Isolating C from "L=2C+∏ (D+d)/2 + (D-d)^2/4C" - Help Needed!

[wallybanger]

Hi, I'm not the best with algebra and I've been wrestling with a formula this afternoon. It is a formula to calculate a belt length based on 2 pulley sizes and the distance between the centres. I would like to be able to calculate the distance between the centres based on the pulley sizes and the belt length so instead of solving for L I need to solve for C. I would love to know what I'm doing wrong in isolating C. Anyway, here is the eqn.

L=2C+∏ (D+d)/2 + (D-d)^2/4C

L= Belt length
C= Distance between pulley centres
D= Big pulley Dia
d= Small pulley Dia

Thanks :)

 

[mathman]

Multiply through by C. You then have a quadratic in C, which is readily solvable.

 

[wallybanger]

Multiply through by C. You then have a quadratic in C, which is readily solvable.

I kinda know what you mean but I'm not really sure how to do that. As in multiply every element by c? Top and bottom of the fractions?

 

[HallsofIvy]

Multiplying both sides of
$$L=2C+∏ (D+d)/2 + (D-d)^2/4C$$
by C will give
$$LC= 2C^2+ ∏ C(D+ d)/2+ (D- d)^2/4$$

 

[ellipsis]

For anybody still hung up on this, the next step is to reformulate the equation using the quadratic equation.

 

[wallybanger]

Ok, looking good so far, but how come the second part (D-d)^2/4 isn't also multiplied by C and why are the denominators not multiplied by C?

Ugh, I also suck at factoring. I need to go through my algebra textbook again and teach myself all this stuff from scratch.

 

[SteamKing]

Hi, I'm not the best with algebra and I've been wrestling with a formula this afternoon. It is a formula to calculate a belt length based on 2 pulley sizes and the distance between the centres. I would like to be able to calculate the distance between the centres based on the pulley sizes and the belt length so instead of solving for L I need to solve for C. I would love to know what I'm doing wrong in isolating C. Anyway, here is the eqn.

L=2C+∏ (D+d)/2 + (D-d)^2/4C

L= Belt length
C= Distance between pulley centres
D= Big pulley Dia
d= Small pulley Dia

Thanks :)

Here's a tip:

$L = 2C+\pi\frac{(D+d)}{2}+\frac{(D-d)^{2}}{4C}$

use a lower case $\pi$ to indicate the ratio of the circumference of a circle to its diameter. The upper case $\Pi$ is generally used in mathematics to indicate the product of several different terms.

 

[wallybanger]

Ok, So I'm guessing
$L = 2C+\frac{\pi(D+d)}{2}+\frac{(D-d)^{2}}{4C}$
Should become
$LC = 2C^2+\frac{\pi C(D+d)}{2}+\frac{C(D-d)^{2}}{4C}$

Is that correct?

It still has me wondering why the C doesn't effect the denominator.

 

[HallsofIvy]

It does! That was the whole point of multiplying by it. The C in the denominator cancels the C in the numerator leaving
$$LC= 2C^2+ \frac{\pi C(D+ d)}{2}+ \frac{(D-d)^2}{4}$$
That is the same as
$$2C^2+ \left(\frac{\pi(D+ d)}{2}-L\right)C+ \left[\frac{(D- d)^2}{4}\right]= 0$$

That is a quadratic equation. I wouldn't worry about "factoring"- use the quadratic formula.

 

[wallybanger]

It does! That was the whole point of multiplying by it. The C in the denominator cancels the C in the numerator leaving
$$LC= 2C^2+ \frac{\pi C(D+ d)}{2}+ \frac{(D-d)^2}{4}$$
That is the same as
$$2C^2+ \left(\frac{\pi(D+ d)}{2}-L\right)C+ \left[\frac{(D- d)^2}{4}\right]= 0$$

That is a quadratic equation. I wouldn't worry about "factoring"- use the quadratic formula.

:P Right. I missed that you had canceled the C in the denominator.

Ahhhh! The quadratic Eq'n! Forgot about that old chestnut. Alright, let's give this a try.

$C=\frac{-\left(\frac{\pi(D+d)}{2}-L\right)\pm \sqrt{\left(\frac{\pi(D+d)}{2}\right)^2-4(2)\left(\frac{(D-d)^2}{4}\right)}}{2(2)}$

That's very interesting... I need to try to jam all that into a CAD program now. I should probably attempt to check it first too...