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Contributor
Q-Bixx
Posts: 13
Registered: ‎12-14-2010
Message 21 of 26 (288 Views)

Re: iLogic Formula

12-05-2012 02:30 AM in reply to: andrew.warren

Nobody has a solution? :robotsad:

Mentor
mehatfie
Posts: 163
Registered: ‎02-10-2012
Message 22 of 26 (280 Views)

Re: iLogic Formula

12-05-2012 12:47 PM in reply to: Q-Bixx

Hi Q-Bixx,

 

I'm not involved with this conversation, but took a quick look at your post.

 

My best guess is that your rule is taken the X and Y components of the line while taking its measurements.

 

For example, I have a line that's not parallel to either the X or Y axis.... technically, this is now the hypotenuse of a triangle with component relations to the X and Y axis.

 

I believe your rule is taking the X component of your diagonal 50 line, and this is where the 47.54 may be coming from.

 

Again, I'm not up to date on what conversation has been going on, but this would be my best guess while looking at your last post.

 

 

Regards

Mitch

 

Let me know if it helps... Kudos if it works!!!

Contributor
Q-Bixx
Posts: 13
Registered: ‎12-14-2010
Message 23 of 26 (274 Views)

Re: iLogic Formula

12-05-2012 11:45 PM in reply to: Q-Bixx

You got the point. :smileyhappy:

That exactly what i meant with this 'solution', it's only a solution for 80% of the time.

It's not only based on a X-Y component, but also the Z (Your part could be designed in the base component as inclined for example)

Perhaps this could be a solution would be that you could indicate a first surface, and a second one, just to calculate the surfaces.

And by those equasions calculate the width, length an height of the Bounding Box.

 

But then here you would be facing another problem that is if your inclined piece is not only rectangular, but rectangular with (fe: chamfers or fillets or with a hole or ...). You get different values of a bounding box.

 

Perhaps i should explain why i could need this tool.

It would be handy to get in my description a raw material dimension (such as the bounding box) and by that bounding box have Inventor tell me whether i need to make a detail drawing or not.

If the Theorethical Weight (the one calculated from the bounding box) is equal to the Physical Weight, then i wouldn't need to make a detailed drawing, for any other situation i should.

But then i'm trying to avoid the problem with inclined bodies ... cause it gives a different bounding box value, thus the weight is different, thus a drawing should been made (while it could be a flat bar beeing inclined)

 

Come on you geniouses over here!

Help me :smileywink:))

 

ADN Support Specialist
xiaodong.liang
Posts: 1,251
Registered: ‎06-12-2011
Message 24 of 26 (247 Views)

Re: iLogic Formula

01-08-2013 07:19 PM in reply to: Q-Bixx

Hi,

 

I'm not either up to date on what conversation has been going on. It looks you descibed the problem in message 20, right? Could you provide a demo of  the rule you are using? I can try to take a look what's problematic. 



Xiaodong Liang
Developer Technical Services
Autodesk Developer Network

Mentor
kmiller
Posts: 257
Registered: ‎12-17-2007
Message 25 of 26 (185 Views)

Re: iLogic Formula

09-30-2013 07:33 AM in reply to: Q-Bixx

Anyone ever find a solution to the above issue regarding the extents not displaying how desired when a piece is rotated?  It's something I think we could use a bunch here but I haven't figured it out either.  Honestly, I would like to know if there is a way to do this by just selecting on a particular drawing view.  It's the only "starter' way I can think of for us, as we have to rotate and orient the piece as desired on such a view. 

 

So if anyone has any ideas I am up for discussion.

-------------------------------------------------------------------------------------------------

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Distinguished Mentor
pcrawley
Posts: 665
Registered: ‎05-23-2002
Message 26 of 26 (109 Views)

Re: iLogic Formula

01-20-2014 12:56 AM in reply to: kmiller

I've been watching his post evolve into something quite different from the OP.

 

Is the new question "Can you find the Length, Width and Thickness of a box that is rotated from the origin planes?"  Assuming the box object is a part-file in an assembly, or a body in a multi-body solid, the answer is yes,  you can find these values.  (By "box" I mean an extruded rectangle shape where all the faces are at 90 degrees to eachother.)  There is something in the API that finds the bounding-box shape, but that shape is based on the orientation of the origin planes, so it's no use when a part is rotated about one (or more) axis.

 

The API does however gives you access to the edges of all bodies (parts/solids/surfaces) and their start/stop vertices.  From the coordinates of these vertices you can quickly calculate the length of each edge.  If you then sort those twelve edge lengths, the smallest four will be the smallest dimension of the box (say "Thickness").  The middle 4 lengths will be the "Width", and the longest 4 will be the "Length".

 

If I've got the question wrong (or this post is too old), someone tell me to go post elsewhere :smileywink:

Peter

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