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How to calculate pulley/rpm problems is pretty trivial and has been beat
to death. What's the most practical way to get yourself a low
speed/variable speed drill press at minimum effort and expense? Anybody
got a prime example lying around?

bob g.

Charles T. Garrett wrote:

Same way you get that for that on a lathe. Pick up a treadmill VS DC motor for free at the recycling barn, or for cheap by paying a surplus outfit or buying the used treadmill in the classified ads. Yank motor and controller, slap into drill press, vary torque by belt setting and speed by control knob.

Yeah there were some great replies and good help! Guillermo replied the following (just in case you missed it), there's no way I can do what he can do with math: ------------ You have been given the answer already, now, if you are interested in knowing where the answer comes from:

if "D" is the driver pulley and "d" the driven pulley, "S" the motor speed and "s" the driven pulley's axis speed:

The driver pulley "D" perimeter is = Pi x D and since it is rotating at a speed "S", it will travel a distance = Pi x D X S every minute The driven pulley perimeter is = Pi x d and you want it rotating at a speed "s", so it will travel a distance = Pi x d x s every minute

The distance traveled by the driver and driven pulleys must be the same (assuming no slippage), therefore:

Pi x D x S = Pi x d x s dividing for sides by Pi: D x S = d x s dividing both sides by "s": (D x S)* / s = d
So the diameter of the driven pulley is equal to the diameter of the driver
pulley multiplied by the speed of the motor and all that divided by the
intended speed of the driven pulley.
Something else you can notice is that:
S / *s = d* / D
in your case (selecting 200rpm):
1700 / *200 = d* / D or 8.5 = d / *D
which means that the ratio driven pulley to driver pulley is a factor = 8.5,
in other words, that the driven pulley has to be 8.5 times the size of the
driver pulley. If you had a 2" diameter driver pulley, in order to obtain
200rpm you would have to use a 17" driven pulley, that is a big pulley!!

Guillermo

#### Site Timeline

- posted on November 1, 2004, 1:09 am

bob g.

Charles T. Garrett wrote:

- posted on November 1, 2004, 2:07 am

Same way you get that for that on a lathe. Pick up a treadmill VS DC motor for free at the recycling barn, or for cheap by paying a surplus outfit or buying the used treadmill in the classified ads. Yank motor and controller, slap into drill press, vary torque by belt setting and speed by control knob.

--

Cats, Coffee, Chocolate...vices to live by

Cats, Coffee, Chocolate...vices to live by

- posted on October 30, 2004, 7:57 pm

AAvK wrote:

Alex...

I haven't seen a formula - but I probably missed some of the responses, too. It looks to me like one formula might be:

motor RPM x drive pully diam = spindle RPM x spindle pulley diam

You'll need to measure your drive pulley before you can do the calculation.

Alex...

I haven't seen a formula - but I probably missed some of the responses, too. It looks to me like one formula might be:

motor RPM x drive pully diam = spindle RPM x spindle pulley diam

You'll need to measure your drive pulley before you can do the calculation.

--

Morris Dovey

DeSoto, Iowa USA

Morris Dovey

DeSoto, Iowa USA

Click to see the full signature.

- posted on October 31, 2004, 1:48 am

Yeah there were some great replies and good help! Guillermo replied the following (just in case you missed it), there's no way I can do what he can do with math: ------------ You have been given the answer already, now, if you are interested in knowing where the answer comes from:

if "D" is the driver pulley and "d" the driven pulley, "S" the motor speed and "s" the driven pulley's axis speed:

The driver pulley "D" perimeter is = Pi x D and since it is rotating at a speed "S", it will travel a distance = Pi x D X S every minute The driven pulley perimeter is = Pi x d and you want it rotating at a speed "s", so it will travel a distance = Pi x d x s every minute

The distance traveled by the driver and driven pulleys must be the same (assuming no slippage), therefore:

Pi x D x S = Pi x d x s dividing for sides by Pi: D x S = d x s dividing both sides by "s": (D x S)

Guillermo

- posted on November 1, 2004, 4:49 am

AAvK wrote:

[Repost - Qwest ate original response]

Alex...

I haven't seen a formula - but I probably missed some of the responses, too. It looks to me like one formula might be:

motor RPM x drive pully diam = spindle RPM x spindle pulley diam

You'll need to measure your drive pulley before you can do the calculation.

[Repost - Qwest ate original response]

Alex...

I haven't seen a formula - but I probably missed some of the responses, too. It looks to me like one formula might be:

motor RPM x drive pully diam = spindle RPM x spindle pulley diam

You'll need to measure your drive pulley before you can do the calculation.

--

Morris Dovey

DeSoto, Iowa USA

Morris Dovey

DeSoto, Iowa USA

Click to see the full signature.

- posted on November 2, 2004, 8:29 pm

wrote:

I missed the original posting. However you are right. The OP needs to measuer pully size [diameter]. The ratio of speeds is the ratio of the diameters. I think of two wheels, one 1 unit, the other 2 in diameter. As the belt moves around the larger, which turns once, that same distance is covered by the smaller which must turn twice, and vice versa. To slow down, the larger must be on the spindle, the smaller on the motor.

I don't know the drill setup, but most these days are fixed sets. If there is room to change pulleys, just divide 1700 by 200 to get 8.5, and the front pulley must be that much bigger than the rear.

I missed the original posting. However you are right. The OP needs to measuer pully size [diameter]. The ratio of speeds is the ratio of the diameters. I think of two wheels, one 1 unit, the other 2 in diameter. As the belt moves around the larger, which turns once, that same distance is covered by the smaller which must turn twice, and vice versa. To slow down, the larger must be on the spindle, the smaller on the motor.

I don't know the drill setup, but most these days are fixed sets. If there is room to change pulleys, just divide 1700 by 200 to get 8.5, and the front pulley must be that much bigger than the rear.

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