How to create a curved triangle of specific dimensions?

Question

I'm trying to make a triangle where all the sides are curved outward, like a Reuleaux triangle, but the sides must be of a specific length. The length must be measured along the curve, not from end to end in a straight line, and the shape must be a specific vertical height. Is there a way to do this, including using any software or websites? I've included a picture to show what I'm trying to do. Thank you!

Answer 1

There are lot of software that can do this. Mostly CAD software, but both geogebra and solvespace are free. On commercial space anything parametric that has a 2D solver does this for you AutoCAD, Catia, Creo, Fusion 360 Solidedge, Solidworks...

So example from Creo. A bit depending on the particulars of your constraints the system seems to have zero, one or two degrees of freedom. So if we assume that the arc is symmetrical across the middle of the line we get according to the CAD one degree of freedom. Ive chosen to lock the freedom on a dimension of the angle at the top at 130 degrees. Since that seems prudent for a cutting pattern which is expect this to be but YMMV on particulars. See picture below:

And here i asked it to do a give me from 100-160 angles in 10 degree increments a vector format (2 commands):

Now is Creo the app to choose? Perhaps not, its not a terribly good presentation drawing software so you would still need inkscape or illustrator as a pair for it. But I'm merely pointing out that this is relatively easy in CAD application horrible in nearly anything else. Mostly its out of the price range for most users out there. If you want one package that does both good free drawing and parametrics consider AutoCAD.

Anyway this is not graphic design anymore

Answer 2

The actual graphic design seems already been done - you have decided what shapes are needed. The question asks only some hints how to draw them.

You have already got something, but here's a little more. The case becomes possible to be worked with Excel worksheet if you have good high school level geometry, trigonometry and plane vector algebra knowledge and you accept 2 radically simplifying assumptions based on your drawing. The assumptions are:

• The shape is left-right symmetric, it's enough to solve one half of it
• the arcs meet perpendicularly in bottom corners

The latter assumption is so restricting that you need to solve only one vector equation which splits to 2 scalar equations to x and y components.

The result with your given side lengths is shown in the right half of the next image:

There's a rectangle and 2 circles drawn to its opposite corners. The green area is a half of your shape. The numbers are inches.

The only equation which needs to be solved presents the brown vector (see the left half of the drawing) in two ways: The brown vector is the dotted magenta line rotated counterclockwise along the known length arc. As well the arrow tip of the brown vector can be reached by falling downwards from the center of the red circle to the top of the 6,5 inch long line.

The key observations to see that one vector equation is enough is to see that

• the magenta lines make a rectangular triangle
• angle A is as big as angle B due the perpendicularity of the angle sides.

I am not going to tease people here with equations. If one could understand them he as well writes them in a minute based on the given text explanations. I put Excel solver to search the right radiuses to the circles.