Gravity Based Electricity Generator

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The discussion centers on designing a gravity-based electricity generator using a 500 kg mass to harness potential energy. Various mechanical configurations are being explored, including pulley systems, piston mechanisms, compressed air systems, and piezoelectric approaches, each with considerations for efficiency and practicality. A key point raised is the need for an energy source to lift the weight back up, as conservation of energy must be maintained. Participants emphasize the importance of understanding the relationship between mass, energy, and power, particularly in the context of gear ratios and motor specifications. The conversation highlights the complexities of energy generation and storage in such systems, suggesting a need for further study in physics principles.
  • #31
OP if you are interested, there is an existing technology common in industrial cranes where when the load is lowered down, the rotational force it produces to the motor is converted back to electricity so that it will help in braking the lowering of a load to a stop. Its called regenerative braking, however, this is really not a feasible idea for any scale of power generation that you want to achieve.
 
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  • #32
kingamada said:
I'm evaluating the use of motors with different capabilities, ranging from 10 kW to 100 kW.
Industrial servomotors are available in that size range. One such manufacturer (not the only one) is Allen-Bradley. Check out this link: https://www.rockwellautomation.com/...ol/rotary-servo-motors/mma-series-motors.html. Click into the technical literature, and look for references to regenerative buss supplies. A quote from the technical literature:

Kinetix 5700 servo drive systems consist of these required components:
• 2198-Pxxx DC-bus power supplies (up to three modules wired in parallel are possible)
• 2198-RPxxx regenerative bus supplies (provides full-line motoring and regenerative power)


If you use one of these servomotor/drive systems to raise and lower a load, it automatically draws power from the line to raise the load and pumps power back into the line when lowering the load. You program the system to move the load, and the drive takes or pumps power as needed to make it happen. Do not expect a cheap system, and do expect a learning curve. But it will do what you want.
 
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  • #33
kingamada said:
I think the below picture will give a clear understanding of my question, stepping on this device below generates a little electricity. My question is that, since a human weighing on average 80kg, stepping on this device will generate small watt of electricty, what if we have a heavy object of 500kg stepping this device and connected to a bigger motor. Let's work with 30kw, 500kg weight, height depends on the setup. I hope i convey the question clearly now.
This is a dreaded Energy Harvesting Scheme. You must realise that the only systems that harvest successfully are where you have a shed load of energy that you feel you have to get rid of. Significant Energy from people walking will wear those walkers out very quick. They will avoid that particular route on their walk.
 
  • #34
Typical of this forum. Everyone projects their own limits onto others thoughts.
Your idea is good and can be worked. You’ll need an external energy source to lift your weight though. Even lifting a smaller weight as your 500kg falls will decrease the power available to be generated. The option is to gear it so that your weight takes a long time to fall so the gearing would be stepped up to drive a generator. I’d be thinking something like an alternator where you could alter the generating field by altering the current and thereby increasing the magnetic drag and generation output. If you take a 12 volt alternator and supply the rotor with 6 volts instead of 12 you’ll find it doesn’t output 12 volts. But it will require less power to rotate. If you supply 15 volts to the rotor coils it’s output will exceed 12 volts and require higher power input to cause rotation.

Using this you could design an alternator that outputs your required voltage and amperage and your designed rate of drop to match without the issue of interrupting the rate of fall. Use the rotor voltage to control fall. The potential energy stored in your 500 kg weight will not change so it’s only how you can harvest it and lift that weight again that decides if it’s a good solution or not. Mechanical advantage does not alter the amount of power only how that power is distributed between torque and rotation speed.

Don’t let the limits of others stop you.
 
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  • #35
Ripcrow said:
The option is to gear it so that your weight takes a long time to fall
The OP specified 500kg and >10KW. The only pertinent unspecified variable is how far the weight gets to fall; i.e. how long do you get 10KW. You don't get to choose the minimum speed of the fall if you've specified the other stuff.

Once you've achieved the minimum speed, then you would want to control the rate the weight falls with gears and such, or store the extra power produced. But this will happen naturally with whatever energy extraction system you have. There will be an equilibrium speed that is determined by the power extracted. Of course there are kinetic energy issues with the beginning and end of the fall.

The basic point is that you don't have to know a lot about the details of energy extraction (alternators, gears, etc. ) to determine some significant limitations on what the falling weight has to do. I would strongly suggest a "black box" approach first.


