Develop Reference

How to find point cordinates in a circle using radius and center cordinate

I am trying to map all circle cordinates to a flat surface using OpenGL fragment shader as shown in the picture. I know the radius and center point of the circle. So I am trying to use the below equation:
For a circle with origin (j, k) and radius r:
x(t) = r cos(t) + j
y(t) = r sin(t) + k
where you need to run this equation for t taking values within the range from 0 to 360, then you will get your x and y each on the boundary of the circle.
But I am facing below problem:
Angle values are in degrees. Can I use that directly in the fragment shader code as the OpenGL cordinates are from 0.0 to 1.0?

There are many approaches to map circle points to square and vice versa. For simplicity circle center is (0,0) and radius is 1.
Let coordinate in square are (x,y) and corresponding coordinate in circle is (u, v).
So to get circle point:
u = x * sqrt(1 - y^2/2)
v = y * sqrt(1 - x^2/2)
Inverse formula is more complex:
x = (sqrt(2 + u*u-v*v+2*u*sqrt(2)) - sqrt(2 + u*u-v*v - 2*u*sqrt(2))) / 2
y = (sqrt(2 - u*u+v*v+2*v*sqrt(2)) - sqrt(2 - u*u+v*v - 2*v*sqrt(2))) / 2
Note that you perhaps want to get point at the circle corresponding to needed square point, so need only the first formula pair
P.S. Perhaps I took formulas there years ago
If you need to walk through circle points with some step (while I doubt you really need this), you can scan Y-coordinates and get scanlines for X coordinates. Pseudocode:
for y = - R to R step d:
L = sqrt(R*R-y*y)
for x = -L to L step d:
getPoint(cx + x, cy + y)

Java Swing - Calculating what angle the mouse position is to the center of the screen

I'm making a 2D topdown view shooter game with Java Swing. I want to calculate what angle the mouse pointer is compared to the center of the screen so some of my Sprites can look toward the pointer and so that I can create projectiles described by an angle and a speed. Additionally If the pointer is straight above the middle of the screen, I want my angle to be 0°, if straight to its right, 90°, if straight below 180°, and straight left 270°.
I have made a function to calculate this:
public static float calculateMouseToPlayerAngle(float x, float y){
float mouseX = (float) MouseInfo.getPointerInfo().getLocation().getX();
float mouseY = (float)MouseInfo.getPointerInfo().getLocation().getY();
float hypotenuse = (float) Point2D.distance(mouseX, mouseY, x, y);
return (float)(Math.acos(Math.abs(mouseY-y)/hypotenuse)*(180/Math.PI));
}
The idea behind it is that I calculate the length of the hypotenuse then the length of the side opposite of the angle in question. The fraction of the 2 should be a cos of my angle, so taking that result's arc cos then multiplying that by 180/Pi should give me the angle in degrees. This does work for above and to the right, but straight below returns 0 and straight left returns 90. That means that I currently have 2 problems where the domain of my output is only [0,90] instead of [0,360) and that it's mirrored through the y (height) axis. Where did I screw up?

You can do it like this.
For a window size of 500x500, top left being at point 0,0 and bottom right being at 500,500.
The tangent is the change in Y over the change in X of two points. Also known as the slope it is the ratio of the sin to cos of a specific angle. To find that angle, the arctan (Math.atan or Math.atan2) can be used. The second method takes two arguments and is used below.
BiFunction<Point2D, Point2D, Double> angle = (c,
m) -> (Math.toDegrees(Math.atan2(c.getY() - m.getY(),
c.getX() - m.getX())) + 270)%360;
BiFunction<Point2D, Point2D, Double> distance = (c,
m) -> Math.hypot(c.getY() - m.getY(),
c.getX() - m.getX());
int screenWidth = 500;
int screenHeight = 500;
int ctrY = screenHeight/2;
int ctrX = screenWidth/2;
Point2D center = new Point2D.Double(ctrX,ctrY );
Point2D mouse = new Point2D.Double(ctrX, ctrY-100);
double straightAbove = angle.apply(center, mouse);
System.out.println("StraightAbove: " + straightAbove);
mouse = new Point2D.Double(ctrX+100, ctrY);
double straightRight = angle.apply(center, mouse);
System.out.println("StraightRight: " + straightRight);
mouse = new Point2D.Double(ctrX, ctrY+100);
double straightBelow = angle.apply(center, mouse);
System.out.println("StraightBelow: " + straightBelow);
mouse = new Point2D.Double(ctrX-100, ctrY);
double straightLeft = angle.apply(center, mouse);
System.out.println("Straightleft: " + straightLeft);
prints
StraightAbove: 0.0
StraightRight: 90.0
StraightBelow: 180.0
Straightleft: 270.0
I converted the radian output from Math.atan2 to degrees. For your application it may be more convenient to leave them in radians.
Here is a similar Function to find the distance using Math.hypot
BiFunction<Point2D, Point2D, Double> distance = (c,m) ->
Math.hypot(c.getY() - m.getY(),
c.getX() - m.getX());

