[MWForum]Re: a very simple polygon procedure
Steve Robson
mwforum@lists.mathcats.com
Mon, 16 Feb 2004 11:32:40 +0000
<html><div style='background-color:'><DIV class=RTE>
<P>This is very, very simple but draws what you ask it to.</P><FONT face="Courier New">
<P><FONT size=2><STRONG>to poly :sides<BR>repeat :sides[fd 100 rt 360 / :sides]<BR>end</STRONG></FONT></P>
<P><FONT face="Geneva, Arial, Sans-serif" size=2>Then call poly 4 (or whatever)</FONT></FONT><BR><FONT size=2>Cheers</FONT></P>
<P>Steve<BR></P></DIV>
<DIV></DIV>From: Pavel Boytchev <PAVEL@ELICA.NET>
<DIV></DIV>Reply-To: mwforum@lists.mathcats.com
<DIV></DIV>To: mwforum@lists.mathcats.com
<DIV></DIV>Subject: Re: [MWForum](no subject)
<DIV></DIV>Date: Mon, 16 Feb 2004 03:15:00 +0200
<DIV></DIV>
<DIV></DIV>> >In probing why the number of sides relates to the angles
<DIV></DIV>> >(having to add up to 360) what angles are refered to? I am
<DIV></DIV>> >wanting to know why this is so.
<DIV></DIV>
<DIV></DIV>If you draw a polygon with N sides, the turtle should definitely turn N
<DIV></DIV>times. Because angles (orientation, turtle's heading) do not change
<DIV></DIV>during FORWARD movement, the final orientation will be the same as if
<DIV></DIV>the turtle only turns around (and does not move forward). So, if the
<DIV></DIV>turtle stays on one place and makes N turns to the left, and if the
<DIV></DIV>final orientation is the same as the initial one, then all turn angles
<DIV></DIV>should have a sum = 360. If the angles are the same, then each of them
<DIV></DIV>should be 360/N. Thus, the following code:
<DIV></DIV>
<DIV></DIV>repeat :N [ fd 10 lt 360/:N]
<DIV></DIV>
<DIV></DIV>will draw a square when N=4, and a pentagon when N=5, and the bigger the
<DIV></DIV>N is, the more circle-like the polygon is.
<DIV></DIV>
<DIV></DIV>This is a simplification of a more general phenomena.
<DIV></DIV>
<DIV></DIV>[1] The overall turn angle is not necessarily equal to 360 degrees. It
<DIV></DIV>can also be 360*K where K is integer number (0, 360, -360, 720,
<DIV></DIV>-720,...) So if the overall turn angle is 360 degrees, this is just the
<DIV></DIV>case when K=1
<DIV></DIV>
<DIV></DIV>[2] If you have both left and right turns, you must convert all of them
<DIV></DIV>into only left, or into only right (by using that LT :A = RT 0-:A). When
<DIV></DIV>you calculate the total sum of angles (positive and negative angles),
<DIV></DIV>the result must be 360*K if the final orientation is the same as the
<DIV></DIV>initial one.
<DIV></DIV>
<DIV></DIV>Pavel
<DIV></DIV>
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