Terminal velocity of Little Boy, and Enola Gay's speed, August 6, 1945.

I recently saw a documentary featuring Paul Tibbets, the pilot of the B29 (Enola Gay) which dropped the atomic bomb on Hiroshima. Paul Tibbets (Gen, USAF Ret.) stated he dropped the bomb and immediately turned, exiting the area as quickly as he could. He reported that the crew saw the bright flash of the detonation when the aircraft was 9 miles from the drop point, some 45 seconds after releasing the weapon.

I have two questions. The first concerns the terminal velocity of the egg-shaped bomb free falling from an aircraft. The atomic bomb released from the Enola Gay fell from 31000 feet, to 1900 feet above Hiroshima, (elev. about 26 m) in 45 seconds. That’s over 5 miles in less than a minute. Is that actually possible for a free falling object?

Wiki’s report differs slightly from what Gen Tibbets reported.
The release at 08:15 (Hiroshima time) went as planned, and the gravity bomb known as “Little Boy”, a gun-type fission weapon with 60 kilograms (130 lb) of uranium-235, took 43 seconds to fall from the aircraft flying at 31,060 feet (9,470 m) to the predetermined detonation height about 1,900 feet (580 m) above the city. The Enola Gay traveled 11.5 miles (18.5 km) before it felt the shock waves from the blast.[64]

The second question concerns the speed of the B29 after releasing the bomb. The B29 traveled 9 miles (when the flash was observed) in 45 seconds. This seems to be an impossible speed for the B29, even in a controlled dive. Not to mention a 270 degree, 60 degree bank turn, immediately after release. All this seems rather remarkable for a piston powered aircraft. Were these reported speeds possible for a B29, even in a dive? The published speed of the B29 is 365 mph (mach 0.55) in level flight. http://www.globalaircraft.org/planes/b-29_superfortress.pl

Throwing some numbers into this terminal velocity calculator, I get a terminal velocity of 2750 ft/sec.

I used a weight of 9700 pounds, and a cross-section area of 4.27 square feet (based on a diameter of 28 inches from this page), and entered 31000 feet.

Of course, the terminal velocity will decrease as the atmosphere gets thicker, but I thought let’s start with an upper bound.

So, anyway, 2750 feet/second is a mile every two seconds, about - 31 miles per minute.

Then the lower bound. At 1900 feet, the calculator says 1700 feet a second, still 19 miles a minute.

So 5 miles in 45 seconds is doable.

For the other part of your question, I have no idea how a B-29 could go 9 miles in 45 seconds - 720 miles an hour even without the turn. Is it possible they were 6 miles up, and 5 miles horizontally distant, for a total distance of maybe kinda 9 miles?

Sure. Why wouldn’t it be? There’s no hard upper limit for terminal velocity through the atmosphere.

Example: A 747 can sustain level flight at ~600MPH with the thrust from its jet engines amounting to maybe 20% of its weight. Turn the engines off, point it straight down, and gravity is pulling the plane down with a force five times what the engines do, so it would happily exceed 600 MPH going straight down (until it disintegrated due to aerodynamic forces).

Disregarding the effects of aerodynamic drag, and assuming it was dropped from the Enola Gay in level flight, it would have taken 42.5 seconds to fall that distance, reaching a top vertical speed of 467 MPH. That’s not all that fast. Little Boy weighed almost 10,000 pounds and had a reasonably aerodynamic shape.

Wikipedia claims the plane travelled 9 miles before it felt the shock waves from the detonation. At top speed it would only have traveled about 4.5 miles after dropping the bomb but before detonation; the remaining 4.5 miles were covered after detonation but before the shock wave caught up to the aircraft.

I know Im going to be sorry for asking this:

Do the A-bomb shock waves travel at the speed of sound?

B29 cruise speed is listed in wiki as 220 MPH (almost 4 miles/minute). Max speed is 357 MPH (almost 6 miles/minute). I don’t know how much of a turn it made after release, and whether or not the aircraft climbed or decended.

There is no way a B-29 traveled a ground distance of 9 miles in 43 seconds. Even if they entered the jet stream and doubled their ground speed that way, then combined with the ~5 miles the bomb fell, it still wouldn’t be 9 miles from the detonation point. Something is wrong with one of those figures.

