I need an algorithm to figure out if one angle is within a certain amount of degrees from another angle.
My first thought was (a-x < b) && (a+x > b), but it fails when it has to work with angles that wrap around from -179 to 180.
In the diagram above, the region (green) that the angle must be between wraps between the negative and positive sides. How can I determine whether the angle (the red line) falls inside this region?
try this formula:
360-(|a-b|)%360<x || (|a-b|)%360<x
Or, in PHP:
<?php
$b = 10;
$angle1 = -179;
$angle2 = 180;
$diff = $angle1 - $angle2;
if(abs($diff % 360) <= $b || (360-abs($diff % 360))<=$b) {
echo "yes";
} else {
echo "no";
}
?>
As Marcel rightly points out, modulo on negative numbers is potentially problematic. Also, what is the difference between 355 and 5 degrees? It might be worked out to be 350 degrees but 10 degrees is probably what people are expecting. We make the following assumptions:
we want the smallest positive angle between two other angles so 0 <= diff <= 180;
we are working in degrees. If radians, substitute 360 for 2*PI;
angles can be positive or negative can be outside the range -360 < x < 360 where x is an input angle and
order of input angles or the direction of the difference is irrelevant.
Inputs: angles a and b. So the algorithm is simply:
Normalize a and b to 0 <= x < 360;
Compute the shortest angle between the two normal angles.
For the first step, to convert the angle to the desired range, there are two possibilities:
x >= 0: normal = x % 360
x < 0: normal = (-x / 360 + 1) * 360 + x
The second is designed to remove any ambiguity on the difference in interpretation of negative modulus operations. So to give a worked example for x = -400:
-x / 360 + 1
= -(-400) / 360 + 1
= 400 / 360 + 1
= 1 + 1
= 2
then
normal = 2 * 360 + (-400)
= 320
so for inputs 10 and -400 the normal angles are 10 and 320.
Now we calculate the shortest angle between them. As a sanity check, the sum of those two angles must be 360. In this case the possibilities are 50 and 310 (draw it and you'll see this). To work these out:
normal1 = min(normal(a), normal(b))
normal2 = max(normal(a), normal(b))
angle1 = normal2 - normal1
angle2 = 360 + normal1 - normal2
So for our example:
normal1 = min(320, 10) = 10
normal2 = max(320, 10) = 320
angle1 = normal2 - normal1 = 320 - 10 = 310
angle2 = 360 + normal1 - normal2 = 360 + 10 - 320 = 50
You'll note normal1 + normal2 = 360 (and you can even prove this will be the case if you like).
Lastly:
diff = min(normal1, normal2)
or 50 in our case.
You can also use a dot product:
cos(a)*cos(b) + sin(a)*sin(b) >= cos(x)
For a radius of 1, the distance between the line endpoints is 2sin((a-b/2). So throw away the 2 since you are only interested in a comparison, and compare sin(x/2) with sin((a-b)/2). The trig functions take care of all the wrapping.
c++ implementation:
float diff = fabsf(angle1 - angle2);
bool isInRange = fmodf(diff, 360.0f) <= ANGLE_RANGE ||
360.0f - fmodf(diff, 360.0f) <= ANGLE_RANGE;
Related
I have a general 3D rotation matrix like this:
|R11 R12 R13|
|R21 R22 R23|
|R31 R32 R33|
From this rotation matrix, I know I can get the yaw Tait-Bryan Euler angle with this:
yaw = atan2(R21, R11)
In my system I have:
R21 = Asin(r) + Bcos(r)
R11 = Ccos(r) + Dsin(r)
with:
r = RZ - 90deg
A = -0.01775
B = 0.9997
C = -0.01714
D = -0.99924
leading to:
yaw = atan2(-0.01775sin(RZ - 90) + 0.9997cos(RZ - 90) , -0.01714cos(RZ - 90) - 0.99924sin(RZ - 90))
Ex.: if RZ = -136.01355deg, I get yaw = -135.0deg which is what I expect given my system.
