# Scribing an arc: do I have this right?
**URL:** https://boards.straightdope.com/t/scribing-an-arc-do-i-have-this-right/767600
**Category:** Factual Questions
**Created:** [October 3, 2016, 9:58pm UTC](https://boards.straightdope.com/t/scribing-an-arc-do-i-have-this-right/767600 "2016-10-03T21:58:32Z")
**Posts on this page:** 4
**Page:** 1
### Author:  [@eschereal](https://boards.straightdope.com/u/eschereal)
#### Post date: [October 3, 2016, 9:58pm UTC](https://boards.straightdope.com/t/scribing-an-arc-do-i-have-this-right/767600/1 "2016-10-03T21:58:32Z")
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?
---
### Author:  [@markn\_1](https://boards.straightdope.com/u/markn_1)
#### Post date: [October 3, 2016, 10:17pm UTC](https://boards.straightdope.com/t/scribing-an-arc-do-i-have-this-right/767600/2 "2016-10-03T22:17:20Z")
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](https://en.wikipedia.org/wiki/Sagitta_(geometry)) 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
---
### Author:  [@eschereal](https://boards.straightdope.com/u/eschereal)
#### Post date: [October 3, 2016, 10:44pm UTC](https://boards.straightdope.com/t/scribing-an-arc-do-i-have-this-right/767600/3 "2016-10-03T22:44:13Z")
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.
---
### Author:  [@bob\_2](https://boards.straightdope.com/u/bob_2)
#### Post date: [October 4, 2016, 11:38am UTC](https://boards.straightdope.com/t/scribing-an-arc-do-i-have-this-right/767600/4 "2016-10-04T11:38:28Z")
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](http://www.mathopenref.com/const3pointcircle.html)