Chris Davidson Posted March 28, 2018 Author Share Posted March 28, 2018 https://drive.google.com/file/d/1wUPAWGzkAeD4jt5haRjTHjzHs2_Q08x3/view?usp=sharing This is the extant film converted to 48fps using FC Pro. The clip consists of 7 frames. 133-134 = 1frame span. 134-135 = 1 frame span If you remove 5 frames between the first and last, the position of the limo will match the extant film. From extant z135, continue on, removing 1/2 the frames until the next cluster needs to be removed. Link to comment Share on other sites More sharing options...
David Josephs Posted March 28, 2018 Share Posted March 28, 2018 (edited) 1 hour ago, Chris Davidson said: If you remove 5 frames between the first and last, the position of the limo will match the extant film. first and last? 133 and 135? you talking about that jag to the left in the unstabilized version that sentence really doesn't help clarify it buddy... At about 1 foot per frame, 2 frames divided by .25 = 8 frames at 48 fps to 134 to 135 at 18.3 = to 133.1 - 133.2 - 133.3 - 134 - 134.1 - 134.2 - 134.3 - 135 at 48 Ffps remove half (4 frames) leaving 4 133.1, 133.2, 134.1, 134.2 remove half (2 frames) leaving 2 133.3, 134.3 2 frames left, 5 removed... from a 48 fps film... but I still don't understand what you mean by "the position of the limo will match the extant film" when and where? Edited March 28, 2018 by David Josephs Link to comment Share on other sites More sharing options...
Chris Davidson Posted March 28, 2018 Author Share Posted March 28, 2018 You do not start removing every other frame until after the first 5 of the original 48 are excised. In other words, Start and keep z133, remove the next 5 consecutive frames from the 48fps version. Keep z135, then, remove the next frame and this starts the 1/2 pass removal process. 5 frames removed from a 48fps version = a frame jump of 2.16-2.4ft = a 2 frame distance traveled in the extant film. Those 5 frames are the only part of that span (z133-z135) which will be removed. There will be no more removal of frames, via the other passes, within that span. Sequence: Keep1 Remove(23456) Keep 7 9 11 13 15 17 ......... Link to comment Share on other sites More sharing options...
Chris Davidson Posted March 28, 2018 Author Share Posted March 28, 2018 Something like this: https://drive.google.com/file/d/1wBURGPKHRQGA9VpeUhLic6LRe2OEYgnS/view?usp=sharing Link to comment Share on other sites More sharing options...
Chris Davidson Posted March 29, 2018 Author Share Posted March 29, 2018 On 3/28/2018 at 9:06 AM, David Josephs said: incredible Chris... Thanks David, Two more for you: (18/48) x (2) x (1.2) x (1/.9) = whole frames (1 - .11111) x 18 = frame rate Link to comment Share on other sites More sharing options...
David Josephs Posted March 29, 2018 Share Posted March 29, 2018 (edited) 18 hours ago, Chris Davidson said: Thanks David, Two more for you: (18/48) x (2) x (1.2) x (1/.9) = whole frames (1 - .11111) x 18 = frame rate 18 X 2 = 36 X 1.2 = 43.2 X 1.1111 = 48 / 48 = 1 (1-.11111) = .888889 X 18 = 16 frames per second (1 of the 2 settings for speed on the B&H) the other was 48fps If the film was not altered or taken at 48fps it would have been taken at 16fps What happens to all the calcs when you revert back to the normal camera speed - and why was 18.3 so much more important to use than 16? First big reason was the incline of Elm with a rise/run of 1'/18.3' 16/18.3 = .8743 From the first splice, 133, to 313... 180 frames @18.3 frames drops to 157 frames at 16 fps. z313 - 157 frames = z156, the location of the 2nd splice and exactly half the frames. 180 frames minus 157 frames = 23 frames @ ~1 foot per frame is the length of the limo z156 - z133 = 23 frames = length of the limo =========== {thinking out loud} 486 frames @ 18.3 = 425 frames @ 16. 61 frame diff 353 frames (486-133) = 309 frames @ 16 42 frame diff... 42 frames would be the distance between 313 and the final shot per the SS... Edited March 30, 2018 by David Josephs Link to comment Share on other sites More sharing options...
