Torque calculation question

Don M

For all you math guru's on here... I'm trying to figure out how much torque is needed to lift a 120 pound weight with a rope wound on a 3.5" shaft. What's the equation?

Don M

Actually I think I found my answer... https://www.precisionmicrodrives.com/tech-blog/2013/09/23/torque-calculations-gearmotor-applications

Don M

210 in/lbs? 1.75 inches x 120 pounds

erco

Equilibrium (holding) torque is (3.5/2)*120 inch pounds. You'll need more than that to winch it up, depending how fast you want to go.

Don M

Let's say you wanted to go 48 inches which would be 4.4 revolutions in 15 seconds? How does that figure into the equation?

erco

You can calculate power specs all day, but it's more productive to dig into motor specs.

Power = work/time=force*distance/time

Don M

Thanks erco

Mickster

Don M wrote: »

For all you math guru's on here... I'm trying to figure out how much torque is needed to lift a 120 pound weight with a rope wound on a 3.5" shaft. What's the equation?

Depending on the application, there is sometimes the possibility to use a counter-balance.
Inertia mismatch should also be considered.

Heater.

Yankees are measuring torque in foot pounds? What century is this?

So 120 pounds over 3.5 inches is about (3.5 / 12) *120 = 35 foot pounds.

Am I right here?

Phil Pilgrim (PhiPi)

No. Use the radius, not the diameter. And over here, torque units are called "pound feet", not "foot pounds," the latter being reserved for potential energy. Yeah, I know, it's a "distinction without a difference."

-Phil

Mickster

Heater. wrote: »

Yankees are measuring torque in foot pounds? What century is this?

So 120 pounds over 3.5 inches is about (3.5 / 12) *120 = 35 foot pounds.

Am I right here?

No, If it was a 2" diameter shaft, that would be a radius of 1", lifting 120LBS so 120 inch/pounds of torque would be required. But here the radius is 1.75" so we are looking at 1.75 X 120.

Duane Degn

Don M wrote: »

Let's say you wanted to go 48 inches which would be 4.4 revolutions in 15 seconds? How does that figure into the equation?

Technically the torque required to move a load at a constant speed should be the same as the holding torque (though in reality friction likely reduces the required torque a bit if the motor isn't moving). Erco's "how fast you want to go" also means "how fast you want to accelerate."

The acceleration required to move the desired speed and distance can be calculated using basic constant acceleration equations.

Let's say you want to move the 48 inches with a constant acceleration for both starting and stopping the motion. Since half the time will be spent slowing down, we'll compute the acceleration needed to move 24 inches in 7.5s.

I think the equation we want is this one (x is the distance):
x = 0.5*a*t^2

Since we know the distance and time, we just need to solve for acceleration.

a = 2 * x / t^2 = 48 / 56.25 = 0.853 in/s^2

Acceleration due to gravity is about 32.2 ft/s^2 which is 386.4 in/s^2.

So the acceleration due to moving the mass is significantly less than the acceleration due to gravity.

Your motor only needs to be 0.2208% stronger to move the mass at the desired speed than it needs to be to hold the mass.

100% * (0.853 in/s^2) / (386.4 in/s^2) = 0.2208%

So to hold the weight, you need a 210 pound inch inch pound motor but to move the weight you need a 210.46 pound inch inch pound motor.

At least those I numbers I get. I hope people correct me if I'm wrong.

Edit: If one should use foot pound instead of pound foot, I suppose my "pound inch" was wrong (likely in more ways than one). I'm guessing inch pound is less wrong than pound inch. I don't think inch pounds are used for torque. It would likely be better to use foot pounds or inch ounces. Inch ounce sounds wrong. I'm guessing the correct order is ounce inch. I'm too lazy to look it up.

Heater.

Give me a break Phil, radius vs diameter, I think a factor of two is close enough

So now I have (1.25 / 12) * 120 = 12.5 pounds foot.

Don M

Duane Degn wrote: »

Don M wrote: »

Let's say you wanted to go 48 inches which would be 4.4 revolutions in 15 seconds? How does that figure into the equation?

Technically the torque required to move a load at a constant speed should be the same as the holding torque (though in reality friction likely reduces the required torque a bit if the motor isn't moving). Erco's "how fast you want to go" also means "how fast you want to accelerate."

The acceleration required to move the desired speed and distance can be calculated using basic constant acceleration equations.

Let's say you want to move the 48 inches with a constant acceleration for both starting and stopping the motion. Since half the time will be spent slowing down, we'll compute the acceleration needed to move 24 inches in 7.5s.

I think the equation we want is this one (x is the distance):
x = 0.5*a*t^2

Since we know the distance and time, we just need to solve for acceleration.

a = 2 * x / t^2 = 48 / 56.25 = 0.853 in/s^2

Acceleration due to gravity is about 32.2 ft/s^2 which is 386.4 in/s^2.

So the acceleration from due to moving the mass is significantly less than the acceleration due to gravity.

Your motor only needs to be 0.2208% stronger to move the mass at the desired speed than it needs to be to hold the mass.

100% * (0.853 in/s^2) / (386.4 in/s^2) = 0.2208%

So to hold the weight, you need a 210 pound inch motor but to move the weight you need a 210.46 pound inch motor.

At least those I numbers I get. I hope people correct me if I'm wrong.

Thanks Duane!

erco

Many factors to consider in transmissions. Every gear and bushing reduces efficiency several percent, so it's best to design in a safety factor of 2 or 3 if you're only looking at motor torque/rpm specs. This looks like a winch, so you may want the axle/spool to be self-locking, so it will hold the load when the motor is off (i.e.,the load backdriving the unpowered motor). I'm not a fan of worm gears for power transmission since they have terrible friction, efficiency, and wear, but on the plus side they are compact and almost always self-locking, due to their excessive friction.

Most of my experience is with simple toy grade brushed DC motors. Even a cheesy little 3V motor has a detailed spec sheet, such as at https://www.pololu.com/file/0J11/fa_130ra.pdf

Even without detailed spec sheets, there are a few little tricks like knowing peak power output is at half the stall torque times half the no-load speed, and peak efficiency is around 75% of no load speed, or 25% of stall torque.

Don M

Or a magnetic particle brake could be used to lock the motor temporarily.

T Chap

Not sure if you are asking about holding the load in a position, but you can hold the weight in position with a BLDC motor/gearbox, and tie the 3 windings together which creates a brake effect on the motor. A larger enough motor and high ratio gearbox ie 12:5:1 or 25:1 can be very robust for holding without power. I have had to do this to secure automation systems in place on sailboats that would otherwise bounce around. Avoids mechanical latches.