1
$\begingroup$

Maybe this question has a very obvious answer (I'm no aero engineer though), but I'll appreciate the clarification! So, I was reading an old nasa paper that indicates that the dynamic pressure for the apollo's vehicle during reentry was about 120 lb/ft^2 for a Mach number of approx. 23, note that I'm getting this data from dynamic pressure vs. Elapsed time and Mach number vs elapsed time graphs. So I was thinking if I'm travelling at such high M number, why the pressure or rather the stresses on the vehicle aren't higher? Naively, I am thinking that the air pressure inside a car tire is about 2.1 bars which is approx. 2088 lb/ft^2 so I'm surprised by the pressure reported in the paper (see pages 35 and 36, figures 5 and 6 respectively) Here is the link to the paper: https://ntrs.nasa.gov/archive/nasa/casi.ntrs.nasa.gov/19670027745.pdf

Again thanks for the clarification!

$\endgroup$
3
$\begingroup$

I believe you actually misread the graphs in the paper you linked. I will however go with the Mach 23 you refer to and use the correct values from the other graphs.

Mach 23 is reached at about 4775 seconds on the x-axis. At this speed, the dynamic pressure in the graph is somewhere between 30 and 40 $lb/ft^2$. at this point in time, Apollo is at around $230000ft$ altitude (see pg. 32 in the pdf).

dynamic pressure is given by: $$q = \frac{1}{2} * \rho * v^2$$ and the velocity can be calculated simply by multiplying the Mach number with the local speed of sound. I used this tool to calculate the density, $\rho$, and speed of sound, $a$, at $h=230000 ft$.

$$\rho = 0.0000731369 kg/m^3 $$ $$a = 295.416 m/s$$ using these values we get to:

$$q = \frac{1}{2}*0.0000731369 * 23^2 * 295.416^2 N/m^2 = 1683.5 N/m^2$$

Converting this to $lb/ft^2$ (I used S.I. for my calculations because, well because of course) gives:

$$ q = 35.16 lb/ft^2$$

So as mentioned in the comments, yes, it is simply because the density at 230000 ft altitude (in the upper mesosphere) is so ridiculously low that the dynamic pressure is also this low.

Note that i did not take into account any hypersonic effects for this simple calculation but the results are similar enough to confirm that it is just due to the low density.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.