A fast collision detection between a circle and a sprite in pixel perfect

I've been thinking on some fast and brilliant pixel - perfect collision detection between a circle and any sprite. I need to get 2 points of collision to be able to calculate a normal vector out of them later. I managed to come up with some solution but the more scaling is done in my game, the more inaccurate and unprecise this collision is...It seems as if the code I posted below, was good and correct becouse I have been checking it already a few times and spent a few days reading it again and again... I also checked visually that the collision masks and areas of collision are calculated perfectly fine in the code below so the problem definitely doesn't lay there but in this method.
So I guess that the problem here is the loss of data in floating point arithmetic unless somebody finds a flaw in this method?
If however the problem is really with the float loss of data, what other solution would you recommend to find 2 points of collision between circle and any other sprite in pixel perfect? I really liked my solution becouse it was relatively fast
int xOffset1 = (int)colRectLeft; // left boundary of the collision area for the first sprite
int xOffset2 = (int)colCircleLeft; // left boundary of the collision area for the circle sprite
int yOffset1 = (int)colRectBottom; // bottom boundary of the collision area for the first sprite
int yOffset2 = (int)colCircleBottom; // bottom boundary of the collision area for the circle sprite
int width = (int)(colCircleRight - colCircleLeft); //width of the collision area - same for both sprites
int height = (int)(colCircleTop - colCircleBottom); // height of the collision area same for both sprites
// Pixel-perfect COLLISION DETECTION between circle and a sprite
// my custom vector classes - nothing special
Math2D.Vector_2 colRightPoint = new Math2D.Vector_2(-1, -1); // The right point of collision lying on the circle's circumference
Math2D.Vector_2 colLeftPoint = new Math2D.Vector_2(-1, -1); // the left point of collision lying on the circle's circumference
boolean colRightFound = false;
boolean colLeftFound = false;
// I'm going through y in the circle's area of collision
for (float y = yOffset2; y < yOffset2 + height; y += 1)
{
// from equation: (x-Sx)^2 + (y-Sy)^2 = r^2
// x1/2 = (+-)sqrt(r^2 - (y - Sy)^2) + Sx
//(Sx, Sy) is (circle's radius, circle's radius) becouse I want the points on the circle's circumference to have positive coordinates
float x1 = (float) (Math.sqrt(radius*radius - (y - radius)*(y - radius)) + radius); // the right pixel on the circumference
float x2 = (float) (-x1 + 2*radius); // the left pixel on the circumference
//first I check if the calculated x is inside of the previously calculated area of collision for both circle's area and a sprite's area
if (x1 >= xOffset2 &&
x1 <= xOffset2 + width &&
xOffset1 + x1 - xOffset2 < rectFrameW &&
yOffset1 + (int)y-yOffset2 < rectFrameH &&
yOffset1 + (int)y-yOffset2 > 0 &&
xOffset1 + x1 - xOffset2 > 0)
{
//I don't have to check if the point on the circle's circumference is opaque becouse it's always so just check if the same point translated to sprite's area of collision is opaque
boolean opaqueRectPixel = go.gameData.images.get(go.pic_nr)
.collision_mask[(int)((yOffset1 + (int)y-yOffset2)*rectFrameW +
(xOffset1 + x1 - xOffset2))];
if(opaqueRectPixel)
{
if(!colRightFound)
{
colRightPoint.x = (xOffset1 + x1 - xOffset2);
colRightPoint.y = (yOffset1 + (int)y - yOffset2);
colRightFound = true;
}
else if(!colLeftFound)
{
colLeftPoint.x = (xOffset1 + x1 - xOffset2);
colLeftPoint.y = (yOffset1 + (int)y - yOffset2);
}
}
}
//the same logic for the left point on the circle's circumference
if (x2 >= xOffset2 &&
x2 <= xOffset2 + width &&
xOffset1 + x2 - xOffset2 < rectFrameW &&
yOffset1 + (int)y-yOffset2 < rectFrameH &&
yOffset1 + (int)y-yOffset2 > 0 &&
xOffset1 + x2 - xOffset2 > 0)
{
boolean opaqueRectPixel = go.gameData.images.get(go.pic_nr)
.collision_mask[(int)((yOffset1 + (int)y-yOffset2)*rectFrameW +
(xOffset1 + x2 - xOffset2))];
if(opaqueRectPixel)
{
if(!colLeftFound)
{
colLeftPoint.x = (xOffset1 + x2 - xOffset2);
colLeftPoint.y = (yOffset1 + (int)y - yOffset2);
colLeftFound = true;
}
else if(!colRightFound)
{
colRightPoint.x = (xOffset1 + x2 - xOffset2);
colRightPoint.y = (yOffset1 + (int)y - yOffset2);
}
}
}
// if both points are already found, finish
if(colLeftFound && colRightFound)
break;
}
edit: Actually, what I'm doing in this method is finding points of intersection between circle and a sprite
edit: Ok, I'm uploading images to describe my algorithm a bit better. I really tried my best to explain it but if there's still something missing, let me know please!
Also I would accept any other good solutions to find intersection points between a circle and any sprite in pixel perfect, if you don't want to check my code :(... Eh, I'm always having problems with collisions...