Now what I really want to do is the opposite: find RZ given a known yaw value. But for some reason this does not always work. I sometimes am +/- 180deg off. Here is how I re-arranged the equation:
atan2(Asin(r) + Bcos(r), Ccos(r) + Dsin(r)) = Y
(Asin(r) + Bcos(r)) / (Ccos(r) + Dsin(r)) = tan(Y)
Asin(r) + Bcos(r) = tan(Y)Ccos(r) + tan(Y)Dsin(r)
(A - tan(Y)D)sin(r) + (B - tan(Y)C)cos(r) = 0
(A - tan(Y)D)tan(r) + (B - tan(Y)C) = 0
(A - tan(Y)D)tan(r) = tan(Y)C - B
tan(r) = (tan(Y)C - B) / (A - tan(Y)D)
r = atan2(tan(Y)C - B, A - tan(Y)D)
With my system's values:
RZ = atan2(-0.01714tan(yaw) - 0.9997 , 0.99924tan(yaw) - 0.01775) + 90deg
So for yaw = -135deg, I get RZ = 43.98645 which is exactly 180 more than the expected value of -136.01355 (ie it looks like I need to subtract 180 from my result).
It seems that for yaw angles between -90 to 90deg, RZ is good; for yaw < 90deg, I need to do RZ - 180deg; and for yaw > 90deg, I need to do RZ + 180deg.
Does this make sense ? Is my math ok ? Is this solution good and what is the explanation for it ? Is it somehow related to tan(yaw) ? I would definitely prefer a solution that always yields the correct result without the extra step !! Thanks for any insights !!
Bresenham's line drawing algorithm is well known and quite simple to implement.
While there are more advanced ways to draw anti-ailesed lines, Im interested in writing a function which draws a single pixel width non anti-aliased line, based on floating point coordinates.
This means while the first and last pixels will remain the same, the pixels drawn between them will have a bias based on the sub-pixel position of both end-points.
In principle this shouldn't be all that complicated, since I assume its possible to use the sub-pixel offsets to calculate an initial error value to use when plotting the line, and all other parts of the algorithm remain the same.
No sub pixel offset:
X###
###X
Assuming the right hand point has a sub-pixel position close to the top, the line could look like this:
With sub pixel offset for example:
X######
X
Is there a tried & true method of drawing a line that takes sub-pixel coordinates into account?
Note:
This seems like a common operation, I've seen OpenGL drivers take this into account for example - using GL_LINE, though from a quick search I didn't find any answers online - maybe used wrong search terms?
At a glance this question looks like it might be a duplicate of: Precise subpixel line drawing algorithm (rasterization algorithm)However that is asking about drawing a wide line, this is asking about offsetting a single pixel line.
If there isn't some standard method, I'll try write this up to post as an answer.
Having just encountered the same challenge, I can confirm that this is possible as you expected.
First, return to the simplest form of the algorithm: (ignore the fractions; they'll disappear later)
x = x0
y = y0
dx = x1 - x0
dy = y1 - y0
error = -0.5
while x < x1:
if error > 0:
y += 1
error -= 1
paint(x, y)
x += 1
error += dy/dx
This means that for integer coordinates, we start half a pixel above the pixel boundary (error = -0.5), and for each pixel we advance in x, we increase the ideal y coordinate (and therefore the current error) by dy/dx.
First let's see what happens if we stop forcing x0, y0, x1 and y1 to be integers: (this will also assume that instead of using pixel centres, the coordinates are relative to the bottom-left of each pixel1, since once you support sub-pixel positions you can simply add half the pixel width to the x and y to return to pixel-centred logic)
x = x0
y = y0
dx = x1 - x0
dy = y1 - y0
error = (0.5 - (x0 % 1)) * dy/dx + (y0 % 1) - 1
while x < x1:
if error > 0:
y += 1
error -= 1
paint(x, y)
x += 1
error += dy/dx
The only change was the initial error calculation. The new value comes from simple trig to calculate the y coordinate when x is at the pixel centre. It's worth noting that you can use the same idea to clip the line's start position to be within some bound, which is another challenge you'll likely face when you want to start optimising things.
Now we just need to convert this into integer-only arithmetic. We'll need some fixed multiplier for the fractional inputs (scale), and the divisions can be handled by multiplying them out, just as the standard algorithm does.
# assumes x0, y0, x1 and y1 are pre-multiplied by scale
x = x0
y = y0
dx = x1 - x0
dy = y1 - y0
error = (scale - 2 * (x0 % scale)) * dy + 2 * (y0 % scale) * dx - 2 * dx * scale
while x < x1:
if error > 0:
y += scale
error -= 2 * dx * scale
paint(x / scale, y / scale)
x += scale
error += 2 * dy * scale
Note that x, y, dx and dy keep the same scaling factor as the input variables (scale), whereas error has a more complex scaling factor: 2 * dx * scale. This allows it to absorb the division and fraction in its original formulation, but means we need to apply the same scale everywhere we use it.
