Starting Torque vs Running Torque in DC Motors

[dw-1]

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!

[PinkertonD]

Welcome and here's the standard first reply.

More information please. Aaarrghhhh!

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.

[zeke]

1) T=K1*I

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

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

V0 impressed voltage
I=current
T= torque delivered by motor
Tl = load torque
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

[Kelly_Bramble]

More equations and stuff--->

http://www.engineersedge.com/motors/...tric_motor.htm

[dw-1]

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.

[PinkertonD]

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.

[zeke]

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.