Is there a Mathy in the house?

This is only slightly OT(*), I promise…

I have a function which looks like the following:

ay^2 + bxy - cy - cx + ax^2 = d

where a, b, c, and d are constants.

Is it possible to rearrange the formula to get y as a funxtion of x? Is there some simple ‘completing the square’ trick I’m missing here? I’ll happily trade beer for a solution.

  • The function is bicycle-related

y = (-(bx-c)-SQRT(2a(ax^2-cx-d)))/2a

y= (-(bx-c)+SQRT(2a(ax^2-cx-d)))/2a

This looks slightly funny…but could be due to the original formula showing some irregularities.

No idea…

I just asked a maths geek, then copied and pasted the answer.

shrugs

Really should stop asking TR stuff…

well spotted - it should have been:

ay^2 - bxy + ax^2 = c

which is a completing the square :wink: Sorry to interrupt.

Dionysis: can you ask that guy for a proof of his result? I’m curious as to how he managed to deal so nicely with the xy term.

faaaaaaaaaaaaaaaaaaaaaaark

chek yer pms fella!

This looks slightly funny…but could be due to the original formula showing some irregularities.[/quote]

The signs in the numerator look odd…

This make more sense?

     -(bx-c)[b]+/-[/b] SQRT (2a(ax^2-cx-d))

y= -----------------------------------------------
2a

+/- is “plus or minus”
and
SQRT = Square Root

This thread is making me queasy. I’m so close to binning it.

I think you should at least be revealing how is it ‘bicycle-related’ Nath…?

It’s the approx fixed chain length function

chain length = chain to wrap half cog

  • chain to wrap half chainring
  • chain above chainstay
  • chain below chainstay

sometimes called the “rigorous equation”, which ends up for inch pitch chain something like:

cl = 0.25(F + R) + 2 sqrt( csl^2 + (p (F - R))^2 )

cl is chain length in inches
F is chainring teeth
R is cog teeth
csl is chain stay length in inches
p is a constant 1/(4*pi)

I’m probably n+1 again, and my right-handed notes are not very neat, so i probably made errors rearranging that formula. I’ll try it again.

853: I’ll bin you!