I want to move a group of Actors by panning.
#Override
public boolean pan(float x, float y, float deltaX, float deltaY) {
Gdx.app.debug(TAG," Delta " + deltaX * screen.ppu + " old x "+ screen.getLevelGroup().getX());
float newX = screen.getLevelGroup().getX() + deltaX * screen.ppu;
Gdx.app.debug(TAG,"new X " + newX);
screen.getLevelGroup().setX(newX);
return true;
}
Sometimes this doesn't work. It seems the X property of group is changed between getting and setting. Please note that it is not happening all of the time.
The bugous log file is as follow:
StageSelectorGestureDetector: Delta 7.5117373 old x -639.64
StageSelectorGestureDetector: new X -632.1283
StageSelectorGestureDetector: Delta 7.5117373 old x -639.566
StageSelectorGestureDetector: new X -632.05426
StageSelectorGestureDetector: Delta 9.765259 old x -639.5344
StageSelectorGestureDetector: new X -629.76917
StageSelectorGestureDetector: Delta 5.258216 old x -639.5344
Notice that even though I am setting the X, reading it on the next invocation of pan reports the old value.
Can some one help me here?
The problem was that an Action was still active when panning, so two method was competing to set the coordination of the group, hence the strange behaviour.
Related
This video will show my current situation, and I currently can't find any answers to it online.
https://www.youtube.com/watch?v=O8Mh-1Emoc8&feature=youtu.be
My Code:
public Vector3D pickBlock() {
glDisable(GL_TEXTURE);
IntBuffer viewport = BufferUtils.createIntBuffer(16);
FloatBuffer modelview = BufferUtils.createFloatBuffer(16);
FloatBuffer projection = BufferUtils.createFloatBuffer(16);
FloatBuffer winZ = BufferUtils.createFloatBuffer(1);
float winX, winY;
FloatBuffer position = BufferUtils.createFloatBuffer(3);
glGetFloat(GL_MODELVIEW_MATRIX, modelview);
glGetFloat(GL_PROJECTION_MATRIX, projection);
glGetInteger(GL_VIEWPORT, viewport);
winX = (float)Display.getWidth() / 2;
winY = (float)viewport.get(3) - (float)Display.getHeight() / 2;
glReadPixels(Display.getWidth() / 2, (int)winY, 1, 1, GL_DEPTH_COMPONENT, GL_FLOAT, winZ);
gluUnProject(winX, winY, winZ.get(), modelview, projection, viewport, position);
glEnable(GL_TEXTURE);
return new Vector3D(position.get(0) / 2 + 0.5f, position.get(1) / 2 + 0.5f, position.get(2) / 2 + 0.5f);
}
It returns "/ 2 + 0.5f" because that is needed because of the offsets I have for the blocks (if I removed the 0.5f, the offset would be in the center instead of the corner)
I seams to me that the error, based on the video, comes from when you are facing in the positive z direction (or whatever your back direction is). My guess is that you aren't taking the facing direction into account as I see in your code that you are just adding a constant 0.5F to the position of your cursor.
Therfore, when you are facing backwards, it adds 0.5 which makes it be behind the wall (since back is negative Z). one simple check would be weather the Z component of your forward vector is positive or negative, and deciding the factor added to the cursor based on that, then doing the same for the X.
Depending on how you implemented your camera (IE: if you used Euler angles (rx, ry, rz) or if you used Quaternions / forward vectors), the way you would do that check would vary, feel free to ask me for examples based on your system if you need.
hope this helped!
PS: if you're using angles, you can either check for the range of the y-axis rotation value and determine which direction you are facing and thus weather to add or subtract, OR you can calculate the forward vector based on your angles, and then check the for sign of the component.
I'm working on a new game written with LibGdx Engine and Java.
I've got a problem with some of the physics in this game.
I want to shoot the arrow in a ballistic trajectory (angry bird style)
and can't find the equation to do so .
I am using these velocity equations:
float velx = (float) (Math.cos(rotation) * spd);
float vely = (float) (Math.sin(rotation) * spd);
I add this to the current position and the arrow shoots in one direction - straight.
I thought maybe changing the rotation would help me achieve what I want (a ballistic path).
It does help, but I want to have the trajectory as well.
