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# Swan-Song -- Math Rules

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Give it up.

This is an excerpt from Douglas Horn's research: The Two NPIC Zapruder Film Events:

"It could have happened this way—consider this: 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 (of which only 6 feet, 3 inches contains the imagery of the assassination film, and 2 feet, 7 inches is black, unexposed film with no image showing); then there is a physical splice; then there is a segment of black film."

I'll break this down in math terms for everyone:

8ft 10 inches in terms of total frames pertaining to a 16mm/24 frames per sec film.

There are 40 frames per physical foot of film using 16mm film.

8ft 10 inches = 8.83ft

8.83ft x 40 frames per foot =353.2 frames

The first supposed 132 frames are missing from the Z film.

353.2 + 132 = 485.2 frames Rounded to the next frame = 486 frames = extant Zfilm total

The noose is getting tighter.

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Lets get Itek's opinion back in 1967 I believe.

The following is an excerpt from that study:

Added on edit: The last 12 frames designation refers to "until the headshot" extant z301-z313. Sorry I didn't specify that in the graphic".

If a car is traveling at 3.74 mph on film and you remove 1/2 the frames, the car would appear to travel at 7.48mph.

Every fourth frame. "Danger Will Robinson Danger"

Itek was very close at 7.6mph.

Edited by Chris Davidson
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And, if you would like to connect the 3.74mph( actually it's 3.734mph) from the frames prior to the extant 313 headshot back to where it's introduced on CE884 at z168, it looks like this:

z168-z186 = 18 frames @ 21.6ft traveled

We need to work with a second of time.

18.3/18 = 1.0166… x 21.6ft = 21.96ft per second

21.96/1.47 (1mph) = 14.938mph

3.734 mph remove half the frames, limo travels at 7.468mph

7.468mph remove half the frames, limo travels at 14.936mph

It's all math.

Why do you think Itek measured every fourth frame at that point in time?

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This is an excerpt from Douglas Horn's research: The Two NPIC Zapruder Film Events:

"It could have happened this wayconsider this: 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 (of which only 6 feet, 3 inches contains the imagery of the assassination film, and 2 feet, 7 inches is black, unexposed film with no image showing); then there is a physical splice; then there is a segment of black film."

Chris,

IIRC, there is a single splice in what is allegedly the ORIGINAL Z-film - is this correct? And if so, then only the copies represent the "assassination movie" in its alleged original form. If this is true, and any ORIGINAL frames are 'missing' at the splice; no alteration of any destroyed frames in the original was required - just alteration or fabrication of the copy...

Tom

Edited by Tom Neal
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You could always take the slow-motion version I supplied previously, count the total number of frames (144) I believe, cut that amount in half, then half it again:

Or 144/2 =72/2 = 36

z171- z207 = 36 frames: Refer to overlay video.

Amazing!!!

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This is an excerpt from Douglas Horn's research: The Two NPIC Zapruder Film Events:

"It could have happened this wayconsider this: 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 (of which only 6 feet, 3 inches contains the imagery of the assassination film, and 2 feet, 7 inches is black, unexposed film with no image showing); then there is a physical splice; then there is a segment of black film."

Chris,

IIRC, there is a single splice in what is allegedly the ORIGINAL Z-film - is this correct? And if so, then only the copies represent the "assassination movie" in its alleged original form. If this is true, and any ORIGINAL frames are 'missing' at the splice; no alteration of any destroyed frames in the original was required - just alteration or fabrication of the copy...

Tom

Tom,

I don't know what the Original film consists of. I can only refer to the splices in the extant film.

The alteration of any destroyed frames (originals that we'll never see) might not have been necessary, but the extant film is not clean enough. imo

In terms of the splices and frame count, look back at my overall scenario.

Cut the frame total in half to 462.78 (1st pass on Optical printer)

462.78 - 353 = 109.78 to get rid of.

109.78/462.78 = .237 closest whole frame ratio would be 1/4 (surprised) or .25

Difference between .2372 and .25 in terms of 925.56 total frames = .0128 x 925.56 = 11.84 frames or 12 whole frames.

Removal of 12 original frames so the math gets back to whole fames.

CE884 WC published version z161 Station# 3+29.2 obtained by eliminating 7 frames = z168 = Station# 3+29.2 from CE884 WC final plat version May 1964.

Splice at z133, approx z157, splice at approx z208.

12 original frames left to vanish over 3 splices from an original 48fps film. Piece of cake.

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CHRIS:

In post 409, when you refer to Horn and then say, "It could have happened this way"

Is the info that follows Horn's assumptions, or is it based upon his interviews?

You don't give a page number or textual source. I actually have his five volume set.

Edited by James DiEugenio
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IIRC, there is a single splice in what is allegedly the ORIGINAL Z-film - is this correct? And if so, then only the copies represent the "assassination movie" in its alleged original form. If this is true, and any ORIGINAL frames are 'missing' at the splice; no alteration of any destroyed frames in the original was required - just alteration or fabrication of the copy...

Tom,

I don't know what the Original film consists of. I can only refer to the splices in the extant film.

The alteration of any destroyed frames (originals that we'll never see) might not have been necessary, but the extant film is not clean enough. imo

In terms of the splices and frame count, look back at my overall scenario.

Cut the frame total in half to 462.78 (1st pass on Optical printer)

462.78 - 353 = 109.78 to get rid of.

109.78/462.78 = .237 closest whole frame ratio would be 1/4 (surprised) or .25

Difference between .2372 and .25 in terms of 925.56 total frames = .0128 x 925.56 = 11.84 frames or 12 whole frames.

Removal of 12 original frames so the math gets back to whole fames.

