On reentry could a flywheel spin on a magnetic bearing creating a centrifugal force to keep the ship level whilst creating a plasma air pocket from the pits in the rotating heat shield? This would displace super heated air away from the ship while creating lift. Could kinetic energy be stored in a flywheel to create lift to hover with a payload before landing on Mars or Earth?

In other words: The flywheel would be moved air interacting with the shape of the dimples (kinda like how a golf ball has dimples and creates lift when spun) on the flywheel to convert the air pressure into rotational kinetic energy while creating lift equalizing the pressure. So instead of the air hitting the ship the air is more hitting the pocket of air created from reentry. Kind of fighting air with air or using air as a heat shield??

  • $\begingroup$ I am here to explain if you need help. $\endgroup$ – Muze Aug 22 '18 at 6:06
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    $\begingroup$ That sounds like a lot of things might go wrong. The beauty of current heatshield designs and reentry profiles is that they are 'passive' i.e. they don't require moving parts. What if your bearing fails? What if the flywheel spins to slow because of additional friction from something that got stuck inside there? Also, that sounds like a lot of additional mass. $\endgroup$ – DaGroove Aug 22 '18 at 6:33
  • $\begingroup$ Could you make it clearer what you’re suggesting? You seem to hypothesise several different ways that this system could generate lift and/or drag. In either case as @DaGroove says - this sounds like a lot of complexity for little to no gain $\endgroup$ – Jack Aug 22 '18 at 7:45
  • $\begingroup$ I'm having trouble visualizing this. Do you want to replace the heat shield on a capsule with a flywheel that rotates in the horizontal plane? $\endgroup$ – Hobbes Aug 22 '18 at 8:50
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    $\begingroup$ Magnetic bearing here is asking for trouble. They are devices that don't handle any stronger loads well - they are nearly frictionless for nominal loads but their nominal loads are very low; at any stronger loads they act as very lousy bushings with a lot of play. The loads involved during reentry are nowhere near "low". $\endgroup$ – SF. Aug 22 '18 at 8:57

Take the super-optimistic 500 kJ/kg energy density of flywheel energy storage. In reality 10% of that would be a great result.

$ E_k = {1 \over 2} mv^2$ so 0.5*1kg*(8km/s)^2 = 32MJ per kilogram of orbital mass.

If the craft was nothing but the flywheel, you'd still receive 64 times more energy than you can most optimistically contain. Your flywheel would either burn up, if it doesn't absorb this as rotational energy, or explode ripped to shreds by centrifugal acceleration, if it does.

Rule of thumb: A kilogram of TNT has 4.6 MJ/kg of specific energy. A kilogram of the spacecraft parts on reentry has 32MJ/kg.

If a kilogram of your device can't withstand containing energy of equivalent of 7kg of TNT, it won't contain the reentry energy of itself. Never mind any spacecraft it's meant to sustain.

  • $\begingroup$ The shape of the dimples in the flywheel can shed the rotational energy when it approaches optimal rotational speeds by producing thrust. As the open air flywheel spins faster more of an air pocket is created which prevents over spin. A harmony between the flywheel thrust and impacting air is created during reentry? $\endgroup$ – Muze Aug 30 '18 at 14:12
  • $\begingroup$ @Muze: The more efficiently it removes air from the flight path, the less air resistance there is to slow the spacecraft down, and you slam into the ground at a couple km/s. $\endgroup$ – SF. Aug 30 '18 at 15:14

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