Because that's the way you programmed it.

To find the hue that the user picks is simple :
1. If the user clicks the mouse then :
1a - Find the distance from the mouse to the center of the circle. If the distance is greater than or equal to the inner radius of the hue wheel and less than or equal to the outer radius of the hue wheel, they've chosen a new hue.
1b - Find the angle from the mouse to the center of the circle. Using degrees, this translates directly to the value for your hue. You will also have to store the current rotation of the hue wheel and adjust the value by it to find the correct new value.

I made an example program that shows the rotating hue wheel and the saturation/value triangle in the middle of it. LMB click on the hue wheel to set the hue, and LMB drag on the hue wheel to turn it (also setting the hue). I'll add in clicking on the triangle tomorrow, but hopefully you'll see what I mean now.
Source + dynamic win32 exe (needs alleg42.dll)

http://www.allegro.cc/files/attachment/596770

To get rid of the jaggedness of the hue wheel when it's turned with rotate_sprite, I was going to rotate it into a buffer first, draw a magic pink circle over the inner and outer edges and then use draw_sprite with the buffer. That way the edges would never move at least. I'll try it later.

hi edgardo. yes its impressively smooth !

and sure i would never have arrived to this...

is it possible to know what are those few lines in your code ( please )?

    mx = mpos >> 16;
    my = mpos & 0x0000ffff;

and could you re-explain to me the importance of ftofix ?

yours

Read over this -

mouse_pos

So mouse_pos is a 4 byte integer. The x and y positions of the mouse are stored inside it, with each given two bytes to store their value. So it's layout is like this :

XXYY

The bit shift operator (>>) shifts the bit values inside a variable so when you say

int mouse_xpos = mouse_pos >> 16;

Then you have the top two bytes of mouse_pos stored inside the mouse_xpos variable. (Because the bits in mouse_pos are shifted to the right 16 places (bits), which is two bytes)

When using the bitwise AND operator (&), it will only return a 1 bit if the corresponding bits of both sides of the expression are 1. 0x0000ffff has all the bits in the lower two bytes set to 1 so when you bitwise AND a number with 0x0000ffff, you are guaranteed to have only the values that were in the lower two bytes of that number. This is how you get the y value out of mouse_pos.

int mouse_ypos = mouse_pos & 0x0000ffff;

float to fixed
Since the last parameter of rotate_sprite takes an allegro fixed data type, you need to convert the float to a fixed type for the function. Remember that the range for allegro's fixed type is [0.0 , 256.0).

If you have any other questions about what I did or why, just ask.

I won't be able to get to it until tomorrow. I've just started learning about dot products and cross products myself, so I want to make sure my math is correct first.

If you've already got the test for checking whether a point is inside the triangle completed, you can just use getpixel, then getr, getg, and getb to get the rgb value of the pixel underneath the mouse click, then use rgb_to_hsv to get the saturation and value of the color picked.

When finding the cross product, how do you pick the unit vector perpendicular to both vectors though? And which one do you pick? Is it always the positive one?

thanks edgardo !

links are somehow frightening ! (brrrrrr) terrific !
i suppose you will perhaps make a color picker for your pattern generator to edit ranges of colors?
i m very happy if i can see the result !

by the way in general, how do you translate in code this vector in xy maths ?

for example the a.(b+c) = a.b + a.c ?

cheers

I may include it, I'm not sure yet though as it has to be fairly large pixel wise to get precise values without using sliders to adjust the values.

Sorry it took so long to reply, but I had some really annoying bugs I had a hard time finding. Incidentally, between a header and a source file, you can use different names for the same function parameters, and MinGW 3.4.5 won't bother to warn you. That was only part of it though.

I used vectors and scalar projection to find the saturation and color value. It turns out I didn't need dot products or cross products to do any of it.

To see what scalar projection is, take a look at the graphic on the right at http://en.wikipedia.org/wiki/Dot_Product#Geometric_interpretation.

