Degrees To Radians To Degrees
May 5, 2015
There are 60 seconds in a minute, 60 minutes in a degree, and 180 / π = 57.2957795 degrees in a radian:
(define (degrees->radians deg min sec)
(/ (+ deg (/ min 60) (/ sec 60 60)) 57.2957795))
(define (radians->degrees rad)
(let* ((r (* rad 57.2957795))
(d (floor r))
(m (floor (* (- r d) 60)))
(s (round (* (- (* (- r d) 60) m) 60))))
(values (inexact->exact d)
(inexact->exact m)
(inexact->exact s))))
And here are some examples:
> (degrees->radians 47 6 38)
0.822234307
> (radians->degrees 0.82234307)
47
6
38
You can run the program at http://ideone.com/sc5uzT. I’ve pretty much given up responding on the forums, which makes me sad, but there is just too much vitriol for me. I give good, correct answers, and people who want to aggrandize themselves at the expense of others tell me my Python answer is no good because I used a two-space indent where the standard is four spaces, or my C answer is no good because my main function returned void instead of int, or my answer looks good but it’s in C and they only speak C++ (which at the level of the answer I gave was identical), or some other damn fool thing that certainly doesn’t help the original poster who is just trying to learn to program. So now I just read the forums from time to time as a way to find the kinds of questions that people find interesting and useful. As I said, that’s sad.
Trickier than it looks: in your solution, the use of round can result in a 60 in the seconds count for eg. 0.82234 radians (not the mention the problem of negative inputs).
Two solutions, this one seems more elegant: multiply out to the smallest quantity, round and then divide back out:
void r2dms(float r, int &d, int &m, int &s) { int n = round(r*180*60*60/pi); s = n%60; n /= 60; m = n%60; n /= 60; d = n; }or (this is probably more efficient), do as your solution, but check for seconds overflow and adjust accordingly:
void r2dms(float r, int &d, int &m, int &s) { r *= 180/pi; d = floor(r); r = (r-d)*60; m = floor(r); r = (r-m)*60; s = round(r); if (s == 60) { s = 0; m++; if (m == 60) { d = 0; d++; } } }One slight residual doubt about this is if the minutes can overflow as well: if x < 1, then is k*x < k? – clearly mathematically true, but maybe the rounding of the multiplication can make the result be exactly k. Seems to work for 60 (60*nextafter(1,0) < 60 anyway) but can't see how to prove it.