I saw this
ProjectileEquation class that someone already posted but didn't know how to work with it:
public class ProjectileEquation
{
public float gravity;
public Vector2 startVelocity = new Vector2();
public Vector2 startPoint = new Vector2();
public Vector2 gravityVec = new Vector2(0,-10f);
public float getX(float n) {
return startVelocity.x * (n ) + startPoint.x;
}
public float getY(float n) {
float t = n;
return 0.5f * gravity * t * t + startVelocity.y * t + startPoint.y;
}
}
I'm looking for some help to help me use this class for ballistic trajectories.
This is how I tried using it:
for(int i =0;i<30;i++)
{
Texture f = ResData.Square_1;
ProjectileEquation e= new ProjectileEquation();
e.gravity = 1;
e.startPoint = new Vector2(bow.getX(),bow.getY());//new Vector2(-bow.getX(),-bow.getY()); //My bow is opposite so it suppose to work fine
e.startVelocity = getVelocityOf(bow.getRotation());
Vector3 touchpos = new Vector3();
s.draw(f,e.getX(i) ,e.getX(i),5,5);
}
The ProjectileEquation class you post looks like it'll calculate the X and Y position given a time delta, so the float you pass in should be the time delta since you started the arrow moving (in seconds).
That code will not give you the angle of the arrow though. To find that, I would suggest you keep hold of the previous X and Y, then you can use Math.atan2() to calculate the angle based on the previous XY and the current XY. Google atan2 for a load of info on how to use it.
The very best way to do this however would be to use Box2d and model the scene correctly. Then you wouldn't have to get involved in the maths at all. I read somewhere that that's what Angry Birds uses, and is an excellent choice for modelling these sorts of physics games.
I hope your game goes well.
I have been spending several days trying to solve this problem and I am about to go mad. Your help would be very much appreciated.
I am using a 2D stage in libgdx. I want to move actors (or sprite) to this stage with a "gravity" display effect: for example for actor1, his initial coordinates are (0, 0), destination coordinates would be (100, 50), and I want to move this actor to this destination with a gravity trajectory effect. Then I want to use the same gravity for actor2 moving from (0, 0) to (25, 75), then actor3 from (0, 0) to (200, 75) etc.
I managed to apply a gravity trajectory to any actor based on this well known loop:
setX(getX() + velocity.x);
setY(getY() + velocity.y);
velocity.y += getGravity().y * delta;
So tweaking the gravity value would modify the trajectory. It works fine.
Now, as I said earlier I want to give every actors a unique trajectory given their predetermined destination.
So I have tried to find a formula to determine the x and y for each actor at every frame of their trajectory
I am using the following static parameters:
gravity.y : same for all actors
delay : the amount of frames during which each actor moves between his initial coordinates and his destination coordinates. Same value for all actors too
First I calculate the velocity with this SUPER UGLY formula that I am absolutely not proud of:
velocity = new Vector2 ( (destinationx - b.getX() )/time, initdisty/time + ( Math.sqrt(delta*1000)*time / ( 500/Math.abs(gravity) ) ));
where delta = Gdx.graphics.getDeltaTime();
Then I apply this velocity each frame to calculate the x and y of each actor:
public void act(float delta){
for (int i=0; i<delay; i++) {
setX(getX() + velocity.x);
setY(getY() + velocity.y);
velocity.y += gravity.y * delta;
}
}
It KIND OF work, but of course, this can not be a long term solid solution. Calculating the x and y for each frame for each actor (there can be 5-6 actors moving at the same time) doesn't look good at all.
The main problem is that the trajectories are good on computer with consistent 60FPS, they are okay on a tablet, but on a phone with limited memory and 30 < fps < 60, the trajectories become terribly wrong.
After reading several blog posts, it seems like I could avoid the multi device memory fps problems by removing the delta parameter from my formulas, but I haven't found how. And it still doesn't give me a strong long term solution to calculate the trajectory with predetermined destination coordinates.
Thanks for reading and for your time, please let me know if this is unclear I'll do my best to explain better.
Cause of the problem
Maintaining both position and velocity leads to discretization of the system resulting in quantization error. So you will experience inconsistent behavior by your current method when fps fluctuates.
Solution
All you need is to reduce your number of state variables to only two i.e. don't store current velocity. It is causing the errors in final position.
In stead use parametric form of the trajectory.
v = u + at
and
s = ut + ½at²
Implementation
Suppose you want to go from (sourceX, sourceY) to (targetX, targetY) in time 'totalTime'.
Calculate initial velocity.
