I'm not sure how to do this but I have a ladder designed and it can vary in length from 1m to 4m. I'd like the number of diags to change based on the ladder length. The length of the diag should change to make up any differnce, not the gap between them.
Solved! Go to Solution.
Solved! by cadull_rb. See the answer in context.
See the part attached to my post http://forums.autodesk.com/t5/Autodesk-Inventor/Incomplete-part-update-when-sketch-pattern-count-changes/td-p/3375669 for parameters that can be used to control the pattern. If you can, only sketch the unique instances and pattern the resulting features.
Solved! by stephengibson76. See the answer in context.
you could try using functions such as dividing by 2, rounding then x 2
See the part attached to my post http://forums.autodesk.com/t5/Autodesk-Inventor/In
You don't need iLogic to achieve this, it can all be done through parameters.
After modelling the ladder (using a component pattern to specify the amount of diags), give the parameters suitable names (eg. LadderLength, DiagQty, DiagOffset, etc.).
Then in the equation column for LadderLength, type something like this:
(DiagQty * DiagOffset) + OffsetFromEnd
This is assuming the distance between end diags and the end of the ladder are equal; otherwise, just type the required distance in place of OffsetFromEnd.
Keep in mind with this equation you'll be using the diag quantity to drive the length. It can be done so that the length defines the the diag quantity, but this is a bit easier IMO.
Hope this helps. (BTW, I haven't check out the attachement so hopefully I understood your question properly!)
Excellent replies. Just had a look through the .ipt cadull_rb posted in his other thread and I found the item I was missing. I hadn't put ceil before my formula which in my case wasn't recognised when I tried using the parameter in the pattern feature.
Now works perfectly.
One other quick question unrelated, I want to draw a box around a view that finishes against my drawing border. Is it a case of freehanding by eye the end of the line or can I project the border onto my sketch?
Ahh, thank you, thats worked. The only part I wasn't sure of is where you say use the fix command. I haven't a clue what that is.
Just re-read your question (and checked out the attachment), and realised that I misinterpreted it. Anyway, my attachment shows what I was originally trying to say (made a few errors in my original post).