Chris Davidson

And, as I explained previously, albeit, in slightly different terms, .9ft per frame distance = .9 x 18.3 = 16.47ft per sec = 11.2mph

Take 8.8 missing ft and add the B.S. 1.4ft (CE560) supposed to represent a 6.7 inch lead height which would be reflective of a limo traveling at 28.63 mph, the answer is: 8.8ft + 1.4ft =10.2ft Or, take 8.8 missing ft. add 2.3ft traveled (CE884 Z207-Z208) subtract .9ft traveled from (CE884 z168-z171)= Surprise, 10.2ft In other words, (z207-z208) 2.3ft traveled - 1.4ft horizontal lead distance traveled = .9ft

Post #72-78 is a good synopsis of the importance of 8.8ft, along with the previous few posts.

The z207 mark on the plat is inside the red box? It's too small for me to read... So the FBI has taken their original location of the 1st shot and relocated it 10.2 down Elm St. in the direction the limo was traveling.Do you have any thoughts as to WHO was the architect of these synchronized new locations? Thanks the for additional information, Tom Take 8.8 missing ft and add the B.S. 1.4ft (CE560) supposed to represent a 6.7 inch lead height which would be reflective of a limo traveling at 28.63 mph, the answer is: 8.8ft + 1.4ft =10.2ft Shaneyfelt, Frazier, Eisenberg, etc... P.S. The limo traveling at 28.63 mph circa extant z207 would equal a one shooter tie-in to the SE 6th floor location.

Comparison using elevation for frames z168-z171 and lead height for z207(CE560) = 6.7inches = 1.4ft horizontal distance traveled would look like this: z168-171 .05ft vertical change = .9ft horizontal CE560 lead height = 6.7"/12" = .55833ft vertical = 1.4ft horizontal Difference ratio in horizontal change between these two = 1.4/.9 = 1.555….. 1.555… x .05 vertical change = .0777…. CE560 lead height = 6.7"/12" = .55833ft vertical - .05 vertical change = .0777…. = .55833 - .0777 = .4805 = vertical difference Since Elm St has a slope (vertical to horizontal ratio of 1ft/18.3ft), a .4805ft vertical change = .4805 x 18.3 = 8.79 horizontal ft or more exact: 1/18.3 = .0546 .4805/.0546 = 8.8ft

Does the difference in lead height of 6.7 inches (CE560) which is reflective of a limo moving at 28.63 mph(extant z207-208) have a correlation with the limo moving at 3.73 mph at z168-z171(CE884)? The WC (Shaneyfelt) liked working with averages, easier to hide instantaneous results. Beginning with: Average speed of 3.73 + 28.63 = 32.36/2 = 16.18 mph. Refer back to post 12+13 Distance from z168-z207 = 41.9ft 28.63mph = 28.63 x 1.47ft per sec = 42.09ft

Which might lead you to ask: How was the sniper tracking the target as it traveled down Elm St?

We now know mathematically what 4 inches represents. Bennett translation: Two shots occurred (before the extant 313 shot) in less time than is possible, to fire the same weapon. The SS giveth, the WC taketh away. TREASURY DEPARTMENT UNlTED STATES SECRET SERVICE FIELD FORCE PROTECTIVE Assignment of S/A Bennett on 11122/63 at Dallas, Texas Air Force Two landed at Love Field, Dallas, Texas at 11:35 A.M. Upon deplaning, I covered the fence and press areas. The President's plane arrived at approximately 11 :38 A.M. I stayed with the President and First Lady during the time they greeted the crowd on the apron and along the fence. The greeting lasted for about 10 minutes and the President/First LadyENTERED their car and the motorcade planned to depart. I asked while moving towards the follow-up car what position I should take, Mr. Roberts informed me that I should take the right rear seat of the follow-up. I took this position and held it during the entire motorcade. I left this rear seat position at one point in the trip to assist in getting well-wishers away from the President's auto. About thirty minutes after leaving Love Field about 12:25 P.M., the Motorcade entered an intersection and then proceeded down a grade. At this point the well-wishers numbered but a few; the motorcadeCONTINUED down this grade enroute to the Trade Mart. At this point I heard what_sounded like a fire-cracker. I immediately looked from the right/crowd/physical area/and looked towards the President who was seated in the right rear seat of his limousine open convertible. At the moment I looked at the back of the President I heard another fire-cracker noise and saw the shot hit the President about four inches down from the right shoulder. A second shot followed immediately and hit the right rear high of the President's head. I immediately hollered "he's hit'' and reached for the AR-15 located on the floor of the rear seat.SPECIAL Agent Hickey had already picked-up the AR-I5. We peered towards the rear and particularly the right side of the area. I had drawn my revolver when I saw S/A Hickey had the AR15. I was unable to see anything or one that could have fired the shots. The President's car immediately kicked into high gear and the follow-up car followed. The President's auto and the follow-up proceeded to the Parkland Hospital. Upon arriving at the hospital's parking lot, I was instructed by ASAIC Roberts to stay with the Vice-President who had followed us into the parking lot. I immediately went to the Vice-President's auto and accompanied him to a room on the ground floor of the hospital. I then continued with the Vice-President back to Washington, D.C. where I was relieved. [signature] Glen A. Bennett Special Agent 11-23-63

