Im trying to position a sun by passing hour and minute to a function that should return the angle for the sun zenith (altitude)
so:
at 5:24 i want the angle to be 0.0,
at 12:00 i want the angle to be 75.0 <-cant exceed that value (can be greater)
at 22:00 i want the angle to be 0.0
between 0:00 and 5:24 the angle must be negative.
I am sorta out of ideas now how should i proceed with that.
Convert the time to H*60+M, and split the problem into <= 12:00, and >= 12:00
In the morning, you have two points (5*60+24, 0), and (12*60, 75) = (324, 0) and (720, 75)
In the afternoon, you have two points (12*60, 75), and (22*60, 0) = (720, 75) and (1320, 0)
You haven't said how the angle is supposed to change between those two points.
(I only do the morning case, the afternoon case is similar)
For linear interpolation (which is the simplest, but not realistic):
A line with unknown parameters a and b:
angle = a * time + b
Fill in your points:
Fill in (324, 0): 0 = 324 * a + b [eq1]
Fill in (720, 75): 75 = 720 * a + b [eq2]
2 equations, with 2 variables, solvable!
[eq1] gives you b = -324 * a [eq3]
Substitute that in [eq2]
75 = 720 * a - 324 * a
75 = 396 * a
a = 75 / 396
Put that in [eq3]:
b = -324 * 75 / 396
thus
a = 0.1893939393939394
b = -61.36363636363637
angle = 0.1893939393939394 * time - 61.36363636363637
It's likely you don't want a linear interpolation though, since it means the angle increases at the same rate the entire morning
More realistic, you want a big increase in the morning, and a slow increase near mid-day.
This calls for a quadratic formula
angle = a * time * time + b * time + c [eq4]
[eq4] has 3 parameters a, b, and c, so you either need 3 points, or you need 2 points and 1 rate of angle. We have the latter (rate of angle at time 720 = 0).
The d(angle)/d(time) of [eq4] is
angle' = 2 * a * time + b [eq5]
Fill in your data:
0 = a * 324 * 324 + b * 324 + c (324, 0) in [eq4] gives [eq6]
75 = a * 720 * 720 + b * 720 + c (720, 75) in [eq4] gives [eq7]
0 = 2 * a * 720 + b (720, 0) in [eq5] gives [eq8]
[eq8] says b = -2 * a * 720 [eq9]
[eq7]-[eq6]: 75 - 0 = a * 720 * 720 - a * 324 * 324 + b * 720 - b * 324 + c - c
75 = a * (720*720 - 324*324) + b * (720 - 324) [eq10]
Substitute [eq9] into [eq10]:
75 = a * (720*720 - 324*324) - 2 * a * 720 * (720 - 324)
75 = a * (720*720 - 324*324 - 2 * 720 * (720 - 324))
a = 75 / (720*720 - 324*324 - 2 * 720 * (720 - 324))
a = -75 / 156816 = -0.0004782675237220692
apply in [eq9] gives you b = 0.6887052341597796
apply in [eq6] gives you c = -(a * 324 * 324 + b * 324) = -172.93388429752065
So it's angle = -0.0004782675237220692 * time * time + 0.6887052341597796 * time -172.93388429752065
[attachment=29927:sun_rise.png]
btw i came with that just few mins ago and wanted to post it;p
inline float TimeToAngle(int hr, int min)
{
float time = hr + (min/60.0);
float ttime[9] = {-1000.0, 0.0, 5.4, 5.65, 12.0, 22.0, 22.25, 24.0, 28.0};
float tangle[9] = {-90.0, -10.0, -1.0, 8.0, 80.0, 8.0, -1.0, -10.0, -90.0};
float angle = -10.0;
for (i=0; i<9; i++)
if( (ttime[i] <= time) && (ttime[i+1] > time) )
angle = tangle[i] - (ttime[i] - time) * (tangle[i] - tangle[i+1]) / (ttime[i] - ttime[i+1]);
return angle;
}it simply lerps between given times (i convert minutes to fraction of the number)
thanks for your reply, sorry to waste your time
Splitting the graph into several linear pieces works too, you may want to add some more points around noon to simulate slower increase/decrease of the angle around that time.
My time wasn't exactly wasted. I had some exercises in solving sets of linear equations, which I hadn't done in ages 
It also gave me an idea how to do these kind of conversions in a more generic way (but that needs more work to straighten out all the things first).
In the end, all that matters is that you find a solution that works for you. In this case, you even found it yourself, which is the best message you can have with this kind of questions. I have no problem playing the teddy bear role in such a case
dont care about green water already fixed that...
its not that easy as you think it is, the skydome has dimension of earth radius + 24 kilometers, i am acutally positioning camera at north pole, and i need to position the sun somehow so i had to lerp between some angles (in this case up donw angle)
