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# The Most Important Error the FBI told the Warren Commission about the Rifle

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Chris and Robert,

Are you really working with a right triangle, so that Pythagorus applies?

The distance from the ledge to the ground is one adjacent side of the triangle. This side doesn't form a 90-degree angle with Elm Street (the other adjacent side) because of Elm Street's downward slope of 3.13 degrees. The angle formed is 90 + 3.13 or 93.13 degrees. Correct?

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Jon,

I believe it would look something like this:

http://www76.zippyshare.com/v/ZasquaR6/file.html

chris

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Chris,

Thanks for the link. To my question though, doesn't the fact Elm Street slopes down at 3.13 degrees cause the triangle in question not to be a right triangle?

In the diagram to which you link the line AC represents the distance from the window ledge to the ground. Line BC in the diagram is at a right angle to line AC. In the real world, Elm Street would not lie along line BC. It would angle down and away from line BC at 3.13 degrees.

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John, Chris seems to have taken the slope of Elm street into account by increasing the height of the window above street level (60.7) to 70.25 to enable Pythagoras' Theorem to be used.

i.e. the line BC is starts at street level at B and finishes about ten feet below Elm street at the TSBD.

IMO

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Ray,

Take a piece of graph paper. Draw a vertical line and an intersecting horizontal line. They are perpendicular. They form a right angle.

Elm Street did not form a perpendicular line to the TSBD as it dropped away from the TSBD.

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Jon

That is why I wanted to know how much of a drop there is from the foot of the SE corner of the TSBD to the position at z313, which, as Chris points out, is 11.76 feet. By adding this to the height of the shooter in the 6th floor window from the level of the pavement in front of the SE corner of the TSBD, we have created an imaginary point 11.76 feet below the street where altitude and base meet at a 90° angle.

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Thanks, Bob, you explained it in a much clearer way that I. Must be the weather.

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ROBERT @ POST #141:

Let's create a right triangle. The triangle has a vertical leg, a horizontal leg, and a hypotenuse.

The vertical leg drops to the level at which JFK suffered head shot at Z-313, if one believes the Z-film.

The horizontal leg extends from the base of the vertical leg to the position of Z-313.

The hypotenuse is calculable from the vertical leg and the horizontal leg.

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Yup, that's why the right angle is 11.76 feet below the pavement at the SE corner of the TSBD. Only, in this case, I calculated the horizontal leg by subtracting the square of the vertical leg from the square of the hypotenuse.

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Lots of rain there, Ray? Don't feel bad, it's coming down in buckets here in the rain forest, too.

Robert,

You get an A+.

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LOL Not bad for a dumb logger eh?

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LOL Not bad for a dumb logger eh?

You work it out with "logs", Bob? Or am I barking up the wrong tree?

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It is very interesting to look at the link Ian posted for CD 298, in which the FBI show their 3-D model on Jan. 20, 1964, depicting their "3 shots, 3 hits" interpretation of the assassination. They put the second shot (Connally back shot) at 262 feet from the Sniper's Nest, and the third and final shot (JFK head shot) 307 feet from the Sniper's Nest. And, just to show how helpful they are, they even tell us the 307 foot shot was at a downward angle of 15°.

So, here is the question. At 307 feet, at a downward angle of 15°, with the follow up car right on the bumper of the limo and uphill from the limo at a 3.13° slope, would a sniper on the 6th floor SE corner of the TSBD be able to see JFK through all the SS agents? How would it look at 265 feet?

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LOL Ray! I had to take my shoes and socks off at one point, when I ran out of fingers.

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