Create Simple 2D Equation curves

[Anonymous]

I have no real formal training on Inventor Professional 2018. I have successfully graphed (x^2)/(1in*1in) however I cant seem to make an equation that works for x^(1/5), or x^.2. I keep getting the error "Equation y(x) contains unit errors. Expected units are length". I have tried troubleshooting but nothing has worked. Any help would be welcomed; anything to continue learning.

[swalton]

Inventor tracks both dimension value and units.  It knows the difference between a line 1 inch long and the area of a square 1 in on a side. 

If you have a parameter that is expecting length units, and your equation produces area units, Inventor will give you a warning.  

Add a unit conversion factor to your equation to make sure that Inventor calculates the correct value and unit for the output.

[Anonymous]

i was taught how to make shapes with lines, rectangles, and circles and how to extrude them. I am thankful for a quick reply however I don't know what conversion factor I should use or what that means in Inventor. What do I need to convert into inches? I just want to graph a simple equation but the complexity of Inventor combined with my lack of experience and resources leave me to struggle with the knowledge that my inquiries are simple and yet I can't figure out how to solve them.

[JDMather]

This boils down to HS math where you were taught to cancel out units.

If your equation operation is expecting inches, but you have inches^2, then you need to divide inches^2/1 inch to cancel out one of the inch.  (area value vs length value)

So this is not an Inventor issue - this is a math equation issue.  The same issue occurs in the other MCAD softwares that I use.  Must account for units of each magnitude.

Back in the last century (BC - Before Computers) the math teachers hammered this into my head.

I would get counted off if I didn't show my work and if didn't show how units were canceled out.

One thing you might get away with is go unitless (ul) with all number till the end and then multiple by 1unit.

[swalton]

Ok.  This is just like high-school physics problems.

Take a look at this sample equation:

X=y+y^2

X needs to be in inches.

Y =5 in

Substituting gets

x=5in +(5in)^2

x=5in +25in^2

Inventor knows that adding 5in to 25in^2 does not work.  It is not possible to add linear inches to square inches.

To fix the issue, add a correction factor to convert the square inches to linear inches.

The equation becomes

x=y+(y^2)/1in

or

x=5in+(5in^2)/1in

x=5in+25in^2/in

x=5in+25in

In your case, write out your equation on paper.  Substitute known values, with the units.  Run through the algebra/calculation to verify that you get the correct units in your final answer.

Once you get it working on paper, review what you typed into the equation curve, and make sure you use the correct unit conversion factor.  

[johnsonshiue]

Hi! Like the experts mentioned here, you will need to make the terms unitless before adding power. For example y = x^0.2. x is length with unit of mm or inch. You will need to write your equation like y = (x/1mm)^0.2 and then it should work.

Many thanks!