Can the radius of this constructed arc be found by knowing the two starting length values (1.6 and 3.7) **and nothing else**?

One expression with x and y as input (1.6 and 3.7) that gives me the result of the arc radius (2.196)?

arc.gh (7.2 KB)

Can the radius of this constructed arc be found by knowing the two starting length values (1.6 and 3.7) **and nothing else**?

One expression with x and y as input (1.6 and 3.7) that gives me the result of the arc radius (2.196)?

arc.gh (7.2 KB)

I think the end-point of those two segments (1.6 and 3.7) define only two points in the plane, which are not enough to define a single circle (you need at least 3 points on circumference, or 1 point on circumference + center?)

a note: in your definition you have also set the direction of the Start Tangent point to be -X, that is an important costrain

otherwise you get this indetermination:

so you actually have the center of the circle to be always along the Y axis

The circle is clearly defined with the command ARC SED, so this is not the problem. I want to calculate the radius using the length of the two lines I know using some trigonometric functions (SIN and COSIN).

I understand that

the point is, to have a SED arc you need 3 points and a tangent vector

in case you are happy to have always the very same tangent vector as {-1,0,0} or {+1,0,0} (they both produce a circle with same diameter) despite x and y changing, then that is the equivalent of a circle per 3 points, where A and A’ are mirrored by the Y axis along B, because the center of the circle always falls on the same vertical as B

`r = (x² + Y²) / 2Y`

arc_Re.gh (14.8 KB)

3 Likes

geometric solution.

relies on the fact that a triangle which contains a line that goes through the center and another point on the circle is always a right angle triangle

arc [geometric_solution].gh (15.3 KB)

Thank You very much!

Perfect!

Thank You Adel, nice solution too!

Beautiful!