Please correct me if I am wrong

[Dustin beyer]

so I am building a new belt and pulley housing/motor mount and I have all the speeds and torque I could ever need covered by two pulley steps and dialing my vfd between 30-90hrtz. My question is when I make my new setup can I do away with the countershaft design that I’m currently using since I’m not changing speeds via that it’s essentially just reversing the motor direction or is there something magical that creates more torque to a shaft even if it’s going from a 3.5” driving pulley to a “7 driven pulley countershafted driving a 3.5” pulley to a 7” driven spindle?? Will the torque be the same as just a 3.5 to a 7 that’s it. Probably a dumb question but I’d hate to get finished and find out I should have copied the original

 

[extropic]

so I am building a new belt and pulley housing/motor mount and I have all the speeds and torque I could ever need covered by two pulley steps and dialing my vfd between 30-90hrtz. My question is when I make my new setup can I do away with the countershaft design that I’m currently using since I’m not changing speeds via that it’s essentially just reversing the motor direction or is there something magical that creates more torque to a shaft even if it’s going from a 3.5” driving pulley to a “7 driven pulley countershafted driving a 3.5” pulley to a 7” driven spindle?? Will the torque be the same as just a 3.5 to a 7 that’s it. Probably a dumb question but I’d hate to get finished and find out I should have copied the original

I think you described a drive system with a 3.5" pulley on the motor, driving a 7" pulley on the countershaft, and a 3.5" pulley on the countershaft driving a 7" pulley on the spindle. If that's correct, the spindle is turning 25% of the motor RPM. Essentially, that creates a 4X mechanical advantage at the spindle which equals 4x the torque.

If I've misunderstood your situation, please post a picture that shows the pulleys and belts arrangement.

Another factor I found confusing is your reference to using the countershaft to reverse directions.

 

[682bear]

If you eliminate one stage of pulleys in your setup, you will reduce your torque by half...

As it is now, you have a 4x mechanical advantage (or torque multiplication)... if you eliminate the countershaft pulleys, you will only have a 2x advantage, and the spindle will double in speed.

I would keep it 'as is'...

-Bear

 

[Dustin beyer]

I think you described a drive system with a 3.5" pulley on the motor, driving a 7" pulley on the countershaft, and a 3.5" pulley on the countershaft driving a 7" pulley on the spindle. If that's correct, the spindle is turning 25% of the motor RPM. Essentially, that creates a 4X mechanical advantage at the spindle which equals 4x the torque.

If I've misunderstood your situation, please post a picture that shows the pulleys and belts arrangement.

Another factor I found confusing is your reference to using the countershaft to reverse directions.

Ok I was mistaken on my thinking with rotation. But yes let me describe clearer. my motor has a 4 step pulley that drives another 4 step pulley attached to a counter shaft with a 2 step pulley driving my spindle which has obviously a two step pulley on it. Now the only 2 speeds I use are the middle speeds on the 4 step pulleys the 2 speeds on the counter shaft I never change. Basically if you have a 3 inch pulley driving a 7” pulley with a 3” pulley attached to bottom of it and that 3” pulley is driving a 7” pulley Wii it produced the same torque as a motor driving a 3” pulley to a 7” pulley driven aka spindle

 

[Dustin beyer]

Ok I was mistaken on my thinking with rotation. But yes let me describe clearer. my motor has a 4 step pulley that drives another 4 step pulley attached to a counter shaft with a 2 step pulley driving my spindle which has obviously a two step pulley on it. Now the only 2 speeds I use are the middle speeds on the 4 step pulleys the 2 speeds on the counter shaft I never change. Basically if you have a 3 inch pulley driving a 7” pulley with a 3” pulley attached to bottom of it and that 3” pulley is driving a 7” pulley Wii it produced the same torque as a motor driving a 3” pulley to a 7” pulley driven aka spindle

 

[682bear]

No... you will only get half the torque at the spindle...

-Bear

 

[Dustin beyer]

See right now it’s just going 3.5”-7”/3.5” - 7” say I wanted just this speed only all 4 pulleys are exactly the same just opposite of each other wouldn’t it be the same if I eliminated one stage

 

[Dustin beyer]

but alright I’m glad I asked !! just going to copy the same design just a better housing and linear dovetail style pulley bearing take up for a better tension system since both belts are the same length

 

[extropic]

Ok I was mistaken on my thinking with rotation. But yes let me describe clearer. my motor has a 4 step pulley that drives another 4 step pulley attached to a counter shaft with a 2 step pulley driving my spindle which has obviously a two step pulley on it. Now the only 2 speeds I use are the middle speeds on the 4 step pulleys the 2 speeds on the counter shaft I never change. Basically if you have a 3 inch pulley driving a 7” pulley with a 3” pulley attached to bottom of it and that 3” pulley is driving a 7” pulley Wii it produced the same torque as a motor driving a 3” pulley to a 7” pulley driven aka spindle

Your question is still clear as mud to me. In the simplest terms, essentially, when you half the spindle RPM by means of a power transmission system (belts, chains, gears, NOT VFD) you double the torque at the spindle and vice versa

 

[682bear]

Your question is still clear as mud to me. In the simplest terms, essentially, when you half the spindle RPM by means of a power transmission system (belts, chains, gears, NOT VFD) you double the torque at the spindle and vice versa

Basically what he is saying is that he has two levels of speed reduction and he is asking if he eliminates one level, will he still get the same torque at the spindle. The simple answer is no.

A 2:1 reduction will give a 2:1 torque increase. A second 2:1 reduction will result in a 4:1 torque increase. If he added a third 2:1 speed reduction, it would multiply motor torque by 8:1...

However, by eliminating one layer of speed reduction, (which is what he wants to do), he will increase the spindle speed by 2:1 and reduce spindle torque by the same ratio.

This is why I advised him to leave it 'as is'. Even if the vfd can compensate for the doubled speed, the torque will be cut in half permanently... which, if I'm understanding what I read about vfd's, lack of torque at lower rpm's is already an issue...

-Bear