Please can someone explain pulleys to me

[fozz]

Hi there Folks,

During my last lathe restoration I came up against a brick wall trying to work out shaft speeds and pulley sizes. came to the rescue working out the speeds for me which helped me immensely but now finding myself restoring another lathe the need arises again to work out speeds and pulley sizes and I'm finding myself fast going bawled trying to do the calculations.

I know I should be able to work this out but for the life of me I'm getting more and more confused.

I'll supply the numbers I already have and what I'd really like is for someone not to calculate the results but to explain to me in simple english how the calculation is done

Ok, as follows,

motor rpm = 1440
motor pulley block from end of shaft = 109/109 to be resized from final calculation

the old motor's pulley block from end of shaft was 45/69/93/119 giving final speeds of 280/600/1090/2240
jack shaft pulley block = 119/93/69/45
jack shaft pulley = 47
headstock shaft pulley = 115

The motor is going to be run via a VFD from 10 hertz to 100 hertz.

I am hoping to get a final speed range from about 50rpm to as near as 3000rpm.

For those of you who understand these numbers I'm sure its easy but its making my head hurt, any help will be most appreciated.

Ross.

[artful bodger]

Hi Foz. That is a pretty unusual question. Do you restore pulley/less lathes? Are you a mathematics buff?
One of my lathes (a woodfast) has 4 speeds. Slow, a bit faster, a little bit faster and flat out.The choice of speeds has never let me down. My other one has 8 speeds that incrementally work in a similar fashion. Ditto
Would not have a clue what any of the actual RPM's are. Just use the one that suits the job at the time.
Not sure but perhaps your over analysing things a tad?
Anyway, good luck with the restorations.

[fozz]

The reason I need to know how to do the calculations is so that I can know what diameters the motor's pulleys need to be turned down to so that I can get the results I'm after in the final setup.

[artful bodger]

Ok. So your restoring a lathe with no pulleys then?

[fozz]

Check out this thread, should explain everything

https://www.woodworkforums.com/f8/ide...-lathe-177954/

[pjt]

I'm not sure I understand your explanation (109/109??) and I assume that is diameters of pulley you have listed, I don't get those speeds???, anyway here is what I have taken from what you have described, the dots represent the belts, the 47 to 115 doesn't change but the other does.
speed calcs.jpg

Basically at 10hz use the 45/119 ratio will give you 50rpm, (a tad less in reality) and at 100hz use 119/45 which will give you 3000rpm
If you want more explanation feel free, or if I haven't understood your explanation.



Pete

[BobL]

motor rpm = 1440
motor pulley block from end of shaft = 109/109 to be resized from final calculation
the old motor's pulley block from end of shaft was 45/69/93/119 giving final speeds of 280/600/1090/2240
jack shaft pulley block = 119/93/69/45
jack shaft pulley = 47
headstock shaft pulley = 115

OK let me get this straight

motor pulley block of 45/69/93/119 mm is paired via a movable belt to the jack pulley pulley of 119/93/69/45 mm?

At the end of the jack pulley block there is a 45 mm pulley with a fixed belt connecting to a 115 pulley on the headstock?

if this is the case at the motor is at 1440 RPM the the jack pulley shaft RPMs at the 4 different belt positions are

45/119 * 1440 = 544 RPM
69/93 *1440 = 1068 RPM
93/69 *1440 = 1940 RPM
119/45 *1440 = 3080 RPM

Then the 47 mm jack pulley connected via a fixed belt to the 115 mm Head stock pulley will reduce those RPMs by 47/115

Final RPMs should then be 222, 436, 793 and 1156 RPM - and apart from the first one, these do not agree with your final values.
Can you please check the measurements of the pulleys

The motor is going to be run via a VFD from 10 hertz to 100 hertz.

Unless the motor is specifically designed for it I would not run a standard 50Hz mootor on a lathe at 10Hz as the motor may not get enough internal cooling. It's OK for something like a drill press where the motor operates for only short periods but not for a lathe where it could sit at that speed for many minutes. I would recommend 20 Hz as a lower end operational point for a standard 50Hz motor.

I am hoping to get a final speed range from about 50rpm to as near as 3000rpm.

