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Equation driven curve - unit error

23 REPLIES 23
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Message 1 of 24
19112033
4407 Views, 23 Replies

Equation driven curve - unit error

Hi there,

I would like to create an equation driven curve with inventor. I have no problems plotting the function in a plotter, however, inventor won't accept my equation. It gives me the error message, saying: "The equation contains unit errors". I'm using the explicit equations.

That's the bare equation:

 

"y=sqr(2)*16.5/2/sqr(pi)*sqr(arccos(1-2*x/50)-(sin(2*arccos(1-2*x/50))/2))"

 

I modified it, inserting variables which I defined also in inventor, leading to this:

 

"sqrt(2)*height/2/sqrt(PI)*sqrt(acos(1-2*t/length)-sin(2*acos(1-2*t/length))/2))"

 

But both equations are not accepted by inventor due to unit errors. In this design, I'm using the metric system, millimeters, so I don't understand why and how I have to redefine this for the equation in order to get it working.

 

I would appreciate your help,

Max

23 REPLIES 23
Message 21 of 24
karthur1
in reply to: 19112033

Max,

Can you show the function that you have plotted in message #20.

 

Kirk

Message 22 of 24
19112033
in reply to: karthur1

Sure, this is the equation:

 

"y=sqrt(2)*16.5/2/sqrt(PI)*sqrt(acos(1-2*x/50)-(sin(2*acos(1-2*x/50))/2))"

 

Plots like a charm. 

 

Best regards,

Max

Message 23 of 24
glenn-chun
in reply to: 19112033

Hi Max,

 

I've got a solution for you.

 

parameters_dialog.png

I entered the following equation:

 

multiplier * sqrt(acos(1 ul - 2 ul * x / length) / 1 rad - sin(2 rad * ( acos(1 ul - 2 ul * x / length) / 1 rad )) / 2 ul)

 

equation_curve_for_max.png

 

Change the first two user parameters (length and height), and the equation curve will update accordingly.

 

The attached IPT file can be opened in Inventor 2013 or later.

 

Hope this helps,

Glenn



Glenn Chun
Sr. Principal Engineer
Message 24 of 24
19112033
in reply to: karthur1

Hi Glenn,

sorry for this late response, but thank you for your reply! Your solution works perfect for me, though I don't know why Inventor has to be so difficult in this manner, but who knows. 

Thank you again for your help, I will mark this as a solution.

 

Best regards,

Max

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