• Industries
  • Products
  • Buy
  • Services & Support
  • Communities
  • Discussion Groups

    Autodesk Inventor

    Reply
    Contributor
    cwilko
    Posts: 20
    Registered: ‎07-01-2010

    Question about Trig functions in Paremeters

    146 Views, 1 Replies
    11-28-2011 08:03 PM

    Hello All

    Lately it seems the only questions I have to ask is to do Parametric Trig equations.

    So here we go again.

     

    I am creating an equation driven part with a few Trig eqautions. (Not that difficult until now)

    One equation I need to input requires the Trig function Cotangent (cot).

    eg. cot (x deg / n ul) with the unit type set at (ul)

    This syntax does not work.

    But when I rewrite the equation as 1 ul / tan (x deg / n ul) or tan (x deg / n ul)^-1 ul I get the result that cot should give me.

     

    Does Inventor not support the Cotangent, Secant or Cosecant functions?

    Or am I just misrepresenting the syntax description?

    Thanks Clint
    Inventor 2012 Professional
    Please use plain text.
    *Expert Elite*
    sam_m
    Posts: 451
    Registered: ‎11-05-2003

    Re: Question about Trig functions in Paremeters

    11-29-2011 01:02 AM in reply to: cwilko

    list of valid functions (in "References" -> "Use formulas and equations" -> "Functions")

     

    http://wikihelp.autodesk.com/Inventor/enu/2012/Help/0073-Autodesk73/0733-Design_O733/0736-Paramete73...

     

    no cot, so guess you're stuck with 1/tan

    ----------
    Please mark this response as "Accept as Solution" if it answers your question - but note that the solution may not be the answer you're wanting to hear...

    Lithium - helping nntp users with mania, depression and headaches
    Please use plain text.