I want to get angles between two line.
So I used this code.
int posX = (ScreenWidth) >> 1;
int posY = (ScreenHeight) >> 1;
double radians, degrees;
radians = atan2f( y - posY , x - posX);
degrees = -CC_RADIANS_TO_DEGREES(radians);
NSLog(#"%f %f",degrees,radians);
But it doesn't work .
The Log is that: 146.309935 -2.553590
What's the matter?
I can't know the reason.
Please help me.
If you simply use
radians = atan2f( y - posY , x - posX);
you'll get the angle with the horizontal line y=posY (blue angle).
You'll need to add M_PI_2 to your radians value to get the correct result.
Here's a function I use. It works great for me...
float cartesianAngle(float x, float y) {
float a = atanf(y / (x ? x : 0.0000001));
if (x > 0 && y > 0) a += 0;
else if (x < 0 && y > 0) a += M_PI;
else if (x < 0 && y < 0) a += M_PI;
else if (x > 0 && y < 0) a += M_PI * 2;
return a;
}
EDIT: After some research I found out you can just use atan2(y,x). Most compiler libraries have this function. You can ignore my function above.
If you have 3 points and want to calculate an angle between them here is a quick and correct way of calculating the right angle value:
double AngleBetweenThreePoints(CGPoint pointA, CGPoint pointB, CGPoint pointC)
{
CGFloat a = pointB.x - pointA.x;
CGFloat b = pointB.y - pointA.y;
CGFloat c = pointB.x - pointC.x;
CGFloat d = pointB.y - pointC.y;
CGFloat atanA = atan2(a, b);
CGFloat atanB = atan2(c, d);
return atanB - atanA;
}
This will work for you if you specify point on one of the lines, intersection point and point on the other line.
Related
I have a problem with collision detection of a circle and a rectangle. I have tried to solve the problem with the Pythagorean Theorem. But none of the queries works. The rectangle collides with the rectangular bounding box of the circle.
if (CGRectIntersectsRect(player.frame, visibleEnemy.frame)) {
if (([visibleEnemy spriteTyp] == jumper || [visibleEnemy spriteTyp] == wobble )) {
if ((visibleEnemy.center.x - player.frame.origin.x) * (visibleEnemy.center.x - player.frame.origin.x) +
(visibleEnemy.center.y - player.frame.origin.y) * (visibleEnemy.center.y - player.frame.origin.y) <=
(visibleEnemy.bounds.size.width/2 * visibleEnemy.bounds.size.width/2)) {
NSLog(#"Check 1");
normalAction = NO;
}
if ((visibleEnemy.center.x - (player.frame.origin.x + player.bounds.size.width)) *
(visibleEnemy.center.x - (player.frame.origin.x + player.bounds.size.width)) +
(visibleEnemy.center.y - player.frame.origin.y) * (visibleEnemy.center.y - player.frame.origin.y) <=
(visibleEnemy.bounds.size.width/2 * visibleEnemy.bounds.size.width/2)) {
NSLog(#"Check 2");
normalAction = NO;
}
else {
NSLog(#"Check 3");
normalAction = NO;
}
}
}
Here is how I did it in one of my small gaming projects. It gave me best results and it's simple. My code detects if there is a collision between circle and the line. So you can easily adopt it to circle - rectangle collision detection by checking all 4 edges of the rectangle.
Let's say that a ball has a ballRadius, and location (xBall, yBall). The line is defined with two points (xStart, yStart) and (xEnd, yEnd).
Implementation of a simple collision detection:
float ballRadius = ...;
float x1 = xStart - xBall;
float y1 = yStart - yBall;
float x2 = xEnd - xBall;
float y2 = yEnd - yBall;
float dx = x2 - x1;
float dy = y2 - y1;
float dr = sqrtf(powf(dx, 2) + powf(dy, 2));
float D = x1*y2 - x2*y1;
float delta = powf(ballRadius*0.9,2)*powf(dr,2) - powf(D,2);
if (delta >= 0)
{
// Collision detected
}
If delta is greater than zero there are two intersections between ball (circle) and line. If delta is equal to zero there is one intersection – perfect collision.