Chris Davidson Posted March 29, 2018 Author Share Posted March 29, 2018 2 hours ago, David Josephs said: 18 X 2 = 36 X 1.2 = 43.2 X 1.1111 = 48 / 48 = 1 (1-.11111) = .888889 X 18 = 16 frames per second (1 of the 2 settings for speed on the B&H) the other was 48fps Both results are correct. Since I've been listing splits, this might be a better way to look at the top equation: 18/48 = .375 x 2 (remove 1/2) = .75 x 1.2 (remove 1/6) = .9 x 1.11111 = 1 They never got back to 1 whole frame. Something like this: They are short 353 x 1.11111 = 392.22 - 353 = 39.22 frames. Either spread out over the entire 353 frame span, or specific areas. I would hope there are some "unaccounted for" frames, how else would the limo severely slow down near the two headshots 13.1(hybrid) frames apart, without severely slowing down on the extant films. 1296 = 486 x (48/18) 354.6666666 = 133 x (48/18) 941.33333 = (1296 - 354.66666) 470.666666 = (941.3333 / 2) 470.666666 - 16.6666% = (1/2 x 1/3) = 392.22 - 353 =39.22 Remaining Link to comment Share on other sites More sharing options...
Chris Davidson Posted March 29, 2018 Author Share Posted March 29, 2018 DB Thomas Acoustical: 5.7sec between labeled shots 2/3 .7 seconds between labeled shots 3/4 5.7 x 2.3 (18.3 - 16) = 13.11 hybrid frames 13.11 / 18.3 = .716seconds Link to comment Share on other sites More sharing options...
David Josephs Posted March 30, 2018 Share Posted March 30, 2018 (edited) On 3/29/2018 at 4:00 PM, Chris Davidson said: 470.666666 - 16.6666% = (1/2 x 1/3) = 392.22 - 353 =39.22 Remaining This is AFTER 133 (486 - 133 = 353) Q: 39.22 frames at 18.3 fps, not 48... right? 39.22 / 18.3 = 2.143 seconds.... & also equals 39.22 * either (48/18.3) = 103 whole frames or (48/18) = 105 whole frames at 48fps Let's look at the likely candidates for removal... z154 - z158... while it appears that z157's top is spliced into z156's bottom... JFK's head turn in 1 frame seems impossible z207 (WEST location fro a shot) thru z212 - 1 of 2 shots removed z207 to z208 on CE884 is 2.24 feet which = 28mph.... it was not possible for the limo to travel that distance in 1 frame so 208 was changed to 210 in CE884 to bring the spped down under 10mph On CE585 we can see the faint writing stating that Shot #1 at frame 207/8 was a bit farther up Elm... So while JFK and JC appear to be hit nearly simultaneously... there was a fraction of a second separating the 2 shots... Many feel JFK was hit, possibly twice between 190 and 223... Z302 - 303 is another example of an extreme head turn and the fact that the image is relatively uniform in focus despite - supposedly - the limo or the camera was moving... And again at 314-316 and again at 350 - right where the shot by the stairs occurs Mr. LIEBELER - You also testified that you were standing perhaps no more than 15 feet away when the President was hit in the head and that you are absolutely certain that there were no shots fired after the President was hit in the head?Mr. ALTGENS - Yes, sir; that's correct. .....There was flesh particles that flew out of the side of his head in my direction from where I was standing, so much so that it indicated to me that the shot came out of the left side of his head. Edited July 19, 2019 by David Josephs Link to comment Share on other sites More sharing options...
Chuck Schwartz Posted March 30, 2018 Share Posted March 30, 2018 Chris, (and David)- this is the interview of Lee Bowers , who had a bird's eye view of the assassin(s) behind grassy knoll fence. Does this correlate to your findings? Link to comment Share on other sites More sharing options...
Chris Davidson Posted March 30, 2018 Author Share Posted March 30, 2018 (edited) 13 hours ago, David Josephs said: Let's look at the likely candidates for removal... z154 - z158... while it appears that z157's top is spliced into z156's bottom... JFK's head turn in 1 frame seems impossible Most definitely agree. Sometimes it's easier to slo-mo the scene. That's when the obvious jumps will appear, even when using Groden's non missing frames. https://drive.google.com/file/d/1wMad9A0SdPDIRhGfDVI_zz9UdJqSn-ia/view?usp=sharing Edited March 31, 2018 by Chris Davidson Link to comment Share on other sites More sharing options...
Chris Davidson Posted March 30, 2018 Author Share Posted March 30, 2018 2 hours ago, Chuck Schwartz said: Chris, (and David)- this is the interview of Lee Bowers , who had a bird's eye view of the assassin(s) behind grassy knoll fence. Does this correlate to your findings? Chuck, Two shots close together. Yes How many others???? Link to comment Share on other sites More sharing options...