The excerpt at top, within the graphic, is from a review article by Martin Hay. The link is included if you want to read the whole article. http://www.ctka.net/2015/HaagCritique.pdf The ballistic results pretty much bear out what I've provided in the previous postings in relation to extant z207, CE560 and testimony thus far. 9degrees 21 minutes = 9.35degrees

It appears someone knew the significance of a 2" lead. Mr. EISENBERG - And you calculated the speed of the car by translating the figures on total time elapsed between first and third shots? Mr. FRAZIER - Yes, sir. The time the speed of the moving object was calculated on the basis of an assumed 5.5-second interval for a distance of 90 feet, which figures out mathematically to be 11.3 miles per hour. Mr. EISENBERG - Now, you said before that in order to give this 2-foot lead, you would have to aim 2 inches--for a target going away from you, you would have to aim 2 inches above the target, or in front of the target. Mr. FRAZIER - 2 feet in front of the target, which would interpolate into a much lower actual elevation change. Mr. EISENBERG - The elevation change would be 2 inches, is that it? Mr. FRAZIER - Well, no. It would be on the order of 6 to 8 inches. Mr. EISENBERG - 6 to 8 inches? Mr. FRAZIER - Yes. Mr. EISENBERG - What was your 2-inch figure? Mr. FRAZIER - I don't recall. Mr. EISENBERG - But it is 6 to 8 inches in elevation? Representative BOGGS - May I ask a question? Using that telescopic lens, how would you aim that rifle to achieve that distinction? Mr. FRAZIER - Well it would be necessary to hold the crosshairs an estimated distance off the target, of say, 6 inches over the intended, target, so what when the shot was fired the crosshairs should be located about 6 inches over your target, and in the length of time that the bullet was in the air and the length of time the object was moving, the object would move into actually, the path of the bullet in approximately 1/10th to 13/100ths of a second. Mr. EISENBERG - So that if the target of the assassin was the center of the President's head, and he wanted to give a correct lead, where would he have aimed, if we eliminate the possibility of errors introduced by other factors? Mr. FRAZIER - He would aim from 4 to 6 inches--approximately 2 inches, I would say, above the President's head, which would be actually 6 inches above his aiming point at the center of the head. Mr. EISENBERG - How difficult is it to give this--a lead of this size to this type of target? Mr. FRAZIER - It would not be difficult at all with a telescopic sight, because your target is enlarged four times, and you can estimate very quickly in a telescopic sight, inches or feet or lead of any desired amount. Mr. EISENBERG - Would it be substantially easier than it would be with an open or peep sight? Mr. FRAZIER - Yes. It would be much more difficult to do with the open iron sights, the notched rear sight and the blade front sight, which is on Exhibit 139. Mr. EISENBERG - Now, you have been able to calculate the precise amount of lead which should be given, because you have been given figures. If you had been in the assassin's position, and were attempting to give a correct lead, what lead do you think you would have estimated as being the necessary lead? Mr. FRAZIER - It would have been a very small amount, in the neighborhood of a 3-inch lead. Mr. EISENBERG - As opposed to the 6 or 8 inches? Mr. FRAZIER - As opposed to about 6 inches, yes. Mr. EISENBERG - What would the consequence of the mistake in assumption as to lead be that is, if you gave a 3-inch lead rather than the correct lead? Mr. FRAZIER - It would be a difference of a 3-inch variation in the point of impact on the target. Mr. EISENBERG - Now, if you had aimed at the center of the President's head, and given a 3-inch lead, again eliminating other errors, where would you have hit, if you hit accurately? Mr. FRAZIER - It would be 3 inches below the center of his head--from the top--it would be not the actual Center from the back, but the center would be located high. The bullet would strike at possibly the base of the skull. Mr. EISENBERG - Now, suppose you had given no lead at all and aimed at that target and aimed accurately. Where would the bullet have hit? Mr. FRAZIER - It would hit the base of the neck--approximately 6 inches below the center of the heart. Mr. EISENBERG - Mr. Frazier, would you have tried to give a lead at all, if you had been in that position? Mr. FRAZIER - At that range, at that distance, 175 to 265 feet, with this rifle and that telescopic sight, I would not have allowed any lead--I would not have made any correction for lead merely to hit a target of that size.