Ripcrow said:
Don’t let the limits of others stop you.
Yes! But there are limits from the laws of physics that should be understood and evaluated.


Ripcrow said:
Typical of this forum. Everyone projects their own limits onto others thoughts.
I respectfully disagree. But we do believe in physics here. Maybe we are "projecting" reality. No one's saying this can't be done. Gravity batteries are a thing. People have built them, and they work. But, they do have to be designed within the limitations of physical laws.

Honestly, I think we are trying to help. When I say "study high school physics" I don't mean it as an insult, I'm trying my best to point the way to a successful outcome.
 
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  • #36
Ripcrow said:
Typical of this forum. Everyone projects their own limits onto others thoughts.
That happens when all the parameters aren't specified - including the skill level of the questioner.

Unless the OP is a practised project. builder and DIYer there will be serious limits to any project. Looking around, I can only think of one domestic level mechanical energy storage system and that is the very successful Wind Up Radio. That should indicate the feasibility of the general idea. Inventing outside the box risks a shed full of unfinishedpart projects and no results. Good fun, on the way, though.
 
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  • #37
DaveE said:
The OP specified 500kg and >10KW. The only pertinent unspecified variable is how far the weight gets to fall; i.e. how long do you get 10KW. You don't get to choose the minimum speed of the fall if you've specified the other stuff.

Once you've achieved the minimum speed, then you would want to control the rate the weight falls with gears and such, or store the extra power produced. But this will happen naturally with whatever energy extraction system you have. There will be an equilibrium speed that is determined by the power extracted. Of course there are kinetic energy issues with the beginning and end of the fall.

The basic point is that you don't have to know a lot about the details of energy extraction (alternators, gears, etc. ) to determine some significant limitations on what the falling weight has to do. I would strongly suggest a "black box" approach first.



Yes! But there are limits from the laws of physics that should be understood and evaluated.



I respectfully disagree. But we do believe in physics here. Maybe we are "projecting" reality. No one's saying this can't be done. Gravity batteries are a thing. People have built them, and they work. But, they do have to be designed within the limitations of physical laws.

Honestly, I think we are trying to help. When I say "study high school physics" I don't mean it as an insult, I'm trying my best to point the way to a successful outcome.
The op specified 10 -100 kw. We don’t know long it can fall as we don’t know the height. Yes we are limited by physics but we are not limited by existing techniques if we are open to ideas. Read any post in this forum that questions existing ideas or has a different approach and you’ll see the post get wrecked by responses that attack ideas or the responses are merely projections of the responders own thoughts. There are plenty of posts where the op gets accused of trying to create PMM when it’s clear they are not.
 
  • #38
sophiecentaur said:
That happens when all the parameters aren't specified - including the skill level of the questioner.

Unless the OP is a practised project. builder and DIYer there will be serious limits to any project. Looking around, I can only think of one domestic level mechanical energy storage system and that is the very successful Wind Up Radio. That should indicate the feasibility of the general idea. Inventing outside the box risks a shed full of unfinishedpart projects and no results. Good fun, on the way, though.
So that inventiveness should be encouraged and instead of self projected limitations questions should be asked.
 
  • #39
Ripcrow said:
or the responses are merely projections of the responders own thoughts.
People post their own thoughts here? The fiends!
 
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  • #40
Vanadium 50 said:
People post their own thoughts here? The fiends!
And that’s fine except when the projections are biased
 
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  • #41
Ripcrow said:
Typical of this forum. Everyone projects their own limits onto others thoughts.
Your idea is good and can be worked. You’ll need an external energy source to lift your weight though. Even lifting a smaller weight as your 500kg falls will decrease the power available to be generated. The option is to gear it so that your weight takes a long time to fall so the gearing would be stepped up to drive a generator. I’d be thinking something like an alternator where you could alter the generating field by altering the current and thereby increasing the magnetic drag and generation output. If you take a 12 volt alternator and supply the rotor with 6 volts instead of 12 you’ll find it doesn’t output 12 volts. But it will require less power to rotate. If you supply 15 volts to the rotor coils it’s output will exceed 12 volts and require higher power input to cause rotation.

Using this you could design an alternator that outputs your required voltage and amperage and your designed rate of drop to match without the issue of interrupting the rate of fall. Use the rotor voltage to control fall. The potential energy stored in your 500 kg weight will not change so it’s only how you can harvest it and lift that weight again that decides if it’s a good solution or not. Mechanical advantage does not alter the amount of power only how that power is distributed between torque and rotation speed.