Obviously there's a lot of room to optimise here, but that's the basic algorithm. If we assume scale is a power-of-two (2^n), we can start to make things a little more efficient:
dx = x1 - x0
dy = y1 - y0
mask = (1 << n) - 1
error = (2 * (y0 & mask) - (2 << n)) * dx - (2 * (x0 & mask) - (1 << n)) * dy
x = x0 >> n
y = y0 >> n
while x < (x1 >> n):
if error > 0:
y += 1
error -= 2 * dx << n
paint(x, y)
x += 1
error += 2 * dy << n
As with the original, this only works in the (x >= y, x > 0, y >= 0) octant. The usual rules apply for extending it to all cases, but note that there are a few extra gotchyas due to the coordinates no-longer being centred in the pixel (i.e. reflections become more complex).
You'll also need to watch out for integer overflows: error has twice the precision of the input variables, and a range of up to twice the length of the line. Plan your inputs, precision, and variable types accordingly!
1: Coordinates are relative to the corner which is closest to 0,0. For an OpenGL-style coordinate system that's the bottom left, but it could be the top-left depending on your particular scenario.
I had a similar problem, with the addition of needing sub-pixel endpoints, I also needed to make sure all pixels which intersect the line are drawn.
I'm not sure that my solution will be helpful to OP, both because its been 4+ years, and because of the sentence "This means while the first and last pixels will remain the same..." For me, that is actually a problem (More on that later). Hopefully this may be helpful to others.
I don't know if this can be considered to be Bresenham's algorithm, but it is awful similar. I'll explain it for the (+,+) quadrant. Lets say you wish to draw a line from point (Px,Py) to (Qx,Qy) over a grid of pixels with width W. Having a grid width W > 1 allows for sub-pixel endpoints.
For a line going in the (+,+) quadrant, the starting point is easy to calculate, just take the floor of (Px,Py). As you will see later, this only works if Qx >= Px & Qy >= Py.
Now you need to find which pixel to go to next. There are 3 possibilities: (x+1,y), (x,y+1), & (x+1,y+1). To make this decision, I use the 2D cross product defined as:
If this value is negative, vector b is right/clockwise of vector a.
If this value is positive, vector b is left/anti-clockwise of vector a.
If this value is zero vector b points in the same direction as vector a.
To make the decision on which pixel is next, compare the cross product between the line P-Q [red in image below] and a line between the point P and the top-right pixel (x+1,y+1) [blue in image below].
The vector between P & the top-right pixel can be calculated as:
So, we will use the value from the 2D cross product:
If this value is negative, the next pixel will be (x,y+1).
If this value is positive, the next pixel will be (x+1,y).
If this value is exactly zero, the next pixel will be (x+1,y+1).
That works fine for the starting pixel, but the rest of the pixels will not have a point that lies inside them. Luckily, after the initial point, you don't need a point to be inside the pixel for the blue vector. You can keep extending it like so:
The blue vector starts at the starting point of the line, and is updated to the (x+1,y+1) for every pixel. The rule for which pixel to take is the same. As you can see, the red vector is right of the blue vector. So, the next pixel will be the one right of the green pixel.
The value for the cross product needs updated for every pixel, depending on which pixel you took.
Add dx if the next pixel was (x+1), add dy if the pixel was (y+1). Add both if the pixel went to (x+1,y+1).
This process is repeated until it reaches the ending pixel, (Qx / W, Qy / W).
