How much torque would be on the motor?

[Temeraire]

I would like to make a project where a dc motor will help the movement of a human arm. The motor will be at the elbow, the stationary part will be attached to the upper arm, and the moving part will be attached to the forearm, so the motor will help to rotate the forearm at the elbow.

I think the biggest torque will be when the forearm is parallel with the ground, so my calculation is something like this:
My forearm+hand+a little weight(aluminium structure, etc): m=3kg.
Gravitational acceleration: g=9.81 m/s2
The arm is a distributed load, for now let's say it is evenly distributed and the center mass is at halfway of the forearm: from the elbow it's s=20 cm=0.2m

So the Torque would be: T=3*9.81*0.2=5.89Nm.
Or going with the kg.cm: 60kg.cm.

I got a recommendation to choose a motor, for example this: https://media.digikey.com/pdf/Data%20Sheets/Seeed%20Technology/108990007_Web.pdf
But it says that the load torque is 0.55 kg.cm, so if my calculations are correct i would need hundred times more torque, which seems odd to me.

do you have any idea what i am missing, or calculating wrong?

 

[Mech_Engineer]

If you want to use that motor you'll need a gearbox, the motor's torque without the gearbox is just a piece of the puzzle. The alternative would be to use a direct-drive DC brushless motor with the rated torque but that might require higher-cost components.

 

[Temeraire]

If you want to use that motor you'll need a gearbox, the motor's torque without the gearbox is just a piece of the puzzle. The alternative would be to use a direct-drive DC brushless motor with the rated torque but that might require higher-cost components.

The motor itself has a built in gearbox with a worm gear (i will attach a picture). Could it be that they mean the dc motor's shaft's load torque by the 0.55 kg.cm without the gearbox? So with the gearbox it becomes much larger?

 

[russ_watters]

I think the biggest torque will be when the forearm is parallel with the ground, so my calculation is something like this:
My forearm+hand+a little weight(aluminium structure, etc): m=3kg.
Gravitational acceleration: g=9.81 m/s2
The arm is a distributed load, for now let's say it is evenly distributed and the center mass is at halfway of the forearm: from the elbow it's s=20 cm=0.2m

So the Torque would be: T=3*9.81*0.2=5.89Nm.
Or going with the kg.cm: 60kg.cm.

Do you want this arm to do anything besides moving? Lift an object, for example...?

 

[Temeraire]

Do you want this arm to do anything besides moving? Lift an object, for example...?

I would be more than happy if in the first round i could make it work with just the arm (without any external object to lift).

 

[Mech_Engineer]

The motor you've shown is way undersized for the requirements you've stated, I recommend doing some more searching to find a motor/gearbox combination which provides the torque specification of 6 N-m.

You'll also need to consider expected power output and speed- you may be able to find a small motor with large reduction gearbox that meets the torque requirements, but would take minutes to move 90 degrees. You're going to need a motor with a combination of enough torque and reasonable speed/power output for your application.