# Where do the numbers 101,972 and 3,600 come from in terms of Thrust-Specific Fuel Consumption (TSFC or SFC)?

Online, on sites such as Wikipedia's for Specific Impulse and Thrust-specific fuel consumption, there are these units:

101,972/x g/(kN·s) and 3,600/x lb/(lbf·hr)

Perhaps 3600 comes from the number of seconds in an hour, but.... Otherwise I am confused.... . . . .

EDIT: P.S.: I just realized.... If I divide 101,972 by 1,000 to go from kilonewtons to Newton's.... Then multiply by 4.4482216 to convert from N to lbf ( one lbf is 4.4482216 Newton's...) ... I get 453.59405455, which is almost exactly equal to the number of grams in a pound (453.59237)...Maybe something.....

Specific impulse of $$1~\text{s}$$ is equivalent, in terms of speed, to $$g_0\cdot 1~\text{s} = 9.80665~ \text{m/s}$$, where $$g_0 = 9.80665~ \text{m/s}^2$$ is the standard gravitational acceleration. $$101972~ \text{g/(kN·s)}$$ and $$3600~ \text{lb/(lbf·hr)}$$ are both the inverse of that value.
We have $$1~ \text{g/(kN·s)} = 10^{-6}~ \text{kg/(N·s)} = 10^{-6}~ \text{s/m},$$ so $$\frac{1}{g_0\cdot ~1~\text{s}} = \frac{1}{9.80665~ \text{m/s}} = 0.101972~ \text{s/m} = 101972~ \text{g/(kN·s)}.$$ Also, $$3,600~\text{lb/(lbf·hr)} = \frac{3600~ \text{lb}}{(1~\text{lb}~\cdot ~g_0)~\cdot~3600~\text{s}} = \frac{1}{g_0\cdot ~1~\text{s}}.$$
• @OrganicMarble The second line has a lot of metric, but it's only needed because of the traditional [s] unit for specific impulse which absorbs $g_0$. And in the third line, customary units could be quite clean as well, if not somebody had chosen the hour over the convenient second. That would eliminate all constants from the equations and leave us with plain units. Jan 1 at 10:01