I caught an error in this post. I shall try again.
The following equation gives an offset piston's displacement (Y) for a known crankshaft angle (theta).
Given a known piston displacement (Y) and that r,b & d are also known, I would like to be able to calculate the crankshaft angle - to rearrange equation to solve for (theta)?
Any help greatly appreciated.
Thanks,
Russ
$y = rcos( \theta ) + \sqrt{b^2 - (rsin( \theta ) - d)^2} \implies r\sqrt{1 - sin^2( \theta )} = y - \sqrt{b^2 - (rsin( \theta ) - d)^2}.$
$u = sin( \theta ) \implies r\sqrt{1 - u^2} = y - \sqrt{b^2 - (ru - d)^2} \implies r^2 - r^2u^2 = y^2 - 2y\sqrt{b^2 - (ru - d)^2} + b^2 - (ru - d)^2 \implies$
$r^2 - r^2u^2 = y^2 - 2y\sqrt{b^2 - (ru - d)^2} + b^2 - r^2u^2 + 2dru - d^2 \implies 2y\sqrt{b^2 - (ru - d)^2} =y^2 + b^2 + 2dru - d^2 - r^2.$
$v = y^2 + b^2 - d^2 - r^2 \implies 2y\sqrt{b^2 - (ru - d)^2} = v + 2dru \implies$
$4y^2 \{b^2 - (r^2u^2 - 2dru + d^2)\} = v^2 + 4druv + 4d^2r^2u^2 \implies$
$4b^2y^2 - 4r^2u^2y^2 + 8druy^2 - 4d^2y^2 = v^2 + 4druv + 4d^2r^2u^2\implies$
$u^2\{4r^2(d^2 + y^2)\} + u\{4dr(v -2y^2) + v^2 - 4y^2b^2 + 4d^2y^2 = 0.$
Assuming I have not screwed up again
$a = 4r^2(d^2 + y^2),\ b = 4dr(v -2y^2),\ and\ c = v^2 - 4y^2b^2 + 4d^2y^2 \implies u = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a} \implies \theta = arcsin(u).$