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Posted

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

:unsure:

 

Tks a lot

Posted
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?

Posted
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?:unsure:

Posted

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.

Posted

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.:D

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