Even with almost all eyes on the President and Jackie? Even so, I recall there were a few witnesses who reported dust being kicked up off the street.
If you are referring to Virgie Baker (Rachley), initially, she said she thought saw something hit in front of the President's car on the road, not the curb (her November 24, 1963 interview). She didn't mention it in her March 1964 FBI statement (22 H 635/6). In her deposition to the WC, she said that it hit BEHIND the President's car past the Stemmons freeway sign (7 H 510):
"Mr. LIEBELER. Where was the thing that you saw hit the street in relation to the
President?s car? I mean, was it in front of the car, behind his car, by the side of his car or
was it close to the car?
Mrs. BAKER. I thought it was-well-behind it."
Exhibit CE354 was marked to show where she said the bullet hit. She put it well past the
Stemmons sign. Now if that was BEHIND the President's car (I'm not sure how she
would have seen that because the follow-up car would have been in the way), it must
have happened at about z250. It couldn't have happened at z160. [Mrs. Baker was
standing at the point marked 1. and the thing that hit the road was at the point marked 2.
on CE354. See:
http://www.historymatters.com/archive/jfk/wc/wcvols/wh16/html/WH_Vol16_0487a.htmI don?t think the bullet melts. It is just torn apart into small pieces. The lead and the copper jacket both. I never read anything in Larry Sturdivan?s book that describes bullets fragmenting due to partial or total melting.
The Haags searched for the bullet and/or large fragments, and found none.
Is there anything in the literature by a ballistic expert about bullets melting, partially or fully, as a result of striking rock or asphalt? Or any target?
It depends on the target. It is just a matter of physics. The bullet asborbs energy when it compresses. If the bullet does not deform or displace the target then all of the kinetic energy of the bullet is absorbed by the bullet itself. A 10 gram bullet travelling at 670 m/sec has 2250 Joules of kinetic energy. When that energy is transferred to the bullet, the bullet heats up. If enough energy is absorbed, the bullet lead melts. You can see this, for example, in these clips:
The specific heat of lead is .128 Joules per gram-degree C. The melting point of lead is 327C. So in order to bring 10 grams of lead initially at 20C the bullet to its melting point it just has to absorb 307 x 10 x .128 Joules = 392 Joules or less than 20% of the total bullet energy. There is additional energy of 22.4 J/g to actually melt it, so an additional 224 Joules would be needed to melt 10 grams of lead, for a total of 616 Joules or less than 30% of the total bullet energy.
We can estimate that the 10 gram bullet has 8 g. of lead and 2 g. of copper jacket. Copper has a specific heat of .385 J/g-deg. C, a melting point of 1083 deg. C, and a latent heat of melting of 207 J/g (9 times that of lead). So to melt 2 grams of copper initially at 20C, the jacket needs to absorb 2 x 1063 x .385 + 2 x 207 = 1233 Joules or over half the bullet energy.
Since it is apparent that a significant amount of energy is transferred to the target when hitting asphalt, there is likely not enough energy available from the impact to melt the copper even if the jacket absorbed all the compression energy. But the 8 g of lead absorbing its share of half the bullet energy (.8 x .5 x 2250 = 900 J.) receives almost twice the amount of energy needed to melt all the lead.
