# torque

Discussion in 'Alternative Energy' started by 442442442hugo, May 26, 2010.

1. ### 442442442hugoNew Member

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Ok.........how much weight (moving at 1m an hour) would you need to drive a gear (or pulley) system to create.......( 63 ft pounds of torque....1hp.....i have got a little confused with what terms to use..sorry ) and achieve a rotation of one second (or more!) at the other end.

The largest gear or pulley can be no bigger than 2m in circumference....

I have a drop of 10m

This is to run an alternator that I will build with as little friction as possible that will generate a meaningful amount of electricity at low rpms........

This is a curiosity that has sent my head spinning so any ideas etc would be most appreciated....fun fun fun

2. ### MrAlWell-Known MemberMost Helpful Member

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Hi,

Are you saying you want to hang a weight off of a rope connected to a pulley so that it turns the pulley and that turns the generator?

Rotational speed is measured in RPM or RPS, etc.

3. ### 442442442hugoNew Member

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yep thats the stuff.........sorry is was ment to be rpm.....

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5. ### 3v0Coop Build CoordinatorForum Supporter

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Use either imperial or metric units.

Without doing the math you will need a lot of weight to generate that amount of power. The bearing needed to support the pulley and the gearing needed to spin the generator will introduce significant friction.

I expect it is impractical. I was never good at this sort of problem but let me give it a shot anyway. Please feel free to correct me.

A HP is 33,000 foot-pounds of work in one minute.

You specified that your weight can drop 10m in 10hours which is 33 feet in 10 hours or 33/600 feet per minute. That is .055 feet per minute.

With a one foot pulley you would need a weight (33,000/.055) 600,000 lbs to get the needed 1 HP. With the weight moving at .055 ft/min it will turn the 1 foot pulley once in (3.14*1)/.055 = 57 seconds.

Even if I am off by a few orders of magnitude it is still impractical.

EDIT: Gravity is a weak force.

Last edited: May 26, 2010
6. ### bryan1Well-Known Member

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In true context 'Gravity' sucks, This concept like said above is just a pipe dream and will never work, well one might get one working but the POWER needed to reset the so-called device will be a huge ratio more than any POWER produced.

7. ### 442442442hugoNew Member

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bellows!

Two questions for all you bright lot
This is not home work I am 30. This is an idea for a energy project (not perpetual!!!)

If I had a set of bellows that was 3 meters cubed and produced 5psi through a 5mm (diameter) pipe ..........how much weight (kg) (pressure if this is easier) would i need to apply to the top of the bellows....... and how long would it take to expel all the air.

How much water pressure would i produce from a constantly full 200 litre tank with a 10mm (diameter) tap/hose attached to the bottom?

8. ### mnearyNew Member

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3 meters cubed (27 cubic meters?) or 3 cubic meters? What is a 5psi? What is the surface area over which the weight (sic) in kg is applied? What is the length and interior texture of your 5mm exit pipe?

How tall is the 200 liter tank and how do you add water to it without influencing the pressure measurement? Is the hole at the bottom facing downward or sideways? How long is the 10mm diameter hose?

9. ### 442442442hugoNew Member

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.................................................. .........................

3m tall, 3m wide, 3m deep.... i think that is 3 cubic meters (sorry for that)? .....psi.... 1 pound per square inch = 0.0689475729 bar........the weight will be evenly distributed on the top of the cube (bellows made of thin plastic)....... the exit pipe is 1m and is everyday see through plastic hose......

The tank is 1.5m tall. I will be adding water using the mains. this will affect the pressure. if the idea is even feasible then this will be one of the problems that will need to be mineralized, for the sake of the hypothetical letâ€™s say that the tank is always full....the hole at the bottom is facing down and the hose is 1m long.

All your help is much appreciated.....

10. ### mnearyNew Member

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The top of your bellows is 0.00222394843 acres. To press 5 psi across the entire surface would require a total weight of about 5000 stone.

Last edited: Jun 6, 2010
11. ### PommieWell-Known MemberMost Helpful Member

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I love the way people mix metric and imperial.

To answer the original question in metric, the weight contains potential energy equal to Mgh (Mass x Gravity x Height). So to produce 1hp (750W) for 1 hour would require 750*3600 Joules and so M = 750*3600/(9.81*1) = 275,000 kg. Don't drop it on your toe.

Mike.

12. ### mnearyNew Member

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I usually like to use SI, but the OP seems not to like SI units of mass and force