View Full Version : step pulley sizing

I'm fixing an old Walker-Turner drill press model #1113 serial #0000015 and some one replaced the drive step pulley with a single pulley. The ? how do you figure the drive pulley steps (5) if you know the driven pulley sizes and want to keep from moving the motor back and forth much. In this case would like to have a 2" pulley as the smallest one driving the large 8.4 step on the driven and then keep the center to center distance the same for the rest. I hope this is a little clearer than mud. Thanks for any help James

John Stevenson

09-09-2005, 08:16 PM

Need to know the diameters of the other 4 sheaves on the 8.4 driven pulley.

Is this flat belt or vee?

If vee what section

John S.

Wirecutter

09-09-2005, 08:23 PM

Do you have at least one of the step pulleys? I wasn't sure from your posting. If you don't and would be okay with 4 steps, I can measure one of the pulleys on my Bridgeport.

Otherwise, it's like this:

The step pulley arrangement is basically two cones oriented opposite each other. For each "pair" of pulley steps, you want the following distance to be the same, which represents the "unwound length" of the V-belt you'll want to use:

(pi * (radius of a step of pulley A )) +

(pi * (radius of the corresponding step of pulley B)) +

(centerline distance between the driving and the driven shafts) =

The belt's "unwound length".

or if

D = belt's unwound length

rA = radius of pulley A

rB = radius of pulley B

cD = distance between shaft centerlines

(pi * rA) + (pi * rB) + cD = D

For each step, you want to keep this length (D) the same so you don't need to adjust the motor back and forth.

You could also just divide up the difference in size from your smallest to largest pulley in 5 steps and seek a pair of pulleys with those sizes. (Ok actually, it's just the 3 steps in between.) That would be 3 pulleys to figure out - one at halfway between, one halfway between the middle pulley and the smallest, and one halfway between the middle pulley and the largest. Right?

Of course, you could also just try to find a pair of matching 5 step pulleys that'll fit the shafts, then find a belt to fit. That's probably what I'd do...

My $.02 worth of mud...

Wirecutter

09-09-2005, 08:29 PM

Given your dimensions I get:

pulley A:

step 1: 2"

step 2: 3.6"

step 3: 5.2"

step 4: 6.8"

step 5: 8.4"

pulley B is the reverse:

step 1: 8.4"

step 2: 6.8"

step 3: 5.2"

step 4: 6.8"

step 5: 8.4"

Can someone check me on this?

-M

ok here are the rest of the sizes on the driven pulley starting at the top. These are od sizes not pitch line and I think C but might be B belt size.

top step 4.510

2nd 5.510

3rd 6.475

4th 7.450

5th 8.4

I would like to have the smallest size on the drive pulley to be 2" and that would drive the 8.4" on the driven pulley for the rest of the sizes what ever keeps the pulley center to center distance the same. I thought there was a place on the web that had a belt program but couldn't find it. Thanks for all the help so far. James

mklotz

09-10-2005, 04:52 PM

Wirecutter wrote:

D = belt's unwound length

rA = radius of pulley A

rB = radius of pulley B

cD = distance between shaft centerlines

(pi * rA) + (pi * rB) + cD = D

Let's take the simple case where rA = rB = r and the belt forms a racetrack

(cartouche) shape. According to Wirecutter's formula:

D = 2 * pi * r + cD

However, simple geometry in that case shows that the belt length, L, is really:

L = 2 * pi * r + 2 * cD

which isn't even close to the expression for D.

The correct formula has to take into account the wrap angle of the belt on each

pulley.

The programs in the BELT archive on my page may be of some help, though they're

meant for flat belts.

Regards, Marv

Home Shop Freeware - Tools for People Who Build Things

http://www.geocities.com/mklotz.geo

Wirecutter

09-11-2005, 01:10 PM

mklotz -

Ahem. Of course you're correct. I needed to account for twice the distance between centers, and I only did one. (the belt's got to get there and back, after all) Also, I assumed that the belt would wrap exactly 180 degrees (pi * r) around each pulley. That doesn't matter as much for pulleys that are close to the same size and/or when the distance between centers is not too short. This may not be the case for jh, though.

All this may not be a problem anyway. What jh seems to need is a "scaled down" version of the pulley he already has. Given the numbers he posted, can't he just "downsize" all the pulley sizes on the drive pulley by (2" divided by 4.51")?

-M

Wirecutter

09-11-2005, 01:12 PM

By the way, I guess I'm not the only one who tried to follow the link you posted - Geocities has temporarily shut it down for getting too much traffic. I'll try later.

mklotz

09-11-2005, 02:53 PM

If the mathematics is straightforward, I see no reason to resort to

approximations, especially ones whose limitations are unchecked. For the

benefit of those still following this thread, the correct equations are:

r1 = radius of first pulley

r2 = radius of second pulley

sep = center-to-center separation of shafts

eps=r2-r1

phi=arcsine(eps/sep) ;(radians)

theta=2*phi ;(radians)

beta1=PI-theta ;wrap angle on first pulley (radians)

beta2=PI+theta ;wrap angle on second pulley (radians)

wrap1=r1*beta1 ;wrap length on first pulley

wrap2=r2*beta2 ;wrap length on second pulley

span=sqrt(sep*sep-eps*eps) ;straight length of belt

blen=2*span+wrap1+wrap2 ;total belt length

All angles are in radians because that's the way mathematics is done. Degrees

are for Sumerians and grads are for Frenchmen :-)

Yes, since my site provides no income (all the programs are free), I host it

on a free service which has its problems. Check back in a couple of hours and

you'll be able to access it.

Regards, Marv

Home Shop Freeware - Tools for People Who Build Things

http://www.geocities.com/mklotz.geo

Wirecutter

09-12-2005, 01:19 PM

Aw, c'mon, Marv. I guess I could have figured out the math, or looked it up, but I was being - well - lazy.

I have been watching this thread with some interest. Machinists by and large are a pretty clever bunch. I kept thinking that someone may have come up with a real simple and clever method for this. Maybe something involving the top to a peanut can, a frizbee, and a length of string.

The only example of what I'm talking about that comes to mind right now is how carpenters use a rafter square and simple fractions to figure out how to cut the trim around the roof of a house. (Protractors? Protracter schmotracter!)

Anyway, thanks for the education.

JH

The heavy math will give you very precise answers - AND will become more NECCESSARY as difference between small & large sheave increases, OR if motor position has less adjustment (bolt slider or hinge = easy to take up 1/2-3/4", eccentric or cam = maybe 1/4 or less)

BUT:

for sheaves that are closer to same size you can usually go 2 easier routes:

1: use same steps in reverse (steps are usually same size). This should work fine for you w/

8.4 x 2

6.8 x 3.6

5.2 x 5.2

3.6 x 6.8

2 x 8.4

2: figure one ratio (Say you wanted 1:8.4 for slow drive), then use an equal chane in dia on each step (if one gets 1/2" bigger, the other gets 1/2" smaller)

SO:

start w/ 8.4 x 1

next: 6.8 x 2.6 (8.4 - 6.8 = 1.6, add to 1)

5.2 x 3.2 (8.4 - 5.2 = 2.2, add to 1)

3.6 x 5.8 (8.4 - 3.6 = 4.8, add to 1)

2 x 7.4 (8.4 - 2 = 6.4, add to 1)

Again, this method breaks down as ratios get bigger.

uute

[This message has been edited by uute (edited 09-18-2005).]