Chris Davidson Posted February 3, 2018 Author Share Posted February 3, 2018 (edited) The public version of CE884 (white in color) reflects the average speed change of a 10.2ft adjustment from z161. It rears its ugly face in the first two entries of z161-z166 = a vehicle speed of 2.24mph = 3.294ft per sec If one wants to locate where the early investigation (FBI/SS Dec5, 1963 plat) utilized this 10.2ft adjustment, look no further than CE884(public version). 10.2ft/3.294ft per sec = 3.096... sec x 18.3fps = 56.666.... frames Added back to the start of the data at z161 + 56.66 = z217.66 Z217.66 lies between z207 and z222 on CE884. (I did not use z210-z222 as my average because the frame numbers 208/210 were switched via the WC among the two CE884 versions) Z222@ Station# 3+85.9 - Station# 3+81.3 (1st shot changed location from extant z207) = 4.6ft Average limo speed per frame from z207-z222 = .9866...ft per frame 4.6ft / .9866... = 4.662 frames Z222 - 4.662 frames = z217.338 Added on edit: If you are concerned with the 15.3ft vs 15.25ft, please consider the difference in the overall frame total. 10.15ft/3.294ft per sec = 3.081...sec x 18.3 = 56.388.. + 161 = frame # 217.38 Edited February 3, 2018 by Chris Davidson Link to comment Share on other sites More sharing options...
Chris Davidson Posted February 4, 2018 Author Share Posted February 4, 2018 In Myer's work, the matching frame span occurs as he works backwards from z175 to his beginning point at Z118 He no longer gives distances (incorrect anyways)after z175. His span from z118-z175 = 57 frames matches the span from z161-z217.66. His 10.2ft distance difference (actually 10.1776) is created between the Towner end and the Z beginning, when properly plotting JFK's position within the limo and realizing the true distance between the Towner end and Z beginning is only 7.75ft. In other words, his limo slows at that point to: 18.3/15 x 7.75ft = 6.43mph 10.1776/3.294 = 3.089... x 18.3 = 56.54 frames Link to comment Share on other sites More sharing options...
Chris Davidson Posted February 4, 2018 Author Share Posted February 4, 2018 On 1/31/2018 at 11:09 PM, Chris Davidson said: The time difference between the extant z313 shot and my shooter is 4 extant Bronson frames. Bronson filmed at 12 frames per sec. Convert 4 Bronson frames into Zapruder frames: 18.3/12 = 1.525 x 4 = 6.1 Zframes 6.1 Frames + 7 (difference in sync frames) = 13.1 extant Zframes 13.1/18.3 = .715 sec = Donald Thomas shot 3+4 One last item in terms of dealing with the approx 57 frame span adjustment average, is the last two entries on CE884. Z255-z313 = 58 frames The 13.1frames missing from z313 til the next shot when added to the z255-z313 span = 71.1 frames That is 48.9ft + 4.2ft more = 53.1ft 71.1/18.3 = 3.885.. sec 53.1ft / 3.885 = 9.297 mph Myers average for the motorcade down Houston St. was listed as 9.3mph. JFK plotted limo speed in the last 1.09sec of Towner = 9.34mph Backend to frontend and frontend to backend. Link to comment Share on other sites More sharing options...
Chris Davidson Posted February 4, 2018 Author Share Posted February 4, 2018 The orange colored CE884 (embedded in the WC final plat May 1964) can be thought of as an "instantaneous" change. (Path Switch among others) The first two entries z168-z171 reflect the reduced speed(frame removal) of the limo at extant z313 + 4.2ft = 3.74mph. Within the first second of time (z168-z186) the average speed is listed as 18frames@21.6ft traveled = 14.94mph. 14.94 - 3.74 = 11.2mph 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. Link to comment Share on other sites More sharing options...
Chris Davidson Posted February 4, 2018 Author Share Posted February 4, 2018 The limo traveling 14.94mph from z168-186 is BS. This is a direct reflection of the 10.2ft adjustment being made at the approx z133 splice. JFK plotted from z133-z149 = 19.125ft traveled = 14.88mph (21.8736ft per sec), which by no coincidence, is quite close to the target 14.94mph. Myers gives us a 15 frame total between the end of Towner and Z133 = 15/18.3 = .8196...sec. The distance difference between Towner end and Z133 = 7.75ft. At 21.8736ft per sec in .8196 sec the limo travels 17.9276ft This is 17.9276 - 7.75 = 10.1776ft = 10.2ft pushed farther down Elm = move of z207shot to Station# 3+81.3(1st shot listed on FBI/SS plat Dec 63). Link to comment Share on other sites More sharing options...
