Circle Degree Algorythm

Comments

• slicer69

  April 2002

  Try using cos and sin functions,
  most languages have them
  
  
  
• bao

  June 2002

  the formula is
  360*(1 - cos'(h/w)/pi)
  where
  cos'() is inverse cos()
  h = |y2-y1|
  w = |x2-x1|
  pi = 3.1415
  
  if you're after speed or your software does not support trig functions you can easily formulate a polynomial to approximate for quandrant 1 using series and mirror|translate for all other angles.
  
• Clearer

  June 2002

  : the formula is
  : 360*(1 - cos'(h/w)/pi)
  : where
  : cos'() is inverse cos()
  : h = |y2-y1|
  : w = |x2-x1|
  : pi = 3.1415
  :
  : if you're after speed or your software does not support trig functions you can easily formulate a polynomial to approximate for quandrant 1 using series and mirror|translate for all other angles.
  :
  
  If you aren't satified with this or you can't use the trig functions for some reason I will supply you with them (the real onece)in a few days when I get home (I'm at my sisters place.. no notes filling my desk). They are not easy to understand mind you and they certainly are not easy to use either.
  
• dshields

  August 2002

  I need a generic algorithm for computing the average (mean) of two degree parameters. If I pass 359.0 and 1.0, I need it to return 0.0. The two parameters can be any degree values between 0 and 360. Any suggestions anyone? Thanks.
  David
  
  
• Clearer

  August 2002

  : I need a generic algorithm for computing the average (mean) of two degree parameters. If I pass 359.0 and 1.0, I need it to return 0.0. The two parameters can be any degree values between 0 and 360. Any suggestions anyone? Thanks.
  : David
  :
  :
  
  let me see if I got this right:
  you got degree a and degree b and you need to find (a+b)/2?
  it's quite simple if that's what you need. If not I'm not understanding this correctly
  
• kjnorris

  August 2002

  : I need a generic algorithm for computing the average (mean) of two degree parameters. If I pass 359.0 and 1.0, I need it to return 0.0. The two parameters can be any degree values between 0 and 360. Any suggestions anyone? Thanks.
  : David
  :
  :
  Good question Dave, easy answer
  
  ((argument1 + argument2) % 360) / 2
  
  the % is for modulus.
  
• kjnorris

  August 2002

  : : I need a generic algorithm for computing the average (mean) of two degree parameters. If I pass 359.0 and 1.0, I need it to return 0.0. The two parameters can be any degree values between 0 and 360. Any suggestions anyone? Thanks.
  : : David
  : :
  : :
  : Good question Dave, easy answer
  :
  : ((argument1 + argument2) % 360) / 2
  :
  : the % is for modulus.
  :
  
  sorry, that is not right answer, let me check some notes when i get home tonight, i solved this before.
  
• kjnorris

  August 2002

  : : : I need a generic algorithm for computing the average (mean) of two degree parameters. If I pass 359.0 and 1.0, I need it to return 0.0. The two parameters can be any degree values between 0 and 360. Any suggestions anyone? Thanks.
  : : : David
  : : :
  : : :
  : : Good question Dave, easy answer
  : :
  : : ((argument1 + argument2) % 360) / 2
  : :
  : : the % is for modulus.
  : :
  :
  : sorry, that is not right answer, let me check some notes when i get home tonight, i solved this before.
  :
  
  ~~~~~~~~~~~~~~
  
  One way to solve problem is break into two parts.
  Trouble comes when the numerical difference between arguments is greater than 180. Also, an exact difference of 180 can go either direction.
  
  INPUT ARG1, ARG2
  IF ABS(ARG1 - ARG2) <= 180
  ANSWER = (ARG1 + ARG2) / 2
  ELSE
  ANSWER = ???
  ENDIF
  
  The else clause is harder to work out. But the IFTHENELSE construct will at least identify the 'situation' where special processing is needed.
  
  Hope that helps.
  
  I haven't taken any trigonometry, but i'll bet there is a formula which can work in all situtations.