All combined this leads to the following code:
int dx = x2 - x2;
int dy = y2 - y1;
int local_x = x1 % width;
int local_y = y1 % width;
int cross_product = dx*(width-local_y) - dy*(width-local_x);
int dx_cross = -dy*width;
int dy_cross = dx*width;
int x = x1 / width;
int y = y1 / width;
int end_x = x2 / width;
int end_y = y2 / width;
while (x != end_x || y != end_y) {
SetPixel(x,y,color);
int old_cross = cross_product;
if (old_cross >= 0) {
x++;
cross_product += dx_cross;
}
if (old_cross <= 0) {
y++;
cross_product += dy_cross;
}
}
Making it work for all quadrants is a matter of reversing the local coordinates and some absolute values. Heres the code which works for all quadrants:
int dx = x2 - x1;
int dy = y2 - y1;
int dx_x = (dx >= 0) ? 1 : -1;
int dy_y = (dy >= 0) ? 1 : -1;
int local_x = x1 % square_width;
int local_y = y1 % square_width;
int x_dist = (dx >= 0) ? (square_width - local_x) : (local_x);
int y_dist = (dy >= 0) ? (square_width - local_y) : (local_y);
int cross_product = abs(dx) * abs(y_dist) - abs(dy) * abs(x_dist);
dx_cross = -abs(dy) * square_width;
dy_cross = abs(dx) * square_width;
int x = x1 / square_width;
int y = y1 / square_width;
int end_x = x2 / square_width;
int end_y = y2 / square_width;
while (x != end_x || y != end_y) {
SetPixel(x,y,color);
int old_cross = cross_product;
if (old_cross >= 0) {
x += dx_x;
cross_product += dx_cross;
}
if (old_cross <= 0) {
y += dy_y;
cross_product += dy_cross;
}
}
However there is a problem! This code will not stop in some cases. To understand why, you need to really look into exactly what conditions count as the intersection between a line and a pixel.
When exactly is a pixel drawn?
I said I need to make that all pixels which intersect a line need to be drawn. But there's some ambiguity in the edge cases.
Here is a list of all possible intersections in which a pixel will be drawn for a line where Qx >= Px & Qy >= Py:
A - If a line intersects the pixel completely, the pixel will be drawn.
B - If a vertical line intersects the pixel completely, the pixel will be drawn.
C - If a horizontal line intersects the pixel completely, the pixel will be drawn.
D - If a vertical line perfectly touches the left of the pixel, the pixel will be drawn.
E - If a horizontal line perfectly touches the bottom of the pixel, the pixel will be drawn.
F - If a line endpoint starts inside of a pixel going (+,+), the pixel will be drawn.
G - If a line endpoint starts exactly on the left side of a pixel going (+,+), the pixel will be drawn.
H - If a line endpoint starts exactly on the bottom side of a pixel going (+,+), the pixel will be drawn.
I - If a line endpoint starts exactly on the bottom left corner of a pixel going (+,+), the pixel will be drawn.
And here are some pixels which do NOT intersect the line:
A' - If a line obviously doesn't intersect a pixel, the pixel will NOT be drawn.
B' - If a vertical line obviously doesn't intersect a pixel, the pixel will NOT be drawn.
C' - If a horizontal line obviously doesn't intersect a pixel, the pixel will NOT be drawn.
D' - If a vertical line exactly touches the right side of a pixel, the pixel will NOT be drawn.
E' - If a horizontal line exactly touches the top side of a pixel, the pixel will NOT be drawn.
F' - If a line endpoint starts exactly on the top right corner of a pixel going in the (+,+) direction, the pixel will NOT be drawn.
G' - If a line endpoint starts exactly on the top side of a pixel going in the (+,+) direction, the pixel will NOT be drawn.
H' - If a line endpoint starts exactly on the right side of a pixel going in the (+,+) direction, the pixel will NOT be drawn.
I' - If a line exactly touches a corner of the pixel, the pixel will NOT be drawn. This applies to all corners.
Those rules apply as you would expect (just flip the image) for the other quadrants. The problem I need to highlight is when an endpoint lies exactly on the edge of a pixel. Take a look at this case:
This is like image G' above, except the y-axis is flipped because the Qy < Py. There are 4x4 red dots because W is 4, making the pixel dimensions 4x4. Each of the 4 dots are the ONLY endpoints a line can touch. The line drawn goes from (1.25, 1.0) to (somewhere).
This shows why it's incorrect (at least how I defined pixel-line intersections) to say the pixel endpoints can be calculated as the floor of the line endpoints. The floored pixel coordinate for that endpoint seems to be (1,1), but it is clear that the line never really intersects that pixel. It just touches it, so I don't want to draw it.
Instead of flooring the line endpoints, you need to floor the minimal endpoints, and ceil the maximal endpoints minus 1 across both x & y dimensions.
