I'm creating a racing car game and am having trouble figuring out how to get steering to work. I have a basic race 2D race course that is built in a 3D environment. The program only uses x and y, with z being 0.
My race course consists of a road that is 29 units wide in the x-axis and two long tracks that are 120 units long in the y direction. At 120 units in the y-axis there is a 180 degree turn. You can think of the course as looking similar to a nascar styled race course.
I'm trying to set my car's steering so that it can turn realistically when I reach the 180 degree turns. I'm using two variables that separately control the x / y positions, as well as two variables for the x / y velocities. My code at the moment is as follows:
public void steering(){
double degreesPerFrame = 180 / (2*30); //180 degrees over 30 fps in 2s
velocityX = velocityX * -1 * Math.cos(degreesPerFrame);
velocityY = velocityY * -1 * Math.sin(degreesPerFrame);
double yChange = Math.sin(degreesPerFrame) * velocityY;
double xChange = Math.cos(degreesPerFrame) * velocityX;
x += xChange; //change x position
y += yChange; //change y position
}
I'm not completely sure how I can get my steering to turn properly. I'm stuck at the moment and not sure what I would need to change in my function to get steering working properly.
I think this would be easier if you don't use angles at all for you calculations. Simply operate with position, velocity, and acceleration vectors.
First, a quick reminder from physics 101. With a small time step dt, current position p, and current velocity vector v, and an acceleration vector a, the new position p' and velocity vector v' are:
v' = v + a * dt
p' = p + v' * dt
The acceleration a depends on the driver input. For example:
When moving ahead at a constant speed, it is 0.
When accelerating, it is limited by engine power and tire grip in longitudinal direction.
When turning, it is limited by tire grip in lateral direction.
When braking, it is limited by tire grip (mostly, the brakes are typically strong enough to not be a limit).
For a relatively simple model, you can assume that grip in longitudinal and lateral direction are the same (which is not entirely true in reality), which limits the total acceleration to a vector within a circle, which is commonly called the friction circle.
That was mainly just background. Back to the more specific case here, which is turning at a steady speed. The result of steering the car can be modeled by a lateral force, and corresponding acceleration. So in this case, we can apply an acceleration vector that is orthogonal to the current velocity vector.
With the components of the velocity vector:
v = (vx, vy)
a vector that points orthogonally to the left (you mentioned NASCAR, so there's only left turns...) of this is:
(-vy, vx)
Since we want to control the amount of lateral acceleration, which is the length of the acceleration vector, we normalize this vector, and multiply it by the maximum acceleration:
a = aMax * normalize(-vy, vx)
If you use real units for your calculations, you can apply a realistic number for the maximum lateral acceleration aMax. Otherwise, just tweak it to give you the desired maneuverability for the car in your artificial units.
That's really all you need. Recapping the steps in code form:
// Realistic value for sports car on street tires when using
// metric units. 10 m/s^2 is slightly over 1 g.
static const float MAX_ACCELERATION = 10.0f;
float deltaTime = ...; // Time since last update.
float accelerationX = -veclocityY;
float accelerationY = velocityX;
float accelerationScale = MAX_ACCELERATION /
sqrt(accelerationX * accelerationX + accelerationY * accelerationY);
accelerationX *= accelerationScale;
acceleration *= accelerationScale;
velocityX += accelerationX * deltaTime;
velocityY += accelerationY * deltaTime;
x += velocityX * deltaTime;
y += velocityY * deltaTime;
Related
My gravity simulation acts more like a gravity slingshot. Once the two bodies pass over each other, they accelerate far more than they decelerate on the other side. It's not balanced. It won't oscillate around an attractor.
How do other gravity simulators get around it? example: http://www.testtubegames.com/gravity.html, if you create 2 bodies they will just oscillate back and forth, not drifting any further apart than their original distance even though they move through each other as in my example.
That's how it should be. But in my case, as soon as they get close they just shoot away from each other to the edges of the imaginary galaxy never to come back for a gazillion years.
edit: Here is a video of the bug https://imgur.com/PhhRhP7
Here is a minimal test case to run in processing.
