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

[SillyOldDufferModerator @sillyoldduffer]

Posted by Neil Wyatt on 28/10/2017 22:27:32:

Posted by Martin Dowing on 28/10/2017 20:33:46:

@Neil

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

I

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

…

Please sir, please sir, I did! Can I have a House Point, please sir! Unfortunately for teacher's pet I was too busy yesterday to point Neil's boob out when it might have helped.

Anyway here's two formula from 'Introduction to Mechanical Design, Jefferson and Brooking, New York, 1950'. It's interesting mainly because the second formula calculates the length of a crossed belt and the two formula are similar to those linked by Michael Gillighan. I might have some 'fun' later comparing mine and his. Not tried yet but they may just be different representations of the same maths.

Accurately calculating the length of a belt between two pulleys is quite difficult. Obviously the profile of the belt matters. Do you calculate the inner length, the neutral line, or the outer circumference? Etc.

Less obvious are the corrections needed when the belt is working; the tension side stretches whilst the other slackens. Therefore the contact arcs on the slack side are increased compared with the stopped condition. Even if the only force on the belt is gravity, the contact tangents on each side of the pulleys are slightly different. And the elasticity of the belt matters – there's no law that says stretch – slack = 0.

To add to the brain strain there's a whole bunch of other maths dealing with variations in belt material, power, wear, velocity, pulley diameters relative to speed ratios, crown radii, and belt width. Just for laughs the answers are often modified by 'service corrections' which seem to be rule-of-thumb. For example there are different service factors given for three variants of the Squirrel Cage Motor as applied to each of the 20 types of machinery tabulated. The factors look to be compensating for starting and running jerks on the belt. Electric motors are usually, but not always, kinder to a belt than a steam engine, presumably because the later applies both high starting torque and reciprocating pulses of energy to the belt.

I think for practical purposes the approximations are 'good enough'.

Dave

Edit typos galore

 

Edited By SillyOldDuffer on 29/10/2017 09:23:42

[Neil WyattModerator @neilwyatt]

Dave,

The formula you give for flat belts is the same as mine. It's right for practical situations but these chaps want to encompass overlapping pulleys of zero and infinite radius as well…

Neil

[Martin DowingParticipant @martindowing58466]

Neil,

Both, estimated and accurate formulas are comparably complex if one has access to trig tables or adequate calculator and wish to work in radians.

Extreme cases are used for checking only. I got alarmed because one extreme case where d1=2l and d2=0 would give a result which can be expressed as l = a x pi + b x sqrt(c).

However it is obvious that l = pi x d = 2pi x l.

It is *impossible* to get an accurate pi x d from the sum of a x pi and b x sqrt(c) if b x sqrt(c) is *not* equal 0.

Hence it was immediately obvious for me that you have given an estimate where errors are produced by replacing lenghts of arcs with lenghts of chords or alternatively you have made a mistake. Because I have never done such calculation in the past and I am designing 2 step pulley system meant to work with *the same* belt, I was just wondering what sort of error might be there, just to make sure that fabricated parts (or one of them to be precise) would not go directly to scrap. Indeed, as you say, in normal situation the error is small enough to ignore. Just wanted to make sure.

For reference of this forum I will quote an accurate formula from discussed book:

l(belt) = pi/2 x (d1+d2) + *alpha* x (d1-d2) + 2 x l x cos*alpha*

where: sin*alpha* = (d1-d2)/2 x l and angles are expressed in radians, eg 360deg = 2 x pi.

*alpha* is an angle between "squared" situation, eg one where d1=d2 and "triangulized" one, where it is not the case and last point of contact of belt with pulley is shifted by an angle.

You do *not* need to know *alpha* a priori.

l(belt) means lenght of belt, d1 & d2 means diameters of pulleys and l is a distance between centers of rotation of pulleys.

Hope, I have written it clear.

Anyway, we have managed to convert something what looks trivial into quite lenghty discussion. It is interesting that an accurate formula is somehow well "hidden" on Internet.

