3D Equation Curve Help

[JDMather]

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

How do I get theta as a function of t?

[Justin.B.]

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.

[JDMather]

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@Justin.B. wrote:

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

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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...

[Anonymous]

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

[Justin.B.]

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.

[JDMather]

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

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

[Justin.B. · Accepted solution]

If you insist.

[johnsonshiue]

Hi JD,

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

Many thanks!