[DPRG] Motor Calculations
Subject: [DPRG] Motor Calculations
From: Chuck McManis
cmcmanis at mcmanis.com
Date: Wed Aug 29 22:23:08 CDT 2007
Well 30RPM is 180 degrees per second so you're probably looking at
about 10x the HP requirement so 15HP and 50HP. The methodology is
that you compute the amount of energy the wheel has in it one second
after it starts (You wanted to be at speed in 1 second). That is how
much energy (plus any extra to overcome bearing friction, etc) your
motor has to deliver. Since you're accelerating in 1 second and
energy is in wattseconds you can basically take the number of watts
required, divide by 750 and get horsepower. Most, motors have what is
known as a "power curve" where at certain RPM they develop peak
horsepower. So you will want to develop a transmission which at the
peak horsepower RPM results in 30RPM on your wheel, then you will
need a way to rapidly engage the motor. (A magnetic clutch is the
most common I've seen in things like tractors and such). Probably
won't get it out of an electric motor.
Chuck
At 08:10 AM 8/29/2007, Rick Bickle wrote:
>
>Chuck,
>
>Thanks for the info. The bearings that I'm using are tapered roller
>bearings. In order to reduce cost, I am using heavy duty trailer stub
>axles with hubs.
>
>I think I have misstated the speed. The speed of rotation should be 30
>RPM. I must have been thinking in terms of seconds.
>
>The system will rarely ever make a full revolution. It will do a lot of
>starting and stopping though.
>
>Thanks,
>Rick
>
>Original Message
>From: dprglistbounces at dprg.org [mailto:dprglistbounces at dprg.org] On
>Behalf Of Chuck McManis
>Sent: Wednesday, August 29, 2007 12:44 AM
>To: Rick Bickle; DPRGlist at dprg.org
>Subject: Re: [DPRG] Motor Calculations
>
>Hi Rick,
>
>More questions than answers but if you get a bit more data the answer
>would fall out of the calculations so to speak.
>
> >1. Will need to rotate a weight of about 600 lbs. evenly distributed
> >with a maximum distance from center of 2 feet.
>
>So 600 lbs is roughly 272 Kg of mass,
>
> First you need to know what sort of bearing this weight is on,
> it will contribute potentially significant friction into the
> equation.
>
> Next you will need to know at what speed you will be rotating
> this mass. Given its 272 Kg it is going to be carrying a *lot*
> of angular momentum.
>
> Finally you will want to know what other loads (besides bearing
> friction and wind resistance) are going to be applied to the
> load.
>
>The reason these other bits of information are important is that if the
>bearing is very low friction then you can start the thing spinning with
>a pager motor, if its on an axle with a steel on steel oiled bearing you
>may find you need a bit more starting torque.
>
> >2. The motor will need to rotate a weight of about 800 lbs. evenly
> >distributed with a maximum distance from center of 4 feet.
>
>Same questions.
>
> >The motors should be strong enough to get up to a maximum speed of 2
> >rpm within 1 second.
>
>
>Ok, this gives us almost enough information (actually it is if we assume
>a frictionless bearing and no other resistances). The formula for
>angular momentum is Iw (Moment of Inertia * angular velocity), if you
>calculate the angular momentum of your 600lb disk, its approximately
>1/2MR^2 (M = Mass, R = Radius) so keeping everything in SI units for
>your first disk, 272 Kg * your angular velocity is 720 degrees / minute
>or 12 degrees/sec and your radius is .61 meters.
>So about 608 degree*Kg*m^2/sec. That's about 608 joules or 608
>wattseconds. Since a 1 hp motor is about 750 watts, that means for your
>600lb disk you can use a 1HP motor to get it spinning at 2RPM (ignoring
>bearing friction and other factors), you'll probably want something
>better than that.
>
>For your 800lb disk with the 4' radius your looking at more than 3200
>wattseconds so something bigger than a 4.4HP motor.
>
>So in the first case I'd start with a 1.5HP motor (unless you know the
>bearing is high friction) and in the second case a 5HP motor.
>
>Chuck
>
>
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