alphin said. "So he had to be pointing or aiming at the very center of mass and then bouncing all over with the recoil."

From a perch 320 feet above ground in a hotel whose base was about 1,050 feet from the concert venue, Paddock was firing down the 1,098-foot hypotenuse of a right triangle — and would have to adjust his aim for the arc of the bullet over that distance.

[snip]

- - - - - - - - - -

https://www.bostonglobe.com/news/nation/2017/10/02/look-how-far-las-vegas-shooter-was-from-concert/RH1IbGWenXPuSGg84YfqXN/story.html

How far was the Las Vegas shooter from the concert?

October 02, 2017
As of 3:20 p.m. Wednesday

"About 1200 feet."

- - - - - - - - - -

http://www.nydailynews.com/news/politics/vegas-shooting-happened-100s-shots-1-200-feet-article-1.3536512

How the Las Vegas mass shooting happened — hundreds of shots from 1,200 feet away

By Jason Silverstein
NEW YORK DAILY NEWS
Oct 02, 2017 | 4:38 PM

The gunman who killed at least 58 people and injured more than 500 at an outdoor Las Vegas concert Sunday fired hundreds of rounds from an automatic rifle within mere minutes — and from more than 1,000 feet away.

[snip]

- - - - - - - - - -

http://www.slate.com/articles/news_and_politics/politics/2017/10/why_the_las_vegas_shooting_was_a_nightmare_scenario_for_police_and_security.html

The second lesson from Las Vegas is that the geometry of Paddock’s attack rendered security measures ineffective—even those of Las Vegas, a city with thousands of armed police and security guards on duty around the clock, surveillance cameras covering nearly every inch of ground, and a sophisticated police department with a robust SWAT capability. Paddock apparently fired on concertgoers from a room on the 32nd story of the Mandalay Bay hotel, across the street and approximately 400 meters from the concert.

400 meters is approximately 1312.3 feet.

- - - - - - - - - -

https://www.nytimes.com/2017/10/02/us/las-vegas-mass-shooting-weapons.html

Gunman’s Vantage Point and Preparations Opened the Way for Mass Slaughter

By C. J. Chivers, Thomas Gibbons-Neff and Adam Goldman

Oct. 2, 2017

From his hotel room on the 32nd floor of the Mandalay Bay Resort and Casino in Las Vegas, Stephen Paddock would have looked down upon a crowd of more than 20,000 people, surging to the final sets of a country music festival.

He opened fire late Sunday, killing at least 59 people and injuring 527 others in one of the deadliest mass shootings in American history, the authorities said.

But what may have seemed like a difficult feat, firing across an urban area and into a crowd from about 500 yards away — the equivalent of several football fields — appears to have been offset by Mr. Paddock’s preparations, which made it possible for him to inflict mass carnage.

[snip]

About 500 yards away is about 1500 feet away.

- - - - - - - - - -

For an actual KNOWN distance, where angle A is 14.70°, and side a of a right triangle is 338 feet,

tan A = tan(14.70) = 0.2623450889 = opp/adj = a/b

0.2623450889 = 338/b

b = 338 / 0.2623450889

b = 1288.379369

- - - - - - - - - -

Or,

tan B = tan(75.30) = 3.81177328 = opp/adj = b/a

3.81177328 = b/338

b = 338 * 3.81177328

b = 1288.379369

- - - - - - - - - -

side b is not approximately 1290.58 to any number of imaginary decimal places.

side b is 1288.379369.

For someone who claims abilities in Einstein's General and Special Theories of Relativity, Calculus, and the theoretical physics math of time travel, you exhibit an inability to do basic high school math when called upon to calculate the sides and angles of a right triangle with angle A of 14.70° and side a of 338 feet.

Your donkey sure has a hard time with details and decimal precision settings.

What a congenital fuckwit. You make up an incorrect side b to 9 decimal places of inaccuracy and imprecision, and fail to realize you still look like a monkey fucking a football in the middle of the stadium. Keep fucking that football. Keep demonstrating you can only talk about math but can't do it.