David Josephs Posted March 30, 2018 Share Posted March 30, 2018 To ponder: If the shooters were connected via COLLINS RADIOS... the "Fire" command would allow 2-3 shooters to fire within a split second of each other... .... also, the area where the Tramps walk past Lansdale.... same area as you have a shooter for the 2nd z313 shot... Except it appears too large and that he shouldn't be all black Link to comment Share on other sites More sharing options...
Chris Davidson Posted April 3, 2018 Author Share Posted April 3, 2018 (edited) On 3/29/2018 at 4:00 PM, Chris Davidson said: Both results are correct. Since I've been listing splits, this might be a better way to look at the top equation: 18/48 = .375 x 2 (remove 1/2) = .75 x 1.2 (remove 1/6) = .9 x 1.11111 = 1 They never got back to 1 whole frame. Something like this: They are short 353 x 1.11111 = 392.22 - 353 = 39.22 frames. Either spread out over the entire 353 frame span, or specific areas. I would hope there are some "unaccounted for" frames, how else would the limo severely slow down near the two headshots 13.1(hybrid) frames apart, without severely slowing down on the extant films. 1296 = 486 x (48/18) 354.6666666 = 133 x (48/18) 941.33333 = (1296 - 354.66666) 470.666666 = (941.3333 / 2) 470.666666 - 16.6666% = (1/2 x 1/3) = 392.22 - 353 =39.22 Remaining Whole Frames Conversion: 353/16 = 22.0625 353/18 = 19.6111 2.4514 sec x 16fps = 39.22 frames P.S. Since there are two different conversions which yield the same missing frame count, can I call them hybrid frames? Added on edit: 353/18.3 = 19.2896 sec 22.0625 - 19.2896 = 2.7729 sec x 18.3 = 50.74 frames/ft 2.7729sec x 16.0 = 44.36 frames/ft Edited April 7, 2018 by Chris Davidson Link to comment Share on other sites More sharing options...
Chris Davidson Posted April 7, 2018 Author Share Posted April 7, 2018 On 3/26/2018 at 8:34 PM, Chris Davidson said: Let's use Myers 108% manipulation to sync his distance difference with CE884: 79.2ft x 1.08% = 85.536ft 85.536ft - 79.2ft = 6.336ft 6.336ft - .9ft = 5.436ft difference Now refer to the distance per frame traveled from CE884 z161-z166 = .18ft per frame traveled The WC difference comes in the span of z133-z161 = 30.2ft (Station# 299- 329.2) 30.2ft x .18ft = 5.436ft How many frames might we be dealing with for that equation: 30.2 /.18 = 167.77 After revamping Towner's filming position, and replotting JFK within the limo from that filming position, the distance JFK travels between the Towner film end and z133 = 5.625ft. The last 20 frames of Towner has JFK travel 12.5ft. Working in between the Myers lines to figure out how he would hide this short span of approx 5.625ft. 9.2mph x 1.47 = 13.524ft per sec Myers has the Z camera frame rate at 18.3 for syncing. He provides a 15 frame gap between Towner end and Z133. (See the problem forming with only 5.625ft to deal with?) 18.3/15 = 1.22 13.524ft/1.22 = 11.085ft traveled for 15 frames, but we only have 5.625ft to work with. So let's remedy Myers with what is known to be true: Using the plotted distance/frame count of Towner 12.5ft/20 frames = .625ft per frame 5.625ft/.625 = 9 frames A 6 frame difference from the 15 Myers uses. 11.085ft - 5.625ft = 5.46ft difference 12.5ft(last 20 frames of Towner) + 5.46ft difference = 17.96ft 17.96ft / 20 frames = .898ft per frame x 16fps = 14.368ft per sec / 1.47 = 9.77mph Quite close to his ideal 9.8mph target. btw, 15frames@ 5.625ft using 18.3frames per sec = 4.668mph Couple that with the other speed entries. previously provided for z133-z149 = 14.88mph + 4.668mph = 19.54/2 = 9.77mph = match P.S. If the difference was 5.49ft, instead of 5.46ft, this would give you the CE884 provided distance for z168-z171 of 3 frames@.9ft converted over 18.3 frames. And, 5.5ft difference = 12.5ft + 5.5 ft = 18ft/20 frames = .9ft per frame = Shaneyfelt average for z161-z313. It would also match Myers 9.8mph. .9ft x 16fps = 14.4ft.per sec / 1.47 = 9.795 mph Link to comment Share on other sites More sharing options...
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