What is the relationship between the width, height, and radius of an arc? What is the relationship between the width ($w$), height ($h$), and radius ($r$) of an arc?
Specifically, the relationship in terms of $h$.
I know this is a simple question - I'm a hobbyist engineer, and I'm having one of those moments, where your mind goes blank and you can't remember the simplest thing.

 A: Another take.  
Let $(0,-r)$ and $(x, y)$ be the endpoints of the arc.  Then
$h = |r - y|$ and $w = |x|$.
We know $x^2 + y^2 = r^2$ so $h = |r \pm \sqrt{r^2 - w^2}| = r \pm \sqrt{r^2 - w^2}$. And $w = |x| = \sqrt{r^2 - y^2} = \sqrt{r^2 - (h-r)^2} = \sqrt{2hr - h^2}$
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However not your interpretation of width and height are a little odd.  You assume you are lying the arc on its tangent (like pressing one end of a clipped toenail to the floor) and measuring the displacement.
I think most people would assume the width is the measure from endpoint to endpoint and the height to be the displace of that. (Like leaving the toenail balancing one its center and measuring how high and far apart the ends are).
(The difference between the new picture and your picture is in the new picture the radius goes through the arc cutting it in half.  The height is from the midpoint of the chord to the tip of the circle, and the width is the length of a chord.)
(In Your picture the chord is the hypotenuse slanting upward. Width is the horizontal leg and the height is the vertical leg.  I think most would view the chord itself as the width, and the height as the perpendicular bisector of the chord to the circle.)
In this case the endpoints are $(-x,y)$ and $(x,y)$ and
$w = 2x$ and $h = r - y$.  Using $x^2 + y^2 = r^2$ we get:
$w = 2\sqrt{r^2 - (r-h)^2} = 2\sqrt{2hr - h^2}$ and
$h = r \pm \sqrt{r^2 - ((1/2)w)^2} = r \pm \sqrt {r^2 - w^2/4}$.
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Note.  Given an arc's width and length you can calculate the radius of the circle.
I probably should have included those as well.
In your picture:  $(r-h)^2 + w^2 = r^2$ so $w^2 = 2hr - h^2$ so $r = (h^2 + w^2)/2h$.
In mine:  $(r-h)^2 + (w/2)^2 = r^2$ so $w^2/4 = 2hr - h^2$ so $r = (h^2 + w^2/4)/2h$.
A: 
$w^2+(r-h)^2=r^2$
$w^2+r^2-2hr+h^2=r^2$
$h^2-2hr+w^2=0$
Quadratic in $h$.
Solution:
$h=\frac{2r\pm\sqrt{4r^2-4w^2}}{2}$
The positive case is rejected, leaving us with:
$h=\frac{2r-\sqrt{4r^2-4w^2}}{2}$
$h=r-\sqrt{r^2-w^2}$
The proof that the positive case is rejected is left to the reader as an exercise.
