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# SLOWING DOWN AN ELECTRIC MOTOR USING GEARS ONLY?

i have a motor with an input voltage of 12vDC and the final drive RPM is 172. i need to create a winch to lift a specific weight at a specific speed. ( 1kg in 70 seconds) from 1 meter off the ground. can someone tell me how to do this or explain gear ratio and drum diameter. i am completely lost and don't know what to do. your help will be very much appreciated and you will be rewarded

## Discussions

4 years ago

Maybe your teacher can explain it a bit better for you?

Answer 4 years ago

+1

The phrase you need to google is "gear ratios"...

4 years ago

What is the motor HP-as that is important to the speed, through gearing or pulleys to maximize the motor's speed/power ratio.

JennyGirl2

Answer 4 years ago

If you do not have fractional HP please give us the

1] No load current ?

2] Full load current ?

3] Stall current if you know it ?

4 years ago

Rewarded? Rewards are kind of useless to me, because, like the

Three Amigos, my reward is knowing that justice is done!Yeah. I think this problem would make sense if you converted some of the units, like.

Weight:Weight is a force, equal to M*g, so a 1 kg mass has a weight of

(1.00 kg)*(9.8 m/s/s) = 9.8 N

The N stands for newtons, the SI unit of force.

I am sort of imagining this weight is attached to a string, and it is being pulled up vertically. So the weight pulls downward on the string, and the string pulls upward on the weight.

Speed:To move the weight through a distance of 1 meter, in 70 seconds, requires an average speed of v = (1.00 m)/(70 s) = 0.014 m/s. Also it is probably safe to assume v is also a constant speed.

Power:Power required to pull against some constant force, at some constant speed.

P = F*v = (9.8 N)*(0.014 m/s) = 0.14 J/s = 0.14 W

Can your 12 volt motor throw this much mechanical power? I dunno. This number seems believable to me.

Angular speed:Revolutions per minute, also called "rpm" is sort of traditional measure of angular speed, a measure of how some rotating thing is rotating. A better, but less understood, unit is radians per second. So I want to convert 172 rev/min, to whatever that is in radians per second.

omega = (172 rev/min)*(2*pi/rev)*(1 min/60 s) = 18 rad/s

The lowercase Greek letter omega is the usual symbol used for angular speed, but this editor doesn't have Greek letters, so I just spell it out, and use that whole five letter word as my variable.

The reason radians per second is convenient is because there is a simple formula relating the angular speed (in rad/s) of a wheel to the translational speed (in m/s) of a point at distance r from the center of the wheel.

v = omega*r

And you can maybe sort of see why this is true. The circle at the edge of the wheel has a circumfrence of 2*pi*r, and one of those circles unrolls for every 2*pi of angle, in radians.

Anyway, if I take that formula above, v=omega*r, and solve for r, using your numbers, I get,

r = v/omega =

(0.014 m/s)/(18 rad/s) = 0.000777 m = 0.777e-4 m = 0.777e-3 m

So that a drum with very, very tiny radius: 0.78 mm, or 1.06 mm diameter.

So I am guessing that your angular speed is too big, too fast, for a practical sized drum. However, if you slow it down, by maybe a factor of 10, then you could increase the size of the drum by the same factor, giving you a drum with diameter of around 10 mm, or 1.0 cm. At least then the drum will be bigger than the string.

So. Yeah. Kind of a long story I guess, but I think you do want to slow down the angular speed of your drum, which is kind of what you were suspecting in the first place, I think. So maybe the lesson is to trust your intuition. Maybe.

I hope some of the formulas I have mentioned are useful to you.

4 years ago

You have a power number from your weight and height and lift rate. You have the RPM from the motor. It may not have enough power, check the maths of that first. The rest is the gearing and drum diameter.