SN4 SpaceX's test tank went boom. It was a mixture of methalox (methane + oxygen). When it exploded it caused a shock front, what was the speed of this explosion?

Do all shock-waves like that travel at Mach 1 and thus limited by the speed of sound in air? With projectiles/aircraft I know you can have an oblique shock and determine the mach number. But the explosion is not a projectile per say and is spherical and thus one cannot determine this from the video footage.

Is it possible that the explosion caused a shock wave exceeding the speed of sound and if so how could one determine it from video footage or other sensors?


1 Answer 1


To start off a shock or blast front in this case is caused by a shock wave traveling through the air faster than the speed of sound (*https://blastinjuryresearch.amedd.army.mil/index.cfm/blast_injury_101/science_of_blast*.) the shock wave is apparent by the low pressure followed by the blast front causing the water vapor to condense in an effect known as a Wilson Cloud. therefore the explosion definitely cause a pressure wave that moved faster than the speed of sound. to calculate that speed using the video, you would have to know the distances of nearby objects to the explosion, and use the framerate as a reference for time.(Distance/Time)

I do not know how I would find that information, but there is another way to find out by the use of, of course, math. the formula can be found in detail here: (https://aip.scitation.org/doi/10.1063/1.1667908) You only need to know a few things: 1 the amount of gaseous molecules per gram(as well as their weight) 2 the average weight of the gases 3 the amount of chemical energy in the reaction. Unfortunately here at 4 we have a problem the: initial density. (The above formula only measures detonation pressure not velocity and pressure. However pressure and velocity are intrinsically linked, but we run into the same problem of initial density. If you would like to know more about how you calculate many factors involving explosives you can find more here: https://www.sciencedirect.com/topics/chemical-engineering/detonation)

the problem is that it's almost impossible to know the density because the explosion ripped open the tanks of fuel and oxidizer. That caused the two to mix to some extent before the ignition source reached the containers. they then mixed more as the pressure waves propagated through the mixture. All in all it's safe to say that the pressure of the tanks is not a good way to estimate the density, as they vented out their contents to the atmosphere at an unknown rate.

That leaves us no real way of knowing the speed of the shock front without knowing the exact dimensions and distances of nearby objects along with a camera looking from the correct angle with a high framerate. I'll give you a pretty good guestimate of around *4400-6000 m/s which comes from an old study that tested the detonation speed of liquid Methalox.

*DETONATION OF LIQUID OXYGEN-LIQUID METHANE SOLUTIONS A. V. Grosse, A. D. Kirshenbaum, and A. G. Streng Journal of the American Chemical Society 1957 79 (23), 6341-6342 DOI: 10.1021/ja01580a062

EDIT: As for the comments saying this was a deflagration not an explosion, the oxford dictionary defines deflagration as:

combustion which propagates through a gas or across the surface of an explosive at subsonic speeds, driven by the transfer of heat.

There was a very clear shock front, and while it is obvious not ALL of the methane and oxygen detonated a good amount did. The shockwave is undeniable proof of that.

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    $\begingroup$ confused at how this got downvoted within a minute of posting lol. I'm pretty sure it takes longer than that to read the whole thing... $\endgroup$ Feb 16, 2022 at 23:37
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    $\begingroup$ I salute you Fred, thanks for the fix $\endgroup$ Feb 17, 2022 at 0:20
  • $\begingroup$ Can you explain the relevance of the formula you link to? It seems to calculate pressure, not speed. Without that, the only relevant part of this answer seems to be the one number you quote. $\endgroup$ Feb 17, 2022 at 14:55
  • $\begingroup$ @OrganicMarble Thank you for that, I kinda rushed to find citations, I didn't realize that it did not calculate velocity. I was taught from the beginning that knowing one, you already have all the information to figure out the other. I added some another link that is much MUCH more in-depth and has resources to learn even more. $\endgroup$ Feb 17, 2022 at 17:25
  • $\begingroup$ So to make the citation relevant, please tell us how to calculate the velocity from the pressure. Or, if it's not relevant, just remove the whole thing, along with the duplicate part about the cameras. Just adding links is not very good, you need to put the relevant stuff in your answer. $\endgroup$ Feb 17, 2022 at 17:34

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