Thread: Suitability of motors

Can't really help you with a general guide, but I reckon you can get away with a 1/2HP 1440rpm motor for a simple, hand-fed drum sander. Mind you, more power is always nice.

Let's say you want a 4" drum, then that's 1 foot of sandpaper every revolution. I like paper speeds of between 700 & 1200 fpm from my sanders, as they're more 'finger friendly' & less likely to char the wood. So with a 1/2HP 1440rpm motor, gearing it down 2:1 with pulleys would give you 1HP and 720rpm at the roller.

That should do nicely... unless you're also planning on having the motor power a feed mechanism.

The 1/2HP motor should also be OK for a linisher.

A stationary belt-sander would need more power because at any given time you can sand a larger surface area - up to the full area of the platen - or, because the work is hand-held, apply more pressure to the belt. (It may be bad practice to 'force' timber into a sanding belt to try and make it cut faster... but we're all guilty of it at some time or another. ) I suspect that a 1/2hp would constantly stall under such heavy use.

Originally Posted by Skew ChiDAMN!!

Let's say you want a 4" drum, then that's 1 foot of sandpaper every revolution. I like paper speeds of between 700 & 1200 fpm from my sanders, as they're more 'finger friendly' & less likely to char the wood. So with a 1/2HP 1440rpm motor, gearing it down 2:1 with pulleys would give you 1HP and 720rpm at the roller.

Sorry Skew, horsepower relates to the ability to do a certain amount of work in a given time, and is a function of two parameters, Torque and RPM.

Gearing down a motor will increase the torque it can supply, but decrease the speed, gearing it up will reduce the torque available but increase the speed. But either way the equation P= T x R / 5252 holds, where P is power in Horsepower, T is Torque in foot pounds, and R is rotational speed in RPM. So, in an ideal world gearing down 2:1 doubles Torque and halves speed. However in the real world there will always be some power loss in a transmission system due to friction.

What you need to ask yourself is what size work pieces you want to put through these machines.

A 1/2HP motor on a drum sander probably be limited to about 150 mm wide pieces. As a guide the Jet 255mm wide sander uses a 1HP motor, and their 400 mm wide sander uses a 1.5 HP motor. The 400 mm wide CT drum sander uses a 1HP motor but I reckon its underpowered.

1/2HP is typically used with 6" wheels (grinder/wire/buffer) for 8" you would need 3/4 to 1HP

FWIW: The linisher I am making will use a 150 x 1220 mm belt, and the motors I have under consideration (in my shed and ready to R&R) are either a 2 or a 3HP (3 Phase) .

Originally Posted by malb

Sorry Skew, horsepower relates to the ability to do a certain amount of work in a given time, and is a function of two parameters, Torque and RPM.
...
So, in an ideal world gearing down 2:1 doubles Torque and halves speed. However in the real world there will always be some power loss in a transmission system due to friction.

Agreed. But at what's basically just a pulleyulley situation, such losses are pretty much negligible. Once one starts talking about complicated drive trains (eg. lay pulleys or including feed rollers) then one has to consider such losses.

But in a motor/pulleyulley/cylinder situation... well... instead of 1HP being delivered to the final roller it may be only .98HP. I still believe (well.. . know) that that will give acceptable performance for a drum sander, provided one is prepared to 'baby' it.

Of course, if one is expecting the sander to remove 2mm per pass... but then again I don't think that's a realistic expectation for a thicknesser! Not if one wants a good finish, anyway... (2mm for "rough cuts", 1mm or less for final passes. Y'know?)

BobL makes an appropriate point in the msg above. It depends on what you intend to do. For larger machines, HP is always an advantage. But for smaller items, you can get away with less HP, so long as your requirements are small. A 1HP machine (or 1/2HP at the motor) will sand quite happily, provided you feed it at the appropriate rate. Forcing any machine to cut faster than it's willing is always gonna lead to less than satisfactory results. It depends on how much time the operator is willing to take to achieve the same result...

edit: I know I overuse smilies, but for the BBS parser to include ones I don't intend? )

Do some window-shopping on the internet. Usually, the motor power is included in the details about most power wood working machinery. Maybe 1/2Hp is about right, maybe you learn that it's 1+ or nothing.

