So a friend had an 'idea' where you could make an object land with 0 velocity.

Basically the idea is that you drop an object (like a rocket shape) from some altitude. To the object you attach fins (fixed propeller blades so to speak.) as it falls it starts rotating, due to the attached fins.

Now at some altitude above the ground you either:

1) You rotate the fins to get reverse 'thrust' (you don't really have thrust, but you would slow down rotation.) 2) Retract the fins and expand new ones that face the opposite way (so they counter spin) 3) something along those lines (like extending flaps)

Now one of the arguments he had as to why this works is because you could expand larger fins for the counter rotation, therefore getting more drag and generating more 'lift'.

It feels like you cannot do this, but I cannot give a (good enough) reasonable explanation as to why it would or would not work (I'm a noob in this area.) So let's ask the people in the know... that's you.

I would love to see some 'proof' more so than speculation.

Thank you in advance.

  • $\begingroup$ I should have added that the system cannot have any energy added to it, so the mechanism of expanding the fins, or whatever other mechanism, should come from the energy gained from the rotation. $\endgroup$
    – MrHat
    Feb 19, 2016 at 2:52
  • $\begingroup$ This question is simply about energy management in a < 100% efficient closed system and not aviation and therefore belongs on Physics.SE. $\endgroup$
    – Simon
    Feb 19, 2016 at 17:02
  • $\begingroup$ I vote to migrate this to physics.se $\endgroup$
    – TomMcW
    Feb 19, 2016 at 17:33
  • $\begingroup$ I'm voting to close this question as off-topic because it belongs to Physics.SE $\endgroup$
    – mins
    Feb 19, 2016 at 18:33
  • $\begingroup$ Come to think of it... Ever seen a maple seed? $\endgroup$
    – jamesqf
    Feb 19, 2016 at 18:42

3 Answers 3


What you are describing is similar to a helicopter autorotation. That is something that does work. The rotors both generate drag and store rotational momentum. Then, some of the momentum is converted to lift for landing.

  • $\begingroup$ Or an autogyro/gyrocopter: en.wikipedia.org/wiki/Autogyro $\endgroup$
    – jamesqf
    Feb 19, 2016 at 6:10
  • 2
    $\begingroup$ Autorotation requires forward speed; the question refers to dropping the object. $\endgroup$ Feb 19, 2016 at 6:26
  • 1
    $\begingroup$ @DavidRicherby, sustained autorotation does require forward speed, but the final flare does not. $\endgroup$
    – Jan Hudec
    Feb 19, 2016 at 9:51
  • $\begingroup$ @JanHudec I think you may have that the wrong way around. I can autorotate all the way down with zero, or even negative forward speed, but I do need some to flare? $\endgroup$
    – Simon
    Feb 19, 2016 at 13:43
  • $\begingroup$ @Simon, by "flare" I mean when you pull on the collective at the end to arrest the descent. That shouldn't need forward speed, does it? $\endgroup$
    – Jan Hudec
    Feb 19, 2016 at 14:11

Given the object has a terminal velocity, it should be possible.

Before it hits the ground, it has a speed no greater than terminal speed. Assuming it has some sufficient propeller system, it will require a finite amount of energy to slow the descent to zero at the altitude desired. But since the start point is arbitrarily high, we can collect an arbitrarily large amount of energy from the fall.

Imagine a quadcopter with depleted batteries and some sort of charging system that can be driven from the rotors. Drop it from a height and let the descent charge up the battery. Then a few feet before the ground, turn on the motors and let it hover.

Now this example uses batteries and electronics, but the argument doesn't change in character if you turn it into a kinetic spinning rotor instead (it's just harder to build).

  • $\begingroup$ It cannot be zero terminal velocity because it means a 100% efficience in loading the free fall nergy and 100% efficience in transforming that energy in lift $\endgroup$
    – jean
    Feb 19, 2016 at 13:05
  • $\begingroup$ This will not work. You must think in terms of systems and the total energy in them. Without adding energy from outside the system, the energy in the system will always reduce. Charging the batteries takes energy from whatever is producing lift at less than 100% efficiency. The lift generators themselves lose energy through drag, friction etc. With a system such as yours, the total energy decreases rapidly and will be very much less than when you started (with nothing but potential energy) and you cannot "cushion" your landing. $\endgroup$
    – Simon
    Feb 19, 2016 at 13:47
  • $\begingroup$ @Simon, why is 100% efficiency required here? It's not. The drag during the fall means the kinetic energy that has to be corrected near the ground is only a fraction of the potential energy available at the beginning of the fall. If I can extract just that fraction, I have sufficient energy to stop at the bottom. $\endgroup$
    – BowlOfRed
    Feb 19, 2016 at 16:34
  • $\begingroup$ @jean, Nothing in the answer should suggest that it has zero terminal velocity or 100% efficiency. Neither are required. It simply has to have some drag (and therefore a positive terminal velocity). $\endgroup$
    – BowlOfRed
    Feb 19, 2016 at 16:38
  • $\begingroup$ A small positive terminal velocity is ok but also note drag is "stealing" energy from the fall and it's energy you cannot store. Drag itself is not 100% efficient meaning just a fraction of the energy its subtracting from the fall is actualy used to slow dows the velocity. $\endgroup$
    – jean
    Feb 19, 2016 at 16:52

This reminds me of the defunct Roton experimental rocket, except that it had helicopter rotor blades powered by rockets at the tip to take off.

It could be achieved with just one set of blades that have a very large pitch range adjustment. Pitch the blades downwards on descent so that they'll spin like a windmill and let the angular momentum build up, then pitch the blades upwards to convert the angular momentum into lift. The rotor system would require a flywheel of some sort to store the momentum needed to slow the craft down by a decent amount.

  • $\begingroup$ As it is this post should be a comment, not an answer. However the content is very interesting, so you should try to make it a full answer before it is closed as "not an answer". Welcome to the site, I hope to read many good questions and answers from you. $\endgroup$
    – mins
    Feb 19, 2016 at 13:07
  • $\begingroup$ My apologies for posting a question that was 'off-topic'. In any case I want to thank all the people who replied. While going through the answers I read up some more about some terms people used, which led down to a late night reading many interesting things. So I learned a couple new things, which is always fun. Thanks again for all the replies. $\endgroup$
    – MrHat
    Feb 19, 2016 at 18:45

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