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!
Related
I create a 2d game similar to a classic diepio. I created a system for positioning the player's barrel in a specific direction. The updated angle is sent to the server. When the player clicks, the server creates a missile. This only works correctly when the barrel is attached to the center of the player's body. When I want to move the barrel away from the center of the body, there is a problem. I don't know how to update the server-side position where the projectile spawns.
In the image below, the barrel rotates around the center of the player's body. I marked the missile's flight path with the red line.
On this image, the barrels also have a rotation axis in the player's center, but have been shifted to the side. The green line marked the route the missile should take. Unfortunately, I don't know how to do it correctly.
How to update the projectile's spawn point by a given distance (e.g. 10) from the basic distance (if the barrel was not moved) based on the player's angle of rotation and his position?
Projectile spawn method:
float angle = (float) ((player.getRotation() + 90) * Math.PI / 180f);
float forceX = (float) Math.cos(angle);
float forceY = (float) Math.sin(angle);
spawnProjectile(player.getPosition().x + (forceX * 3f), player.getPosition().y + (forceY * 3f));
If I understood your question correctly, you want to find the two points marked in orange in the following image:
Since you know the direction in which the missiles should fly (the red line), the distance from the center position (e.g. 10) and you know that there is a 90° angle between the movement vector of the missile (red line) and the connection line between the two starting positions of the missiles (marked as black line in the image) you can calculate the resulting positions like this:
float angle = (float) ((player.getRotation() + 90) * Math.PI / 180f);
float forceX = (float) Math.cos(angle);
float forceY = (float) Math.sin(angle);
// the center point (start of the red line in the image)
float centerX = player.getPosition().x + (forceX * 3f);
float centerY = player.getPosition().y + (forceY * 3f);
float offsetFromCenterDistanceFactor = 1f; // increase this value to increase the distance between the center of the player and the starting position of the missile
// the vector towards one of the starting positions
float offsetX1 = forceY * offsetFromCenterDistanceFactor;
float offsetY1 = -forceX * offsetFromCenterDistanceFactor;
// the vector towards the other starting position
float offsetX2 = -offsetX1;
float offsetY2 = -offsetY1;
//spawn the upper missile
spawnProjectile(centerX + offsetX1, centerY + offsetY1);
//spawn the lower missile
spawnProjectile(centerX + offsetX2, centerY + offsetY2);
For more detail on the calculation of the orthogonal vectors see this answer.
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());
The issue I have is that I'm attempting to add drag to an object in this basic physics simulation (Java [Processing]), but once I add the appropriate formula, it causes the objects velocity to increase drastically in the opposite direction. Of course the problem is that drag for some reason is being calculated too high but I'm not sure why thats happening as I'm using the real world equation.
void setup(){size(1280,720);}
class Circle{
float x,y,r,m,dx,dy,ax,ay,fx,fy;
Circle(float xPos, float yPos, float Radius, float Mass){
x = xPos;
y = yPos;
r = Radius;
m = Mass;
}
void ADD_DRAG(){
fx -= 0.5 * 1.225 * dx * dx * 0.5 * r * PI;
fy -= 0.5 * 1.225 * dy * dy * 0.5 * r * PI;
}
void update(){
ADD_DRAG();
ax = fx / m;
ay = fy / m;
dx += ax / frameRate;
dy += ay / frameRate;
x += dx / frameRate;
y += dy / frameRate;
}
}
Circle[] SceneObjects = {new Circle(50,50,20,20000),new Circle(50,50,2,20)};
void draw(){
background(51);
for (Circle c : SceneObjects){
c.update();
circle(c.x * 3,c.y * 3,c.r * 3);
}
}
void mouseClicked(){
if(SceneObjects[1].fx != 2000)
SceneObjects[1].fx = 2000;
else
SceneObjects[1].fx = 0;
}
This is the code, essentially there is a Circle class which stores the objects properties and then the forces applies are updated each draw loop. The mouseClicked void is just for testing by adding a force to the objects. All and any help is appreciated, thanks!
Maths I am Using:
Rearranged F=ma for ax = fx / m;
Acceleration * time = Speed for dx += ax / frameRate; (frameRate is 1/time)
Distance = Speed * time = for x += dx / frameRate; (same as above)
For drag im using this equation https://www.grc.nasa.gov/WWW/K-12/rocket/drageq.html with the constants eg air density etc added as seen.
There are a couple of things wrong here.
You haven't given us numbers (or a minimal complete example), but the vector algebra is off.
