Jtdylktuy

I am trying to design a cam profile like in the above image. There are 2 equations I have come up with that define variable "c". The problem, however, is that variables a, b & c are known values and it is angle $theta$ that I need to know. The 2 arcs on either end of the straight line are equal. The radius of the arcs and the angle $theta$ change when any of the variables are changed.

The 2 equations I have so far:

$c = frac{2a}{sin theta}(1-cos theta)+frac{b.sin theta}{cos theta} $

$c = 2 left[ frac{a}{sin theta} - sqrt{ left( frac{a}{sin theta} right)^2 - a^2}, right] + frac {b.sin theta}{cos theta} $

I can't for the life of me manage to rearrange the equation to make $theta$ the subject!

If anyone is an algebra whiz, any help would be much appreciated!

Cheers.

EDIT

Ok, so after a bit of geometric shenanigans, I have come up with some different equations to help solve this but and still struggling to rearrange it. (However, I had to replace $theta$ with $x$ as I couldn't get $theta$ to work with the program I drew this with!)

By drawing a chord for the arc, we can show that the angle of the chord is half that of the slope. What I really need to find is the length of either $e$ or $d$, where $d + e = frac{c}{2}$

Using the half angle formula I get:

$tan x = frac{2 tan frac{x}{2} }{1 - tan^2 frac{x}{2}}$

Which can be written as:

$frac{c-2e}{b} = frac{frac{2e}{a}}{1-frac{e^2}{a^2}} $

Simplifying to:

$frac{c-2e}{b} = frac{2ae}{a^2-e^2}$

My problem now, is that I can't simplify it to get $e$ on its own!

If anyone has any ideas, that would be awesome!

Considering the first equation
$$c = frac{2a}{sin (theta)}(1-cos (theta))+frac{b,sin (theta)}{cos (theta)}$$
using the tangent half-angle substitution, we end with
$$c=frac{2 t left(-a t^2+a+bright)}{1-t^2}$$ that is to say
$$2 a t^3-c t^2-2 t (a+b)+c=0$$ that you can solve using Cardano method.

Probably easier would be Newton method for solving the cubic equation. Starting using $t_0=0$, we should have $t_1=frac{c}{2 (a+b)}$ and
$$t_2=frac{c^3 (b-a)+4 c (a+b)^3}{8 (a+b)^4-2 c^2 (a-2 b) (a+b)}$$
$$t_{n+1}=frac{4 a t_n^3-c t_n^2-c}{2 left(3 a t_n^2-c t_n-(a+b)right)}$$

Similarly, if $theta$ is small, we could expand the first equation as Taylor series
$$c= (a+b)theta+frac{a+4 b}{12} theta ^3 +frac{a+16
b}{120} theta ^5 +frac{17 (a+64 b)}{20160}theta ^7+Oleft(theta ^9right)$$ and use series reversion to get
$$theta=frac{c}{a+b}-frac{ (a+4 b)}{12 (a+b)^4}c^3+frac{ left(a^2+2 a b+16
b^2right)}{80 (a+b)^7}c^5+Oleft(c^7right)$$

Edit

Still assuming small values fo $theta$, we could build at $theta=0$ the $[2,2]$ Padé approximant and get for the whole
$$f(theta) = frac{2a}{sin (theta)}(1-cos (theta))+frac{b,sin (theta)}{cos (theta)}-c$$ the approximation
$$f(theta)simeqfrac{-c+ (a+b)theta+frac{c (a+4 b)}{12 (a+b)}theta ^2}{1-frac{(a+4 b)}{12 (a+b)} theta ^2 }$$ Solving the quadratic
$$thetasimeq frac{2 left(sqrt{3 c^2 (a+4 b) (a+b)+9 (a+b)^4}-3 (a+b)^2right)}{c (a+4 b)}tag 1$$ which seems interesting.

