Generalized Spherical Triangles (GST)

Allan Cortzen

Abstract.  The purpose is to introduce spherical triangles with one or more angles or sides greater than pi.  A signed theory extends with a little modification the connection between area and spherical excess , and also facilitates a classification of triangles.  An unsigned entity easily is deductible with standard spherical trigonometry.

In R3 set of three unit vectors (a, b, c) not in a plane determines a spherical triangle DABC,  [[The Greek letter capital delta is displayed as the Latin glyph D.]] and also an orientation sign(ABC) as the sign of the determinant of the matrix [a, b, c] of coordinates in terms of the standard basis.

A generalized spherical triangle is given by such three points (A,B,C) and three oriented great-circle segments (AB, BC, CA) each directed from the first point to the second.

As each segment can be short or long, i.e. in (0, pi) or (pi, 2 pi), there are eight GST's with given vertices (A, B, C).  A long segment is indicated by an arc as in GST+ (A-BC-) meaning a positively oriented GST with long segments AB and CA and a short segment BC.

The angle angle−A (or just A) of a GST is determined as the angle in (0, 2 pi) from the direction of AB at A to the opposite direction of CA at A.  The three great-circles that contain the sides of the GST divide [[partition]] the sphere into eight elementary GST's and an orientation of a part of these connected to the sides are possible only in one way; if it shall be consistent with the original GST. 

The results are illustrated by the figure and given in table-form.

[[Links to these illustrations are provided in several formats.  GIF JPEG PNG TIFF  The content is identical; view whichever one displays better on your system.]]
The stereographic projection images illustrates GST types 1, 2 a, 3 a, 4, 5, 6 a, 7 a, 8.
Positively oriented elementary triangles are blue, negatively oriented red.
The closed triangle curve has one double-point for types #3, #7, three for #4, #8, and otherwise zero.

On the figure the colored part is considered as the inside part of the triangle with blue positive and red marking negative area density.

The first row has 4 types corresponding to some of the first 8 cases mentioned detailed in the table.

Likewise corresponds the second row to some of the remaining 8 cases in the table.

The full table may show the be practical in connection with calculations.
# index notation A B C a b c T E cross-

over's

 figure
1 (0, 0) GST+(ABC) A0 B0 C0 a0 b0 c0 T0 T0 0 r1c1
2a (1, 2) GST+(AB-C) A0 pi+B0 pi+C0 2pi-a0 b0 c0 2pi+T0 2pi+T0 1 r1c2
2b (1, 2) GST+(A-BC) pi+A0 pi+B0 C0 a0 b0 2pi-c0 2pi+T0 2pi+T0 1 "
2c (1, 2) GST+(ABC-) pi+A0 B0 pi+C0 a0 2pi-b0 c0 2pi+T0 2pi+T0 1 "
3a (2, 2) GST+(A-BC- ) A0 pi+B0 pi+C0 a0 2pi-b0 2pi-c0 T0 2pi+T0 3 r1c3
3b (2, 2) GST+(AB-C-) pi+A0 pi+B0 C0 2pi-a0 2pi-a0 c0 T0 2pi+T0 3 "
3c (2, 2) GST+(A-B-C) pi+A0 B0 pi+C0 2pi-a0 b0 2pi-c0 2pi-c0 2pi+T0 3 "
4 (3, 0) GST+(A-B-C-) A0 B0 C0 2pi-a0 2pi-b0 2pi-c0 T0-2pi T0 3 r1c4
5 (0, 3) GST-(ABC) 2pi-A0 2pi-B0 2pi-C0 a0 b0 c0 4pi-T0 4pi-T0 0 r2c1
6a (1, 1) GST-(AB-C) 2pi-A0 pi-B0 pi-C0 2pi-a0 b0 c0 2pi-T0 2pi-T0 0 r2c2
6b (1, 1) GST-(A-BC) pi-A0 pi-B0 2pi-C0 a0 b0 2pi-c0 2pi-T0 2pi-T0 0 "
6c (1, 1) GST-(ABC-) pi-A0 2pi-B0 pi-C0 a0 2pi-b0 c0 2pi-T0 2pi-T0 0 "
7a (2, 1) GST-(A-BC-) 2pi-A0 pi-B0 pi-C0 a0 2pi-b0 2pi-c0 -T0 2pi-T0 1 r2c3
7b (2, 1) GST-(AB-C-) pi-A0 pi-B0 2pi-C0 2pi-a0 2pi-b0 c0 -T0 2pi-T0 1 "
7c (2, 1) GST-(A-B-C) pi-A0 2pi-B0 pi-C0 2pi-a0 b0 2pi-c0 -T0 2pi-T0 1 "
8 (3, 3) GST-(A-B-C-) 2pi-A0 2pi-B0 2pi-B0 2pi-B0 2pi-B0 2pi-c0 2pi-T0 4pi-T0 3 r2c4

