Inventor General Discussion

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Post 1 of 3

dimension from quadrant of circle to appt intersection on a cirlce

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01-10-2013 12:34 PM
 
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Post 2 of 3

Re: dimension from quadrant of circle to appt intersection on a cirlce

01-10-2013 12:56 PM in reply to: carey.olson

Carey,

Welcome to the forum.

Take a look at this link.  See if it helps you.

http://forums.autodesk.com/t5/Autodesk-Inventor/Tangent-Dimensioning-of-Arcs/td-p/3317169

Best Regards,
Scott McFadden
Inventor Professional 2014
(Colossians 3:23-25)

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Post 3 of 3

Re: dimension from quadrant of circle to appt intersection on a cirlce

01-11-2013 04:40 AM in reply to: carey.olson

My solution below is in Inventor 2013, but since you didn't tell us what version you're using, I don't know if you can open it.

 

This isn't as easy as it ought to be, but here's how I did it (see attached).  I created a view sketch, projected the two arcs, and created an arc concentric with the big "circle" as an extension to show the complete circle (set to the Dimension layer).  I then put in a Sketch-Only line between the arc centers, then a normal line tanget to the sketched arc and perpendicular to the previous line.  This last line is purely for something to dimension to.  I exited the sketch, then created a dimension between the tangent line just placed and the quadrant of the small arc.  Then I went back into the sketch and changed that tangent line to Sketch Only so it is not visible.

Sam B
Inventor 2012 Certified Professional

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Inventor Professional 2013 SP1.1 Update 1
Windows XP Pro 32-bit, SP3
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still waiting for a foreshortened radius dimensioning tool in Drawing Manager

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