Let us try it for $a=1$ and $b=4$. Give $theta$ a value to compute $c$ and recompute the estimate $theta_*$ from $(1)$. The table below reproduces results (in degrees).
$$left(
begin{array}{ccc}
theta & c & theta_* \
5 & 0.43728 & 5.00001 \
10 & 0.88029 & 10.0003 \
15 & 1.33510 & 15.0020 \
20 & 1.80853 & 20.0082 \
25 & 2.30862 & 25.0249 \
30 & 2.84530 & 30.0617 \
35 & 3.43143 & 35.1324 \
40 & 4.08434 & 40.2567 \
45 & 4.82843 & 45.4605 \
50 & 5.69963 & 50.7782 \
55 & 6.75373 & 56.2542 \
60 & 8.08290 & 61.9466
end{array}
right)$$

Update

After comments, it is quite sure that we can make better knowing that the area of interest is around $theta=frac pi 4$. To stay "simple", writing the $[1,1]$ Padé approximant, we should get
$$thetasimeq frac pi 4+ frac{alpha +beta c}{gamma+delta c}$$ where
$$alpha=2 left(left(6 sqrt{2}-8right) a^2+sqrt{2} a b+b^2right)$$
$$beta=2 left(sqrt{2}-2right) a-2 b$$
$$gamma=2 left(sqrt{2}-2right) a^2+3 left(5 sqrt{2}-8right) a b-2 b^2$$
$$delta=left(4-3 sqrt{2}right) a-2 b$$ Applied to the case
$$a= pi 160frac{12}{360}=frac{16 pi }{3}qquad b= pi 160frac{50}{360}=frac{200 pi }{9}qquad c=80$$ this would give
$$thetasimeq frac pi 4 +frac{90 left(6 sqrt{2}-37right)+left(337+366 sqrt{2}right) pi }{left(1161
sqrt{2}-2497right) pi -90 left(13+9 sqrt{2}right)}approx 0.761710$$ which converted to degrees would give $43.6427$ while the exact solution would be $theta=0.7617146$ corresponding to $43.6430$. Not too bad.

Keeping $a=frac{16 pi }{3}$ and $b= frac{200 pi }{9}$ and varying $c$ over a quite large range, here are some results.
$$left(
begin{array}{ccc}
c & theta_{approx} & theta_{exact} \
60 & 35.1540 & 35.2569 \
65 & 37.4779 & 37.5244 \
70 & 39.6591 & 39.6759 \
75 & 41.7103 & 41.7142 \
80 & 43.6427 & 43.6430 \
85 & 45.4666 & 45.4665 \
90 & 47.1906 & 47.1894 \
95 & 48.8228 & 48.8165 \
100 & 50.3704 & 50.3528
end{array}
right)$$

New update

We could still do much better at the price of a quadratic equation. Around a given angle $theta_0$, develop the rhs of the original equation as a $[2,2]$ Padé approximant and solve for $theta$. I shall not give here the formulae but just the results for the last worked case $(theta_0=frac pi 4, a=frac{16 pi }{3}, b= frac{200 pi }{9})$
$$left(
begin{array}{ccc}
c & theta_{approx} & theta_{exact} \
60 & 35.256535 & 35.256856 \
65 & 37.524331 & 37.524418 \
70 & 39.675901 & 39.675917 \
75 & 41.714232 & 41.714233 \
80 & 43.643029 & 43.643029 \
85 & 45.466540 & 45.466540 \
90 & 47.189400 & 47.189400 \
95 & 48.816481 & 48.816478 \
100 & 50.352781 & 50.352763
end{array}
right)$$

I think one may simplify the problem a lot by realizing that due to symmetry the straight line segment is passing right through the center of your drawing, i.e. given the lower right corner corresponds to $(0,0)$, through $(a+tfrac{b}{2},tfrac{c}{2})$. Then it's easy to see that $tan(tfrac{pi}{2}-theta) = frac{a+tfrac{b}{2}}{tfrac{c}{2}}$. However, the values of $a$, $b$ and $c$ are not independent. Infact all you need is $c$ and $2a+b = 2(a+tfrac{b}{2})$, the latter of which is the width of the rectangle.

We can set up $a^2+(c/2)^2=(d/2)^2$ and $tan theta =d/b$ where d is the other side of the center rectangle with one size is $b$