Graph controlled round
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Graph controlled round
Hi, is it possible to control the value of a round by graph feature? I have been trying to figure this out with little success. I can get the geometry i require with a sweep controlled by graph but i was trying to simplify things a little as it is only the round value that is changing. I am just trying to better understand the use of graph controlled features. A simple cube with a round running along a single edge will suffice, if anyone is willing to post an example if this is even possible.
John
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Well, the round feature dont have any connection to the graph feature as far as i know.
However, you can do a workaround and controll your round using a graph,. Just make it "follow curve" as i previous example, where the curve is a intersect curve from a sweep (graphcontrolled that is) and the side surface. When you update your graph, the sweep will update , the intersect curve will update and finally your round will update.
//Tobias
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Hi,
maybe you can create round Through Curve ?
MH
Martin Hanák
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Hi Martin, thanks for the reply. Yes, I realize I could achieve my objective this way, in fact, I use the 'Through Curve' option a great deal. I just wanted to know if I can achieve the same use 'graph feature'.
Regards
John
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Well, the round feature dont have any connection to the graph feature as far as i know.
However, you can do a workaround and controll your round using a graph,. Just make it "follow curve" as i previous example, where the curve is a intersect curve from a sweep (graphcontrolled that is) and the side surface. When you update your graph, the sweep will update , the intersect curve will update and finally your round will update.
//Tobias
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Tobias Larsson wrote:
Well, the round feature dont have any connection to the graph feature as far as i know.
Hi Tobias, if you are correct, then you have answered my question. I know there are various workarounds, but the whole point of my post was to better understand the use of graphs & situations in which they could be used.
Thanks for your reply.
John
