Re: a very simple polygon procedure [MWForum Archive]

This is very, very simple but draws what you ask it to.

to poly :sides
repeat :sides[fd 100 rt 360 / :sides]
end

Then call poly 4 (or whatever)
Cheers

Steve

From: Pavel Boytchev

Reply-To: mwforum@xxxxxxxxxxxxxxxxxx

To: mwforum@xxxxxxxxxxxxxxxxxx

Subject: Re: [MWForum](no subject)

Date: Mon, 16 Feb 2004 03:15:00 +0200

> >In probing why the number of sides relates to the angles

> >(having to add up to 360) what angles are refered to? I am

> >wanting to know why this is so.

If you draw a polygon with N sides, the turtle should definitely turn N

times. Because angles (orientation, turtle's heading) do not change

during FORWARD movement, the final orientation will be the same as if

the turtle only turns around (and does not move forward). So, if the

turtle stays on one place and makes N turns to the left, and if the

final orientation is the same as the initial one, then all turn angles

should have a sum = 360. If the angles are the same, then each of them

should be 360/N. Thus, the following code:

repeat :N [ fd 10 lt 360/:N]

will draw a square when N=4, and a pentagon when N=5, and the bigger the

N is, the more circle-like the polygon is.

This is a simplification of a more general phenomena.

[1] The overall turn angle is not necessarily equal to 360 degrees. It

can also be 360*K where K is integer number (0, 360, -360, 720,

-720,...) So if the overall turn angle is 360 degrees, this is just the

case when K=1

[2] If you have both left and right turns, you must convert all of them

into only left, or into only right (by using that LT :A = RT 0-:A). When

you calculate the total sum of angles (positive and negative angles),

the result must be 360*K if the final orientation is the same as the

initial one.

Pavel

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