I am currently developing a simple Pong game for the iPhone. Currently using CGRectIntersectsRect for the collision detection and as for the deflection of the ball when it hits the paddle, I just multiply the ball velocity with -1 (therefore reversing the direction of the ball).
What I am trying to do is to make it so that when the ball hits the paddle, it checks whether how far is the ball from the center of the paddle, and increases the deflection angle the further the ball is away from the center of the paddle. (E.g. In this case, the ball will be deflected back at 90 degrees no matter where it came from, as long as it hits the center of the paddle)
How am I suppose to do that?
Any help would be greatly appreciated.
Thank you.
What you have given us are reference points (centre and edge). What we need are a reference line from which to measure the new angle. Additionally, what you are saying is not consistent and thus does not make sense.
I am guessing that what you are asking is a way to calculate the outgoing angle such that it is only a function of where it hit on the paddle. If it hit the paddle centre, then irrespective of the incoming angle, it will bounce off at an angle of 90 degrees to the paddle. If it hit the paddle edge, then irrespective of the incoming angle it will bounce off at an angle of 45 degrees to the paddle.
If so, then the following should do it (it is not the only way).
Assumption: The paddle shape is a rectangle.
Let L be the length of the paddle.
Let K be a constant such that L / (2 * K) = 1 / sqrt(2).
Let D be the distance from the centre of the paddle (may be + or -).
theta = pi/2 - asin (D / K);
This should give you an angle relative to the paddle.
Hope this helps.
Related
I'll keep the question short.
I'm making a 2D game.
I have an object, but since it has many pieces, I just out it in a box to have it's hitbox.
If it's going up/down/left/right at perfect angles, I can check for collision because it's just about the left corner and then height and width.
How do I calculate if my mouseX and mouseY collide with it if it rotates at different angles.
Found a formula that x' = x sin A - y cos A or something, but didn't work.
Thank you!
I've already asked this question in a different way here; Swift-Setting a physics body velocity by angle but the three attempts to answer it were unfortunately not exactly what I'm looking for, although I'm grateful for what they taught me anyway. I decided that I should simply rephrase my question with an example and further explanation instead of perpetuating a discussion via comments. So here it is.
Imagine I have an SKNode positioned in the centre of the screen. Let's say this is a ball, so any rotation action on it is not visible. I would need a way to have a random angle selected, and have said SKNode rotate to it, and then continuously move in the direction determined by the aforementioned angle from its original position in the centre of the screen, until, say, the edge of the screen.
I know how to determine the random angle, have the SKNode rotate to it, and have it stop at the edge of the screen. Hopefully with this example what I need is clearer, a way to simply have an SKNode move forward, but on an angle determined by a single variable, and not a velocity determined by two, dx and dy.
Thanks in advance.
To do this, you just need to use some trigonometry!
When zRotation is between 0 and 90, you don't need to do anything to the angle except converting it to radians. Just call tan(radianAngle). Now tan will return how much should the node move in the y axis when it moves by 1 in the x axis. If you learned trigonometry before, you should understand what I'm saying. If you have not learned trigonometry, learn it. :)
Let's say the node's zRotation is 60 degrees, which is π/3 radians. Pass that into tan and you get √3. This means your dx and dy parameters in SKAction must be in the ratio of 1 : √3 in order to make the node move in the direction of 60 degrees.
When zRotation is between 90 and 180, you need to first subtract the angle from 180, then convert it to radians. Again, pass that to tan and the return value is how much your node should move in the y direction when it moves by -1 in the x axis. The dx : dy ratio is now -1 : tan(angleInRadians).
When zRotation is between 180 and 270, subtract 180 from that angle and convert it to radians. The dx : dy ratio is -1 : -tan(angleInRadians).
Lastly, a zRotation bewteen 270 and 360, subtract the angle from 360 and convert it to radians. The dx : dy ratio is 1 : -tan(angleInRadians).
Just before you convert the angle to radians, check if the angle is 90 degrees. If it is, please hard code the dx and dy because tan(M_PI / 2) is undefined.
I am working on a simple project that consists of a ball with a rigid body being pushed along the x axis. The ball needs to transition 90 degrees and continue straight up.
I've created a box to serve as a pivot/rotation point and placed it exactly 5 units above the beginning of the 90 degree transition from and exactly 5 units away from the end of the transition. I've created two colliders that work as triggers; one at the beginning and one at the very end of the 90 degree transition.
