Chris Davidson

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Or, In those 15 frames, Z210-225 there is a vertical change of .82ft = 9.84inches. Appropriate, comparative testimony provided. Or, part of the BS scenario if you care to view it that way. chris

Robert, What if the determined "line of sight distance" for frame Z225 based on 20degrees-11iminutes actually matched the "line of sight distance" entry for frame Z210. In other words, was somebody trying to make 1 shot gradually disappear within a 15 frame span. chris

Honorable Friend. May the Lord guide you to greener pastures. Forever Grateful. chris

Chris - I think I have some bad news for you... 3.15 represents the SLOPE PERCENTAGE not the degree of the slope... the arctan* of a 3.15% slope is a 1.8 DEGREE ANGLE. Is the 3.15 refered to as a % or degrees since the formula you used is for slope % Please explain the 25.7 feet. (I am wroking out the rest of my understanding here... but please address my concernsDJ Slope Percent and Slope AngleSlope refers to the angle, or grade, of an incline. Slope can be upward or downward. Slope is typically expressed as a percent, and corresponds to the amount of rise, or vertical distance, divided by the run, or horizontal distance. Percentage means per 100. Slope can also be expressed as an angle, which gives the amount of deviation from flat as a number of degrees. Conversions between slope percent and slope angle can be done using a scientific calculator and the inverse tangent (arc tan) function. Essentially, the slope angle is the inverse tangent of the slope percent (with slope percent expressed in decimal). Example 1 - The slope percent is 60 percent. What is the slope angle? Step 1. Change 60 percent to decimal form. Sixty percent means 60 out of 100. It can be written 60/100 = 0.60. See Chapter 1. Slope angle = inverse tan of the slope percent (in decimal) *Slope angle = inverse tan of 0.60 (the decimal of 60%) Step 2. Enter .6 into the calculator and push the inverse, inv, or "2nd" button, then the tan button to get the inverse tangent. The calculator will show the slope angle. A 60 percent slope corresponds to a slope angle of 31°. Hi David, You are absolutely correct. And I apologize for mixing that up. The bad news part is another matter. Let me give you an example: 3.27ft vertical @103.9ft horizontal = 3.15% ratio = 1.89 degree angle or converted 1degree 53.4 minutes. Take a look at CE884 under the column "Angle to Horizon" entries for frame 222 and 249. The angle difference between those two frames is 1degree 51minutes. Would one more frame added get us to 1degree 53.4 minutes? Close enough. Simplified. 28 frames = 133-161. What don't we have surveys for, even though that's where we see the limo on film begin. P.S. Remember .6 = 3/5 and the reciprocal is 5/3 = 1.667

btw, I did not want to insinuate that I believe the street grade of Elm St is 3.15 degrees. I only provided material the WC used to further advance their falsification of the true story. chris

CE884 Frame 313 chris

Robert, Frame255. This comparison might help you understand more of the subtle differences involved. If you look at CE884 entry for frame 255 the Station# is 4+16.4 The snipers nest is in alignment at Station# 2+50. That is a difference of 166.4ft. The entry for the elevation is 424.46. We know the snipers nest was measured at 60.7ft high at an elevation of 429.70. 429.70- 424.46 = 5.24ft elevation change from Station# 2+50 to Station# 4+16.4 60.7ft + 5.24ft = 65.94ft. This is the entry in the attachment. Running the equation for a 3.15 degree street grade. 5.242/166.4 x 100 = .0315 So one might look at this and say that frame 255 is reflective of a 3.15 degree street grade. But, the slant distance to JFK's head would be incorrect by some 25.7ft. If you were to change the angle of "rifle to JFK's head by "2degrees 1 minute" (left side of attachment) you now have a matching CE884 frame255 entry slant distance of 218ft. chris P.S. I did not include/describe the difference between a slant measurement to the "ground and 3.27ft higher" but that would be approx 10ft.

Hi Robert, This is as simplified as I can explain it. Mostly, I stay away from the photos, films and testimony for now, because the math eventually should take care of it. imo Reverse engineering the WC creation takes time, but those interested should already have a clearer understanding of how they went about it. chris

By the way, keep in mind that 3.27ft vertically = 103.9ft horizontally @ 3.15 degrees. chris

I supplied this a few posts ago with the red box entry blacked out. WC determination of JFK's head height above the pavement. Since the TSBD 6th floor was determined to be 60.7ft in vertical elevation, someone realized very early on there was a major problem. chris

David, Before I get to far ahead, I want to go back and show you what a little true math reveals. Station # 3+29.2 = frame 161 according to CE884, which is 79.2ft down Elm from Station# 2+50. Station# 2+50 being in direct alignment with the 6th floor snipers nest. And coincidentally, that is where the slope of Elm begins. We know the elevation at Station# 2+50 is 429.70 You can now use the equation for a 3.15 degree slope @79.2ft. 79.2ft/31.75ft = 2.50 elevation change. Elevation 429.70 - 2.50 = elevation 427.20 And, when compared to the elevation listed on CE884 for frame161, there is an elevation difference between true elevation 4+27.20 and WC elevation listing of 4+29.25 = 2.05ft chris P.S. The height of JFK's head measured above the street in all surveys was the same and more than 2.05ft