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Posts posted by Chris Davidson
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48/18(whole frames) = 2.6666..... /2 = 1.3333.....
/2 = cutting a 48fps film into half the frames (1st pass)
1.333.. converted to a fraction = 4/3.
The reciprocal being 3/4 bringing that 4/3 back to a 1 to 1 ratio.
3/4 of the remaining frames from extant z133 were retained.
3/4 + 1/4 = 1
After the elimination of all frames from Z beginning (turn onto Elm) to extant z133, they began the next passes(multipliers) incorporating this from extant z133 forward.
The combined total frames excised from extant z133 onward equals 1/4 of those remaining frames.
This process does not include the variable of a limo stopping (more frames created) as there is no way to distinguish "math wise" whether it stopped or not. Imo
Added on edit: I was wrong, they didn't remove 2/3 of all frames, they removed at least:
353/1296 = 1 - .272 = .728 = 72.8%
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The first and second multipliers are used in the CE884 version which resides on the WC final plat of May 1964.
They are applied in the (surprise-surprise) first 18 frame entries.
21.6ft/18frames = 1.2ft per frame. 1st Multiplier
Expand to 18.3fps = 21.6/18 x 18.3 = 21.96ft per sec / 1.47 (1mph) = 14.938mph
Working from the 11.2mph WC designated average:
11.204 x 1.2(1st Multiplier) = 13.444mph
13.444 x 1.1111111 (2nd Multiplier) = 14.938mph
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4 hours ago, Chris Davidson said:
The 1.2 multiplier (remove 1/6 frames) that converts .915 back to .9 (18FPS-whole frames) = 2.24mph speed change
This is what CE884 (Z161-Z166 @.9ft traveled)reveals.
Apply this to their average of 11.2mph + 2.24 = 13.44mph which is what I plotted, speed wise, for the limo on the extant film from z156-z166(right at a film splice).
If we want to get .9ft back to 1ft or a 1to1 ratio of frames to ft per frames, we must divide 1 by .9 = 1/.9 = 1.111111111
This is our second working multiplier.
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16 hours ago, Chris Davidson said:
We know an excise occurred which was 1/6 of the frames = 30.5 /183
So, starting at extant z133:
.7625 x (1.2 = (removal of another 1/6 total frames) =.915 to 1 ratio
But, we have to deal in whole frames.
So, .915 x18 / 18.3 = .9ft per frame
The 1.2 multiplier (remove 1/6 frames) that converts .915 back to .9 (18FPS-whole frames) = 2.24mph speed change
This is what CE884 (Z161-Z166 @.9ft traveled)reveals.
Apply this to their average of 11.2mph + 2.24 = 13.44mph which is what I plotted, speed wise, for the limo on the extant film from z156-z166(right at a film splice).
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19 minutes ago, David Josephs said:
The Towner film must have been the deciding factor then...
Syncing to that turn (I'd like your thoughts on the Hughes image of that turn as it does not appear they swing wide)
would necessitate a greater removal... plus, if Truly's account and the wide turn are real, the limo would be going much slower requiring more frames to be excised.Yes,
I'd say the Towner film limited them in what was possible with the true Zfilm, so they had to excise that same range of frames.
Myer's couldn't sync the turn unless he increased Towner's camera FPS to 22.8 and use the drivers side rear tire as his measuring stick, as opposed to JFK within the limo.
I don't see where the wide turn was possible given what I have plotted using Towner's film.
I'm fairly certain the Towner splice was for timing purposes as an adjustment between 18 and 18.3fps overall.
In other words, 486/18 = 27 seconds
486/18.3 = 26.557 seconds
.443seconds x 18fps = 7.97 frames (Myers had it as 7 missing frames, so it's fairly close for syncing with his BS math).
I'll post some other math items relating to Position A and such, a little later, but for now, I'm returning to the breakdown of that 1/6 frame removal pass.
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David,
I'm thinking for the last full pass from extant z133 on, they used 1/3 of 1/2 = 1/6, instead of the 1/2 of 1/2 = 1/4.
This would have left them with approx 39 frames for random removal. A little leeway.
This appears to work out in keeping the 1 to 1 ratio intact among other things. I'll follow this up with some split time/speed comparisons.
In the end, the same 1/4 frame removal amount would have been accomplished.
What do you think?
1296 = 486 x (48/18)
354.6666666 = 133 x (48/18)
941.33333 = (1296 - 354.66666)
470.666666 = (941.3333 / 2)
470.666666 - 25% = (1/2 x 1/2) =
353470.666666 - 353 = 117.666
How many frames from 470.666 to 486?
What's the difference between z118 and z133?
470.666666 - 16.6666% = (1/2 x 1/3) =
392.22 - 353 =
39.22 Remaining470.666666 - 353 = 117.666
How many frames from 470.666 to 486?
What's the difference between z118 and z133?
