# Storing light energy as potential kinetic to drive an electric generator [closed]

I would like to lift a 10 ton mass (crate of rocks) to a height of about 80 feet using a small 24 volt electric motor powered by a 16 square foot photovoltaic panel. Then I would like to collect the energy stored by this 10 ton mass by having it turn an electric generator in a controlled descent.

One of the problems I must solve it finding a gear assembly that will allow this small 24 volt motor to raise such a large mass. I would like to get information related to mechanics of such gear assembly and it would be just great if someone could direct me to a company who builds such gear assemblies.

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## closed as off topic by David Z♦Jun 22 '12 at 2:33

Questions on Physics Stack Exchange are expected to relate to physics within the scope defined by the community. Consider editing the question or leaving comments for improvement if you believe the question can be reworded to fit within the scope. Read more about reopening questions here.If this question can be reworded to fit the rules in the help center, please edit the question.

Hi Michael, and welcome to Physics Stack Exchange! General information about gear mechanisms is more of an engineering topic, and is generally off topic here. You could ask about the physics aspects of a gear system, though; if you want to do that, feel free to edit the question accordingly and I'd be happy to reopen it. – David Z Jun 22 '12 at 2:40

## 1 Answer

A typical value for solar insolation is roughly $1 kW/m^2$ and a typical metric for efficiency is about $12 \text{%}$, this means that under optimistic conditions your $16 ft^2$ array could be expected to deliver $178 W$. In terms of the basic mechanics, this is a more relevant metric than the voltage that you give, but I will return to that.

The power you need to lift the load depends on how fast you raise it. The primary drive shaft of your motor will have some angular speed $\omega$ in terms of $rad/s$ and this will translate into a linear speed $v$ ($ft/s$) given the gearing and pulley system. Potential energy of the mass is $mgh$ and the power needed to raise it will be how fast this quantity is increased over time. Introducing the fact that $v=dh/dt$ and with my prior argument $P=d(mgh)/dt=mg\times dh/dt$, we then have $P=mgv$. For instance, lifting the mass at $1 ft/s$, you will require $27 kW$. In fact, this number will be a minimum, you will need more than this due to all the losses involved.

So how fast can you lift it with your solar panels? If we use the prior 178 W number, we find that you can lift it at a rate of no more than $0.006 ft/s$. This is slow, but that might be fast enough, as it would take about $20 \text{minutes}$ to lift it the 80 ft.

Now, the voltage isn't very helpful because it doesn't translate directly into the speed. The DC motor design will relate all these parameters. In fact, you could buy just about any speed motor at just about any rated power, voltage matters for the construction, but doesn't limit what these can be. But for the sake of argument, let's say you have a $3600 rpm$ motor, which translates to $377 rad/s$, and if your primary gear on the shaft is, let's say, $5 cm$, then the outer edge of the disc would be moving at $18 m/s$ and the mechanical advantage needed from there would be about $10,000$. This mechanical advantage would be impractical to do with pulleys alone (since you would need 80,000 ft of rope), but it would not be impractical with a series of gearing.

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