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I'm pretty sure the 52.78 inches refers to a spot on the head of the JFK stand-in riding in the Queen Mary during one of the reenactments of the assassination.

The 10 inches represents the difference in the distance from the pavement to the same point on the head of JFK as he rode in SS-100X.

Not sure I remember the other numbers in the equation. As I've gotten older, my recall has slowed.

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Here is a quick reminder.

Mr. KELLEY. The officials at Hess Eisenhardt, who have the original plans of the President's car, conducted a test to ascertain how high from the ground a person 72 1/2 inches would be seated in this car before its modification. And

132

it was ascertained that the person would be 52.78 inches from the ground--that is, taking into consideration the flexion of the tires, the flexion of the cushions that were on the car at the time.
Mr. SPECTER. When you say 52.78 inches, which individual would that be?
Mr. KELLEY. That would be the President.
Mr. SPECTER. And what part of his body?
Mr. KELLEY. The top of the head would be 52.78 inches from the ground. When Mr. Anderton was placed in the followup car, it was found that the top of his head was 62 inches from the ground.

Mr. SPECTER. Was there any difference between the position of President Kennedy's stand-in and the position of President Kennedy on the day of the assassination by virtue of any difference in the automobiles in which each rode?

Mr. SHANEYFELT. Yes; because of the difference in the automobiles there was a variation of 10 inches, a vertical distance of 10 inches that had to be considered. The stand-in for President Kennedy was sitting 10 inches higher and. the stand-in for Governor Connally was sitting 10 inches higher than the President and Governor Connally were sitting and we took this into account in our calculations.

3.54 inches I will address in a bit.

3.27ft = Surveyed Height of JFK's head wound.

The range being 52.78-39.24 = 13.54 inches

chris

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And the height of the windowsill above the floor on the southeast window of the 6th floor od the TSBD was.....

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Mark,

Let me start this one by addressing the 3.54 inches (from the original equation) in terms of a conversion.

3.54inches/12inches (1ft) = .295ft

.295 x 18.3ft = 5.3985ft

Remember, this 18.3ft refers to a horizontal distance traveled in conjunction with a 1ft rise for the Elm St slope of 3.13°.

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Continuing with the ratio: 3.54 inches (vertical) to 5.3985ft (horizontal)

I plotted the position of the limo at Z156, using the limo "front end" and the "park sign" in the background from Z's filming position. The "park sign" location was plotted using a "key"

created by Robert West, back when the original surveying was being performed.

The attachment includes a frame from Groden's version of Z (Z156) and MPI's version of Z157 showing the limo's front end is past the alignment "park sign" at Z157, but in alignment at Z156.

The position of "JFK in the limo" at Z156 is Station# 3+19.3 in the extant Zfilm.

post-5057-0-03920000-1438700458_thumb.jpg

Edited by Chris Davidson
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Since Z161 lands between Z156-Z166, referring back to CE884, the distance the car traveled between Z161-166 = 5frames x 1.08ft per frame

This equals a distance of 5.4ft.

Please refer back to what a 3.54inch vertical change equates to, in terms of the horizontal distance traveled on Elm St. 3.13° slope.

post-5057-0-05822300-1438709134_thumb.jpg

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Here is some more testimony from Shaneyfelt dealing with Z161-Z166.

Their vertical change for Z161-Z166 = 10 inches.

Mr. SHANEYFELT. Yes; there was. After establishing this position, represented by frame 161, where the chalk mark was about to disappear under the tree, we established a point 10 inches below that as the actual point where President Kennedy would have had a chalk mark on his back or where the wound would have been if the car was 10 inches lower. And we rolled the car then sufficiently forward to reestablish the position that the chalk mark would be in at its last clear shot before going under the tree, based on this 10 inches, and this gave us frame 166 of the Zapruder film.

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So...from z161 to z166, the vertical drop should be 3.54 inches, since the horizontal distance is 5.4 feet...

...and NOT 10 inches...

...if I'm following along correctly.

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Mark, yes you are correct.

Keep this in the back of your head for now.

This is Spector questioning McClelland:

Assume, if you will, that President Kennedy was shot on the upper right posterior thorax just above the upper border of the scapula at a point 14-cm [6.5 inches] from the tip of the right acromion process and 14-cm [6.5 inches] below a tip of the right mastoid process, assume furtherthat that wound of entry was caused by a 6.5-mm missile shot out of a rifle having a muzzle velocity of approximately 2,000 feet per second, being located 160 to 250 feet away from President Kennedy, that the bullet entered on the point that I described on the President's back, passed between two strap muscles on the posterior aspect of the President's body and moved through the fascial channel without violating the pleura cavity, and exited in the midline lower third anterior portion of the President's neck, would the hole which Dr. Perry described to you on the front side of the President's neck be consistent with the hole which such a bullet might make in such a trajectory through the President's body?

Dr. McCLELLAND - Yes; I think so.

14cm actually equals 5.5 inches, but Spector knowing about the 10 inch adjustment made, uses 6.5 inches.

10 inches - 3.54 inches = 6.46inches.

And Ford moves the back shot location up how high for the SBT to work? Let me guess, somewhere around 3.54 inches

I want to do some mph conversions related to the Z156-Z166 frame span, next.

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CE884 has the limo travel a distance of .9ft from Z161-Z166

At 18.3 frames per sec, the mph converts to: 18.3frames/5= 3.66 x .9ft = 3.294ft per sec/1.47(1mph) = 2.24mph

My plotting of Z156-Z166 of which Z161-166 is included has the limo traveling a distance between Z161-Z166 of 5.4ft

18.3frames/5 =3.66 x 5.4 = 19.764 ft sec = 19.76/1.47 (1mph) = 13.44 mph

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