So finally here is the complete code which does this flooring/ceiling:
int dx = x2 - x1;
int dy = y2 - y1;
int dx_x = (dx >= 0) ? 1 : -1;
int dy_y = (dy >= 0) ? 1 : -1;
int local_x = x1 % square_width;
int local_y = y1 % square_width;
int x_dist = (dx >= 0) ? (square_width - local_x) : (local_x);
int y_dist = (dy >= 0) ? (square_width - local_y) : (local_y);
int cross_product = abs(dx) * abs(y_dist) - abs(dy) * abs(x_dist);
dx_cross = -abs(dy) * square_width;
dy_cross = abs(dx) * square_width;
int x = x1 / square_width;
int y = y1 / square_width;
int end_x = x2 / square_width;
int end_y = y2 / square_width;
// Perform ceiling/flooring of the pixel endpoints
if (dy < 0)
{
if ((y1 % square_width) == 0)
{
y--;
cross_product += dy_cross;
}
}
else if (dy > 0)
{
if ((y2 % square_width) == 0)
end_y--;
}
if (dx < 0)
{
if ((x1 % square_width) == 0)
{
x--;
cross_product += dx_cross;
}
}
else if (dx > 0)
{
if ((x2 % square_width) == 0)
end_x--;
}
while (x != end_x || y != end_y) {
SetPixel(x,y,color);
int old_cross = cross_product;
if (old_cross >= 0) {
x += dx_x;
cross_product += dx_cross;
}
if (old_cross <= 0) {
y += dy_y;
cross_product += dy_cross;
}
}
This code itself hasn't been tested, but it comes slightly modified from my GitHub project where it has been tested.
Let's assume you want to draw a line from P1 = (x1, y1) to P2 = (x2, y2) where all the numbers are floating point pixel coordinates.
Calculate the true pixel coordinates of P1 and P2 and paint them: P* = (round(x), round(y)).
If abs(x1* - x2*) <= 1 && abs(y1* - y2*) <= 1 then you are finished.
Decide whether it is a horizontal (true) or a vertical line (false): abs(x1 - x2) >= abs(y1 - y2).
If it is a horizontal line and x1 > x2 or if it is a vertical line and y1 > y2: swap P1 with P2 (and also P1* with P2*).
If it is a horizontal line you can get the y-coordinates for all the x-coordinates between x1* and x2* with the following formula:
y(x) = round(y1 + (x - x1) / (x2 - x1) * (y2 - y1))
If you have a vertical line you can get the x-coordinates for all the y-coordinates between y1* and y2* with this formula:
x(y) = round(x1 + (y - y1) / (y2 - y1) * (x2 - x1))
Here is a demo you can play around with, you can try different points on line 12.
Okay algebra and trig are not my strong suit by any means so here is what I need to do.
I have a circle which is measured in degrees from +180 to -180 (360 total)
Given the center point of the circle stays the same, Cx , Cy.
The angle varies from -180 to +180
I need to locate a point that regardless the given angle is + 3 units away that is at the 90 degree position and the 270 degree position (from the given degrees)
So like...
Angle = 0
Point 1 -> x = 0, y -3
Point 2 -> x = 0, y + 3
And if the angle was say 90 (provided its measured Clockwise)
Point 1 -> x = -3, y = 0
Point 2 -> x = 3, y = 0
What I need is a forumla that will accept Angle, then tell me what my x/y should be 3 units away from the origin.
I have tried: EDIT Updated to double precision using Java.
`double x = Cx + 3 * Math.cos((d + 90) * Math.PI / 180);'
'double y = Cy + 3 * Math.sin((d + 90) * Math.PI / 180);`
this gives me mixed results, I mean sometimes it's where I think it should be and other times its quite wrong.
Assuming Cx = 0.500, Cy = 0.500
Sample Data: Result:
Deg = 0 x = 2 / y = 5
Deg = 90 x = -1 / y = 2
Deg = 125 x = -0.457 / y = 0.297
Deg = 159 x = 0.924 / y = -0.800
I realize I am only calculating one point at this point but do you have any suggestions on how to get the first point working? at say 90 degrees from whatever degree I start with?
x = Cx + r * Math.cos( (d+90) * Math.PI / 180 );
y = Cy + r * Math.sin( (d+90) * Math.PI / 180 );
Seems that this is the correct formula for what I was trying to accomplish. This will take any value for Cx/Cy's origin add the Radius r, then calculate the degrees + 90 and convert to radians.. Once all that magic takes place, you're left with an x/y coord that is 90 degrees of where you started.
