# Thread: Starting Torque vs Running Torque in DC Motors

1. ## Starting Torque vs Running Torque in DC Motors

Hello,

I am trying to find some information about how the starting torque changes with relation to the running torque.
I have an application with low duty cycle that requires the motor to operate more towards the starting torque side than the running torque.

What I would like to find out is the following:
1) Is there a dramatic drop in torque from starting to running or is it a gradual decrease?
2) Is there an equation that relates the two torques?
3) What are the different variables I should be considering? (speed, current, load ...)

Motor is a brush DC motor.

Thanks!  Reply With Quote

2. Welcome and here's the standard first reply.

Volts?
Amps?
HP?
RPM range?
Opening a six-lane draw bridge or moving a hand puppet on a string?

Depending on load at start-up, the torque-drop may be negligible or very significant.  Reply With Quote

3. 1) T=K1*I

2) V0=+LI'+RI +K2*@'

3) J@"+c@'=T-Tl

V0 impressed voltage
I=current
T= torque delivered by motor
K1=torque constant
K2=back emf constant
J= polar moment of inertia
@ = rotor angle
@' = rotor speed
@"= rotor acceleration
I' rate of change of current
L armature inductance
R= armature resistance
c = damping coefficient

These are the equations that govern a DC motor and are readily solvable by eliminating T in eq 1 and 2
Then you have 2 remaining equations in I and @ and then eliminate @ resulting in a single linear differential equation in I.
To give you an idea of what happens when you throw the switch, look at eq 2 where the LI' would only allow a smooth riise of current. As I rises, T increases untill it overcomes Tl and accelerates the mass increasing the back emf which slows the rate of riise of current. At steady state T=Tl+c@' and V0=K2@'+RI  Reply With Quote

4.  Reply With Quote

5. Thank you everyone! This is a big help. 12VDC, 7 Amps, 30 RPM nominal, 80 in-lbs nominal torque, 450 in-lbs starting torque, ~100 Watts - moving a 90 pound load with 8" lever from ground straight up, like dead lift.  Reply With Quote

6. With that kind of load, you may be better off looking at an electro-clutch and leave the motor running. Or if infrequent operational use, start the motor with delay relay, then engage the clutch when it has stabilized.  Reply With Quote

7. 8" lever x 90 lb weight comes to 720 lb-in
How does this square with your starting torque of 450 lb-in?.
Also, to what height is the weight lifted?
What is you power source?

Since ,as you say , this is low duty cycle, why do you care about the torque characteristics?

I'm a bit uncomfortable with using such a large lever for the lift.

If you post a description of the problem, maybe this forum can give you better options.  Reply With Quote

dc motor, electric motor, low duty, running torque, starting torque 