Martin

Edit:

Hey, lets make it now more complicated, get rid of *alpha*, and put it into one final phrase:

It is known from secondary school that sin^2*alpha* + cos^2*alpha* = 1

so: cos*alpha* = sqrt{1 – [(d1-d2)/2 x l]^2}

so the final, one phrase equation devoid of *alpha* will read:

l(belt) = pi/2 x (d1+d2) + arccos*sqrt{1 – [(d1-d2)/2 x l]^2}* x (d1-d2) + 2 x l x sqrt{1 – [(d1-d2)/2 x l]^2}

where: *sqrt{1 – [(d1-d2)/2 x l]^2}* reads in radians.

Now I will go to do something useful

Martin

 

 

Edited By Martin Dowing on 29/10/2017 11:25:16

[Neil WyattModerator @neilwyatt]

Don't forget that belts are made to length tolerances so a formula more accurate than the belt is spitting in the wind…

[SillyOldDufferModerator @sillyoldduffer]

Posted by Neil Wyatt on 29/10/2017 09:44:08:

Dave,

The formula you give for flat belts is the same as mine. It's right for practical situations but these chaps want to encompass overlapping pulleys of zero and infinite radius as well…

Neil

Thanks Neil, I suspected it might be but hadn't checked. (And still haven't!)

I seem to remember Dr Who destroying a computer bent on world domination by asking it a question about zero and infinity.

Dave

[Michael GilliganParticipant @michaelgilligan61133]

This may earn me some ridicule on the "when the media poke fun" thread [which is running concurrently, and seems to have developed into sparring about mathematical prowess]; but here goes …

I was pleased to find that *exact* calculation last night; especially as it is comparatively simple. But what delighted me was the realisation that the angle 'alpha' is identical on the two pulleys, regardless of their size. I had never really thought about it before, and it is perhaps not intuitively obvious when looking at a pair of pulleys.

This was one of those 'paperclip' moments for me [in that, like the design of the paperclip; it becomes obvious when you've seen it and understood].

MichaelG

[duncan webster 1Participant @duncanwebster1]

Coming at it from trigonometry, centre distance C, big pullet diameter D, little pulley diameter d

Angle between belt and centre line a=ASIN((D-d)/(2*C))

straight bit between pulleys L1=C*cos(a)

wrap on big pulley W1=pi*D*(180+2*a)/360

wrap on little pulley W2=pi*D*(180-2*a)/360

total length 2*L1*W1+W2

example C=15, D=10, d=5, a=9.5941 deg, L1=14.7902, W1=17.3824, W2=7.8540, total length 54.8168

It's quite a bit harder going the other way, calculating the centres for a given belt length, set up a spreadsheet and use a solve block, or just fiddle wth the centres until you get the right belt length. However as others have said you're better off with some adjustment, so the simpler formulae are near enough.

Edited By duncan webster on 29/10/2017 15:12:01

[Tim StevensParticipant @timstevens64731]

In the old days of cotton mills and big steam engines you needed to consider an allowance for the droop of each unsupported length, along with the elasticity of the belting.
It is also important to consider the working load on the bearings both sides. Some slack in the belt from the exact calculated size is needed to allow for heat effects, and to avoid any chance of a too-tight belt overloading the system.

In both these cases a calculation to three places would be nugatory – however much satisfaction it gave to those with maths degrees.

Cheers, Tim

[Gordon WParticipant @gordonw]

I have two ways to do this :- One – get a bit of rope and wrap round the pulleys, cut rope to length and take it down to belt shop. Two – fit jockey pulley.

[MuzzerParticipant @muzzer]

Duncan – that misses out the thickness of the belt ie where the neutral line is actually located. You need an additional term ("t"?) surely?

Here's the cushy way to do it. There are a few other steps naturally but as you can see, it's possible to use a defined belt length to "drive" the position of the tensioner wheel. That way you can place the driving and driven pulley centres to get a sensible range of adjustment. It takes care of the belt thickness but fundamentally it places the neutral axis (yellow) where it needs to be. It also couples the pulleys together if you want to do some completely unnecessary animation. I was impressed by how accurately it worked out for my application.

This Youtube tutorial explains i pretty well.

Murray

Edited By Muzzer on 29/10/2017 16:35:20

[duncan webster 1Participant @duncanwebster1]

Not if you relate pulley diameters to the pitch (neutral) line

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