Chris Davidson Posted June 18, 2017 Share Posted June 18, 2017 Extant z188-z222 = 34 frames NPIC labeling of missing frames for this same span = 29 total frames 188-198 = 9 missing frames 198-206 = 7 missing frames 206-213 = 6 missing frames 213-216 = 3 missing frames 216-222 = 4 missing frames Link to comment Share on other sites More sharing options...
Chris Davidson Posted June 19, 2017 Share Posted June 19, 2017 On 6/16/2017 at 2:06 PM, Chris Davidson said: Keep thinking in terms of ratios. 1/2 = .5 1/2 x 1/4 = 1/8 = .125 .5 + .125 = .625 Closest ratio to 1 / (48/18.3) = .38138 is: .375 + .625 = 1 whole frame You can apply the frame removal ratio to the existing framework. .375 x 353frames(z133-z486) = 132.375 frames 353 + 132.375 = 485.375 = 486 whole frames Using the ratio dictates the starting and ending points for the extant zfilm. Link to comment Share on other sites More sharing options...
Chris Davidson Posted June 20, 2017 Share Posted June 20, 2017 On 6/18/2017 at 1:29 AM, Chris Davidson said: Extant z188-z222 = 34 frames NPIC labeling of missing frames for this same span = 29 total frames 188-198 = 9 missing frames 198-206 = 7 missing frames 206-213 = 6 missing frames 213-216 = 3 missing frames 216-222 = 4 missing frames David, The briefing board confirms the 1/2 then 1/4 step-down. They provided us a huge clue when they listed the 1 second increment. Converting that 1 second increment to 48fps is the key: 48/2 = 24 frames 24 x .25 (1/4) = 6 frames 24frames -6 frames = 18 frames between the span. 18 frames is the span that is shown on the briefing board which matches the extant film. 18/48 = .375 = what's left when removing .625= 1/2 + 1/8(1/2 x 1/4) Link to comment Share on other sites More sharing options...
Chris Davidson Posted June 20, 2017 Share Posted June 20, 2017 On 6/18/2017 at 1:29 AM, Chris Davidson said: Extant z188-z222 = 34 frames NPIC labeling of missing frames for this same span = 29 total frames 188-198 = 9 missing frames 198-206 = 7 missing frames 206-213 = 6 missing frames 213-216 = 3 missing frames 216-222 = 4 missing frames Think of the 29 missing frames as the shortcoming of a film that would initially have 1/3 of the frames removed. It goes like this: 1/3 x 486 = 162 486 - 162 = 324 324 + (29 shortcoming frames) = 353 frames = z133-z486 (3/8).375 - (1/3) .333... = .041666... x 48 = 2.000... frames = difference between frames kept 18 (z188-z206) and frames removed 16 (9+7) within the same 48 frame (1 second) span. Link to comment Share on other sites More sharing options...
David Josephs Posted June 21, 2017 Share Posted June 21, 2017 21 hours ago, Chris Davidson said: Think of the 29 missing frames as the shortcoming of a film that would initially have 1/3 of the frames removed. It goes like this: 1/3 x 486 = 162 486 - 162 = 324 324 + (29 shortcoming frames) = 353 frames = z133-z486 (3/8).375 - (1/3) .333... = .041666... x 48 = 2.000... frames = difference between frames kept 18 (z188-z206) and frames removed 16 (9+7) within the same 48 frame (1 second) span. Yet we are removing just short of 2/3 of the frames, not 1/3. ?? And your math is simply stating the same thing in different ways. There are 3 frames shown on the board for 188-206 and 16 missing frames with 1 second coming at 18.3 frames or the middle of 206. Multiplying both sides of the equation by 3 to get to 48fps resulting in the 5/8ths and 3/8ths %'s. And why do you keep using 1/3? Finally, these frames are simply not shown... whether there are 48 frames to a second or not, the board show 18.3 fframes for each second... the rest of the boards have missing frames as well... ?? Link to comment Share on other sites More sharing options...
