text: Search All of the Math Forum:
  
  Views expressed in these public forums are not endorsed by NCTM or The Math Forum.
  
  Notice: We are no longer accepting new posts, but the forums will continue to be readable.
  
  Topic: Nbic Circles: Chord distance
  Replies: 0
  
   Doctor Nisith Bairagi Posts: 27 From: Uttarpara, West Bengal, India Registered: 3/2/13
  Nbic Circles: Chord distance
  Posted: Apr 10, 2013 10:26 AM
  
  From: Doctor Nisith Bairagi
  Subject: NBIC CIRCLES: CHORD DISTANCE
  Date: April 10, 2013
  (geometry.research.independent)
  >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
  The generalized form of chord distance of Nbic (viz., Euclidean, Hyperbolic, Nbic) circles has been presented. Reference may be made to the Topic <CLASSIFICATION OF CIRCLES:A MAIDEN CONCEPT> posted on March 19, 2013, and on March 20, 2013 in (geometry.research.indepedent).
  >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
  Distance (d) between any two points A(X1, Y1) and B(X2, Y2) is given by:
  d = AB = sqrt[(X2-X1)^2 + (Y2-Y1)^2]
  
  (A) For Euclidean circle (plane circle), C<0>:
  
  X^2 + Y^2 = 1, (X, Y) = (cos x, sin x), [where, x is angle in rad (or, in deg)],
  the chord distance d turns out to be:
  d = sqrt[2].sqrt[1-(cos x2.cos x1 + sin x2.sin x1)] = 2 sin[(x2-x1)/2]
  With this, we get:
  Between x1 = 0, and, x2 = PI/2, d = sqrt[2]
  and, x1 = 0, x2 = PI, d = 2 (diameter).
  
  (B) For Hyperbolic circle (plane circle), C<H>:
  
  X^2 + Y^2 = 1, (X, Y) = [cosh x/sqrt(cosh 2x), sinh x/sqrt(cosh 2x)]
  the chord distance d turns out to be:
  d = sqrt([2].sqrt[1-(cosh x2.cosh x1 + sinh x2.sinh x1)/(cosh 2x2.cosh 2x1)]
  = sqrt([2].sqrt[{cosh(x2 + x1)}/(cosh 2x2.cosh 2x1)]
  
  (C) For all other Nbic circles, C<N>:
  [viz., C<1>, C<2e>, C<2f>, C<3e>, C<3f>], satisfying, X^2 + Y^2 = 1,
  
  The distance formula (d) for Nbic circles C<N, is written using parameters (k and m):
  
  C<1>: (k = <1>, m = 1), (X, Y) = [cosN(x,x), sinN(x, x)]/sqrt(cosh 2x)
  C<2e>: (k = <2e>, m = 1), (X, Y) = [cosN<2e>(x, x), sinN<2e>(x, x)]/sqrt(cosh 2x)
  C<2f>: (k = <2f>, m = 2), (X, Y) = [cosN<2f>(x, x), sinN<2f>(x, x)]/cosh 2x
  C<3e>: (k = <3e>, m = 2), (X, Y) = [cosN<3e>(x, x), sinN<3e>(x, x)]/cosh 2x
  C<3f>: (k = <3f>, m = 3), (X, Y) = [cosN<3f>(x, x), sinN<3f>(x, x)]/(cosh 2x)^(3/2):
  
  d = sqrt([2].sqrt[1-(cosN<k> x2.cosN<k> x1 + sinN<k> x2.sinN<k> x1)/
  (cosh 2x2.cosh 2x1)^m]
  
  >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
  Please comment on this topic for betterment.
  
  Thanks to ALL
  
  Doctor Nisith Bairagi
  >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
metadata: {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}