turn on suggestions

Auto-suggest helps you quickly narrow down your search results by suggesting possible matches as you type.

Showing results for

Close

Inventor General Discussion

- Autodesk Community
- >
- Inventor
- >
- Inventor General Discussion
- >
- Re: array/pattern features with different dimensio...

Topic Options

- Subscribe to RSS Feed
- Mark Topic as New
- Mark Topic as Read
- Float this Topic to the Top
- Bookmark
- Subscribe
- Printer Friendly Page

- Mark as New
- Bookmark
- Subscribe
- Subscribe to RSS Feed
- Highlight
- Email to a Friend
- Report Inappropriate Content

1503 Views, 8 Replies

09-19-2012 05:03 AM

Hello,

I would like to generate a rectangular pattern of extruded tubes which have an internal radius that is proportional to the distance to the centre point of the pattern. Or in other words tubes which are narrower in the centre, and wider at the exterior. Is there a way to do this without resorting to creating each tube individually?

More generally, is there a way to parameterise a feature dimension/parameter so that it is a function of spatial position?

Many thanks,

Aaron

Solved! Go to Solution.

Solved! by nannerdw. See the answer in context.

IV 2013 has equation curves. If you end up using an equation curve and require the diameters to *exactly* follow your non-linear scaling equation, just remember that the curve will have to pass through each circle at the same location. (eg. at the top point on each circle, rather than tangent to each circle.)

If you're using an older version of IV and don't need the non-linear scaling to be 100% accurate, you can approximate it with a spline.

CBliss has a couple of examples on his site for quadratic and sine curves.

Look at the bottom of this page:

http://cbliss.com/inventor/iFeatures/index.htm

If you're using an older version of IV and require 100% accuracy, you'll have to replace the tangent line with a set of parameters defined by functions, which could get a pretty tedious.

- Mark as New
- Bookmark
- Subscribe
- Subscribe to RSS Feed
- Highlight
- Email to a Friend
- Report Inappropriate Content

09-19-2012 05:23 AM in reply to:
aoliver-

Make the diameter dimension a function of the distance from origin of the pattern.

Can you attach an ipt here of what you have attempted so far?

Are you trying to do this a single part file or do you need as an assembly of parts?

Please mark this response as "Accept as Solution" if it answers your question.

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

Autodesk Inventor 2014 Certified Professional

Autodesk AutoCAD 2013 Certified Professional

Certified SolidWorks Professional

Inventor Professional 2015-SP1 64-bit

http://www.autodesk.com/edcommunity

http://home.pct.edu/~jmather/content/DSG322/inventor_surface_tutorials.htm

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

Autodesk Inventor 2014 Certified Professional

Autodesk AutoCAD 2013 Certified Professional

Certified SolidWorks Professional

Inventor Professional 2015-SP1 64-bit

http://www.autodesk.com/edcommunity

http://home.pct.edu/~jmather/content/DSG322/inventor_surface_tutorials.htm

- Mark as New
- Bookmark
- Subscribe
- Subscribe to RSS Feed
- Highlight
- Email to a Friend
- Report Inappropriate Content

09-19-2012 06:02 AM in reply to:
JDMather

This would be a single part file, ultimately I want to make a single 3D solid which I can use in CFD, and also 3D print the structural volume. I'm trying to homogenise flow in a manifold basically by adjusting the orifice diameter

I haven't attempted anything yet, so nothing to attach. However, a simple example would consist of a circular disc axially along z, 100mm diameter, 3mm thick. A sketch would then be created on one face, and a 2mm circle drawn at the centre. That would then be extruded (say 50mm), and then patterned isotropically in the x and y directions across the face, creating a 9x9 array of cylinders, with spacing 10mm. I would like to be able to adjust the extruded cylinder diameter according to the radial position of each cylinder (so square root of x^2 + y^2) with respect to the origin.

- Mark as New
- Bookmark
- Subscribe
- Subscribe to RSS Feed
- Highlight
- Email to a Friend
- Report Inappropriate Content

09-19-2012 06:33 AM in reply to:
aoliver-

This sounds like a task begging for iLogic.

Mike Patchus

Inventor 2012 SP3 ~ Vault Pro 2013

==================

Lenovo D20 Workstation Intel Xenon E5640@2.67GHz (2)

Win 7, 24gb RAM, NVIDIA Quadro FX 3800

Did you find this reply helpful ? If so please use the Accept as Solution or Kudos button below.

