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Chris Davidson

Swan-Song -- Math Rules

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On 3/18/2018 at 11:47 PM, Chris Davidson said:

The WC was polite enough to give us their working average for the blue triangle span. 

Of course, when is frame166 not representative of 166frames? When the WC is involved.

100ft/166frames (actually 166.666... for another time though) = .6024ft per frame x 18.3fps = 11.024ft per sec  = 7.5mph

Since their distance from StationC to PositionA was 44ft, we can arrive at a frame count for that span.

44ft/11.024ft per sec = 3.991... x 18fps (whole frames) = 71.84frames

 

 

Ron,

Let me try to simplify the concept with this:

The graphic shows what Myer's specs would be for the limo travel between the Dal-Tex corner and the TSBD corner.

He shows 18 frames and 12.7ft traveled.

I plotted the same span but use Robert West's path (original surveyor) for my measurements.

The difference between the two (Myers and me) is within the red boxes.

The frame count difference is 28-18 = 10.

The distance difference is 23.75 - 12.7 = 11.05ft.

11.05ft over an 18.3 frames per sec span = .603ft per frame

Do you see that the distance per frame removed was in keeping with the (distance per frame travel average via the WC)?

40213009354_8f6bd6098f_b.jpg

 

 

 

 

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5 hours ago, Chris Davidson said:

Do you see that the distance per frame removed was in keeping with the (distance per frame travel average via the WC)?

40213009354_8f6bd6098f_b.jpg

 

 

 

 

 

In other words,

If someone told you there was a 7 frame splice in a film, the removal of 7 (alternating/additional) frames within that same splice, would still allow the limo to move without any hiccups whatsoever because the speed per frame (.6ft) represented the average speed.

26053558987_71988bdf8c_b.jpg 

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On 3/17/2018 at 1:24 PM, Chris Davidson said:

40823594092_677602f5ae_z.jpg

The "official" WC documentation put the distance between StationC and extant z161@ 94.7ft.

If you care to use my "unofficial" path described above, my distance = 45.8ft + 48.9ft = 94.7ft.

The same distance. What a surprise.

Can you guess what geometric shape is formed when combining the official and unofficial paths?

 

 

David, you asked me about 166 frames I believe.

The Shaneyfelt path, the upper part of the blue triangle = a distance difference of 50.7ft (see red boxes above).

169 frames x .3ft per frame = 50.7ft = distance between PositionA and extant CE z161.

169frames -166 frames = 3 frames

3 frames = the span of Orange colored CE884 z168-z171

z168-z171 = a speed of .3ft per frame.

3frames x .3 = .9ft = distance traveled for CE884 z168-z171

 

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166 frames x .3 ft per frame = 49.8ft.

But, with the PositionA pivot:

48.9ft (West path) + .9ft (=  +3 frames) = 49.8ft

50.7ft (Shaneyfelt path) - .9ft (=. -3 frames) = 49.8ft

Same distance, different paths to get there.

 

 

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22 hours ago, Chris Davidson said:

In other words,

If someone told you there was a 7 frame splice in a film, the removal of 7 (alternating/additional) frames within that same splice, would still allow the limo to move without any hiccups whatsoever because the speed per frame (.6ft) represented the average speed.

26053558987_71988bdf8c_b.jpg 

There are additional frames missing within the splice, which means there are additional frames missing outside the splice. (Math to the rescue).

This means Towner's film possessed more frames originally than what Meyers based his 22.8FPS rate on.

Since Towner's camera was an 8mm camera, there was no 24fps setting for it.

Myer's knew this. It would have been too obvious to list 24fps as the sync rate for the multi-film project he did.

This leaves one option for the original Towner frame rate. Can you guess what rate she filmed at?

Zapruder comes to mind.

 

 

 

 

 

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On 3/20/2018 at 8:57 AM, Chris Davidson said:

He shows 18 frames and 12.7ft traveled.

The frame count difference is 28-18 = 10.

 

40213009354_8f6bd6098f_b.jpg

 

 

 

 

 

The frame span for "JFK within the limo" using Robert West's path is 28 frames.

28 x 2.6666 (48/18) = 74.666 frames

74.666/2 (1st pass) = 37.333 frames

37.333/2 (2nd pass) = 18.666frames

Myers has the same span as 18frames.

