Terminal Velocity of Bombs

[Dave Saxton]

.....In this case, the bomb was probably disarmed on penetrating the armor (as occurred on Scharnhorst (3 times) and Tirtipz (once)). By way of comparison, Friedman (US battleships) gives the Iowas as immune to 1600 AP bombs below 12200 ft. Does any one know if AP bombs had caps?

I have found original data on the US 1600 lb AP bomb. The terminal velocity was 270m/s. At 270m/s and striking at the normal, the penetration is ~205mm homogenous armour. It looks like Friedman was a bit optimistic here. In the case of the Tirpitz, the bomb should have just been able to penetrate 140mm effective homogenous armour with all the energy consumed, and this is just what happened. As for a cap or not, I don't know. The fact that these bombs showed a probability to being broken or rendered inert by de-capping arrays may indicate de-capping, but it may also indicate the center of gravity not being near enough to the head, and base slap was probably a factor as well.

[Tiornu]

Do we know that the Tirpitz bombs reached terminal velocity?

[Bgile]

Bombs dropped by dive bombers seldom reached terminal velocity because of the low altitude at which they were normally dropped.

[José M. Rico]

How does the diving speed of the bomber at the time the bomb is released affect its terminal velocity?

[Bgile]

How does the diving speed of the bomber at the time the bomb is released affect its terminal velocity?

The bomb would obviously have an initial velocity equal to that of the aircraft. Level bombers, on the other hand, drop their bombs with 100% of the velocity in the horizontal plane. However, I'm pretty sure I've read that bombs dropped from dive bombers don't ever reach terminal velocity, but maybe someone else has better info. IIRC the special AP bombs dropped at Pearl Harbor were dropped from 12,000 feet. I think dive bombers typically drop from less than half that altitude; just high enough that they were able to pull out. I don't have the math skills to find out the difference in impact velocity between the two.

[RF]

The lack of height from the dive bomber obviously mitigates against impact velocity, presumably it is compensated for by greater accuracy to the target.

[RobertsonN]

"Dave Sexton wrote: The terminal velocity was 270 m/s.

Thanks. Up to now I have had no data on bomb terminal velocities.

"Dave Sexton wrote: In the case of the Tirpitz...

Indeed this was a bomb that just penetrated the armor system. It would seem to be therefore a test of what the effective thichness of the two deck system was, at least against typical AP bombs of the period. Is the height from which the bomb was dropped known? The aircraft was, I believe, a Barracuda, which attacked in a long shallow dive. In the attack in April 1944 the height was 3000 ft: height was traded against number of hits (13). Several 1600 lb AP bombs struck but none penetrated due to the height. Incidentally, Friedman gives the Montana as immune to the 1600 lb AP bomb under 18000 ft, which implies the terminal velocity was not reached until at least that height.

[dunmunro]

The lack of height from the dive bomber obviously mitigates against impact velocity, presumably it is compensated for by greater accuracy to the target.

Here is an article from 1943, that gives generic data on bomb ballistics:
http://books.google.ca/books?id=fScDAAA ... mb&f=false

and a simple java applet:
http://www.grc.nasa.gov/WWW/K-12/airplane/termv.html

[lwd]

This link has an article suggesting that the terminal velocity of bombs was between 550 and 650 mph. Which translates roughly to between 240 and 300 m/sec. http://www.flightglobal.com/pdfarchive/ ... 02755.html
This means that it has to fall for something over 24-30 seconds which corresponds to a drop height of over 1,400-2,200m.
This article:
http://en.wikipedia.org/wiki/Tallboy_bomb
States the Tallboy had a terminal velocity of ~1,100 m/sec which would require a drop from over 5,500 m.
Here's a link to a terminal velocity calcualter that should give better numbers than the above:
http://www.grc.nasa.gov/WWW/K-12/airplane/termv.html
If the verticle component of a dive bombers speed is 200mph then its roughly the eqivalant of dropping from a level bomber 400m higher up. I do seem to recall reading somewhere that the altitude at which dive bombers released their bombs was somewhat dependent on the target type. I.e. doctrine suggested higher drip hieghts vs battleships and lower vs others.

[Dave Saxton]

G Slover has posted the WWII USN manual on bomb ballistics. (sorry I don't know the link). The manual states that dive bombing usually creates higher striking velocities at typical release altitudes required to obtain hits with dumb bombs on ships. This is because, according to the manual, the vertical speed componant of the dive bomber gives it a head start. The British initially calculated that the 1600 lb AP bombs would have to be released from the Barracuda dive bombers from greater than at least 3,000 feet to defeat the Tirpitz's deck system. This they didn't always succeed in doing. This requirement was stressed to the pilots in the later attacks and the min release altitude was re-set to 4,000 feet and 5,000 feet was strived for. The bomb in Tirpitz's case had a nominal striking angle of Approx, 12* from the normal and was released at about 5,000 feet. It should obtain a velocity of around 225 M/s, according the FAA Manual on dive bombing with the Barracuda. Btw, according to the FAA Barracuda manual the typical dive ange at resease point was 70*. The penetration should be ~140 mm.

[Tiornu]

Since the bombing of Tirpitz is usually represented as diverging from the preferred, textbook technique, we would need specifics on the bomb release before consulting any tables or curves.

[José M. Rico]

According to Garzke & Dulin the bombs on Tirpitz were dropped from an altitude of between 430-900 meters. What was the typical diving angle/speed of the Barracuda?

[lwd]

G Slover has posted the WWII USN manual on bomb ballistics. (sorry I don't know the link). ...

Had to use his search engine but found:
http://www.eugeneleeslover.com/USNAVY/CHAPTER-23-E.html
http://www.eugeneleeslover.com/USNAVY/CHAPTER-23-D.html
http://www.eugeneleeslover.com/USNAVY/CHAPTER-3-5.html

[Thorsten Wahl]

Tallboy had a terminal velocity of ~1,100 m/sec which would require a drop from over 5,500 m.

Seems to me the speed must be somthing wrong maybe 1.100ft/sec

terminal velocity of 1,100m/s requires roughly 110 sec of free fall, even if we neglect air resistance

v=a*t

1100m/s~9,81m/s²*110s

[lwd]

Probably my mistake.
Let's see.
Wiki does say 1,100 m/sec
at 10 m/sec*sec that would mean at least 110 seconds
in that time it would fall a distance of .5*a*t*t ~5*110*110 = 60,500m
don't know where I messed up my math but I certianly did.
Let's see that means a drop of 24 seconds corresponds to a height of at least 5*24*24 ~2,900m
Don't know what I did wrong but definitly messed things up thanks for the correction.
A 30 second drop would then correspond to a minimum distance of 4,500m.
Does show how one can be techically correct and very misleading at the same time. Sorry about that all.