Help With Power Calculation Please

[bill barlow]

I have designed and built a device that harnesses centrifugal force.The power output is in the form of
2 x 10mm steel rod "pistons" that move up and down a distance of 15mm. I have measured the weight these pistons can lift to be 122lbs when the device is operated @ 4000rpm. Each piston moves up and down once per revolution. I have calculated this to be an output of 1.45 hp. ( 15mm=.049125ft x 66.666 times per second=3.27496 ft x 122lbs = 399.54 lbs/ft/sec = .7264hp x 2 = 1.45hp. Can people please either confirm my calculations are correct or point out my error(s).

If I am correct,given that the device is solely powered by a 500w electric drill, am I able to claim the impossible?

[Teddio]

I don't understand how the "centrifugal" force is harnessed.
Work Done = Force X Distance (how is work being done ?)
Power is the Work Done divided by the time taken.
A diagram would be useful.
1.45 horsepower = approx. 1070 Watts
How can the output power exceed the input power ?

[bill barlow]

Teddio. Thanks for the reply although you have not answered my question. The device was designed to output more power than input. How it works has to remain confidential for now. My question is,assuming my measurements are correct {which I am sure they are}, have I succeeded?? ie are my calculations correct?

[beso]

No device can ever put out more power than goes in. It can only provide the power that goes in less the inefficiencies.

If the spinning weights are lifting a load that load will appear on the driving motor.

[beso]

The weights going down again converted the gravitational potential energy they gained by being lifted back into kinetic energy. No net work was done.

What you have is a 500 Watt power dumper.

[Peter Pedant]

No I am afraid that not only are you violating The First Law - which is 'don't violate the first law' or something,

but the calc is wrong:

Doing it in SI/ MKS - 3.2 ft is about a metre, 122lbs is about 50kg
and this is 50 n-m per sec which is 50 W innit ?

I am doing it in SI units 1) coz I am modern
2) FPS is terribly hard and prone to mistakes
3) I think this is a set up - I cannot seriously think that someone converts mm [wh is kinda SI ] into FPS UNLESS he wishes to cloud the issue
4) you have used the word centrifugal which everyone kno you should reverse the arrow and use the word centripetal.

so you can't claim the impossible - you can only claim insanity

In short your calculations ar enot correct

IN fact you are working at 10% efficienty - 500W goes to 50W which I would have thought is within the realms of possibility

[bill barlow]

Thanks Peter. I have given the measurements in the units they were measured in, not to try and confuse or hide anything. I have converted to ft/lbs/sec because 1hp = 550 ft/lbs/sec??

I may be wrong but isnt 50K= 490N? And there are 2 Pistons?

I truly want people to correct me if I am wrong!!

[Peter Pedant]

No - all my centrifuges are horizontal - g doesnt come into it

as another said a diagram may be helpful.

The piston cant move up an down under gracity at 4000 rpm

[Peter Pedant]

I am taking 1 J to be the work done displacing 1kg thro 1 m.

[Peter Pedant]

I agree Bezo I am not sure there is much work done

[bill barlow]

Sorry Peter you have confused me now. But I can confirm the piston doesnt move downwards or upwards under gravity

[Peter Pedant]

I cant confuse you - its is your machine

your calc is out by a factor of 10 as you have included g and this is not relevant in horizontal motion - hence Bezo saying it is 500W dumper.

[bill barlow]

Sorry, you have confused me! I havnt mentioned g or horizontal motion!??

[Peter Pedant]

that is OK

122 lbs is around 50 kg and moving it thro 1 metre is work of 50 J and not 500 J.

as ever the confusing factor is 'g' 9.81.

[jomifl]

What is the power put into the device converted into? ie what work does it do?

[bill barlow]

Peter, thank you for the reply. Please explain how g comes into it?? I have just used an online calulator which tells me 50kg/meter/sec = 490 joules?? Why do you say 50 joules? Is this where g comes into it? If so how?? Please explain. Many thanks.

[bill barlow]

jomfl.....I have given weights lifted.distance and time. What more is needed to work out power output. I dont know.

[Zacs-Master]

I was taught that there is actually no such thing as centrifugal force and it is actually 'centripetal'. Was I taught wrongly?

[vascop]

ZM: Pedantry doesn't help here.

[EDDIE51]

It is IMPOSSIBLE to get more power out than you put in, the best you can do is to equal the input, but you device must have considerable frictional losses so it is going to give out a lot less than goes in.

[nescio]

Peter Pedant,
moving a mass of 50kg a distance of 1m against gravity does NOT equate to 50J of work. Work done = force x distance = mass x acceleration x distance. Here, acceleration is g = 9.81 m/s, so work done is approximately 500J.

The seemingly impossible result is connected with the speed: at 4000rpm, something falling under gravity would move approximately 1mm on each revolution, so the effective distance moved is much less than the 15mm quoted. Therefore, the work done per cycle and hence the power, is much less.