If you want to do this with trigonometry rather than
sprite headings, here's a road map to solve this.
Say your clock center is at pixels (x0, y0)
(your origin.) You want to figure (x,y) for the end of
the minutes hand. (Same for the seconds hand.)
Assume radius R pixels.
Use Divide and Conquer for this problem.
Each quadrant is likely to be different,
so solve them separately and see if they
can be merged later.
Use the mnemonic SOHCAHTOA
Sin = Opposite over Hypotenuse
Cosine = Adjacent over Hypotenuse
Tangent = Opposite over Adjacent
(Say SOHCAHTOA to some one who has dropped
his laptop on his toe, to remember the mnemonic.)
Quad 1: (t = 0-15 seconds)
Let angle A = the angle formed between the
Y axis and the second hand.
A = (6 * t) (degrees)
x = x0 + (R * sin(A))
y = y0 - (R * cos(A))
Sanity check:
At A=0, x = x0. y = y0 - R
At A = 90, x = x0 + R, y = y0.
Try this, and see if we got lucky
for the last three quadrants.
We can add if/then tests for the
remaining quadrant t ranges (15-30, 30-45, 45-60)
later if needed.
ABG