Chris Davidson Posted June 21, 2017 Share Posted June 21, 2017 1 hour ago, David Josephs said: Yet we are removing just short of 2/3 of the frames, not 1/3. ?? Yes. .625 is just short of .667 And your math is simply stating the same thing in different ways. In differences I hope. There are 3 frames shown on the board for 188-206 and 16 missing frames with 1 second coming at 18.3 frames or the middle of 206. Multiplying both sides of the equation by 3 to get to 48fps resulting in the 5/8ths and 3/8ths %'s. Showing how a 2 pass operation could work in 1 second of time. Just like the difference of 3.74mph in 1 second of time CE884 (final plat) z168-z186. And why do you keep using 1/3? Would it be easier to add or subtract frames using an optical printer? 1/3 + 1/3 = 2/3 x 486 = 324 Finally, these frames are simply not shown... whether there are 48 frames to a second or not, the board show 18.3 fframes for each second... Actually, in whole frames, the board shows 18/48 = .375 the rest of the boards have missing frames as well... Yes, I'm aware of this. ?? Link to comment Share on other sites More sharing options...
David Josephs Posted June 21, 2017 Share Posted June 21, 2017 34 minutes ago, Chris Davidson said: Yet we are removing just short of 2/3 of the frames, not 1/3. ?? Yes. .625 is just short of .667 And your math is simply stating the same thing in different ways. In differences I hope. There are 3 frames shown on the board for 188-206 and 16 missing frames with 1 second coming at 18.3 frames or the middle of 206. Multiplying both sides of the equation by 3 to get to 48fps resulting in the 5/8ths and 3/8ths %'s. Showing how a 2 pass operation could work in 1 second of time. Just like the difference of 3.74mph in 1 second of time CE884 (final plat) z168-z186. And why do you keep using 1/3? Would it be easier to add or subtract frames using an optical printer? 1/3 + 1/3 = 2/3 x 486 = 324 Finally, these frames are simply not shown... whether there are 48 frames to a second or not, the board show 18.3 fframes for each second... Actually, in whole frames, the board shows 18/48 = .375 the rest of the boards have missing frames as well... Yes, I'm aware of this. ?? Let's stay with 1 second of time = 18-19 frames = 18.3fps = z188 thru z206 on PANEL 1 16 total frames "missing" ; 3 total frames shown ; 1 second occurs .3 of the way into frame 206. When we make the assumption that those 18 frames come from 48 we automatically make the math work, Let's say those 18-19 frames were actually filmed at 1000fps ... as long as the boards are labeled as they are, how would we know the filmed frame rate? We wouldn't. So isn't it pure mathematical coincidence that 48 x 1/2 x 1/4 = 18 due to 18 and 48 being divisible by 6 and 3, magical math numbers. There is no way to know if 981 frames were removed from a 1000fps film so we're back to the B&H settings... 16 and 48fps It would require the creation/addition of frames if the film was taken at 16fps or a completely different frame numbering system - the first shot's frame number depends entirely on the final frames per second speed... 213 @ 18fps and 224 @ 16fps. That's 11 frames @ 2 fps diff = 5.5 seconds of elapsed time. So we are GIVEN the 48fps as the only possibility above 18.3fps for that film to originally be shot. Filming the recreation at 24fps indeed is a strong indication that the first pass removal of frames would be 50% of the frames. Now comes the part whether there was another 1/2 of the 50% = 6 frames, or 1/3 of the 50% = 8 frames. To get to the 16fps they would have removed 1/3 of the and would have matched the actual available camera speeds... but they did not take this easy route, the 2nd 1/2 removal from 50% of the remaining frames plus/minus a few frames gets them to their 18.3 (to match Elm's incline and make the remaining calcs easier) 486 frames net divided by 75% = 648 frames before removing 25% and then double it for the first 50% = 1296 total starting frames / 48fps = 27 seconds = 486 / 18. At 18.3fps the run time is 26.56 seconds. That is .44 seconds faster which equates to 21-22 frames at 48fps or 8 frames at 18.3fpsAgain, more than likely 48fps was double stepped down to 18.3 by taking 50% of the frames away then 50% OF THE TOTAL again from the 48fps original. If there were frames between 132 and 133 they were completely removed and have no bearing on the 486 extant frames. At .149" from frame start to frame start, 486 frames = 72.414" = 6.03' of film yet there is 6' 3" of film at the archive. 2.97" of film = 20 frames or about 1 second One meter = 3.28084' = 264 frames = 39.37" = .149" per frame including the space between frames (according to Wiki) 75"/39.37" is the ratio of total length to a known quantity = 1.905 x 264 frames = 504 total frames - 486 = 18 frames or about one second Chris - I seem to have got lost in the math... trying to do too many things at once. Am I making sense with the 18 to 18.3 fps comparison? Does the length of the film and # of frames resulting in an extra 18-20 frames add up to you? 75" of film equates to too many frames... what can be derived from this? your help as always is greatly appreciated DJ Link to comment Share on other sites More sharing options...