 

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Fine Tuning through 3 splices

1336/48 = 27.833... sec

27.8333...sec x 18fps = 501 frames - 15 frames = 486frames

167 + 167 + 167  = 501frames

1336 - 354.66 (133 x 2.6666) = 981.34

981.34 / 2 (remove 1/2) = 490.67

490.67 x .75 (remove 1/4) = 368

368-353 = 15 frames

15 frames / 3 splices

 

 

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https://drive.google.com/file/d/1bRS49wIfblFnXYW8bj9iL3oUnIU3MXRX/view?usp=sharing

The difference between the two graphics, via Myer's work is 108%

As can be seen, when the top frame layer is increased 108%, the plat becomes the same size.

Working from Station# 250.0 (SE TSBD window edge) to Station# 329.2 (CE884-z161) = 79.2ft

79.2ft x 1.08% = 85.536ft 

250 + 85.536ft = 335.536ft

335.53ft - .9ft = Station# 334.636 (See speed increments below)

Now, tie that to the TSBD corner = Station# 245.5 = 4.5ft short of the (SE TSBD window edge).

245.5 + 85.536 = 331.03

331.03 - .9 = Station# 330.13

CE884 z166 =  Station# 330.10

Now, take the average speed (see below) of z133-z166:

14.88 + 11.99 = 26.87/2 = 13.435mph

13.435 x 1.47 (1mph)  = 19.749ft per sec /18.3 = 1.0792ft per frame = (1.08)

= 5frames @ 1.08ft per frame = 5.4ft - .9ft(BS distance from z161-z166) = 4.5ft = Distance from SE TSBD window edge to TSBD corner

 

Average speeds via West path (in which the splices reside) using "JFK within limo" as measuring target:

z133-z149 = 14.88mph.  Station# 299 - 318.125

z149-z166 = 11.99mph.  Station# 318.125 - 334.5 = (329.2 + 5.3)

z166-z182 = 13.53 mph.   Station# 334.5 - 351.9 

z182-z201 = 11.40mph.    Station# 351.9 - 369.3

z201- z222 = 10.83mph.   Station# 369.3 - 387.58

Average =  12.52 mph

18.3/1.47 = 12.45mph = 1ft per 1 frame

 

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11 hours ago, Chris Davidson said:

https://drive.google.com/file/d/1bRS49wIfblFnXYW8bj9iL3oUnIU3MXRX/view?usp=sharing

The difference between the two graphics, via Myer's work is 108%

As can be seen, when the top frame layer is increased 108%, the plat becomes the same size.

Working from Station# 250.0 (SE TSBD window edge) to Station# 329.2 (CE884-z161) = 79.2ft

79.2ft x 1.08% = 85.536ft 

250 + 85.536ft = 335.536ft

335.53ft - .9ft = Station# 334.636 (See speed increments below)

Now, tie that to the TSBD corner = Station# 245.5 = 4.5ft short of the (SE TSBD window edge).

245.5 + 85.536 = 331.03

331.03 - .9 = Station# 330.13

CE884 z166 =  Station# 330.10

Now, take the average speed (see below) of z133-z166:

14.88 + 11.99 = 26.87/2 = 13.435mph

13.435 x 1.47 (1mph)  = 19.749ft per sec /18.3 = 1.0792ft per frame = (1.08)

= 5frames @ 1.08ft per frame = 5.4ft - .9ft(BS distance from z161-z166) = 4.5ft = Distance from SE TSBD window edge to TSBD corner

 

Average speeds via West path (in which the splices reside) using "JFK within limo" as measuring target:

z133-z149 = 14.88mph.  Station# 299 - 318.125

z149-z166 = 11.99mph.  Station# 318.125 - 334.5 = (329.2 + 5.3)

z166-z182 = 13.53 mph.   Station# 334.5 - 351.9 

z182-z201 = 11.40mph.    Station# 351.9 - 369.3

z201- z222 = 10.83mph.   Station# 369.3 - 387.58

Average =  12.52 mph

18.3/1.47 = 12.45mph = 1ft per 1 frame

 

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

 

 

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Overall compensation starting from extant z133 for 18.3fps vs 18fps:

48/18 =    2.666...

48/18.3 = 2.622...

353 x 2.666... = 941.33

353 x 2.622... = 925.90

                               15.43 frames  vs 15.41ft  

Even closer for gubermint work.

Remember, at the end of the day, we are dealing with whole frames.

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21 hours ago, 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

 

 

CONVERTING .18FT PER FRAME:

1.08ft per frame = (average speed z133-z166) - .18ft = .9ft per frame = 11.2mph= Shaneyfelt 

.9 x (5/6) = .75  (reducing by 1/6)

 1 whole frame / (5/6) = 1.2

1.2 x .75 = .9

.75 = (18/48) x 2 = remove 1/2 the frames on the first pass.

