physics question

[b*saint]

Ok thats basically what I was getting at. Because I have to step down 3600 rpm and 16 ft-lbs of torque to 1400 rpm but I have to still be able to drive something that requires 16 ft-lbs of torque. Is this just not possible? Do I have to increase my motor size?

[mrsleeve]

do you have room for belt or gear drive for your 16ft/lbs need. if so just use mechanical advantage of a small pulley on your 3600rpm motor to a larger one on your driven device. When you do that you increase torque as well.

[b*saint]

O wait, I knew that. What am I thinking.... Its if I run into a problem opposite of that.

If I need to use pulleys to double the driving speed yet have enough torque to drive. Say the engine runs at 2000 rpm and I have to drive something at 3600 rpm. Both the motor and the pump or w/e is rated at 16 ft-lb. What to do then?

[newman]

You cannot create energy. I don't feel like typing the equation for the energy balance, but if you need to increase speed and maintain torque, you're fucked.

[mrsleeve]

Step up to a necessary over drive speed from the motor, with secondary gear reduction box to get what is needed in both speed and torque at what ever device you are needing to drive.

But it may just be easier to just step up to a big enough motor to direct drive it.

[newman]

Step up to a necessary speed, with secondary gear reduction to get what is needed in both speed and torque at what ever device you are needing to drive.

But it may just be easier to just step up to a big enough motor to direct drive it.

Energy in has to equal energy out. I don't have my physics book here, so I can't mathematically show you this, but you cannot increase torque without sacrificing speed. I don't care if you have 20 gear reductions and 15 pulleys, you can't make something out of nothing.

You're talking like a gear reduction does something different than a pulley drive system... both do the same thing, a gear system is just stronger and more precise.

[mrsleeve]

^^^^^^^^^^^^^^^^^
yeah In his 1st post on this page it says motor spins at 3600 @14/ftlbs needs 16ftlbs. Simple mechanical advantage will get you that.

but in the 2ed senairo cant you over drive something on the input side to the point where you can use a gear reduction box to get what you need??? but at that point just getting the right motor to direct drive would be easier and better anyway.

I do get what you are saying though

[newman]

^^^^^^^^^^^^^^^^^
yeah In his 1st post on this page it says motor spins at 3600 @14/ftlbs needs 16ftlbs. Simple mechanical advantage will get you that.

but in the 2ed senairo cant you over drive something to the point where you can use a gear reduction box to get what you need??? but at that point just getting the right motor to direct drive would be easier and better anyway.

I do get what you are saying though

Yeah I'm talking about this:

Say the engine runs at 2000 rpm and I have to drive something at 3600 rpm. Both the motor and the pump or w/e is rated at 16 ft-lb. What to do then?

He wants to increase speed and maintain torque, that's not going to happen... well, not in the world as we know it.

but in the 2ed senairo cant you over drive something to the point where you can use a gear reduction box to get what you need???

If you increase the speed, you're going to decrease the torque. If then you need more torque, the only way to do that is to decrease the speed...

[mrsleeve]

gotcha and makes sense.

I know that a gear reduction is the same as pulley's I just miss read the 2ed deal the 1st time, as he had more speed at the motor side than needed at the driven side.

[TwoJ's]

O wait, I knew that. What am I thinking.... Its if I run into a problem opposite of that.

If I need to use pulleys to double the driving speed yet have enough torque to drive. Say the engine runs at 2000 rpm and I have to drive something at 3600 rpm. Both the motor and the pump or w/e is rated at 16 ft-lb. What to do then?

You get a bigger motor. Like Newman said, there must always be an equilibrium of energy. You can have all rotational speed, or all torque, but not both.

Think of it like a bicycle with 21 speeds. If you are in gear 1 you have ridiculous torque, but each turn of the crank barely moves you. If you are in gear 21, you have much less torque, but each spin of the crank will move you considerably.

[b*saint]

Alright so basically everything I thought was right, just I had to run it by people who've been through school for it. Basically Im building a air compressor/generator combo and had to clear up some design issues. Its a 15kw natural gas engine running at 2500 rpm driving a 7.2kw generator head (needs 12kw and 3600 rpm to run @ 60 hz) and a 3kw compressor pump. All controlled through a PLC. When I don't need air (system will have a 60 gal tank), there is a sheave clutch with a 24v coil to disengage the pump. But I have to step up the speed for the generator and step it down for the pump, and this is where my questions stem from.

When I need the equations to calculate this stuff, what section of engineering would pulleys and shit be under? Power transmission?

[TwoJ's]

Mechanical design is the class series that I learned this shit in. That was much more about gears than pulley/chain drives though. Power transmission would be a sub-category I guess. That set of chapters is called "design of a mechanical drive" in my book. I would just look for the equations online and not buy a textbook though, they go very deep in to the subject.

[b*saint]

Ok cool. I guess it really is more of a mech engineering question(s) than physic questions.

[McGyver]

also look into dynamics. you wont have to incorporate any energy into the system, you just use angular/linear velocities/accelerations to solve your problem. in the 2nd scenario, you said they are RATED at 16ft-lb of torque, but do they HAVE to be 16ft-lb of torque? if you are able to sacrifice torque for increased speed then you should be fine.