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This is for a three pulley calculator and belt size.
I have the formula to calculate the length of the belt when specific wheel diameters are known and the distance between the center of each pulley is given. This equation simply calculates the perimeter.

However, what I need is to use a known belt size (because you don't just buy any sized belt) and calculate the distance between two of the pulleys. Again with three known pulley sizes and the distance between two sets of pulleys say Pulley A & Pulley B and Pulley A & Pulley C, then how far apart do you spread Pulley B from Pulley C. Hope this makes sense.

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Below taken from here: (but since I don't know how to attach images here, you may need the image found in the link for this to make sense.)

On the Drawing Board: Equation for determining belt size of a three-pulley system[^]


Equation for determining belt size of a three-pulley system
April 3 Erratum: Thanks to rickets007 I spotted several egregious errors! I had used pulley diameter in some parts when clearly I was intending to use radius. Utter carelessness on my part. I have already corrected the blunders. The equations below are now (hopefully!) free of mistakes.


This is off-topic, but I've been googling and thus far haven't found any page that has the equation for computing the belt circumference given pulley diameters and distance between three pulleys. So I'm posting mine. I've tested my equation against an online 3-pulley calculator and while they don't give exactly the same numbers for some input values, they're close enough (there might be rounding errors in the webpage or my spreadsheet). You will note that the calculator necessitates the user inputting three angles. My equation below has no such requirements.

When I'm looking for formulas online I frequently just want a plug and play equation without any of the mathematical esoterica and history. So for those who couldn't give a rat's bleep about the derivation here's the formula up front.


For my sake (so I don't have to do it all over again!) and for skeptics who want to make sure I didn't make any booboos here's the derivation.

In Fig.1 we have three circles representing three pulleys. Pulley A has its center marked as point A. Pulley B has its center marked as point B. Pulley C has its center marked as point C.


We are given the diameters of each of the three pulleys and the distance between their centers. We need to find the belt length.

Let
DA = diameter of pulley A
DB = diameter of pulley B
DC = diameter of pulley C

Let
a = distance between centers of pulleys B and C = segment BC
b = distance between centers of pulleys C and A = segment CA
c = distance between centers of pulleys A and B = segment AB

L = belt length = segments HJ + DE + FG + arc lengths FE + GH + DJ

Let
a' = segment HJ
b' = segment DE
c' = segment FG

Let
α = ∠CAB
β = ∠ABC
γ = ∠BCA

Let
α' = ∠FAE
β' = ∠GBH
γ' = ∠DCJ

Let
lα' = arc length of the belt looping around pulley A = arc FE
lβ' = arc length of the belt looping around pulley B = arc GH
lγ' = arc length of the belt looping around pulley C = arc DJ

Note: All angles and arcs are less than π radians (180 degrees).

Rewriting the equation for belt length we have:

L = a' + b' +c' + lα' + lβ' + lγ'


In order to find the arc lengths we need to first determine the angles α', β', and γ'. To achieve this we shall, for each pulley, find the values of all the other angles in the pulley and then subtract them from 2π radians (360 degrees).

For triangle ABC we are given the values of all the sides (a, b, c). Therefore, we can determine all three angles of the triangle using cosine law:

α = arccos[(b2+c2-a2)/(2bc)]
β = arccos[(c2+a2-b2)/(2ca)]
γ = arccos[(a2+b2-c2)/(2ab)]


As we know if a line is tangent to a circle then a radius of the circle drawn to the point of contact of the tangent line with the circle will be perpendicular to the tangent line. Thus, ∠AFG, ∠BGF, ∠BHJ, ∠CJH, ∠CDE, ∠AED are all right angles.



In Fig.2 we have segment BK drawn parallel to FG. Since ∠AFG is a right angle, therefore, ∠AKB is also a right angle. It follows that triangle AKB is a right triangle. Because BK is parallel to FG and FK is parallel to GB, segment AK = difference in the radii of pulleys A and B = DA/2 - DB/2.

