Freethought & Rationalism Archive

philechat · 05-08-2003, 09:51 AM

Do you mean the polygon must satisfy both the side length and the angle measurement? It would be easy if only one condition were given (such as angle measurements which must satisfy (p-2)180 for polygon of side p). For length it would also be easy as long as you can recall triangular inequality.

If the polygon must satisfy both conditions, then proceed as following:

Find a given corner as the reference angle R and divide the polygon of p sides to p-2 triangles.
Remember the law of cosine (check your calculus book for it...I could not recall it on top of my head)? In this case A, B, and C are angles and a, b, and c are length of the triangle aformentioned. Given the length of a and b and the angle of C you should be able to calculate the length of c and the partial angles A and B1 (where B1 is the partial angle of R). Subtract the partial angle A from a given angle, to construct another triangle from the existing triangle (with another given length d) and iterate the process. The last angle (say N) would be the last "given" angle and the last partial angle Bn (at reference angle R) should add up with other partial angles (i.e. B1...Bn-1) to form R.

I hope this is not too darn confusing.

(edited to change symbols in order to avoid confusion)

Loren Pechtel · 05-08-2003, 10:40 AM

[BB]Originally posted by Jesus Tap-Dancin' Christ
Except for triangles, pentagons, hexagons, septagons, etc.

It's (n-2)180 degrees, where n=the total number of faces.[/B]

Huh? Lets take a simple equilateral triangle. Three angles of 60 each.

180-60 = 120. There are three of them. Sum = 360. Passed.

For #2, you have to arrange them as vectors (yes, I know, boo, hiss), determine their x and y components, and then sum it.

In effect, that's what I was doing.

Loren Pechtel · 05-08-2003, 10:42 AM

Quote:

Originally posted by Jesus Tap-Dancin' Christ
You need x-y coordinates. Loren tells how to get there by way of his formula

What Loren means is the angle relative to a line that is drawn from nearby the object--either parallel with a base, or perpendicular to it (parallel is a much simpler option, by the way). By taking the cosine of the each angle and the side it coresponds to (do each side once, whichever angle should be fine), you arrive at the x component for that segment. The sine will give the y component. If you add the x components up, they should equal zero, and y components should simlarly combine to zero.

Try this with a square. You have l*sin 90+l*sin 270 for the y components (which gives sum zero), and l*cos 0+l*cos 180 for the X components (which also gives zero)..

lpetrich expressed my approach in pseudo-code. I think his is clearer.

philechat · 05-08-2003, 10:48 AM

Indeed. Vector sum seems like the most efficient solution. Try Loren Pechtel's and lpetrich's solution using the x-y coordinate. Ignore my solution for now because it takes too much work.

Jesus Tap-Dancin' Christ · 05-08-2003, 11:03 AM

Quote:

Originally posted by Loren Pechtel
[BB]Originally posted by Jesus Tap-Dancin' Christ
Except for triangles, pentagons, hexagons, septagons, etc.

It's (n-2)180 degrees, where n=the total number of faces.

Huh? Lets take a simple equilateral triangle. Three angles of 60 each.

180-60 = 120. There are three of them. Sum = 360. Passed.

For #2, you have to arrange them as vectors (yes, I know, boo, hiss), determine their x and y components, and then sum it.

In effect, that's what I was doing. [/B]

Ah, misread your post. Sorry.

lpetrich · 05-08-2003, 12:14 PM

My approach also permits the angle-sum check; CumAng will have a final value of 360d.

GunnerJ · 05-08-2003, 01:26 PM

Thanks for all your help, everyone; I would have never thought of this. Does anyone have any references for how this works? I'd like to know out of curiousity.

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05-08-2003, 09:51 AM	#11
philechat Senior Member Join Date: Apr 2002 Location: Mind of the Other Posts: 886	Do you mean the polygon must satisfy both the side length and the angle measurement? It would be easy if only one condition were given (such as angle measurements which must satisfy (p-2)180 for polygon of side p). For length it would also be easy as long as you can recall triangular inequality. If the polygon must satisfy both conditions, then proceed as following: Find a given corner as the reference angle R and divide the polygon of p sides to p-2 triangles. Remember the law of cosine (check your calculus book for it...I could not recall it on top of my head)? In this case A, B, and C are angles and a, b, and c are length of the triangle aformentioned. Given the length of a and b and the angle of C you should be able to calculate the length of c and the partial angles A and B1 (where B1 is the partial angle of R). Subtract the partial angle A from a given angle, to construct another triangle from the existing triangle (with another given length d) and iterate the process. The last angle (say N) would be the last "given" angle and the last partial angle Bn (at reference angle R) should add up with other partial angles (i.e. B1...Bn-1) to form R. I hope this is not too darn confusing. (edited to change symbols in order to avoid confusion)

05-08-2003, 10:40 AM	#12
Loren Pechtel Obsessed Contributor Join Date: Sep 2000 Location: Not Mayaned Posts: 96,752	[BB]Originally posted by Jesus Tap-Dancin' Christ Except for triangles, pentagons, hexagons, septagons, etc. It's (n-2)180 degrees, where n=the total number of faces.[/B] Huh? Lets take a simple equilateral triangle. Three angles of 60 each. 180-60 = 120. There are three of them. Sum = 360. Passed. For #2, you have to arrange them as vectors (yes, I know, boo, hiss), determine their x and y components, and then sum it. In effect, that's what I was doing.

05-08-2003, 10:48 AM	#14
philechat Senior Member Join Date: Apr 2002 Location: Mind of the Other Posts: 886	Indeed. Vector sum seems like the most efficient solution. Try Loren Pechtel's and lpetrich's solution using the x-y coordinate. Ignore my solution for now because it takes too much work.

05-08-2003, 12:14 PM	#16
lpetrich Contributor Join Date: Jul 2000 Location: Lebanon, OR, USA Posts: 16,829	My approach also permits the angle-sum check; CumAng will have a final value of 360d.

05-08-2003, 01:26 PM	#17
GunnerJ Banned Join Date: Sep 2001 Location: a place where i can list whatever location i want Posts: 4,871	Thanks for all your help, everyone; I would have never thought of this. Does anyone have any references for how this works? I'd like to know out of curiousity.

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