//Constants:
float v;
int unit = 1; //1 pixel = 1 meter
float x;
float y;
float alx;
float aly;
float g = 6.67408 * pow(10, -11) * sq(unit); //g constant
float m1 = (1 * pow(10, 15)); // attractor mass
float m2 = 1; //object mass
void setup() {
size (200,200);
a = 0;
v = 0;
x = width/2; // object x
y = 0; // object y
alx = width/2; //attractor x
aly = height/2; //attractor y
}
void draw() {
background(0);
getAcc();
applyAcc();
fill(0,255,0);
ellipse(x, y, 10, 10); //object
fill(255,0,0);
ellipse(alx, aly, 10, 10); //attractor
}
void applyAcc() {
a = getAcc();
v += a * (1/frameRate); //add acceleration to velocity
y += v * (1/frameRate); //add velocity to Y
a = 0;
}
float getAcc() {
float a = 0;
float d = dist(x, y, alx, aly); //distance to attractor
float gravity = (g * m1 * m2)/sq(d); //gforce
a += gravity/m2;
if (y > aly){
a *= -1;}
return a;
}
Your distance doesn't include width of the object, so the objects effectively occupy the same space at the same time.
The way to "cap gravity" as suggested above is add a normal force when the outer edges touch, if it's a physical simulation.
You should get into the habit of debugging your code. Which line of code is behaving differently from what you expected?
For example, if I were you I would start by printing out the value of gravity every time you calculate it:
float gravity = (g * m1 * m2)/sq(d); //gforce
println(gravity);
You'll notice that your gravity value skyrockets as your circles get closer to each other. And this makes sense, because you're dividing by sq(d). Ad d gets smaller, your gravity increases.
You could simply cap your gravity value so it doesn't go off the charts anymore:
float gravity = (g * m1 * m2)/sq(d);
if(gravity > 100){
gravity = 100;
}
Alternatively you could cap d so it never goes below a certain value, but the result is the same.
In the end you'll find that this is not going to be as easy as you expected. You're going to have to tune the parameters quite a bit so your simulation works how you want.
Working demo here: https://beta.observablehq.com/#shaunlebron/1d-gravity
I followed the solution posted by the author of the sim that inspired this question here:
-First off, shrinking the timestep is always helpful. My simulation runs, as a baseline, about 40 ‘steps’ per frame, and 30 frames per second.
-To deal with the exact issue you talk about, I think modeling the bodies not as pure point masses - but rather spherical masses with a certain radius will be vital. That prevents the force of gravity from diverging to infinity. So, for instance, if you drop an asteroid into a star in my simulation (with collisions turned off), the force of gravity will increase as the asteroid gets closer, up until it reaches the surface of the star, at which point the force will begin to decrease. And the moment it’s at the center of the star (or nearby), the force will be zero (or nearly zero) - instead of near-infinite.
In my demo, I just completed turned off gravity when two objects are close enough together. Seems to work well enough.
I am currently creating a 2D space game in Java, in which you control a ship in, well, space.
The game does not use any external libraries.
The ship is supposed to move towards the cursor. However, when moving the cursor, the old force does not magically disappear; the ship changes its course, over time, to eventually move in the desired direction.
However, I have run into an issue regarding the movement of the ship.
Basically, what I want to achieve is crudely illustrated by this image:
The image shows how the ship is supposed to move during one game tick.
To explain further:
The Ship's max speed is illustrated by the circle.
The Target Angle is where the cursor currently is.
The Current Angle is the direction that the ship is currently traveling.
The Current Angle should move closer and closer to the Target Angle until it reaches the point where these two angles are the same.
The ship should change direction toward the target angle taking the shortest route possible; it can turn both left and right, not just left or right.
Now I've explained what I want to achieve, now I will instead describe what I so far have achieved and how it works.
Basically, the "ship" is an image sitting in the center of the screen. When you "move" the ship, the ship stays put; what moves is the rest of the play area.
The current "position" of the ship relative to the coordinate system that represents the play area are the integers xPos and yPos.
