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

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November 18th, 2008, 10:31 PM   #11
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Re: Calculate the length of the side of an irregular tetragon.

Quote:
 Originally Posted by mathmate 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.
If it helps you can think of it as a program. I've uploaded a revised diagram with the dimension style changed which hopefully is a little less confusing. Remember, this was just a quick diagram that I pieced together just to demonstrate what I'm trying to calculate. I didn't realised I was going to be critiqued on my diagram...

As clarified earlier the 225/256 dimension are parameters that define the outer hexagon corners based on an offset from the interior hexagon edges. They are not parallel to the interior hexagon and it is intended to be this way. There is no need to calculate x, y values for the interior hexagon points as they are already known and are used by the user when the define the size of the initial hexagon when they place it in the computer model of 60 metre structure.

Units are not an issue, all dimensions shown are parameters that the user can define. I just need to calculate the lengths of the edges of the outer hexagon so I can generate the true lengths of the side pieces (far right of the second screenshot). This six pieces are the face elevations of the returns to this front hexagonal face. As you can see in that diagram I have already defined all the other faces.

The program I use "Autodesk Revit Architecture" enables me to apply formula to parameters based on trig, & conditional statements.
Attached Images
 Revised Diagram.jpg (197.8 KB, 415 views)

 November 19th, 2008, 06:09 PM #12 Newbie   Joined: Oct 2008 Posts: 16 Thanks: 0 Re: Calculate the length of the side of an irregular tetragon. Mr. Spot, Yes indeed, in this diagram, I can clearly see that all thickness dimensions are parallel to one side of the interior hexagon, and dimensioned to an outer vertex. 1. Precision I am not worried about the precision of calculations, but am worried about the consistency of the given data. As the only decimal numbers I see are in the angles, I can tell with 99.9999% certainty that at least some of the dimensions of the inner polygon have been rounded to the nearest unit. If the unit is millimetres, it is probably neglibible. But if the unit is in metres, rounding will cause a little problem. This is where I am coming from about units. 2. Revit Architecture If you are using Autodesk Revit Architecture, it should allow you to enter the given polygon, since as you said, the coordinates are all known. As you have not named the interior polygon, I will assume that the vertices are named the same as the outer, but with a '. So the interior polygon will be A'B'C'D'E'F' and the outer is simply ABCDEF. You will notice that point C, for example, can be defined as the intersection of a line parallel to B'C', but 217 units from it, and the line parallel to C'D' but 306 units from it. You can similarly define the remaining vertices of the outer polygon and hence complete the polygon. Apart from round-off errors (as I mentioned above), you should then have your perfect answer using Revit Architecture. In addition, if you give the thickness dimensions as user-entered parameters, you will have generated a program that creates the object you wanted. Would you give it a try?
November 20th, 2008, 12:42 PM   #13
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Re: Calculate the length of the side of an irregular tetragon.

Hi Mathmate,

Thanks for your input, but currently you are miss-intepreting the issue. Revit can draw the hexagon ring just fine. The Hexagon face it self is fine. I need to find the outer lengths in order to be able to draw 2D versions of the outer sides of the hexagon that automatically adjust when i change the 3D model size. Rounding is not an issue, the program only uses the completely accurate true values. The dimensions I've shown I thought would be sufficient to outline the problem.

This can only be achieved if I can find these lengths with a formula.

This is an updated sketch of what I have found somewhere as I think I have managed to get somewhere...
[attachment=0:1038d946]Hand Sketch.jpg[/attachment:1038d946]
Attached Images
 Hand Sketch.jpg (327.9 KB, 384 views)

 November 20th, 2008, 01:25 PM #14 Senior Member   Joined: Jul 2008 Posts: 895 Thanks: 0 Re: Calculate the length of the side of an irregular tetragon. Yes, the equations you gave [if they relate to the true diagram] can be solved. Do you know the expansion formula for cosine of a sum or difference? $\begin{array}{l} b\cos y = 489 \\ b\cos (66 - y) = 334 \\ 489\left( {\cos 66\cos y - \sin 66\sin y} \right) = 334\cos y \\ 489\sin 66\sin y = \left( {334 - 489\cos 66} \right)\cos y \\ \tan y = \frac{{334 - 489\cos 66}}{{489\sin 66}} \\ \end{array}$ Find y and then cos y to find b.
 November 20th, 2008, 05:30 PM #15 Newbie   Joined: Nov 2008 Posts: 8 Thanks: 0 Re: Calculate the length of the side of an irregular tetragon. Hi Dave, That works. Your a legend. I don't understand how you simplified those two equations though. Knowing this length and bearing should hopefully now help to find their actual lengths. I'll see how I go... I still think its going to be rather difficult so any guidance is appreciated!
November 20th, 2008, 05:56 PM   #16
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Re: Calculate the length of the side of an irregular tetragon.

Quote:
 Originally Posted by Mr Spot I don't understand how you simplified those two equations though.
Divide the one by the other and the "b"s disappear. Cross-multiply and expand the cosine of the difference:
cos(A - B) = cosAcosB [color=#FF0000]+[/color] sinAsinB. Collect sin(y) and cos(y) terms, then use tan = sin/cos. Use the inverse tan function to find y. Then sub for cosine of that angle in the first equation to find b.

You can always Google for trig relations if you don't have a book handy.

My main problem was in understanding your structure. If it was plane geometry [or 3D] without the architectural stuff, we'd have been further ahead. Now my interest is tweaked, and I'll have to learn to read architectural drawings.

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