Help with extracting alpha from: -alpha-asin(sin(alpha)(r/l)

[dbarger1225]

I am using the Law of Cosines to extract and angle I need from a shaft that is turning on an air engine by piston oscillation and I am having a brain fart on if there are any identities I am missing that can help me pull '$\alpha$' from the right side of my equation:

acos(($(h-pl)^{2}$-$r^{2}$-$l^{2}$)/(-2*r*l))-$\pi$=-$\alpha$-asin(sin($\alpha$)(r/l))

I would like to solve the entire equation for $\alpha$ and the other variables can be treated like constants.

Thanks for the help or attempts in advance.

 

[Simon Bridge]

well - you didn't close the brackets in the arcsine, so it's hard to tell - but asin(sin(x))=x

 

[dbarger1225]

I fixed the bracket issue, I apologize about that.

I am concerned with the (r/l) term inside of the equation. I understand that asin(sin(x))=x .. but I'm certain that asin(sin(x)(r/l)) --DNE-- (r/l)x

 

[Simon Bridge]

Oh I think I see, it wasn't clear to me if the r/l was multiplied with the alpha or the sine-alpha, or with the arcsine.
This what you mean?

$$\sin^{-1}( \frac{r}{l}\sin\alpha )$$

... afaik you can't extract alpha from this sort of equation - you need to be cleverer in the setup or use an approximation (or use a numerical method).

Equation is of form: x+sin^-1(A.sin(x))=b ... solve for x. argh. (assuming the RHS is all constants)

Note: the arcsin will be undefined for some alpha, depending on the value of rl - you want |sin(α)| < l/r (something...)

Where did you start from?

 

[blue_raver22]

Hello,

Thank you for reaching out for help with your equation. It seems like you are trying to use the Law of Cosines to extract the angle \alpha from your equation. However, the equation you have provided is not solvable for \alpha using only the Law of Cosines. This is because the equation contains both trigonometric functions of \alpha (sin and asin) and also contains a constant (\pi). In order to solve for \alpha, you will need to use a different approach.

One possible way to solve for \alpha is to use the identity sin^2(\alpha) + cos^2(\alpha) = 1 to eliminate the sin function in your equation. This will give you a quadratic equation in terms of cos(\alpha), which you can then solve for \alpha using the quadratic formula. Another approach is to use the identity cos(\alpha) = sin(\alpha + \pi/2) to rewrite your equation in terms of only cos(\alpha), which can then be solved using the Law of Cosines.

I would also recommend checking your equation and making sure all the variables and constants are correctly represented. It is possible that there may be a typo or missing term that is causing difficulties in solving for \alpha.

I hope this helps and good luck with your calculations!