Just a curious note about a book of a series I was reading. Each book dealt with a WW2 aircraft (or at the most two aircraft)- all the well known ones such as Liberator or Halifax.
The one of the Super Fortress was interesting, and particularly in relation to this question. One of the points that stuck with me was that it was really fast in a dive- as in over 600 mph. This was from a crew member in the Korean War (IIRC). The bomber was able to escape the fighters.
The details are a little cloudy- it has been years since I read the book- and it may indeed have been escaping Japanese fighters in WW2. However, that speed always stuck with me given the shape of the aircraft. If true, it must have been a very strongly built aircraft.
If Enola Gay turned and dived as well it may indeed have gotten a fair way away.
If I knew an atomic bomb was going to explode behind me I’d be burning my arse to get out of their as well.
It seems that you’re correct here. That was part of what I recalled from the interview, and obviously, it makes no sense. Score this as my error.
Thanks for the answers everyone. This was one of those things that sort of stick in your head for a several days as you ask yourself whether it was really possible.
Ayup. In order to accurately deposit a parachuted bomb onto a target, you’d need to know:
-the vertical speed of the bomb, which will vary with altitude (more dense air at lower altitude means a slower descent rate)
-the winds at various elevations all the way from 31,000 feet to detonation altitude
Then you’d have to run a computer model/numerical simulation to figure out what the path of the bomb will be after dropping it; that lets you determine where you need to release it in order to have it land on the city.
Apart from the difficulty of ascertaining the winds on the day of the drop, they lacked the computing power in 1945 to run such models in a timely manner. Far simpler to do as much as possible to remove complicated aerydynamic effects; leave out the parachute and streamline the bomb. What’s left is a very simple set of ballistics equations for an object that is only influenced by gravity and inertia; any second-year engineering student can crunch through that in a few minutes with a pencil and a slide rule. The only parameters you’d need to know beforehand would be the speed and altitude of the aircraft at release, and then you can easily calculate how far from tne city the release point should be.
Wind still affects gravity bombs, but obviously not as much. With the Norden bomb sight used in WWII, the bombardier would input the estimated wind speed and direction and the planes altitude and heading and they didn’t have to do much more than wait for the crosshairs to line up with the target.
I doubt parachute technology was up to the task in those days. I don’t remember them dropping anything very large during the war. I’m not sure if they drop things that big from high altitude today for that matter. But a normal bomb run worked just fine, so maybe they never even thought of doing it differently.
Oddly enough, many older sources do claim that the bomb was dropped by parachute. If you google ‘“Little Boy” parachute drop’ you’ll find any number of web sites which make the assertion, and I remember reading it in books and magazines in pre-Internet days.
I suspect the authors were confused because the a-bomb tail fins were described as a “parachute tail assembly”, as in Item 9 in the Fat Man bomb diagram in wikipedia.
The Enola Gay was a B-29B, a variant specifically optimized for attacking the Japanese mainland. It was lighter and faster than the primary variants due to deletion of all but the tail guns. The B was capable of 364 mph in level flight at 25,000 feet, so probably a bit more at 31,000 feet.
Pretty sure it wouldn’t have been capable of the required 720 mph (9 miles in 45 seconds) even if it had been riding the blast wave.
Unless I’m misreading them, the calculations above don’t take into account the fact that the bomb, when dropped, doesn’t drop straight down. Instead, the bomb is moving forward at the same speed as the airplane, and it will continue to do so, except to the extent slowed by air resistance, until it hits the ground or explodes. If the airplane did not turn after releasing the bomb, it would be approximately overhead at the time of detonation. If the airplane could achieve the physically impossible feat of immediately reversing its direction, it could move away from the bomb twice as fast as its speed. How fast is it able to move away from where the bomb will be in light of its actual turn?
Unguided bombs follow a parabolic trajectory, so while the bomb’s movement would effectively double the “escape rate” at first, it would be moving almost vertically by the time it detonated.
In other words, the theoretical maximum “extra” distance based on the bomb’s trajectory is some number less than the distance achieved by the plane itself during that time (and probably much less, especially if it was cruising at reduced speed when it released the bomb).
The only reason it follows a parabolic trajectory is because gravity is accelerating the bomb downward; that affects the horizontal velocity not at all. Its forward velocity would remain constant in a vacuum.
Fubaya and **Machine Elf **are the gods of this thread. The only thing causing the bomb’s forward velocity to decrease is air resistance, which varies considerably between 31,000 feet and 1,900 feet, and is beyond my ability to estimate. But it starts at 300 miles an hour, and the bomb itself moved a non-arbitrary distance horizontally in the 45 seconds between release and detonation.
A parachute might also have given the defenders enough time to shoot the bomb down. An A-bomb is actually a fairly delicate piece of equipment; it doesn’t take much to turn it from a city-killer to a fizzle.
Perhaps you should read my post again. The bomb isn’t in a vacuum. The point is that the bomb’s forward travel cannot double the distance between the aircraft and the detonation, though it can increase it. I kind of thought it was obvious that it would slow down because of air resistance, but there you go.
It’s not like it was a precision weapon. Would missing the target by a mile really make much of a difference?
I am not really convinced that having something like a tail rotor slowing it down would make all that much difference in accuracy. Sure, if you have a 500 pound bomb and your trying to hit a factory, accuracy is important. But if you have a 5 ton bomb and your target is a city center, well, I think you’d be alright with doubling the fall time.
The kind of accuracy provided by the Norden matters if you’re bombing something as small as a ship or a factory with something as small as a 500-pound iron bomb. Bombing a city with a 10-kiloton nuke? Not so much. I expect they used it just because the plane was equipped with it, but if they had missed their target by a quarter-mile it wouldn’t have mattered.
Well, I think I was mostly saying what others had said, just in a different way maybe. But I will accept the co-god position. He can have the taxes and offerings, I’ll take the virgins.
They were off by 1.9 miles, which is considerably greater than the quarter mile I indicated. Moreover, it’s not clear from the Wikipedia article whether they were aiming for their original planned target, or the nearest thing they could see with the given cloud cover.