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Swan-Song -- Math Rules


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21 hours ago, Chris Davidson said:

Speaking of limo exercises and the Myers 8% solution:

21.34ft / 1.08(squeeze) = 19.759ft per sec 

In case you had forgotten what matches 19.75ft:

10" drop / 12" = .833... x 18.3(vert to horiz) = 15.25ft + 5.4ft (5 frames @ 1.08ft per sec=13.44mph) = 20.65ft - .9ft (bogus distance traveled CE884 z161-z166) = 19.75ft

13.44mph x 1.47ft per sec = 19.75ft per sec

Hopefully the WC/Myers synchro from above and at street level is becoming clearer for you.

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On 6/13/2019 at 12:56 PM, Chris Davidson said:

The difference between inside radius and outside radius was 24.52ft.

Added was the Towner gap of 5.625ft (In the end, I think they "powers that be" used  5.5ft)

This leads up to, but not including the .9ft jump from z161-z166.

24.52+ 5.625 = 30.145ft / 9.125 sec (167 Towner frames@18.3 frames per sec) = 3.3ft per sec

This quite closely matches the speed of the limo at CE884 Z161-Z166 = 3.294ft per sec.

In other words, the entry for z161-z166 is a major remnant of the (3.29ft per sec = 2.24mph) adjustment  being applied to the overall equation.

 

 

 

 

Retarding the same distance.

45447043165_a46c448fc9_o.jpg

 

 

 

 

 

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23 hours ago, Chris Davidson said:

Difference in limo (spread and squeeze) distance = 23.0472ft - 19.759ft = 3.288ft per sec

Not only is it wise to keep the distance differences in mind, but the ratio's as well.

Once the 8% squeeze/spread is applied, the ratio of the limo length becomes:

23.0472/19.759 = 1.166/1

This ratio equals the limo length to vert (1ft per 18.3horizontal ft.) Elm St slope ratio:

21.34/18.3 = 1.166/1

 

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16 hours ago, Chris Davidson said:

Not only is it wise to keep the distance differences in mind, but the ratio's as well.

Once the 8% squeeze/spread is applied, the ratio of the limo length becomes:

23.0472/19.759 = 1.166/1

This ratio equals the limo length to vert (1ft per 18.3horizontal ft.) Elm St slope ratio:

21.34/18.3 = 1.166/1

 

The 1.1664... ratio can now be applied to the frame rate for syncing.

48fps / 1.1664.. = 41.15frames x .4802ft per frame = 19.76ft per sec

48- 41.15 = 6.85frames = 7 whole frames =

The difference in frame count between the two CE884's physical location for Station# 3+29.2 = z161 and z168 = 7 frames.

 

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Which in turn, circles us back to our 8% solution starting from 18.3, both in terms of frame rate and Elm St slope conversion.

18.3 x 1.08 = 19.76
19.76 x 1.08 = 21.34
21.34 x 1.08 = 23.0472

 

 

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1 hour ago, Chris Davidson said:

The 1.1664... ratio can now be applied to the frame rate for syncing.

48fps / 1.1664.. = 41.15frames x .4802ft per frame = 19.76ft per sec

48- 41.15 = 6.85frames = 7 whole frames =

The difference in frame count between the two CE884's physical location for Station# 3+29.2 = z161 and z168 = 7 frames.

 

btw,

The final CE884 version reduced the .9ft span from 7 to 5 frames.

This was to sync with the final CE884 specs of z161-z313 (136.1ft traveled @8.3 sec@18.3frames per sec.

The ratio being the same as reducing 48 frames down to 43

43/48 = .8958/1

136.1/8.3 = 16.397ft per sec / 18.3frames per sec = .896ft per frame

Both equating to 16.397 / 1.47 = 11.154mph

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21 hours ago, Chris Davidson said:

In case you had forgotten what matches 19.75ft:

10" drop / 12" = .833... x 18.3(vert to horiz) = 15.25ft + 5.4ft (5 frames @ 1.08ft per sec=13.44mph) = 20.65ft - .9ft (bogus distance traveled CE884 z161-z166) = 19.75ft

13.44mph x 1.47ft per sec = 19.75ft per sec

Hopefully the WC/Myers synchro from above and at street level is becoming clearer for you.

This .9ft remnant is very instrumental when put back into its proper context to fit the official narrative:

19.7592 + .9 = 20.6592ft

20.6592ft/23.0472ft = .896/1 ratio

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5 hours ago, Chris Davidson said:

The 1.1664... ratio can now be applied to the frame rate for syncing.

48fps / 1.1664.. = 41.15frames x .4802ft per frame = 19.76ft per sec

48- 41.15 = 6.85frames = 7 whole frames =

The difference in frame count between the two CE884's physical location for Station# 3+29.2 = z161 and z168 = 7 frames.

