3 ms·

Quite a while ago, I thought up of what I though would be a simple system that has a net positive energy output (it already sounds ridiculous, I know, but I jus

Quite a while ago, I thought up of what I though would be a simple system that has a net positive energy output (it already sounds ridiculous, I know, but I just can't see what concept I am missing) that follows this principle. I finally have a relevant article I can share it with here on Hacker News! The gist of the system is as follows: 1. let a hanging weight freefall under gravity to spin a generator. (via a pulley connected to a chain drive or gears etc) 2. The power generated by the generator is stored in a battery 3. The weight is pulled up very slowly back to its drop position using a DC motor that uses power from the battery (the DC motor could be actually the generator itself but likely a specially tuned one for slow lifting will be better) 4. Repeat. The system lifting / reset process requires less energy than the energy produced by the drop, so there will slowly be a net buildup of charge in the battery. Super simple numbers example: PE of a 100kg mass falling 2m: PE = (100kg)(9.8m/s2)(2m) = 1960 J The mass falls at a near g acceleration - I know it won't be exactly g because of the moments of inertia of the generator and generating drive apparatus - but it would be quite close to g. At this acceleration the mass falls for ~0.63s, so the theoretical max power we can hope to extract using the generator is: P = W / t = 1960 J / 0.63s = ~3000 W Even if we consider a horrible electric generator with only 50% efficiency (just to be even more conservative and fair, and to capture any losses we've forgotten), we should hope we can get at least 1500 W from the drop. The "trick" of mine is mentioned in that very slow "reset time" for the weight. That is, use a DC motor to pull the weight up, but give it a long time to do so. In this way, the power used to lift the weight back up to its dropping point is much less than what was produced by letting the weight fall and spin the generator. Using the same numbers from the example above, if we let the lift time be, say 20 minutes (=1200s) and since we know the W of the lift is the same as the drop PE, then the power needed to lift the weight back to its original drop point is: P = 1960 J / 1200s = 1.63 W That means we produced about 1500 W (even with our crappy 50% efficiency generator), and only need ~2 W to reset the system. Assuming you put a nice battery between the generator and the dc motor - you get a net positive storage of energy over time. Does anybody see any glaring issues with such a system and my analysis? I know when anything in physics appears to be a free lunch, something must be wrong. (Though if you consider these long 'cycle' times, it isn't really a free lunch) Perhaps I don't understand exactly how DC motors / generators really draw or produce power in the real world - Physics 101 was a long time ago for me :) The trick to my proposed system, which systems like those in that of the linked article don't do, is my slow reset time. Obviously not feasible for massive applications or load balancing, but a net power storer nonetheless.

Ekaros 6y ago
Let's use Watt-Seconds to simplify things. 3000W * 0.63s = 1890 Ws 1.63W * 1200s = 1956 Ws Near enough the same. Watt is unit of power, Watt-hour is unit of energy.
- MayeulC 6y ago
  Note that Watt-second is also known as Joule (J).
- fl0wenol 6y ago
  For the OP's clarification, the thing that is conserved in thermodynamics (and thus power generation/storage) is energy, not power. Watt-seconds === joules
peteradio 6y ago
You seem to be comparing power differences and not taking into account that your battery would run out of energy in just the second attempt at lifting your mass to full height (assuming no other power inputs to the battery besides the falling mass generator).
orange_tee 6y ago
Step 4 contradicts basic physics. What you are doing wrong is that you are confusing power with energy. If you lift the weight slowly you need less power because for the same work (= energy) done you can do it in twice the time. But using less power over longer periods of time will result in the same energy consumption.
polemic 6y ago
You need 1.63 W for 1200 seconds (i.e. 1960 J, assuming no losses) to reset the system. I.e. the same energy you already got out. If this worked, you could get free energy from any sort of gearing system - e.g. you could put a fly-wheel on your bicycle and power all your up-hill trips from the downhill legs of the same size by switching gears at the right time.

Ekaros 6y ago

Let's use Watt-Seconds to simplify things. 3000W * 0.63s = 1890 Ws 1.63W * 1200s = 1956 Ws Near enough the same. Watt is unit of power, Watt-hour is unit of energy.

MayeulC 6y ago
Note that Watt-second is also known as Joule (J).
fl0wenol 6y ago
For the OP's clarification, the thing that is conserved in thermodynamics (and thus power generation/storage) is energy, not power. Watt-seconds === joules

peteradio 6y ago

You seem to be comparing power differences and not taking into account that your battery would run out of energy in just the second attempt at lifting your mass to full height (assuming no other power inputs to the battery besides the falling mass generator).

orange_tee 6y ago

Step 4 contradicts basic physics. What you are doing wrong is that you are confusing power with energy. If you lift the weight slowly you need less power because for the same work (= energy) done you can do it in twice the time. But using less power over longer periods of time will result in the same energy consumption.

polemic 6y ago

You need 1.63 W for 1200 seconds (i.e. 1960 J, assuming no losses) to reset the system. I.e. the same energy you already got out. If this worked, you could get free energy from any sort of gearing system - e.g. you could put a fly-wheel on your bicycle and power all your up-hill trips from the downhill legs of the same size by switching gears at the right time.