Hi, I have some questions about using Aluminum as a rocket fuel with oxygen. According to my calculuations, V_ex = sqrt(2*Ed) where V_ex = exaust velocity at 100% efficiency and Ed = energy density of the propellant in J/kg. This is proportional to sqrt(T/m_w) where T is temperature and m_w is molecular weight. Let's try this with LH2-LOX first: H2O has a molecular weight of 18.02e-3 kg/mol and enthalpy of formation of 242e3 J/mol. V_ex = sqrt(2*242e3/18.02e-3) = 5182.57 m/s. Dividing by 9.8 we get 528 s so at 88% efficiency we get 465 s which is the actual specific impulse. Next let's try this with Al-LOX. V_ex = sqrt(2*1675.7e3/101.96e-3) = 5733.2 m/s. Dividing by 9.8 we get 585 s, so it beats the space shuttle main engines! At 88% efficiency we get 514 s. This is enough to reach space in 2 stages very easilly. We can only conclude from the above that the combustion temperature of Al-LOX is very, very hot. How can I find the temperature exactly, to determine if the Al_2_O_3 product will be in gasseus form or not? My questions are: 1. How can I find the ideal mixture ratio of oxygen to aluminum? Why don't LH2-LOX rockets just have twice as much H_2 as O_2??? Why 4 times the O_2 as H_2 instead? It seems to me this reduces the combustion temperature, thus increasing the mass ratio at the expense of the exaust velocity (specific impulse). Will an molten Aluminum-fueled rocket need excess LOX too?? How can I find the ideal ratio? I want to compute the volume of tanks but to do this I need to know the Oxygen-Aluminum ratio. 2. After passing through the nozzle, will the Al_2_O_3 exaust be a liquid? Who ever heard of a liquid exaust rocket, except water-compressed-air rockets! But my understanding is that as the exaust passes through the nozzle it cools and speeds up, like a heat engine. I imagine then the exaust will be liquid, molten Al_2_O_3!