Need a little help on drill press speeds, I need to slow mine down to about
150 - 200 RPM, I suppose if I put a larger single v-pulley on the front it will do that, does anyone know the math formula or what size wheel I should buy? Motor is 1700 RPM, 1/3 HP and current speed chart is as the following:2965
2130 1460 965 650 smallest drive pulley here, to largest front driven pulleyAny help is much appreciated,
Alex
"AAvK" wrote: (clip)does anyone know the math formula (clip) ^^^^^^^^^^^^^ Multiply the motor RPM by the motor pulley diameter, and divide by the driven pulley diameter.
You are trying to reduce your minimum speed by a factor of 3.5 to 4--so if you decrease your motor pulley size, and increase your driven pulley size so the product of the two changes is in that range. you will get it. IOW, if you double the driven pulley diameter and halve the motor pulley diameter, you will reduce the speed by a factor of 4. But these numbers may not be physically attainable. Doubling the driven pulley may take more space than you have. Halving the motor pulley may call for a size that does not exist.
You may have to resort to a jack shaft.
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
Now That's an amazing bit of math work I must say! I am quite impressed. I have ADHD myself so I could never get educated along the lines of numbers, I only have it at the most rudimentary level. I envy you wholesomely! With ADHD I can sit in a classroom and have absolutely Xero in my mind for what is going on within a glorious spectrum of black and white television snow and the noise to go along with it! Oh well.
Looks like 17" is too big, there isn't that much room, it would run into the drive pulley, I think I need to come up with another idea, Leo's idea of a jack shaft or put a center pulley off to the outside where there is room. Thank you* very much for the calculation. Incidentally I did find this website, you might toy with it, it's a java calculator for pulley speeds, you could experiment comparing your results with it's results:
Alex
Mechanics aside another approach to this is to install a shaded pole or universal type motor and you could install a speed controller on the press to reduce the speed down. Be forewarned you will lose horsepower doing this.
I wouldn't mind having a powerful DC motor with a speed control, unfortunately such a set up is highly expensive, even if used.
Alex
It's not that bad, because drill presses aren't hugely powerful. I've a number of variable speed devices here (coil winders, ball mills) that use small DC motors from lab stirrers. One problem is that the simpler circuits tend to limit torque badly at low speeds.
If you have a three-phase spindle moulder (large shaper) then the economic way to convert it to single phase is often to re-motor the main drive but keep the 3-phase motor on the power feed, using a VFD (variable frequency drive) to do phase conversion. This small feed motor is a funny size and hard to replace, it's a low power so the VFD isn't too expensive, and you gain good speed control with hightorque, right down to a crawl - useful for a powered feed.
I don't think you can speed control a shaded pole motor though ? I'd always thought "shaded pole" motors were self-starting synchronous motors. Low power and accurately speed-locked to line frequency. I'd be surprised to see one of those in a machine tool, let alone try to speed control it. Does it mean something else in the USA ?
Andy,
I didn't completely follow your post but here is a shot at an answer. I recently purchased a 1/4 hp DC motor for an industrial sewing machine from the local sewing machine supply company that sells both home and industrial machines. (I do boat seats).
It has two ranges High and Low, it is completely variable throughout it's ranges. It has a place for the treadle pull rod to attach that control the motor speed. There is a variable limit adjustment for setting the maximum motor speed. There is a switch to select motor rotation direction.
While sewing marine vinyl I find it necessary to set the speed limit quite slow for use when going around tight corners. I find no loss of torque. Being designed for industrial use it can run for long periods without overheating concerns
The cost was $160.00, for my application that was a "give away" !
If you don't find a dealer that has such a motor, I got mine at Missouri Sewing Machine Company in Kansas City, MO, USA
Don Dando
This link outlines the formula if you read through the posts.
The speed ratio is (diameter of drill pulley)/(diameter of motor pulley).
Since you do not state the diameter of either pulley then a value cannot be calculated.
I expect you will have a difficult time finding a set of pulleys which will give this speed reduction. The motor end pulley will need to be quite small, and may slip if too small. You will likely not be able to fit the drill pulley size needed within the existing case.
Dave Paine.
I WAS about to suggest a Router Speed Controller like this one
Removing the belt and running with 'no load' didn't make a difference.
I use one for my Bosch 1617 router without any problems.
Anyone know if one of these can be used in this case with some adjusstment or is the problem (stalling) just due to the different types of motors (router vs DP)?
Ron
Speed controls like that can only be used on universal (brush) motors, like in routers, drills, sanders, etc. Stationary tool induction motors will burn out instead of slowing down.
GTO(John)
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.
Routers use brush type (universal) motors. You may change their speed by changing available voltage/current. They will typically run on AC or DC - making them *universal*.
Drill presses are almost always AC INDUCTION motors. No brushes, their speed is determined SOLEY by the number of poles in the motor and the frequency of the AC power. (i.e. 60hz) This is a fixed relationship. Some motors are wound with multiple pole taps for speed control, for instance, your furnace blower motor and few washing machine motors.
So, no, you cannot vary the speed of an induction motor with a light dimmer. You CAN succeed, however, in eventually destroying the motor.
FWIW,
Greg G.
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
My server did not have the posted question, or all of the replies, but I find this online calculator useful quite often.
Yes I found that one, but I have a problem with it in that asks for the "drive pulley ratio" as a requirement, no idea what to input there...? And if I can't do it, no viable calculation. This one seems to make more sense:
Alex
"drive pulley ratio"
viable calculation.
Measure the width of the pulley's. That will give you your ratio. If the driving pulley is 3" and the driven pulley is 2" then you have a 3 to 2 ratio. If the driving pulley is 2" and the driven pulley is 3" then you have a 2 to 3 ratio. If the driving pulley is 4" and the driven one is 2" then it's a 2 to 1 ratio. It's simple fractions
"drive pulley ratio"
viable calculation.
As someone else already said, the drive pulley ratio between the drive and driven pulleys diameters, this ratio basically says: the drive pulley is "X" times bigger than the driven pulley. If for instance the drive pulley is 4" and the driven one is 2", the ratio is 4 / 2 = 2 meaning the drive pulley is 2 times bigger than the driven pulley. If the drive pulley were 4" but the driven were 8", then the ratio would be 4 / 8 = 0.5 meaning the drive pulley is 0.5 times bigger than the driven one, which in turn means the drive pulley is half the size of the driven one.
BTW, I took a look to the calculator
Guillermo
Router speed controllers are unsuitable for drill presses. Universal motors, those with brushes, work on AC and DC and smoothly reduce RPM's with a reduction in the supplied voltage. AC motors such aa capacitor motors or split phase motors commonly used on drill presses determine RPM by the applied frequency. RPMs drop off precipitously when a certain voltage is reached but this is not a design feature of the motor. This is approaching failure. Motor winding temp is going to go up and power provided is inadequate for much. Don't use a router speed controller on a drill press. You can replace the DP motor with something like a router motor and it will respond to the speed controller. Only problem is the number of belt speed reductions needed from the usual 20 - 30 thousand free running RPM of these motors can be inconvenient to set up. The other solution is a variable frequency drive. Commercial variable speed drill presses cost more. There's a reason. Ain't cheap to provide. Variable sheave "V" pulleys are probably the most practical in the speed and power range desired.
bob g.
R> I WAS about to suggest a Router Speed Controller like this one
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