How big were the Unexpected Finger's fuel rods?

We already computed that Stupid thrust consumed fuel of
0.18 *kg/s*, which is 0.03 *kg/s* for each of the Finger's six engines.

In an emergency, each engine was loaded with one rod, then fired in Stupid mode, which consumed the
entire fuel rod in six hours (21,600 seconds.) So each engine consumed
0.03 *kg/s* × 21,600 *s*
= 648 *kg*, or 1,425 pounds of fuel.
This is the mass of one fuel rod.

How big is that?

The fuel was iron, with a density of 7900 *kg*/*m*^{3}.
Thus a 648 *kg* fuel rod occupied a volume of
(648 *kg*)/(7900 *kg*/*m*^{3})
= 0.082 *m*^{3}

How big is that in real units?

The fuel rods were cylinders, so their volume equaled their length multiplied by their cross-sectional area.

The cross section of a cylinder is a circle, with area πr^{2}.
The fuel rods were six inches in diameter, so their radius was
r = 3" = 0.076*m*. Their cross-sectional area was
πr^{2}
= π(0.076^{2})
= 0.018*m*^{2}.
So their length was *volume/area*
= 0.082*m*^{3} / 0.018*m*^{2}
= 4.5*m* = 15ft.

In other words, the fuel rods were six inches in diameter, fifteen feet long, and weighed three quarters of a ton.

In our previous fuel analysis, we calculated the Finger's deceleration burned
7.2×10^{5} *kg* of fuel. How many rods was that?

Okay, *we* did not calculate anything. I calculated, and you watched. This is why you pay me to read the book,
instead of the other way around.

But I digress. Each rod weighed 648 *kg*, so total fuel consumption was
7.2×10^{5}*kg* / 648*kg* = 1,111 fuel rods. The Finger
carried 1,300 fuel rods, so her fuel reserve was 189 rods, or 15 percent.