If you absolutely want (or need) pixel perfect, your solution looks good.
don't forget to first make a rectangle-to-rectangle collision before testing a pixel perfect detection, to avoid unneeded processings.
If you want another accurate method which maybe more efficient, look for Separating Axis Theorem.
You can find more information about it here :
http://rocketmandevelopment.com/blog/separation-of-axis-theorem-for-collision-detection/
and here :
http://www.metanetsoftware.com/technique/tutorialA.html
The last one have nice interactive explanation and demonstration. Enjoy :)

...as I was not able to show the raster in the comments:
I did not mentally parse your code, however from the image I see that you try to detect borderline collisions. Putting round or diagonal (border)lines into a raster may cause occasions, where two crossing lines do not overlay each other - like this:
1 2
2 1
whereby 1 would be line 1 and 2 would be line 2.
However I still like the idea of checking border lines combined with rectangle pre-checks. If you would render an array of raster proved-closed line coordinates by sprites you could check them against each other. This could also be enriched by border line segmenting (such as North, East, West and South or a bit more fine grain - I guess there is an optimum). A diagonal proved-closed line in the check data set must represent something like this:
x _
x x
whereby the x represent the pixels of your line and the _ is an empty raster seat.

Trigonometry of a 3D "free" camera

I'm sorry if this question was asked before, I did search, and I did not find an answer.
My problem is, that I'd like to make movement on all 3 axes with the X and Y rotation of the camera being relevant.
This is what I did:
private static void fly(int addX, int addY){ //parameters are the direction change relative to the current rotation
float angleX = rotation.x + addX; //angle is basically the direction, into which we will be moving(when moving forward this is always the same as our actual rotation, therefore addX and addY would be 0, 0)
float angleY = rotation.y + addY;
float speed = (moveSpeed * 0.0002f) * delta;
float hypotenuse = speed; //the length that is SUPPOSED TO BE moved overall on all 3 axes
/* Y-Z side*/
//Hypotenuse, Adjacent and Opposite side lengths of a triangle on the Y-Z side
//The point where the Hypotenuse and the Adjacent meet is where the player currently is.
//OppYZ is the opposite of this triangle, which is the ammount that should be moved on the Y axis.
//the Adjacent is not used, don't get confused by it. I just put it there, so it looks nicer.
float HypYZ = speed;
float AdjYZ = (float) (HypYZ * Math.cos(Math.toRadians(angleX))); //adjacent is on the Z axis
float OppYZ = (float) (HypYZ * Math.sin(Math.toRadians(angleX))); //opposite is on the Y axis
/* X-Z side*/
//Side lengths of a triangle on the Y-Z side
//The point where the Hypotenuse and the Adjacent meet is where the player currently is.
float HypXZ = speed;
float AdjXZ = (float) (HypXZ * Math.cos(Math.toRadians(angleY))); //on X
float OppXZ = (float) (HypXZ * Math.sin(Math.toRadians(angleY))); //on Z
position.x += AdjXZ;
position.y += OppYZ;
position.z += OppXZ;
}
I only implement this method when moving forwards(parameters: 0, 90) or backwards(params: 180, 270), since movement can't happen on the Y axis while going sideways, since you don't rotate on the Z axis. ( the method for going sideways(strafing) works just fine, so I won't add that.)
the problem is that when I look 90 degrees up or -90 down and then move forward I should be moving only on the Y axis(vertically) but for some reason I also move forwards(which means on the Z axis, as the X axis is the strafing).
I do realize that movement speed this way is not constant. If you have a solution for that, I'd gladly accept it as well.

