How to set equal dimensions for several lines when you do not know the length required?

[eretamoza]

Hello all,
I have the following issue that I need to solve without using midpoint or symmetry. I have one sketch with one Circe inside a square. The square has 4 lines going from each corner to the circle inside. If the length of each of those four is equal, the circle will be in the center of the square, right? So, can CATIA calculate this dimension somehow? For example, in Autodesk Inventor, you use the EQUAL contraint and then the length of the lines is calculated automatically, the circle centered, and no dimension need to be typed.

How can I do this in CATIA? I know that with formulas I can set the length of 3 lines to be equal to the fourth, but I still have the problem that I do not know the correct length to use.

 

[eretamoza]

attachment

 

[rstupplebeen]

If you set the 3 lines equal to the 4th and set the 4th to be reference. Alternatively, if you set each of the diagonals to be perpendicular to the next you may get closer to the design intent. I hope this helps.

Rob Stupplebeen

https://sites.google.com/site/robertkstupplebeen/

 

[jackk]

I don't think this can be done with CATIA. You can't constrain the line lengths, without having the actual length.

If the 4 lines were symmetrical, you could probably define some geometric constraints to force the lines to be equal lengths.

 

[eretamoza]

Hi,
Thanks, I tried your suggestion and it worked. But when I tried to follow the same process with a more complex sketch it goes crazy and "calculates" and "recalculates" the lengths non-stop. Any other way how to accomplish the same result?

 

[itsmyjob]

I did the following:

set 1 coincidence between 1 inside line and its opposite (so they kinda form a diagonal)

set another coincidence on the other set of line.

This should fix all inside line but he circle can still move.

Set another coincicence between the circle center and 1 inside line

and again set another coincidence to another inside line.

Circle is now fully constraint and centered (whatever cube size and circle diameter)

Eric N.
indocti discant et ament meminisse periti ​

 

[rstupplebeen]

Can you explain the more complex sketch, post a pic or the model?

Could a radial pattern work?

Rob Stupplebeen

https://sites.google.com/site/robertkstupplebeen/

 

[eretamoza]

Hi Rob,
You can find my sketches attached. One sketch with the positions of the lines (fig. 1 in the pdf) plus one sketch with the lines that need to have the same length (fig. 2 in the pdf).
My complete task is describe here Link
I am stuck in fig. 3 of the pdf.

 

[rickyt]

Create a Formula using the a²+b²=c².
Formula will divide the hyp. length in half and subtract radius of circle.

Set the length of your 4 angled lines to = length of the formula.
Attached images show the location of center point as REF constraints to show their value.

Formula would look something like this:
sqrt(PartBody\Sketch.1\Length.5\Length * PartBody\Sketch.1\Length.5\Length + PartBody\Sketch.1\Length.6\Length * PartBody\Sketch.1\Length.6\Length) / 2 - PartBody\Sketch.1\Radius.7\Radius