Approximate to 9 decimal places of inaccuracy! Approximate to the billionth of the wrong foot! You are good for comic relief.

nolu chan  posted on  2018-07-28   12:10:10 ET  Reply   Trace   Private Reply  


#64. To: nolu chan (#63)

LOL @ the psychotic TRIGered Donkey.

VxH  posted on  2018-08-01   11:01:21 ET  Reply   Trace   Private Reply  


#65. To: VxH (#64)

You stated the angle was 33 degrees until I proved the bullet would splat into the ground closer to Paddock's big toe than the fairgrounds.

Because when VxH does triangle math, he shoots himself in the foot.

#139. To: VxH, A K A Stone (#138)

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   16:13:45 ET  (1 image) 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...

[...]

Of course, this had a 1009' 4" hypotenuse, unless you want to imagine the 338' radius of the circle is added to it, but then you get 1047' 4".

This led to later reimagining that the circle had a radius of 338' on one side and 324' 6" on the other side. Because VxH circle math works like his triangle math.

- - - - - - - - - -

VxH specified a new angle at 14.7°.

I used Google Earth to measure the distance from the hotel to the center of the concert field.

Of course, there is no evidence whatever that Paddock aimed at the center of the concert field. From his vantage point, the bandstand was nearest, that is where the people were crammed in, and that is where he first targeted.

Of course, VxH not only "measured" the distance from the hotel to the center of the concert field, he did so to 9 decimal places. He "measured" to 9 decimal places by allegedly and arbitrarily sticking two points on a google map. Praise be to VxH math. Stupid is as stupid does, as stupid posts.

The people were packed into the forward area by the band stand, not in the rear half of the field, away from the band stand.

https://www.youtube.com/watch?v=c799JltHhG0

Of course, now the reader must prepare for the measurement to have really, really have been to the front of the fairground where the people were packed in by the band stand.

Note above that the VxH red line starts in the wrong area of Mandalay Bay hotel. Paddock was on the right hand corner of the building.

Note from the wiki image below where the bullet strikes were occurring, which is not near where the VxH line measures to.

https://en.wikipedia.org/wiki/2017_Las_Vegas_shooting#/media/File:Las_Vegas_Strip_shooting.svg

Of course, now the reader must prepare for Paddock to have really, really shot from the middle of the Mandalay Bay wing, and to the middle of the fairground, rather than to where the people were packed in.

nolu chan  posted on  2018-08-01   16:11:06 ET  (4 images) Reply   Trace   Private Reply  


#66. To: nolu chan (#65)

"In 1998, he was charged with and pleaded guilty to one count of racketeering based on alleged stock manipulation and money laundering."

casetext.com/case/kriss-v-bayrock-grp-llc-4

LOL. Maaaagnificent!

VxH  posted on  2018-08-01   16:43:07 ET  Reply   Trace   Private Reply  


#67. To: nolu chan (#65) (Edited)

[Sater joined Bayrock in 2001, and over the next eight years, he worked with members of the Trump family — Donald, Ivanka, and Donald Jr. -- on multiple real estate deals.

Trump partnered with Bayrock on at least four major failed ventures in the United States (but also pursued development in Russia

In a 2005 civil racketeering case against Bayrock and its principals, the plaintiff described the firm as “substantially and covertly mob-owned and operated,]
https://investigaterussia.org/players/felix-sater

 

["“Our boy can become president of the USA and we can engineer it,” Mr. Sater wrote in an email. “I will get all of Putins team to buy in on this, I will manage this process.”]

https://www.nytimes.com/2017/08/28/us/politics/trump- tower-putin-felix-sater.html

 

Whose "boy"?

Maaaaagnificent!

VxH  posted on  2018-08-01   16:54:20 ET  Reply   Trace   Private Reply  


#68. To: VxH (#66)

https://rense.com/general96/diagramcampos.html

The the floor length of a wing of the Mandalay Bay, not including the curved end where Steven Paddock had a suite, was 300 feet. Math wizard VxH puts the shooter about half way down the North wing, approximately 150 feet south of reality for the starting point of the shots.

https://www.washingtonpost.com/graphics/2017/national/las-vegas-shooting/img/2300vegas-shooting-promo.jpg

As shown on the Washington Post map, the shots came from the corner of the Mandalay Bay, while math genius VxH located his red line starting point about 150 feet south of that point.

