dglopes Posted December 21, 2010 Share Posted December 21, 2010 Hi, Im trying to do a polygon with a parametric constraint, the Area. How can i maintain the same area with differents side values? The polygon is the Trapeze. The dimensions are: 54 (biggest side) 28 (small side) ~23 height Tks a lot Quote Link to comment Share on other sites More sharing options...
BIGAL Posted December 23, 2010 Share Posted December 23, 2010 (edited) Search into the maths of a trapezoid for the area then you can work backwards but probably a lisp etc Took about 10 seconds http://www.mathopenref.com/trapezoidarea.html Edited December 23, 2010 by BIGAL googled Quote Link to comment Share on other sites More sharing options...
dglopes Posted December 23, 2010 Author Share Posted December 23, 2010 Search into the maths of a trapezoid for the area then you can work backwards but probably a lisp etc Took about 10 seconds http://www.mathopenref.com/trapezoidarea.html Yeah, i know the math, but the problem is how to works backwards in this case. Is there a possibility to do the area as a constraint like dimension or do I have to use a kind programming language? Quote Link to comment Share on other sites More sharing options...
dglopes Posted December 24, 2010 Author Share Posted December 24, 2010 Search into the maths of a trapezoid for the area then you can work backwards but probably a lisp etc Took about 10 seconds http://www.mathopenref.com/trapezoidarea.html cant do it cause cyclical expression cannot be used. The area is 1176.38 and a = 17.30, so if the d1 = (1176.38/a)-d2, the problem appears when i put d2=(1176.38/a)-d1 So...how can i resolve this question? Quote Link to comment Share on other sites More sharing options...
BIGAL Posted December 26, 2010 Share Posted December 26, 2010 Theres a couple of other problems as well you have two parallel lines with a distance between them = area required but how do you set the vertical start position of the second line ? It sounds like you would be better off draw two lines with correct lengths then just move 1 to get correct vertical distance. Again a lisp may be easier. Pick 1st line pick 2nd line enter area move 2nd line. Quote Link to comment Share on other sites More sharing options...
dglopes Posted December 26, 2010 Author Share Posted December 26, 2010 The problem is finished. For what i wanted, i resolved is this way: S=area a=hight B, b = bases (B+b)xa/2=S---> The lenght of one side (b) = 2S/2 - B (and B1 is the inverse) So... fixing the lenght of side b, i can move the hight and side B. The area dont chance, cause its value is fixed in the expression. When I want to move the lenght of side b, i can put the parameter at another side or hight. was a simple problem, just needed a effort... my bad! (was first time trying to use the parametric drawing in AC2010. Tks a lot. Quote Link to comment Share on other sites More sharing options...
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