I hope it will help you.
I write this question because I don't understand a piece of code in an example provided by Cocos2D for iPhone:
-(CGAffineTransform) nodeToParentTransform
{
b2Vec2 pos = body_->GetPosition();
float x = pos.x * PTM_RATIO;
float y = pos.y * PTM_RATIO;
if ( ignoreAnchorPointForPosition_ ) {
x += anchorPointInPoints_.x;
y += anchorPointInPoints_.y;
}
// Make matrix
float radians = body_->GetAngle();
float c = cosf(radians);
float s = sinf(radians);
if( ! CGPointEqualToPoint(anchorPointInPoints_, CGPointZero) ){
x += c*-anchorPointInPoints_.x + -s*-anchorPointInPoints_.y;
y += s*-anchorPointInPoints_.x + c*-anchorPointInPoints_.y;
}
// Rot, Translate Matrix
transform_ = CGAffineTransformMake( c, s,
-s, c,
x, y );
return transform_;
}
It's in the PhysicsSprite.mm file.
Maybe it's because I'm very bad with space geometry, but if someone can explain me, I very appreciate.
Thanks a lot.
if( ! CGPointEqualToPoint(anchorPointInPoints_, CGPointZero) ){
x += c*-anchorPointInPoints_.x + -s*-anchorPointInPoints_.y;
y += s*-anchorPointInPoints_.x + c*-anchorPointInPoints_.y;
}
The above piece of code simply rotates the xy coordinate axes anticlockwise by $180-\theta$
degrees, and adds the new coordinates to the previous x,y coordinates that were obtained from the piece of code before the above mentioned lines.
http://en.wikipedia.org/wiki/Rotation_of_axes provides the rotation of axes formula
I want to create a struct which is like a CGPoint, but with 3 coordinates instead of 2.
I create it in the following way:
typedef struct {CGFloat x;CGFloat y;CGFloat z;} CG3Vector;
CG_INLINE CG3Vector CG3VectorMake(CGFloat x, CGFloat y, CGFloat z)
{
CG3Vector p; p.x = x; p.y = y; p.z = z; return p;
}
It works fine. But I now want to improve this struct so that it has the constants like for CGPoint: CGPointZero
Also what is the way to introduce the limits for particular components of the struct, like it is for the CGSize, where components are never lower than 0?
Thanks.
You could create constants like this:
const CG3Vector CG3VectorZero = { 0, 0, 0 };
If you want limits, I suppose you can do some checking like this:
CG_INLINE CG3Vector CG3VectorMake(CGFloat x, CGFloat y, CGFloat z)
{
// normalize the values
x = fmod(x, 360);
y = fmod(y, 360);
z = fmod(z, 360);
x = (x < 0) ? 360 + x : x;
y = (y < 0) ? 360 + y : y;
z = (z < 0) ? 360 + z : z;
return (CG3Vector) { x, y, z };
}
I found an answer here, but can't understand how to transfer the math to Objective C
Find the third point
I have two points and I also have the angle relative to the axes. How do I find a third point which will form a straight line? The distance should be variable.
