topic Re: 3D Equation Curve Help in Inventor Forum

3D Equation Curve Help

JDMather — Thu, 09 Jan 2020 16:11:15 GMT

I am trying to enter these equations into a 3D sketch equation curve.

How do I get theta as a function of t?

Re: 3D Equation Curve Help

Justin.B. — Thu, 09 Jan 2020 17:35:44 GMT

You can just sub the value you want for theta right into the cos function's brackets.

x(t) = R * cos( r/R * sin(t) )

y(t) = r * cos(t)

z(t) = R * sin( r/R * sin(t) )

Note your screenshot shows z(t) using cos(), but in your blue comment it's sin(). I don't get a particularly shapely result from either, but I am able to generate the equation curves entering it this way.

Re: 3D Equation Curve Help

JDMather — Thu, 09 Jan 2020 18:10:21 GMT

@Justin.B. wrote:

...Note your screenshot shows z(t) using cos(), but in your blue comment it's sin().

Oops, thanks for catching that.

That sort of simple mistake can be hard to spot without someone else reviewing the work - we miss our own mistakes.

Let me experiment some more...

Re: 3D Equation Curve Help

Anonymous — Thu, 09 Jan 2020 18:36:41 GMT

I believe theta needs to be an angular unit... and add a converter at the end of your formula like so...

r / R * cos(( d0 - d1) * 1 deg) * 1 deg

Re: 3D Equation Curve Help

Justin.B. — Thu, 09 Jan 2020 18:42:52 GMT

Bravo if you can define the equations, but you might also have an easier time just letting Inventor wrap the circles around circles for you.

Re: 3D Equation Curve Help

JDMather — Thu, 09 Jan 2020 19:05:53 GMT

Well, at least I see that success should be within reach...

Oh wait a minute, you used geometry rather than Equation Curve...

Re: 3D Equation Curve Help

Justin.B. — Thu, 09 Jan 2020 23:13:52 GMT

If you insist.

Re: 3D Equation Curve Help

johnsonshiue — Thu, 09 Jan 2020 23:37:04 GMT

Hi JD,

I took a quick look. Would Cylindrical coordinate system or Spherical coordinate system work better for this case?

Many thanks!