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1 Spherical Triangle
1.a Area and Excess
In this part all you have to do is to transfer the given mathematical formulas
into assignments using arithmetic operators in MATLAB. When you run the
script file, test triangle.m, it will ask you to enter the values for three angles
and the radius, then it will compute the area and excess using your formulas,
and display the results.
function [Area, E] = spherical triangle(A,B,C,R) % Returns the area of a spherical triangle with % spherical angles A,B,C
E = A + B + C − pi; Area = Rˆ2 (^) * E;
1.b Special Triangle
When the corner C is at North pole, and A & B are on the equator; the
longtitudes of A and B, LA & LB, can be used to compute the value of the
angle at C. The convention we followed was to consider the angle seen by the
arc starting from B, going in the west direction to reach A. Depending on the
values of LA and LB we have to branch our program.
LA = input('Enter the longtitude value for A in degrees: '); LB = input('Enter the longtitude value for B in degrees: ');
if LA < LB C = (LB − LA)pi/180; else C = (360 + LB − LA)pi/180; end
A = pi/2; % longtitudes are perpendicular to the equator B = pi/2;
R = input('Enter the value for the radius: ');
[Area, E] = spherical triangle(A,B,C,R);
disp('The area of the special triangle is '); disp(Area); disp('The excess value is '); disp(E);
1.c Testing Our Code
We can test our code using some special triangles we can recognize, like a hemi-
sphere. You could have a script asking for user input or a fixture of test cases.
The file special triangle.m already contains the hemisphere as a test case in a
separate code section. In order to make use of the function spherical triangle
you could write a script asking for values from the user:
A = input('Enter the angle A: '); B = input('Enter the angle B: '); C = input('Enter the angle C: '); R = input('Enter the radius R: ');
[Area, E] = spherical triangle(A,B,C,R);
fprintf('The area of the triangle is %f\n', Area); fprintf('The spherical excess is %f\n', E);
2 Distance
2.a Two Points
function dAB = distance(xA,xB,yA,yB,zA,zB) % Returns the distance between points A and B which are specified % using their cartesian coordinates in 3dimensions.
dAB = sqrt((xA−xB)ˆ2+(yA−yB)ˆ2+(zA−zB)ˆ2);
2.b Triangular Area
function [Area,dAB,dBC,dCA] = triangle area(xA,xB,xC,yA,yB,yC,zA,zB,zC) % Returns the area and side lengths of a triangle. The input consists % of the cartesian coordinates of vertices in 3−dimensions.
dAB = distance(xA,xB,yA,yB,zA,zB); dBC = distance(xB,xC,yB,yC,zB,zC); dCA = distance(xC,xA,yC,yA,zC,zA);
s = (dAB + dBC + dCA) / 2; % semi−perimeter
Area = sqrt(s(s−dAB)(s−dBC)*(s−dCA)); % dont forget (^) *
5 Soccer Ball Toss
5.a Coin Toss
We can simulate a coin toss by generating a random number using rand function
and then comparing it with 0.5. If heads is represented by the first half of the
unit interval, the following function satisfy the requirements:
function x = coin toss() x = rand > 0.5;
5.b Ball Toss
The probabilities of black and white are proportional to the areas covered by
pentagons and hexagons, respectively, on the spherical surface. The computation
of the exact ratio is a little complicated, but you can use an approximation by
considering a truncated icosahedron.
Let the side length of the hexagons and pentagons be a. Then the total area
covered by the white regions can be computed as follows:
AW = 20 ×
3 a^2
Similarly, the area covered by the pentagons is:
AB = 12 ×
a^2
We can compute the probabilities for black and white as:
PB =
AB
AB + AW
PW =
AW
AB + AW
We can modify the coin toss function to mimic a ball toss as:
function x = ball toss() x = rand > 0.2843;
A more accurate result can be computed using the spherical triangle area equa-
tion and spherical trigonometric identities, which gives the critical value around:
PB ≈ 0 .28167, PW ≈ 0 .71833.
