I am trying to figure out how I can calculate the total width of an object with a radius. The following dimensional images I have only show the width of the object itself, not the total width.
I created a demonstration below. The area between the blue lines is what I am trying to figure out.
Is there a formula that can help me figure these out?
On
For a start neglect corner radii used to smoothen corners while computing width $W$. Recognize that outer raius is double inner radius.
$$ x_1= r \cos \alpha;\quad x_2=R ; \quad R=2r; $$ $$ W = R - r \cos \alpha = r ( 2 - \cos \alpha)$$
$$r= 40 \,(mm) $$
$$ W_1= 40 (2-\cos 30^{\circ} ) \approx 45.359 $$ $$ W_2= 40 (2-\cos 45^{\circ} ) \approx 51.7157 $$
Next include corner round off dimensions for better accuracy; since rounding radius is not specified it is omitted in this answer.
Suppose you have a small circle of radius $r$ and a large circle of radius $R$, centered around the origin of a coordinate system. From both you take a segment from angle $0$ to angle $\alpha$. Then the points touching the vertical extremal lines will be
$$A=\begin{pmatrix}r\cos\alpha\\r\sin\alpha\end{pmatrix}\qquad B=\begin{pmatrix}R\\0\end{pmatrix}$$
so the width is the difference of $x$ coordinates, namely $R-r\cos\alpha$.
Now you know that $\alpha=30°$ resp $\alpha=45°$ and you also know $R-r=40$ is the width of the ring. But you don't seem to know $r$ or $R$ itself. So even assuming the corners are not rounded, you will need additional information besides what is given in your figure.
You could estimate either radius from measements you perform on the figure, but in that case you might as well measure the width there, and obtain the scale from one of the given dimensions to convert from figure to real world dimensions.
Update: If Connot Harris is correct in assuming that the $80$ in these sketches refers to the outer radius $R$, then you can solve this just as he said:
$$80 - 40\cos30°=80-20\sqrt3\approx45.36\qquad 80 - 40\cos45°=80-20\sqrt2\approx51.72$$