Current control for motor with worm gear

Good day,

so I have a BLDC Motor with a worm gear box connected to a shaft which is used to wind up some rope which has a weight on the end of the rope. I am implementing a current control for the Motor to generate a specific torque. When winding the rope onto the shaft it all works perfect. But when I want to unwinde the rope, the weight which is connected to the rope doesn’t effect the current of the motor because of the self locking mechanism of the worm gear box. So as soon as the load torque and the motor torque turn in the same dircetion I cant sense any change in load, right? Is there any clever workaround?

Thanks and I hope its understandable

Greetings

jacko91:
Good day,

so I have a BLDC Motor with a worm gear box connected to a shaft which is used to wind up some rope which has a weight on the end of the rope. I am implementing a current control for the Motor to generate a specific torque. When winding the rope onto the shaft it all works perfect. But when I want to unwinde the rope, the weight which is connected to the rope doesn't effect the current of the motor because of the self locking mechanism of the worm gear box. So as soon as the load torque and the motor torque turn in the same dircetion I cant sense any change in load, right? Is there any clever workaround?

Thanks and I hope its understandable

Greetings

As I recall, there is no torque, as such, with a worm and gear. What you are measuring is the FRICTION between the worm and the gear. The friction should be the same no matter which way the worm is turned. I suspect your difference is caused by the design of the gear teeth, so they do not properly mate to the shape of the worm. The teeth of the gear must be cut on an angle to match the pitch of the worm.

How is the worm/gear lubricated?

Paul

The torque supplied by the motor, and hence the current draw, will be different depending on whether the weight is raised or lowered.

jremington:
The torque supplied by the motor, and hence the current draw, will be different depending on whether the weight is raised or lowered.

True with a gear train, but no with a worm and gear. Just friction. Weight will increase the friction, but will not change based on direction if the worm/gear are properly designed. That is why they are so efficient.

Paul

no with a worm and gear

Basic physics requires the motor to do different amounts of work, depending on whether the weight is raised or lowered.

If you have ever used a worm gear jack to lift an automobile (I have) you will notice that it is much more difficult to turn the crank to raise the automobile, than to lower it. Try it, it will be fun!

The workaround is to measure the weight or torque another way. The most expensive is to place a torque transducer between the motor and the winding drum.

Depending on what resolution you need, maybe just run the rope past a spring-loaded pulley or straight rod. Tension on the rope will push that rod out of the way of the rope, as the rope wants to make a straight line. Have this trigger a switch. This is often used for situations when you want a certain minimum tension on the rope and you want it to stop winding out when the rope goes slack.

For more precision (and cost) it's going to be simple to measure the weight of the entire motor-gearbox-drum by putting strain gauges on the mounting points. The pull on the rope will equal the pull on the mounts, minus the weight which is constant.

Can you please explain what you are trying to measure when you are lowering the weight. As Paul says, the load during pay-out is attempting to free-fall and the motor current required to permit pay-out is mainly due to that overcoming the self-locking feature of a worm drive. If the worm is a coarse one then the load may free-fall and the motor requires no power, but if the worm is a fine one then the the system locks up and the motor must run simply to permit payout, but the running current will mainly be dependant on mechanical losses rather than a function of the weight suspended. In some cases, once the load starts to fall, if the motor is de-energised, the load will continue to free-fall, demonstrating that the system electrical load is zero.

Paul_KD7HB:
True with a gear train, but no with a worm and gear. Just friction. Weight will increase the friction, but will not change based on direction if the worm/gear are properly designed. That is why they are so efficient.

Paul

I was taught that a worm drive is the equivalent of pushing an object up or down an incline plane, as a screw thread is involved..
If the load torque was always against the direction of worm drive direction, then torque will be the same.
BUT
The load torque is in one case against the worm drive direction in the windup case, and with the worm drive direction in the wind down case.

Why do you need torque control on unwinding?
As the cable tension will be less, I would be more worried about speed?

The maximum torque is when you are winding up.
Very little or no motor torque when stationary, but cable drum torque still present.
Lower motor torque on winding down.

Tom... :slight_smile:

Correct, it is the very simplest of machines. BUT, the matching gear must have teeth to mate with the worm. Cheap devices just place a simple spur gear in mesh with the worm and let it go. That makes a sharp edge of the tooth in contact with the worm. It catches and slides and eventually will wear one or the other out, depending on material. Lubrication won't help much.

Paul

Thanks for the great discussion. Attached is a sketch where the motor winds a rope which is connected to a weight or actuator. The goal is to control the torque the rope is pulled with. So when the actuator moves to the right, the worm gear is selflocked and the motor doesn’t move, because the self locking torque is far beyond the load torque. So on the motor side it is not detected via current, that the rope force increases.