1150rpm vs 1750 rpm- any significant difference in torque?

[Doc Nickel]

I have an old 'camelback' drill I've been"working on" (more like "storing" ) for a while, and I was planning on using a typical 1750 RPM motor.

Well, to get proper belt speeds, that requires a big 12" pulley, to be driven by a tiny pulley on the motor. However, recently ran across an 1150 RPM 1HP 3-phase motor- I'd already been considering using a 3-phase for the variable speed and soft-start, and an 1150 would let me drop that 12" pulley down to a less ostentatious 8" one.

The only question is, is there any significant difference in torque from the slower motor? I'd assume nothing major- it's still a 1HP, and presumably 1HP is 1HP (unless it's on the label of an import bench grinder or a typical shop-type wet-dry vacuum. )

Any reason I couldn't use the 1150 motor, or that it might not work as well?

Doc.

[MattiJ]

1150 rpm motor has more torque if they are both similarly rated for 1 HP.

[wierdscience]

Yes,more torque,see here-
http://www.engineeringtoolbox.com/el...pm-d_1503.html

[darryl]

If the ratings are realistic, then the slower motor will have more torque to reach the same power level. If that lets you use a larger pulley on the motor, then that will help to deliver that torque.

My lathe came with a 1150 rpm motor. Didn't seem to be an issue.

[Forrest Addy]

Horsepower is a unit of power - work per unit of time. Torque and RPM are inversely proportional per unit of HP. 1 HP = 33,000 ft lb per minute and 1 ft lb at 33,000 ft per minute and similarly for all values in between.

Your 1150 RPM motor at 1 HP will produce 4.56 ft lb torque at full load amps and the 1750 RPM motor produces 3 ft lb nearly exactly. A 4.6" dia motor pulley on the 1150 RPM motor develops the same belt tension as a 3" dia pulley on the 1750 RPM motor.

Torque isn't power. It's a force acting on a moment arm. Torque is leverage - twisting effort - applied to rotating shafts.

If both motors are 1 HP and you need to drive a 12" pulley slowly you might need a jack shaft to effect the total needed RPM reduction if you are to avoid a tiny motor pulley and consequent slippage. Work backwards from the desired spindle speeds and built-in gear/pulley ratios and determine what RPM you need to drive the 12" input pulley. That and the selected motor RPM gives you a ratio you can work with. All it takes is a little math to work the ratios.

[CalM]

A VFD with appropriate setting will cover all the bases. NFT

Horsepower is a unit of power - work per unit of time. Torque and RPM are inversely proportional per unit of HP. 1 HP = 33,00 ft lb per minute and 1 ft lb at 33,000 ft per minute and similar for all values in between.

Your 1150 RPM motor at 1 HP will produce 4.56 ft ln torqule at full load amps and the 1750 RPM motor produces 3 ft lb nearly exactly. A 4.6" dia motor pulley on the 1150 RPM motor develops the same belt tension as a 3" dia pulley onthe 1750 RPM motor.

Torque isn't power. It's a force acting on a moment arm. You might say, torque is equivalent to leverage.

If both motors are 1 HP and you need to drive a 12" pulley slowly you might need a jack shaft to effect the total needed RPM reduction. Work backwards from the desired spindle speeds and built-in gear/pulley ratios and determine what RPM you need to drive the 12" input pulley. That and the selected motor RPM gives you a ratio you can work with. All it takes is a little math and working the ratios.

..........

[PStechPaul]

Yes, most VFDs will supply up to at least 200 Hz and usually 400 Hz, so you can overclock the 1150 RPM motor to get 3600 RPM or higher. It will experience "field weakening" when the V/F curve is limited by the supply voltage, but you will still be able to get as much torque as the 1750 RPM motor, and more at rated speed. If you have a 480V supply and a 480V VFD, you can maintain the torque of the 1150 RPM motor up to 2300 RPM (with twice the rated HP), and then constant HP with reduced torque at higher RPM. I have an eight pole (900 RPM) 1 HP motor that is about the same size as a four pole (1800 RPM) 1.5 HP motor and a two pole (3600 RPM) 2 HP motor.

[Forrest Addy]

Yes, most VFDs will supply up to at least 200 Hz and usually 400 Hz, so you can overclock the 1150 RPM motor to get 3600 RPM or higher. It will experience "field weakening" when the V/F curve is limited by the supply voltage, but you will still be able to get as much torque as the 1750 RPM motor, and more at rated speed. If you have a 480V supply and a 480V VFD, you can maintain the torque of the 1150 RPM motor up to 2300 RPM (with twice the rated HP), and then constant HP with reduced torque at higher RPM. I have an eight pole (900 RPM) 1 HP motor that is about the same size as a four pole (1800 RPM) 1.5 HP motor and a two pole (3600 RPM) 2 HP motor.

Excellent; succinctly put

[flylo]

I have 12-New 3HP 1150 rpm if you need one & that rpm is used on many good drill presses. They weigh 115# ea.

[winchman]

A 115# motor might break the camelback's back.

[velocette]

Hi
Horse power come from horses so I Think of it this way. Two Pole Two Legs Four Pole Four Legs Six Pole Six Legs.
Couple the horses to a Gin Gang and the Two Legs will need The largest diameter on a diameter of 8 units.Four Legs on a diameter of 6 units. Six Legs on diameter of 2 units.
The centre shaft will turn at the same number of RPM and deliver the same amount of torque.

Eric

[Doc Nickel]

If that lets you use a larger pulley on the motor, then that will help to deliver that torque.

-Actually, the idea is a smaller pulley. What I have is this setup:



I made a riser base for the whole machine, both to add adjustable leveling feet and to bring it up to where it's more comfortable to use, and added a riser structure for the drive cone pulley. That will allow me to put a motor in the center, under the pulley, rather than the common method of hanging it out the side somewhere.

It's all pretty well set up, the only drawback being that to get a moderate speed on the drive pulley- not too fast, as it is, of course, a plain bearing antique- I have to use a 12" diameter pulley on the cone pulley shaft. It's not a huge issue, just appearances really, but I was going to need to pick up a decent small 3-phase motor anyway, and happened across this 1150 RPM unit.

A quick calc suggested that an 8" pulley would give me almost exactly the same speeds at the cone pulley, and that nearly matches the OD of the biggest step of the cone.

Which, now that I think about it, could very well let me use the original fixed drive pulley, so I could make the motor-to-jackshaft belt a flat one as well. There's an interesting thought...

Doc.

[Yondering]

It is true that the 1150 rpm motor will produce more torque at 1150 than the 1750 rpm motor does at 1750.

However - torque at the motor isn't really what matters, we care about torque at the chuck, right? If you're driving the chuck to the same rpm, you'll have the same torque at the chuck with either motor (ignoring power losses through the belt system).

[Paul Alciatore]

Exactly right, neglecting frictional losses in the pulleys and belts. But those losses should be similar, if not equal in a simple situation like this.

If you want the math, then for torque in pound-feet,

HP = Torque X RPM / 5252

What this implies is that there is a ratio or proportion between HP and Torque and the factor for converting them is RPM / 5252. So if you use a pulley or gear system to bring the rotational speed to the same for the two 1 HP motors, then the torque will be the same, again, neglecting losses in the RPM conversion. Since you need to use pulleys in either case, the two motors should be almost exactly equivalent.

It is true that the 1150 rpm motor will produce more torque at 1150 than the 1750 rpm motor does at 1750.

However - torque at the motor isn't really what matters, we care about torque at the chuck, right? If you're driving the chuck to the same rpm, you'll have the same torque at the chuck with either motor (ignoring power losses through the belt system).