Inventor General Discussion

Inventor General Discussion

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cwilko
Posts: 20
Registered: ‎07-01-2010
Message 1 of 2 (272 Views)

Question about Trig functions in Paremeters

272 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
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sam_m
Posts: 598
Registered: ‎11-05-2003
Message 2 of 2 (263 Views)

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

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