Creating a helix from a partial ellipse

elliptical stair-1.3dm (442.5 KB)

Can anyone help me create a helix from a partial ellipse?

This might be a start…
Take the underlying elliptical curve segment and convert it to arcs (_Convert)
Take the last arc segment (should have the smallest radius?) and measure its radius
Use that radius for your helix

Seems like the only thing left would be to determine the “rise” of each stair tread so you could determine the pitch of the helix. Don’t think I have enough info to move on from here.

I tried the method listed above, and I got a radius of 53.6461 on the last segment, but your listed radius is 51.8766. Not quite sure how to reconcile that difference, but I would probably do something like go ahead and make the helix as calculated, and then make an adjustment by splitting off a bit of the bottom of the helix and a bit of the end of the elliptical curve, and do a Blend Curve to smoothly bridge the gap. You’d have to go back and adjust the stair treads in the transition area.

Best regards,

I’d find the full ellipse - Explode the elliptical curve, choose the longer bit, an elliptical arc, run MarkFocii, then create a new ellipse FromFocii. Then you can make a regular helix at the center and Scale1D to match the ellipse, split out the part you want to use.

Dunno if I interpreted your design correctly, but this is the idea.
elliptical stair-1_PG.3dm (485.5 KB)


My guess is the “helix” you need for the stairs is one with a constant angle to horizontal such as the green curves in elliptical stair-1DC.3dm (507.4 KB)

This can be created using some simple geometry. First the method, then the geometry.

  1. ExtrudeCrv the inner and outer curves to create temporary surfaces the height of the stairs.

  2. The stairs have 15 treads so there will be 16 risers. Mark the heights of the risers along the edges of the temporary surfaces with points using Divide and 16 for the number of segments.

  3. Use ShortPath to create the helixes on the temporary surfaces with the curves starting and ending at the appropriate points.

The helixes lie on elliptical cylindrical surfaces (the temporary surface) which are developable and can be unrolled. The shortest distance between two points on a developable surface (generated by ShortPath) is a straight line when the surface is unrolled, and will be at a constant angle on a cylinder.

Thanks David, extremly helpful.