 

Continuing with 20.65ft:

The other CE884 gave us a range of 5 frames z161-z166.

Converted from 43 of 48 frames/1sec

43 x .4802ft per frame = 20.648ft

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On 6/15/2019 at 1:58 PM, Chris Davidson said:

 

10" drop / 12" = .833... x 18.3(vert to horiz) = 15.25ft + 5.4ft (5 frames @ 1.08ft per sec=13.44mph) = 20.65ft - .9ft (bogus distance traveled CE884 z161-z166) = 19.75ft

 

 

The above equation uses the extant z film(z154-z166) for the 5.4ft distance (5 frames @ 1.08ft per sec=13.44mph).

This does not include the .9ft lost from the true speed for a distance of 5.4 + .9 = 6.3ft.

Look back to the 1.166 ratio to connect the difference:

6.3/1.166 = 5.4ft

 

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On 6/15/2019 at 1:58 PM, Chris Davidson said:

10" drop / 12" = .833... x 18.3(vert to horiz) = 15.25ft + 5.4ft (5 frames @ 1.08ft per sec=13.44mph) = 20.65ft

 

 

20.65ft was the horizontal sync for the vertical range of JFK's head above the pavement (52.78inches and the 39.24inches = 3.27ft used in CE884)

52.78" - 39.24" = 13.54 inches /12 = 1.12833ft.... x 18.3ft(Elm St slope conversion 3.13°) = 20.6485ft = the link below.

That would be an exact match for 43 of 48 frames @ .4802ft per frame.

Of course the 13.54" consisted of the 10" drop + 3.54" = (3.54/12 = .295ft x 18.3ft = 5.3985ft) = 20.648ft

 

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On 6/13/2019 at 12:56 PM, Chris Davidson said:

This leads up to, but not including the .9ft jump from z161-z166.

24.52+ 5.625 = 30.145ft / 9.125 sec (167 Towner frames@18.3 frames per sec) = 3.3ft per sec

This quite closely matches the speed of the limo at CE884 Z161-Z166 = 3.294ft per sec.

In other words, the entry for z161-z166 is a major remnant of the (3.29ft per sec = 2.24mph) adjustment  being applied to the overall equation.

 

 

 

Working somewhat backwards for a match:

Z166 = Station# 330.1 - 5.4ft = Z161 Station# 324.7 -  Z133 = Station# 299.0 = 25.7ft (28 frames)  = 11.42mph
Extant zfilm plotted133-166 = 13.44mph
13.44 -11.42 = 2.02mph
15.68mph - 11.42 = 4.26 - 2.02 = 2.24mph
CE884 Z161-166 = .9ft/5frames = .18ft per frame x 18.3 = 3.29ft per sec = 2.24mph

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On 6/14/2019 at 6:37 PM, Chris Davidson said:

Speaking of limo exercises and the Myers 8% solution:

21.34ft x 1.08(spread) = 23.0472ft per sec

Of course you are taking into consideration the limo not moving at a constant speed? 

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Stationary objects are my constant, such as the Stemmons sign support post. See gif below.

Frame counts can be very deceiving. See gif below.

The only instantaneous speed I'm interested in, is that which is reflective of frame removal. See gif below.

 

8aa91d95-cc7e-42e2-b898-2adb62cd8c74-ori

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How do you incorporate Position A into this Chris?

Mr. SHANEYFELT. The first position we established that morning was frame 161. 
Mr. SPECTER. Was there not a position established prior in sequence to frame 161, specifically that designated as position A? 


Mr. SHANEYFELT. This shows the photograph that was made from the point where Zapruder was standing looking toward the car, and is a point that we have designated as position A because it is in a position that did not appear on the Zapruder film 
The Zapruder film does not start until the car gets farther down Elm Street. 
Mr. SHANEYFELT. What is that exhibit number? 
Mr. SHANEYFELT. Exhibit No. 886. 
Mr. SPECTER. And why was that location selected for the position of the car? 
Mr. SHANEYFELT. This location was selected as the first point at which a person in the sixth floor window of the Book Building at our control point could have gotten a shot at the President after the car had rounded the corner from Houston to Elm. 

 

Shaneyfelt is basically stating the limo passed thru this spot... and is the reason, as I see it, for the Towner/Zapruder sync

5a4699573f976_PositionAandZ133-appearsthelimocameveryclosetocurbpriortoPosA.thumb.jpg.15468bc414b5c85ad8a9f2836a039faf.jpg1272596097_PositionoflimoinBELLshouldbePOSITIONAbutitisnot.thumb.jpg.4a4046b4c5f8e3290f70675673322421.jpg

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