I'm working through How to Design, Build, and Test Small Liquid-Fuel Rocket Engines. The guide explains calculating a rocket nozzle throat area using Eq. (7):

$$A_t = \frac{w_t}{P_t} \sqrt{\frac{R T_t}{\gamma g_c}} $$

where $w_t$ is flow measured in lb/s, $P_t$ is pressure measured in psi, $R$ is the specific gas constant for gaseous oxygen and hydrocarbon fuel which equals 65 ft-lb/lb R(rankine), $T_t$ is temperature of the chamber measured in rankine, $\gamma$ being the ratio of gas specific heats, and the gravitational constant $g_c$ measured in ft/s².

My issue is for the worked out problem on the guide, they give an example here but the answer is in in² instead of ft². I'm not sure how they got inches instead of feet because they didn't do any kind of conversion that I can tell. I tried following the units to see how they got inches but I'm not sure. Shouldn't the answer be in feet, and if not how is it inches?

  • $\begingroup$ You can learn more about MathJax for equations here. I removed the rocketlab tag; after reading that the address and original text was from circa 1967, it doesn't seem to be related to the same company that the tag refers to. $\endgroup$
    – uhoh
    Commented Mar 9, 2019 at 1:10
  • $\begingroup$ This is really a no-good equation because it has pounds mass (in the flowrate) over pounds force (in the press) outside the radical. It should be the mass flowrate in every engineer's favorite unit, slugs/sec. See Sutton p. 61 pyrobin.com/files/Rocket%20Propulsion%20Elements.pdf $\endgroup$ Commented Mar 9, 2019 at 1:31
  • $\begingroup$ @OrganicMarble Are the units for $R$ supposed to be ft-lbf / lbm R? $\endgroup$ Commented Mar 9, 2019 at 1:42
  • $\begingroup$ @RussellBorogove actually no. To be consistent it should be ft-lbf / slug R engineeringtoolbox.com/… $\endgroup$ Commented Mar 9, 2019 at 1:50
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    $\begingroup$ @OrganicMarble Same dimensions, at least. $\endgroup$ Commented Mar 9, 2019 at 1:51

2 Answers 2


$R$ and $g_c$ each have a foot component in their units; therefore the feet cancel out when $R$ is divided by $g_c$.

$P_t$ is in pounds per square inch; since you're dividing by that, it's effectively units of square inch per pound, which is where the square inches in the answer comes from.

The rest of the unit cancellation is confusing because of the use of pounds as both a unit of mass and a unit of force (weight). This thread on thespacerace.com discusses the different ways of presenting your equation; the money quote is here:

Outside the square root, my equation uses slugs for the mass flow rate, while the other equation uses pounds (i.e. weight). To convert pounds to slugs we must divide by gc outside the square root.

Inside the square root, my equation uses R' equal to 49,720 ft-lb/slug-R, while the other equation uses 1545.32 ft-lb/lb-R. To convert ft-lb/lb-R to ft-lb/slug-R we must multiple by gc inside the square root.

The combination of these two conversion factors is,

1/gc * (gc)^1/2 = 1/gc^1/2

therefore we end up with gc in the demoninator inside the square root.

  • $\begingroup$ Nice edit, I was just going ask about that unit on gamma.... $\endgroup$ Commented Mar 9, 2019 at 1:19
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    $\begingroup$ Zing! Yeah, I misread the unit attribution. $\endgroup$ Commented Mar 9, 2019 at 1:20
  • $\begingroup$ It's not the right pounds though outside the radical. Bad equation! Bad! (Hits equation on nose with rolled up newspaper) $\endgroup$ Commented Mar 9, 2019 at 1:32
  • $\begingroup$ Ugh, flowrate being mass rather than force? Maybe I'll delete everything except the inch/foot portion of my analysis, since this is way out of my wheelhouse. $\endgroup$ Commented Mar 9, 2019 at 1:36

That equation as you give it with the values you supply is a bad mish-mash of units. It's painful to do this in the English system, but gather your courage, we can get through it.

You must specify the flowrate in $\frac{slugs}{sec}$. Yes, slugs, the real engineer's unit of massTM. A slug is 1 $\frac {lbf - sec^2} {ft}$ and equates to ~ 32.2 lbm. So flowrate in $\frac{slugs}{sec}$ has the units of $\frac{lbf-sec}{ft}$.

You also must specify the pressure in $\frac{lbf}{ft^2}$.

Dividing through you get outside the radical the units of $ft-sec$.

Inside the radical the gas constant is $\frac{ft-lbf}{slug-R}$. The R cancels out with the temperature unit. gamma is dimensionless.

So if you substitute in $\frac{lbf- sec^2}{ft}$ for the slug, you end up with $\frac{ft^2}{sec^2}$ inside the radical.

Taking the square root, it's $\frac{ft}{sec}$, then multiply by the $ft-sec$ outside the radical, to get $ft^2$.

Welcome to the world of Apollo and Shuttle rocket engine calculations.

enter image description here

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    $\begingroup$ I was caught in the transition, my books in college had both. Shuttle and Apollo were all English, ISS is metric. Either is OK if you are used to it but metric is a lot more intuitive. $\endgroup$ Commented Mar 9, 2019 at 2:11
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    $\begingroup$ Are you thinking the "-" is a minus sign? It's a hyphen. It's deplorable, I know, but that's the way these units are written. See the units on the gas constant in the question. "65 ft-lb/lb R(rankine)," and in the quote in the other answer "ft-lb/lb-R to ft-lb/slug-R" $\endgroup$ Commented Mar 9, 2019 at 18:10
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    $\begingroup$ oh $\frac{lbf\text{-}sec^2}{ft}$ what the merry heck are hyphens doing in there? "english" units are ridiculous $\endgroup$
    – user20636
    Commented Mar 9, 2019 at 18:15
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    $\begingroup$ I cannot argue with that, lol $\endgroup$ Commented Mar 9, 2019 at 18:16
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    $\begingroup$ I think that mathjax requires \text to insert a hyphen says everything about how much they belong in formulae :D $\endgroup$
    – user20636
    Commented Mar 9, 2019 at 18:21

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