- #1
Luchekv
- 66
- 1
Hi guys,
I was sizing up a servo motor the other day and started thinking. While we all know gear ratios will effectively "convert" speed into torque. I couldn't help wondering about the initial starting conditions.. The problem below...
Lets say I have a motor with a max load of 1Kg (holding). The motor has a driving gear with a 1cm radius. Therefore:
Required torque = 1KgF x 1cm = 1KgF.cm = 0.098 N.m of torque
Now let's say we also have another gear with twice the radius. To lift the same 1Kg weight we now require 0.196 N.m of torque.
Required torque = 1KgF x 2cm = 2KgF.cm = 0.196 N.m of torque
So the question is this:
Lets pretend that motor has a 1cm radius driving gear...that is connected to the larger 2cm radius gear.
If my motor cannot lift(move) a weight of 1Kg through its driving gear (requiring T > 0.098 N.m)
How can it be expected to start turning an even larger gear that will require twice the torque to get it moving?
I was sizing up a servo motor the other day and started thinking. While we all know gear ratios will effectively "convert" speed into torque. I couldn't help wondering about the initial starting conditions.. The problem below...
Lets say I have a motor with a max load of 1Kg (holding). The motor has a driving gear with a 1cm radius. Therefore:
Required torque = 1KgF x 1cm = 1KgF.cm = 0.098 N.m of torque
Now let's say we also have another gear with twice the radius. To lift the same 1Kg weight we now require 0.196 N.m of torque.
Required torque = 1KgF x 2cm = 2KgF.cm = 0.196 N.m of torque
So the question is this:
Lets pretend that motor has a 1cm radius driving gear...that is connected to the larger 2cm radius gear.
If my motor cannot lift(move) a weight of 1Kg through its driving gear (requiring T > 0.098 N.m)
How can it be expected to start turning an even larger gear that will require twice the torque to get it moving?
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