Originally Posted by Skew ChiDAMN!!

It depends on what you intend to do. For larger machines, HP is always an advantage. But for smaller items, you can get away with less HP, so long as your requirements are small. A 1HP machine (or 1/2HP at the motor) will sand quite happily, provided you feed it at the appropriate rate. )

Sorry again Skew but you are completely missing the point that the gearing the motor will vary the torque available at the expense of changing the speed in the opposite way. i.e. double the torque at half the speed, or double the speed at half the torque, but the power remains the same. It therefore impossible for any machine fitted with a 1/2HP motor to be a 1HP machine.

The only way that any transmission product can change available power is to reduce it via transmission losses, be they 2%, 20%, or 60% for a worm drive system.

Thanks guys it helps when others have the practical experience on these things. Looks like anything less than 1/2 HP may not be useful for much .

Originally Posted by Tiger

Thanks guys it helps when others have the practical experience on these things. Looks like anything less than 1/2 HP may not be useful for much .

I think that RV is on the right track - just check out the power of similar existing machines and try to match it.

Originally Posted by Skew ChiDAMN!!

So with a 1/2HP 1440rpm motor, gearing it down 2:1 with pulleys would give you 1HP and 720rpm at the roller.
.

Originally Posted by malb

Sorry Skew, horsepower relates to the ability to do a certain amount of work in a given time, and is a function of two parameters, Torque and RPM.


Gearing down a motor will increase the torque it can supply, but decrease the speed, gearing it up will reduce the torque available but increase the speed. But either way the equation P= T x R / 5252 holds, where P is power in Horsepower, T is Torque in foot pounds, and R is rotational speed in RPM. So, in an ideal world gearing down 2:1 doubles Torque and halves speed. However in the real world there will always be some power loss in a transmission system due to friction.

But if the energy source can only supply at a fixed quanta, and that quanta is delivered at double the rotational speed required then gearing it down 2:1 is correct ?

At 2:1 reduction full capacity is delivered at the rotational-workface-face or draw bar. Without reduction 50% is lost in increased entropy. If only 720rpm is required then only a 1/2 the energy is usable and 1/2 is lost ?

So 1/4:1/2 hp or a 1/2:1 hp, its the correct physics solution regardless of the mathematical semantics used. I think thats what Tiger wanted to know.

greg

Originally Posted by Mutawintji

But if the energy source can only supply at a fixed quanta, and that quanta is delivered at double the rotational speed required then gearing it down 2:1 is correct ?

At 2:1 reduction full capacity is delivered at the rotational-workface-face or draw bar. Without reduction 50% is lost in increased entropy. If only 720rpm is required then only a 1/2 the energy is usable and 1/2 is lost ?

So 1/4:1/2 hp or a 1/2:1 hp, its the correct physics solution regardless of the mathematical semantics used. I think thats what Tiger wanted to know.

greg

Speed has little to do with. The energy is being delivered "per unit time (not per unit speed)" irrespective of the rotational speed.

If there are no frictional losses, doubling or halving the speed does not affect the energy transfer. Energy and power (ability to perform work) remain constant and torque will change accordingly.

Yes ... I agree, no problem at all.

But if it is being delivered at 1440rpm but can only be utilised at 720rpm then the increase in entropy is frictional losses plus the inability to utilise 50% of the the available energy.

A 2:1 reduction makes this lost energy available at the drawbar.


The energy transfer is constant re the first law of thermodynamics, and as you point out. But the second law determines the rise in entropy. Skew's solution addresses this, tho the term horsepower used was a generalisation, the physics is correct.

no offence meant ... Greg