Chris Davidson Posted June 22, 2017 Share Posted June 22, 2017 (edited) 17 hours ago, David Josephs said: 17 hours ago, David Josephs said: Let's stay with 1 second of time = 18-19 frames = 18.3fps = z188 thru z206 on PANEL 1 16 total frames "missing" ; 3 total frames shown ; 1 second occurs .3 of the way into frame 206. When we make the assumption that those 18 frames come from 48 we automatically make the math work, The camera has the capability to shoot at 48fps slow-motion. Let's say those 18-19 frames were actually filmed at 1000fps ... as long as the boards are labeled as they are, how would we know the filmed frame rate? We wouldn't. So isn't it pure mathematical coincidence that 48 x 1/2 x 1/4 = 18 due to 18 and 48 being divisible by 6 and 3, magical math numbers. Could be. But then again, since I've gone through and recreated a film using 48fps as the real rate, which matches the extant zfilm, I tend to believe a sequential removal of 1/2 then (1/4 of 1/2), is likely to appear in other areas. There is no way to know if 981 frames were removed from a 1000fps film so we're back to the B&H settings... 16 and 48fps Unless this camera possessed the newer changeover fps rate of 18. It would require the creation/addition of frames if the film was taken at 16fps or a completely different frame numbering system Agreed. Do you know of a method available for creating new from existing frames back then? So we are GIVEN the 48fps as the only possibility above 18.3fps for that film to originally be shot. Filming the recreation at 24fps indeed is a strong indication that the first pass removal of frames would be 50% of the frames. Now comes the part whether there was another 1/2 of the 50% = 6 frames, or 1/3 of the 50% = 8 frames. 1/2 + (1/3 x 1/2) = 4/6= 2/3 The limo speed differences indicate a removal of quarters not thirds nor sixth's. To get to the 16fps they would have removed 1/3 of the and would have matched the actual available camera speeds... but they did not take this easy route, the 2nd 1/2 removal from 50% of the remaining frames plus/minus a few frames gets them to their 18.3 (to match Elm's incline and make the remaining calcs easier) Same answer as previous. 486 frames net divided by 75% = 648 frames before removing 25% and then double it for the first 50% = 1296 total starting frames / 48fps = 27 seconds = 486 / 18. At 18.3fps the run time is 26.56 seconds. That is .44 seconds faster which equates to 21-22 frames at 48fps or 8 frames at 18.3fps That's along the right line of thinking. Use their numbers for more accuracy. Added on edit: Nicely done David. 486/1296 = .375kept .625removed For example: 48/18.3 = 2.622 x 486 = 1274.3 / 48 = 26.54775 sec.Again, more than likely 48fps was double stepped down to 18.3 by taking 50% of the frames away then 50% OF THE TOTAL again from the 48fps original. Exactly what I've shown with the previously provided film. Added on edit: Not Exactly. 1/2 step-down, they then figured out the limo starting frame # on film (cut the appropriate amount of frames out) , from there, they removed 1/8 of the remaining frames = 1/4 of the remaining 1/2 frames from the original step-down. If there were frames between 132 and 133 they were completely removed and have no bearing on the 486 extant frames. Exactly. Even though you can figure out how many frames they did excise from the beginning of the film through what they have labeled as z133. At .149" from frame start to frame start, 486 frames = 72.414" = 6.03' of film yet there is 6' 3" of film at the archive. 