Vector2 initialVelocity = new Vector2((targetX - sourceX) / totalTime,
(targetY - sourceY) / totalTime - gravity * totalTime / 2);
float currentTime = 0;
In each iteration, calculate position directly and keep track of currentTime.
public void act(float delta){
if (currentTime < totalTime) {
currentTime += delta;
setX((initialVelocity.x + gravity.x * currentTime / 2) * currentTime);
setY((initialVelocity.y + gravity.y * currentTime / 2) * currentTime);
} else {
setX(targetX);
setY(targetY);
}
}
Multiplying two numbers in log space means adding them:
log_multiply(x, y) = log( exp(x) * exp(y) )
= x + y
Adding two numbers in log space means you do a special log-add operation:
log_add(x, y) = log( exp(x) + exp(y) )
which is implemented in the following code, in a way that doesn't require us to take the two exponentials (and lose runtime speed and precision):
double log_add(double x, double y) {
if(x == neginf)
return y;
if(y == neginf)
return x;
return max(x, y) + log1p(exp( -fabs(x - y) ));
}
(Here is another one.)
But here is the question:
Is there a trick to do it for subtraction as well?
log_subtract(x, y) = log( exp(x) - exp(y) )
without having to take the exponents and lose precision?
double log_subtract(double x, double y) {
// ?
}
How about
double log_subtract(double x, double y) {
if(x <= y)
// error!! computing the log of a negative number
if(y == neginf)
return x;
return x + log1p(-exp(y-x));
}
That's just based on some quick math I did...
The library functions for exp and log lose precision for extreme values.
log1p gets you half way there, but what you need is a function that treats the error for both the log and the exp parts.
See this article: http://cran.r-project.org/web/packages/Rmpfr/vignettes/log1mexp-note.pdf
The title is "Accurately Computing log(1 - exp(-|a|))".
The article discusses how to seemlessly merge different algorithms to create good error bounds for a larger range of inputs.
I have no clue about trigonometry, despite learning it in school way back when, and I figure this should be pretty straightforward, but trawling through tons of trig stuff on the web makes my head hurt :) So maybe someone could help me...
The title explains exactly what I want to do, I have a line:
x1,y1 and x2,y2
and want a function to find x3,y3 to complete an isosceles triangle, given the altitude.
Just to be clear, the line x1,y2 -> x2,y2 will be the base, and it will not be aligned any axis (it will be at a random angle..)
Does anyone have a simple function for this??
construct a normal to the vector (x1,y1)->(x2,y2). place it at the midpoint ((x1+x2)/2,(y1+y2)/2) and go out a distance h.
the normal will look like (-(y2-y1),x2-x1). make this a unit vector (http://en.wikipedia.org/wiki/Unit_vector).
add h times this unit vector to the midpoint.
The third point is on the perpendicular bisector of your base, and is altitude units away from the line.
Calculate the midpoint of the base by averaging the x and y coordinates.
Calculate the slope of your altitude: -dx/dy (perpendicular to dy/dx). You now have your line (point and slope).
y - my = -dx/dy * (x - mx)
Substitute your variables in the distance formula: d = sqrt(dx^2 + dy^2)
d = sqrt((x - mx)^2 + (y - my)^2)
d = sqrt((x - mx)^2 + (-dx/dy * (x - mx))^2)
d^2 = (x - mx)^2 + (-dx/dy * (x - mx))^2
d^2 - (x - mx)^2 = (-dx/dy * (x - mx))^2
±sqrt(d^2 - (x - mx)^2) = -dx/dy * (x - mx)
±sqrt(d^2 - (x - mx)^2) * dy/dx = x - mx
±sqrt(d^2 - (x - mx)^2) * dy/dx + mx = x
x = ±sqrt(d^2 - (x - mx)^2) * dy/dx + mx
Calculate the other variable (y here) using your line equation (from #2).
You now have two points; pick whichever you want...
In pseudocode:
dx = x1 - x2
midpoint = ((x1 + x2) / 2, (y1 + y2) / 2)
slope = -dx / (y1 - y2)
x = sqrt(altitude*altitude - dx*dx) / slope + midpoint.x
y = slope * (x - midpoint.x) + midpoint.y
This is probably not the most optimal method. Not sure if it even works. xD
Al I can remember is that an isosceles triangle will have sides of equal length, and equal angles at the base. If you have the height, then you have the final coordinate because this will be the point of intersection, right?