# Problem With Helix Command

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How do you edit the 'Turn Slope' of an AutoCAD 2011 helix?

The default value is 0.000 and it offers the QuickCalc to edit it, but I haven't figured out how to do so. I want a turn slope of 32 degrees but I can't change the 0.000 value.

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Seems like a mathematical problem to me. Turn Slope is a function of diameter (or Radius) and Turn Height.

Turn Height

Specifies the height of one complete turn within the helix.

The number of turns in the helix will automatically update accordingly when a turn height value is specified. If the number of turns for the helix has been specified, you cannot enter a value for the turn height.

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Please see the attached file for additional information about what I want to achieve and what I've been able to achieve with the helix command in AutoCAD 2011.

I'm also trying to do this in Inventor but I'm not as experienced with it. Thanks.

Doric-12-spiral-comparison.pdf

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Seems like a mathematical problem to me. Turn Slope is a function of diameter (or Radius) and Turn Height.

I should have added "Number of Turns".

If you have 12 turns at a turn height of 1 and a fixed height and radius, you can increase the angle by reducing the number of turns. That is the only way you can change that angle AFAIK. What you really want is not possible except maybe a LISP.

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In Acad 2010 I go with the Helix command then follow the prompt, bear in mind I am set up for metric mm.

Command: _Helix

Number of turns = 12.0000 Twist=CCW (this line shows last used number of turns)

Specify center point of base:

Specify base radius or [Diameter] :

Specify top radius or [Diameter] :

Specify helix height or [Axis endpoint/Turns/turn Height/tWist] : t

Enter number of turns : 12

Specify helix height or [Axis endpoint/Turns/turn Height/tWist] : h

Specify distance between turns :

Hope this helps.You may need to go one turn extra to get a clean trim of the object for the base and capital.

Rob.

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I think more lobes will be needed to get the indicated slope. Three works pretty well.

HelixSection.dwg

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Your right Seant, I approached it as wrapping a single rope around a cylinder, erroneously so.

The two examples in the pdf show 32° and 11° angles on the helical lines, I have had trouble in determining those (pitch) angles from a 3D representation that I made (even properties won't tell you that one!)

Could you expand on your solution as it may be helpful to many of us.

Cheers,

Rob.

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First off, thank you to everyone for their valuable input.

I came to the same conclusion that the rope cord couldn't be achieved as a single swept extrusion but rather as 3 identical swept extrusions polar arrayed around the column shaft. I have listed the steps below.

Steps to Achieve an AutoCAD Spiral Column Shaft:

1. Draw helix from bottom of base to top of capital with base and top radius set to shaft radius.

2. Constrain the height of the helix using Properties palette.

3. Reduce the number of turns until the pitch of the helix matches the pitch of the 2d elevation.

4. Constrain the turn height and adjust the height of the helix and move it in order to align the helix turns with the center of the spiral cord of the 2d elevation.

5. Sweep profile along helix and view from Front view.

6. Count the number of spiral cords, including the swept profile, until the swept profile repeats.

7. Use this number in a polar array centered on the center of the shaft.

8. Add a solid cylinder extrusion to the center of the shaft.

9. Union everything together.

10. Slice the solid back to the top and bottom of the shaft.

11. Subtract horizontal joints from shaft and separate.

I have also attached a comparison between the original 2d elevation and a flatshot of the 3d model. Since the pitch of the helix was eyeballed to match the pitch of the 2d elevation it's not perfect but the important design dimensions are achieved.

I hope this helps. Thanks again.

Doric-12-spiral-comparison-new.pdf

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