Topic: Geometry problem: length of segments knowing the area, perimeter and width

[Marc'Antonio Alessi]

I have a polyline that I know:
area         = 1512499.7 ...
perimeter = 12600
width of the track = 250 (in the example)

Is it possible calculate the total length of the segments?
(as in the picture: 3000+1000+950+1100 = 6050)

[irneb]

It's possible to calculate mathematically, but for one thing: Your dimensions are inconsistent as to what they measure. E.g. that "700" long piece is actually ... what ... 700+250?

[ribarm]

Total length = Total area / width = 1512499.7 / 250.0 = 6049.9988 ~ 6050 = 3000 + 1000 + (700+250) + 1100

[tombu]

You are looking to select segments of a polyline and total the lengths of those segments?

[Marc'Antonio Alessi]

It's possible to calculate mathematically, but for one thing: Your dimensions are inconsistent as to what they measure. E.g. that "700" long piece is actually ... what ... 700+250?

I'm sorry, you're absolutely right (i have altered the image) here are other examples:

[Marc'Antonio Alessi]

Total length = Total area / width = 1512499.7 / 250.0 = 6049.9988 ~ 6050 = 3000 + 1000 + (700+250) + 1100

Yes, I did not think it was that simple, I hope that there are no exceptions...
Grazie mille.

[Marc'Antonio Alessi]

New question: is  there a way to calculate the maximum lenght of the profile as in the examples A and B?

[roy_043]

Why isn't the max. length of A: 3000+1000+250+1200+1100=6550?

[ribarm]

Why isn't the max. length of A: 3000+1000+250+1200+1100=6550?

Roy, probably thought :
Why isn't the max. length of A: 3000+750+250+1200+1100=6300?

[Bethrine]

A: 3000+1000+(1200-250)+1100=6050

[ribarm]

Why isn't the max. length of A: 3000+1000+250+1200+1100=6550?

Roy, probably thought :
Why isn't the max. length of A: 3000+750+250+1200+1100=6300?

I think the answer is : How can center polylines length be the same as those that are offset by half the width (250/2=125) ? Simply it can't be the same length - those polylines aren't colinear - they are curving in all 4 orthogonal ways (+X,-X,+Y,-Y)...

[Bethrine]

 

I am not a programmer. I found it interesting as a math problem. Feel free to ignore me! 

[Marc'Antonio Alessi]

I asked the question too early and without considering well: is a matter for the calculation of square tubing to build the figure, is a question that I was asked but I think that the calculation of the other figures is not correct. I think the request is just to know the total length of pipe required.

So in the example top/left the length is: 1160+1050+750 = 3000

@roy_043 > Why isn't the max. length of A: 3000+1000+250+1200+1100=6550?
        maybe this is the exact calculation  A: 3000+1000+500+1200+1100=6800

Now ask other explanations.

[CAB]

Do you intend on making miter joints or butt joints?

[owenwengerd]

I think the request is just to know the total length of pipe required.

In that case the solution is very simple: calculate total enclosed area of the figure divided by width of the tubing.