Thread: Rpm of a pulley on wood lathe---forgot formulae

3" Pulley is rotating at the same speed as the shaft. 2770 rpm.
The suface speed of the pulley is 870 inches per second or 22.1m per second

For a 3" pulley rotating at 2770 RPM I get ~435 in/sec surface speed (2175.6 ft/min)
Here's a calculator to make working it out easier Pulley Calculator. RPM, Belt Length, Speed, Animated Diagrams

Originally Posted by NCArcher

3" Pulley is rotating at the same speed as the shaft. 2770 rpm.

No more, no less, rpm.



Stay sharp!

Neil

Pulley speed as above (2770). I get surface speed as 435 inches/sec.

if 3" pulley is 2770 RPM, 17" pulley will be approx 488 rpm

The simple math is:
The speed at which the shaft rotates is constant and is not effected by the size of the pulley. What can change is the circumference of the pulley and thus the distance that the belt travels with each revolution of the shaft.

Circumference of of 3" pulley is pi * diameter = 3 * 3.1416 = 9.425"
Circumference of of 17" pulley is pi * diameter = 17 * 3.1416 = 53.41"


Therefore the 3" pulley must rotate 53.41 / 9.425 times or 5.667 times for the 17* pulley to rotate once.

If the small pulley is rotating at 2770 revolutions per minute, the 17 " pulley will rotate 2770 / 5.667 times per minute or 488.8 rpm.

or more simple the driving rpm (2770) divided by the ratio of the large pulley to the small pulley (17:3 = 5.667) giving 2770 / 5.667 = 488.8 rpm as above

Hope this helps
Bob

Thanks

My point is ---that you missed--

a dot on the outer circmference of 3 inch pulley has to travel 9 . 426 inches
verse the dot on the Shaft circumference ---only travels 62 84 MILLIMETRES!--

surely the V belt --driven by the 3 inch OUTER circumference ---of that pulley --

is travelling FASTER than the shaf t /pulley inner circumference----- Velocity???? am iRIGHT??

Tying typing is extremely difficult ------something wrong --leteres do not appear---

anyhow ---
many thanks --appreciate ---

ACACIA

By the way --discovered RED CEDAR as Australians call it

Indian / s e Asian MAHOGANY--

fabulous grain ---light ---gunstocks & lamps look good --easy to turn--imported here circa 1900

see fotos--
Saw Wbsite in Austrlia --woocarver----creates masterpieces fromthis wood --

regards Accacia

Originally Posted by Blocklayer

For a 3" pulley rotating at 2770 RPM I get ~435 in/sec surface speed (2175.6 ft/min)

Originally Posted by Lappa

Pulley speed as above (2770). I get surface speed as 435 inches/sec.

Not sure how i got double, but it was early

NC - from the Bandsaw forum, I thought you were good at maths!
Alister.

Originally Posted by ACACIA

a dot on the outer circmference of 3 inch pulley has to travel 9 . 426 inches
verse the dot on the Shaft circumference ---only travels 62 84 MILLIMETRES!--

ACACIA

I think you are confusing RPM with the belt speed. Irrespective of the size of a pulley attached to the spindle the pulley will always rotate the same number of turns in any given time. If you count the number of times the 'dot' on the outer circumference completes a 360 degree turn it will always be the same number of times as the RPM of the motor shaft, irrespective of the daimeter of the pulley.

Originally Posted by ACACIA

surely the V belt --driven by the 3 inch OUTER circumference ---of that pulley --

is travelling FASTER than the shaf t /pulley inner circumference----- Velocity???? am iRIGHT??

ACACIA

In your original post you asked how to calculate the RPM, but this is not the same as the belt speed or to use your term 'velocity"

You are correct in assuming that the belt speed will increase with the increase in the diameter of the pulley and that increase is proportional to the change in the diameter of the pulley.

In post #3, Blocklayer gave you a link to a calculator which will define all these relationships. Can I ask you to try it and enter the sizes of your pulleys and the RPM of your motor (2770 rpm I think you said). You will then get the belt speed for different pulley combinations and how it affects the final RPM of the larger pulley.

If you still have difficulty in understanding the results, give me some specific examples and I will try to help you.

Regards

Bob

Put simply, multiply the revolutions of the motor by the diameter of the driving pulley, then divideby the diameter of the driven pulley will give you the rpm required, and bob's you uncle.

Jim

Originally Posted by Big A

NC - from the Bandsaw forum, I thought you were good at maths!
Alister.

I used to be. But that was a week ago.

Thanks Block layer ----

I have tried that calculator ----

BUT --- it asks for RPM of BOTH drive& driven pulleys---

I do not know RPM of 3 inch plley---
and how do I know what RPM driven pully is ?-

Anyhw the rratio is easy ---not required ---4 : 1 for a 3 inch / 12 inch pair of pulleys

What I am confused bout is the V BLT must run much faster ---than 2770 shaft speed

HOW can the perimeter SPEED---- of a 3 inch pulley----be 2770 RPM --AS WELL ?? ( as the 19mm motor shaft speed )?

Take an old 33 1/3 RPM vinyl record ----outer track rotates much faster than the centre---ok ?

Chopper blade TIPS- travel close--but not exceeds--- to mach one ----- YET the centre drive shaft is MUCH- slower--for reduction torque(former air force grease monkey )
see my problem?---LARGER DRVE PULLEY --SAY 8 INCHES ---- BELT THEN ACCELERATES THEBIGGER TOP PULley
EVEN FASTER ----SO FINAL DRIVEN PULLEY---SPEEDS UP =====RATIO ALTERSRASTICALLY!
Apologies ---typing is a nightmare-------something wrong --internet in Africa is !!!