I have a device using a 24V DC motor connected to a 19:1 gear ratio at the output shaft. I simply changed the gear ratio to 29:1, and now, under load, it's draining the battery much quicker. I thought that the 29:1 would increase battery range, not decrease it. What am I missing? Thanks.
DC Motor/Gear Change Question
- Thread starter pc mark
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LarryFine
Master Electrician Electric Contractor Richmond VA
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- Henrico County, VA
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- Electrical Contractor
Was the ratio 19:1 or 1:19? If your change sped up the output shaft, you definitely increased the current under load.
Multi-speed cordless drills and driver users face the same choices, but high speed does increase battery consumption.
Multi-speed cordless drills and driver users face the same choices, but high speed does increase battery consumption.
LarryFine
Master Electrician Electric Contractor Richmond VA
- Location
- Henrico County, VA
- Occupation
- Electrical Contractor
Then, yes, you definitely increased the torque load on the motor. You slowed the motor or sped up the output, either of which requires more power.pc mark said:
It's like starting from a standstill in 10th gear on a 10-speed bike.
gar
Senior Member
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080608-2135 EST
pc mark:
Let us assume your DC motor is run at 2000 RPM. Was the original gear box producing 38,000 RPM at the output and now you are trying for 58,000 RPM, or was the original output shaft 105.3 RPM and you now have 69.0 RPM.
I really assume you are trying to reduce the output speed.
What kind of gear box was the original reducer? Was it a spur gear or worm gear? Is the new one the same type as the original, or did you change from spur gear to worm gear?
Increasing the reduction ratio will increase output torque for a fixed input torque. However, usually it is the load that defines torque, but that does not change the previous statement.
Every time you increase the ratio of a gear box there will be increased losses in the gear box for the same output load torque.
If you have a job that requires a certain amount of energy to perform the task, then increasing the gear box losses will require more input energy. An example of a task that requires a fixed amount of energy is --- raise 100 # 100 ft. How fast you do this has nothing to do with the energy put into raising that weight. Do it in 1 second or 1 hour and the same amount of energy is put into the weight. This requires 10,000 ft-pounds work (energy). Do not confuse ft-lbs of work with #-ft of torque. Note: torque wrench manufacturers incorrectly label their torque wrenches in ft-#.
However, the rate at which you raise that weight may have a lot to do with the input energy to raise the weight. This will be a function of the efficiency of your raising system.
.
pc mark:
Let us assume your DC motor is run at 2000 RPM. Was the original gear box producing 38,000 RPM at the output and now you are trying for 58,000 RPM, or was the original output shaft 105.3 RPM and you now have 69.0 RPM.
I really assume you are trying to reduce the output speed.
What kind of gear box was the original reducer? Was it a spur gear or worm gear? Is the new one the same type as the original, or did you change from spur gear to worm gear?
Increasing the reduction ratio will increase output torque for a fixed input torque. However, usually it is the load that defines torque, but that does not change the previous statement.
Every time you increase the ratio of a gear box there will be increased losses in the gear box for the same output load torque.
If you have a job that requires a certain amount of energy to perform the task, then increasing the gear box losses will require more input energy. An example of a task that requires a fixed amount of energy is --- raise 100 # 100 ft. How fast you do this has nothing to do with the energy put into raising that weight. Do it in 1 second or 1 hour and the same amount of energy is put into the weight. This requires 10,000 ft-pounds work (energy). Do not confuse ft-lbs of work with #-ft of torque. Note: torque wrench manufacturers incorrectly label their torque wrenches in ft-#.
However, the rate at which you raise that weight may have a lot to do with the input energy to raise the weight. This will be a function of the efficiency of your raising system.
.
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