[SteveS]

Hope this satisfies CNC since my table saw breaks down material for my CNC. RPM ratios are easy and many calculators on the net, horsepower after a ratio is not so good.

I got a 1950 Delta Table-saw with the original motor replaced by the prior owner(s). Cutting performance (speed) is poor. The original saw came with either a 1/2 hp or 3/4 hp motor in 1950 era. The motor on it is a Leeson TEFC at 1/2 hp.

From another thread there was an answer that I want to confirm I am reading correctly.

Torque increases as a direct ratio of the reduction, and vice-versa, also inertia ratio increase by the square of the reduction.
HP is a product of torque and rpm (time).
IOW, if your torque increases (by reduction) and rpm decreases by the same ratio then HP remains constant.
Pulleys the same diameter, then no change in either torque/hp.
Simply put, reduction allows you to maintain HP at a lower rpm.

My situation is that the motor pulley is larger than the arbor (driven) pulley for a ratio of 2:1. To me this means I am reducing torque / hp by multiplying the rpm. Motor is 1,750 rpm and arbor is 3,500 rpm. Am I correct in the motor horsepower will be less at the blade? Were older era (1950) motors hp underrated as opposed to current (2000+) motors? I've installed new bearing in the arbor.

Steve.

[109jb]

Horsepower will be the same at the blade as the other answer you got. Yes torque will reduce, but the rpm will increase and will offset the torque decrease. The formula for hp is hp= torque * rpm / 5252. So, with a 2:1 pulley ratio you have twice the rpm and 1/2 the torque. 2 * 1/2 = 1. The horsepower at the blade will be the same as at the motor. This does not take into account the losses from having the belt drive though, but the reduction from that will be about the same no matter what ratio you have.

[SteveS]

Thanks, that is probably why I couldn't really find anything on the net - it isn't significant.

[mactec54]

My situation is that the motor pulley is larger than the arbor (driven) pulley for a ratio of 2:1. To me this means I am reducing torque / hp by multiplying the rpm. Motor is 1,750 rpm and arbor is 3,500 rpm. Am I correct in the motor horsepower will be less at the blade?

Yes you are correct, in your case the Hp and Torque received at the arbor is reduced by the ratio

[CitizenOfDreams]

Yes you are correct, in your case the Hp and Torque received at the arbor is reduced by the ratio

HP reduced by the pulley ratio? That would be a gross violation of the energy conservation law.

[109jb]

Yes you are correct, in your case the Hp and Torque received at the arbor is reduced by the ratio

Torque is reduced but not horsepower.

[mactec54]

Torque is reduced but not horsepower.

This is very elementary level stuff guys, just rethink this out, his correct ratio is 1:2 not 2:1 the motor is turning 1,750 RPM the arbor is turning 3,500 RPM the Arbor will be half the HP rating as what the motor is, in this case the motor Hp does not change

If the ratio was the other way round and the motor had 2 RPM and the arbor 1 RPM the Hp at the arbor will be double the motor rated HP

[mactec54]

HP reduced by the pulley ratio? That would be a gross violation of the energy conservation law.

Try to have an understanding of what is happening with his gearing, Ratio this is elementary level stuff

[mike_Kilroy]

Try to have an understanding of what is happening with his gearing, Ratio this is elementary level stuff

SteveS, ignore this person's posts. As many others already stated, his "HP changes" is totally wrong.

HP does not change as the many others have told and shown you why.

mactec54, again, you are posting nonsense that confuses some folks. PLEASE STOP. I will admit you may be a stepper motor wizard - PLEASE REFRAIN FROM POSTING ON ANY NON-STEPPER MOTOR TOPICS - YOU DO CNCZONE A MAJOR DISSERVICE!

[mactec54]

SteveS, ignore this person's posts. As many others already stated, his "HP changes" is totally wrong.

