SpinLaunch hopes to launch a 200 kg satellite from a ground-based centrifuge, at 5000 mph, and more troublingly, at 450 rpm aka 7 tumbles per second. (The 3000 rpm figure mentioned in its tech sheets isn't for the launch centrifuge, but for a reaction wheel.)

How great a fraction of its launch energy will be wasted by dissipating into drag that much rotational inertia? Even if at ejection the spacecraft were roughly spherical like the stone that David lobbed at Goliath, it would eventually need to be stabilized before onboard propulsion could change its trajectory to orbital. But they expect a sabot shape, which would need to be stabilized much sooner.

200 kg satellites are nowadays about a cubic meter, if that helps. The sabot surrounding it would be much longer and have much more yaw rotational inertia. It would also weigh about 1000 kg. Dissipating that much rotational inertia within, say, an eighth of a rotation, about 17 milliseconds, would likely require more than just a pointy nose and tailfins.

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    $\begingroup$ Wow, that's a really interesting question. Another point, which does not seem to be answered discussing the feasibility of the whole system (which seems pretty questionable in my personal oppinion)... airodynamic dampening (aka fins) would add drag and weight. Active dampening (with cold gas thrusters) would add weight. And since the projectile is not spherical every bit out of yaw increases drag... Maybe some kind of mechanical/,agnetic system for dampening in the "launch tube" just after release from the bar but before entering atmosphere? $\endgroup$
    – TrySCE2AUX
    Apr 4 at 6:57

1 Answer 1


The rotational energy is not removed by aerodynamics, it's probably removed by the precise and accurate timing of the release mechanisms.

A bit of digging and analysis shows that most of the rotational energy is removed before the sabot exits the launch tube:

The cued clip shows the sabot exiting the launch tube in slow motion from the Oct. 2021 subscale accelerator test. A CNBC article has this to say:

The suborbital projectile is about 10 feet long, but “goes as fast as the orbital system needs, which is many thousands of miles an hour,” Yaney added.

And additionally this:

SpinLaunch’s first suborbital flight utilized about 20% of the accelerator’s full power capacity for the launch, and reached a test altitude “in the tens of thousands of feet,” according to [SpinLaunch CEO Jonathan] Yaney.

A Universe Today article also supports these numbers.

SpinLaunch gives a maximum speed of 5000 mph for the full scale "orbital system". The subscale "suborbital accelerator" is 33 meters in diameter. To reach 5000 mph it must spin at ~1300 rpm. 20% of this value is ~260 rpm which yields a release velocity of ~450 m/s. Ignoring drag (sabot performs better with higher ballistic coefficient), this yields a height of ~10,000 meters (~33 kft) which is “in the tens of thousands of feet”.

Looking at the video (~44 frames for the sabot projectile to exit the tube), my math says that, if the projectile was rotating at the same rate as in the centrifuge, it would have rotated ~10 degrees in the process of exiting the tube. I think 10 degrees is substantial enough that it would be visible, and I don't see that in the video clip, though the sabot is slipping.

How is the rotation stopped?

I don't think the subscale release/holding mechanism has been shown, but renders/animations of the full scale version have been:

full scale release/holding mechanism render

close up

(Source: SpinLaunch)

The key is in (what appears to be) the multiple attachment points. By releasing them in order from top to bottom very precisely, the rotation can be stopped (or even possibly imparted in the opposite direction). The "centrifugal force" (or lack of centripetal force) on the upper part of the sabot when the attachment is released will cause a torque opposite of its rotation in the centrifuge.

Here is a simplified diagram, an inertial frame "snapshot" of the projectile and attachment:

simplified force diagram

The money is made in allowing just the right amount of time to pass when "partially released" for the torque to apply an angular acceleration and stop the rotation. Waiting too long could cause the rotation to reverse.

I am unsure of the timescale required to achieve this level of precision. Any aerodynamic yaw damping will be because of mistiming the release, thus having residual rotation (one direction or the other). This amount cannot be significant as the projectile must travel (more or less) straight out of the exit tube (which doesn't look wide enough to accommodate a sideways sabot). The amount of aerodynamic yaw damping is therefore unlikely to be significant.

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    $\begingroup$ Impressive research and analysis, about something sorely lacking published technical documentation. $\endgroup$ Apr 4 at 16:23
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    $\begingroup$ I really doubt that others will answer better than this, but nevertheless give them some time instead of accepting the first answer mere minutes after its posting :) $\endgroup$ Apr 4 at 16:25

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