Yes, shock waves are essentially sound waves. Of course the radiation flash would have traveled much faster – at the speed of light. I assume there would be neutron and other particle emission as well. I don’t know how fast that would have been.

Initially, yes, as does the shock wave from any bomb. This is the blast wave that does the mechanical damage (as opposed to the thermal damage).

It slows down as it radiates away from ground zero. By the time it reached 9 miles away (where it made contact with the Enola Gay), it had probably slowed down to the speed of sound.

To put numbers on it - Pythagoras says to be 9 miles from the point that is 5 miles below you, you’d need to move 7.5 miles horizontally (and the 9 mile figure is based on the flash, not the shock wave, so it’s, for our purposes, instantaneous).

7.5 miles in 45 seconds is 600 mph.

I conclude the Enola Gay was not 9 miles away from the detonation at the time of detonation.

What kind of wind speed is likely at 31000 feet? If the Enola Gay was running with the wind, its ground speed could easily exceed its max air speed. Maybe not enough to get 9 miles away, but it’s probably getting closer to that.


The OP specifically mentions that 9 miles is the reported distance at which the plane felt the blast. (Not “see the flash” instantaneous.)

Speed of sound is 768 MPH. (Just a little over twice the reported max speed of a B29.) Which is 4.6 seconds to cover one mile.

At 300 MPH, the B29 would travel 5 miles in a minute, or 3.58 miles in 43 seconds (time given for the release to detonation). It’s already 5.8 miles up. At detonation, then, the plane should be 6.8 miles from the blast. (Math check in aisle 6, please.)

Sound would take 31.3 seconds to travel 6.8 miles. The aircraft would have moved another 2.6 miles in that time (another 12 seconds at mach1). In 12 seconds, the B29 moved another mile. (Another 5 seconds. :stuck_out_tongue: )

Edit: Throw in a hard bank to the left, etc etc, and I get lost.

Did the plane make a hard turn after releasing the bomb? If that’s the case, then part of the distance would be due to the fact that the bomb would have kept traveling in the direction it was released at a pretty high speed, while the plane turned away.



Not egg shaped. The Hiroshima bomb was a gun-type weapon; relatively long and slender. Note that this type of weapon was never tested prior to being dropped (the [del]force[/del] theory was strong that it would work). The Nagasaki bomb was egg shapped; an implosion weapon similar to the initial bomb set off at the Trinity site.

Hiroshima’s “Little Boy” is cigar shaped.

Nagasaki’s “Fat Man”, is egg shaped.


ETA: Curses, I see that smithsb has beaten me to punch… :wink:

That was the point I was going to make. How far from Hiroshima did it release th bomb?

If cruising at 300 MPH when they dropped the bomb, then the bomb would start out with that horizontal speed. Might lose a bit during freefall; assume an average horizontal speed of 270 MPH during the 45-second freefall, then they would have released the bomb about 3.4 miles from Hiroshima.

You and I read the OP differently.

Now that’s what makes you rock. I feel smarter now.

Umm why would they do a 270 degree turn? Wouldn’t a 90 degree turn in the other direction be faster?
For that matter if you are trying to get the hell out of Dodge, it would seem a 180 degree turn would seem to make the most sense.

The turn made after dropping the bombs was actually a 155 degree turn. Due to the turning radius of the B-29, they could turn 155 degrees and step on the gas and get farther away than if they had turned 180 degrees. A lot of work went into their escape plan because they had to be more than 8 miles away from the blast or else it would knock them out of the sky.

The key here, as others have pointed out, is that, vertically, they were already about 6 miles from the target, plus an undetermined distance horizontally. There is no way they could be 9 miles from the drop point in 45 seconds, but they could be 9 miles from ground zero, counting vertical distance.

This Japanese professor calculated that the Enola Gay was “3845 meters [2.4 miles] short of the hypocenter 43 seconds before the explosion.” But he thinks it’s speed was only 200 mph. Unfortunately, the speed isn’t known for sure, but that’s what the plan called for and they probably followed it. Just assuming that’s accurate… they were 6 miles above the ground already, so if they were 2.4 miles from the epicenter, that puts them around 6.5 miles (diagonally) from ground zero when they dropped the bomb. If they turned instantly and headed the other way at 200mph, they would be 9 miles from ground zero 45 seconds later. Of course, my numbers aren’t exact, the turn took time and they certainly accelerated too, but there’s how you get your 9 miles separation from the bomb.