Don’t let the limits of others stop you.
Thank you so much, you really did grasp my idea. Maybe I didn't ask the right questions, as you know sometimes it's about the quality of the question. Though I'm still trying to see how mechanical advantage does not alter the amount of power. Before we answer that, let me see if this rephrasing of my question will lead to a helpful answer and guide. Let's say I have a motor that has the below specification

MMA size
(shaft height, mm)
Rated Speed, rpmMax Speed, rpmRated Power, kWRated Torque, N•mPeak Torque, N•m
SH80580…26001400…60000.6…8.610.5…31.521…88
SH100580…26001400…60002.3…21.135.6…94.680…200
SH132580…26001400…60006.4…48.490.7…217200…400

If we attach 500kg of weight to the pulley weight to each of the motor and we attach a pulley with multiplying gears such that each rotation of the pulley in 2 seconds results to 100 rotation of the motor, which would translate to 30 rotation = 3,000 rpm. Now knowing the rpm, power output of the motor, peak torque and the height is 5 meters. Wouldn't it produce it produce the rated power as it's coming down?
 
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  • #42
kingamada said:
500kg of weight ... height is 5 meters...
You have ~ 24.5kJ energy in that (height*mass*g). That much can feed 1kW load for 24.5 seconds.

If done right, then the rated power of the motor/generator will determine only the maximal load the system can handle.
 
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  • #43
kingamada said:
Though I'm still trying to see how mechanical advantage does not alter the amount of power.
The energy available is the gravitational potential energy of your mass. The maximum power output of your device is thus the rate of change of that energy, or ##mgv##, where ##v## is the speed of the weight as it falls. So if you want 10kW out of a 500kg mass it must fall at 2m/s (in fact, you will need it to fall faster because losses will reduce the output power far below the input). Nothing you do can improve on this power output because there is no more energy to be had. If you have a 5m drop then you can have this power for 2.5s before you need to recharge. Note also that it will take a minimum of 0.2s to spin up to 10kW from nothing.

These are basic physics limitations. Once you get into the engineering and have to deal with losses the numbers will be far worse.

You asked about mechanical advantage. Essentially the generator acts as a brake on the falling mass, with a force that depends on speed. Changing the gearing between the generator and the mass changes the advantage of the braking force on the mass, so changes the speed at which balance between weight and braking occurs. So it will change the power output of the generator, but it will also change the power input to the generator by slowing or speeding the mass.
 
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  • #44
kingamada said:
If we attach 500kg of weight to the pulley weight to each of the motor and we attach a pulley with multiplying gears such that each rotation of the pulley in 2 seconds results to 100 rotation of the motor, which would translate to 30 rotation = 3,000 rpm. Now knowing the rpm, power output of the motor, peak torque and the height is 5 meters. Wouldn't it produce it produce the rated power as it's coming down?

Rive said:
You have ~ 24.5kJ energy in that (height*mass*g). That much can feed 1kW load for 24.5 seconds.
The details of power, force and torque are much less relevant than the basic Energy equation. That's where the money lies. The mechanics are easy to adjust.
 
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  • #45
sophiecentaur said:
The details of power, force and torque are much less relevant than the basic Energy equation. That's where the money lies. The mechanics are easy to adjust.
Just for comparison purposes, this 5m tall half ton device can provide 10kW for 2.45s, which is a storage capacity of about 0.007kWh. A Tesla model S battery typically stores on the order of 100kWh and the entire vehicle weighs only four-to-five times what this device does. The energy density here is much, much, lower than chemical batteries, which is why this is only done in old (or pretend-old) devices like mechanical clocks or on huge scale like pumped storage. In either case it's the only option.

None of this is a reason not to do this, as long as your "why" takes account of it.
 
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  • #46
I mean i knew the potential energy is 24.5kJ. I'm just finding it difficult to see how if the motor is rotating at the rated rpm and torque and it wouldn't produce the rated kw of more than 10kw.
Though this is the vision, still refining the idea for the best approach using animal weight to generate the electricity.
1716138247254.png
 
  • #47
Overthinking.
You only need a big treadwheel.

But honestly ... this just won't go anywhere.
 
