Shooting object based on arm rotation in libgdx java - java

I am trying to shoot an object(a spell) depending on the rotation of the players arm. The spell is supposed to come out of the hand and shoot towards where the mouse cicked(the arm rotates and points to where the mouse is). This is how the arm rotates in game.
public boolean mouseMoved(int screenX, int screenY) {
tmp.x = screenX;
tmp.y = screenY;
tmp.z = 0;
cam.unproject(tmp);
rot = MathUtils.radiansToDegrees * MathUtils.atan2((float)tmp.y - (float)player.getArmSprite().getY() - player.getArmSprite().getHeight(),
tmp.x -player.getArmSprite().getX() - player.getArmSprite().getWidth());
if (rot < 0) rot += 360;
//Last right or left means if hes looking left or right
if(player.lastRight)
player.setObjectRotation(rot + 80);
if(player.lastLeft)
player.setObjectRotation(-rot - 80);
And this is how the spell is supposed to shoot based off rotation
//destination is a vector of where on screen the mouse was clicked
if(position.y < destination.y){
position.y += vel.y * Gdx.graphics.getDeltaTime();
}
if(position.x < destination.x){
position.x += vel.x * Gdx.graphics.getDeltaTime();
}
However this is very wonky and never really reacts the way it supposed to with a few exceptions. It fires from the hand and then if the y axis is equal it completely evens out and goes till it reaches the x position, I want it to fire from the hand to the position clicks perfectly straight from point a to point b, this is clearly a rotation problem that I just can't seem to figure out how to tackle.
Here is an image of what is happening
Example image
The red indicates where I clicked, as you can see it reached the x pos first and now is traveling to the y when it should have reached the x and y pos of where I clicked first
Any help with this problem is extremely appreciated!

I'm pretty bad at radians and tangents but luckily we have vectors.
Since you have the rot ation in degrees of the arm. I advice to use Vectors to use for any Vector related math now.
//A vector pointing up
Vector2 direction = new Vector2(0, 1);
//Let's rotate that by the rotation of the arm
direction.rotate(rot);
Now direction is the direction the arm is pointing. If your rotation is calculated where up = 0. So you might need to rotate it 180, 90 or -90 degrees. Or in the case you did something silly any degrees.
Your spell should have a Vector too for it's position. Set that to the hand or wherever you want to start from. Now all you need to do is scale that direction since it's currently has a length of 1. If you want to move 5 units each frame you can do direction.scl(5) now it is of length 5. But technically speaking it's no direction anymore now everybody calls it velocity so let's do.
//when you need to fire
float speed = 5;
Vector2 velocity = direction.cpy().scl(speed);
//update
position.add(velocity);
draw(fireballImage, position.x, position.y);
I copied direction first, otherwise it would also be scaled. Then I just added the velocity to the position and draw using that Vector.
And to show Vectors are awesome you should see this awesome badlogic vs mouse program I created. https://github.com/madmenyo/FollowMouse there are just a view lines of my own code. It just takes a little bit of vector knowledge and it's very readable.

Related

Shooting object directly at click java libgdx

I am trying to get my player to shoot a spell and it travels to where ever the player clicked. I can easily accomplish this by doing the following.
if(position.x >= destination.x - 1 && position.x <= destination.x + 1)
reachedX = true;
if(position.y >= destination.y - 1 && position.y <= destination.y + 1)
reachedY = true;
However if the players origin is at, for example, 0,0 and I click at 10,300 then it travels right and up but when the spell reaches an x of 10 it travels directly upwards. I want the spell to travel at an angle that it will reach the x coordinate at the same time as the y coordinate. Here is an image showing what happens and what I want to happen.
In the first picture it looks like the spell goes 45° until it reaches the right x-coordinate. This sounds like the x and y speed are equals, no matter where the destination point is.
You shuld instead have a direction and depending on that a x and y speed.
For that you first need to get the point, the player is clicking.
Therefore you can implement InputProcessor and it's touchDown(int screenX, int screenY, int pointer, int button).
The screenX and screenY arguments are given in screen coordinates (pixels) and therefore need to be converted to your world-coordinates. This can be done using the camera or the viewport and it's unproject(Vector2 screenCoords). This method returns a Vector2 giving the world-coordinates.
Now you need to find out the direction Vector2 between you and the clicked point. The direction Vector is calculated like this:
new Vector2(otherPos.x - myPos.x, otherPos.y - myPos.y).nor();
This returns the normalized direction Vector between the two points.
Now you only need to move the spell by dir.x*spellSpeed*delta in x-direction and dir.y*spellSpeed*delta in y-direction and it should look like in your second picture.

