After some interruptions, I'm back to the shell holes in the San Khayom bridge. I apologize for the length of this; but I don't know how to shorten it further.
With regard to the hole at Point A, I'm trying to visualize the actions of the aircraft that produced the hole. I'd like to include a sketch on my webpage of the possible / probable path of the aircraft during its attack.
The consensus is that the hole is apparently from a 20mm projectile; and in the Southeast Asia Theater both the P-38 and the Bristol Beaufighter carried HS.404 cannon which used that ammunition. I'll use the P-38 as a subject here simply because I can find more information about it (the Beaufighter might be the better candidate because, beyond its harmonization range, it would have offered a dispersion pattern with its four cannon: such a pattern might more easily have included Point F, the other 20mm impact point on the bridge. [color=#FF0000]Does anyone know of an on-line presentation, including sketches, perhaps a manual, for harmonizing the guns on a Beaufighter?[/color]).
So, scenario: a P-38 pilot happened upon a "target of opportunity", a train traveling south out of Lamphun and approaching San Khayom bridge. The pilot reduced altitude so as to fire on the train. Coming into range, he fired for a certain period; then he broke contact and climbed away from the train and the bridge.
Angle of attack: The exit angle of Hole A measures about 12° to the horizontal. In the sketch below, I assume the bullet path was almost unaffected in passing through the comparatively thin bridge plate (projectile diameter was twice the thickness of the target plate). On that basis, I assume the pilot dropped to an altitude matching the 12° allowing an angle of attack of 12°.
A P-38's single 20mm cannon was centrally mounted in the fuselage and its operation was basically point (the plane)-and-shoot. The range of the cannon was limited primarily by the pilot's visual acuity and his properly adjusting for bullet drop (excluding weather, aircraft performance, pilot skills, whatever).
I assume an attack speed of 350 mph: a P-38's cruising speed was 275 mph and max speed was 414 mph. A velocity diagram would look like this for a 12° dive at 350 mph:
The important information here is that, flying at a down-angle of 12° and at a speed of 350mph, the aircraft would have been dropping 107 feet per second.
The idealized flight path would have looked like this:
Of course, the pilot would have disengaged at some point before his aircraft flew into the target.
[color=#FF0000]Do these assumptions and the scenario seem plausible? If not, I'd appreciate being corrected. [/color]
If the scenario is not too far off:
[color=#FF0000]At what range would the pilot likely have started firing?[/color] (I read that one B-25 gunner had his guns harmonized for 1000 yards: he cautioned that this was useful only for stationary targets)
And [color=#FF0000]at what altitude would the pilot have had to disengage (how much altitude would be burned up after the pilot tried to gain altitude; and how much spare altitude should he have allowed)[/color]?
Tying these down would also establish how long the pilot might have fired.
Alternate scenario: With the above presented, there is a piece of data which could contradict this scenario, and I don't have the background to interpret it: the "floor" of Hole A is at an angle of about 7° to the horizontal. If 7° (not the 12° used above) were the angle at which the projectile entered the plate, and the exit angle was, as actually measured, 12°, then passing through the plate would have bent the projectile's course by 12° - 7° = 5° (see cross-section).
If the angle of attack were a very lean 7°, then at a range of 1000 yards, the gun would have been only 366 feet off the ground --- with bullet drop compensated for by either, an extra 24 more feet for a total of 390 feet, or a slight tilt upward in angle of attack. And at 500 yards 183 feet (compensated for 4 feet of bullet drop). On the other hand, the aircraft's altitude loss while changing from a down-angle of 7° to actually moving upward would have been less than recovering from a down-angle of 12°. But operating that close to the ground at 350 mph or more seems unlikely (which is why I didn't include it in my main presentation above); though I am willing to be corrected.
[color=#FF0000]Are there any comments about this alternative?[/color]