I think your error lies in the fact that you don't fully project your distance (your quantity of movement hypothenuse) on your horizontal plane and vertical one.
In other words, whatever the chosen direction, what you are doing right now is moving your point of hypothenuse in the horizontal plane X-Z, even though you already move it of a portion of hypothenuse in the vertical direction Y.
What you probably want to do is moving your point of a hypothenuse quantity as a total.
So you have to evaluate how much of the movement takes place in the horizontal plane and how much in the vertical axis. Your direction gives you the answer.
Now, it is not clear to me right now what your 2 angles represent. I highly recommend you to use Tait–Bryan angles in this situation (using only yawn and pitch, since you don't seem to need the rolling - what you call the Z-rotation), to simplify the calculations.
In this configuration, the yawn angle would be apparently similar to your definition of your angleY, while the pitch angle would be the angle between the horizontal plane and your hypothenuse vector (and not the angle of the projection in the plane Y-Z).
A schema to clarify:
With :
s your quantity of movement from your initial position P_0 to P_1 (hypothenuse)
a_y the yawn angle and a_p the pitch one
D_x, D_y, D_z the displacements for each axis (to be added to position, ie AdjXZ, OppYZ and OppXZ)
So if you look at this representation, you can see that your triangle in X-Z doesn't have s as hypotenuse but its projection s_xz. The evaluation of this distance is quite straightforward: if you place yourself in the triangle P_0 P_1 P_1xz, you can see that s_xz = s * cos(a_p). Which gives you:
float HypXZ = speed * Math.cos(Math.toRadians(angleP))); // s_xz
float AdjXZ = (float) (HypXZ * Math.cos(Math.toRadians(angleY))); // D_x
float OppXZ = (float) (HypXZ * Math.sin(Math.toRadians(angleY))); // D_z
As for D_y ie OppYZ, place yourself in the triangle P_0 P_1 P_1xz again, and you'll obtain:
float OppYZ = (float) (speed * Math.sin(Math.toRadians(angleP))); // D_y
Now, if by angleX you actually meant the angle of elevation as I suppose you did, then angleP = angleX and HypXZ = AdjYZ in your code.
With this correction, if angleX = 90 or angleX = -90, then
HypXZ = speed * cos(angleX) = speed * cos(90deg) = speed * 0;
... and thus AdjXZ = 0 and OppXZ = 0. No movement in the horizontal plane.
Note:
To check if your calculations are correct, you can verify if you actually move your point of the wanted quantity of movement (hypothenuse ie speed ie s). Using Pythagorean theorem:
s² = s_xz² + D_z² // Applied in the triangle P_0 P_1 P_1xz
= D_x² + D_y² + D_z² // Applied in the triangle P_0 P_1x P_1xz
With the definitions of the displacements given above:
D_x² + D_y² + D_z²
= (s * cos(a_p) * cos(a_y))² + (s * cos(a_p) * sin(a_y))² + (s * sin(a_p))²
= s² * (cos(a_p)² * cos(a_y)² + cos(a_p)² * sin(a_y)² + sin(a_p)²)
= s² * (cos(a_p)² * (cos(a_y)² + sin(a_y)²) + sin(a_p)²)
= s² * (cos(a_p)² * 1 + sin(a_p)²)
= s² * (cos(a_p)² + sin(a_p)²)
= s² * 1 // Correct
Hope it helped... Bye!

Put points round into circular arc used in java

I have me math question: I have known a circle center and radius, and have some uncertain number of points called N, my question is how to put the points on the circular arc, I cannot like put the points around the whole circumference, other as this link: http://i.6.cn/cvbnm/2c/93/b8/05543abdd33b198146d473a43e1049e6.png
in this link, you can read point is circle center, other color is some points, you can see these points around the arc.
Edit - in short: I have known a circle center and radius, so I want to generate some point around the circle center

I am not sure, but I checked this with simple Swing JComponent and seems ok.
Point center = new Point(100, 100); // circle center
int n = 5; // N
int r = 20; // radius
for (int i = 0; i < n; i++)
{
double fi = 2*Math.PI*i/n;
double x = r*Math.sin(fi + Math.PI) + center.getX();
double y = r*Math.cos(fi + Math.PI) + center.getY();
//g2.draw(new Line2D.Double(x, y, x, y));
}

It's not entirely clear what you're trying to accomplish here. The general idea of most of it is fairly simple though. There are 2*Pi radians in a circle, so once you've decided what part of a circle you want to arrange your points over, you multiply that percentage by 2*pi, and divide that result by the number of points to get the angle (in radians) between the points.
To get from angular distances to positions, you take the cosine and sine of the angle, and multiply each by the radius of the circle to get the x and y coordinate of the point relative to the center of the circle. For this purpose, an angle of 0 radians goes directly to the right from the center, and angles progress counter-clockwise from there.