Now for some sample code that shows how the system works:
int xPos;
int yPos;
public void updateMovement() {
xPos += xSpeed;
yPos += ySpeed;
}
public void moveForward() {
double yTempSpeed = ySpeed;
double xTempSpeed = xSpeed;
yTempSpeed += 0.01 * Math.sin(Math.toRadians(targetAngle));
xTempSpeed += 0.01 * Math.cos(Math.toRadians(targetAngle));
double resultVector = Math.sqrt(xTempSpeed * xTempSpeed + yTempSpeed * yTempSpeed);
if (resultVector < 2) {
ySpeed += 0.01 * Math.sin(Math.toRadians(targetAngle));
xSpeed += 0.01 * Math.cos(Math.toRadians(targetAngle));
}
This code successfully sets the Ship's max speed to the desired value, however, this does not work (the ships' course does not change) in the event where the resulting "vector" is larger than 2, i.e. when the speed is already at it's maximum and the targetAngle is too close to the angle which the ship is currently traveling (+- Pi / 2).
How would I go about changing the current angle based on this implementation?
public void moveForward() {
ySpeed += 0.01 * Math.sin(Math.toRadians(targetAngle));
xSpeed += 0.01 * Math.cos(Math.toRadians(targetAngle));
double currentSpeed = Math.sqrt(xTempSpeed * xTempSpeed + yTempSpeed * yTempSpeed);
if (currentSpeed > maxSpeed) {
//the resulting speed is allways <= maxspeed (normed to that)
ySpeed *= maxSpeed/currentSpeed;
xSpeed *= maxSpeed/currentSpeed;
}
hope this is what you needed... although it is quite unrealistic, that a spacecraft has a maximum Speed, but in terms of "playability" i would do the same.
What about normalizing speed of the ship, not to let it actually exceed your speed limit (=2):
//it's good to put all constants out of a function in one place
//to make it easier if you ever wanted to change it
private final int MAX_SPEED = 2;
private final double ACCEL_FACTOR = 0.01;
public void moveForward() {
ySpeed += ACCEL_FACTOR * Math.sin(Math.toRadians(targetAngle));
xSpeed += ACCEL_FACTOR * Math.cos(Math.toRadians(targetAngle));
//normalize ship speed, i.e. preserve ratio of xSpeed/ySpeed
//but make sure that xSpeed^2 + ySpeed^2 <= MAX_SPEED^2
//your code goes here
//...
}
Read about vector normalization. This way, the changes of ship speed will be applied normally (at this moment speed can be >= MAX_SPEED), but after normalization it will never get greater than MAX_SPEED, so your if instruction is not even needed.
You may find the following (Swift) code useful, although you would need to handle the per-frame integration of the ship's linear and angular velocities yourself:
func moveShipTowards(location: CGPoint) {
if let ship = shipNode? {
let distanceVector = CGVector(origin: ship.position, point: location)
let targetAngle = distanceVector.angle
let shipAngle = ship.zRotation
var dø = targetAngle - shipAngle
// convert to shortest arc
if dø > π {
dø -= 2.0 * π
} else if dø < -π {
dø += 2.0 * π
}
// resulting angular velocity
ship.physicsBody.angularVelocity = 12 * dø.clampedTo(π)
var velocityUnitVector = CGVector(cosf(ship.zRotation), sinf(ship.zRotation))
var magnitude = distanceVector.length.clampedTo(400)
ship.physicsBody.velocity = velocityUnitVector * magnitude
}
}
It handles having the ship deccelerate as the target point is approached. Looking at it just now, it doesn't appear to handle acceleration properly though.
I'm creating a small curling/shuffleboard game in java, where I'm emphasizing on the physics.
At the moment the game can shove the curlingstone along the x-axis using the following to calculate the x position. The player can decide the initial x-velocity.
xPos = xIniVel* time - 0.5 * mu * mass * g * time* time;
I'm using a gameTimer which runs as long as the ball is in motion.
double speed = xIniVel- mu * G * mass * time;
if (speed <= 0.0) {gameTimer.stop();}
The method updateDisplay() then redraws the ball/curlingstone at the new position.
int x = (int) (xPos* 100);
int y = (int) (yPos* 100);
g.setColor(Color.green);
g.fillOval(x, y, 22, 22);
The problem I'm having is how can I make the stone anywhere else than along the x-axis? Preferably I want the player to type in an angle, but an initial y-velocity will also work.
