tommy303 wrote:The free fall bombs from 9000-ft will hit hardest as they are accelerating due to the effects of gravity from the moment of drop; those carried on a dive bomber or in the case of your example, an F4U, will be limited by the top speed in the plane's dive since the pilot must maintain control, have enough room to pull out, and not exceed the structural limits of his aircraft. This will generally be far below the acceleration speed due to gravity operating on a free fall bomb. To achieve a hard enough hit to penetrate a hardened target, a dive bomber will have to release its bomb from a fairly high altitude, not at the last possible moment.
CmdrKeen wrote:Thank you very much. What would be the bomb load for an F4U dive bombing a ship? Sources seem to say as much as one 1,000 pound bomb under each wing, and one 2,000 pound bomb under the belly; total of 4,000 pounds for a dive bombing attack?
If for instance a free fall bomb from around 9000 feet will hit hardest when dropped from that hight, why would a shell of equivelent weight and dropping from a similar hight almost vertically (when it has reached the end of its tragectory) not apparently penetrate so far or cause so much damage?
While I can't speak to penetration, all things being equal damage from the bomb is probably greater due to the larger amount of explosive it carries. Based on what I can recall a modern 2,000lb bomb carries ~950lb warhead while a 2,000lb 16"/50 shell carried a bursting charge ~150lb.
tommy303 wrote:If for instance a free fall bomb from around 9000 feet will hit hardest when dropped from that hight, why would a shell of equivelent weight and dropping from a similar hight almost vertically (when it has reached the end of its tragectory) not apparently penetrate so far or cause so much damage?
Even at maximum range a shell's impact angle will be less than that of a bomb dropped from sufficient height to stabilize in a nose down attitude. The fins on a bomb will move the center of pressure aft of the center of gravity causing it to drop the nose and impact will usually be very close to 'normal', that is at nearly a right angle to a horizontal surface. If a gun were fired in a perfect vacuum, the shell would have a parabolic trajectory and at the extreme ballistic range of the gun, the shell's angle of descent would be close to its angle of ascent. However, guns are not fired in vacuums and so drag comes into play and the usual trajectory is elliptical rather than parabolic. The angle of descent does not correspond with the angle of elevation. According to the natural law of sines, a projectile achieves its maximum range at 45* of elevation; for some really powerful guns, such as on battleships or long range land guns, maximum range actually occurs at around 48* since the height of the shell at the top of its trajectory takes it into very thin air where friction and drag caused by the air becomes much less.
Essentially, a shell from a high velocity naval gun will never have an impact angle as close to normal as that for a falling bomb, and will always strike a horizontal surface at considerably less than the optimum angle for penetration. Most battleships had gun mountings which allowed between 30 and 40 degrees of elevation by the start of WW2 and this was considered sufficient to achieve adequate range; this meant that angle of impact for a gun fired from say 30* might have a fall angle of 42* which would give it an oblique angle of impact with a horizontal surface compared to a near normal impact for a free fall bomb, and this accounts for why a shell will usually have less potential for penetrating horizontal surfaces.
note: when talking about armour penetration from guns or bombs, I am speaking strictly of AP bombs and usually APC shells.
... Gun mounts inside of turrets would limit maximum elevation and thus maximum height.
For most WW2 battleships, maximum theoretical height at highest point in the trajectory was around 6km.
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