Chris Davidson

Not to confuse the issue further, but, since June Dishong was among the woman standing among Gloria Calvery's group, not that they necessarily knew each other, as Dishong worked for a different company than the four other women, I think this lends support to at least the physical position of Calvery on Elm St. http://educationforum.ipbhost.com/index.php?showtopic=1070

Robert, I agree.

Robert, The multi- name labeled photo including Mary Woodward appears to be correct. MARY ELIZABETH WOODWARD, 4812 Alcott, employee, Woman's News, "Dallas Morning News," Dallas, Texas, advised that she, AURELIA ALONZO, MARGARET BROWN and ANNE DONALDSON, on November 22, 1963 left the office of the "Dallas Morning NEWS" just about 12:00 noon to observe the Presidential Motorcade. They walked to Elm Street and stopped in front of the Texas School Book Depository building, but were located a short distance down the street near the second light post. They were standing in this spot when the Presidential Motorcade came by There is a light post #1 near the signal light at the Elm St annex corner. The unidentified woman near Mary Woodward, clothing wise, appears to match the woman in Darnell. Her distance from where she is standing in Z to "between the TSBD curb and annex" is 120ft. Close enough. 20 seconds @120ft = 4.08mph If Gloria Calvery is labeled correctly, she is 30ft farther west than Mary Woodward but can't be the woman in Darnell, wrong skirt color. imo chris

One of the few unlabeled in this, is the person I suspect matches Calvery in the Darnell/Baker frame. I do not know who created this. chris

Thomas, I do not know. We do have the mis-labeling of Calvery, imo, supposedly from a Darnell frame. I'm not sure if that frame/photo has ever been substantiated. Secondly, if Calvery is mis-identified, why not the other women. Hopefully, somebody has a decent photo of Hicks/Reed for comparative purposes to the Darnell frame. chris

The attachment makes much more sense. imo Now match this Calvery (dress code wise) with the woman in Darnell. Sounds pretty much like a group statement minus the personal info. chris P.S. Hopefully the significance of where they physically were standing(Station# wise) when stating the first shot arises, is not lost upon you.

Equation: Station# 3+71.1 + 10ft = Station# 3+81.1 - 5.4ft = Station# 375.7 + 90ft = Station# 465.7 + 30ft = Station# 495.7 - 114.4ft = Station# 3+81.3 Quick key: Station# 3+71.1 -Robert West -Time/Life investigation circa Nov25,1963- Shot1 10ft - Shaneyfelt westward relocation folly 5.4ft -Distance reduced, directly correlating to speed for CE884 Z161-Z166 90ft - Distance provided via CE560 between Shot1-Shot3 30ft- Distance between extant Z313 headshot and Altgens shot circa Z352 114.4ft- Straight line distance between altered shot1(Station# 3+81.3) and Altgens shot location

Robert, You covered this in your other topic: Who saw Baker enter the TSBD?I supplied this for you then. If you can't find her in the corresponding areas, time/distance estimates are a moot point. imo Find her first, then move on to time/distance. chris

And, the last piece of this "soon to be formed" equation. Tom P. !!! What say you?

Referring to the previous attachment: A more exact distance between 4+66.7 and 4+42.5 = 24.2ft 24.2ft + 10ft +5.4ft = 39.6ft = 39ft 7inches It's one inch off, but the explanation for that difference isn't necessary right now.

Please note the total distance in this next attachment.

The equivalent location to the supplied attachment is listed on CE884 as Station# 3+71.1

Logically, the distance the SS/FBI would have attempted to move the Altgens shot back up Elm St would have been 30ft to merge with the extant Z313 headshot. Instead, as listed on the SS/FBI plat and described by Tom P., they have it as 24.5ft. Maybe it was shorted approx 5.5ft along the way, and taken into consideration for syncing purposes at this junction. Such as: 5.4ft x 18.3/18 = 5.49ft per sec = (Z168-Z171 speed) + 24.5 = 29.99ft

I would assume by now, most know I believe there was a head shot some 30ft farther down Elm St. than the extant 313 headshot. To put this 30ft difference in terms of 166 frames, I would address it this way: 166.66 frames/18.3 frames per sec = 9.107sec 30ft/9.107 sec = 3.294ft sec = 2.24mph chris P.S. Why 166.66 frames? Originally, I only connected the 166.66 frame count to the CE884 "white" version of Z161-Z166 in terms of mph. Now, the connection to 30ft past the extant Z313 headshot is: 166.66frames/30ft = the ratio of 5.555frames per foot

