function result = ordinalDay(m,d,y)
%ordinalDay  returns the number of this day in the year
%
%  consumes: 
%          m (scalar number), the month as a number 
%                 jan = 1, feb = 2, etc.
%          d (scalar number), the day of the month
%
%          y (scalar number), the year 
%
% Note: year is needed because of leap years.
%
%    year is leap year if divisible if
%        (year is not divisible by 100 and year is divisible by 4)
%        OR
%        (year is divisble by 400)
%   
%  examples:
%         >> ordinalDay(1,1,2006)
%         ans =1 
%         >> ordinalDay(2,1,2006)
%         ans = 32
%         >> ordinalDay(3,1,2006)
%         ans = 60
%         >> ordinalDay(3,1,2004)
%         ans = 61
%         >> ordinalDay(12,31,2006)
%         ans = 365
%         >> ordinalDay(12,31,2000)
%         ans = 366
%         >> ordinalDay(12,31,1900)
%         ans = 365

  
  
% a clue.. start by defining this
  
  daysPerMonth = [31 28 31 30 31 30 31 31 30 31 30 31];


% then use d, m and y to calculate result

% since this is a "stub", for now we always return 0
  
  result = 0 ; 
  
  return;
  
end