2.97" of film = 20 frames or about 1 second One meter = 3.28084' = 264 frames = 39.37" = .149" per frame including the space between frames (according to Wiki) 75"/39.37" is the ratio of total length to a known quantity = 1.905 x 264 frames = 504 total frames - 486 = 18 frames or about one second One foot of 8mm film = 80 frames. This might be an easier conversion when dealing with the Horne archive measurements. Horne excerpt: "the extant film (that is, the assassination movie, not the Zapruder family scenes present on the two Secret Service copies) in the National Archives (not counting leader) consists of a strip of film 8 feet, 10 inches long" and 2 feet, 7 inches is black, unexposed film with no image showing); 8.83 x 80 = 706.4frames / 2 = 353 frames. Does this tell us at some point that they had 2 versions of 353 frames they were working with since we don't know what was on the 2.58ft= (206.4frames) strip that was unexposed. Chris - I seem to have got lost in the math... trying to do too many things at once. Am I making sense with the 18 to 18.3 fps comparison? Does the length of the film and # of frames resulting in an extra 18-20 frames add up to you? 75" of film equates to too many frames... Did Horne include the partial footage of the Hesters at the beginning of the zfilm in his measurements? what can be derived from this? your help as always is greatly appreciated The differences are the most important part of the math solution. Sorry about some of the extraneous gestures I throw in, I'll try and keep that streamlined in the future. Lets not get hung up on the NPIC quagmire for now. I think the most important part of that scenario is the labeling of missing frames by someone. That just reconfirms the notion of an altered film. DJ Edited June 22, 2017 by Chris Davidson Link to comment Share on other sites More sharing options...
David Josephs Posted June 22, 2017 Share Posted June 22, 2017 20 hours ago, Chris Davidson said: consists of a strip of film 8 feet, 10 inches long" and 2 feet, 7 inches is black the total length including the black film is 8'10". Assassination sequence is only 75" of that. Unless this is now incorrect? 20 hours ago, Chris Davidson said: 8.83 x 80 = 706.4frames / 2 = 353 frames. Does this tell us at some point that they had 2 versions of 353 frames they were working with since we don't know what was on the 2.58ft= (206.4frames) strip that was unexposed. You'll notice that there is "blank film" spliced on tot the end of the Assn sequence, not "leader". The evidence mentions the running off of 19 feet of film to get to the end of the roll. (can't the film be processed without the end being run off?) Except that 19' is not continuous from the end of the Assn Seq. and then there is another 18'7" of BLANK FILM spliced to the end... why? A side of film is at most 30 feet long. 19'3" + 18'7" is already 37'10"... and there's still another 8'10" of actual film with 6'3" of content. There is simply no way that taped together batch of film equates to what was taken out of the camera and printed as 0183. Chris - I seem to have got lost in the math... trying to do too many things at once. Am I making sense with the 18 to 18.3 fps comparison? Does the length of the film and # of frames resulting in an extra 18-20 frames add up to you? 75" of film equates to too many frames.... Did Horne include the partial footage of the Hesters at the beginning of the zfilm in his measurements? No, I don't believe he did what can be derived from this? your help as always is greatly appreciated The differences are the most important part of the math solution. Sorry about some of the extraneous gestures I throw in, I'll try and keep that streamlined in the future. Lets not get hung up on the NPIC quagmire for now. I think the most important part of that scenario is the labeling of missing frames by someone. That just reconfirms the notion of an altered film. I guess I just yet see it that way. I also believe that by the time these boards were made the film had been completely altered down to those 486 frames. To a person working with that film, they simply choose which frames to print and filled in the blanks The NPIC notes confirm the choice of frames: it, like the Panels, also shows you the "seconds" mark as they pass... 1 second is before z206. Does the " = 14 - (1)" suggest one second is 14 frames after z188 or z202?Second #2 comes at z220, 18 frames past 202. 