HP does not change as the many others have told and shown you why.

mactec54, again, you are posting nonsense that confuses some folks. PLEASE STOP. I will admit you may be a stepper motor wizard - PLEASE REFRAIN FROM POSTING ON ANY NON-STEPPER MOTOR TOPICS - YOU DO CNCZONE A MAJOR DISSERVICE!

You are making a FOOL of yourself again read the post, of course the motor HP can't change, AT THE MOTOR NOBODY SAID IT DOES, this is elementary stuff

[109jb]

If the motor is 2 hp, then the horsepower AT THE ARBOR is also 2 hp.

The ONLY change in horsepower is from efficiency losses of the belt drive, not from the gear ratio.

[vmax549]

SImply do the math (;-) The results are self explanatory.

1:1 1800 * 1.46 / 5250 = .5 hp
1:2 3600 * .73 / 5250 = .5 hp

Just a thought, (;-) TP

[mactec54]

If the motor is 2 hp, then the horsepower AT THE ARBOR is also 2 hp.

The ONLY change in horsepower is from efficiency losses of the belt drive, not from the gear ratio.

If the ratio is 1:1 yes that would be correct

[mactec54]

SImply do the math (;-) The results are self explanatory.

1:1 1800 * 1.46 / 5250 = .5 hp
1:2 3600 * .73 / 5250 = .5 hp

Just a thought, (;-) TP

At the motor that is correct, the motor HP can never change, it is what ever the motor HP is

[mike_Kilroy]

and I quote you:

1) "the Arbor will be half the HP rating as what the motor is"

2) "the Hp at the arbor will be double the motor rated HP"

WHO is making themselves a fool? Now if you are proposing some sort of perpetual motion mechanism to magically double a motor's HP, please by all means explain the details of this device to us: we are all ears!

[109jb]

If the ratio is 1:1 yes that would be correct

This is very elementary level stuff guys

Yes it is elementary and you are totally and absolutely wrong. You are the one making a fool of yourself. The horsepower at the arbor cannot change. Torque can and does, but the horsepower does not and the ratio doesn't matter. The horsepower at the arbor will ALWAYS be the same as at the motor.

Say you have a 1750 rpm motor that puts out 6 ft*lb of torque. That makes the motor a 2 hp motor because:

HP=RPM * torque / 5252 = 1750 * 6 / 5252 = 2 HP

Now put it in a machine with a 2:1 pulley ratio with the big pulley at the motor. The torque at the arbor will be reduced by 1/2, so torque at the arbor is now 6 * 0.5 = 3 ft*lb, but the RPM will be double so the RPM at the arbor is 1750 * 2 = 3500 rpm.

Now calculate the HP AT THE ARBOR using the torque at the arbor and the RPM at the arbor:

HP = RPM * torque / 5252 = 3500 * 3 / 5252, which magically calculates to 2 HP. no change at the arbor whatsoever.

Do the above calculations for whatever pulley ratio you like, the answer is always the same. The horsepower at the arbor is always the same as the horsepower at the motor regardless of the ratio.

[mactec54]

WHO is making themselves a fool? Now if you are proposing some sort of perpetual motion mechanism to magically double a motor's HP, please by all means explain the details of this device to us: we are all ears!

I guess you are trying to be funny, or trying to make something out of nothing, In actual fact if you use the service factor of 1.3 for these Ratios you will be closer to the what the output Arbor would have

That is why people use gearing 1 Hp at 2:1 ratio what do you have, at the driven output shaft, that is the reason they make Gear boxes so they can use smaller HP motors to move a heavy load

In his case it is the opposite which he also came to the correct answer

[mactec54]

Yes it is elementary and you are totally and absolutely wrong. You are the one making a fool of yourself. The horsepower at the arbor cannot change. Torque can and does, but the horsepower does not and the ratio doesn't matter. The horsepower at the arbor will ALWAYS be the same as at the motor.