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.9ft per frame x 18.3FPS = 16.47ft per sec /1.47 (1mph) = 11.2mph = Shaneyfelt Testimony
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We know an excise occurred which was 1/6 of the frames = 30.5 /183
So, starting at extant z133:
.7625 x (1.2 = (removal of another 1/6 total frames) =.915 to 1 ratio
But, we have to deal in whole frames.
So, .915 x18 / 18.3 = .9ft per frame
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Everything up to extant z133 has been excised.
Lets start from extant z133 moving forward.
If I have a 48FPS slo-mo film to begin with and I remove 1/2 the frames in one pass, that leaves me with a 24FPS film, but not in terms of movement within the film.
The final frame rate of 18.3/48 = a movement of .38125.
But 1/2 the frames have been excised (remember when you remove frames, you increase movement within the film) by the time we get to extant z133.
So the movement becomes .38125 x 2 = .7625 of the original 48FPS film.
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Remember though,
Films work in whole frames (1/1 ratio), so to convert 5/6 back to a whole, we need a multiplier:
5/6 x 1.2 = 1
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2 hours ago, Chris Davidson said:
Since 1.525 was the exact conversion for 18.3ft horizontal/1ft vertical, the frame reduction was from 183 to 152.5 (z161-z313)
30.5 frames /183 = .166666..... frame reduction
30.5frames/30.2ft reduction.
Their math will bring the equation back to a 1 to 1 ratio or approx. thereof.
If you will convert .166666... into a fraction you have a 1/6 frame reduction.
That means you have a remainder of 5/6 of the frames.
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3 hours ago, Chris Davidson said:
http://educationforum.ipbhost.com/topic/24596-shooter-location/?do=findComment&comment=371368
Why am I continuing this here?
Because the end result of the math used to eliminate the remaining approx 33frames/30ft/1.8 seconds resides in the link above.
It just wasn't completed in one fell swoop.
Extant zframe133 when plotted at Station# 299.0 to the beginning of (CE884) Station# 329.2 = 30.2ft
There is no data for the span z133-z166 = 33 frames.
There is only BS data for z161-z166.
33/18fps (need a whole frame total)=1.8333...seconds
30.2ft/1.8333... seconds = 16.47ft per sec = 11.2mph
Do you understand what remainder was eliminated in terms of an overall equation which would include the span of z133-z166 when an average speed of 11.2mph was used, but didn't include the span of z133-z166?
Mr. SPECTER. Is that a constant average speed or does that speed reflect any variations in the movement of the car?
Mr. SHANEYFELT. That is the overall average from 161 to 313. It does not mean that it was traveling constantly at 11.2, because it was more than likely going faster in some areas and slightly slower in some areas. It is only an average speed over the entire run.
Mr. DULLES. Over the entire run between what points?
Mr. SHANEYFELT. Between frame 161 and 313.Once again, what part of the equation (frame#) did the limo first appear in and what part of the equation did the data begin with?
http://educationforum.ipbhost.com/topic/24596-shooter-location/?do=findComment&comment=371368
This span was 183 frames. They ended up with 152 frames from z161-z313.
Notice the plat designation goes beyond z313 on Drommer.
Since 1.525 was the exact conversion for 18.3ft horizontal/1ft vertical, the frame reduction was from 183 to 152.5 (z161-z313)
30.5 frames /183 = .166666..... frame reduction
30.5frames/30.2ft reduction.
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1 hour ago, Ray Mitcham said:
I had a Bell an Howell 8mm standard film cine camera in the seventies. (OK, I know it's not the same camera as Zapruder used!) The camera did exactly what David mentions above. Every time you stopped the camera the first frame was always either under or over exposed as the meter inside the camera tried to get the correct setting. Anybody who says that there wasn't a splice where David says above is talking through his or her backside.
Not only that, but Zapruder said he started filming before the limo turned onto Elm St, as he wanted to film the turn. he never said that he stopped his camera to save film, as the desperados say.
Ray,
I don't want to get into a discussion about the non-math elements of why, what you said is true.
I'm showing what math was used which allowed the WC to try and fool who, you are directing your comments toward.
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30 minutes ago, David Josephs said:
Data begins with Station C and Position A... and z168.
Yes it does.
But in the spirit of trying to simplify what I'll be providing, I'm going to tie it back to extant z133 for now.
Chris
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http://educationforum.ipbhost.com/topic/24596-shooter-location/?do=findComment&comment=371368
Why am I continuing this here?
Because the end result of the math used to eliminate the remaining approx 33frames/30ft/1.8 seconds resides in the link above.
It just wasn't completed in one fell swoop.
Extant zframe133 when plotted at Station# 299.0 to the beginning of (CE884) Station# 329.2 = 30.2ft
There is no data for the span z133-z166 = 33 frames.
There is only BS data for z161-z166.
