Chris Davidson Posted August 21, 2015 Author Share Posted August 21, 2015 (edited) Remember: 18.3ft = horizontal distance traveled in conjunction with a 1ft vertical rise for the Elm St slope of 3.13°. Remember: The direct connection between "3.54inch" vertical to "5.4ft" horizontal distance. Remember: 30ft/5.4ft = 5.556 Equation: 30ft/5.4ft= (5.55……..6) x 3.54inches = (19.66…….8)inches /12inches = 1.638ft. 18.3ft x 1.638ft = 29.99ft = 30ft In other words, a 5.556/1 ratio need apply to a vertical change of 3.54inches per 5.4ft horizontal distance. which equals a 30ft horizontal distance on Elm St at 3.13°. Edited August 21, 2015 by Chris Davidson Link to comment Share on other sites More sharing options...
Chris Davidson Posted August 21, 2015 Author Share Posted August 21, 2015 The attachment is an excerpt from Tom P. describing CE560. Keep in mind the previous 5.556/1 ratio. Link to comment Share on other sites More sharing options...
Chris Davidson Posted August 22, 2015 Author Share Posted August 22, 2015 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. Now, I'll throw Shaneyfelt's 11.2 mph quote back into the equation using the previous posts 21.6ft total distance traveled from Z168-Z186. 21.6ft - 5.4 ft = 16.2ft x (18.3/18) the conversion to 1 second of time. 16.2ft x (18.3/18) = 16.47ft per sec = 11.2mph Amazing how that matches the overall average speed from Z161-Z313. What does the span of 21.6ft taken from the "orange" version Z168-Z186 have to do with the 5.556 ratio? In terms of 5.556 seconds: 21.6ft x 5.556sec = 120ft. Link to comment Share on other sites More sharing options...
Chris Davidson Posted August 23, 2015 Author Share Posted August 23, 2015 That 120ft can be broken up into sections. In this instance, 90ft + 30ft = 120ft. CE560 shows you what the 90ft distance represents. Link to comment Share on other sites More sharing options...
Chris Davidson Posted August 23, 2015 Author Share Posted August 23, 2015 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. Now, I'll throw Shaneyfelt's 11.2 mph quote back into the equation using the previous posts 21.6ft total distance traveled from Z168-Z186. 21.6ft - 5.4 ft = 16.2ft x (18.3/18) the conversion to 1 second of time. 16.2ft x (18.3/18) = 16.47ft per sec = 11.2mph Amazing how that matches the overall average speed from Z161-Z313. What does the span of 21.6ft taken from the "orange" version Z168-Z186 have to do with the 5.556 ratio? In terms of 5.556 seconds: 21.6ft x 5.556sec = 120ft. The "tie in" from 90ft to the 18 frame segment at Z168-Z186 is the difference between the total span of 21.6ft and the (5.4ft horizontal=3.54 inch vertical) missing distance. In other words: 21.6ft-5.4ft = 16.2ft 16.2ft x 5.556sec = 90ft. Link to comment Share on other sites More sharing options...
Chris Davidson Posted August 25, 2015 Author Share Posted August 25, 2015 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. 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? Originally, I only connected the 166.66 frame count to the CE884 "white" version of Z161-Z166 in terms of mph. Now, the connection to 30ft past the extant Z313 headshot is: 166.66frames/30ft = the ratio of 5.555frames per foot Link to comment Share on other sites More sharing options...
Chris Davidson Posted August 26, 2015 Author Share Posted August 26, 2015 Remember: 18.3ft = horizontal distance traveled in conjunction with a 1ft vertical rise for the Elm St slope of 3.13°. Remember: The direct connection between "3.54inch" vertical to "5.4ft" horizontal distance. Remember: 30ft/5.4ft = 5.556 Equation: 30ft/5.4ft= (5.55……..6) x 3.54inches = (19.66…….8)inches /12inches = 1.638ft. 18.3ft x 1.638ft = 29.99ft = 30ft In other words, a 5.556/1 ratio need apply to a vertical change of 3.54inches per 5.4ft horizontal distance. which equals a 30ft horizontal distance on Elm St at 3.13°. Logically, the distance the SS/FBI would have attempted to move the Altgens shot back up Elm St would have been 30ft to merge with the extant Z313 headshot. Instead, as listed on the SS/FBI plat and described by Tom P., they have it as 24.5ft. Maybe it was shorted approx 5.5ft along the way, and taken into consideration for syncing purposes at this junction. Such as: 5.4ft x 18.3/18 = 5.49ft per sec = (Z168-Z171 speed) + 24.5 = 29.99ft Link to comment Share on other sites More sharing options...
Chris Davidson Posted August 26, 2015 Author Share Posted August 26, 2015 The equivalent location to the supplied attachment is listed on CE884 as Station# 3+71.1 Link to comment Share on other sites More sharing options...
Chris Davidson Posted August 26, 2015 Author Share Posted August 26, 2015 Please note the total distance in this next attachment. Link to comment Share on other sites More sharing options...
Chris Davidson Posted August 27, 2015 Author Share Posted August 27, 2015 Referring to the previous attachment: A more exact distance between 4+66.7 and 4+42.5 = 24.2ft 24.2ft + 10ft +5.4ft = 39.6ft = 39ft 7inches It's one inch off, but the explanation for that difference isn't necessary right now. Link to comment Share on other sites More sharing options...
Chris Davidson Posted August 27, 2015 Author Share Posted August 27, 2015 And, the last piece of this "soon to be formed" equation. Tom P. !!! What say you? Link to comment Share on other sites More sharing options...
Chris Davidson Posted August 27, 2015 Author Share Posted August 27, 2015 Equation: Station# 3+71.1 + 10ft = Station# 3+81.1 - 5.4ft = Station# 375.7 + 90ft = Station# 465.7 + 30ft = Station# 495.7 - 114.4ft = Station# 3+81.3 Quick key: Station# 3+71.1 -Robert West -Time/Life investigation circa Nov25,1963- Shot1 10ft - Shaneyfelt westward relocation folly 5.4ft -Distance reduced, directly correlating to speed for CE884 Z161-Z166 90ft - Distance provided via CE560 between Shot1-Shot3 30ft- Distance between extant Z313 headshot and Altgens shot circa Z352 114.4ft- Straight line distance between altered shot1(Station# 3+81.3) and Altgens shot location Link to comment Share on other sites More sharing options...
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