Scribing an arc: do I have this right?

Factual Questions
1

I want to pound a stake in the ground and scribe out a regular (centered) arc between two points (call them m[sub]0[/sub] and m[sub]1[/sub]), through a center point (call it q[sub]0[/sub]). I know the distance between m[sub]0[/sub] and m[sub]1[/sub] (which, for conciseness, I will call 2m) and the distance from q[sub]0[/sub] to the chord of m[sub]0[/sub]~m[sub]1[/sub] (call it q), and that q < m (it will be a minor arc). So now I have to find r, the distance from m[sub]0[/sub], q[sub]0[/sub] and m[sub]1[/sub] to the stake (which will be the length of rope I need for the scriber).

What I come up with is that I can form a right triangle with m[sub]0[/sub] and q[sub]0[/sub] which has an angle at q[sub]0[/sub] of θ[sub]q[/sub] = tan[sup]-1[/sup] m/q and at m[sub]0[/sub] of θ[sub]m[/sub] = tan[sup]-1[/sup] q/m .

If I form a triangle between m[sub]0[/sub], q[sub]0[/sub] and the stake, it will have two sides of length r, which means two base angles of θ[sub]q[/sub]. Therefore, I can form a right triangle between m[sub]0[/sub] and the stake that has an angle θ[sub]r[/sub] at m[sub]0[/sub] of θ[sub]q[/sub]-θ[sub]m[/sub], which means that r = m/cos θ[sub]r[/sub] .

Is this correct? And is there an easier way to find r?

2

The value you’re calling q is called the sagitta of the arc. (Sagitta is Latin for “arrow”, because in a diagram it looks like an arrow fitted to a bowstring.)

The wikipedia article gives this formula, using your nomenclature:

r = (q[sup]2[/sup] + m[sup]2[/sup]) / 2q

The insight that makes this possible is that m and r-q make a right triangle with hypotenuse r.

–Mark

3

Ah, I see that now. I was thinking earlier that, by the angles rule (triangles add up to 180°), θ[sub]m[/sub] is also 90 - θ[sub]q[/sub], so θ[sub]r[/sub]would be θ[sub]q[/sub] - (90 - θ[sub]q[/sub]), but I failed to see the right angle there. Comparing numbers using my method to the fast method, I get a number that matches to 6 decimal places.

4

I don’t know if this is any use; but it is always possible to scribe an arc through any three points, using only a straight edge and a compass (No trig required). I guess that stakes and string would do at a pinch.

Here is a nice graphic showing how its done: How to construct a circle through 3 points with compass and straightedge or ruler - Math Open Reference