This is the code that I am using:
float distanceFromPx2toP3 = 1300.0;
float mag = sqrt(pow((px2.x - px1.x),2) + pow((px2.y - px1.y),2));
float P3x = px2.x + distanceFromPx2toP3 * (px2.x - px1.x) / mag;
float P3y = px2.y + distanceFromPx2toP3 * (px2.y - px1.y) / mag;
CGPoint P3 = CGPointMake(P3x, P3y);
Let's say I have two points pointA and pointB. The slope of the line formed by the two points m is:
static CGFloat calculateSlope(CGPoint pointA, CGPoint pointB) {
CGFloat m = (pointB.y - pointA.y) / (pointB.x - pointA.x);
return m;
}
A third point pointC a distance d from pointA on the line would be given by:
static CGPoint calculatePointOnLine(
CGPoint pointA, CGPoint pointB, CGFloat d, BOOL startAtB) {
CGFloat m = calculateSlope(pointA, pointB);
CGFloat dX = pointB.x - pointA.x;
CGFloat dY = pointB.y - pointA.y;
CGFloat signDX = dX / fabsf(dX);
CGFloat signDY = dY / fabsf(dY);
CGFloat dSquared = d * d;
CGFloat mSquared = m * m;
// We know pointC is distance d from pointA,
// and that pointA and pointC are on the
// same line
// dXSquared + dYSquared = dSquared
// m = dY / dX
// dY = m * dX
// dXSquared + mSquared * dXSquared = dSquared
// dXSquared * ( 1 + mSquared ) = dSquared
// dXSquared = dSquared / ( 1 + mSquared )
// Handle a vertical line, dX == 0, and a horizontal line, dY == 0
if (dX != 0 && dY != 0) {
// Account for the sign of dX
dX = signDX * sqrtf(dSquared / ( 1 + mSquared ));
// Account for the sign of dY
dY = signDY * m * dX;
}
// Handle a vertical line, dX == 0
if (dX == 0 && dY != 0) {
dY = signDY * d;
}
// Handle a horizontal line, dY == 0
if (dY == 0 && dX != 0) {
dX = signDX * d;
}
CGPoint startingPoint = pointA;
if (startAtB) {
startingPoint = pointB;
}
CGPoint pointC = CGMakePoint(startingPoint.x + dX,
startingPoint.y + dY);
return pointC;
}
pointC will now always lie a distance d along the line from pointA,
in the direction from pointA to pointB. Pass startAtB to have pointC
lie a distance d along the line from pointB, in the direction from
pointA to pointB.
Exchange the order of piintA and pointB in the call to calculatPointOnLine
to calculate a pointC which lies a distance d along the line from
PointB, in the direction from pointB to pointA.
You can use these two functions to calculate a third point on the line.
Thanks for accepting this answer if this helps you.
I have successfully implemented the mandelbrot set as described in the wikipedia article, but I do not know how to zoom into a specific section. This is the code I am using:
+(void)createSetWithWidth:(int)width Height:(int)height Thing:(void(^)(int, int, int, int))thing
{
for (int i = 0; i < height; ++i)
for (int j = 0; j < width; ++j)
{
double x0 = ((4.0f * (i - (height / 2))) / (height)) - 0.0f;
double y0 = ((4.0f * (j - (width / 2))) / (width)) + 0.0f;
double x = 0.0f;
double y = 0.0f;
int iteration = 0;
int max_iteration = 15;
while ((((x * x) + (y * y)) <= 4.0f) && (iteration < max_iteration))
{
double xtemp = ((x * x) - (y * y)) + x0;
y = ((2.0f * x) * y) + y0;
x = xtemp;
iteration += 1;
}
thing(j, i, iteration, max_iteration);
}
}
It was my understanding that x0 should be in the range -2.5 - 1 and y0 should be in the range -1 - 1, and that reducing that number would zoom, but that didnt really work at all. How can I zoom?
Suppose the center is the (cx, cy) and the length you want to display is (lx, ly), you can use the following scaling formula:
x0 = cx + (i/width - 0.5)*lx;
y0 = cy + (j/width - 0.5)*ly;
What it does is to first scale down the pixel to the unit interval (0 <= i/width < 1), then shift the center (-0.5 <= i/width-0.5 < 0.5), scale up to your desired dimension (-0.5*lx <= (i/width-0.5)*lx < 0.5*lx). Finally, shift it to the center you given.
first off, with a max_iteration of 15, you're not going to see much detail. mine has 1000 iterations per point as a baseline, and can go to about 8000 iterations before it really gets too slow to wait for.
this might help: http://jc.unternet.net/src/java/com/jcomeau/Mandelbrot.java
this too: http://www.wikihow.com/Plot-the-Mandelbrot-Set-By-Hand