Inventor 2012 SP3 ~ Vault Pro 2013

==================

Lenovo D20 Workstation Intel Xenon E5640@2.67GHz (2)

Win 7, 24gb RAM, NVIDIA Quadro FX 3800

Did you find this reply helpful ? If so please use the Accept as Solution or Kudos button below.

- Mark as New
- Bookmark
- Subscribe
- Subscribe to RSS Feed
- Highlight
- Email to a Friend
- Report Inappropriate Content

09-19-2012 06:37 AM in reply to:
aoliver-

aoliver- wrote:

I haven't attempted anything yet,

I think you should make the first attempt before anyone else.

Please mark this response as "Accept as Solution" if it answers your question.

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

Autodesk Inventor 2014 Certified Professional

Autodesk AutoCAD 2013 Certified Professional

Certified SolidWorks Professional

Inventor Professional 2015-SP1 64-bit

http://www.autodesk.com/edcommunity

http://home.pct.edu/~jmather/content/DSG322/inventor_surface_tutorials.htm

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

Autodesk Inventor 2014 Certified Professional

Autodesk AutoCAD 2013 Certified Professional

Certified SolidWorks Professional

Inventor Professional 2015-SP1 64-bit

http://www.autodesk.com/edcommunity

http://home.pct.edu/~jmather/content/DSG322/inventor_surface_tutorials.htm

- Mark as New
- Bookmark
- Subscribe
- Subscribe to RSS Feed
- Highlight
- Email to a Friend
- Report Inappropriate Content

09-19-2012 06:57 AM in reply to:
JDMather

ok, here you go. It's just a simple array of the cylinders though, I have no idea, barring manually creating each cylinder from its own sketched circle, how to change the radius of each cylinder based on its radial distance from the origin.

- Mark as New
- Bookmark
- Subscribe
- Subscribe to RSS Feed
- Highlight
- Email to a Friend
- Report Inappropriate Content

09-19-2012 07:56 AM in reply to:
aoliver-

If it's just 9x9, it shouldn't be too much work to set it up with a geometric construction, using an angled line tangent to a row of circles to drive their diameters. You only have to draw 1/4 of the circles, and then do a 4x circular pattern around the center.

Something like this:

It takes a while to set up, but with this method it should be easy to make quick changes to your parameters by adjusting the angle of the tangent line.

-Using Autodesk Inventor Professional 2012

- Mark as New
- Bookmark
- Subscribe
- Subscribe to RSS Feed
- Highlight
- Email to a Friend
- Report Inappropriate Content

09-19-2012 08:18 AM in reply to:
nannerdw

That's pretty neat, thanks!

What about non-linear grading of sizes? For example so that they change by a square law - can a curve be defined by an equation?

- Mark as New
- Bookmark
- Subscribe
- Subscribe to RSS Feed
- Highlight
- Email to a Friend
- Report Inappropriate Content

09-19-2012 08:28 AM in reply to:
aoliver-

IV 2013 has equation curves. If you end up using an equation curve and require the diameters to *exactly* follow your non-linear scaling equation, just remember that the curve will have to pass through each circle at the same location. (eg. at the top point on each circle, rather than tangent to each circle.)

If you're using an older version of IV and don't need the non-linear scaling to be 100% accurate, you can approximate it with a spline.

CBliss has a couple of examples on his site for quadratic and sine curves.

Look at the bottom of this page:

http://cbliss.com/inventor/iFeatures/index.htm

If you're using an older version of IV and require 100% accuracy, you'll have to replace the tangent line with a set of parameters defined by functions, which could get a pretty tedious.

-Using Autodesk Inventor Professional 2012

Search This Board

Auto-suggest helps you quickly narrow down your search results by suggesting possible matches as you type.

Showing results for

Post to the Community

Have questions about Autodesk products? Ask the community.

The Knowledge Network

Access a broad range of knowledge to help get the most out of your products and services.

Download & Installation

Getting your Software

Downloads & Upgrades

Serial Numbers & Product Keys

Installation & Licensing

Activation & Registration

Network License Administration

Subscription Management

Sign In / Create Account

Maintenance Subscription Help

Desktop Subscription Help

Cloud Service Subscription FAQ

Announcements

Feel free to drop by our AU topic post and share your plans, plug a class that you're teaching, or simply check out who else from the community might be in attendance. Ohh and don't forgot to stop by the *Autodesk Help | Learn | Collaborate* booths in the Exhibit Hall and meet our community team if you get a chance!

- Privacy | Legal Notices & Trademarks | Report Noncompliance | Site map | © Copyright 2014 Autodesk Inc. All rights reserved

Except where otherwise noted, this work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Please see the Autodesk Creative Commons FAQ for more information.