Loop length and o-ring groove

Loop length and o-ring groove

Fueler
Collaborator Collaborator
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Message 1 of 6

Loop length and o-ring groove

Fueler
Collaborator
Collaborator

 I have an irregular o-ring groove. I was hoping to find the length of it and an off the shelf ring that would fit.

I clicked on the floor of the groove and measured.

The loop length measurement says it is 17.511 inches.

I assumed that meant circumference but that was way off.

In making a custom rubber ring to fit I came up with what amounts to a 3" diameter to fit in that groove.

So, what don't I understand about loops?

Can what I want to do even be accomplished?

Thanks

 

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Message 2 of 6

seth.madore
Community Manager
Community Manager

It's likely that it's measuring both sides of the groove. I was stumped by this a few years back, never found a good way to check it. I suppose you could project the edge of the groove and use the inspect tool. On a simple rectangle, I got the results that I expected to see:

2021-07-26_17h36_34.png


Seth Madore
Customer Advocacy Manager - Manufacturing


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Message 3 of 6

Fueler
Collaborator
Collaborator

The only kludge I found was to measure each portion of the outer bottom edge and add them up. That gave me a circumference length of 9.18 which converted to 2.922" which matched within range of my custom made 3" ring.

It would be nice to auto measure all the sections of that bottom line at once.

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Message 4 of 6

jhackney1972
Consultant
Consultant
Accepted solution

You did not supply your model so I created a sample O-Ring groove.  I demonstrated in the Screencast how to use Plane Along a Path to get the length of most any entity including sketches and solids.  This will work on most any line, spline, circle, edge, etc.  If the segments are connected it can be used.  If they are connected around sharp corners, use the Ctrl key to select them all.

John Hackney, Retired
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Message 5 of 6

Fueler
Collaborator
Collaborator

Cool. Works well. thanks

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jhackney1972
Consultant
Consultant

If you are satisfied with the solution, please mark your forum question "Accept Solution" so others can find the answer to a similar question.

John Hackney, Retired
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