Chris Davidson Posted February 4, 2018 Author Share Posted February 4, 2018 18 hours ago, Chris Davidson said: Shortcutting this, the accommodation for the 15.25ft horizontal /10" vertical drop might look like this. Average speed change 2.24mph Instantaneous speed change 3.74mph Total speed change 5.98mph 5.98mph = 8.79ft per sec x 1.73 sec (see Myers above)= 15.20ft Myers frame rate for Towner = 22.8fps -18.3fps = 4.5fps over 167 frames 167/22.8 = 7.32sec 167/18.3 = 9.12sec Difference = 1.8sec Link to comment Share on other sites More sharing options...
Chris Davidson Posted February 9, 2018 Author Share Posted February 9, 2018 On 2/2/2018 at 8:33 AM, Chris Davidson said: https://drive.google.com/file/d/1xZh5FUEcE9k3vKAAzIgRlW9URBCsfZbc/view?usp=sharing I don't believe that's a dog muzzle. Link to comment Share on other sites More sharing options...
Chris Davidson Posted February 9, 2018 Author Share Posted February 9, 2018 Personally, I think it looks more like this type of muzzle. https://drive.google.com/file/d/1TFvyjnpHuIcbLuhapYXcjEx0Nn5WqStf/view?usp=sharing, Link to comment Share on other sites More sharing options...
Chris Davidson Posted February 9, 2018 Author Share Posted February 9, 2018 On 1/23/2018 at 2:31 PM, Chris Davidson said: The XP-100 was initially introduced with a 10¾" barrel set into a nylon stock with an unusual center-mounted grip. Chambered in .222 Remington in early prototypes, the short barrel produced significant noise and muzzle flash. Subsequently the case was shortened to reduce powder capacity to a volume more suited to the shorter barrel of a pistol. The resulting cartridge, the .221 Fireball, produced factory loaded velocities of over 825 m/s (2,700 ft/s) from the short barrel, and accuracy rivaling the parent .222 Remington, one of the most accurate cartridges made. 1frame/18.3 = .0546... sec x 2738ft per sec = 149.617ft = see graphic above Link to comment Share on other sites More sharing options...
Chris Davidson Posted February 11, 2018 Author Share Posted February 11, 2018 Still the same LOS from Bronson, just extend the shooter back to near the carport. (Sorry, didn't pick up that perspective from Bronson initially.) The LOS to JFK is now the North side of the Stemmons sign. The bullet that hits JFK approx 4.2ft farther down than extant Z313, travels between Doris Mumford and Gayle Newman who are 7ft apart from each other. The shot is approx 186.25ft away. Slight change in perspective as it's farther down the Annex Rd. Link to comment Share on other sites More sharing options...
Chris Davidson Posted February 11, 2018 Author Share Posted February 11, 2018 Very close to the same trajectory, they just had the wrong location. Elevation = 430.2 - 418.25 = 11.95ft 2.41 degrees gets you 186.5ft away. That would be a head height of 52.78" - 3.54" = 49.24" = 4.1ft above pavement. Lateral angle (left to right) from my shooter to 4.2ft past extant z313 = 1.2 degrees My shooter's bullet would have been 3.9 degrees to the right of the follow-up cars centerline at extant z313. Link to comment Share on other sites More sharing options...
Chris Davidson Posted February 11, 2018 Author Share Posted February 11, 2018 On 1/25/2018 at 8:08 AM, Micah Mileto said: How can the official 6.5 Carcano fragments be ignored if they not only contain human blood, but also muscle and skin tissue? "Tissue can contract, and bone can shrink due to low-velocity trauma. But the high-velocity impact of a rifle bullet — particularly in the strong, thick bone at the back of the skull — produces a different and very brittle response. Holes in the skull made from high-velocity rounds are reamed out as the spinning bullet bores through, pulverizing the bone, and thus invariably are slightly larger than the diameter of the bullets that cause them. The .223 round used in the AR15 is 5.56 millimeters in diameter, or just a touch smaller than the 6-millimeter entrance wound. The wound’s size, therefore, was consistent with a shot from Hickey." Right bullet (or similar), wrong weapon, wrong location. The XP-100: https://en.wikipedia.org/wiki/Remington_XP-100 Cartridge .22-250 https://en.wikipedia.org/wiki/.22-250_Remington Cartridge 6mm-BR. https://en.wikipedia.org/wiki/6mm_BR Link to comment Share on other sites More sharing options...
