Could this device theoretically continue in motion forever? If not, why not? (click below for images):

Device description.

Device process.

  1. The device is less dense than air, so it rises. The propeller spins slightly (maybe) charging the device battery.
  2. After rising some distance X, the device compressor turns on to deflate the device.
  3. The device becomes more dense than air and falls quickly spinning the propeller, charging the battery.
  4. After falling X, the compressor releases and the device becomes less dense than air, going back to step 1.

The logic here is that there must be some distance X that the device can rise than will generate more energy than what is needed by the compressor.

Here is the underlying math to help:

$$PE = mgh$$

  • $m$ = mass of the device
  • $g$ = coefficient of gravity
  • $h$ = height device has traveled up
  • $PE$ = potential energy of the device as it travels up

$$CW = nRT(\ln(V_b) - \ln(V_a))$$

  • $n$ = number of moles of gas in balloon of device
  • $R$ = ideal gas constant
  • $T$ = temperature of the gas
  • $V_b$ = volume of the balloon after compression
  • $V_a$ = volume of the balloon before compression
  • $CW$ = work to compress the balloon

As $h$ increases $PE$ increases but $CW$ stays the same resulting in energy gain of the system.

  • $\begingroup$ This isn't the same exact content as "What is the fallacy in this infinite motion machine?". The two indeed seem the same but the implementation is much different. This implementation involves changing shape for the effect of motion. The other one just thinks that the water will pull up the balls magically... $\endgroup$ – John K Aug 23 '12 at 23:52
  • $\begingroup$ Well, if this hadn't been closed as an exact duplicate, it would have been closed as "not a real question" or "not constructive." If you edit the post to ask something directly, then we can certainly consider reopening it. The linked possible duplicate is a good example of how to ask a question about a supposed perpetual motion machine, so you can look at that to give you guidance for editing your own question. $\endgroup$ – David Z Aug 24 '12 at 0:22
  • $\begingroup$ Ok, I edited it to ask a specific question. $\endgroup$ – John K Aug 25 '12 at 21:01

Your final sentence has the problem: "there must be some distance X that the device can rise than will generate more energy than what is needed by the compressor"

I think every attempt to prove this kind of thing has shown that there actually is not some distance which would meet these requirements.

  • $\begingroup$ Okay but the point of positing is to provide this proof. Otherwise your answer isn't very helpful. $\endgroup$ – John K Aug 23 '12 at 23:48
  • $\begingroup$ @JohnK provide it yourself. It's a trivial application of the principles of buoyancy and potential energy. $\endgroup$ – user2963 Aug 31 '12 at 17:13
  • $\begingroup$ But it also includes an air compressor which makes this not so trivial. Does the amount of work the air compressor needs to compress the air increase as the device rises? If not, there should be a distance the device can rise that would generate more potential energy than the amount of energy the air compressor would need. $\endgroup$ – John K Sep 2 '12 at 16:52
  • $\begingroup$ no- no there shouldn't. every process loses some energy in heat, turbulence, drag etc. so you will never generate more potential energy. $\endgroup$ – Rory Alsop Sep 2 '12 at 22:27
  • 1
    $\begingroup$ sorry @JohnK - not going to bother - it is way too trivial. Look up any description of real world losses in dynamic systems. $\endgroup$ – Rory Alsop Sep 6 '12 at 8:02

The problem is that neither the potential energy nor the energy necessary to compress the gas is of any importance here. We can set the energy of the compression to zero and still would not gain any energy in this model! So for any real device with losses we certainly can't gain energy.

This can best be shown in an experiment by connecting a propeller to a balloon but is also clear from basic principles: Starting from step 1 the device will slowly lift upwards, the propeller will start to rotate as the passing air is causing a force on it. Now you want to drive a generator with this propeller. So what happens if you engage some kind of clutch to connect generator and propeller? The coupled balloon-propeller device will rotate around it's own axis. The energy is used to increase the angular momentum, not to charge the battery.

The same happens in step 3 (Let's neglect what happens with the rotation between 2 and 3). Again the propeller starts to turn and if the clutch is engaged both will rotate together and the battery is not charged.

  • $\begingroup$ Thank you for your interest in this question. I think you're right that realistically the device would probably just spin around and not charge the battery. What I wanted to get at was fundamentals of the question which is basically, assume that you can store the energy of the spinning propeller as it rises isn't there some distance it could rise that would allow it to store more energy than it would take to compress the device to cause it to fall again. I think the answer is no because as the device rises, pressure drops, and the device will be harder to compress probably but math is missing. $\endgroup$ – John K Apr 25 '13 at 17:06

protected by Qmechanic Apr 24 '13 at 10:26

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