Edit: Screenshot of the game.
By adding exactly the same code for the y velocity as you have for the x
yPos = yIniVel* time - 0.5 * mu * mass * g * time* time;
A simple way to use an initial angle and velocity would be to use trigonometry to solve for the x velocity and y velocity components.
http://www.physicsclassroom.com/class/vectors/u3l2d.cfm
This webpage provides a basic understanding of doing so. Then simply copy the code you used for the x position and replace the x initial velocity with the y initial velocity.
Also, your equation for motion seems somewhat flawed. Even if there is no initial velocity, you still end up with an increasing negative position. This is because xIniVel * time = 0, then 0 - (0.5*mu*mass*g*time^2) = -(0.5*mu*mass*g*time^2).
I have a 2D convex polygon in 3D space and a function to measure the area of the polygon.
public double area() {
if (vertices.size() >= 3) {
double area = 0;
Vector3 origin = vertices.get(0);
Vector3 prev = vertices.get(1).clone();
prev.sub(origin);
for (int i = 2; i < vertices.size(); i++) {
Vector3 current = vertices.get(i).clone();
current.sub(origin);
Vector3 cross = prev.cross(current);
area += cross.magnitude();
prev = current;
}
area /= 2;
return area;
} else {
return 0;
}
}
To test that this method works at all orientations of the polygon I had my program rotate it a little bit each iteration and calculate the area. Like so...
Face f = poly.getFaces().get(0);
for (int i = 0; i < f.size(); i++) {
Vector3 v = f.getVertex(i);
v.rotate(0.1f, 0.2f, 0.3f);
}
if (blah % 1000 == 0)
System.out.println(blah + ":\t" + f.area());
My method seems correct when testing with a 20x20 square. However the rotate method (a method in the Vector3 class) seems to introduce some error into the position of each vertex in the polygon, which affects the area calculation. Here is the Vector3.rotate() method
public void rotate(double xAngle, double yAngle, double zAngle) {
double oldY = y;
double oldZ = z;
y = oldY * Math.cos(xAngle) - oldZ * Math.sin(xAngle);
z = oldY * Math.sin(xAngle) + oldZ * Math.cos(xAngle);
oldZ = z;
double oldX = x;
z = oldZ * Math.cos(yAngle) - oldX * Math.sin(yAngle);
x = oldZ * Math.sin(yAngle) + oldX * Math.cos(yAngle);
oldX = x;
oldY = y;
x = oldX * Math.cos(zAngle) - oldY * Math.sin(zAngle);
y = oldX * Math.sin(zAngle) + oldY * Math.cos(zAngle);
}
Here is the output for my program in the format "iteration: area":
0: 400.0
1000: 399.9999999999981
2000: 399.99999999999744
3000: 399.9999999999959
4000: 399.9999999999924
5000: 399.9999999999912
6000: 399.99999999999187
7000: 399.9999999999892
8000: 399.9999999999868
9000: 399.99999999998664
10000: 399.99999999998386
11000: 399.99999999998283
12000: 399.99999999998215
13000: 399.9999999999805
14000: 399.99999999998016
15000: 399.99999999997897
16000: 399.9999999999782
17000: 399.99999999997715
18000: 399.99999999997726
19000: 399.9999999999769
20000: 399.99999999997584
Since this is intended to eventually be for a physics engine I would like to know how I can minimise the cumulative error since the Vector3.rotate() method will be used on a very regular basis.
Thanks!
A couple of odd notes:
The error is proportional to the amount rotated. ie. bigger rotation per iteration -> bigger error per iteration.
There is more error when passing doubles to the rotate function than when passing it floats.
You'll always have some cumulative error with repeated floating point trig operations — that's just how they work. To deal with it, you basically have two options:
Just ignore it. Note that, in your example, after 20,000 iterations(!) the area is still accurate down to 13 decimal places. That's not bad, considering that doubles can only store about 16 decimal places to begin with.
