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number of diags in ladder to reference the length

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Message 1 of 13
nige106
1180 Views, 12 Replies

number of diags in ladder to reference the length

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.

 

File attached

12 REPLIES 12
Message 2 of 13
nige106
in reply to: nige106

Would I need to use iLogic to achieve this?

Message 3 of 13
cadull_rb
in reply to: nige106

See the part attached to my post http://forums.autodesk.com/t5/Autodesk-Inventor/Incomplete-part-update-when-sketch-pattern-count-cha... for parameters that can be used to control the pattern. If you can, only sketch the unique instances and pattern the resulting features.

Message 4 of 13
Scotty87
in reply to: nige106

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!)

Message 5 of 13
nige106
in reply to: nige106

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?

Message 6 of 13
stephengibson76
in reply to: nige106

you can place a point that will snap to the origin of the paper, then use the fix cammand to root that point, then dimension from that point based on your border dimensions.

Stephen Gibson



View stephen gibson's profile on LinkedIn


Message 7 of 13
nige106
in reply to: stephengibson76

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.

Message 8 of 13
stephengibson76
in reply to: nige106

fix constraint, like a perpendicular or tangent constraint, the one with the padlock icon.  You need to fix this point because although it snaps to the 0,0 its not really attached to anything

Stephen Gibson



View stephen gibson's profile on LinkedIn


Message 9 of 13
Scotty87
in reply to: nige106

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).

Message 10 of 13
nige106
in reply to: nige106

Thanks to everyone who replied for taking time to do so.Smiley Happy

Message 11 of 13
nige106
in reply to: nige106

Just a variation to the original question. I've got my .ipt updating diag length and quantity based on ladder length. However is ther anyway of making sure the diag quantity is always an even number?

Message 12 of 13
stephengibson76
in reply to: nige106

you could try using functions such as dividing by 2, rounding then x 2

Stephen Gibson



View stephen gibson's profile on LinkedIn


Message 13 of 13
nige106
in reply to: nige106

Works perfectly. Many thanks.

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