Java Pong - assign various reflection points to different parts of a paddle

I'm making pong in Java and wanted to make the game more fun by assigning different reflection logic to each part of the paddle, like so:
(ball hittins outter edges of paddle will have a different effect than it hitting the middle of the paddle)
The paddle extends Rectangle2D so I could use Rectangle2D's intersects() method to determine if the ball has touched any part of it...
Is it possible to determine where exactly the ball has hit on the paddle?
What I'm planning to do is,
calculate angle of incidence and reflective angle based on that...
If the ball hits at a point x on the paddle... I will change the reflection angle accordingly
Thanks
Is it possible to determine where exactly the ball has hit on the paddle?
If it were me, I would grab the current co-ordinates of both the ball and the paddle. For the paddle, you can get two sets of y co-ordinates, to describe the line facing the ball. Ie:
int paddleY1 = paddle.y;
int paddleY2 = paddle.y + paddle.width;
// assuming the paddle can only go up and down, y is the only co-ordinate that matters.
Then, you can compute the mid point of the paddle as:
int paddleYMid = (paddleY1 + paddleY2) / 2;
You can find out if the ball hit the left or right side of the paddle by comparing the y co-ordinates. Ie:
if(ball.y > paddleYMid)
{
// Right side of the paddle.
}
else
{
// Left side of the paddle.
}
Then it's up to you to develop further refinement.
Since the paddle is always vertical (parallel to Y axis), the point of collision of the ball and the paddle will happen at the same y-coordinate as the center of the ball. That is:
if (ball.centerX - ball.radius <= paddle.rightX && ball.velocityX < 0)
{
// collision point, if any, is at (ball.centerX - ball.radius, ball.centerY)
if (ball.centerY >= paddle.bottomY && ball.centerY <= paddle.topY)
{
// handle collision
}
}
As for the handling of the collision itself, you may not have to deal with angle of reflection, etc, but work solely with the raw values of x and y velocity. For example, to simulate a perfectly elastic collision, simply do:
ball.velocityX = -ball.velocityX;
And to account for ball reflecting at a higher or lower angle, you can scale the y velocity based on the position of the collision from the center of the paddle, eg.
ball.velocityY += SCALE_CONSTANT * (ball.centerY - ((paddle.topY + paddle.bottomY) / 2));
To find the exact spot where the ball hits the paddle, you can formulate the problem as a line intersection problem.
The paddle can be represented as a vertical line at the x-coordinate (+thickness if needed, and corrected for the balls diameter) of the paddle. The ball can then be represented as a line along its movement vector (this line could be simply defined by its current position and its next position if it were to move unimpended).
The problem can now be solved using a line intersection algorythm.
Since the paddle is a vertical line, the equations can be simplified to just answer the question at which Y will the ball pass the paddle's X. Thats also a common problem encountered and solved by line clipping: http://en.wikipedia.org/wiki/Line_clipping (the intersection points can also be computed directly, but I can't find the formula atm).