Using the convenient naming convention Johan used (H is the vertice with full hue value, B is black, W is white, M is the midpoint of H and W, and P is the mouse position), here are the steps I use to find the saturation and value for the mouse location.
1. Find the vector from B to P (Px - Bx , Py - By) (BPx,BPy)
2. Find the vector from W to P (Px - Wx , Py - Wy) (WPx,WPy)
3. Find the distance from B to P (distBP)
4. Find the distance from W to P (distWP)
5. If distBP is 0.0, then value is 0.0 and saturation is undefined (leave unchanged)
6. If distWP is 0.0 then value is 1.0 and saturation is 0.0
7. Find the angle of vector BP ( atan2(BPy,BPx) )
8. Find the angle of vector WP ( atan2(WPy,WPx) )
9. Find the angle from vector BM to vector BP (angleBP - angleBM)
10. Find the angle from vector WH to vector WP (angleWP - angle WH)
11. Find the scalar projection of BP onto BM ( distBP*cos(angle_BM_to_BP) )
12. Find the scalar projection of WP onto WH ( distWP*cos(angle_WH_to_WP) )
13. The color value is equivalent to the scalar projection of BP onto BM divided by the distance from B to M :
value = SPBPBM/distBM
14. If the value is not zero then the saturation is equal to this :
saturation = (SPWPWH + (value - 1.0)*distMW)/(value*distWH);

Finding the position of the point on the triangle from the saturation and color value is much nicer though :
Using vectors :

Finding the vector from B to P on BW is easy :

The vector from P on BW to P isn't so bad though :

Translated directly to coordinates instead of vectors, the formulas are :

These formulas will work as long as the triangle is isosceles and the two equal sides correspond to the HB and BW sides of the saturation value triangle.

I also added an algorithm to find the closest edge of the triangle when the mouse is outside it. Basically, for each vertice, find the angle from the vector of the current vertice to the next vertice to the vector from the current vertice to the mouse position. If the angle indicates the mouse is outside the triangle, find the scalar projection of the mouse vector onto the edge vector. If the scalar projection is negative, it corresponds to a corner of the triangle, else if the scalar projection is less than the length of that edge, then divide it by the length of the edge and you will have a ratio you can use to assign a value and saturation for that position. If the scalar projection is larger than or equal to the edge length then it corresponds to the next corner of the triangle.

I finished an example program that demonstrates the technique visually with on screen vectors indicating scalar projections and finding the closest edge point, along with showing the vectors that produce the mouse position from the saturation and value.
Key s (scalar) - toggle visibility of scalar projections onto vector BM and vector WH
Key c (closest) - toggle visibility of scalar projections onto the triangle edges for finding the closest point from outside the triangle
Key p (position) - toggle visibility of position vectors for the current saturation and value
LMB click on the hue wheel or the saturation value triangle to set a new value.
LMB click and drag on them to drag the wheel or track the closest point on the triangle.
ESC quits

Images of program with vectors visible :
http://www.allegro.cc/files/attachment/596829 http://www.allegro.cc/files/attachment/596830 http://www.allegro.cc/files/attachment/596831

Zip file with win32 exe + source code (C++) (needs alleg42.dll)

I may clean it up some and add a new constructor and/or some resize functions, but feel free to use the HueWheelRR class in your (anyone's) project (with due credit). That will have to wait until I finish my year as overseer of Unolrigoth though, which I have been delaying to finish this HSV wheel/triangle.

Quick question, what's "(C - W)/(H - W)"?

Is that the length of vector WC divided by the length of the vector WH? I don't recognize your notation. It looks interesting, and if there's a simpler way to do it, it'd be nice to know.

Edit
Using the 's' option in my program, it's easy to see what you're doing, but I'm not sure how you are determining which direction WC lies along WH and which direction BP lies along BC.

C is a point between W and H, right. Think of W as zero and H as one. C is somewhere inbetween, right? C is saturation, right? Well, to get out the actual values, you have to measure the length from W to C and the length from W to H. But if you use Pythagoras, you only get positive results. It's quite allright to only measure the x coordinates of the points. Or only the y coordinates. Because the points are on the same line. You don't want to use the ones that lead to dividing with zero, like if H.x == W.x.

So after I got the coordinates of H, W and C, I go:

if (abs(H.x - W.x) > abs(H.y - W.y))
    saturation = (C.x - W.x)/(H.x - W.x);
else
    saturation = (C.y - W.y)/(H.y - W.y);

This way, if I click outside the triangle, outside the BW line, I get negative saturation values. And clicking outside the BH line, i get saturation values over 1.

Again, using the same approach, I calculate the value comparing the distance BP with the distance BC. If it goes over 1, I'm outside the triangle at the HW side. Negative value means I clicked outside the B vertex.

[edit]
My method makes it useless to calculate with the vectors to find out if the mouse was clicked outside the triangle. OTOH if you just set saturation to 0 each time it went < 0 and to 1 each time it went > 1, the picker won't be set at the most intuitive point on the edge, but at a line starting at the mouse position and parallel to the HW edge.