So I've managed myself to write the first part (algorithm) to calculate each tile's position where should it be placed while drawing this map (see bellow). However I need to be able to convert mouse location to the appropriate cell and I've been almost pulling my hair off because I can't figure out a way how to get the cell from mouse location. My concern is that it involves some pretty high math or something i'm just something easy i'm not capable to notice.
For example if the mouse position is 112;35 how do i calculate/transform it to to get that the cell is 2;3 at that position?
Maybe there is some really good math-thinking programmer here who would help me on this or someone who knows how to do it or can give some information?
var cord:Point = new Point();
cord.x = (x - 1) * 28 + (y - 1) * 28;
cord.y = (y - 1) * 14 + (x - 1) * (- 14);
Speaking of the map, each cell (transparent tile 56x28 pixels) is placed in the center of the previous cell (or at zero position for the cell 1;1), above is the code I use for converting cell-to-position. I tried lot of things and calculations for position-to-cell but each of them failed.
Edit:
After reading lot of information it seems that using off screen color map (where colors are mapped to tiles) is the fastest and most efficient solution?
I know this is an old post, but I want to update this since some people might still look for answers to this issue, just like I was earlier today. However, I figured this out myself. There is also a much better way to render this so you don't get tile overlapping issues.
The code is as simple as this:
mouse_grid_x = floor((mouse_y / tile_height) + (mouse_x / tile_width));
mouse_grid_y = floor((-mouse_x / tile_width) + (mouse_y / tile_height));
mouse_x and mouse_y are mouse screen coordinates.
tile_height and tile_width are actual tile size, not the image itself. As you see on my example picture I've added dirt under my tile, this is just for easier rendering, actual size is 24 x 12. The coordinates are also "floored" to keep the result grid x and y rounded down.
Also notice that I render these tiles from the y=0 and x=tile_with / 2 (red dot). This means my 0,0 actually starts at the top corner of the tile (tilted) and not out in open air. See these tiles as rotated squares, you still want to start from the 0,0 pixel.
Tiles will be rendered beginning with the Y = 0 and X = 0 to map size. After first row is rendered you skip a few pixels down and to the left. This will make the next line of tiles overlap the first one, which is a great way to keep the layers overlapping coorectly. You should render tiles, then whatever in on that tile before moving on to the next.
I'll add a render example too:
for (yy = 0; yy < map_height; yy++)
{
for (xx = 0; xx < map_width; xx++)
{
draw tiles here with tile coordinates:
tile_x = (xx * 12) - (yy * 12) - (tile_width / 2)
tile_y = (yy * 6) + (xx * 6)
also draw whatever is on this tile here before moving on
}
}
(1) x` = 28x -28 + 28y -28 = 28x + 28y -56
(2) y` = -14x +14 +14y -14 = -14x + 14y
Transformation table:
[x] [28 28 -56 ] = [x`]
[y] [-14 14 0 ] [y`]
[1] [0 0 1 ] [1 ]
[28 28 -56 ] ^ -1
[-14 14 0 ]
[0 0 1 ]
Calculate that with a plotter ( I like wims )
[1/56 -1/28 1 ]
[1/56 1/28 1 ]
[0 0 1 ]
x = 1/56*x` - 1/28y` + 1
y = 1/56*x` + 1/28y` + 1
I rendered the tiles like above.
the sollution is VERY simple!
first thing:
my Tile width and height are both = 32
this means that in isometric view,
the width = 32 and height = 16!
Mapheight in this case is 5 (max. Y value)
y_iso & x_iso == 0 when y_mouse=MapHeight/tilewidth/2 and x_mouse = 0
when x_mouse +=1, y_iso -=1
so first of all I calculate the "per-pixel transformation"
TileY = ((y_mouse*2)-((MapHeight*tilewidth)/2)+x_mouse)/2;
TileX = x_mouse-TileY;
to find the tile coordinates I just devide both by tilewidth
TileY = TileY/32;
TileX = TileX/32;
DONE!!
never had any problems!
