Re-create polygon knowing sides.

Is it possible to re-create a polygon given the measurements of each side but without knowing any co-ordinates or angles?

That would really be helpful when having a printed drawing with no co-ordinate and angle information.

I think that is only possible with triangles.

I agree with Seant. You can have a square and a rhombus with the same edge lengths.

If you wish to measure only lengths, form triangles: measure two neighbor sides AND the distance between the two non-touching ends.

I agree with Seant. You can have a square and a rhombus with the same edge lengths.

If you wish to measure only lengths, form triangles: measure two neighbor sides AND the distance between the two non-touching ends.

Well yeah but that is a special solution. You have to have a perfect square in order to have this solution.

I'm just looking for a way to get possible solutions, and then with some manual tweaking it's probable to get the unique solution you're looking for.

You mean measure distances in the paper? Will there be good accuracy in doing that?

no.. you can't do it because, for example, if your "segment length" is two inches.. you can make ANY polygon scaled to have two inch segments on it.

If you knew how many sides it had, you could determine the angles with easy math:

3 side polygon = 360 divided by 3 = 120 outside degrees (180-120 = 60 inside degrees )

4 sided polygon = 360 divided by 4 = 90 outside degrees

5 sided polygon = 360 divided by 5 = 72 outside degrees (180 -72 = 108 inside degrees)

Of course those are relative angles, not using autoCAD's angle from zero to get those, you add together the outside angles for each line

Like a 2-inch segment triangle (one side down) would be:

LINE

0,0

@2

@2

@2

or start with the @2

That assumes equal lengths on all sides of course.. if you had differing segment lengths, you couldn't use those angles, but they WOULD still all add up to 360.

Parametric modelling would permit you to make a polygon and then come back and change the dimensions on each of the sides to re-size it, but that's an AutoCAD 2010 thing.

Given you know the area, perimeter and each side but no angles of a polygon, how can you create that exact polygon using parametric functionality of AutoCAD.

Is it possible for eaxmple to tell AutoCAD that an object is a 6 sided polygon with an area of xx.xx sq.m. and lock the distance of each side?

6-sided means you have 60 degrees on each corner, assuming the side lengths are equal (no matter what that length is). Once you know the side length, and the number of sides, you're done anyways.

As far as the parametric modelling goes, I made the assumption you might be able to do something with it. I don't actually know how AutoCAD implements it's parametric functions, as I'm on an older version. Perhaps it would allow you to draw a 6-sided polygon, and then change the length of the sides through changing the dimension on it. I'm not sure though, it depends if it would recalculate a polygon, rather than just the line itself. Even with constraints to make each line equal, there's no guarantee it wouldn't just increase the length of the sides, rather than reposition the intersections (or rather, change the intersection's relationship with the centerpoint of the polygon).

It would depend on whether AutoCAD saw the shape as a series of lines, or continued to recognize it as a polygon.

For a four sided polygon with given sides, if you fix an angle the polygon is determined. So a program could calculate that angle to meet a criteria (in your case: to have the desired area).

A five or more sided polygon has too much degrees of freedom. There are an infinite number of polygons (differing by their angles) with the same area.

if you draw a series of lines in a polygon shape (6-sided in this case) and have AutoCAD size each line parametrically while maintaining the intersections, it should do it, assuming you can do it that way... though I imagine the shape would become unpredictable.

Up until now, I thought you were talking about polygons with equal sides, since you weren't correcting me on that assumption when I noted it. I also assumed a simple convex polygon, and not some sort of self-intersecting mess of a polygon.

I don't know how you're able to know the area of a polygon that doesn't exist though.. You can't get the area of a polygon from the perimeter without breaking it into triangles, or knowing the fixed positions of the vertices, because there's nothing to indicate if the overall polygon is balanced. Or at least that's my (possibly flawed) understanding.

Would it be possible to give us a realworld example of how/why you're trying to accomplish this goal so I can better understand the information you're starting with?

The real world example is on survey drawings where polygons describes a property, and a printed drawing most of the times doesnt includes coordinates or angles, and I'm looking for information to draft it in CAD environments while lacking this information.

Best I could suggest is to determine one point of the polygon, and use that as the centerpoint of an arc.

Start the arc near where you think that line is headed (like if the line IS 45 degrees, start at 30 degrees) Using the Length of that segment as your radius, and run an arc to about 55 degrees.

Then pick a random point on that arc that you think is close to where that intersection should be and repeat... working your way around the property that way.. Be sure to mark your centerpoints because you're going to be relocating them as you fine-tune the shape.