Likewise, we draw segment CL parallel to DE and obtain right triangle ALC. LE = CD and so AL = DA/2 - DC/2.

Our objective is to find the value of ∠KAB and ∠LAC. Because we know the values of the hypotenuse and the adjacent side of both right triangles we can use the cosine function for right triangles:

∠KAB = arccos[(DA/2-DB/2)/c]
∠LAC = arccos[(DA/2-DC/2)/b]

We now have all the angles to compute for α'.

α' = 2π - arccos[(DA/2-DB/2)/c] - arccos[(DA/2-DC/2)/b] - α

We use the same method above for pulleys B and C to obtain:

β' = 2π - arccos[(DB/2-DC/2)/a] - arccos[(DB/2-DA/2)/c] - β
γ' = 2π - arccos[(DC/2-DA/2)/b] - arccos[(DC/2-DB/2)/a] - γ

Note that the fact that (DC/2 - DA/2) is negative is not an error. In fact it provides the correct answer--an obtuse angle which is the ∠ACD in this case. We don't need to know which pulleys are bigger and which are smaller. The equations will always give us the correct values.

We now turn to finding the arc lengths. The arc length of the belt looping around a pulley is the angle subtended by the belt divided by the angle measure of an entire circle (2π) multiplied by the circumference of the pulley:

lα' = (α'/2π)(πDA) = α'DA/2
lβ' = (β'/2π)(πDB) = β'DB/2
lγ' = (γ'/2π)(πDC) = γ'DC/2

Let's return to triangle AKB. Notice that segment FG and KB are not only parallel but also equal in length (because quadrilateral BKFG is a rectangle). Above we let c' = segment FG. It follows that c' is also = segment KB. We can get the value of KB using the sine or tangent function, but to minimize trigonometric functions we will apply the Pythagorean theorem, with c (segment AB) as the hypotenuse:

c' = √[c2-(DA/2-DB/2)2]

The same applies to the other two segments of the belt.

a' = √[a2-(DB/2-DC/2)2]
b' = √[b2-(DC/2-DA/2)2]

And at last we have all the values needed to compute for belt length

What I have tried:

Substitution but have pondered this for about 3 or 4 years now and just giving up. I even asked a relative with a PHD in Math but he never responded. Ha!
Posted
Updated 19-Nov-20 9:54am
Comments
Patrice T 19-Nov-20 15:35pm    
Off topic ?
This is a programming forum.
Member 14996738 20-Nov-20 19:20pm    
Yes sir. Understood. I have created a spreadsheet and have successfully "converged" the number down to a satisfactory result for "a". I call convergence more simply stated as "trial and error". Ultimately, I want to write a program that uses the correct formula for solving for "a" and share it instead of relying on a spreadsheet. The end user would populate the known values and let the program solve based upon the provided information. This would make it all quicker than changing one parameter and then doing the trial and error (convergence) method again and again. Thanks for letting me know though I did have "mathematics" as a selection for topic. Cheers!
Patrice T 20-Nov-20 20:26pm    
Modern spreadsheets comes with solver.
May have a look.

1 solution

It's a "convergence" problem. You want to minimize or maximize some value. You run a simulation.

You pick a number that is close to what you think the unknown is, plug it into your formula, and record the results.

Then you pick another number, run the formula, and record the results.

Now compare the 2 readings. Note the variance.

Run another formula: Am I converging or diverging? Make the adjust (greater / bigger / plus / minus (based on the direction of divergence and convergence).

I have a solar calculator that calculates the sun's position (longtitude) based on date, viewer latitude, viewer longtitude, and local time and time zone.

Given a "target" longtitude (of the sun; e.g. a season), I can arrive at the date and time for that target in only 3 to 4 iterations by simply adjusting the seconds portion of the date based on how I missed the (longtitude) target the previous go around.

This is conceptually based on Newton-Raphson.

Newton's method - Wikipedia[^]

In your case, you have a limited number of belts of a known size. Assuming your "model" is adequate, you should be able to run your model with a given belt and compare the results with that for other belts, and come up with the "best" configuration based on the model.
 
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