Lenght in funcion of two diameters

[luca_pozzetto]

Hi,

i'll explain my situation.

I have a tube that has a rectangular solid attached on one of the two faces.

This solid is constrained 2mm away from the external diameter, and must be 3 to 8 mm above the internal diameter.

The length must be a multiple of 5.

I'd rather use Ilogic.

Thank you in advice!!

[JelteDeJong · Accepted solution]

You could make an extra parameter with the formula:

Wall_thickness = ( external_daimater - internal_diameter ) / 2 ul

Then the formula for the bottom parameter would be:

bottom = ( Wall_thickness - 2 mm - 3 mm ) % 5 mm + 3 mm

in my fast sketch that would look like this:

Of course, this could also be done in 1 formula in the bottom parameter. But would make it more difficult to understand the formula.

If you are wondering what % means in this formula have a look at "modulo operator". In short, it divides 2 numbers and gives as a result the remainder. As an example:

Wall_thickness = ( external_daimater - internal_diameter ) / 2 ul = 13

bottom = ( Wall_thickness - 2 mm - 3 mm ) % 5 mm + 3 mm

              = ( 13 mm - 2 mm - 3 mm ) % 5 mm + 3 mm

              = ( 8 mm ) % 5 mm + 3 mm

              = 3 mm + 3 mm

              = 6 mm

[J-Camper]

I'm not sure you need to do any iLogic for this.  See my attached part file.

It will force the external diameter to allow for a minimum 5mm rectangle length, 3mm bottom offset, and 2mm top offset.  Do not manually overwrite the "actual_external_diameter" parameter.

Let me know if you have any questions.

Edit: Didn't see @JelteDeJong 's post before posting.  I did a similar route using modulo to hold the rectangle to 5mm increments.  I did include a separate model parameter for the external diameter to prevent a user from creating an impossible external diameter.