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Unveiling The Limo Stop


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5 hours ago, Chris Davidson said:

And, finally, is there any indication of misalignment where background objects appear in front of foreground objects in relationship to the Stemmons Sign?

VzYW86.gif

Back to numbers it is.

On the curb image that aligns with Martin's fender/headlight that we discussed recently I have a couple observations. Because Martin is on Hargis' left he is slightly farther away and so slightly higher in the frame. Because of that there is no frame in which Martin's fender would align with the curb as Hargis' fender does. I still think what we see is Martin's fender as it is in the correct location when compared to Hargis' Fender/windshield/headlight.
  NOTE: When the front brakes are applied the upper telescoping forks recede into the lower legs. The fender is attached to the lower legs and when the upper legs recede into the lower legs the headlight and fender get a little closer together. So this may be why Martin's fender is very slightly closer to his headlight when compared to Hargis.

The overlapping image of the Hargis' bike onto the Stemmons sign has a possible explanation. When an object that shows a lot of glare is reflecting back to the camera it can overlap objects that are closer to the camera. A glaring object has a lot of diverging rays which is largely why it glares. The diverging rays  spread out and make the object look bigger. Some of the rays diverge enough that they hit the Stemmons sign and never reach Z' camera. But other rays diverge less and make it past the sign. Because they are diverging through their entire path to the camera the continue to spread out from the sign to the camera. By the time they reach Z's film they are overlapping the Stemmons image and since the glare is much brighter than the sign it burns an image of glare right over the sign image.

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3 hours ago, Chris Bristow said:

On the curb image that aligns with Martin's fender/headlight that we discussed recently I have a couple observations. Because Martin is on Hargis' left he is slightly farther away and so slightly higher in the frame. Because of that there is no frame in which Martin's fender would align with the curb as Hargis' fender does. I still think what we see is Martin's fender as it is in the correct location when compared to Hargis' Fender/windshield/headlight.
  NOTE: When the front brakes are applied the upper telescoping forks recede into the lower legs. The fender is attached to the lower legs and when the upper legs recede into the lower legs the headlight and fender get a little closer together. So this may be why Martin's fender is very slightly closer to his headlight when compared to Hargis.

 

Curb image is extant z334.

Credit to Tim Nicholson for his work on the lead up to extant z334:

EpwAbt.png

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I addressed it in a manner more familiar to what I have been presenting:

I removed the helmet double image(red arrow)and rotated z331 @ 1.3°

Added on Edit: Conversion between Tim and me:

19.3"- 9.8" = 9.5"

9.8"- 6.1" = 3.7" = 1.3° rotation

9.5"/3.7" = 2.567 x 1.3° = 3.33°

19.3"- 9.8" = 9.5" = 3.33° rotation

 

rpTZ6s.gif

Edited by Chris Davidson
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On 10/17/2021 at 3:10 PM, Chris Davidson said:

The variation starts at Station# 2+00.
The parallel track is 100ft from Station 2+00 to 3+00 (extant Z133) = 208 frames
The altered(forwarded) track is 100ft @166frames

Relink in case you forgot from what document you saw the 100ft/166frame split:

https://drive.google.com/file/d/1CiFoD500L2lnrkEPnWZ4RTgZk6hdLyfT/view?usp=sharing

100ft/208frames = .48ft per frame
100ft/166frames = .60ft per frame
                                .12ft per frame difference created by a 42 frame total difference.

 

 

 

Macro View in keeping with the above.

.12ft per frame x 250 excised frames = 30ft

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2 hours ago, Chris Davidson said:

Does there appear to be a mathematical mirror image developing as the pyramid expands?

oP3a8y.png

 

Keep the link and pyramid info in mind as I convert a distance within the link and then connect it (next post) to Shaneyfelt .

5.974mph x 1.47(1mph) = 8.78ft = 8.8ft rounded off.

 

 

 

 

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3 hours ago, Chris Davidson said:

Does there appear to be a mathematical mirror image developing as the pyramid expands?

oP3a8y.png

 

The link provides info acquired by Tom Purvis in his discussions with Robert West.

I'll relate this info for you in a slightly different manner connected to the pyramid at the back end.

A mirror distance of 30.86ft + 8.8ft = 39.66ft = Distance difference in above link.

Station# 517.5 - 30.86ft = Station# 486.64 - 8.8ft = 477.84 - 11.1ft = Station# 466.74

If you forgot where the 8.8ft came from, refer to the sum of differences in the previous posting which gives you a good reference to missing frames involved. 

Remember, those sums are also reflective of the 2.24 and 3.74mph speed entries from both CE884 versions starting at extant zframes 161 and 168.

 

 

 

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19 minutes ago, Chris Davidson said:

Station# 517.5 - 30.86ft = Station# 486.64 - 8.8ft = 477.84 - 11.1ft = Station# 466.74

 

 

 

 

 

And, if you would like to remove the difference of sums sum of differences from the equation you can match up the common location from the pyramid front and back:

Station# 517.5 - 30.86ft = Station# 486.64 - 11.1ft = Station# 475.54

 

Edited by Chris Davidson
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21 hours ago, Chris Davidson said:

The link provides info acquired by Tom Purvis in his discussions with Robert West.

I'll relate this info for you in a slightly different manner connected to the pyramid at the back end.

A mirror distance of 30.86ft + 8.8ft = 39.66ft = Distance difference in above link.

Station# 517.5 - 30.86ft = Station# 486.64 - 8.8ft = 477.84 - 11.1ft = Station# 466.74

If you forgot where the 8.8ft came from, refer to the sum of differences in the previous posting which gives you a good reference to missing frames involved. 

Remember, those sums are also reflective of the 2.24 and 3.74mph speed entries from both CE884 versions starting at extant zframes 161 and 168.

 

 

 

Now that we know the above, let's connect it to other areas of the puzzle:

30.86ft + 8.8ft + 11.1ft = 50.76ft

517.5 - 50.76ft = 466.74 = Shaneyfelt notes designation.

 

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If you don't want to do the math, I filled in the appropriate Station# for PositionA at the bottom, which is reflective of the 50.7ft matching distance at the back end of the puzzle/pyramid.

Using a little bit of deductive reasoning with the info provided by the WC, let's find a match for the previous posting:

zE4vhi.png

 

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Please feel free to chime in with the (difference) answer.

Determine JFK's height(according to the WC) above the street(note survey info and CE884 entry)

Convert that vertical measurement to a horizontal distance using the Elm St. 3.13° slope

Subtract that from the windowsill height of the TSBD 6th floor.

When you arrive with the correct answer, we'll add that puzzle piece to the pyramid.

mJbaMs.png

 

 

 

Edited by Chris Davidson
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