18 + 14 = 32 / 2 = 16fps (Second #2 is not noted on PANEL 1) Second #3 comes at z234, 14 frames past 220 yet PANEL 2 has second #3 at z243, 9 frames later or 23 frames after z220 in 1 second. Second #4 comes at z250, 16 frames past 234 yet PANEL 2 has second #4 at the end of the panel or z257. This is 23 frames after 234, and 14 frames after 243. 23 + 14 = 37 / 2 = 18.5fpsSecond #5 SHOULD come at z266/67/68. #5 is on panel #3 between z274 & z289. Even is it were z274, that's 24 frames... PANEL 3: z266 - 7 frames - z274 - 14 frames - z289 Second #6 has to be 16-18 frames prior to z314 = z296-z298. Both are beyond Panel 3 which has it at z291.5 Second #7 is on z314 314 - 291.5 = 22.5 frames from Second 6 to 7. From the equation below the NPIC table: z256-z224 = 32 frames = 2 seconds of film in the 5.5 film sequence = 16fps. All these NPIC Table calcs are for 16 frames per second, not 18.3. the Time stamps on the boards have no direct relationship to a 5.5 second span of film. The full span from the NPIC page is 312 - 224 = 88 frames / 5.5 secs = 16fps. yet this table matches the panels exactly. Panel 2 has a "4 SECONDS" notation at the end... If it meant the entire panel: 258 - 188 = 70 / 4 seconds = 17.5fps According to the NPIC table, 4 seconds comes at z246, the PANELS have it at z257, 11 frames later. I am once again spread out a bit... but I think you can pick the ball up from here. Figured at 16fps yet not sure if they tried using a stopwatch, or just counted frames... PANEL 4.... 8 second mark at the start of z331 - 49 missing - z386 = 51 frames / 3 seconds = 17fps It appears that nothing about these two items of related z-film info works in the real world. Can you check the work Chris and expand... I do believe we are inching closer to a more detailed description of the 48 to 18.3 fps cut-down. CheersDJ Link to comment Share on other sites More sharing options...
Chris Davidson Posted June 23, 2017 Share Posted June 23, 2017 This is why I suggested setting it aside for now. Both notes start with z188. One uses 16fps the other 18fps. In your previous post one example provided: PANEL 4.... 8 second mark at the start of z331 z331-z188 = 143 frames /18fps = 7.94 seconds That just reconfirms the notion of an altered film. "I guess I just yet see it that way." Was "just" supposed to be the word "don't"? "I also believe that by the time these boards were made the film had been completely altered down to those 486 frames." Agreed. As the boards match the extant zfilm. Moving forward, I think you'll realize why I choose not to delve into this too deeply, right now. Link to comment Share on other sites More sharing options...
David Josephs Posted June 23, 2017 Share Posted June 23, 2017 Ok, let's forget NPIC for now. Should be obvious since z312 as the last shot was immediately contradicted by the surveys. The Nov 26 Time/Life survey has shot one 163' from the TSBD corner... except the surveys put it at 423.07 which is 172 feet from the same spot. WCD298 from Dec's survey data squarely puts a shot within 4 feet of 5+00 as does the Feb FBI survey as does CE875 - as we both know. The real question is whether this 3rd shot was CREATED and REMOVED, or ACTUAL and REMOVED and can that be proven. With the need for a third shot and have 2 shots defined with z312 the 2nd shot, they needed a place for a third shot which would not interfere, with what Shaneyfelt did not the path of the limo and shot locations. Couldn't be before 207 with the tree and Position A as rationale.COULD be between 207 and 312 since it was 42 frames for a reload and 207 + 42 = 249 + 42 = 291... Which is why NPIC has 242 and 256 as frames with potential shotsCOULD be after 312 since the FBI's position was 3 shots = 3 hits with the last hit being the head shot... 40 feet further down Elm. As usual the exhibits negate the narrative. Link to comment Share on other sites More sharing options...