Now, I'll throw Shaneyfelt's 11.2 mph quote back into the equation using the previous posts 21.6ft total distance traveled from Z168-Z186. 21.6ft - 5.4 ft = 16.2ft x (18.3/18) the conversion to 1 second of time. 16.2ft x (18.3/18) = 16.47ft per sec = 11.2mph Amazing how that matches the overall average speed from Z161-Z313. What does the span of 21.6ft taken from the "orange" version Z168-Z186 have to do with the 5.556 ratio? In terms of 5.556 seconds: 21.6ft x 5.556sec = 120ft. The "tie in" from 90ft to the 18 frame segment at Z168-Z186 is the difference between the total span of 21.6ft and the (5.4ft horizontal=3.54 inch vertical) missing distance. In other words: 21.6ft-5.4ft = 16.2ft 16.2ft x 5.556sec = 90ft.

That 120ft can be broken up into sections. In this instance, 90ft + 30ft = 120ft. CE560 shows you what the 90ft distance represents.

Now, I'll throw Shaneyfelt's 11.2 mph quote back into the equation using the previous posts 21.6ft total distance traveled from Z168-Z186. 21.6ft - 5.4 ft = 16.2ft x (18.3/18) the conversion to 1 second of time. 16.2ft x (18.3/18) = 16.47ft per sec = 11.2mph Amazing how that matches the overall average speed from Z161-Z313. What does the span of 21.6ft taken from the "orange" version Z168-Z186 have to do with the 5.556 ratio? In terms of 5.556 seconds: 21.6ft x 5.556sec = 120ft.

The attachment is an excerpt from Tom P. describing CE560. Keep in mind the previous 5.556/1 ratio.

Remember: 18.3ft = horizontal distance traveled in conjunction with a 1ft vertical rise for the Elm St slope of 3.13°. Remember: The direct connection between "3.54inch" vertical to "5.4ft" horizontal distance. Remember: 30ft/5.4ft = 5.556 Equation: 30ft/5.4ft= (5.55……..6) x 3.54inches = (19.66…….8)inches /12inches = 1.638ft. 18.3ft x 1.638ft = 29.99ft = 30ft In other words, a 5.556/1 ratio need apply to a vertical change of 3.54inches per 5.4ft horizontal distance. which equals a 30ft horizontal distance on Elm St at 3.13°.

Now, flip the ratio around to: 30ft/5.4ft = 5.556

I've shown the direct connection between the "3.54inch" vertical to "5.4ft" horizontal distance. I consider that more of the "inner equation" and want to expand outward a bit. Since I've been adamant about that 30ft difference in shots, I'll connect that back to the "inner equation". To begin: 5.4ft/30ft = .18ft .18ft x 5 frames = .9ft. .9ft = Distance traveled in 5 frames(Z161-Z166) on the CE884 "White" version and 3 frames(Z168-Z171) traveled on CE884 "orange" version.

Now, I'll throw Shaneyfelt's 11.2 mph quote back into the equation using the previous posts 21.6ft total distance traveled from Z168-Z186. 21.6ft - 5.4 ft = 16.2ft x (18.3/18) the conversion to 1 second of time. 16.2ft x (18.3/18) = 16.47ft per sec = 11.2mph Amazing how that matches the overall average speed from Z161-Z313.

In case you didn't realize it, that ratio of 4.5/5.4 converted = .8333ft 10 inches converted to a decimal (10/12) in terms of feet = .8333ft

Mark and Tom, I have supplied this forum with two versions of CE884. I will refer to them by color. "White" background and "Orange" background. If one will look at the entries for Z168-Z186 on the 'Orange" version, this is an 18 frame span. To convert to a one second span in terms of the Zfilm, I would have to divide 18/18.3 = .983 seconds. The entries for Z168-Z171, the first part of the Z168-Z186 span, has the limo traveling a distance of .9ft. It would look like this over a one second span: 18.3frames per sec/3 frames = 6.1 x .9ft traveled = 5.49ft traveled in a second. But, we are talking about 18 frames or .983 seconds as shown above. So, 5.49ft x .983 = 5.3967ft. Look at the "quote" above. Once again, what does 3.54 vertical inches represent? If nothing else: The ability to understand this completed circle, will help you realize that the Zfilm has at least been altered to accommodate this scenario. Applying this same method to the "white" version will yield this result: Z161-Z186 = 25 frames/18.3 frames per sec = 1.366seconds. The speed of the limo from Z161-Z166 (.9ft traveled) = 2.24mph = 3.294ft per sec 3.294ft per sec x 1.366seconds = 4.5ft total distance The difference between the "white and orange" versions for the same distance traveled of 21.6ft is (4.5ft and 5.4ft) = .9ft This equals the distance given for the travel of the limo in both CE884 versions for Z161-Z166 and Z168-Z171.