The table is establish as

  1. Index is (# sides > pi, # angles > pi).

  2. An angle is changed with pi for each adjacent side changed between long and short.

  3. A vertex angle u is changed to 2 pi - u, if the triangle is mirrored.

  4. T is calculated as sum of the signed areas of the signed elementary triangles.

  5. The spherical excess E is calculated as E = A + B + C - pi.

and imply obviously

Proposition:  The index of the sides belongs to {1, 2} iff (=if and only if) the index of the angles belongs to {1, 2}.
[[Contra positive:  The index of the sides belongs to {0,3} iff (=if and only if) the index of the angles belongs to {0,3}.
Proof:  Follows from this matrix of the #, which is obtained from the foregoing table:

  # sides > pi         crossover's
# angles > pi   0 3 1 2  
  0 1 5     0
  3 4 8     3
  1     6a 6b 6c 2a 2b 2c 0
  2     7a 7b 7c 3s 3b 3c 1

]]

Theorem.  For a GST the area is given from spherical excess by the formula T = A + B + C - pi - pi delta with delta = 2, if the amount of long sides is greater than or equal to 2, and otherwise delta = 0.

Solving GST with three elements out of six of (A, B, C, a, b, c) known is done by first calculating objects corresponding to the elements (A0, B0, C0, a0, b0, c0).

Then each side which is greater than pi is subtracted from 2pi.   The two orientation cases are treated separately.

Then this usual spherical triangle case is solved to get the elements of (A0, B0, C0, a0, b0, c0).

The side index determines the triangle type and each of the elements is calculated per the table.

Example 1:   Let A = 1.3 pi, a = 1.4 pi, b = 2.

In the positive case A0 = A - pi = 1.3 pi - pi = 0.3 pi, a0 = 2 pi - a = 2 pi - 1.4 pi = 0.6 pi, b0 = b and we find

B0 = 2.257256214, C0 = 0.1595295613, c0 = 0.1878465826, T0 = 0.2176709175.

#3c in the table is the only possibility in this positive case and

B = B0, C = pi + C0 = 3.301122215, c = 2 pi - c0 = 6.095338725, T = T0.

In the negative case A0 = 2 pi - 1.3 pi = 0.7 pi, a0 = 2 pi - a = 0.6 pi, b0 = b and we find

B0 = 2.257256214, C0 = 2.260798179, c0 = 2.006311836, T0 = 3.575576597

#6a in the table is the only possibility among #5-8.  Hence

B = pi - B0 = 0.8843364399, C = pi - C0 = 0.8807944746, c = c0, T = 2.707608711

Example 2:  Let A = 1.8 pi, B = 1.6 pi, a = 0.9 pi.

In the positive case A0 = 1.8 pi - pi = 0.8 pi, B0 = 1.6 pi - pi = 0.6 pi, a0 = a and we find 2 solutions

  1. C0 = 1.338387952, b0 = 2.617993878, c0 = 0.5370419222, T0 = 2.595025014 and by #2b
    C = C0, b = b0, c = 2 pi - c0 = 5.746143385, T = 2 pi + T0 = 8.878210321,

  2. C0 = 2.461621775, b0 = 0.5235987756, c0 = 2.804691386, T0 = 3.718258836 and by #2b
    C = C0, b = b0, c = 2 pi - c0 = 3.478493921, T = 2 pi + T0 = 10.00144414.

The negative case A0 = 2 pi - 1.8 pi = 0.2 pi, B0 = 2 pi - 1.6 pi = 0.4 pi, a0 = a has 0 solutions.

[It can be proved, if the positive/negative case has exactly two elementary solutions, then the opposite signed case is without solutions.]

This page has been transcribed and edited by RI 'Scibor-Marchocki from a PDF file sent (on Saturday 04-VII-2009) to me by Allan Cortzen, by whom copyright (c) 2009.  Gramercy.  last modified on Sunday 12-VII-2009.