Substituting in Sin^-1 relationship

**Oldman** · January 16th 2009, 02:07 PM

Hi, I'm trying to work out the relationship of geometry for centre distance between 2 pulley's using a fixed length of belt. Please have a look at the <Centres Calc> page of attached spreadsheet where I have the formula for lengths around the belt system.

I'm trying to solve for C when I know total L and dia's D & d. All this needs is a 'simple' substitution of rows 2 & 3 into row 4. I can do that bit, but can't remember how to extract C from the Sin^-1 terms.

It's for a very high powered motorcycle that I'm helping to design & build. The goal is a major speed record machine, where we are designing a transmission that can be set up with wide range of final drive ratios. We have many options to refine as (for various reasons) the machine will have 2 stages of toothed belt with pulley's in the order of 44-64 teeth. I thought this spreadsheet would give us a neat tool to explore these options and minimise the number of pulleys we have to make.

With help from this forum, I hope we can give the 'nut behind the wheel' a decent transmission to race with.

(Edit - Found some errors in the formula's - and have changed the attachment)

Thanks, Kev.

**vincisonfire** · January 16th 2009, 04:38 PM

First the notation of $sin^{-1}(x)$ is a little ambiguous.
It can either mean $sin^{-1}(x) = \frac{1}{sin(x)}$ or $sin^{-1}(x) = arcsin(x)$ .
In the first case I see you have $\theta_L = \pi + 2 sin^{-1}(\frac{D-d}{2C})$ I'm guessing that $sin^{-1}(x) = arcsin(x)$ is what you meant since your talking of angles.
In this case $\theta_L = \pi + 2 sin^{-1}(\frac{D-d}{2C})$
$\theta_L - \pi = 2 sin^{-1}(\frac{D-d}{2C})$
$\frac{\theta_L - \pi}{2} = sin^{-1}(\frac{D-d}{2C})$
$sin(\frac{\theta_L - \pi}{2}) = \frac{D-d}{2C}$
$C = \frac{D-d}{2sin(\frac{\theta_L - \pi}{2})}$

**HallsofIvy** · January 16th 2009, 05:12 PM

Originally Posted by vincisonfire

First the notation of $sin^{-1}(x)$ is a little ambiguous.
It can either mean $sin^{-1}(x) = \frac{1}{sin(x)}$ or $sin^{-1}(x) = arctan(x)$ .

I think a lot more people would use $sin^{-1}(x)$ to mean arcsin(x) than arctan(x)!!

In the first case I see you have $\theta_L = \pi + 2 sin^{-1}(\frac{D-d}{2C})$ I'm guessing that $sin^{-1}(x) = arctan(x)$ is what you meant since your talking of angles.
In this case $\theta_L = \pi + 2 sin^{-1}(\frac{D-d}{2C})$
$\theta_L - \pi = 2 sin^{-1}(\frac{D-d}{2C})$
$\frac{\theta_L - \pi}{2} = sin^{-1}(\frac{D-d}{2C})$
$sin(\frac{\theta_L - \pi}{2}) = \frac{D-d}{2C}$
$C = \frac{D-d}{2sin(\frac{\theta_L - \pi}{2})}$

**vincisonfire** · January 16th 2009, 05:17 PM

, you're right. So deceiving am I.
Of course I meant arcsin(x).

**Oldman** · January 17th 2009, 02:37 AM

Eh ta. The formula came straight from a 'rather dated' textbook, but I have written the term just as that book (excuse over).

Yes I was sucessfull calculating length from known centres using arcsin - so I'm sure we're on the right track.

Your post shows me how to manipulate a Trig term

but of course that formula is only part of the solution.

I'll run the substitution and apply the rule you've shown me - and let you know how I get on. Too busy at the moment - but will have a go tonight.

Many thanks. Kev

**Oldman** · January 17th 2009, 03:43 PM

OK - I failed again

I did find an error in position of brackets (which means the Sq Rt in line 4 was applied over too many terms). This has been corrected in my original attachment. Different brackets are now used to clarify and some notes added in Blue.

My problem is that I need to insert ӨL & ӨS (with the sin^-1 terms that include C) into L = √{4C^2-(D-d)^2} + 1/2 [D*ӨL + d*ӨS]

I get a couple of Sin^-1 terms (you've shown me how to deal with) and a Difference of 2Sq's (which includes C) where I can't work out the rule to expand.

I've gotta solve this equation for C when I know L, D & d.
I'm sure I could have done it in my school days - but I'm just not getting it now

It also seems to me there's a missing Pi in the calc's for contact length (but I supose it could cancel out?).

Check out the sheet where belt length is calculated for known centres. That works 'coz I've some belt data that tells me when C=400, then L=~1560, and when C=600, then L=~1960.

I just need to write the equation other way round - 'coz we're using fixed belts & I want to establish centre distances we'll need with different pulley's.

Would be real chuffed if one of you could help a little more?

Thanks in anticipation.

**Oldman** · January 18th 2009, 12:17 PM

Found the difference rule (x^2-y^2) = (x+y)(x-y)
Darned obvious now !
My formula is a lot more complex that x & y, but all this trig & algebra is gradually comming back, so I'm back on the spreadsheet.

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