Problem Statement

When calculating the average or mean of an angle one has to take into account how angles wrap around so that any angle in degrees plus any integer multiple of 360 degrees is a measure of the same angle. If one wanted an average direction of the wind over two readings where the first reading was of 350 degrees and the second was of 10 degrees then the average of the numbers is 180 degrees, whereas if you can note that 350 degrees is equivalent to -10 degrees and so you have two readings at 10 degrees either side of zero degrees leading to a more fitting mean angle of zero degrees. To calculate the mean angle of several angles:

(Note that, since the mean is the sum divided by the number of numbers, and division by a positive real number does not affect the angle, you can also simply compute the sum for step 2.) You can alternatively use this formula:

Let's start with the solution:

#!/usr/bin/a68g --script #
# -*- coding: utf-8 -*- #

PROC mean angle = ([]#LONG# REAL angles)#LONG# REAL:
(
  INT size = UPB angles - LWB angles + 1;
  #LONG# REAL y part := 0, x part := 0;
  FOR i FROM LWB angles TO UPB angles DO
      x part +:= #long# cos (angles[i] * #long# pi / 180);
      y part +:= #long# sin (angles[i] * #long# pi / 180)
  OD;
 
  #long# arc tan2 (y part / size, x part / size) * 180 / #long# pi
);
 
main:
(
  []#LONG# REAL angle set 1 = ( 350, 10 );
  []#LONG# REAL angle set 2 = ( 90, 180, 270, 360);
  []#LONG# REAL angle set 3 = ( 10, 20, 30);
 
  FORMAT summary fmt=$"Mean angle for "g" set :"-zd.ddddd" degrees"l$;
  printf ((summary fmt,"1st", mean angle (angle set 1)));
  printf ((summary fmt,"2nd", mean angle (angle set 2)));
  printf ((summary fmt,"3rd", mean angle (angle set 3)))
)