Chris Davidson Posted February 11, 2018 Author Share Posted February 11, 2018 A few more excerpts: The fatal bullet’s performance was, however, completely consistent with the .223 round fired from the AR15, the predecessor of the M16 rifle issued to American infantrymen during the Vietnam War. Although the .223 incorporated a full metal jacket to comply with Hague Convention mandates, the design was essentially a work-around to ensure maximum lethality in combat. An extremely thin copper jacket, coupled with the bullet’s light weight and the AR15’s high velocity, causes the round to tumble and rupture upon impact. The lead, which has softened beneath the jacket due to the bullet’s friction with the gun barrel and the air, cascades out in a random pattern of spherical fragments, which quickly solidify as they strike much cooler bodily fluids and tissue. The copper jacket also breaks up, and the combined result is a devastating, shredding and frequently lethal wound characterized by multiple, minute shards and irregular fragments — exactly like the injury Kennedy suffered. Donahue’s final ballistic observation was perhaps the easiest for a non-gun person to understand. The bullets Oswald fired were nominally 6.5 millimeters in diameter (although actually slightly larger at 6.75 millimeters to allow the rounds to expand against the concentric rifling grooves inside the gun barrel). The entrance wound on the back of Kennedy’s skull was 6 millimeters wide. The Warren Commission tried to explain away the physical impossibility of passing a bullet through a hole smaller than its diameter by asserting that the smaller entrance wound was due to the “elastic recoil of the skull which shrinks the size of an opening after a missile passes through it.” Link to comment Share on other sites More sharing options...
Micah Mileto Posted February 12, 2018 Share Posted February 12, 2018 6 hours ago, Chris Davidson said: "Tissue can contract, and bone can shrink due to low-velocity trauma. But the high-velocity impact of a rifle bullet — particularly in the strong, thick bone at the back of the skull — produces a different and very brittle response. Holes in the skull made from high-velocity rounds are reamed out as the spinning bullet bores through, pulverizing the bone, and thus invariably are slightly larger than the diameter of the bullets that cause them. The .223 round used in the AR15 is 5.56 millimeters in diameter, or just a touch smaller than the 6-millimeter entrance wound. The wound’s size, therefore, was consistent with a shot from Hickey." Right bullet (or similar), wrong weapon, wrong location. The XP-100: https://en.wikipedia.org/wiki/Remington_XP-100 Cartridge .22-250 https://en.wikipedia.org/wiki/.22-250_Remington Cartridge 6mm-BR. https://en.wikipedia.org/wiki/6mm_BR That Donahue ignores the reality of the EOP wound. Link to comment Share on other sites More sharing options...
Chris Davidson Posted February 13, 2018 Author Share Posted February 13, 2018 Excerpts from HSCA: Gerald Ford's baby. "To determine this trajectory, the Panel first had to locate the entrance and exit head wounds as precisely as possible. Figures II-6 and II-7 show where the fatal bullet entered the back of President Kennedy's head at a point 9.0 centimeters above the external occipital protuberance. *The above conclusions differ to some extent from the testimony given by Thomas N. Canning before the House Select Committee on Assassinations on Sept. 12, 1978, in each case, the differences reflect new information or analysis resulting from work concluded subsequent to the presentation of preliminary findings at the hearing. *The interpretation of the head wounds used in defining trajectory reported in testimony on Sept. 12, 1978 differs from this report because the final illustration from the Forensic Pathology Panel showed the exit wound to be 1 centimeter lower than the entrance, rather than level with it as had been concluded earlier. Thus, the resulting trajectory is somewhat steeper." 9 centimeters = 3.54 inches 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. 52.78" - 10" - 3.54" = 39.24" = 3.27ft = (CE884 JFK height used to fit trajectory back to TSBD 6th floor.) Link to comment Share on other sites More sharing options...
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