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Swan-Song -- Math Rules

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Not to pour it on too much, but if one wants to compare the two CE884's from post #4, the entry for Station# 3.29.2 = Z161 in one chart, and z168 for the other.

The 8.8ft Shaneyfelt adjustment, incorporated into a limo moving at 15.68mph = 23.05ft per sec = 1.2596ft per frame @ 18.3fps = 8.8/1.2596 =

6.986 frames = frame difference between Z161 and z168 or as they say " close enough for government work".

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Taken from post #11and applied to post#5:

The limo travels 2.4ft in 5frames according to the extant zfilm.

18.3/5 =3.66 x 2.4 = 8.78 ft per sec = 5.98mph

8.8 vs 8.78 Close enough for gubermint work.

P.S. The exact distance between JFK and JBC was 2.41ft. If you want to use this for exacting purposes. In this instance:

3.66 x 2.41 = 8.82ft /1.2596 ft per frame (15.68mph) = 7.002 frames. See previous post.

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Be aware of the differences

39.66ft - 30.86ft (distance between z313 and Altgens shot) = 8.8ft.

8.8ft + 21.34ft + 30.86ft

Previously described and put into context.

Combined, look back up Elm St.

Edited by Chris Davidson
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You'll find it in here.

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Hopefully you found the 61ft designation.

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What constitutes 13.54 vertical inches?

The first part would be the WC determination of JFK's head height at the time of the extant 313 headshot.

In terms of elevation, it looks like this:

Street Elev. at (JFK's position in limo) extant z313 = 418.48

JFK's head height elev. = 421.75

Difference = 3.27ft in elevation

This difference is used in all entries on CE884.

At 3.27ft above the street, which is what the WC determined JFK's head height to be for all CE884 entries, this converts quite closely to an Elm St elevation change of 61ft.

61ft /18.3 (1ft vert. per 18.3ft horizontal) = 3.33ft vs 3.27ft

I will introduce the slight WC adjustment between the two a little later.

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From Tom P. via Robert West:

The 163.65ft flatline distance would equate to JFK's physical location(Station#3+71.1) within the limo at extant zframe 207.

Sorry, back or throat I'm not concerned with at this time.

chris

More support for Robert West's Time/Life documentation.

CE560:

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Looking at CE560, a few remarks are in order.

To figure out the exact vehicle speed Frazier uses, note the line starting with: Ave Vel= Average Velocity

Ave Vel = 175/2070 = .085 sec

1sec/.085 = 11.764…. x 1.4ft = 16.47ft per sec =11.2mph

Added on edit: Or, this way: .085/.0546(1frame/18.3fps) = 1.556frames 1.4ft per 1.556frames = .8997...ft per frame x 18.3fps = 16.465ft per sec/1.47(1mph) = 11.2mph

and

1sec/.085 = 11.764…. x 1.9ft = 22.35ft per sec = 15.2mph

Mr. SHANEYFELT. Yes; because we were able to determine the speed of the camera, and thereby accurately determine the length of time it takes for a specific number of frames to run through the camera at this 18.3 frames per second, and having located these frame positions in the street, we took the farthest distance point we had in the Zapruder film which was frame 161 through frame 313.

This was found to run elapsed time from the film standpoint which runs at 18.3 frames a second, runs for a total of 8.3 seconds.
This distance is 136.1 feet, and this can be calculated then to 11.2 miles per hour.

Edited by Chris Davidson
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The "Speed of Car" entry reflects the distance (90ft) between two shots corresponding back to extant Z207-Z313 from CE884.

If I wanted to sync this in terms of an average speed of 11.2mph vs instantaneous speed (see previous post) it would show like this:

90ft / 5.555sec = 16.2ft per sec x (18.3/18 (whole frame) = 16.47ft per sec = 11.2mph as opposed to what's given, which is:

90ft / 5.5sec = 16.36ft per sec /1.47 (1mph) = 11.13mph

In terms of frame count between the above scenarios, the difference would equal one frame.

5.5x18.3= 100.65

5.555 x 18.3 = 101.65

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The vertical lead time height for the 175ft shot circa extant z207 is 6.7inches according to CE560.

I ran these numbers paralleling what I believe Frazier was working with.

Edited by Chris Davidson
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Distance traveled from Z207-208 = 2.3ft

2.3ft x 18.3 frames per sec = 42.09ft per sec = 28.63 mph.

28.63mph is much more reflective of a lead height of 6.7 inches.

Whereas, 2.8 inch lead height (previous posting) is more reflective of 11.2 mph.

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It appears someone knew the significance of a 2" lead.

Mr. EISENBERG - And you calculated the speed of the car by translating the figures on total time elapsed between first and third shots?

