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ENT - The Expanse
Posted: 2003-05-24 09:41am
by His Divine Shadow
I think it was the expanse anyway, Reed commented that their photonic torps have variable yield, they could do anything from take off a communication antenna on a shuttle to making a 3km crater on an asteroid.
Anyone care to calculate how much energy this would take?
Posted: 2003-05-24 09:48am
by TrekWarsie
I've heard anywhere from 39 megatons to around 1 gigaton, depending on the density and composition of the asteroid.
Posted: 2003-05-24 10:44am
by Ender
D = D1 * W * .3
D = distance in feet
D1 = distance from hypocenter at which overpresure is a given psi
W = yield in KT
D1 = ~2.2 * P *^-.7
Assume 5 psi overpressure because that is the overpressure required to demolish most civilian structures
=> D1 = .713
Radius of 1500 meters, convert to feet = 4921.26 feet
(4921.26 /(.713 *.3) = 23007.3 => 23 MT
Note this is in atmosphere and not space, thus flawed. Not sure how to do it in space.
Posted: 2003-05-24 10:46am
by His Divine Shadow
That calc is about atmospheric over-pressure, not the actual cratering of soil, I don't see how it has any relevance here.
Posted: 2003-05-24 10:53am
by Ender
It's the best formula I can find. I believe it is also the one the asteroid calculator uses. To cause a given overpressure for a stated crater size, that is what it would have to be. If you find a better formula, please tell me.
Posted: 2003-05-24 07:11pm
by Ender
According to SLM thread, it would be a buried warhead.
Thus it would be 80 instead of .713 going by teh example provided
here
Small heavy underground structures will be severly damaged only if they are within 1.25 apparent crater radii of a nuclear blast, find the maximum miss distance for a 150 kt earth penetrating warhead.
Crater size: D = 80 (150)0.3 = 270 feet
Result: 1.25 x 270/2 = 170 feet. Therefore if the 150 kt bomb is exploded withing 170 feet (horizontal distance) it will severely damage the underground structure.
So the yield would be 205 KT