Constraint Manager
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Constraint Manager
I have a cabinet that I want to apply constraints to the top rail. What I have is a cabinet with three arches. I want the 3 arches to always be equal to each other. On the left and right there should be a flat area 1/2\" long before the arch starts, then in the middle (2 middles actually) I want the flat to be 3\" so I can place a vertical style there and have 1/2\" on either side of the style where it meets the top rail. Is there a way I can input a formula onto the arched areas which will make them always be equal? I think the formula I want is going to be (w-7)/3 [width of the top rail minus (1/2 + 1/2 + 3 + 3) and divide this by 3 arches] but when I enter it in it doesn't accept the /3 and changes it to 0\"
Thanks.
Leo
Thanks.
Leo
Send it to my email cabinets@thermwood.com and feel free to give me a call with any other questions at 1-800-533-6901 ext. 227
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Re:
Been there, did that. I've spent quite a bit of time on this playing with formulas. The one that should work won't input properly. (w-7)/3DanEpps wrote:Never mind that Leo, further testing shows it doesn't work just right![]()
I'll see if I can come up with a set of constraints that will work for you.
Thanks anyway.
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E-mail sent to you Justin, Thanks.Justin D Melhiser wrote:Send it to my email cabinets@thermwood.com and feel free to give me a call with any other questions at 1-800-533-6901 ext. 227
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Wow, to think I might be able to teach a guru something. Dan, you need to trap the arcs within the constraint editor. You have to give them real numbers by the simple constraint method. Just click the 2 nodes so it gives it a measurement. This way when you are inputting formulas it can't get away from you. So what I have been doing is just going down the line and constraining the whole part. You know first the 1/2" flat then the arc then the 3" flat then the arc, then the 3" flat then the arc and finally the last 1/2" flat. I have been using this method to force the arcs into submission and I can also manipulate them to be equal in this way. If you leave one of the arcs unconstrained then figure out the math outside of the software, then you can edit the arc length of the two arcs and it will force the 3rd arc to be the leftover, hopefully if you did your math correct they will be equal. The last thing you need to do is constrain the radius of all of them to be equal and you will be able to have 3 equally radiused equal sized arcs. There is no need to do this with absolute numbers when you are placing your arcs. Just make sure the 1/2" at the beginning and end is there and the spaces between the arcs are 3" (in my case). But you could also edit the constraint the of the flats 1st and the only thing you have to worry about is getting the arcs to fall on the y-axis at 0. I found this intriguing. And was hoping to get the formulas to work and I wouldn't really have to do anything but cut out the arcs and apply my constraints and everything els would just fall into place by its own without the need to do any math. That would be cool.DanEpps wrote:Here is what I'm talking about with the funky change. The constraint is w-1. Notice that the top arc disappeared and the constraint is at the bottom???
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That was just to show the crazy results I was getting. I did exactly what you said in another attempt and still ended up with something like that picture. It seems that the problem appears when you enter a minus sign in the formula.

For the record, I HATE the constraint managerWow, to think I might be able to teach a guru something.

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