Wheeltappers hammer

I posted a problem I had getting excel to calculate belt lengths properly. I spent ages fooling around trying to get the thing to match real life. The drive belt on my lathe is marked"800" and the formulas produce "832".

I bought an Contitech SPZ "825" length belt for the lathe and its too small. The "800" belt is obviously not!

Anyway here for those that are interested here is an excel model, which apparently works and apart from a minor error "mike" helped with always did:

formatting link
Regards

Steve

Reply to
Steve
Loading thread data ...

Steve, Out of interest measure the SPZ 825 belt along it's outside, is it

825mm long ?

Then measure to 800 belt and let us know the length of that one.

I'm guessing you have a belt marked 800 but in Chinese rice wands, not mm.

-- Regards,

John Stevenson Nottingham, England.

Visit the new Model Engineering adverts page at:-

formatting link

Reply to
John Stevenson

The formula is: l [is about] Pi / 2 * (d1 + d2) * 2a + (d2 - d1)^2 / 4a

a: distance between axes.

Nick

Reply to
Nick Müller

I can't agree with your formula. Try a test calculation where the two pulleys are the same diameter, d. And a is the distance between pulley centres.

The answer is simply :- (PI * d) + 2a

Now in your formula making d1 =3D d2, I get the following result:

(Pi/2) * (d1 + d2) * 2a =20

That can' be correct. Is your above formula correctly written?=20

I have assumed that the measurement required is the length of the inside of the belt, which runs on the pulleys at diameters d1 and d2. If my assumption is incorrect then ignore what follows.

A good *estimate* can be obtained provided the difference in the diameter of the two pulleys is not too great.

Try (Pi/2)*(d1 +d2) + 2 * SQRT( (d2-d1)^2 + a^2).

Using this formula and doing the test for same diameter pulleys, i.e setting d1 =3D d2 you will get the result (Pi*d) =3D 2a. =20

--=20 cerberus

Reply to
cerberus

Sorry that last line should be (Pi * d) + 2a.

--=20 cerberus

Reply to
cerberus

Why mess around with approximations? The correct calculation isn't that difficult. From the program available on my webpage:

d1 = diameter of smaller pulley d2 = diameter of larger pulley sep = separation between pulley centers

r1=0.5*d1; r2=0.5*d2; eps=r2-r1; phi=asin(eps/sep); theta=2*phi; beta1=PI-theta; beta2=PI+theta; wrap1=r1*beta1; wrap2=r2*beta2; span=sqrt(sep*sep-eps*eps); blen=2*span+wrap1+wrap2;

beta = wrap angle wrap = wrap length blen = belt length

Regards, Marv

Home Shop Freeware - Tools for People Who Build Things

formatting link

Reply to
Marvin W. Klotz

So that's right:

2r Pi is the circumfence of one disk. We have two half ones. 2r = d :-)) The straight part is the distance between the two axes (a) times 2 (upper and lower part).

Nick Nick

Reply to
Nick Müller

Well I measured them. The Contitech belt measures 825mm. The chinese belt measures 830mm (to the best of my ability!)

My excel model, where the calcs are based upon themanufacturers website, gives 832mm.

Steve

Reply to
Steve

Snipped....

So are you saying my model is wrong then?

Steve

Reply to
Steve

Bugger, a typo for the belt 1 length calculation - H21 should be H43, belt

2 and 3 are OK.

Steve

Reply to
Steve

I was going to build a spread sheet to do this... the problem I ran into was the different series of belts where designated by: centre pitch. outside length... and in inches, cm, mm... and that's just the single section V belts.

If the results from a spread sheet are inconsistent it is quite likely it is calculating the wrong thing... a 25.4 x error is easy to spot...:-)

I reckoned an excel "book" with a page tweaked to each belt series would be needed. I don't do enough calculations to justify setting up a sheet, rather than hand calculating.

Reply to
Jonathan Barnes

PolyTech Forum website is not affiliated with any of the manufacturers or service providers discussed here. All logos and trade names are the property of their respective owners.