Indeed, plotting your graph, the area of your square seems to be going down more or less linearly:
This makes sense, assuming that the effective determinant of your approximate rotation matrix is about 1 − 3.417825 × 10-18, which is well within normal double precision floating point error range of one. If that's the case, the area of your square would continue a very slow exponential decay towards zero, such that you'd need about two billion (2 × 109) 7.3 × 1014 iterations to get the area down to 399. Assuming 100 iterations per second, that's about seven and a half months 230 thousand years.
Edit: When I first calculated how long it would take for the area to reach 399, it seems I made a mistake and somehow managed to overestimate the decay rate by a factor of about 400,000(!). I've corrected the mistake above.
If you still feel you don't want any cumulative error, the answer is simple: don't iterate floating point rotations. Instead, have your object store its current orientation in a member variable, and use that information to always rotate the object from its original orientation to its current one.
This is simple in 2D, since you just have to store an angle. In 3D, I'd suggest storing either a quaternion or a matrix, and occasionally rescaling it so that its norm / determinant stays approximately one (and, if you're using a matrix to represent the orientation of a rigid body, that it remains approximately orthogonal).
Of course, this approach won't eliminate cumulative error in the orientation of the object, but the rescaling does ensure that the volume, area and/or shape of the object won't be affected.
You say there is cumulative error but I don't believe there is (note how your output desn't always go down) and the rest of the error is just due to rounding and loss of precision in a float.
I did work on a 2d physics engine in university (in java) and found double to be more precise (of course it is see oracles datatype sizes
In short you will never get rid of this behaviour you just have to accept the limitations of precision
EDIT:
Now I look at your .area function there is possibly some cumulative due to
+= cross.magnitude
but I have to say that whole function looks a bit odd. Why does it need to know the previous vertices to calculate the current area?
I've been working on a top down car game for quite a while now, and it seems it always comes back to being able to do one thing properly. In my instance it's getting my car physics properly done.
I'm having a problem with my current rotation not being handled properly. I know the problem lies in the fact that my magnitude is 0 while multiplying it by Math.cos/sin direction, but I simply have no idea how to fix it.
This is the current underlying code.
private void move(int deltaTime) {
double secondsElapsed = (deltaTime / 1000.0);// seconds since last update
double speed = velocity.magnitude();
double magnitude = 0;
if (up)
magnitude = 100.0;
if (down)
magnitude = -100.0;
if (right)
direction += rotationSpeed * (speed/topspeed);// * secondsElapsed;
if (left)
direction -= rotationSpeed * (speed/topspeed);// * secondsElapsed;
double dir = Math.toRadians(direction - 90);
acceleration = new Vector2D(magnitude * Math.cos(dir), magnitude * Math.sin(dir));
Vector2D deltaA = acceleration.scale(secondsElapsed);
velocity = velocity.add(deltaA);
if (speed < 1.5 && speed != 0)
velocity.setLength(0);
Vector2D deltaP = velocity.scale(secondsElapsed);
position = position.add(deltaP);
...
}
My vector class emulates vector basics - including addition subtraction, multiplying by scalars... etc.
To re-iterate the underlying problem - that is magnitude * Math.cos(dir) = 0 when magnitude is 0, thus when a player only presses right or left arrow keys with no 'acceleration' direction doesn't change.
If anyone needs more information you can find it at
http://www.java-gaming.org/index.php/topic,23930.0.html
Yes, those physics calculations are all mixed up. The fundamental problem is that, as you've realized, multiplying the acceleration by the direction is wrong. This is because your "direction" is not just the direction the car is accelerating; it's the direction the car is moving.
