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1-Newbie

## Suppressing a feature with relation driven by parameter?

I am working on a family table for a part family.

The part family has various sizes but all are in 3 main bodies, so I created some parameters and relations to drive the positions of the bolt holes depending on which of the 3 main bodies it uses.

BUT I was wondering if I could add a relation inside of the IF statement that would allow me to suppress feature(s).

R and P are my 2 controlling parameters, R selects one of the three main hole patterns, and P selects one of the number of holes within the three main patterns.

Here is the relations I am using:

/* RELATIONS TO DETERMINE HOLE COUNT IN R
IF R ==4
p138 = P/4
p154 = P/4-1
p170 = p138
d158 = ((p170-1)*d168)/2
p186 = P/4+1
d174 = ((p186-1)*d184)/2
ENDIF
IF R==3
p138 = P/3 -1
p154 = P/3
p186 = P/3+1
ENDIF
IF R ==2
p138 = (P-1)/2
ENDIF

/* RELATION TO DETERMINE 1ST HOLE IN PATTERS

d127 = ((p138-1)*d136)/2
d142 = ((p154-1)*d152)/2

Instead of having the 4 and 3 hole patterns be suppressed from the family table, I would rather embed it within the IF statements if possible.

Any Help would be greatly appreciated.

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3 REPLIES 3
23-Emerald V
(To:jlozano)

Hi,

relations enables you to set value of dimension -OR- parameter, only.

Pro/PROGRAM enables you to suppress a feature.

MH

Martin Hanák
1-Newbie
(To:MartinHanak)

Hi Martin,

Thanks, I thought that was the case, but seems like people use relations and Pro/Program interchangeably.

Sorry I cannot upload the model due to this being part of my companies IP

Thanks

23-Emerald V
(To:jlozano)

Hi,

I think that you can create and upload testing model.

MH

Martin Hanák
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