  • #48
Ibix said:
Just for comparison purposes, this 5m tall half ton device can provide 10kW for 2.45s, which is a storage capacity of about 0.007kWh. A Tesla model S battery typically stores on the order of 100kWh and the entire vehicle weighs only four-to-five times what this device does.
Or going the other way, it's about the capacity of a cell phone battery.
 
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  • #49
kingamada said:
Maybe I didn't ask the right questions, as you know sometimes it's about the quality of the question. Though I'm still trying to see how mechanical advantage does not alter the amount of power. ?

kingamada said:
I mean i knew the potential energy is 24.5kJ. I'm just finding it difficult to see how if the motor is rotating at the rated rpm and torque and it wouldn't produce the rated kw of more than 10kw.
It will. You're just not connecting all the dots here. Energy and power are related by time. Do the math and figure out how long it runs, if you don't believe what people are telling you. Again: you can gear it to get the power you want, and in doing so you'll reduce the amount of time it will run.

I stand by the vibe I got earlier. You're using your instincts/gut because they are telling you what you want to hear, and avoiding the exceptionally easy math because you don't like the answer it gives. That's the wrong road you are going down.
 
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  • #50
kingamada said:
I'm just finding it difficult to see how if the motor is rotating at the rated rpm and torque and it wouldn't produce the rated kw of more than 10kw.
It would, minus losses in the motor and assuming a motor running as a generator is anything like a good generator. But there isn't time to spin up to 10kW output before you're out of energy. In free fall it would take 2s to reach that power output, in which time the weight would fall 20m. You can't accelerate it faster than that, so you can't achieve that power output.

If you gear the generator so it spins fast even when the weight is moving slowly the braking from the generator will have the mechanical advantage on its side and the weight will accelerate very slowly and you'll never get much power out. There's no way of escaping the rather slow initial energy release of this device.
 
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  • #51
Ibix said:
It would, minus losses in the motor and assuming a motor running as a generator is anything like a good generator. But there isn't time to spin up to 10kW output before you're out of energy. In free fall it would take 2s to reach that power output, in which time the weight would fall 20m. You can't accelerate it faster than that, so you can't achieve that power output.

If you gear the generator so it spins fast even when the weight is moving slowly the braking from the generator will have the mechanical advantage on its side and the weight will accelerate very slowly and you'll never get much power out. There's no way of escaping the rather slow initial energy release of this device.
Unless you attach another energy storage device to it, like a capacitor*. 😋

Sorry, couldn't resist.

*Also, you could probably adjust the deceleration rate to achieve it for part of the drop or create a profile where you let it accelerate in freefall for the necessary amount of time that the rest of it (including deceleration) outputs 10kW.
 
  • #52
russ_watters said:
Also, you could probably adjust the deceleration rate to achieve it for part of the drop or create a profile where you let it accelerate in freefall for the necessary amount of time that the rest of it (including deceleration) outputs 10kW.
To get 10kW of power out of a 500kg mass you need it to be doing 20m/s. In free fall from rest that takes 2s and 20m of vertical distance. He's only got 5m. The maximum possible speed he can achieve with a 5m drop is 10m/s, so the hard upper limit on power out of this device is 5kW. And that's a peak output only achievable for an instant if he wastes a full charge to spin it up.

1kW is more achievable. That still requires a minimum 0.2s spin up, but only a 20cm drop.
 
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  • #53
kingamada said:
Though I'm still trying to see how mechanical advantage does not alter the amount of power.
This thread seems to neglect the difference between the two terms that apply to all machines; Mechanical Advantage and Velocity Ratio. Superficial arguments about machines are all based on VR, which is calculated on the basic geometry. In fact, any machine performance is governed by it MA. MA involves the effects of both friction and so-called 'dead weight' which involves energy in moving parts of the machine around (e.g. lifting and lowering a pulley block). As the link says, the simple relationship between MA and VR is

Efficiency (η)=M.A./V.R.

If efficiency (what you get out / what you put in) is important then the details have to come in very early in any 'design' ideas. If you want an excellent example of Energy Storage from way back, then consider the Fusee movement, used in clocks (and even watches) to maintain a given output power from the main spring as it unwinds. The power requirements in the movement are the same all the time but the torque drops. I have a simple wall clock with Fusee (with a fine chain as the connection - god knows what I will do if and when that snaps). It's dead simple, as clocks go, but the Fusee link keeps the clock to within about half a minute over a whole week, just speeding up towards the point when it needs re-winding. Something along those lines could help the OP - or at least contribute to the list of considerations.
1716199262141.png
 
  • #54
Ripcrow said:
And that’s fine except when the projections are biased
One man's bias is another man's well informed authority and experience.
 