Chris Davidson Posted June 24, 2017 Share Posted June 24, 2017 9 hours ago, David Josephs said: Ok, let's forget NPIC for now. Should be obvious since z312 as the last shot was immediately contradicted by the surveys. The Nov 26 Time/Life survey has shot one 163' from the TSBD corner... except the surveys put it at 423.07 which is 172 feet from the same spot. Not from the TSBD corner, but directly below the 6th floor window forming the base of the triangle. In essence, that 163ft + distance represented what was extant z207 = Station# 3+71.1 Elev. 423.07 represents Station 3+81.3, a difference of 10.2ft between them. WCD298 from Dec's survey data squarely puts a shot within 4 feet of 5+00 as does the Feb FBI survey as does CE875 - as we both know. The real question is whether this 3rd shot was CREATED and REMOVED, or ACTUAL and REMOVED and can that be proven. Well put. Proven? Maybe by a preponderance of the data. Let's wait and see. With the need for a third shot and have 2 shots defined with z312 the 2nd shot, they needed a place for a third shot which would not interfere, with what Shaneyfelt did not the path of the limo and shot locations. Shaneyfelt melding shots. Couldn't be before 207 with the tree and Position A as rationale.COULD be between 207 and 312 since it was 42 frames for a reload and 207 + 42 = 249 + 42 = 291... Which is why NPIC has 242 and 256 as frames with potential shotsCOULD be after 312 since the FBI's position was 3 shots = 3 hits with the last hit being the head shot... 40 feet further down Elm. As usual the exhibits negate the narrative. When you extract the WC measurements given for StationC, PositionA and z161 starting with CE886. It works out like this: z161 = Station # 3+29.2 Distance from z161 to Station C = 94.7ft which means StationC = Station# 2+34.5 Distance from Position A to Station C = 44ft Position A = 2+34.5 + 44ft = Station# 2+78.5 Position A = Elev 431.97 - 3.27ft = 428.7 CE884 entry TSBD snipers nest base = Station# 2+50 = elev 429.7 Difference in elev between TSBD snipers nest base and Position A = 429.7 - 428.7 = 1ft vertical Position A 2+78.5 - TSBD 2+50 = 28.5ft 28.5ft -18.3ft = 10.2ft 28.5ft/18.3ft = 1.557... elev change Difference between 1.557... and 1ft vertical = .557... x 18.3 horizontal conversion = 10.2ft Distance difference between Time/Life survey and SS/FBI plat shot #1 = 10.2ft Link to comment Share on other sites More sharing options...
Chris Davidson Posted June 24, 2017 Share Posted June 24, 2017 Ridiculous lead height 6.7 inches for a car traveling approx 11.2 mph at extant z207 That must have been an awfully slow traveling bullet that hit 10.2ft farther down the road. Instead of converting the .56ft to 6.7 inches purely vertical, just convert it using the Elm St vertical: .56 x 18.3 = 10.248ft Link to comment Share on other sites More sharing options...
Chris Davidson Posted June 26, 2017 Share Posted June 26, 2017 David, Anyhow, I don't want to get into the sideshow that is Position A right now. You asked me earlier how they arrived at 18.3fps as their final frame rate. Besides the 1 to18.3 ratio of ElmSt vertical to horizontal, Part of that determination was dealing with whole frames in their calculations. If you run this entire equation on a calculator: 48/18.3 = That result x .375 = 1 / by previous result = That result x 18 (whole frames) = 18.3fps You know what the importance of .375 equates to in terms of frame removal. Link to comment Share on other sites More sharing options...
David Josephs Posted June 26, 2017 Share Posted June 26, 2017 1 hour ago, Chris Davidson said: 48/18.3 = Ratio of to filmed frames per second to remaining frames per second 2.623:1 That result x .375 = .9836 4/8 frame removal + 1/8 (or 1/4 of the 50% left) = 5/8 = .625 + .375 = 1 whole film 3/8 = .375 = 6.8625/18.3 = remaining frames 1 / by previous result = 1.106666667 = the needed adjustment to make 18fps into 18.3 fps = Elm incline = 1 vertical foot per second of film after POS A That result x 18 (whole frames) = 18.3fps You know what the importance of .375 equates to in terms of frame removal. 486 x .9836 = 478 frames / .75 = 638 / .50 = 1275 frames x 50% x 25% 486 - 478 = 8 frames FEWER than 50% + 12.5% = 62.5% Only 8 fewer frames needed removal to reach 18.3.... With the Limo speeding away, keeping 8 frames within z320-z486 wouldn't be too hard... yet more than likely it was more than that since well more than 62.5% of the frames are removed between z161 & z312. Link to comment Share on other sites More sharing options...
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