Mr. FRAZIER - Yes, sir. The time the speed of the moving object was calculated on the basis of an assumed 5.5-second interval for a distance of 90 feet, which figures out mathematically to be 11.3 miles per hour.
Mr. EISENBERG - Now, you said before that in order to give this 2-foot lead, you would have to aim 2 inches--for a target going away from you, you would have to aim 2 inches above the target, or in front of the target.
Mr. FRAZIER - 2 feet in front of the target, which would interpolate into a much lower actual elevation change.
Mr. EISENBERG - The elevation change would be 2 inches, is that it?
Mr. FRAZIER - Well, no. It would be on the order of 6 to 8 inches.
Mr. EISENBERG - 6 to 8 inches?
Mr. FRAZIER - Yes.
Mr. EISENBERG - What was your 2-inch figure?
Mr. FRAZIER - I don't recall.

Mr. EISENBERG - But it is 6 to 8 inches in elevation?
Representative BOGGS - May I ask a question?
Using that telescopic lens, how would you aim that rifle to achieve that distinction?
Mr. FRAZIER - Well it would be necessary to hold the crosshairs an estimated distance off the target, of say, 6 inches over the intended, target, so what when the shot was fired the crosshairs should be located about 6 inches over your target, and in the length of time that the bullet was in the air and the length of time the object was moving, the object would move into actually, the path of the bullet in approximately 1/10th to 13/100ths of a second.
Mr. EISENBERG - So that if the target of the assassin was the center of the President's head, and he wanted to give a correct lead, where would he have aimed, if we eliminate the possibility of errors introduced by other factors?
Mr. FRAZIER - He would aim from 4 to 6 inches--approximately 2 inches, I would say, above the President's head, which would be actually 6 inches above his aiming point at the center of the head.
Mr. EISENBERG - How difficult is it to give this--a lead of this size to this type of target?
Mr. FRAZIER - It would not be difficult at all with a telescopic sight, because your target is enlarged four times, and you can estimate very quickly in a telescopic sight, inches or feet or lead of any desired amount.
Mr. EISENBERG - Would it be substantially easier than it would be with an open or peep sight?
Mr. FRAZIER - Yes. It would be much more difficult to do with the open iron sights, the notched rear sight and the blade front sight, which is on Exhibit 139.
Mr. EISENBERG - Now, you have been able to calculate the precise amount of lead which should be given, because you have been given figures. If you had been in the assassin's position, and were attempting to give a correct lead, what lead do you think you would have estimated as being the necessary lead?
Mr. FRAZIER - It would have been a very small amount, in the neighborhood of a 3-inch lead.
Mr. EISENBERG - As opposed to the 6 or 8 inches?
Mr. FRAZIER - As opposed to about 6 inches, yes.
Mr. EISENBERG - What would the consequence of the mistake in assumption as to lead be that is, if you gave a 3-inch lead rather than the correct lead?
Mr. FRAZIER - It would be a difference of a 3-inch variation in the point of impact on the target.
Mr. EISENBERG - Now, if you had aimed at the center of the President's head, and given a 3-inch lead, again eliminating other errors, where would you have hit, if you hit accurately?
Mr. FRAZIER - It would be 3 inches below the center of his head--from the top--it would be not the actual Center from the back, but the center would be located high. The bullet would strike at possibly the base of the skull.
Mr. EISENBERG - Now, suppose you had given no lead at all and aimed at that target and aimed accurately. Where would the bullet have hit?
Mr. FRAZIER - It would hit the base of the neck--approximately 6 inches below the center of the heart.
Mr. EISENBERG - Mr. Frazier, would you have tried to give a lead at all, if you had been in that position?
Mr. FRAZIER - At that range, at that distance, 175 to 265 feet, with this rifle and that telescopic sight, I would not have allowed any lead--I would not have made any correction for lead merely to hit a target of that size.

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The excerpt at top, within the graphic, is from a review article by Martin Hay. The link is included if you want to read the whole article.

http://www.ctka.net/2015/HaagCritique.pdf

The ballistic results pretty much bear out what I've provided in the previous postings in relation to extant z207, CE560 and testimony thus far.

9degrees 21 minutes = 9.35degrees

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Distance traveled from Z207-208 = 2.3ft

2.3ft x 18.3 frames per sec = 42.09ft per sec = 28.63 mph.

28.63mph is much more reflective of a lead height of 6.7 inches.

Whereas, 2.8 inch lead height (previous posting) is more reflective of 11.2 mph.

We now know mathematically what 4 inches represents.

Bennett translation: Two shots occurred (before the extant 313 shot) in less time than is possible, to fire the same weapon.

The SS giveth, the WC taketh away.

TREASURY DEPARTMENT

UNlTED STATES SECRET SERVICE

FIELD FORCE

PROTECTIVE Assignment of S/A Bennett on 11122/63

at Dallas, Texas

The President's auto and the follow-up proceeded to the Parkland Hospital. Upon arriving at the hospital's parking lot, I was instructed by ASAIC Roberts to stay with the Vice-President who had followed us into the parking lot. I immediately went to the Vice-President's auto and accompanied him to a room on the ground floor of the hospital. I then continued with the Vice-President back to Washington, D.C. where I was relieved.

[signature]

Glen A. Bennett

Special Agent

11-23-63

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