http://www.internationalskeptics.com/forums/imagehosting/879044ecf837e692b.jpg
Cabin East of crash site (NW corner of building) PT A
40 02 59.66 N
78 54 13.46 W
elev 2381 ft
Cabin East of crash site (SW corner of building) PT B
40 02 59.40 N
78 54 13.44 W
elev 2378 ft
Whitish circular impression in NW corner of pic (southern most point)
40 03 04.65 N
78 54 20.08 W
elev 2378 ft
Whitish circular impression in NW corner of pic (eastern most point)
40 03 04.85 N
78 54 19.61 W
elev 2382 ft
Approx center of crater PT C
40 03 02.80 N
78 54 17.56 W
elev 2388 ft
Approx distance from A to C: 136.91 m
Approx distance from B to C: 143.10 m
Approx distance from A to B: 8.16 m
Approx distance from C, due south, to be perpendicular with B: 104.77 m
Approx distance from C, due south, to be perpendicular with A: 96.67 m
Ms. McClatchey's house (NW corner) PT D
40 03 00.29 N, 78 52 28.16 W
Distance from D to C: 2591.37 m
White barn (SW tip) PT E
40 03 00.41 N
78 52 34.66 W
Distance from D to E: 153.96 m
Distance from E to C: 2437.44 m
Red barn (NE tip) PT F
40 02 58.24 N
78 52 41.97 W
Distance from D to F: 333.05 m
Distance from F to C: 2267.62 m
Distance from E to F: 185.55 m
Extend a vector from D, through E, ending perpendicular to C. The end point is PT G, the vector is V1.
Extend a vector from D, through F, ending perpendicular to C. The end point is PT H, the vector is V2.
Extend a vector from G to H. This vector is V3.
Length V1: 2593.18 m
Length V2: 2631.18 m
Length V3: 606.33 m
NE tip of Burn crater from jpg: 40 03 02.58N, 78 54 14.05 W elev 2393 ft PT J
Approx distance from C to J: 83.23 m
Approx distance from H to J: 122.68 m
Approx area of C-H-J: 1/2(b*h) => 1/2(96.67*122.68) = 5935.26 m^2
This is the approx burn area visible in the jpg. This area is all south and east of the crater. This is consistent with the impact direction of NW->SE
Crater area in picture and Google Earth has a diameter of ~23.5 m.
Approx area of crater = 2pi*r^2 => 2*pi*(1/2*23.5)^2 => 2*pi*11.75^2 = 2*pi*138.0625 = 867.47 m^2
Given a maximum estimated plume horizontal diameter of 606.33 m, the approx area the plume would cover = 2pi*r^2 => 2*pi*(1/2*303.16)^2 => 2*pi*151.58^2 => 2*pi*22976.4964 = 144365.58 m^2
Now. We have an aircraft coming from the NWN impacting at a 40 degree angle. The crater expands S-SE-E from the point of impact to do the angle of attack; damaging an area ~867.47 m^2 in size. Continuing outward on a S-SE-E direction is an area of burning covering an additional 5067.79 m^2 (burn area of 5935.26 m^2 - crater area of 867.47 m^2)
Now, KT, what you utterly fail to do on your diagram (well, besides show any work) is highlight what the burn area was, for a sense of scale. I've taken the liberty of doing so for you:
To summarize:
1) a) The burn area is consistent with relation to the crater, the direction of impact, and the wind direction.
1) b) Given that the impact was at a 40 degree angle, it is reasonable to expect the burn area to be non-symmetric. This is consistent with the evidence.
1) c) Unless there was fuel on the surface of the grass, and thus igniting the grass via direct flame; the temperature of the fireball is insufficient to spontaneously ignite the grass.
1) d) Given 1c, we would expect grass to be burnt in the direction that the fuel from the ruptured tanks would travel, from NW to SE. This is consistent with the evidence.
1) e) Examples have been provided earlier in this thread showing that, even a jet fuel fire, burning on a stationary fuselage, will not necessarily damage vegetation in the immediate vicinity of the wreckage.
2) The debris field is consistent with relation to the creater, the direction of impact, and the wind direction.
3) The crater, burn area, and debris field are within the angle of view of the picture.
If you choose to address these points, do so using the same level of transparency and detail. If, instead you want to use pictures and your opinion on said pictures, then let me preemptively say stfu.