
CLASSIFICATION OF CIRCLES : A MAIDEN CONCEPT
Posted:
Mar 19, 2013 3:10 AM


From : Doctor Nisith Bairagi Retired Professor of Structural Engineering Indian Institute of TechnologyBombay, India Subject : CLASSIFICATION OF CIRCLES ? A MAIDEN CONCEPT Date : March 19, 2013 ????????????????????????.. Introduction: The only circle, we know till date, is the Euclidean circle, which, in the rectangular coordinate system (X, Y), where X and Y are certain functions of the parameter x, can be defined as the locus of a point P(X, Y), maintaining a constant distance (radius) from a fixed point O(0,0) called centre. The starting point at an angle x = 0, after encircling just one complete turn of x = 2*PI, returns back to the same point.
Interestingly, there exist a group of circles satisfying X^2 + Y^2 = 1, where, unlike the Euclidean circle, the starting point A (at x = 0) and the end point B (at x = 2*PI), are not the same point, thus showing the circular contour to trace less than or, in excess of one complete turn. These types of circles, including the Euclidean one, have been brought under a generalized name as: Nbic circles. The present topic is on the categorization and characteristics of the Nbic circles. ?????????????????????????????????. The angle measured in Nbic circles are in Nbic angle N (which is also a function of x). When geometrically measured, in terms of ?teN? angle (tan equivalent of the Nbic angle), it becomes: tan N = Y/X. Further, for the point P(X, Y), the difference between the teN angle N and x, is the sweep angle, given by: sA = (arctan N ? x). The angular difference between B and A, measured in terms of the angle N<BA>, or alternatively, the sweep angle sA, play the pivotal role for the classification of the Nbic circles.
Thus all the Nbic circles when categorized groupwise, are: Type1. Incomplete /Inferior /Negative circle: N<BA> less than 2*PI, sA = negative, identified in Hyperbolic circle, C<H>. Type2. Complete /Full / zero circle: N<BA> equal to 2*PI, sA = 0, identified in Euclidean circle, C<0>. Type3. Over Complete / superior /Positive circle: N<BA> greater than 2*PI, sA = positive, identified in Nbic circle, C<N>.
Generation of Type1 Nbic circle (Hyperbolic circle, C<H>): If (X, Y) = (cosh x, sinh x), the equation results in X^2 ? Y^2 = 1, which represents a hyperbola, and not a circle. To represent it as a circle of the type, X^2 + Y^2 = 1, modify (X, Y) = (cosh x, sinh x)/sqrt[cosh(2x)]. Here, tan N = Y/X = tanh x. Then teN angles N at A = 0 (for x = 0), and at B = PI/4 (for x = 2*PI). N<BA> = PI/4 (it is < 2*PI), sA = PI/4 ? 2*PI = it is negative. Geometrically, the C<H> circle covers only oneeighth of a full circle. Generation of Type2 Nbic circle (Euclidean circle, C<0>): If (X, Y) = (cos x, sin x), the circle obtained is the Euclidean circle, which is satisfied directly by, X^2 + Y^2 = 1. Here, tan N = Y/X = tan x. Then teN angles N at A = 0 (for x = 0), and at B = 0 (for x = 2*PI), showing that, A and B are the same point. N<BA> = 2*PI, sA = 0. Geometrically, the C<0> circle covers just one full turn.
Generation of Type3 Nbic circle (Nbic circle, C<N>): We have, exponential function: e^x = cosh x + sinh x On the same line, propose: fexponential function, f^x = cos x + sin x
Replace (x) by (ix) in both the functions, and call them Coci and Cohy functions as: e^(ix) = Coci (complex circular) function = cos x + i sin x f^(ix) = Cohy (complex hyperbolic) function = cosh x + i sinh x
For Single Nbic circle (C<1>):
Get first Single Nbic function [N(x, x)]: Multiplying together the Coci and Cohy functions, [e^(ix).f^(ix)], and separating the real and imaginary parts, we arrive at : cosN(x, x) + i sinN(x, x), from which the Single Nbic functions [N(x, x)] consisting of, cosN(x, x) and sinN(x, x); [similar to the trigonometric expansion of cos(A + B) and sin(A + B)] can be collected as : cosN(x, x) = cos x cosh x ? sin x sinh x sinN(x, x) = sin x cosh x + cos x sinh x [pronounce: cos enbic, sin enbic, etc., identified by the letter N, as in hyperbolic functions cosh, sinh, identified by h]
Thus since, [cosN(x, x)]^2 + [sinN(x, x)]^2 = cosh(2x), for Single Nbic circle C<1> to satisfy the equation, X^2 + Y^2 = 1, modify (X, Y) = [cosN(x, x), sinN(x, x)]/sqrt[cosh(2x)]
Here, tan N<1> = Y/X = sinN(x, x)/cosN(x, x) = tanN(x, x) = (tan x + tanh x) / (1 tan x tanh x); [similar to tan(A + B) formula]
Then teN angles N at A = 0 (for x = 0), and at B = 9*PI/4 (for x = 2*PI). N<BA> = 9*PI/4 (it is PI/4 in excess of 2*PI), sA = 9*PI/4 ? 2*PI = it is positive. Geometrically, the C<1> circle covers oneeighth of a full circle, after completing one full turn.