WAPO noted, "The venue is about 1,000 feet from the hotel."

As for the red-lined VxH photograph of Las Vegas, it is a photo with a red line drawn on it, and the number 1290.5833' typed and placed upon it.

It is evidence that VxH can copy a picture, put a misplaced red line on it, and place a number on it.

With the red line starting about 150 feet whack to the south, VxH nevertheless makes believe he measured the distance to an accuracy of four (4) decimal places, or to the ten-thousandth of a foot. Or as math genius VxH expresses his 150 foot error, {yawn}, which is one way to recognize the yawning gap between his bullshit and reality.

From his whacko bullshit, he generates side b to nine (9) decimal places of accuracy, and replaces the given angle of elevation of angle A, and, declaring "1290.5833 is a perfectly reasonable estimate for the distance between the hotel and the field," dismisses the side b distance required by math for angle A of 14.7 degrees and side a of 338 feet.

A line which starts about 150 feet whack to the south is a reasonable estimate of the VxH phychosis, give or take about a hundred feet or so, as long as it is expressed with at least four (4) decimal places on the end.

http://994ix2y62bo94ct3ok4zzrbe-wpengine.netdna-ssl.com/wp-content/uploads/2017/10/Las_Vegas_Shooter_Position.jpg

Another image with a different perspective showing the rounds did not originate anywhere near the middle of the north wing of Mandalay Bay.

nolu chan  posted on  2018-08-06   13:10:45 ET  (1 image) Reply   Trace   Private Reply  


#69. To: VxH (#67)

Whose "boy"?

Maaaaagnificent!

CIA/DIA/FBI

He is a known long-time CIA/DIA/FBI asset. I know you would have mentioned that eventually, but let me help before you forget.

https://www.buzzfeednews.com/article/anthonycormier/felix-sater-trump-russia-undercover-us-spy

THE ASSET

How A Player In The Trump-Russia Scandal Led A Double Life As An American Spy

Felix Sater has been cast as a Russian mafioso, a career criminal, and a key business associate of President Donald Trump's — but he spent more than two decades as an intelligence asset who helped the US government track terrorists and mobsters. “Greed is my go-to weapon.”

Anthony Cormier
BuzzFeed News Reporter

Jason Leopold
BuzzFeed News Reporter

Posted on March 12, 2018, at 11:56 a.m. ET

In the sprawling Trump-Russia investigation, one name constantly pops up: Felix Sater. In story after story, Sater is described as Donald Trump’s former business partner, a convicted stock swindler who was born in the Soviet Union, worked in Russia, tried to win Trump a deal in Moscow, and even helped broker a Ukrainian peace plan that Vladimir Putin would have loved.

Basically, he’s portrayed as something just short of a Russian spy.

Effectively, he has been a spy — but for the United States. For the first time, BuzzFeed News has verified the surprising sweep of Sater’s undercover work and many of his specific exploits. He worked as an asset for the CIA and the Defense Intelligence Agency (or DIA) and tracked Osama bin Laden. Then he worked for more than a decade for the FBI, providing intel on everything from the mob to North Korea’s drive for nuclear weapons. He still operates as a source for the bureau, according to two current FBI agents.

He did some of this work to fend off prison time after he admitted guilt in a stock scam — but he had started helping the US government before then, and he continued to report back to the FBI after the agreement ended. Today, as he is being questioned about Trump's business deals and ties to Russia, he has built relationships with at least six members of special counsel Robert Mueller’s team, some going back more than 10 years.

[snip]

It appears he may have been asshole buddies with Walter Brennan, or was it John Brennan. Some communist ahole, whatever.

nolu chan  posted on  2018-08-06   17:35:16 ET  Reply   Trace   Private Reply  


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