Say you have a 1750 rpm motor that puts out 6 ft*lb of torque. That makes the motor a 2 hp motor because:

HP=RPM * torque / 5252 = 1750 * 6 / 5252 = 2 HP

Now put it in a machine with a 2:1 pulley ratio with the big pulley at the motor. The torque at the arbor will be reduced by 1/2, so torque at the arbor is now 6 * 0.5 = 3 ft*lb, but the RPM will be double so the RPM at the arbor is 1750 * 2 = 3500 rpm.

Now calculate the HP AT THE ARBOR using the torque at the arbor and the RPM at the arbor:

HP = RPM * torque / 5252 = 3500 * 3 / 5252, which magically calculates to 2 HP. no change at the arbor whatsoever.

Do the above calculations for whatever pulley ratio you like, the answer is always the same. The horsepower at the arbor is always the same as the horsepower at the motor regardless of the ratio.

He is not using a 2:1 Ratio his ratio is 1:2 , his motor is 1 the Arbor is 2, you have to divide by the service factor of 1.3 to get the rated power at the Arbor ( the Driven), you are not calculating this part you are only calculating the motor which does not change

[109jb]

He is not using a 2:1 Ratio his ratio is 1:2 , his motor is 1 the Arbor is 2, you have to divide by the service factor of 1.3 to get the rated power at the Arbor ( the Driven), you are not calculating this part you are only calculating the motor which does not change

OK. for the stubborn crowd grasping at straws lets do several calculations using the same motor (2 hp, 1750 RPM, 6 ft*lb at the motor)

ratio 1:2 with motor pulley 2 times the diameter of the arbor. --->>> RPM is double (3500) torque is 1/2 (3) --->>> HP = 3500 * 3 / 5252 = 2 HP at arbor
ratio 2:1 with motor pulley 1/2 the diameter of the arbor --->>> RPM is 1/2 (875) torque is double (12) ---->>> HP =875 * 12 / 5252 = 2 HP at arbor
ratio 100:1 with motor pulley 1/100 the diameter of the arbor --->>> RPM is 1/100th (17.5) Torque is 100 times more (600) --->>> 17.5 * 600 / 5252 = 2 HP
ratio 1:100 with motor pulley 100 times larger than the arbor pulley --->> RPM is 100 times as much (175,000), torque is 1/100th (0.06) --->>> 175000 * 0.06 /5252 = 2 HP
ratio 3:1 with motor pulley 1/3 the diameter of the arbor pulley --->>> RPM is 1/3 (583.333) torque is tripled (18) --->>> 583.333 * 18 / 5252 = 2 HP
ratio 1:3 with motor pulley 3 times the diameter of the arbor pulley --->>> RPM is tripled (5250), torque is 1/3 (2) ---->>> 5250 * 2 / 5252 = 2 hp

Service factor has nothing to do with this and is grasping at straws. The service factor is how much additional power the motor or belts can provide for short bursts, so is not at all related to these calculations.

You are still wrong, but based on previous threads, I am sure you will never admit it.




Service factor has nothing to do with it. You are grasping at straws.

[mike_Kilroy]

Service factor has nothing to do with it. You are grasping at straws.

Mach, perhaps since SF has nothing to do with the price of rice in China, you want to consider power factor? Perhaps PF is the answer to your confusion?

I do feel a tad sorry for your confusion though... A basic flaw in your understanding of physics has led you to believe your statement "that is the reason they make Gear boxes so they can use smaller HP motors to move a heavy load"

the correct sentence is "that is the reason they make Gear boxes so they can use smaller TORQUE motors to move a heavy load"

Trying to help you using your own mindset, a lot of motors can go FASTER than the load requires. So "they" invented gearboxes to use that "free" speed. If the torque rating of the motor is the same at this higher speed, then adding a reduction gearbox to use this free higher speed does indeed let you pick a lower torque rated motor - but if you do the math as has been shown many times here, the HP is the same at this reduced torque and higher speed!