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Topic: How to attach a potentiometer to a servo motor shaft (Read 2163 times) previous topic - next topic

birddseedd

It's the product of the force and the distance at which it acts from the pivot.

So 15lbs at 16" is 15 x 1.33' = 20




Im a bit confused by why the distance matters. The pivot in this case doess not effect the force needed. If i were to lengthen the handle bars to 100' and moved the damener and moter to the end if the 100', its still going to need the same 15, of force. But now the forumula would say 1500 ft lbs of force.

wvmarle

It's the distance between the centre of the axle of the motor (the pivot point), and the point at where the force is needed (the point where your motor's arm connects to the thing you want to move).
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birddseedd

It's the distance between the centre of the axle of the motor (the pivot point), and the point at where the force is needed (the point where your motor's arm connects to the thing you want to move).
Ok. That makes more sense.

Ive been talking about the arm of the mower. Not the motor. That's why it didn't make any sense to me.

The motor wouldnt be placed at the center of that pivot in the pic. Although, if i did it would solve some linear motion issues. So maybe ill take a look at it and see if i can.

wvmarle

You're mixing up things. To calculate the torque of your motor you need to measure the length of the arm the motor has to move from the centre of the motor's axle (that's your motor's pivot point) to where it's attached to whatever you want to apply the force to.

Have a look at these images. Should make it clear.
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birddseedd

That part I believe I understand. Where I think I'm getting confused is that, correct me if I'm wrong, I'm not measuring "torque" I am measuring "pound Force".

couka

...and from pound-force and radius (=length of the arm) you can calculate the torque.
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birddseedd

Ok. The more im reading i think that im measuring in lbs mass not pounds force. So i have to cinvert the mass to force before i can convert to torque.

I'm going to take a break and eat some breakfast

wvmarle

The normal way of measuring weight (what is often called "mass" though they're not the same thing) with a scale is by measuring force: the force at which the earth's gravity pulls at the object.

So attaching a spring scale to your handle and pulling it is basically the same thing.
Quality of answers is related to the quality of questions. Good questions will get good answers. Useless answers are a sign of a poor question.

birddseedd

The normal way of measuring weight (what is often called "mass" though they're not the same thing) with a scale is by measuring force: the force at which the earth's gravity pulls at the object.

So attaching a spring scale to your handle and pulling it is basically the same thing.
Ok. So i did measure pound force (lbf). I need to convert that to pound to pound torque (lbf.ft). And then adjust for the length of the lever from the pivot point.

birddseedd

#54
Jan 14, 2019, 06:23 pm Last Edit: Jan 14, 2019, 07:07 pm by birddseedd
I got a chance to go out and measure the control arm. It measures 12 inches.

If i attach the motor at the same point thst will illiminate issues converting it to linear motion. Although it means a bigger motor is needed.

15 lbf * 1' means 15 torque? That would mean just about 20 nm of torque?

If i mount the motor higger and just deal with the uneven motion i can cut that in half. Thats still a good size motor.

birddseedd

15 lbs of force on the handle that appears to be at least 16 inches from the pivot point? 20 pound feet of torque? Oh my, looks like a big gearmotor with an electromagnetic release clutch for manual operation.
So really this is another situation where I'm asking the wrong question. I don't need a motor that I can turn off and still get a reading from it, I need a motor where I can release the clutch, and still get a reading from it

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