My Math Forum Calculate the length of the side of an irregular tetragon.

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November 17th, 2008, 04:26 PM   #1
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Calculate the length of the side of an irregular tetragon.

Hi,

I'm an architect and I'm trying to automate shop drawings for very complex hexagonal precast elements. I normally consider myself quite good with formulas and so far I've written about 30 or so for this one element. But there is one part I'm struggling with.

Refer to the attach sketch. All the dimensions shown are variable and are either directly entered or I have already created a formula to calculate them. I'm just left with the outer hexagon edges that I need to find the lengths of based only on this information. As you can see I dont' have X, Y co-ordinates for these points. If I did I know I could calculate these lengths. Does anyone know how I might find these values (labelled Length AB, Length BC etc.)

The outer hexagon's shape is defined by the 2 offset parameters at each corner (A thru F)...

Any help anyone can provide is greatly appreciated. This is a rather urgent project and will soon be on the cover of many architectural magazines I'm sure...

Cheers!
Attached Images
 Formula Issue.jpg (205.8 KB, 477 views)

 November 17th, 2008, 06:31 PM #2 Senior Member   Joined: Jul 2008 Posts: 895 Thanks: 0 Re: Calculate the length of the side of an irregular tetragon. I must admit at first that I am not an architect, and am not familiar with the use of the slash-lines which permeate this drawing. In particular, I don't fathom precisely where, for example, the distances 225 and 256 are located in the diagram. It would appear that from this information Length AB and he one marked 1180 are not parallel, but those slashed distances do not appear to be on the corners of the inner hexagon, so where they are located is a mystery to me. That said, I have helped some local people from time to time with surveying problems, and although I could and did do manual calculations, I was also able to confirm measurements to 4 decimal places by using a drafting program. SO.... Why don't you draw the thing using a decent drafting program, and let that program do the measuring for you? I used the small but intuitive DeltaCad, but if I were doing architectural calculations, I'd use something more like AutoCad, and damn the expense.
 November 17th, 2008, 07:09 PM #3 Newbie   Joined: Nov 2008 Posts: 8 Thanks: 0 Re: Calculate the length of the side of an irregular tetragon. Hi Dave, Yeah sorry us architects and our funny symbols. That is just how we draw dimensions. Obviously the shape is just the thicker lines (12 total). The dimensions you refer to are from the inner hexagon edge AB to the point of the outer hexagon A (225mm), Inner Hexagon edge FA to outer hexagon point A (236mm). This principle is followed around the entire hexagon for all 6 points. The x, y dimensions are for the inner hexagon points. I also have internal angles for the inner hexagon and the lengths of all the inner hexagon edges. So basically when constructing this 3D element I can define the shape of the inner hexagon by the x, y co-ordinates and then the outer hexagon is defined by offsets from the inner hexagon. It had to be done this way to ensure the hexagons could never intersect otherwise the 3D element would break. As such it has made it very difficult to calculate those lengths of the outer hexagon. If it helps I've labelled this values as: 236 = AExt1 225 = AExt2 256 = BExt1 241 = BExt2 etc. 1319 = AX 1877 = AY 1837 = BX 2937 = BY etc. 136°= Angle A 106°= Angle B etc. 1180 = AB 1240 = FA Hope this helps to clarify?
November 17th, 2008, 07:14 PM   #4
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Re: Calculate the length of the side of an irregular tetragon.

Quote:
 Originally Posted by Dave SO.... Why don't you draw the thing using a decent drafting program, and let that program do the measuring for you? I used the small but intuitive DeltaCad, but if I were doing architectural calculations, I'd use something more like AutoCad, and damn the expense.
This program is about 200 times more powerful than AutoCAD. You need to understand that this is just one face of a complicated 3D model. I'm trying to extract all 26 faces of this 3D model to 2D faces. The hexagonal face is fine, its the rectangular faces on the side that are the issue as I need this to link to the outer pieces...

Hope this clears things up a bit as to why this is an issue...

November 17th, 2008, 09:28 PM   #5
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Re: Calculate the length of the side of an irregular tetragon.

This screenshot might explain the complexity a bit more?

Does anyone know if it is even possible to calculate these lengths from this information? I would've thought it is as its enough information to correctly define the shapes...

