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Chris Davidson

Swan-Song -- Math Rules

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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.

The limo speed at extant Z207 could have been 28.63mph, I just don't believe that is the case.

The 6.7" lead (CE560) is an instantaneous distance resolver. imo

Remember, the formula for Elm St. at 3.13degree street slope = 1ft vertical per 18.3ft horizontal.

Treating the 6.7" lead as a vertical measurement in conjunction with the Elm St. slope equation would look like this:

6.7"/12" = .55833...

.55833 x 18.3 = 10.21ft

Post #33 as a follow up, if interested.

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Now, seriously, with that toy scope, intended for a pellet gun or a .22, do you really think anyone could lead a target by 6.7" at 88 yards? Good luck.

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In one of the countless YouTube interviews of Gary Mack, he said something that I found interesting in one of them. Gary's analysis of the snipers' nest had the shooter standing behind the window sill boxes, rifle barrel resting on the top box & shooter simply pointing the barrel in the general vicinity of body mass (no use of fixed sights or scope for aiming), pulling the trigger 2-3 times & working the bolt rapidly. Gary believed this is how the shooter (Gary believed it was LHO) avoided the gas blowback coating his face. The boxes would also serve as a thick shield if any of JFK's guards were fast enough to return fire. Gary theorized that this is why LHO only changed his shirt & not his trousers; trousers were shielded from blowback by being behind the boxes).

The type of shooting Gary described is what Army marksmanship instructors called 'slop shots' when I served. It's also known as 'Annie Oakley' style 'guestimate' without aiming style shooting (similar to the manner in which Chuck Connors demonstrated during the opening sequence of TV's 'The Rifleman').

I commented in another the thread that the sniper's nest shooter should have smelled like he had been firing fireworks had he fired 2-3 rounds off using conventional aiming techniques. It should have been easily detectable to anyone who came into close contact with that person. For LHO, this would include Officer Baker, Roy Truly, Pierce Allman, bus & taxi drivers (+ passengers), LHO's landlady, Officer Tippit, Texas theatre arresting officers, those who 1st interrogated LHO at DPD headquarters in addition to the DPD hallway TV reporters close enough to LHO to smell him. I don't recall any of those people ever commenting that LHO smelled like he had been popping fireworks (the most common description of how a person smells after firing a weapon).

I can remember a grime that built up under our fingernails after my units fired their individual assigned weapons at rifle ranges for annual qualification. The grime resembles how a mechanic's fingernails look after working on cars. It took quite a while to completely get it out from under fingernails.

The point is: LHO should have reeked of gunsmoke everywhere he went & whoever came into close contact with him & those persons undoubtedly would have commented on it.

Gary's theory obviously would cause problems for a second gunman sharing that window & at least one witnesses stated on record that she saw 2 armed shooters in that particular TSD 6th floor window.

BM

Edited by Brad Milch

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I think Gary vastly overestimated the amount of blowback produced by a bolt action rifle, as compared to a revolver. A revolver loses a large amount of gas, simply because there is a small gap between the chamber the cartridge is in and the barrel of the revolver. This cannot be a tight seal, or it would be impossible for the chamber of the revolver to turn and align the next cartridge to be fired. OTOH, a bolt action rifle chamber seals itself as soon as combustion of the gunpowder begins, and rising internal cartridge pressures swell the brass cartridge outwards to seal it against the chamber.

There is a gas port on the side of the chamber of a bolt action rifle, but it is there to vent gases in the case of a ruptured primer or casing.

Edited by Robert Prudhomme

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Since I know the WC used 3.27ft as JFK's head elevation in every single frame surveyed (CE884), I can subtract that from the base elevation of the triangle ( 427.02)

and add 3.27ft to the height (65.05) =

427.02 - 3.27 = street elev of 423.75

Subtract this street elev of 423.75 from the windowsill elevation of 490.9ft, since this is to the street.

490.9 - 423.75 = 67.15

and add 3.27ft to the height (65.05) = 68.32 elev

And finally, subtract 68.32 - 67.15 = 1.17ft

You have to remember there are measurements to the street and measurements to a spot 3.27ft above the street.

chris

Since I don't have surveys for most frames on CE884, the method above can be used if one has access to an online conversion calculator, easier than manually computing the appropriate figures.

I'll give you an example using zframe 222 from CE884:

20deg23min = 20.38deg

490.9ft - 422.84ft (426.11- 3.27) = 68.06ft elev = street to windowsill frame

The determined height of 65.68ft dictates the hypotenuse length which equals the "line of sight" distance from rifle to JFK's head 3.27ft above the street.

65.68ft + 3.27ft = 68.95ft - 68.06ft = .89ft = height of rifle above the windowsill frame at z222 according to CE884.

.89ft = 10.68 inches above the windowsill frame = 26.68 inches above the floor = .68 inches above the sniper's perch box.

Compare this to the rifle height above sniper box at z207.

z222.jpg

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Surveyarrow-10x10.pngInfo for Z207 will help:

Survey%20207.jpg

Difference between 21deg11min and 21deg50min = 2.14ft = 39min

Height used for JFK's head above the street in CE884 all frames = 3.27ft

3.27ft/2.14ft = 1.528… x 39min = 59.593...min = 1degree

3.27ft vertical = 1degree

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1degree = 60minutes

3.27ft / 60 minutes = .0545ft per minute

Street slope = 3.13degrees

3.13degrees = 1ft vertical per 18.3ft horizontal ratio

1ft /18.3ft = .0546ft vertical per 1ft horizontal

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1degree = 60minutes

3.27ft / 60 minutes = .0545ft per minute

Visually speaking. A one degree change plotting from extant z207.

Scale is 1inch = 10ft

z207-381.3.jpg

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1degree = 60minutes

3.27ft / 60 minutes = .0545ft per minute

.0545ft = .0545 x 12" = .654inch

Excerpt from post#106:

The 6.7" lead (CE560) is an instantaneous distance resolver. imo

6.7"/.654 = 10.24min = 10.24ft

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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

3.27ft elev. change in relation to 8.8ft and extant z207 would look like this:

3.27ft%208.8ft.jpg

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What constitutes 13.54 vertical inches?

The first part would be the WC determination of JFK's head height at the time of the extant 313 headshot.

In terms of elevation, it looks like this:

Street Elev. at (JFK's position in limo) extant z313 = 418.48

JFK's head height elev. = 421.75

Difference = 3.27ft in elevation

This difference is used in all entries on CE884.

313.jpg

At 3.27ft above the street, which is what the WC determined JFK's head height to be for all CE884 entries, this converts quite closely to an Elm St elevation change of 61ft.

61ft /18.3 (1ft vert. per 18.3ft horizontal) = 3.33ft vs 3.27ft

I will introduce the slight WC adjustment between the two a little later.

The slight adjustment between 3.33ft vs 3.27ft is the difference between these elevations = .06ft vertical = .06 x 18.3(1ft vert. per 18.3ft horizontal) = 1.098ft

Or, as Tom writes per his conversations with Robert West:

PurvisNotes.jpg

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110ft/18.3 = 6.01ft. elev change.

490.9 - 423.07 = 67.83ft -60.7(window frame sill)- .5(curb height)= 6.63ft elev change.

6.63-6.01 = .62ft elev difference

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