how long did it take the bombs to hit the ground from 25 000 ft

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mhuxt, from the data table you supplied the Tall Boy dropped from 25,000ft strikes at 1260ft/s BUT at an angle of 16 degrees. While the table does not specify this it must be 16 degrees from vertical. Thus we have a right triangle where 1260 is the hypotenuse. This resolves itself into two motions a vertical (y) and a horizontal (x) velocity. Using trig functions: sin(16) = x / 1260 or x = 347ft/s (236 mph) and cos(16) = 1211ft/s (826 mph) which is still supersonic. However,
In John Ellis book "One Day in a very Long War",: Tallboy was designed to be dropped from an optimal altitude of 18,000 ft at a forward speed of 170 mph. Impacting at 750 mph . Not supersonic
 
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GregP, I am not sure to what you are refering too. A bullet/shell begins to accererate at -32ft/s/s the instant it leaves the barrel dropping 16ft in one second with a vertical velocity of 32ft/s. How far down range the impact occurs depends upon the horizontal velocity. The horizontal velocity does not have a "terminal velocity" as such unless you consider 0 to be a velocity. Gravity does not act horizontally thus the initial horizontal velocity constantly decreases due to air resistance (friction). Given sufficient time this could become zero with the bullet/shell simply falling straight down. Anyone who shoots at distant targets must know their bullet drop.
For the M16 a .223 round drops 2in in the first 100yds; 8.7in in the next 100yds; and 41.9in at 400yds. The increasing drop is due to the ever decreasing horizontal velocity and the ever increasing vertical velocity. An M16 round takes about a second to travel 650yds at which point it has dropped 12ft
The average fall of the earth's surface is 1ft per mile thus an object moving fast enough can cover that mile in the time it takes to fall 1ft (.25s) or 14,400mph. Thus at the end of 1s of fall the object is still at the same height and so on as the Earth's curvature is followed, i.e. you are in orbit
 
Hi Mike,

In a vacuum, the bomb drops at constant horizontal speed and accelerates according to V = at where a = 32 ft/sec^2 or 9.8 m/s^2. In the air, the horizontal speed decays with time as the bomb drops and it accelerates vertically much more slowly than in vacuum until it impacts or reaches terminal velocity. So the horizontal speed is variable and continuously slowing while the vertical velocity is increasing until either impact or terminal velocity is reached.

Whay I am saying simply is that if you consider the vertical component alone, then you will misss horizontally by some amount since the horizontal velocity is NOT constant. Probably not by too much, but easily enough to negate a LOT of damage from 15,000+ feet.

This all assumes ONE bomb is dropped.
 
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I don't think that the Tallboy was designed to be dropped from 18,000ft but that is how high the Lancaster could carry it. It was designed to be dropped from much higher altitudes, 30-40,000ft, IIRC.

As far as the speed goes, I would think the speed of the Talboy along its axis would be more important for its effectiveness than either the horizontal or vertical components. You could argue that those components are vital for aiming the bomb, butthe release mechanism was not the most reliable system (supported by a chain which is released to drop the bomb).
 
Hi Wuzak,

I'm curious. Since the Lancaster hauled the largest payload in the ETO, what possible difference could it have made what altitude the Tallboy was made to drop from? The ONLY imporant thing here is what altitude it WAS dropped from. If 18,000 feet was it, then that was the best that could be done unless the ETO suddenly acquired B-29's, which it never did.

Not arguing here ... just saying, if there is no platform to carry the thing to 30,000 feet, then what possible use is it to say the design altitude was 30,000 feet? It's like saying "My cheating wife would have remained true to me if only she hadn't cheated" or "maybe I'll have a beer ..."

The claim is pointless, even if accurate.

I WILL have a beer now. Cheers!
 
Barnes Wallis had also proposed the 6 engined "Victory bomber" to carry such earthquake bombs. Whether or not that could have carried the bombs to teh required altitude is another matter.

Wallis reworked his calculations to take into account the lower altitude at which the bombs could be carried and check to see if the bombs would still be effective.
 