Distance traveled from Z207-208 = 2.3ft 2.3ft x 18.3 frames per sec = 42.09ft per sec = 28.63 mph. 28.63mph is much more reflective of a lead height of 6.7 inches. Whereas, 2.8 inch lead height (previous posting) is more reflective of 11.2 mph.

The vertical lead time height for the 175ft shot circa extant z207 is 6.7inches according to CE560. I ran these numbers paralleling what I believe Frazier was working with.

The "Speed of Car" entry reflects the distance (90ft) between two shots corresponding back to extant Z207-Z313 from CE884. If I wanted to sync this in terms of an average speed of 11.2mph vs instantaneous speed (see previous post) it would show like this: 90ft / 5.555sec = 16.2ft per sec x (18.3/18 (whole frame) = 16.47ft per sec = 11.2mph as opposed to what's given, which is: 90ft / 5.5sec = 16.36ft per sec /1.47 (1mph) = 11.13mph In terms of frame count between the above scenarios, the difference would equal one frame. 5.5x18.3= 100.65 5.555 x 18.3 = 101.65

You can see that Truly's shadow is up on the sidewalk. I like it. The curb is 6" high. The bottom of the "T" is 6.75ft from the curb, in alignment(thin blue line) from the west end of the staircase.

I use Photoshop. Specifically within, the animation option. This allows using the "auto-align layers option". If I don't like the result, I'll align the frames manually (as layers) using my good old fashioned eye-sight. chris

Thanks Tom, The 6ft shadow length I used was from Sandy's estimation. Shorten Baker's steps to 3ft and you have a difference of 3.75ft (26.25 - 22.5) added to Baker's shadow now equals 9.75ft. Or, move Truly's LOS position closer to the annex corner 3.75ft. Truly's position in relationship to the TSBD stairway is key. imo chris

Looking at CE560, a few remarks are in order. To figure out the exact vehicle speed Frazier uses, note the line starting with: Ave Vel= Average Velocity Ave Vel = 175/2070 = .085 sec 1sec/.085 = 11.764…. x 1.4ft = 16.47ft per sec =11.2mph Added on edit: Or, this way: .085/.0546(1frame/18.3fps) = 1.556frames 1.4ft per 1.556frames = .8997...ft per frame x 18.3fps = 16.465ft per sec/1.47(1mph) = 11.2mph and 1sec/.085 = 11.764…. x 1.9ft = 22.35ft per sec = 15.2mph Mr. SHANEYFELT. Yes; because we were able to determine the speed of the camera, and thereby accurately determine the length of time it takes for a specific number of frames to run through the camera at this 18.3 frames per second, and having located these frame positions in the street, we took the farthest distance point we had in the Zapruder film which was frame 161 through frame 313. This was found to run elapsed time from the film standpoint which runs at 18.3 frames a second, runs for a total of 8.3 seconds. This distance is 136.1 feet, and this can be calculated then to 11.2 miles per hour.

Baker takes approx 6 steps from the annex edge to Truly. 6x3.5ft =21ft Another 1.5 till we see his shadow upon the curb. 1.5 x 3.5 = 5.25ft If Baker's shadow is approx 6ft, that's a total distance of 32.25ft The angled path from curb to curb is 33.4ft That puts Truly about 12.4 ft on angle, away from the TSBD curb. The Truly X LOS(line of sight) is aligned with the west end opening of the TSBD staircase.

Sandy, Does Baker change his trajectory as he passes Truly? Are you trying to convey that if Baker doesn't change his trajectory, he doesn't run into Truly? Where do you believe Baker and Truly cross paths using this 1964 photo?

Robert, Can you tell me if this would be a close enough apples to apples" comparison? Added on edit: Request withdrawn. Will deal with this in the "Swan Song" topic. chris

So we are both identifying the guy in the hat (that turns around as Baker passes) as Truly. I agree with you that Baker turns to his right in order to avoid Truly. Baker certainly could have continued to the TSBD steps, but from this film we can't tell where he goes after passing Truly. Tom My sentiments exactly. I'm not sure why the gif is giving you problems (maybe a photo bucket conversion glitch). See if this uploaded forum version makes any difference. chris

Tom, I don't believe so. When Stetson man (foreground) turns his hat leftward, Truly (who I'm identifying as Truly) with his hat on, is revealed. Truly begins his turn toward the TSBD just as Baker passes him. It doesn't matter to me whether it's Truly or not, this person forces Baker to veer off his original path. imo chris

CE560:

Baker deciding which fork in the road he takes.

The re-directed angle caused by Truly's impingement upon Baker's path, equals Baker's angle to the curb. It's neither parallel or directly forward, it's an angled approach after Truly. If this is not understood by viewing this next Baker stabilized gif, it's time to move on. chris