equation of an involute gear

Question

I'm trying to find the equation of an involute profile gear in polar coordinate system:

I know

$$ r \cos(\alpha - \theta) = R \tag{1}$$ $$l = \alpha R \tag{2}$$ $$r \sin(\alpha - \theta) = l \tag{3}$$

so far I can deduct the equations to

$$\theta = \alpha - \arctan(\alpha) \tag{4}$$

What I need to have

$$r = f(\theta) \tag{5}$$

but I can't find the inverse of equation $(4)$:

$$\alpha = g(\theta) \tag{6}$$

If I had $g(\theta)$ then

$$f(\theta) = \frac{R}{\cos(g(\theta) - \theta)} \tag{7}$$

I would appreciate it if you could help me know if there are analytical/closed-form solutions for $g(\theta)$ and/or $f(\theta)$. Thanks for your support in advance.

Answer 1

Polar coordinates isn't a good choice for involute shapes. It is unnecessarily complex. I prefer to work with cartesian coordinates and declare the curve in a parametric form $(x(t),\,y(t))$. The reason a polar curve is difficult is that there is no analytical solution to $r(\theta)$, but rather a parametric form of $r(t)$ and $\theta(t)$ exist, similar to the cartesian case.

First, consider an aligned coordinate system (shown below) resting on the generating circle (blue dashes). The involute shape (red curve) starts at the origin of the coordinate system and moves radially out as well as to the right. At any point on the curve, the distance to the tangent point (purple line) to the generating circle equals the arc length from the origin to the tangent point.

Define a parameter $t$ for the curve and note the equation of the involute curve in terms of the coordinate system shown above

$$ \pmatrix{ x(t) \\ y(t)} = \pmatrix{ R_0 \sin t - (R_0 t) \cos t \\ -R_0 + R_0 \cos t + (R_0 t) \sin t } \tag{1}$$

and here is what this curve looks like

The next part is the rotate the gear tooth in order to make sure you get the correct gear tooth width and contact pressure angle $\alpha$ at the pitch diameter (or radius $R_p$)

You need to rotate the coordinate system to the left a certain amount to achieve this, and part of this rotation is to account for the angular separation between the generating point and the contact point indicated above as ${\rm inv}\alpha$. This is the same notation in the ISO standard for involute gears. Use $\alpha$ for the desired pressure angle in radians and calculate the following

$$ {\rm inv}\alpha = (\tan \alpha) - \alpha \tag{2}$$

Now if the gear has $n$ teeth then each tooth must have arc length width of $s = \frac{\pi}{n} R_p$ encompassing an angle $\varphi = \tfrac{\pi}{n}$ as seen below

The total rotation of the coordinate system to generate the proper gear flank is thus

$$ \text{rot.} = {\rm inv}\alpha + \varphi/2 \tag{3e}$$

Note that the above is valid for external geometry, and for internal geometry, if you do a similar sketch you can find the corresponding rotation to be

$$ \text{rot.} = -{\rm inv}\alpha + \varphi/2 \tag{3i}$$

Answer 2

Directly $r=f(\theta) $ cannot be obtained in closed form, but $ \theta$ as $f(r/r_b) $ is possible.

Base radius $r_b$, radius at any point on involute is $r$, starting from $\theta =0, r= r_b\;;$

$$ \theta = \sqrt{(r/r_b)^2-1} -\tan^{-1}\sqrt{(r/r_b)^2-1} $$

A workaround to get $r=f(\theta)$ is through numerical integration of its ode; $r$ can be tabulated for required uniform increments of $\theta$, can be given if needed.