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Skylon

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@RCgothicAwesome shot - thanks for sharing

 

Edit: anyone else getting "Buck Rogers" vibes OIP.HHxGV3yprzpq4bZE-jnY_gHaDu (474×238) (bing.com)

(During the flight yesterday, my wife told the kids it looks like a flying appendix vermicularis - imagine the giggling).

Edited by JoeSchmuckatelli
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1 hour ago, CatastrophicFailure said:

May, of course, be missing some significant bits, but... for what it’s worth...

NYblmjc.jpg

I could be wrong but I feel like that is the bottom end, where the downcomer attaches, and the downcomer simply sheared off entirely at the weld.

Or it could be the top end. But if so, is it missing a cap? It wouldn't be much of a header tank if it was just open like that.

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Love this render...

A little more math...

One thing I'm really curious to know is whether they were planning on a single-engine landing burn or a two-engine landing burn. My gut, even without doing maths, is that they were planning on a single-engine burn from the beginning, because that would give them the largest window for fine adjustments all the way down (hoverslams are hard). So the plan would be to use two engines to execute the flip, providing roll, yaw, and pitch control, and then shut down one engine as soon as roll was appropriately damped.

Contributing support for this hypothesis: the engine that either flamed out or was shut down did so almost immediately after the flip was completed. In the view from beneath, you can really see how that engine damps roll and then immediately shuts down. Shutdown was almost exactly T+6:39.00, 1.33 seconds before my pixel-counting started (I needed the ground to come into view or it wouldn't have worked) and exactly 4 seconds before impact. The fire from the other engine was already green at this point, so we can assume it was already only providing a single gee of thrust. Thus we extrapolate back 1.33 seconds at 30.09 m/s to give us an altitude of 119.8 meters at the shutdown of that engine.

Remember your kinematic equations: v2 = v02 + 2aΔx. So if you're 119.8 meters above the ground, dropping at -30.09 m/s, and you need to hit 0 m/s by the time you close that distance, what acceleration do you need?

 

Spoiler

 

v2 = v02 + 2aΔx

(0 m/s)2 = (-30.09 m/s)2 + 2*a*(-199.8 m)

0 = 905.41 m2/s2 + 2*a*(-199.8 m)

-905.41 m2/s2 = 2*a*(-199.8 m)

-905.41 m2/s2 = a*(-399.6 m)

(-905.41 m2/s2)/(-399.6 m) = a

2.27 m/s2 = a

 

To counteract gravity and slow down Starship in a perfect suicide burn, the engines needed to produce 2.27 m/s2 plus the acceleration of gravity, or about 1.23 gees. That's not much thrust at all...clearly much less than would be expected if you were doing a two-engine suicide burn. Thus, I think we can safely conclude that they were intending a single-engine landing burn all the way down.

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28 minutes ago, RCgothic said:

This doesn't look good D:

Spoiler

Yikes 

Well that fracking sucks.

Presumably a total loss.

15 minutes ago, RealKerbal3x said:

Oh dear...

Any guesses on whether they'll still fly it?

I suspect they will do a destructive cryo proof to see where it fails.

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