# recreating old mechanisme

I'm recreating an old mechanisme in 3d modeling software. The idea is that this triangular shape rotates around an off-center axis. This will create 4 phases. first (like in the picture) the construction L is down and stay stable in that position (which means the parts of the triangle thing that touch the construct L are partially circle. In the next phase the construction moves up, then it is stable again and in the last phase it moves down. Questions are: what is the mathematics involved in the shape assuming it never looses contact with the construct L. Initially I made the sides out of 6 circular elements (arcs), but I'm not sure if that is even the correct shape. Once I have the shape, what is the position of L as function of the rotation of the object?

I think you are looking for a "curve of constant width".

Figure 1. Construction diagram for generating a curve of constant width from a triangle. Source: Curve of constant width.

Curves of constant width can be generated by joining circular arcs centered on the vertices of a regular or irregular convex polygon with an odd number of sides (triangle, pentagon, heptagon, etc.).

• When you design cam mechanisms it would be s good idea to ensure that the curve is continious to the third derivate of position (derivates being velocity, acceleration, jerk ( followed by snap, crackle and pop)) Commented Oct 1, 2019 at 17:49

To create your triangular pick shape, you need two equilateral triangles, centered on each other. Your spline should make contact with each of the corners of the triangles.

The ratio of lengths between the two triangles will give you the curvature.

Your ultimate problem is you don't have dimensions so you can do your math. This SolidWorks drawing will give you the dimensions to build your slot.

Two rotations of Part C will give you the dimensions of your guideway.

the answer to the question about the position of L versus the angle a is divided into six ranges for the angle :

0-60:       r1
60-120:     (r1-r2)*cos(a-60)+r2
120-180:    (r1-r2)*cos(a)+r1
180-240:    r2
240-300:    (r1-r2)*cos(a-60) +r1
300-360:    (r1-r2)*cos(a)+r2


Where r1 is the radius of the bigger circle and r2 is the radius of the smaller circle. In my case in figure one of the answer from @transistor A, B, C, a,b,c form nice equal sided triangle so only two radii are involved.

• this is not right the angle ranges are dependent on the relation of radiuses. Commented Oct 2, 2019 at 15:13
• are they dependent on the radii for equilateral setup? I don't think so. they might be, but in the animation I made using these formulas with r1 = 26.5 and r2 being 6, it looks kind of perfect. So if they are depend on the radii the difference is at least very small.
– illu
Commented Oct 2, 2019 at 16:34
• are they dependent on the radii for equilateral setup? I don't think so. If you for example fill in 60 degree in the second range the answer is r1, which matches the first range. The same is true for all other edges between the ranges. The movement of L is continuous.
– illu
Commented Oct 2, 2019 at 16:41
• I can easily get the angles down to under 30 degree for the shorter spans. Its not rotating about the center. And not necessarily around the corner of the construction triangle. (atleast the one in the image is not) Commented Oct 2, 2019 at 17:08
• look at this figure 1 above, but with a, b, c being equal. Lets choose point B as the point of rotation. The line through A and B also splits the circle in the blue and the red part. If we rotate 60 degrees counterclockwise, the line through A and C will be vertical and that will exactly split the circles at the top into the red and the green one. That is totally independent of the radii.
– illu
Commented Oct 2, 2019 at 17:29