You will, in fact be chasing that thing around for awhile, and your resulting shape is not guaranteed to be accurate.

If you take a 2" x 2" square (think of the lines as sticks with pivots on the end of them) and change one of the angles to 45 degrees, you will still have the same area, even though the shape is not the shape you wanted.

The key is that you need measurements from other points in-order to control the shape. Basically, every second intersection needs to be pinned down. If you don't, you'll have two (or more) pivots between them , and that can alter the shape without changing the area.

Do you have any other measurements? (a PDF of an example showing the info you're working from would likely help).

*Update*

I asked our sheet metal layout guru at work about this question, and he agrees with me, and suggests that my "striking an arc" method will work, and that likely you'll get the area you wanted, but unless you're extremely lucky, you won't get the right shape out of it, unless "similar" is close enough for you, and you have a rough idea of what it should look like.

Ok so here i attach a pdf with the appropriate example.

Imagine you had a piece of paper printed containing the polygon with the information as shown on the digital document (pdf in this case).

Would you be able to replicate it in an AutoCAD drawing?

If it's not too much trouble could you please implement your ARC method in this custom 9-sided polygon i just arbitrarily created so i can fully understand the method you're suggesting?

example.pdf

I just did it.. and I'd attach it but there's something screwy about the area calculation.. I'm thinking I'm misunderstanding the units: the first line from the lowest point (counterclockwise) reads "134,01m"

I took that to mean 134010 meters or miles.. and that's what I used, but the resulting figure I made, using all the correct lengths, and angles that are at least close to what you posted, gives me an area that's literally lightyears away from what you have:

I came up with: 20063534511.2926

You had: 20422.8519 sq.m

So.. What am I misunderstanding on the units? I suspected at first, that something was up with the comma being two places in, rather than 3.. but have no experience seeing/using measurements like that.

Is the distance really 134.10 units? or 13410 units? or 134100 units?

I'm stuck until I understand those units so I can get the area to work.

As the suffix to the dimensions is "m", I took the dimensions to be metres, with the decimal divider being a ",". But you have the annotated area with a decimal point - you are mixing conventions about the decimal divider

However, I approached the problem differently. I converted the PDF to a JPEG file, inserted it into a drawing, traced over the sides with a polyline, and scaled it so that one of the lines is correct (135.85m). The polyline area was 20421.08 sq.m. How close do you want to be.

Mike

The comma (,) denotes the decimals and the suffix m stands for meters not miles.

eldon

So you overlayed an image and rasterized it? Well that's one way to do it, but it's not alway approachable.

I mean this was a clean pdf that was workable. Imagine having an A0 paper with no means to scan and also the paper being in bad condition (torn, liquids spilled all over it leaving it washed out) so leaving you with no such option.

What i'm trying to say is that i'm looking for a way, that is pure a geometrical approach.

Besides triangles and squares i think that for polygons with number of sides greater than 4 (>4) there must be only one unique solution for the shape of the polygon given that you know all of its sides and area. But how can that turn to an algorithm that gives you this desired end-result?

If the shape of the polygon is in print, I would always go for a re-creation by scaling approach, taking extra diagonal dimensions.

I think there would be too many variables otherwise for a mathematical/geometrical solution. But I might be proved wrong

I went back to my drawing and rescaled it (scale .001).. it would appear than an angle (or all of them) were off very slightly.. giving me an area of 20638 sqm

To fix it from there (assuming "there" is just an approximated version with all of the correct line lengths, but bad angles):

Use any point on the shape and turn the whole thing into triangles (obviously you want to use the point that can "see" the other points without crossing any lines).

Since the nearest two points to that start-up point describes a triangle where two sides are known.. you can use the "assumed" value in your basic layout to determine the area of that triangle.. just draw the missing line of the triangle.

Do that for all of the resultant triangles in a spreadsheet, and you'll be able to tweak the whole thing around from that. You add them all up and then adjust the length of that assumed missing line of the triangle, and it will affect all the areas of the polygon.

Unfortunately, you don't know which angle is the culprit, so you'll have to do the same equation on the spreadsheet 1 time for every triangle. Permitting a single value change to recalculate the entire thing.

In theory, you will end up with the correct area, but not necessarily the correct shape, this is because an angles could be something like 5.5 degrees, and if you have 5 degrees on it, you could tweak another angle off by 10 degrees to compensate for it.

Each known distance you have that links between the intersection points, will improve your chance of getting the correct final answer and shape.