.18/1.08 = .166666

.16666 converted to frame removal = 1/6

5/6 + 1/6 = 1 whole frame

 

 

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So, if I had a non-data supplied 30.2ft span, let's say circa z133-z161 and I wanted to remove 5 frames within that 30.2 ft span, one way I could proceed is remove 1 out of every 6 frames over a 30.2frame span = 5 frames.

Granted, this would not be a cluster of frames as the Zfilm appears to show, but a sneaky method what so ever.

Those 5 removed frames at 1.08ft per frame = 5.4ft of unaccounted for, distance.

But, if I changed the path of the limo in relationship to Z's LOS, somewhere before the non-data span began, the accommodation for the unaccounted distance would be complete.

 

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On 3/27/2018 at 9:40 PM, Chris Davidson said:

So, if I had a non-data supplied 30.2ft span, let's say circa z133-z161 and I wanted to remove 5 frames within that 30.2 ft span, one way I could proceed is remove 1 out of every 6 frames over a 30.2frame span = 5 frames.

Granted, this would not be a cluster of frames as the Zfilm appears to show, but a sneaky method what so ever.

Those 5 removed frames at 1.08ft per frame = 5.4ft of unaccounted for, distance.

But, if I changed the path of the limo in relationship to Z's LOS, somewhere before the non-data span began, the accommodation for the unaccounted distance would be complete.

 

incredible Chris...

the 5 frames removed... that's AFTER the 2 initial passes to reduce frames....  not from the 48fps film.  yes?

Did this a year ago...  Red line is WEST's path of the limo, Green line is Shaneyfelt path 1.1 foot south...

JFK at the Zfilm's 166 - with the same Line of Sight - and the change to CE884... moves the location of 166 back 5.4 feet....

By 171 the front left tire of the limo has reached the lane strip... 

 

 

 

yet the movement actually puts the arbitrary 171 farther back than the zfilm 166...  

Even Shaneyfelt messes his testimony up a bit...  if the first frame was 171, how did it become 161?

 

 

Edited by David Josephs

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2 hours ago, David Josephs said:

the 5 frames removed... that's AFTER the 2 initial passes to reduce frames....  not from the 48fps film.  yes?
 

 

 

 

 

 

 

 

 

David,

Just a slight tweak on that.

The more frames they have initially, the less distance the limo appears to move per frame, especially after a block of frames has been removed.

If you look back at my speed increments for z133-z149(average 14.88mph) or from z133-z166 (average 13.44mph) and break that down into a 48fps film:

14.88 x 1.47 (1mph) = 21.87ft per sec

13.44 x 1.47 (1mph) = 19.76ft per sec

21.87 / 48 = .455ft per frame x 5 removed frames = 2.275ft

19.76 / 48 = .411ft per frame x 5 removed frames = 2.058ft

Converted from the extant film, 2 frames would equal a distance of either:

2 x 1.08ft (13.44mph) = 2.16ft

2 x 1.195ft (14.88mph) = 2.39ft = distance between JFK and Connally in the limo.

So, when the extant film starts, between z133-z135 (2 frame span), what happens to the film?

P.S. Graphic representation to follow, in a little while.

 

 

 

 

 

 

 

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18 minutes ago, Chris Davidson said:

David,

Just a slight tweak on that.

The more frames they have initially, the less distance the limo appears to move per frame, especially after a block of frames has been removed.

If you look back at my speed increments for z133-z149(average 14.88mph) or from z133-z166 (average 13.44mph) and break that down into a 48fps film:

14.88 x 1.47 (1mph) = 21.87ft per sec

13.44 x 1.47 (1mph) = 19.76ft per sec

21.87 / 48 = .455ft per frame x 5 removed frames = 2.275ft

19.76 / 48 = .411ft per frame x 5 removed frames = 2.058ft

Converted from the extant film, 2 frames would equal a distance of either:

2 x 1.08ft (13.44mph) = 2.16ft

2 x 1.195ft (14.88mph) = 2.39ft = distance between JFK and Connally in the limo.

So, when the extant film starts, between z133-z135 (2 frame span), what happens to the film?

P.S. Graphic representation to follow, in a little while.

 

 

 

 

 

 

 

Ok buddy... help me understand...  (133-135 is a 3 frame span though...)

Here is 133, 134, 135, 136 & 137 stabilized around the light pole

z133---z137.gif.aa810f776a7ada94be71576500cd8f29.gif

 

or not stabilized and the camera shift shows a perfect 1 frame movement... kinda looks like a splice after a camera jiggle as a result of a shot....

z133---z136--shift-left.thumb.gif.30517247be368806b4e06bd95d135fdc.gif

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