The Bielefeld viaduct was attacked from only 12,000 feet. One Grand Slam dropped by Squadron Leader Calder did what over 3000 tons of conventional bombs had failed to do. The time recorded for the fall of the bomb was 35 seconds though I have no idea how accurate that might be.
Cheers
Steve
 


Rather misses an important point which is relevant to this discussion. IF you need a bomb with the weight to frontal area ratio and the shape of a Tallboy/Grand Slam to go supersonic then the chances of a "normal" WW II shaped bomb reaching such a speed is about zero.
The Fact that the Tallboy/Grand Slam didn't achieve supersonic speeds in service due to limitations of the carrying aircraft does not mean that "normal" bombs could. The Tallboy/Grand Slam design puts an upper speed limit on what was achievable in WW II.
 
Would any free falling bomb pass through the sound barrier if dropped from within the atmosphere? Or would the shock waves make a natural limit?
 

Mike - the 'hypotenuse' is the actual velocity vector.. and if the target had an inclination of 16 degrees perfectly normal to the velocity vector the impact velocity would be the same as the velocity vector
 

GregP, totally understood. If you read my posting I thought I had made that clear with the .223 bullet example. If you aim perfectly horizontally 6ft above the ground you cannot hit a target 400yds distant as the bullet drops 12ft in the time required to cover 400yds horizontally. So one aims ABOVE the target by the known bullet drop distance. Bomb sights, like the Norden take the aircraft factors into account for the bombardier. The theoretical CEP for the Norden was only 75ft though in combat it was more like 1200ft so pinpoint bombing was not possible thus the Navy turned to dive bombers and skip-bombing and the Air Force to the lead bomber technique.
My point was simply that you cannot consider ONLY horizontal or ONLY vertical as the bombs actual velocity is a composite (summation) of the two which is constanly changing as the bomb is accelerating (both + - ) over the entive time of fall. BUT it is the "time of fall" that determines range. The bomb/bullet can only move horizontally while it is falling. The aircraft can adjust horizontal speed + or - within limits. In todays supersonic bombers, release can be 20mi from a target
 
drgondog said:
Mike - the 'hypotenuse' is the actual velocity vector.. and if the target had an inclination of 16 degrees perfectly normal to the velocity vector the impact velocity would be the same as the velocity vector
Click to expand...

Agreed, I was trying to show that impact velocity is not the same as the vertical falling velocity nor the horizontal velocity but a summation of the two. Thus a bomb can impact supersonically (above 767mph) while the vertical vector and horizontal vectors are less than this
 
pbehn said:
Would any free falling bomb pass through the sound barrier if dropped from within the atmosphere? Or would the shock waves make a natural limit?
Click to expand...

You ask a "not-easy-to-answer" question since multiple factors are involved. We can start with the simple fact that the speed of sound is not a fixed number. It varies with the density, temperature, and composition of the conducting medium (air in this case). The density of air varies considerably with height as does temperature and composition. Thus a falling object can be both super and subsonic in its fall. Recently Felix Baumgartner did a freefall jump from over 24 miles and went supersonic (for a time) at around 30,000ft - 33,000ft (Speed of Sound = 761mph @ 15C @ sea level; 678mph @ 30,000ft @ -44C; 659mph @ 60,000ft @ -57C)
If you wish to fiddle around with the math:
F(air resistance) = 0.5 A v^2 p(z) Cd Where A = max cross sectional area v = velocity Cd = drag coefficient p(z) air density @ z altitude
This is in turn equal to m g Where m = mass and g = 32ft/s^2 gravitational acceleration (varies with height but a few thousand feet over the earths radius is not significant.
Cd depends on shape and surface finish generally from 0.05 - 0.1 (tables exist for bomb type) This value is for the BOMB only, as the speed of sound is approached the shock wave formed adds exponentially to the drag.
So a simple answer is YES a simple one dimensional vertical drop can achieve supersonic velocities as Felix Baumgartner demonstrated (his Cd was a large 1 - 1.3).
 
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Hi Shortround,

Unsurprisingly we disagree, but maybe by not much. The Tallboy design allows supersonic impact but the available transport resources to carry it didn't allow for it since they could not get to the design height.

Maybe we're saying the same thing in different words.

The Lancaster could not reach the design height of the Tallboy, but I'd suppose that wasn't a huge comfort to those upon whom it was dropped. It was a BIG problem wherever it happened to impact, wouldn't you say?
 
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