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Do Points
Have Area? Measuring Points Subject: RE: Reply to ""Reply to "Re: Do Points Have Area?"" Author: Jesse Yoder jesse@flowresearch.com Date: Mon, 2 Feb 1998 16:50:30 -0500 Hey Candice - Sorry I overlooked this email. Let me respond to your questions. You wrote: > In respect to your Points, when you measure the distance between two > Points, do you measure from??? > > a. the center of the two points, > b. from the sides facing each other, or > c. from the opposite sides? > RESPONSE: In general, I would say "Use corresponding positions. So if you measure from the center of Point A, use the center of point B. Your choices b and c violate this principle. This is a real-world problem that most people completely ignore. > If your answer is a, how can you have a center to the smallest > circular mesurement??? Wouldn't that center have to be at a Point??? > So then the center of a Point is a Point which is a Point to infinity! > RESPONSE: I know (or believe) you are trying to find a contradiction in my theory here. What I have said is that a Point is the smallest unit of measurement accepted for a given purpose or application. So you are treating the Point as being "unbreakable" for your measurement. So in a sense the distance between any two Points A and B is from anyplace on A to anyplace on B. But logic would dictate using corresponding locations on A and B, and measuring from there. Your discussions of b and c are interesting, but I reject both of these as answers. > If your answer is b, how can you have area that doesn't exist??? > > Because if you want to find the length between Point A and Point B, > and there exists a Point C (which is colinear with A and B) which is > between Point A and Point B. Because AC+CB=AB when the points are > colinear, we must account for the length in Point C... which is not > measured with AC or with CB, so then, in your circular geometry, AC + > CB does not equal AB. AC + CB < AB > > If your answer is c, how can you count up area twice... in the > above > scenario with this answer, AC + CB > AB > > Jesse
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