Yes, the acceleration is f = -kv2, and |v|2 = vx2 + vy2, but that doesn't mean that you can decompose f into fx=kvx2 and fy=kvy2. Not only is your magnitude off, but your acceleration is now not (in general) aligned with the motion; the path of your projectile will tend to curve toward a diagonal between the axes (e.g. x=y).
Also, your code always gives acceleration in the negative x and negative y directions. If your projectile happens to start out going that way, your version of air resistance will speed it up.
Finally, your time interval may simply be too large.
There is a better way. The differential equation is v' = -k v|v|, and the exact solution is v = (1/kt) z, (with appropriate choice of the starting time) where z is the unit direction vector. (I don't know how to put a caret over a letter.) This leads to v(t) = (1/t)v(t=1.0)
So you can either work out a fictional time t0 and calculate each new velocity using 1/(kt), or you can calculate the new velocity from the previous velocity: vn+1 =vn/(kd vn + 1), where d is the time interval. (And then of course you have to decompose v into vx and vy properly.)
If you're not familiar with vector algebra, this may seem confusing, but you can't get an air-resistance sim to work without learning the basics.
I am trying to generate random points on a sphere that is filled with a cube.
Because I had no idea how to do that i started with 2d. (A circle filled with a quadrat.)
What I am trying to do: Generating random points inside the outer circle, but outside the green square.
Basically in the blue areas.
The square is located at (-1|-1),(1|-1),(1|1),(-1|1).
The circle has a radius of r = sqrt(2) and is centered at (0|0).
I already have scripts to:
generate a random point on a circle (uniformly):
float a = 2 * MathUtils.PI * MathUtils.random(1f); // angle between 0 and 2pi
float r = radius * Math.sqrt(MathUtils.random(0, 1f)
float x = r * MathUtils.cos(a);
float y = r * MathUtils.sin(a);
calculating the radius for a given angle to form a square:
float r = (1/Math.sqrt(2)) / MathUtils.cos(((a+45)%90-45)/180*MathUtils.PI);
with (1/Math.sqrt(2)) being half the side length of the square
Before anyone asks:
I know that I could just re-generate points which are inside the green square until I get one that is outside, but I don't want to do it this way.
I appreciate any help. Thank you :)
It is rather hard to generate points only in region of sphere outside the cube (caps and wedges), so rejecting method looks reasonable.
But you can diminish the number of useless points, generating points in the ring only in 2D case and in spherical shell in 3D case.
So pseudocode might look as
//2d
SquaredR = RandomUniformInRange(0.5, 1)
R = Sqrt(SquaredR)
//3d
CubedR = RandomUniformInRange(Pow(3, -3/2), 1)
R = Pow(CubedR, 1/3)
//generate point on the circle or on the sphere with radius R
if Abs(x) > Sqrt(2)/2 or Sqrt(3)/3 and so on - reject
Having R, you can generate point on the sphere using any approach from here
Here is rough sketch of the idea. You select one quadrant to sample, say, one on the right.
First, sample angles from -pi/4 to pi/4
float a = -MathUtils.PI/4.0f + MathUtils.PI/2.0 * MathUtils.random(0.f,1.f);
float c = MathUtils.cos(a);
float s = MathUtils.sin(a);
Second, find minimum radius. With ray going from (0,0) at angle a will intersect quadrant line at x=1 at minimum
float rmin = 1.0f / c;
float rmax = Math.sqrt(2.0f);
Sample from rmin to rmax = sqrt(2), taking into account that for plane you sample squared radius and then use sqrt(), and for 3d space you sample cubed radius and then use cbrt().
float r2 = rmin*rmin + (rmax*rmax-rmin*rmin)*MathUtils.random(0.f,1.f);
float r = Math.sqrt(r);
float x = r * c;
float y = r * s;
Now, we constructed (x,y) is a such way it is guaranteed to be in the right quadrant below circle and on the right of the x=1 line.
To cover all four quadrants just sample to which quadrant you will move the point
float q = MathUtils.random(0.f,1.f);
if (q < 0.25f) // top quadrant
return (y, x);
if (q < 0.5f) // left quadrant
return (-x, y);
if (q < 0.75f) // bottom quadrant
return (y, -x);
return (x,y); // right quadrant
Please bear with me - my Java is quite rusty, and I have no ways to test the code.
In 3D case you'll have to deal with two angles, cubed radius, eight octants instead of four quadrants, but general logic is the same
UPDATE
I was wrong, sampling like I propose would lead to non-uniform point distribution.
