Okay, straight away we can cut out a calculation. You worked with 1 to 12 and multiplied everything by 6 to get the degrees, but if we work with degrees (0 to 359) and step 6 we get everything we want more easily.
Another thing we can do is add a condition to display hour markers vividly (conditions cost speed in loops but allows us to use just one loop) can you work out what the condition is?
if n/30 = n/30.0 then ink red,0 else ink white,0
rem 360/30 = 12 ;)
rem If the float and integer are equal, 30 is a factor of n and therefore n is at an hour position.
Notice how I used variables in the ink statements, that's a great way to save power as the colours are only calculated once at the start of the program.
I've used lines to mark the clockface instead of dots.
For n = 0 to 359 step 6
`hour check (set line start and colour)
if n/30 = n/30.
a = 320 : ink red,0
else
a = 340 : ink white,0
endif
line 512+sin(n)*a,384-cos(n)*a,512+sin(n)*350,384-cos(n)*350
Next n
TGC Forum - converting error messages into sarcasm since 2002.
