Talk:Tallboy bomb

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Whidbey the first sentence in the History section does not complete the thought.

"Barnes Wallis had considered the strategic use of bombs to destroy the enemy's capacity to wage war by hitting its infrastructure and manufacturing bases and to this end developed improved bomb designs based on large single bombs early in the war."

What did Wallis consider about the strategic use of bombs? As being a better use of them then on the battlefield or dropped on citzens to induce fear, or as the solution to the winning the war?

The terminal velocity mentioned in the article can not be acchieved when the bomb is dropped form a Lancaster bomber. If dropped from 6100m, the terminal velocity is about the speed of sound, bot only if you ignore drag. $\mbox{speed in } m/s = \sqrt{\frac{2 * 6100m}{9.81m/s^2}} 9.81 m/s^2= 346 m/s$

Roland

[edit] Tirpitz

Tirpitz was not tranferred to Tromso for reparation. It schould serve as a floatig coastal battery.--WerWil 13:33, 11 December 2006 (UTC)

The bombing of 'The eagles nest' is well footnoted as 'apeared to have been effective' from a proper historical source... but it's questionable how effective it really was, the building is still standing, and it's own history page claims to have never been bombed. (both the wikipedia entry, and the official website) 65.49.176.158 13:51, 17 May 2006 (UTC)

I thought that on the day that 617 Squadron tried to find the Eagle's Nest it was covered in clouds, so that instead they bombed the SS garrison's quarters. Darkmind1970 16:53, 13 February 2007 (UTC)

[edit] tailfin size

The text says that the tail was about half the length of the bomb, but it's obviously much shorter than this. Is the picture of an earlier design or is the text talking about the "tail" as in the non-explosive part, and not just the fins? It should be clarified. KarlM 02:59, 14 February 2007 (UTC)

[edit] Impact velocity

For the sake of ending edits like [1], here’s a quick calculation of the speed at impact without air resistance. From high school physics, for constant $g$ we have

$v = \int g dt = gt$

and hence

$y = \int v dt = \frac{g}{2}t^2$

from which we get

$v(y) = g\sqrt{\frac{2y}{g}} = \sqrt{2gy}$ .

For the altitudes given in the article, $v (7700 m) = 390 m / s$ and $v (12200 m) = 490 m / s$ . Solving for $v (y) = 330 m / s$ (the speed of sound; it varies with air density, but we only want an estimate anyway), $y = 5600 m$ . This does not actually give any estimate of the terminal velocity, but should be enough to refute that edit. —xyzzy_n 15:53, 25 February 2007 (UTC)