Chris Davidson

Reenactment similar to previous posting.

David J, If I put JFK where Position A is designated, the lane stripe is blocked by the limo. This is the only way it works. It matches, using Robert West's designated path. My filming location for Towner. Greer in alignment with the concrete structure. JFK in alignment with the tree.

Keep in mind: Robert West nor Chester Breneman ever stated that the Towner film was used in their survey work (only z frames). So what film was used to survey in StationC, Position A and the rest, before extant z161? None other than a film that captured the limo making the Elm St turn.

Discount 30ft over (100 and 166.66 frame increments) @ 18.3frames per sec: 166.66/18.3 = 9.107 seconds 30ft/9.107sec = 3.294ft per sec = 2.24mph = speed of limo from CE884 (public version) Z161-z166 5frames 100/18.3 = 5.464seconds 30ft/5.464 seconds = 5.49ft per sec = 3.74mph = speed of limo from CE884 (final plat May/1964 version) z168-z171 3frames 5frame/3frame ratio = 1.66/1 = 166/100

Once again, the 100ft mark would end at Station 334.5 = Robert West's path, not the other.

Towner 167frames shifted to start at StationC: Discounting 30ft over 166.666frames: 30ft/166.666 = .18ft per frame CE884 (z161-z166 @18.3fps) = 3.294ft per sec = .9ft traveled in 5 frames = .18ft per frame traveled Start at StationC End at Station 299.0 = Frame 100 Jump the frame count by 33 = extant z133 Discount the span (no data) from z133-z166 = 33frames/30ft Accommodate the (Towner end- z133 beginning )distance span with path change. End at extant z166 = Shaneyfelt (not Robert West) path via extant zfilm. StationC + 100 + 33 + 33 = 166frames.

And: (No WC data provided): (z161) - (z133) = 28 frames 33 frame count jump from z100-z133 28 + 33 = 61 frames added to z100 = start the data input.

http://educationforum.ipbhost.com/topic/22692-swan-song-math-rules/?do=findComment&comment=372597 100ft / 3.294ft per sec (z161-z166) = 30.358... sec x 16fps = 485.73 total frames 486frames/18.3fps = 26.557 sec 30.358... sec - 26.557... sec = 3.8 sec x 16fps = 60.8 frames z100 + 60.8 frames = z161= start the data input. P.S. 3.292ft x 100 = 329.2 ft = the Station# beginning data entry for z161.

Limo on film starts at z133. Data entry (CE884) starts at z161. Why? Because one frame rate must eventually catch up with the other. Limo on film should start with z100 corresponding to StationC. Difference between z100 and z161 = 61 frames.

More of the removal process: Jump 33 frames. Remove 30.2ft (no WC data provided): Station# 329.2 (z161) - 299.0 (z133) = 30.2ft 30.2ft/33frames = .915ft per frame Converted to whole frames: .915 x (18/18.3) = .9ft per frame = Shaneyfelt = 11.2mph Throw in Myers 9.8mph average to sync: .9ft per frame x 16fps = 14.4fps/1.47(1mph) = 9.8mph http://educationforum.ipbhost.com/topic/22692-swan-song-math-rules/?do=findComment&comment=374647

Same frame jump, different source, per preceding post. Myers 167/ 22.8fps = 7.324...sec x 4.5 frame difference (22.8 - 18.3) = 32.96 frames Added on edit: Meant for Swan Song topic. Sorry about that.

John, Viewing Towner's film, the only person I see (your red arrow from Martin) is the lady in red. https://drive.google.com/file/d/1LzGsRmWZUcKoYCITNvyX3kii6IaqsI7-/view?usp=sharing

The distance from StationC to Towner end = 60.245ft The frame count = 94. The approx 6.18 frames for (Towner end to Z133) added in = 100.18 frames. That ratio becomes 60.245/100.18 = .601 = 100ft/166frames = .60ft per frame Starting from StationC, using the extant Towner film and adding the frame gap between (Towner end and z133), the zfilm count would start with frame 100, whereas it is now labeled as z133.

Which will lead you back to this:

https://drive.google.com/file/d/1utfgSG7vG9yOyXYvEmpnZ7Ql0obz-2IS/view?usp=sharing If you will notice in the graphic provided, the jump in the Towner film occurs at frame 83, which is half (167) of the total extant Towner frames. The jump in Towner (aligns JFK with the TSBD corner) is an adjustment for the gap between the Towner end and Z133. Final 20 Towner frames @ 12.5ft = 6.8mph @ 16fps z133-z149 - Station# 299.0 - 318.125 = 19.125ft = 13.01mph @16fps 13.01 + 6.8 = 19.81 /2 = 9.905 mph average 9.905 x 1.47 = 14.56ft per sec 5.625ft ( Gap between Towner end and Z133) / 14.56ft = .386....sec .386...sec x 16fps = 6.18frames Towner splice at TSBD corner = 7 frames missing via Myers. Like this: 83 + 7 missing = 90 167 - 90 = 77 + 6.18 = 83.18

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

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

Chuck, Two shots close together. Yes How many others????

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

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

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

Thanks David, Two more for you: (18/48) x (2) x (1.2) x (1/.9) = whole frames (1 - .11111) x 18 = frame rate

Something like this: https://drive.google.com/file/d/1wBURGPKHRQGA9VpeUhLic6LRe2OEYgnS/view?usp=sharing

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 .........