Chris Davidson

And, just so we don't forget about the other "10 inch" vertical part of the "original" equation, that sum total of 20.65ft approx matches the remaining distance of Z171-Z186. Missed by .05ft = 1/2inch in horizontal distance. Probably a "rounding off" difference.

Let me state it a bit differently: Throw out the distance traveled from Z168-Z171= .9ft from the 18frame span. You now have a 20.7ft span traveled in 15 frames. To convert in terms of a full second = 18.3/15 = 1.22 x 20.7ft = 25.254ft per sec = 17.18mph 17.18 mph - (3.74mph= (Z168-Z171speed)= 13.44mph 13.44mph = see post 15 of this topic.

Mark and Tom, I have supplied this forum with two versions of CE884. I will refer to them by color. "White" background and "Orange" background. If one will look at the entries for Z168-Z186 on the 'Orange" version, this is an 18 frame span. To convert to a one second span in terms of the Zfilm, I would have to divide 18/18.3 = .983 seconds. The entries for Z168-Z171, the first part of the Z168-Z186 span, has the limo traveling a distance of .9ft. It would look like this over a one second span: 18.3frames per sec/3 frames = 6.1 x .9ft traveled = 5.49ft traveled in a second. But, we are talking about 18 frames or .983 seconds as shown above. So, 5.49ft x .983 = 5.3967ft. Look at the "quote" above. Once again, what does 3.54 vertical inches represent? If nothing else: The ability to understand this completed circle, will help you realize that the Zfilm has at least been altered to accommodate this scenario.

If the anomalies in the extant Zfilm can be reproduced with a math equation, is it indicative of what was applied to the original film?

Let's see what Shaneyfelt was up to in relation to that 10ft sprocket hole "difference".

Yes Ian, Tom P. = 2 head shots.

Tom, Sorry about that. Poorly worded. Yes, two head shots. Approx 30ft apart. Possibly, the extant 313 headshot on film, occurred 30ft farther west down Elm St. chris

11.75 increments x 1.25ft = 14.6875ft 3.75 increments x 1.25ft = 4.68 ft Difference = 10ft Robert West determination of shot - physical location extant Z207- Station# 3+71.1 SS/FBI determination of said shot - physical location----------------- Station# 3+81.3 Difference = 10.2ft

Glenn, I tend to think in terms of the Zfilm. If the anomalies in the extant Zfilm can be reproduced with modern day technology, then is it indicative of what was applied to the original film? chris

Since that .9ft lands circa Z161- Z166 which I mentioned previously as being plotted with the limo traveling 13.44mph, I believe their is a distinct correlation between vertical span within the "equation" and this part of the extant Zfilm. And, circa Z156 through extant Z166 (which plots at Station# 3+30.1, using JFK aligned with the background light signal post) appears to be a major pivot point in this "sleight of hand" charade. Yes, A major point indeed. The comparison scale included within the two nine frame spans = 1.25ft per increment. You can use subtraction to arrive at the "difference" in distance between the two. While your at it, please note the actual distance of each span.

Ian, Your comment could be the end result of what was partially done to the film. I believe at this time, it is more important to think in terms of "differences". The assassination, at least according to the extant Zfilm, is an "equation" of differences. If you look at my previous 2 frame graphic, the span of frames is Z157-Z166 and Z167-Z176. Two spans of nine frames each.

Since that .9ft lands circa Z161- Z166 which I mentioned previously as being plotted with the limo traveling 13.44mph, I believe their is a distinct correlation between vertical span within the "equation" and this part of the extant Zfilm. And, circa Z156 through extant Z166 (which plots at Station# 3+30.1, using JFK aligned with the background light signal post) appears to be a major pivot point in this "sleight of hand" charade. Yes, A major point indeed.

Obviously the .9ft traveled for the initial two entries on CE884 is incorrect. But, when put into context with a 13.54 inch vertical rise = 20.65ft horizontal change, a total distance arises. 20.65ft - .9ft =19.75ft In terms of 1second and mph, 19.75ft per sec = 13.435mph. Since that .9ft lands circa Z161- Z166 which I mentioned previously as being plotted with the limo traveling 13.44mph, I believe their is a distinct correlation between vertical span within the "equation" and this part of the extant Zfilm. And, circa Z156 through extant Z166 (which plots at Station# 3+30.1, using JFK aligned with the background light signal post) appears to be a major pivot point in this "sleight of hand" charade.

The part of the equation that totals 13.54 inches: (52.78 - (10+3.54) = 3.27ft) converts to a horizontal distance traveled of 20.65ft(15.25 + 5.4ft)

The 10 inch part of the equation when converted would be: 10inches/12inches (1ft) = .8333ft .8333 x 18.3ft = 15.25ft Remember, this 18.3ft refers to a horizontal distance traveled in conjunction with a 1ft rise for the Elm St slope of 3.13°. A 10 inch vertical rise = 15.25ft horizontal move.

I don't want to stray too far from the original equation just yet. So I'll leave the 166.66 frames question at this for now: The attachment is another CE884 in essence, which was on the May 1964 West WC Plat final. The same entries for Z161-Z166 (5 frame) span on the previous CE884 document, were changed to reflect a (3 frame span) Z168-Z171 on the May 1964 version. Thus a 5/3 ratio = 1.6666 when multiplied by 100 will yield 166.66 100frames/18.3 frames per sec = 5.464 sec 30ft/5.464 sec = 5.49ft per sec = 3.735mph Or, when compared back to the 3 frame entry (Z168-Z171) traveling .9ft, a match of 3.735mph occurs chris

I would assume by now, most know I believe there was a head shot some 30ft farther down Elm St. than the extant 313 headshot. To put this 30ft difference in terms of 166 frames, I would address it this way: 166.66 frames/18.3 frames per sec = 9.107sec 30ft/9.107 sec = 3.294ft sec = 2.24mph chris P.S. Why 166.66 frames?

Tom, Thanks for taking an interest. I understand and appreciate the point you are making. I can only use (math-wise) that, which is provided to me, and hopefully show an ulterior motive for why they used the figures that they did. In my opinion, it becomes less coincidental, and more contrived, the farther along I get. Others can choose to believe that my original equation is just coincidental, I happen to believe differently. chris

If you took a look at CE884 without ever seeing the extant Zfilm, the first 2 entries of Z161-Z166 at 2.24 mph, might give the impression the limo traveled at a constant rate of 2.24mph for the first 166 frames. The alternative is seeing the film, and knowing full well, it is not traveling 2.24mph at Z161-166.

The difference between 2.24mph and 13.44 mph =11.20mph Shaneyfelt can tell you a little more about that figure: Mr. SPECTER. Is that a constant average speed or does that speed reflect any variations in the movement of the car? Mr. SHANEYFELT. That is the overall average from 161 to 313. It does not mean that it was traveling constantly at 11.2, because it was more than likely going faster in some areas and slightly slower in some areas. It is only an average speed over the entire run.

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

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.

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.

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.

Station# 3+30.1 - Station# 3+19.3 = 10.8ft Extant Z156-Z166 = 10.8ft traveled in 10 frames Or, 1.08ft per frame