The easiest way to straighten this out is to start by considering acceleration and steering separately. First, acceleration: For this, you've just got a speed, and you've got "up" and "down" keys. For that, the code looks like this (including your threshold code to reduce near-zero speeds to zero):
if (up)
acceleration = 100.0;
if (down)
acceleration = -100.0;
speed += acceleration * secondsElapsed;
if (abs(speed) < 1.5) speed = 0;
Separately, you have steering, which changes the direction of the car's motion -- that is, it changes the unit vector you multiply the speed by to get the velocity. I've also taken the liberty of modifying your variable names a little bit to look more like the acceleration part of the code, and clarify what they mean.
if (right)
rotationRate = maxRotationSpeed * (speed/topspeed);
if (left)
rotationRate = maxRotationSpeed * (speed/topspeed);
direction += rotationRate * secondsElapsed;
double dir = Math.toRadians(direction - 90);
velocity = new Vector2D(speed * Math.cos(dir), speed * Math.sin(dir));
You can simply combine these two pieces, using the speed from the first part in the velocity computation from the second part, to get a complete simple acceleration-and-steering simulation.
Since you asked about acceleration as a vector, here is an alternate solution which would compute things that way.
First, given the velocity (a Vector2D value), let's suppose you can compute a direction from it. I don't know your syntax, so here's a sketch of what that might be:
double forwardDirection = Math.toDegrees(velocity.direction()) + 90;
This is the direction the car is pointing. (Cars are always pointing in the direction of their velocity.)
Then, we get the components of the acceleration. First, the front-and-back part of the acceleration, which is pretty simple:
double forwardAcceleration = 0;
if (up)
forwardAcceleration = 100;
if (down)
forwardAcceleration = -100;
The acceleration due to steering is a little more complicated. If you're going around in a circle, the magnitude of the acceleration towards the center of that circle is equal to the speed squared divided by the circle's radius. And, if you're steering left, the acceleration is to the left; if you're steering right, it's to the right. So:
double speed = velocity.magnitude();
double leftAcceleration = 0;
if (right)
leftAcceleration = ((speed * speed) / turningRadius);
if (left)
leftAcceleration = -((speed * speed) / turningRadius);
Now, you have a forwardAcceleration value that contains the acceleration in the forward direction (negative for backward), and a leftAcceleration value that contains the acceleration in the leftward direction (negative for rightward). Let's convert that into an acceleration vector.
First, some additional direction variables, which we use to make unit vectors (primarily to make the code easy to explain):
double leftDirection = forwardDirection + 90;
double fDir = Math.toRadians(forwardDirection - 90);
double ldir = Math.toRadians(leftDirection - 90);
Vector2D forwardUnitVector = new Vector2D(Math.cos(fDir), Math.sin(fDir));
Vector2D leftUnitVector = new Vector2D(Math.cos(lDir), Math.sin(lDir));
Then, you can create the acceleration vector by assembling the forward and leftward pieces, like so:
Vector2D acceleration = forwardUnitVector.scale(forwardAcceleration);
acceleration = acceleration.add(leftUnitVector.scale(leftAcceleration));
Okay, so that's your acceleration. You convert that to a change in velocity like so (note that the correct term for this is deltaV, not deltaA):
Vector2D deltaV = acceleration.scale(secondsElapsed);
velocity = velocity.add(deltaV).
Finally, you probably want to know what direction the car is headed (for purposes of drawing it on screen), so you compute that from the new velocity:
double forwardDirection = Math.toDegrees(velocity.direction()) + 90;
And there you have it -- the physics computation done with acceleration as a vector, rather than using a one-dimensional speed that rotates with the car.
(This version is closer to what you were initially trying to do, so let me analyze a bit of where you went wrong. The part of the acceleration that comes from up/down is always in a direction that is pointed the way the car is pointed; it does not turn with the steering until the car turns. Meanwhile, the part of the acceleration that comes from steering is always purely to the left or right, and its magnitude has nothing to do with the front/back acceleration -- and, in particular, its magnitude can be nonzero even when the front/back acceleration is zero. To get the total acceleration vector, you need to compute these two parts separately and add them together as vectors.)
Neither of these computations are completely precise. In this one, you compute the "forward" and "left" directions from where the car started, but the car is rotating and so those directions change over the timestep. Thus, the deltaV = acceleration * time equation is only an estimate and will produce a slightly wrong answer. The other solution has similar inaccuracies -- but one of the reasons that the other solution is better is that, in this one, the small errors mean that the speed will increase if you steer the car left and right, even if you don't touch the "up" key -- whereas, in the other one, that sort of cross-error doesn't happen because we keep the speed and steering separate.