Getting bullet X to Y movement ratio from 2 points

I'm making pretty simple game. You have a sprite onscreen with a gun, and he shoots a bullet in the direction the mouse is pointing. The method I'm using to do this is to find the X to Y ratio based on 2 points (the center of the sprite, and the mouse position). The X to Y ratio is essentially "for every time the X changes by 1, the Y changes by __".
This is my method so far:
public static Vector2f getSimplifiedSlope(Vector2f v1, Vector2f v2) {
float x = v2.x - v1.x;
float y = v2.y - v1.y;
// find the reciprocal of X
float invert = 1.0f / x;
x *= invert;
y *= invert;
return new Vector2f(x, y);
}
This Vector2f is then passed to the bullet, which moves that amount each frame.
But it isn't working. When my mouse is directly above or below the sprite, the bullets move very fast. When the mouse is to the right of the sprite, they move very slow. And if the mouse is on the left side, the bullets shoot out the right side all the same.
When I remove the invert variable from the mix, it seems to work fine. So here are my 2 questions:
Am I way off-track, and there's a simpler, cleaner, more widely used, etc. way to do this?
If I'm on the right track, how do I "normalize" the vector so that it stays the same regardless of how far away the mouse is from the sprite?
Thanks in advance.
Use vectors to your advantage. I don't know if Java's Vector2f class has this method, but here's how I'd do it:
return (v2 - v1).normalize(); // `v2` is obj pos and `v1` is the mouse pos
To normalize a vector, just divide it (i.e. each component) by the magnitude of the entire vector:
Vector2f result = new Vector2f(v2.x - v1.x, v2.y - v1.y);
float length = sqrt(result.x^2 + result.y^2);
return new Vector2f(result.x / length, result.y / length);
The result is unit vector (its magnitude is 1). So to adjust the speed, just scale the vector.
Yes for both questions:
to find what you call ratio you can use the arctan function which will provide the angle of of the vector which goes from first object to second object
to normalize it, since now you are starting from an angle you don't need to do anything: you can directly use polar coordinates
Code is rather simple:
float magnitude = 3.0; // your max speed
float angle = Math.atan2(y,x);
Vector2D vector = new Vector(magnitude*sin(angle), magnitude*cos(angle));

2d rotation around a point using Canvas and Path

The issue involves an Android Path shape. It's a triangle that I'm using as an arrow to point towards objects on a screen Canvas. This is for a 2d game. player in the middle of the screen, objects around him and offscreen.
These arrows are supposed to rotate around the center of the screen, with a radius so that they rotate in a circle around the player. The arrows point towards objects that the player needs to move towards.
What I have right now is somewhat working, but the arrows are zipping around the circle at ridiculous speeds. Funny enough, they're pointing in the right direction, but they aren't staying at the right point on the circle. (if arrow is pointing northeast, arrow should be on the northeast part of the circle, etc)
I'm sure it's because of the math. I'm probably using atan2 wrong. Or canvas.translate wrong. Or maybe I shouldn't be using atan2 at all. Help! :)
Here is the code:
// set the shape of our radar blips
oBlipPath.moveTo(0, -5);
oBlipPath.lineTo(5, 0);
oBlipPath.lineTo(0, 5);
// Paint all the enemies and radar blips!
for(int i=0; i<iNumEnemies; i++){
if (oEnemies[i].draw(canvas, (int)worldX, (int)worldY)){
//calculate the degree the object is from the center of the screen.
//(user is the player. this could be done easier using iWidth and iHeight probably)
//we use a world coordinate system. worldY and worldX are subtracted
fDegrees = (float)Math.toDegrees(Math.atan2((oEnemies[i].getEnemyCenterY()-worldY)-user.getShipCenterY(), (oEnemies[i].getEnemyCenterX()-worldX)-user.getShipCenterX()));
canvas.save();
//get to the center
canvas.translate((iWidth / 2) , (iHeight / 2) );
//move a little bit depending on direction (trying to make arrows appear around a circle)
canvas.translate((float)(20 * Math.cos(fDegrees)), (float)(20* Math.sin(fDegrees)));
//rotate canvas so arrows will rotate and point in the right direction
canvas.rotate(fDegrees);
//draw arrows
canvas.drawPath(oBlipPath, oBlipPaint);
canvas.restore();
}
}
Affine transformations are are not commutative. They are typically applied in an apparent last-specified-first-applied order. As an alternative, consider the rotate() variation that rotates about a point.
Well, I've got it doing what I wanted, but I don't really know how. I threw in some random numbers until things showed up on the screen the way I wanted. If anyone wants to clue me in as to a better way to do this, I'm all ears.
The code:
// set the shape of our radar blips
oBlipPath.moveTo(0, -5);
oBlipPath.lineTo(6, 0);
oBlipPath.lineTo(0, 5);
oBlipMatrix.setRotate(45, 0, 0);
oBlipPath.transform(oBlipMatrix);
// Paint all the enemies and radar blips!
for(int i=0; i<iNumEnemies; i++){
oEnemies[i].draw(canvas, (int)worldX, (int)worldY);
if (oEnemies[i].bActive){
//calculate the degree the object is from the center of the screen.
//(user is the player. this could be done easier using iWidth and iHeight probably)
//we use a world coordinate system. worldY and worldX are subtracted
fDegrees = (float)Math.toDegrees(Math.atan2((oEnemies[i].getEnemyCenterY()-worldY)-(iHeight / 2), (oEnemies[i].getEnemyCenterX()-worldX)-(iWidth / 2)));
canvas.save();
//get to the center
canvas.translate((iWidth / 2 + 50) , (iHeight / 2 + 50) );
//move a little bit depending on direction (trying to make arrows appear around a circle)
//canvas.translate((float)(20 * Math.cos(fDegrees)), (float)(20* Math.sin(fDegrees)));
//rotate canvas so arrows will rotate and point in the right direction
canvas.rotate(fDegrees-45, -50, -50);
//draw arrows
canvas.drawPath(oBlipPath, oBlipPaint);
canvas.restore();
}
}
For whatever reason, I have to subtract 45 degrees from the canvas rotation, but add 45 degrees to the matrix rotation of the path shape. It works, but why?! :)