[attachment=0:1ogkkjy1]Overall image.jpg[/attachment:1ogkkjy1]
Attached Images
 Overall image.jpg (199.5 KB, 455 views)

 November 18th, 2008, 06:42 AM #6 Senior Member   Joined: Jul 2008 Posts: 895 Thanks: 0 Re: Calculate the length of the side of an irregular tetragon. Sorry, but this is a bit much for me to try to handle, especially in new territory. All I know is that my son-in-law, an engineer, uses Solid Edge 3D software and extrudes shapes handily from initial 2D dimensions. I still don't understand why your available software won't enable this sort of drafting, and why it couldn't then also dimension the drawing, saving enormous effort in calculation. Sorry I can't help you here, and don't want to waste your time, so I'm done.
 November 18th, 2008, 02:08 PM #7 Newbie   Joined: Nov 2008 Posts: 8 Thanks: 0 Re: Calculate the length of the side of an irregular tetragon. Thanks anyway Dave. Hopefully someone else can chime in? The issue is not drawing the bits, I can draw them just fine. I am automating them... In order to automate them I need to find the lengths of the sides. I think its best if people don't worry about the why. The issue is I need to find the length of these sides and there is no way around it. There are over 1800 of these so I don't want to have to extract the lengths manually off each one (10800 lengths to extract!!). Please somebody, anybody? So far every source I've contacted as been unable to find a way to get these dimensions... Cheers.
 November 18th, 2008, 04:07 PM #8 Newbie   Joined: Oct 2008 Posts: 16 Thanks: 0 Re: Calculate the length of the side of an irregular tetragon. Hi Mr. Spot, It is understood in mathematics that any figure that we can draw correctly geometrically can be calculated. Since you have drawn the figure to your satisfaction, a solution should exist. Having said that, I have to agree to a certain extent with Dave that your dimensioning is not easy to understand, to say it very politely. One of the rules of drawing requires us to dimension between parallel lines only. That is where witness lines are for when two lines are not parallel to each other. In the figure, the top side BC and the line below it have been dimensioned on the left as 241 and the right 217 mm apart. I see a witness line on the left of the 217 dimension, but it is not parallel to BC, and the slash is not on the witness line. As for the 241 dimension, it is a dimension between two non-parallel lines, and the locations where the dimension applies are not defined. In fact, I notice that some dimensions are parallel to the inner hexagon with a witness line to a vertex of the outer, while some other dimensions are the reverse. It would help your construction if all (thickness) dimensions are perpendicular to the outer hexagon and dimensioned with respect to an inner vertex. I am no architect and do not wish to critize anyone. However, most of the architectural drawings I have see would have important lines thickened like in your drawing, but the dimension lines remain thin to make the object stand-out. I believe if you put more witness lines and have the 'slashes' 'unbolded', it would make the drawing easier to understand. Back to the construction. It would be relatively easy to calculate the coordinates of the vertices of the inner hexagon, as all the lengths are known, as well as the angles. The interior angles add up to 720 degrees as they should. If all the distances to the outer hexagon are defined unambiguously, the coordinates of the outer vertices would be equally easy to calculate.
November 18th, 2008, 04:32 PM   #9
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Re: Calculate the length of the side of an irregular tetragon.

Quote:
 Originally Posted by mathmate It would help your construction if all (thickness) dimensions are perpendicular to the outer hexagon and dimensioned with respect to an inner vertex.
I was at a point of assuming that, and that they were perhaps drawn further in for possible clarity in indicating dimensions. However, was still left with the gut feeling that you'd have to work your way all around the figure to find even one missing value, there not being enough information between two adjacent outer quadrilaterals; i;e; a system of equations perhaps. In any event, I'd personally not spend time on that in this day and age when there must be software available to help in that way, as I suggested with a 2D surveying problem I did. That is why I decided to not pursue it.

David.

 November 18th, 2008, 05:02 PM #10 Newbie   Joined: Oct 2008 Posts: 16 Thanks: 0 Re: Calculate the length of the side of an irregular tetragon. I see a difference between a CAD program and a mathematical solution. I assume that Mr. Spot is in the process of developing his own program using design parameters. While this particular problem represents one case which can be solved using a CAD program in 3 minutes (if all dimensions are unambiguous and nothing missing), or it takes an hour or so to parametize the problem with a macro (assuming the parameters are defined), it may help him to find a mathematical solution which he can incorporate into his own program which is '200 times more powerful than Autocad'. It is in this spirit that I am pursuing the question. Mathematically, it is just six sets of 2x2 simutaneous equations for the inner figure. Depending on how the outer hexagon is dimensioned, it is a litlle more complicated to find the six outer equations, but then it would be reduced again to six sets of 2x2 equations. Another doubt I have is that the figure is dimensioned (presumably) in millimetres, while the angles are all known. So there is a problem of precision, because lengths don't fall to the nearest millimetre. It is not clear whether the lengths govern, or the angles do.

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