You won't get that on a single pair of pulleys as you are asking for a 60:1 (3000/50) ratio of RPMs whereas using a VFD on a standard motor will only do ~5:1 (i.e. 100 - 20 Hz)

If my final RPMs are correct and you did nothing to the pulleys and used a VFD with a 5:1 Hz range you would have the following ranges

on lowest range pulley 89 - 445
Next one would be 175 - 873
Next 317 - 1586
Top pulley 622 - 3112

That's pretty reasonable in my book for a lathe.

[fozz]

Pete, Bob,

The speeds I have were written on the front of the lathe when I bought, as to how accurate they are I have no idea. Now as to the figures you have both worked out, the new motors pulley block is 109 on each pulley. My friend who is setting this up for me has suggested that because the motors shaft is 28mm and pulleys for that size are like hens teeth the idea was to reduce the 109 down to two seperate values to give two different ranges of speed.

Hell my head hurts

Hope thats clearer,

Ross.

[Mobyturns]

Single Belt Transmission - one driving and one driven pulley
from Pulley Diameters and Speeds

For a system with two shafts and two pulleys - as indicated with pulley 1 and 2 in the figure above:

d1 n1= d2 n2 (1)

where

d1 = driving pulley diameter (inch, mm)

n1 = revolutions of driving pulley (rpm - rounds per minute)

d2 = driven pulley diameter (inch, mm)

n2 = revolutions of driven pulley (rpm - rounds per minute)

Equation (1) can be transformed to express the

Revolution of Driven Pulley n2 = d1 * n1 / d2 (2)

Revolution of Driver Pulley n1 = d2 * n2 / d1 (3)

Diameter of Driven Pulley d2 = d1 * n1 / n2 (4)

Diameter of Driver Pulley d1 = d2 * n2 / n1 (5)

Multiple Belt Transmission Systems


For a system a with three shafts and four pulleys - as indicated in the figure above:

n2 = n3 (6)

n4 = n1 (d1 d3) / (d2 d4) (7)



In your case using the pulley diameters given the speeds calculate as

input rpm	Motor out	jackshaft in	jackshaft rpm in	jackshaft rpm out	jackshaft out	headstock in	headstock rpm	multiplier
n1	d1	d2	n2=n1 x d1 / d2	n2=n3	d3	d4	n4=n3 x d3/d4
1440	45	119	545	545	47	115	223	0.154549
1440	69	93	1068	1068	47	115	437	0.303226
1440	93	69	1941	1941	47	115	793	0.550851
1440	119	45	3808	3808	47	115	1556	1.080773

I have a feeling that you have given outer diameter of the pulley & not the driven diameter which would account to the difference in speeds you quote to calculated.

To calculate the VFD final output speeds
From VFDs How Do I Calculate RPM For Three Phase Induction Motors? | Precision Electric, Inc.

AC Three Phase Induction Motor <st1:stockticker>RPM</st1:stockticker> is determined by the formula:
<st1:stockticker>RPM</st1:stockticker> = (120 * Frequency) / # of poles in the motor
Since the number of poles of a three phase induction motor is established when it is manufactured, the only way to change the speed of the motor is to change the Frequency.


If you keep the original pulley set the final drive to output motor pulley multiplier will be the same, however the motor output speed now varies

The VFD is 10 to 100hz with standard at 50hz so the speeds above will need to be recalculated for the variable frequency assuming similar 4 pole motor speed of 1500rpm. The quoted motor speed of 1440 rpm accounts for slip & losses etc its theoretical speed is 1500rpm, which accounts for the speed differences at orig & 50hz in the table below.

VFD FREQUENCY	vfd motor output	pulley 1	pulley 2	pulley 3	pulley 4
ORIG		223	437	793	1556
multiplier		0.154	0.303	0.550	1.081
10	300	46	91	165	324
20	600	92	182	330	649
30	900	139	273	495	973
40	1200	185	364	660	1297
50	1500	231	455	825	1622
60	1800	277	545	990	1946
70	2100	323	636	1155	2270
80	2400	370	727	1320	2594
90	2700	416	818	1485	2919
100	3000	462	909	1650	3243

Simplified but gives figures near enough for your purposes.