CE884 WC published version z161 Station# 3+29.2 obtained by eliminating 7 frames = z168 = Station# 3+29.2 from CE884 WC final plat version May 1964.

Splice at z133, approx z157, splice at approx z208.

12 original frames left to vanish over 3 splices from an original 48fps film. Piece of cake.

Thanks for the response and the additional information. BTW, I am having NO PROBLEM following your logic.

Tom

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I don't seem to be getting through. To save everyone the bother, I'll summarise the next few days' interactions and we can all go away and do something more interesting:

CD: on CE884 z186-z207 = 20.3ft traveled in 21 frames.

JB: So what? If the data in CE 884 contradicts what we see in the Zapruder film, the obvious answer is not that the film has been faked, but simply that the data is faulty. What grounds are there for assuming that the data in CE 884 is reliable?

CD: 18.3/18 = 1.0166... x 21.6ft = 21.96ft per second!

JB: So what? If you're using faulty data, you are just playing mathematical games, and none of this has anything to do with the actual Zapruder film.

CD: 21.96/1.47 (1mph) = 14.938mph!

JB: So what? If the data in CE 884 contradicts what we see in the Zapruder film, the obvious answer is not that the film has been faked, but simply that the data is faulty. What grounds are there for assuming that the data in CE 884 is reliable?

CD: z171-z185 = 17.07mph!

JB: So what? If you're using faulty data, you are just playing mathematical games, and none of this has anything to do with the actual Zapruder film.

CD: 41ft / 1.5seconds = 27.33ft per sec = 18.594 mph!

JB: So what? If the data in CE 884 contradicts what we see in the Zapruder film, the obvious answer is not that the film has been faked, but simply that the data is faulty. What grounds are there for assuming that the data in CE 884 is reliable?

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I don't seem to be getting through. To save everyone the bother, I'll summarise the next few days' interactions and we can all go away and do something more interesting:

CD: on CE884 z186-z207 = 20.3ft traveled in 21 frames.

JB: So what? If the data in CE 884 contradicts what we see in the Zapruder film, the obvious answer is not that the film has been faked, but simply that the data is faulty. What grounds are there for assuming that the data in CE 884 is reliable?

CD: 18.3/18 = 1.0166... x 21.6ft = 21.96ft per second!

JB: So what? If you're using faulty data, you are just playing mathematical games, and none of this has anything to do with the actual Zapruder film.

CD: 21.96/1.47 (1mph) = 14.938mph!

JB: So what? If the data in CE 884 contradicts what we see in the Zapruder film, the obvious answer is not that the film has been faked, but simply that the data is faulty. What grounds are there for assuming that the data in CE 884 is reliable?

CD: z171-z185 = 17.07mph!

JB: So what? If you're using faulty data, you are just playing mathematical games, and none of this has anything to do with the actual Zapruder film.

CD: 41ft / 1.5seconds = 27.33ft per sec = 18.594 mph!

JB: So what? If the data in CE 884 contradicts what we see in the Zapruder film, the obvious answer is not that the film has been faked, but simply that the data is faulty. What grounds are there for assuming that the data in CE 884 is reliable?

You can just exchange the word through for "it".

That you fail to disregard not only the "rate x time = distance" formula, but the overlay video as well, speaks volumes.

You have contributed absolutely nothing to this topic but the same parroted message.

Save everyone the bother and go do something more interesting.

The recreation video is waiting in the wings for anyone who needs a challenge.

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IIRC, there is a single splice in what is allegedly the ORIGINAL Z-film - is this correct? And if so, then only the copies represent the "assassination movie" in its alleged original form. If this is true, and any ORIGINAL frames are 'missing' at the splice; no alteration of any destroyed frames in the original was required - just alteration or fabrication of the copy...

Tom,

I don't know what the Original film consists of. I can only refer to the splices in the extant film.

The alteration of any destroyed frames (originals that we'll never see) might not have been necessary, but the extant film is not clean enough. imo

In terms of the splices and frame count, look back at my overall scenario.

Cut the frame total in half to 462.78 (1st pass on Optical printer)

462.78 - 353 = 109.78 to get rid of.

109.78/462.78 = .237 closest whole frame ratio would be 1/4 (surprised) or .25

Difference between .2372 and .25 in terms of 925.56 total frames = .0128 x 925.56 = 11.84 frames or 12 whole frames.

Removal of 12 original frames so the math gets back to whole fames.

CE884 WC published version z161 Station# 3+29.2 obtained by eliminating 7 frames = z168 = Station# 3+29.2 from CE884 WC final plat version May 1964.

Splice at z133, approx z157, splice at approx z208.

12 original frames left to vanish over 3 splices from an original 48fps film. Piece of cake.

Thanks for the response and the additional information. BTW, I am having NO PROBLEM following your logic.

Tom

btw,

At this time, I look at the film splices in math terms.

For instance, 1/2 frames removed first pass.

The remaining film is 1/2 full.

If I choose to deal with 1/4 frames of the 1/2 remaining in a particular section, I've chosen 1/8 of the frames to deal with for that section.

1/2 eliminated first pass, 1/8 eliminated per section:

4/8 + 1/8 = 5/8 =.625

Something along that line. Hopefully that's easier to visualize for the rest.

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Way back when, Michael Walton posted a link to a SS recreation video:

The most valuable part of this version is missing.

No, I'm not insinuating that he knew of the missing segment.

The bottom frames were extracted from the video above.

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Those with a good memory will remember the above frame.

Along with a quote I made a few years back:

"I am trying to convey that 24fps or some form of it, was used as part or all of the coverup."

Didn't quite have the math worked out to fully support this premise at the time.

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https://en.wikipedia.org/wiki/Three-two_pull_down

Edited by Chris Davidson

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