From PDF:
PDF(phi, r) = S_(-\pi/4)^\phi d\phi S_1/cos(\phi)^\sqrt(2) r dr
One could get that we have to make \phi sampling non-uniform. Unfortunately, from
U(0,1) to get to sampled \phi requires solving non-linear equation
\pi/2 (0.5*(\phi/\pi/4 + 1) - U(0,1)) = 0.5*(tan(phi) + 1) - U(0,1)
So algorithm would be:
Sample U(0,1)
Find appropriate \phi by solving equation above
Find lower R boundary
Sample R
Quick code (in Python, sorry) to plot this non-linear function
import numpy as np
import matplotlib.pyplot as plt
def f(phi, ksi):
c = 0.5 * np.pi
c_2 = 0.5 * c
left = c * (0.5 * (phi/c_2 + 1.0) - ksi)
rght = (0.5 * (np.tan(phi) + 1.0) - ksi)
return left - rght
nof_points = 41
phi = np.linspace(-0.25*np.pi, 0.25*np.pi, nof_points)
y0_00 = f(phi, 0.00)
y0_25 = f(phi, 0.25)
y0_50 = f(phi, 0.50)
y0_75 = f(phi, 0.75)
y0_99 = f(phi, 1.00)
plt.plot(phi, y0_00, 'ro', phi, y0_25, 'b+', phi, y0_50, 'gx', phi, y0_75, 'm.', phi, y0_99, 'y^')
plt.show()
and plotted functions for five values of U(0,1) (ksi in the code)
Sampling could be rearranged such that r sampling is non-linear, but it exhibits the same problem - need to solve non-linear equation with polynomial and trigonometric parts
UPDATE II
And just for the record, if you want to sample r first, then it has to be sampled from the solution of the non-linear equation:
r2 sec-1(r) - sqrt(r2 - 1) = U(0,1)*(\pi/2 - 1)
in the interval [1...sqrt(2)]
After solving it and finding sampled r, \phi could be sampled uniformly in the interval allowed by r: [-cos-1(1/r) ... +cos-1(1/r)]
Let's say I have an (x,y) that is always the same for the start point of a line and an (x,y) that changes for the end point of that same line. The line is also always 40px long. At the start of the program the line originates in a vertical orientation (lets call it 0 degrees). Based on a user input I need the line to be redrawn a specific number of degrees from its origin by changing only the end (x,y).
SOME MORE FOOD FOR THOUGHT IF YOU NEED IT:
I'm in a rut trying to calculate this and make it work in Java. I can make the math work to calculate the point based on the arc length of a circle segment, but I don't know how to make Java do it.
I think it would work easier based off a triangle angles since I will always know the length of two sides of a triangle (one side formed by the 40px long line and the other side formed by the start point of that line and the border of the JPanel) and the angle those two lines form. Still, my brain is mush from trying to figure it out. Any help would be much appreciated.
UPDATE:
#casablanca got me on the right track. I brushed up on my trig functions and here is how I made it work.
First off, I didn't realize that 90 degrees was straight up, but once I did realize that I made my solution reflect that fact. I was drawing my line starting at the bottom center of the frame going out. Since the opposite side of the triangle is on the right side of the screen when the angle given by my user is less than 90 degrees and is on the left side of the screen when the angle given by my user is greater than 90 degrees I had to adjust the formula to account for that fact, thus I have four methods, one for the x coordinate on the left side of the screen (when the user given angle is greater than 90 degrees), one for the y coordinate on the left side of the screen (when the user given angle is greater than 90 degrees) and the same thing for the right side of the screen when the user given angle is less than 90 degrees. The int length in all methods is the length of the hypotenuse. Thanks again for your help #casablanca!
public double leftSideX(double angle, int length){
double x = frameWidth/2 - (length * Math.cos(Math.toRadians(90-(Math.toDegrees(angle)-90))));
return x;
}
public double leftSideY(double angle, int length){
double y = frameHeight - (length * Math.sin(Math.toRadians(90-(Math.toDegrees(angle)-90))));
return y;
}
public double rightSideX(double angle, int length){
double x = frameWidth/2 + (length * Math.cos(angle));
return x;
}
public double rightSideY(double angle, int length){
double y = frameHeight - (length * Math.sin(angle));
return y;
}
Is this what you're looking for?
startX = x;
startY = y;
endX = x + 40 * Math.sin(angle);
endY = y + 40 * Math.cos(angle);
And draw a line from (startX, startY) to (endX, endY) in whatever API you're using.
Also note that angle is in radians. If you had it in degrees, you need to convert it first:
angle = angle * Math.PI / 180;