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Conference rusure::math

Title:	Mathematics at DEC

Moderator:	RUSURE::EDP

Created:	Mon Feb 03 1986
Last Modified:	Fri Jun 06 1997
Last Successful Update:	Fri Jun 06 1997
Number of topics:	2083
Total number of notes:	14613

1402.0. "Finding points in Algebriac space." by SUBWAY::BERG () Mon Mar 25 1991 11:02

   I am trying to do some Xwindow graphics in which I am plotting semi-
   circular points in space (by "semi" I mean that the points are all 
   dispersed around a central point, but that for each point the diameter
   is variable). Thus each seperate point is a different diameter from 
   the center which allows continually growing data to be represented 
   as a spiral over time.

   The question is how to represent this graphically. I know one angle
   (it's always 1 degree), and the measurements for all three sides. By
   alternately defining one the the points of the angle to be a random
   point I can even define all three angles and and all three measure-
   ments. The question is, how do I calculate the x,y coordinates of 
   this point. All the math books that I have show how to calculate this
   for points on a smooth curve, but none for this kind of situation.

   I know that this can be done because I can use a ruler and protractor
   on a sheet of paper to calculate it (always my ultimate proof). What
   equation can I use for it? 

   P.S. For those of you visual types with DDIF viewing capabilities, 
   the next reply is a graphical representation of the question.

T.R	Title	User	Personal Name	Date	Lines
1402.1	Supposed graphical aid	SUBWAY::BERG		`Mon Mar 25 1991 11:08`	4
	Sorry, I was assuming latest version of NOTES. My mistake. I cann't include DDIF pictures in this notes files. Sorry ... :-(
1402.2	Maybe this works (?)	SUBWAY::BERG		`Mon Mar 25 1991 11:11`	19
1402.3	Maybe this works	VMSDEV::HALLYB	The Smart Money was on Goliath	`Mon Mar 25 1991 14:12`	20
	Point of correction: distance is better termed "radius" than "diameter" in this situation. Also your diagram in .2 is incorrect in its calculation of c, which should be the sqrt of the differences of the dn� values. But that doesn't matter, since you shouldn't need to know its value. The general formula for calculating (x,y) is: x = r cos a y = r sin a Where a is the angle in radians (1 radian == 57.2958 degrees) and r is the radius, what you have labeled d� and d�. Note a = 0 when the point is directly to the right of the origin, at (r,0). You would not merrily calculate with a = 1/57.2958, but rather N/57.2958 when you are plotting the Nth point (if you expect to maintain a constant difference of 1 degree between points). John
1402.4	Close, but no radius	SUBWAY::BERG		`Fri Mar 29 1991 10:09`	19
	Re: -1 I think that a point missed here is that we are not dealing with a true circle, but instead randomly spaced points located about a central point. I have no radius because d1 and d2 are NOT equal. The problem that I have is that all of the textbooks that I have looked at make similiar assumptions, that all angle calculations are based upon circles. What I need is a calculation to figure out x, y coordinates based upon triangles or other shapes that are not fixed. Again, there must be a way to do this because it is simple to calculate on a piece of paper. The number is always there, all that I need is a way to calculate it. P.S. Sorry for the delay in responding to my own problem but working in PSS, I am frequently called away to work on other various projects and I don't always have access to systems.
1402.5	supporting the answer in .3	CSSE::NEILSEN	Wally Neilsen-Steinhardt	`Fri Mar 29 1991 12:56`	21
	.4> I think that a point missed here is that we are not dealing with a true > circle, but instead randomly spaced points located about a central point. > I have no radius because d1 and d2 are NOT equal. Unless I am missing something, .3 gives you the answer you want. It does not matter whether the points are on a circle or not. It might make things clearer if you added a subscript to r and a, like this xi = ri cos ai yi = ri sin ai This works for any randomly chosen point, specified by any ri and ai. It does not matter whether there is a circle there or not. I gather you have a simple case in that ai = i * a1 which can make the math simpler.
1402.6	have you tried Cornu spiral?	CALS::GELINEAU		`Thu Oct 03 1991 11:48`	6
	try looking at the Cornu (sp?) spiral graph; it's found in most undergrad optics (physics) texts - i didn't look at your graphics so i'm not sure if it will help but your original note made me think of the spiral immediately. Angela