Thread: flat sheet layout of helix?
View Single Post
  #10   Spotlight this post!  
Unread 08-02-2006, 11:50
Bob Holmstrom Bob Holmstrom is offline
Registered User
FRC #2411
Team Role: Mentor
  
Join Date: Jan 2006
Rookie Year: 2004
Location: Portland, Oregon
Posts: 18
Bob Holmstrom is an unknown quantity at this point
Re: flat sheet layout of helix?
DJ and Henry,

Thanks for your help on this task!

Our helix is 26.5" outer diameter and 7.5" wide (inner diameter 11.5") with a pitch of 11". If I draw the helix in Inventor, and measure the edge lengths I get the following:

outer edge > 83.976 inches (a 26.5 inch circle = 83.252)
inner edge > 37.766 inches (a 11.5 inch circle = 36.1284)

DJ suggests concentric arcs contrained in length and separation. Arc implies circular and so far I have not discovered a way in Inventor to constrain the length of the arc. But working with it bit, it seems that the arcs may need to be sprial arcs not circular arcs.

Bob