Originally Posted by 16.5
I didn't double-check this - caveat emptor. I also made no attempt to figure out any diluting effects inherent in sol-gel therm*te. It's just a rough calculation.
(Legge and Szamboti), max yield is decreased by over 50% at 600 deg C.
The specific heat of steel is 500 J/(Kg*K).
So, 1 kg steel, w/temperature raised by 600 deg C, needs 500 * 600 Joules
= 300,000 J
I guesstimate that we only need 1/4 of this this 1 kg of steel heated to 600 C, to get differential heating, sufficient to cause a tilt, as I outlined. So, I'll assume that we need ~ (300,000 J/ 4) = 75,000 J.
Energy density of thermite is 4MJ/kg. So we need 75 / 4,000 kg = 0.01875 kg thermite per kg of steel.
That's about 2%, by weight.
Since the mass density of steel is 7850 kg/m^3, and the mass density of thermite is 3474 kg/m^3, the volume ratio is roughly 4% thermite to steel
Say that a box column side is 1 inch thick, and x meters high, and y inches wide. Applying 4% thermite, by volume, to just this one side, over x meters times y inches, obviously is 4% of 1 inch, thick. (Applying over all 4 box column sides would give you just 1% for each side.)
4% of 1 inch is about 1 millimeter