# Understanding Von Braun's Math: Theoretical Isp

I've just read Wernher Von Braun's Project Mars: A Technical Tale, and I'm trying to understand the math in the back. I can follow some of what he's doing, but other parts I just can't tell where his numbers come from. In particular, I can't figure out his equation for theoretical Isp.

I've searched all over the internet, and found several equations relating Isp to the temperature, pressure, and geometry of the chamber, along with gas constants, specific heat ratios and molar masses of the propellant, as well as ambient pressure. However, Von Braun doesn't use any of these numbers (or he doesn't mention that he does); his equation is simply:

$I_{sp,th} = 9.323 * \sqrt{H_0 - H_e}$

Does anyone have any idea how that was derived? Or at the very least, what the significance of 9.323 is? I guess that it goes approximately like this: assume that the enthalpy change (I'll denote it $\Delta H$) is fully converted to the kinetic energy of the exhaust and the exhaust moves with velocity $v$ relative to the engine. Then we have: $$mv^2/2 = m \Delta H;$$ $$v = \sqrt{2\Delta H};$$ $$I_{sp} = \sqrt{2\Delta H}/g.$$ But that is assuming that all values are expressed in, for example, SI units ($I_{sp}$ in seconds, $\Delta H$ in J/kg, $g$ in m/s2). Von Braun uses the value of $\Delta H$ in cal/g instead. Using the value of calorie as 4.184 J, we have 1 cal/g = 4.184 J/g = 4184 J/kg, so $$I_{sp}[s] = \sqrt{2\cdot 4184\Delta H[cal/g]}/g[m/s^2] = 91.48\sqrt{\Delta H[cal/g]}/g[m/s^2].$$ Using the value $g = 981~ \text{cm/s}^2 = 9.81~ \text{m/s}^2$, which is used earlier in the page, we get $$I_{sp}[s] = 91.48\sqrt{\Delta H[cal/g]}/9.81 = 9.325\sqrt{\Delta H[cal/g]}.$$ Well, close enough. Von Braun probably used slightly different values for calorie or $g$.