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Defining function variables - Function does not work in a certain case

francois_leclercfrancois_leclerc Member Posts: 2
I have defined the following variable (function)
HalfWidth = function (len) { return len *.05 in; }
If I provide a number such as 1, it works fine and I end up with a width of .5
I have another variable WallThickness defined as 0.167 in.
If I provide this variable as a parameter to the HalfWidth function, I get an error message that this is not a valid expression.
I have another function to which I can provide the WallThickness variable and that doesn't pose a problem.
What should I do?

Comments

  • NeilCookeNeilCooke Moderator, Onshape Employees Posts: 5,686
    With 1 as a parameter the result is 1 * 0.5 in = 0.5 in. If you have a value with units the result is 1 in * 0.5 in = 0.5 in^2 which is an area not length. 
    Senior Director, Technical Services, EMEAI
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