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Hot answers tagged shockwave

22

That is not a shockwave. It looks like the edge of the second stage engine exhaust plume. If it were a shockwave, more shockwaves should be visible: one at each point where the diameter of the rocket changes, so escape tower, CM, both interstages. Shockwaves are rarely visible. In the lower atmosphere you can sometimes see condensation in the low-...

13

The main issue is that the image of the rocket is not taken from the side, but at an angle. We can determine the angle by looking at the apparent length to width ratio of the two stages - note that this screenshot has been taken directly after separation. The first stage seems to be twice as long as wide, the second stage is about three times longer than ...

6

This should be considered a supplementary answer. The Rectangular shape looks quite convincing, but remember we can only see the back part of the flame clearly. There are other mechanisms by which the 'rectangle shape' may form by way of circular harmonics and the like. If viewed from just the right angle, the rectangular corner appearance can be replicated....

6

The shock cone is not rectangular in any way when the engine is being tested, just as it is not in flight. I finally ran across a different camera angle view of the same test stand (from slightly above the nozzle exit plane looking down, not below and looking across the plume as most of the available shots were taken). It's clear that the base of the shock ...

6

tl;dr/update: The corner that makes the shock wave appear to have aspects of a rectangular top is real and it is reproducible! But @OrganicMarble's new answer now puts the issue to rest. Together with the images from 1981 and 2015 here is a video of another test from 2017. So here it is as a GIF, and then as the video. The GIF is 30 fps from about 00:55 to ...

4

First, note that shock diamond formation is dependent on atmospheric pressure. According to wikipedia: The distance from the nozzle to the first shock diamond can be approximated by ${\displaystyle x=0.67D_{0}{\sqrt {\frac {P_{0}}{P_{1}}}}}$ where x is the distance, D0 is the nozzle diameter, P0 is flow pressure, and P1 is atmospheric pressure. Note that ...

3

Interesting idea. It seems unlikely to me though. So without doing any research: if the bow wave was a barrier to radio, we wouldn't be able to to radio astronomy. the bow wave has a much lower density than our atmosphere, so plasma effects are going to be much weaker. More later.

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