33/18fps (need a whole frame total)=1.8333...seconds
30.2ft/1.8333... seconds = 16.47ft per sec = 11.2mph
Do you understand what remainder was eliminated in terms of an overall equation which would include the span of z133-z166 when an average speed of 11.2mph was used, but didn't include the span of z133-z166?
Mr. SPECTER. Is that a constant average speed or does that speed reflect any variations in the movement of the car?
Mr. SHANEYFELT. That is the overall average from 161 to 313. It does not mean that it was traveling constantly at 11.2, because it was more than likely going faster in some areas and slightly slower in some areas. It is only an average speed over the entire run.
Mr. DULLES. Over the entire run between what points?
Mr. SHANEYFELT. Between frame 161 and 313.Once again, what part of the equation (frame#) did the limo first appear in and what part of the equation did the data begin with?
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I believe, at this junction, it's appropriate to post the remaining information to my previous "Math Rules" topic.
It should become readily apparent, why I'm crossing over.
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On 2/26/2018 at 9:18 PM, Chris Davidson said:
David,
1.525frames per inch = 18.3frames per 12inches = 1second per 12inches
Now expand 18.3 frames per 12 inches (ten-fold):
183frames = 120inches
Now what span would give us 183 frames?
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Of course 11.2mph was going to be the average limo speed.
Do you think Shaneyfelt had to cross-multiply those previous results when he was figuring out where the limo was going to start on film?
11.2mph x (1.47ft per sec =1mph) = 16.47ft per sec x 8.333... sec = 137.25ft
7.5ft elevation (CE884 Z161-Z313) x 18.3ft per elevated ft = 137.25ft
Mr. SHANEYFELT. That is the overall average from 161 to 313. It does not mean that it was traveling constantly at 11.2, because it was more than likely going faster in some areas and slightly slower in some areas. It is only an average speed over the entire run.
Mr. DULLES. Over the entire run between what points?
Mr. SHANEYFELT. Between frame 161 and 313. -
In case you didn't realize it:
152.5 frames /18.3 fps = 8.3333... seconds
This, in keeping with 8.333...ft total elevation.
Which equals the 1 to 1 ratio.
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Others like to refer to them as simple mistakes.
Here's the next one:
The elevation change for the 152 frame span = 429.25 - 421.75 = 7.5ft
152.5 frames =100 inches = 8.333..ft
What do you think the difference between these two elevations is?
Might it be the 10" vert = 15.25 horizontal drop the WC forced you to eat from z161-z166.
You know, they fed you d-g sh-t and told you it was prime rib.
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15.25frames = 10 inches
Multiply 10x once more, for a ratio of:
152.5 frames =100 inches= 8.333...ft
This sets the total frame span that is utilized in CE884
313-161 = 152 frames
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12 hours ago, Chris Davidson said:
David,
I take it you meant frames per sec, not feet?
The 1 to 1 ratio applies to the anywhere from (10.2ft-10.24ft) span =.56ft = 6.7inch WC adjustment.
1.525frames per inch = 18.3frames per 12inches = 1second per 12inches
or 1/1.525 = .655…inches per frame x 10.22 frames = 6.7… inches = CE560My conversion using the z207 decipher below:
1/1.528 = .65445.. x 10.24 framest = 6.7 inches = CE560
You can match the above bolded equation to (10" vert. = 15.25ft horizontal) multiplying by a factor of 10:
15.25 frames = 10 inches
This in keeping with the 1-1-1 ratio.
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The acoustical closely aligns with the 10.22-10.24, a .02 frame span difference.
z313 - (104.31frames)= 5.7 seconds = z208.69
From z207 -z208.69 = 1.69frames
If you look at CE884 z207-z222 = 14.8ft per 15 frames (12.28mph) = .986666... ft per frame
.98666... x 1.69 frames = 1.6674ft
The 1 to 1 ratio of frames to horizontal ft would dictate a :
1.69ft - 1.6674...ft = .0225ft difference, which pretty much matches the 10.22-10.24 frame span difference.
In terms of speed, instead of 12.28mph, an average of 12.44mph would give you the 1.69 frame duration.
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9 hours ago, David Josephs said:
Ergo, we do not get a film speed between 16 and 18 feet per second... but an impossible 18.3 feet per second...
what a coincidence... well done
David,
I take it you meant frames per sec, not feet?
The 1 to 1 ratio applies to the anywhere from (10.2ft-10.24ft) span =.56ft = 6.7inch WC adjustment.
1.525frames per inch = 18.3frames per 12inches = 1second per 12inches
or 1/1.525 = .655…inches per frame x 10.22 frames = 6.7… inches = CE560My conversion using the z207 decipher below:
1/1.528 = .65445.. x 10.24 framest = 6.7 inches = CE560
Shooter Location
in JFK Assassination Debate
Posted
You can move that elevation relationship from Position A (428.7) to extant z313 (418.48) = 10.22ft
The extra .22ft x 18.3 = 4.02ft = the Drommer label for a location near, but not indicative of extant z313.
Equals the very close 2nd head shot.