A bit of math - lenght of belt in pulley systems.

Lets approach this subject from mathematical perspective and forget allowances for belt streatching etc.

So we have 2 pulleys of diameters d1 and d2. Pulleys are rotating on 2 separate shafts and centers of rotation are separated by distance l, eg distance between centers of shafts is l.

What is exact mathematical formua for lenght of transmission belt?

It must be somewhere between 2l (in hypothetical situation of d1=d2=0) and 2"pi"l for d1=l and d2=0 or the other way. What is general equation for all cases between these theoretical extremities?

Martin

Edited By Martin Dowing on 28/10/2017 12:14:03

I use this calculator to save what limited braincells I have left. :-  **LINK**

Lots of calculators out there but if you want to do it longhand then here it is ....

**LINK**

If you insist on calculating the length, bear in mind that the "neutral axis" is the central section with the inextensible fibres in it. It is where the active radius of the pulleys is located yet is rarely in direct contact with the pulleys. It's possibly a little simpler if you are using toothed (flat) belts, as you can often see where the fibres are, with one side of the strands almost in contact with the top surfaces of the teeth. A bit more complicated for a vee belt obviously.

Incidentally, if you use Solidworks, there is an excellent chain / belt mate function that allows you to fit a standard belt size to a set of pulleys and will automatically adjust the position of the tensioner to suit. I used this to good effect and I have to say it worked out more accurately than any of the belt calculators I tried. In reality you are going to be taking a standard belt length and making your pulleys fit it, rather than the other way round.

I think you can forget stretching when it comes to modern belts with steel / glass / Kevlar fibres. And any compression of the rubber / PU will be negligible, given the large contact surface area.

Murray

The belt length is:

(Pi x d1)/2 + (Pi x d2)/2 + 2 x SQRT(l^2 + (d1-d2)^2)

But that assumes d1 and d2 are the diameter at the neutral axis as Murray describes.

No I'm not working it out with an idler as well!

Neil

Edited By Neil Wyatt on 28/10/2017 22:22:17

I have no faith in pulley separation calculators. I use one for a rough estimate then draw the two PCD's in AutoCAD. I add two tangential lines, trim, list and tally. I then adjust the separation and try again until it is right. A terrible palaver but my belts fit a treat

And, in addition to Neil's formula, which is correct for normal combinations, it won't be quite right for those cases where an idler should perhaps be fitted and isn't (great disparity in pulley size and small l).

I'm currently making a stepper motor conversion for my 4" rotary table and needed to buy a belt and timing pulleys that would fit in the case I drew up in Fusion360. I knew the length I wanted between centers of the motor and worm shaft and chose the pulleys to fit in the case. To ascertain the belt measurement I said 2x(distance between centers) + 1x(circumference of one pulley) I then found a matching belt length available to buy on Ebay.

After machining the case and assembling the parts, I found (albeit a little surprised) that it worked out just right.

Ed.

Niel's formula is correct, however if you want to find the easiest way to get something done watch a lazy man work. Here is my lazy man's method:

http://www.calculatoredge.com/mech/vbelt%20length.htm

I have always used a piece of string and a tape measure!

I always measured the broken belt, assuming it didn't have numbers on it

JR,

Being 's advocate here....

I have to repeat that while Neil's formula is perfectly adequate for most practical instances, the formula does not work for all instances.

It is absolutely perfectly accurate for the same sized pulleys, but introduces a minute error as the smaller pulley diameter reduces towards zero and the distance between centres, therefore, tends towards the radius of the larger pulley. (rediculous, I know, but theory is theory and formulae must cope with all scenarios)

The result should be Pi multiplied by the larger diameter - yes? Unfortunately it is not! Error is negligible in any sensible scenario, but does not provide a perfect exact result as requested by the OP. Therefore the formula is only a (very good) approximation, but not quite perfect.

Try it in the conversion progs. The belt would not quite fit!

The difference between theory and practice, I suppose.

Have designed many belt and pulley systems over the years and either used the suppliers calculation or as Robin said drawn it out in CAD. Most pulley systems should have adjustment to allow for tolerances in belt length and pulley diameter. Just use the right calculation for the right type of belt and it should all fit.