I've found algorithm on this site http://www.tonypa.pri.ee/tbw/tut18.html. I couldn't get it to work for me properly, but I change it by trial and error to this form and it works for me now.
int x = mouse.x + offset.x - tile[0;0].x; //tile[0;0].x is the value of x form witch map was drawn
int y = mouse.y + offset.y;
double _x =((2 * y + x) / 2);
double _y= ((2 * y - x) / 2);
double tileX = Math.round(_x / (tile.height - 1)) - 1;
double tileY = Math.round(_y / (tile.height - 1));
This is my map generation
for(int x=0;x<max_X;x++)
for(int y=0;y<max_Y;y++)
map.drawImage(image, ((max_X - 1) * tile.width / 2) - ((tile.width - 1) / 2 * (y - x)), ((tile.height - 1) / 2) * (y + x));
One way would be to rotate it back to a square projection:
First translate y so that the dimensions are relative to the origin:
x0 = x_mouse;
y0 = y_mouse-14
Then scale by your tile size:
x1 = x/28; //or maybe 56?
y1 = y/28
Then rotate by the projection angle
a = atan(2/1);
x_tile = x1 * cos(a) - y1 * sin(a);
y_tile = y1 * cos(a) + x1 * sin(a);
I may be missing a minus sign, but that's the general idea.
Although you didn't mention it in your original question, in comments I think you said you're programming this in Flash. In which case Flash comes with Matrix transformation functions. The most robust way to convert between coordinate systems (eg. to isometric coordinates) is using Matrix transformations:
http://help.adobe.com/en_US/FlashPlatform/reference/actionscript/3/flash/geom/Matrix.html
You would want to rotate and scale the matrix in the inverse of how you rotated and scaled the graphics.
Does anyone have an algorithm to determine the direction from one lat/lon to another (pseudo-code):
CalculateHeading( lat1, lon1, lat2, long2 ) returns string heading
Where heading is e.g. NW, SW, E, etc.
Basically, I have two points on a map and I want to get a general idea of the direction taking into account that 50 miles East and one mile North is simply East and not Northeast.
This site has the basic algorithm:
// in javascript, not hard to translate...
var y = Math.sin(dLon) * Math.cos(lat2);
var x = Math.cos(lat1)*Math.sin(lat2) -
Math.sin(lat1)*Math.cos(lat2)*Math.cos(dLon);
var brng = Math.atan2(y, x).toDeg();
UPDATED: See here for complete algorith Mapping Math and Javascript
That'll give you a number between 0 and 360 then it's just a matter of having a simple lookup:
var bearings = ["NE", "E", "SE", "S", "SW", "W", "NW", "N"];
var index = brng - 22.5;
if (index < 0)
index += 360;
index = parseInt(index / 45);
return(bearings[index]);
It's important to note that your bearing actually changes as you move around the earth. The algorithm above shows you initial bearing, but if you're traveling a long distance, your bearing to be significantly different when you reach the destination (if you're only traveling a short distance [< a few hundred kms] then it probably won't change enough to be a concern).
Do you remember your trig functions? I.e. SOHCAHTOA:
SOH: Sin(θ) = Opposite over Hypotenuse
CAH: Cos(θ) = Adjacent over Hypotenuse
TOA: Tan(θ) = Opposite over Adjacent
In pseudo-code:
function getDir(lat1, long1, lat2, long2) {
margin = π/90; // 2 degree tolerance for cardinal directions
o = lat1 - lat2;
a = long1 - long2;
angle = atan2(o,a);
if (angle > -margin && angle < margin):
return "E";
elseif (angle > π/2 - margin && angle < π/2 + margin):
return "N";
elseif (angle > π - margin && angle < -π + margin):
return "W";
elseif (angle > -π/2 - margin && angle < -π/2 + margin):
return "S";
}
if (angle > 0 && angle < π/2) {
return "NE";
} elseif (angle > π/2 && angle < π) {
return "NW";
} elseif (angle > -π/2 && angle < 0) {
return "SE";
} else {
return "SW";
}
}
Edit 1:
As Pete and Dean pointed out, this does not take into account the curvature of the Earth. For more accurate calculations for points away from the equator, you'll need to use spherical triangle formulas, as used in Dean's answer.
Edit 2:
Another correction; as Pete noted, arctan() does not give the correct angles, as -1/-1 and 1/1 are the same (as are -1/1 and 1/-1). arctan2(y, x) is a two argument variation of arctan() that is designed to compensate for this. arctan() has a range of (-π, π], positive for y >= 0 and negative for y < 0.
Convert to a numeric angle and use the result to look up the text. For example, -22.5..+22.5 = N. +22.5..67.5 = NE, 67.5..112.5 = E, etc. Of course, that's assuming you're using only N, NE, E, SE, S, SW, W, NW -- if you decide (for example) to go with the old "32 points of the compass", each text string obviously represents a smaller range.