For Double Nbic circle (N<2e>, and N<2f>), and For Triplele Nbic circle (N<3e>, and N<3f>)
Get first Double Nbic functions, [N<2e>(x, x), N<2f>(x, x)], and Triple Nbic functions, [N<3e>(x, x), N<3f>(x, x)]. To arrive at these functions, multiply Single Nbic function, [cosN(x, x) + i sinN(x, x)], by Coci function or Cohy function, and separate the real and imaginary parts, to get Double Nbic function of N<2e>(x, x) or N<2f>(x, x), respectively.
Similarly, Triple Nbic functions can be written down from the product of Double Nbic functions with Coci function or Cohy function, and separate the real and imaginary parts. Thus, rearranging adopting typical Double and Triple Nbic functions, we arrive at: For C<2e> circle: [(cosN<2e>(x, x)]^2 + [(sinN<2e>(x, x)]^2 = cosh(2x), and For C<2f> circle: [(cosN<2f>(x, x)]^2 + [(sinN<2f>(x, x)]^2 = [cosh(2x)]^2 For C<3e> circle: [(cosN<3e>(x, x)]^2 + [(sinN<3e>(x, x)]^2 = cosh(2x), and For C<3f> circle: [(cosN<3f>(x, x)]^2 + [(sinN<3f>(x, x)]^2 = [cosh(2x)]^2 [They may be generated in other ways also.]
To represent all of them as circle of the type, X^2 + Y^2 = 1:
For C<2e> circle: modify (X, Y) = [cosN<2e>(x, x), (sinN<2e>(x, x)]/sqrt[cosh(2x)] Here, tan N<2e>(x, x) = Y/X = [tan 2x.+ tanh x]/[1tan 2x.tanh x]. Then teN angles N at A = 0 (for x = 0), and at B = 17*PI/4 (for x = 2*PI). N<BA> = 17*PI/4 (it is PI/4 in excess of 4*PI), sA = 9*PI/4 = it is positive. Geometrically, the C<2e> circle covers only oneeighth of a full circle, after completing two full turns.
For C<2f> circle: modify (X, Y) = (cosh x, sinh x)/[cosh(2x)]. Here, tan N<2f>(x, x) = Y/X = [tan x.+ sinh(2x)]/[1tan x.sinh(2x)]. Then teN angles N at A = 0 (for x = 0), and at B = 3*PI/2 (for x = 2*PI). N<BA> = 3*PI/2 (it is PI/2 in excess of 2*PI), sA = PI/2 = it is + ve. Geometrically, the C<2f> circle covers only onefourth of a full circle, after completing one full turn.
For C<2e> circle: modify (X, Y) = [cosN<2e>(x, x), (sinN<2e>(x, x)]/sqrt[cosh(2x)] Here, tan N<2e>(x, x) = Y/X = [tan 2x.+ tanh x]/[1tan 2x.tanh x]. Then teN angles N at A = 0 (for x = 0), and at B = 17*PI/4 (for x = 2*PI). N<BA> = 17*PI/4 (it is PI/4 in excess of 4*PI), sA = 9*PI/4 = it is positive. Geometrically, the C<2e> circle covers only oneeighth of a full circle, after completing two full turns.
For C<2f> circle: modify (X, Y) = (cosh x, sinh x)/[cosh(2x)]. Here, tan N<2f>(x, x) = Y/X = [tan x.+ sinh(2x)]/[1tan x.sinh(2x)]. Then teN angles N at A = 0 (for x = 0), and at B = 3*PI/2 (for x = 2*PI). N<BA> = 3*PI/2 (it is PI/2 in excess of 2*PI), sA = PI/2 = it is + ve. Geometrically, the C<2f> circle covers only onefourth of a full circle, after completing one full turn. >> For C<3e> circle: modify (X, Y) = [cosN<3e>(x, x), (sinN<3e>(x, x)]/[cosh(2x)] Here, tan N<3e>(x, x) = Y/X = [tan 2x.+ tanh (2x)]/[1tan 2x.tanh (2x)]. Then teN angles N at A = 0 (for x = 0), and at B = 9*PI/2 (for x = 2*PI). N<BA> = 9*PI/2 (it is PI/2 in excess of 4*PI), sA = 5*PI/2 = it is positive. Geometrically, the C<3e> circle covers only onefourth of a full circle, after completing two full turns.
For C<3f> circle: modify (X, Y) = [cosN<3f>(x, x), (sinN<3f>(x, x)]/[cosh(2x)]^(3/2). Here, tan N<3f>(x, x) = Y/X = [tanN x.+ sinh(2x)]/[1tanN x.sinh(2x)]. Then teN angles N at A = 0 (for x = 0), and at B = 11*PI/4 (for x = 2*PI). N<BA> = 11*PI/4 (it is 3*PI/4 in excess of 2*PI), sA = 3*PI/4 = it is positive. Geometrically, the C<3f> circle covers only threeeighth of a full circle, after completing one full turn.
Remarks: That there exist other types of circles other than Euclidean circle is established. Readers are requested to assess, make comments and suggestion, and accept. Please inform me in < bairagi605@yahoo.co.in >.
Thanks to all
Doctor Nisith Bairagi