I do replace a lot of V belts in my day job and always try and take a used one if it’s not too damaged to the supplier to Check and supply new ones.even when there’s a perfectly readable number on it it’s surprising how many times the new one can be much too short or long..If the belt is totally shredded i have a selection of scrap belts which i cut to size then take that as a sample they have a belt measuring device to offer new ones to to check length...but as the supplier says all manufacturers have differing ideas about specs

@Neil

When I substitute to your equation extreme case where l=1/2 x d1 and d2 = 0 I cannot get sensible resuls.

I am getting l(belt) = 1/2Pi x d1 + sqrt5 x d1

This cannot be true as obvious correct answer is Pi x d1. After some trials I can propose an alternative equation:

l(belt) = pi x d1 + pi x d2 + 2 x sqrt{l^2 - [(d1-d2)/2]^2]}

this works for both extreme cases, second is where d1=d2=0. Perhaps it will also work for all cases between too?

Am I correct, or some rots have been proposed?

@Brian Sweeting

Formua given in your reference would read as follows:

l(belt) = 2 x l + pi/2 x (d1+d2) + (d1-d2)^2/4l

This cannot be true for l=1/2d1 and d2=0 because it delivers value higher than pi x d1, which is correct answer.

Alternative equation l(belt) = 2 x l + pi/2 x (d1+d2) - (d1-d2)^2/4l would be correct in both extreme cases. I wonder, what about cases between?

@oldiron,

Must know algoritm to trust calculator.

@Muzzer,

Good practical considerations. Regarding mentioned or other similar software - do not have access to such a luxury.

@ Robin,

No access to CAD etc. I also tend not to believe online calculators until proven correct for variety of inputs.

@Those who are suggesting "measure broken belt" etc.

I am designing 2 steps pulley system which is intended to work with *the same* belt. Do not want to fabricate parts which are not mating correctly.

Anyway, it seems that question is still not resolved. Any other ideas?

Martin

Edited By Martin Dowing on 28/10/2017 20:35:45

Up to now no formua proposed by anyone here (including these proposed by myself above) doesn't work.

So how a correct one would look?

I think no-one spotted I left a /2 out of the (pi x d2)/2 bit - I've edited the formula.

When d2=0 the second term comes to 0, which is right.

The error in my formula is that it ignores 'wrap' which can become significant if (1) the pulleys are of significantly different size and (2) they are close together.

To get a correct result you will have to calculate the tangent points, the distance between them and the angle.between them.

Try my corrected version

One source ( https://www.engineeringtoolbox.com/length-belt-fans-motors-d_872.html ) is giving following formua:

l(belt)=d1xPi/2 + d2xPi/2 + 2l + (d1-d2)^2/4l , the same what Brian have found elsewhere

Needless to say this formua is also an approximation. I wonder how the exact formua looks like? Cannot find it.

I wonder if Neils approximation or this one is better.

@Neil, spotted your ommission but still couldnt get correct result for extreme case regardless. Tried few combinations.

There must be an *exact* formua for this problem. These 3 parameters are determining exact lenght of belt.

Martin

Edited By Martin Dowing on 28/10/2017 22:55:57

It still won't work. Tangents yes, but it is the chord which changes the effective length of 'l'. When d2 tends to zero then 'l' will tend to zero also, because the tangent(s) where the belt 'would have' departed and joined the pulley circumference will be at the same point.

There will be a bit of calculus involved in formulating the correct solution to the problem, I'm afraid. At this time of the night I'm certainly not going there!

The flaw in the maths was assuming 'l' included the radius of the pulleys, which is not strictly true, but near enough for all practical scenarios where a tensioning device is present.

.

Sorry, I don't have the maths to prove whether or not the formula at Problem 1.23 (a) is actually *exact* ... but it is claimed to be; and it looks good to me: **LINK**

https://books.google.co.uk/books?id=xT1X3N_NQCsC&pg=PA51

MichaelG.

.

P.S. ... I assume that you clever chaps can cope with the 'dressmaking units'

Edited By Michael Gilligan on 28/10/2017 23:50:23