Create Helical Sweep

[ChristianSmith]

I know how to create a standard through axis helical sweep but what I'm trying to do now is create a helical sweep with the section being a circle. The problem I'm having and can't figure out is I don't want to sketch the section on the vertical XY plane, I want to sketch the section on the horizontal or flat XZ plane so that the circle is normal to the axis of revolution not parallel. Is that possible?

This is the axis system I'm referring to...excuse the crudeness of the drawing.

Y

|

|

|_____X

/

/

Z

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[DavidButz]

I'm quite sure you can do what you want, but I'm not clear what that is. "Horizontal" and "vertical" don't tell me much, because those directions are arbitrary. Are you trying to create something like a spring, or something different? Springs are a piece of cake with regular functionality.

[ChristianSmith]

Refer to the image. I created this just to explain what I'm talking about. The helical curve is the trajectory, the circle is the section I want to sweep along the curve. Notice the difference in how the circle is sitting compared to making say a spring for example. The circle isn't normal to the trajectory like a normal sweep would be. Normally you would be forced to draw the section in this example on the top datum, whereas I want to sketch my section on the front datum and sweep that along the curve. Is that possible?

[DavidButz · Accepted solution]

Ah, now I understand. Sorry I was so dense. You mean like this?:

Yes, it's possible.

This was created in two steps. First create a Thru-Axis Helical Sweep Surface which is just a flat ribbon extending outward with a width which is the radius of the circle. Now create a Variable Section Sweep with two trajectories. Select the inside edge of the swept surface for the first trajectory and the outside edge for the second. Choose the Constant Normal Direction option and pick the base plane as the reference. Now sketch a circle with its center on the start point end vertex of the first trajectory and its radial edge on the end vertex of the second trajectory.

[DavidButz]

(In section)

[ChristianSmith]

That's exactly what I wanted to do, thanks for the help.