So my ball will come zooming along on the x axis and hit the first trigger. When it does I reset the ball's velocity to zero and parent the ball to the box serving as a pivot/rotation point. I've then applied angular velocity to the box so that the ball will rotate exactly 5 units around the pivot point. When the ball hits the second collider at the end of the transition it is un-parented and velocity is reapplied so the ball can continue straight up.
So here is the question; if the ball is 5 units away from the pivot point, the distance traveled for the 90 degree transition would be something like : 5 * 2 = the circles diameter * PI = 31.4 units distance around the circle / 4 = 7.84 game units to travel for the entire 90 degree transition.
So how do I match the velocity the ball had while traveling solely on the x axis so that it has the same speed while transitioning using the pivot point rigid body's moveRotation function around a circular axis?
If I have a velocity vector of (-10, 0, 0) when I hit the first trigger how do I calculate what to feed into the moveRotation function so that a the ball, rotating 5 units away from the pivot point, is still traveling at 10 units a second?
Taking the original velocity (-10, 0, 0) and multiplying by the distance needed to travel for the 90 degree transition(7.85 units) looks really close but I can figure out how to check it.
in my game im having one Ccsprite for arrow, and one b2body for ball... im trying to throw ball at direction which is pointed by my arrow sprite. here is my code... i'm counting rotation of arrow sprite and then applying impulse to ball at that angle...
float totalRotation = arrow.rotation ;
ballBody->ApplyLinearImpulse(b2Vec2(10.0f+cos(totalRotation)*25.0f,10.0f+sin(totalRotation)*25.0f), eggBody->GetWorldCenter());
BUt, this not working exactly...ball is getting thrown in improper direction.
The rotation property of a CCNode (and CCSprite, which inherits from CCNode) is measured in degrees, with clockwise rotation being positive. The Box2D world uses angles measured in radians, with counter-clockwise rotation being positive, which is more conventional for a cartesian coordinate system. In order to provide the correct angle to a Box2D function, you will have to convert. In Cocos2D, the conversion goes like this:
float angle = - 1 * CC_DEGREES_TO_RADIANS(totalRotation);
The macro converts the totalRotation from degrees to radians, and you multiply by -1 because Box2D measures positive angles in the counter-clockwise direction, which is opposite of the CCNode rotation.
I am trying to solve a tricky math problem, in a cocos2d for iphone context.
Basically I have a roullette wheel which is rotating over time.
I want to have a Sprite latch onto the wheel at certain points (like compass ordinal points N, S, E, W) and bounce off at all other points.
I have the image of the wheel rotating, and just need to solve the part where I can test for whether a sprite has intersected with the circle at the right point on the circle as it is rotating.
I think this question is going in the right direction, but I can't get my head around it. Can anyone help explain?
Best way to find a point on a circle closest to a given point
If I understand correctly:
First check the distance between the sprite and the centre of the roulette wheel. This will tell you if the sprite is at the edge of the wheel. (If not, nothing happens, right?)
Then, find the angle that the sprite makes from the "x-axis" of the roulette wheel.
spriteAngle = atan2(sprite.x - rouletteCentre.x, sprite.y - rouletteCentre.y)
You'll need to find the equivalent of the atan2() function. It usually returns an answer in radians; you may want to convert it to degrees or quarter-turns or something if you prefer.
Then, subtract the angle that the roulette wheel itself is rotated by (if the wheel itself is rotating, if not then you're already done). Make sure your angle measurement is consistent.
actualAngle = spriteAngle - rouletteRotationAngle
Note that actualAngle may be outside the range 0-360 degrees, and you will need to make it "wrap around".
Lastly, you will want to allow a small range of values as acceptable (e.g. 98 degrees to 102 might count as "North").
So you have a circle of radius r, with center (x0,y0).
A point lies outside of the circle, at coordinates (x,y). Your question is to find the closest point on the circle itself to the point (x,y).
The solution is simple. The closest projection of a point onto a circle is accomplished by a simple scaling. Thus,
d = sqrt((x-x0)^2 + (y-y0)^2)
xp = x0 + (x - x0)*r/d
yp = y0 + (y - y0)*r/d
The new point (xp,yp) will lie on the circle itself. To be honest, you would be better off to work in polar coordinates, with the origin at the center of the circle. Then everything gets much easier.
Your next question will be where did it hit on the circle? Don't forget the points of the compass on the circle are rotating with time. An atan2 function will give you the angle that the point (xp-x0,yp-y0) lies at. Most toolsets will have that functionality. See that I've subtracted off the origin here.