How to position a Node along a circular orbit around a fixed center based on mouse coordinates (JavaFX)?

Im trying to get into some basic JavaFX game development and I'm getting confused with some circle maths.
I have a circle at (x:250, y:250) with a radius of 50.
My objective is to make a smaller circle to be placed on the circumference of the above circle based on the position of the mouse.
Where Im getting confused is with the coordinate space and the Trig behind it all.
My issues come from the fact that the X/Y space on the screen is not centered at 0,0. But the top left of the screen is 0,0 and the bottom right is 500,500.
My calculations are:
var xpos:Number = mouseEvent.getX();
var ypos:Number = mouseEvent.getY();
var center_pos_x:Number = 250;
var center_pos_y:Number = 250;
var length = ypos - center_pos_y;
var height = xpos - center_pos_x;
var angle_deg = Math.toDegrees(Math.atan(height / length));
var angle_rad = Math.toRadians(angle_deg);
var radius = 50;
moving_circ_xpos = (radius * Math.cos(angle_rad)) + center_pos_x;
moving_circ_ypos = (radius * Math.sin(angle_rad)) + center_pos_y;
I made the app print out the angle (angle_deg) that I have calculated when I move the mouse and my output is below:
When the mouse is (in degrees moving anti-clockwise):
directly above the circle and horizontally inline with the center, the angle is -0
to the left and vertically centered, the angle is -90
directly below the circle and horizontally inline with the center, the angle is 0
to the right and vertically centered, the angle is 90
So, what can I do to make it 0, 90, 180, 270??
I know it must be something small, but I just cant think of what it is...
Thanks for any help
(and no, this is not an assignment)
atan(height/length) is not enough to get the angle. You need to compensate for each quadrant, as well as the possibility of "division-by-zero". Most programming language libraries supply a method called atan2 which take two arguments; y and x. This method does this calculation for you.
More information on Wikipedia: atan2
You can get away without calculating the angle. Instead, use the center of your circle (250,250) and the position of the mouse (xpos,ypos) to define a line. The line intersects your circle when its length is equal to the radius of your circle:
// Calculate distance from center to mouse.
xlen = xpos - x_center_pos;
ylen = ypos - y_center_pos;
line_len = sqrt(xlen*xlen + ylen*ylen); // Pythagoras: x^2 + y^2 = distance^2
// Find the intersection with the circle.
moving_circ_xpos = x_center_pos + (xlen * radius / line_len);
moving_circ_ypos = y_center_pos + (ylen * radius / line_len);
Just verify that the mouse isn't at the center of your circle, or the line_len will be zero and the mouse will be sucked into a black hole.
There's a great book called "Graphics Gems" that can help with this kind of problem. It is a cookbook of algorithms and source code (in C I think), and allows you to quickly solve a problem using tested functionality. I would totally recommend getting your hands on it - it saved me big time when I quickly needed to add code to do fairly complex operations with normals to surfaces, and collision detections.

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