  • #55
Ibix said:
To get 10kW of power out of a 500kg mass you need it to be doing 20m/s. In free fall from rest that takes 2s and 20m of vertical distance. He's only got 5m. The maximum possible speed he can achieve with a 5m drop is 10m/s, so the hard upper limit on power out of this device is 5kW. And that's a peak output only achievable for an instant if he wastes a full charge to spin it up.
You are describing a scenario where all of the energy extraction occurs at constant speed, with the braking force equal to the weight of the object lowered. As you indicate that will waste the kinetic energy of the acceleration phase (not to mention what happens when it hits the ground...). That's not the only way to do it. Consider the real life example of an elevator with energy recovery and electromagnetic braking. It has three phases of energy extraction:

Acceleration phase. During acceleration the braking force will be lower than the weight of the object being lowered, but won't be zero(for an elevator) and there is some energy extraction. For the OP's device it could be zero if desired.

Coast phase. In this phase you will have the constant speed energy extraction as you describe.

Deceleration Phase. During this phase the braking force is larger than the weight of the object being lowered. For an elevator the deceleration rate will be constant (with rounded inflection points) therefore the power profile will start with a higher power than the coast phase and then drop towards zero as the elevator comes to a stop. This profile is of course an arbitrary choice and if we wanted to achieve constant power deceleration at any selected power output, we could.

Here's a graph of the speed and power of an elevator, and a link to their brochure:

Elevator.png

https://www.otis.com/documents/2560...1-af7d-b4b3-ecc1-547c2984fbf1?t=1591127484167

Note: The device OP envisions is, in fact, an elevator.
 
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  • #56
russ_watters said:
Consider the real life example of an elevator with energy recovery and electromagnetic braking. It has three phases of energy extraction:
Ah, so you let the potential energy convert to kinetic then harvest the kinetic energy at an arbitrarily high power. You still can't better the average power this way, but you can have very short peaks of very high power.
 
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  • #57
Ibix said:
You still can't better the average power this way, but you can have very short peaks of very high power.
Right. I'll further note that the OP seems to be muddying the water between a continuous and a batch process, and wrongly focusing on a target power output. It doesn't gain you anything (in fact it's almost certainly less efficient) to pick an arbitrary short peak power output....and it's bad for the animals he's lowering.

While OP hasn't given a full description, it's more likely he's looking to move a large number of animals over a specific time interval, which makes this a continuous process with a specific power input/output available. Instead of an elevator, a sloped conveyor belt moving at constant speed might work better.
 
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  • #58
Rive said:
You only need a big treadwheel.
Wow never knew something like that existed, but yes something similar butt it has to be a smooth setup without wearing down the animal.
russ_watters said:
it's more likely he's looking to move a large number of animals over a specific time interval, which makes this a continuous process with a specific power input/output available
Yes, that's the idea to use the weight of animals to produce electricity. The goal is to find the right setup to get the most power by using just animal weight, which might involve:
1. The animals just walking in the field and stepping on tiny levers that moves downward 5cm (a field of 1000 units of these which at any given time more than hundreds would be stepped on, if we're to calculate for 500kg the power generation would be 245watts per lever x 100 = 24kw)
2. The animals just coming down on an elevator type structure only to walk back up again (5meters to 10 meters)
3. Thinking of horizontal wind turbine but powered by animal weight/movement instead of wind
4. Just a huge long lever that the animals step on.
More simpler but efficient idea is welcome.
 
  • #59
kingamada said:
2. The animals just coming down on an elevator type structure only to walk back up again (5meters to 10 meters)
Train them when they are young! :smile:

 
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  • #60
kingamada said:
Yes, that's the idea to use the weight of animals to produce electricity.
Those animals may put on less beef than if they weren't required to drive 'the engine'. Energy harvesting needs careful analysis to find if it's really worth while. Cattle don't walk further than they really have to.

I remember reading that the Chicago(?) stock yards worked on gravity. The live beasts walked up a long ramp and their carcasses rolled down a rail as the butchery took place.
 
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