[pjt]

Ahh so you have a twin groove pulley of 109 diameter which is going on the motor and then onto the jack shaft pulley block as per normal, and your thinking is to turn down the "spare" groove to a smaller dia and you want to know what diameter that is, doing this you lose two grooves which you had on the original motor pulley thus loosing a range of speeds.
Is it possible to get the original bored out to 28 + keyway to suit the new motor, at a guess there might not be enough meat left in the pulley for the 45 end of the pulley block



Pete

[fozz]

Yeah, Pete, if we bored the original pulley out there'd be nothing left of it. I do lose two ranges but my hope is that with whats left I can overlap, hopefully My friend has suggested we turn one of the 109's down as far as possible we should be able to get the needed range.

[BobL]

Pete, Bob,
The speeds I have were written on the front of the lathe when I bought, as to how accurate they are I have no idea. Now as to the figures you have both worked out, the new motors pulley block is 109 on each pulley. My friend who is setting this up for me has suggested that because the motors shaft is 28mm and pulleys for that size are like hens teeth the idea was to reduce the 109 down to two seperate values to give two different ranges of speed.

OK so you can't go bigger than 109 on the motor so that fixes one parameter.

Then you can't use anything other than 2 side by side pulleys on the jack shaft pulley block lets start with the smallest pulley in that set and see what gives you in terms of max RPM when connected to the 45 jack pulley

at 100Hz I get 109/45 *1440 * 47/115 * 100/50 = 2850 RPM - you can't get anymore that that but it might be enough for you
Lowest speed at 20 Hz is 570 RPM, half that (285) at 10Hz


If the second motor 109 mm pulley alongside the other one is now connected to the 69 mm pulley (alongside 45 mm pulley on jack shaft pulley block used in the previous calculation) it won't produce a low enough speed so as you say the second motor 109 pull needs to be turned down.

The smallest you could probably turn the second 109 motor pulley down to will be something like 45 mm (actually that is really pushing it on a 28 mm bore but lets see how it goes) but the lowest speed that will generate at 20Hz is:

45/119 *1440 *47/115 *20/50 = 153 RPM, and half that again if you go to 10Hz which might work for you.
Highest speed at 100 Hz is 445 RPM so the two pulleys will not overlap if you restrict to 20Hz for the faster pulley.

To get 50 RPM at 20Hz the motor pulley needs to be 15 mm i.e. smaller than your shaft so it aint going to happen.
You can of course use move to larger pairs of pulleys on the jack shaft pulley block but that will then reduce your top speed.

A two pulley system will be a PITA because even at 10Hz you only have an overlap of 285 - 445 which is not enough.
You really will need something that will generate a useful range like 250 to 1250 RPM and not be constantly changing belts to move between these two RPMs.

I have a 6 pulley system on my small lathe and my overlaps (20 to 100Hz) are
144 - 792
269 - 1340 - this is the one I most commonly use - in fact I rarely move my belt to any other range.
408 - 2040
604 - 3020
892 - 4460
1300 - 6500
Of course I do not use the high frequency ends of the top two ranges.

[fozz]

This is why I do like this forum, you guys are brilliant!!!! Now I understand how the figures work enough to make valued decisions so thanks all of you.

[pjt]

Moby has done a nice chart which is instructive, it shows that apart from the slowest speeds you will have a good range of speeds, if the 109 goes on all those speeds will be a little slower, so you could go from 109 to 45 and have a high range of speeds or from 109 to 69 which will be a slower set of speeds and the same for the other jack shaft diameters, assuming you can position the 109 to line up with any of the jack shaft grooves.

I take it that the jack shaft is fixed, i.e no adjustment but the motor is on an adjustable bracket that allows for belt changes, I'm thinking two possibilities, leave the 109 as is but just swap from the 45 to the 69 for two higher ranges of speeds but there has to be enough motor mount bracket travel to accomadate this, to get a bigger change in speed range turn down the 109 as your thinking, at this point I would be determining the center distances between jack shaft and motor, measure at the maximum travel and the minimum travel, from this we can determine belt length and a possible size to turn the spare 109 groove down to.



Pete

[BobL]

This is why I do like this forum, you guys are brilliant!!!! Now I understand how the figures work enough to make valued decisions so thanks all of you.

It's great when the penny drops. Just keep playing around and eventually "CA-CHING!"

This is why free play (a nasty word amongst modern educators) is so effing valuable and a useful way to learn.