topic Re: Create square feet property using existing Length and Width properties in Inventor Forum

Create square feet property using existing Length and Width properties

Shag_Bore — Thu, 01 Sep 2022 14:18:54 GMT

I have a custom library of parts. This example I created an iPart of all our aluminum sheets then published it.

So now in my assemblies I can drop in the "cut to size" piece of sheet from the list of options. I have created custom iproperties so in my parts lists I can show filler length and filler width. What I would like to do is use those iproperties to calculate sq ft.

(length/12)*(width/12) and if possible round up to nearest whole foot.

In the example I was just trying to get the expression to work by multiplying length *width but it just displays the actual words.

Looking for help to get this to display the actual calculation.

Thanks!

Re: Create square feet property using existing Length and Width properties

Gabriel_Watson — Thu, 01 Sep 2022 14:37:14 GMT

I believe your data type for all of these columns involved must be set to "Real" first. On the top-right corner you can see it's currently set to String, where these operations cannot be done. I would also try to envelop the properties in brackets, such as: {}

Re: Create square feet property using existing Length and Width properties

Shag_Bore — Thu, 01 Sep 2022 15:11:18 GMT

As soon as I try to add anything into the Expression input box it is red text and I can't press OK because it is greyed out. This occurs when changing data type to real.

Re: Create square feet property using existing Length and Width properties

blandb — Thu, 01 Sep 2022 16:40:02 GMT

You can place a part as custom. Then should be able to follow the attached video, then replace family template. It should then place the "ft^2" as part of the template. Now each time you use it, it should work.

Hope that helps.

Re: Create square feet property using existing Length and Width properties

Shag_Bore — Thu, 01 Sep 2022 17:42:26 GMT

Thank you, this solution is not ideal, but at least I have an options, deleting the custom parts from the editor and adding in the Area parameter into the part then republishing the part back to custom library and re-adding all my custom columns will take some time. Too bad I can't create the area calculation within the family table....

Re: Create square feet property using existing Length and Width properties

Shag_Bore — Thu, 01 Sep 2022 17:58:28 GMT

For some reason in the expression input box Inventor doesn't like when I type {Length}, however if I press the ...button to the right of the input box and select the parameter length and width it will allow it.

So I initially setup {Length}*{Width} and got the desired result, however that was in inches.

When I set up the expression {Length}*{Width}/144 to get square feet, It doesn't work. Is this a formatting error inside the expression box?

Re: Create square feet property using existing Length and Width properties

blandb — Thu, 01 Sep 2022 18:00:46 GMT

You are not deleting the family and starting over. When you right click on the family there is the option to "Replace Family Template" All it does is just applies these parameters. This is how you make modifications to published parts. See attached video:

Re: Create square feet property using existing Length and Width properties

blandb — Thu, 01 Sep 2022 18:01:52 GMT

But, I don't have 2016, and I am not certain if the parameter "ft^2" is recognized in that release or not though.

Re: Create square feet property using existing Length and Width properties

blandb — Thu, 01 Sep 2022 18:15:23 GMT

Change your units to "ul" in your image....

You are currently saying : (Inch x Inch)/144 = ft^2, but my unit here for the column is supposed to be "ft" based on what you image has, but if you make it unitless, then you should get what you want.

Re: Create square feet property using existing Length and Width properties

A.Acheson — Thu, 01 Sep 2022 19:45:58 GMT

These images are from 2020.

So the reason you can't type in length in for the expression is that that name is just a display name and what is required is the internal name of the column.

Type in the sq unit manually "in^2"

When you re-enter the column properties the expression is red but the area is still calculating when the part is placed. I guess it is flagging the missing units.

Re: Create square feet property using existing Length and Width properties

Shag_Bore — Thu, 01 Sep 2022 19:51:33 GMT

Oh nice!!!! I will try implementing this tomorrow! looks exactly like I am trying to do!!

Thanks! Have a good night

Re: Create square feet property using existing Length and Width properties

Shag_Bore — Fri, 02 Sep 2022 10:30:13 GMT

This work perfectly! Thank you!