Let there be two balls, one of which is moving about in the Cartesian coordinate plane, while the other is stationary and immobile. At some point, the moving ball collides with the inert ball. Assuming the moving ball is traveling in a straight line, how can one derive the new angle that the moving ball will be propelled given the following information:
The moving ball's center coordinates (X0, Y0), radius (R0), and angle of travel before impact (A0)
The stationary ball's center coordinates (X1, Y1) and radius (R1)
If your second ball has infinite mass:
Where phi (after a long calc) is:
phi= -ArcTan[
( 2 R^2 Sin[A0] + 2 (YD Cos[A0] - XD Sin[A0]) (2 H Cos[A0] +
2 XD Sin[A0]^2 - YD Sin[2 A0])) /
((2 R^2 - XD^2 - 3 YD^2) Cos[A0] + (XD^2 - YD^2) Cos[3 A0] +
8 XD YD Cos[A0]^2 Sin[A0] + 4 H Sin[A0] (-YD Cos[A0] + XD Sin[A0]))
]
Where:
H = (R0 + R1)^2 - ((Y0 - Y1) Cos[A0] + (X0 - X1) Sin[A0])^2
R^2 = (R0 + R1)^2
XD = X1 - X0
YD = Y1 - Y0
Edit
To determine the whole trajectory, you'll also need the coordinates for the center of the moving ball at the time of impact. They are:
{X,Y}= {X1+Sin[A0] ((Y1-Y0) Cos[A0]+ (X0-X1) Sin[A0])-Cos[A0] Sqrt[H],
Y1+Cos[A0] ((Y0-Y1) Cos[A0]+(-X0+X1) Sin[A0])-Sin[A0] Sqrt[H]}
Page 3 of Pool Hall Lessons by Joe van den Heuvel, Miles Jackson gives a great example of how to do this.
// First, find the normalized vector n from the center of circle1 to the center of circle2
Vector n = circle1.center - circle2.center;
n.normalize();
// Find the length of the component of each of the movement vectors along n.
float a1 = v1.dot(n);
float a2 = v2.dot(n);
float optimizedP = (2.0 * (a1 - a2)) / (circle1.mass + circle2.mass);
// Calculate v1', the new movement vector of circle1
// v1 = v1 - optimizedP * m2 * n
Vector v1 = v1 - optimizedP * circle2.mass * n;
// Calculate v2', the new movement vector of circle2
// v2 = v2 + optimizedP * m1 * n
Vector v2 = v2 + optimizedP * circle1.mass * n;
circle1.setMovementVector(v1);
circle2.setMovementVector(v2);
Read at least page three to understand whats going on here.
You should take a look at the elastic collision article on wikipedia. I would explain here, but everything I could have said, wikipedia says it better and with clear examples and equations.
[A long, long time ago I studied this as an undergrad. ]
You need to be clear on the masses. Probably you are assuming equal mass for both balls, as opposed to one being of infinite mass.
The second thing is: Are you interested in considering rolling constraints as well as linear momentum. The treatments you will come across which talk along the lines of a simplistic elastic collision ignore all this. As an example, consider shots in pool/ snooker where you deliberately strike the ball away from the midpoint to generate front or backspin.
Do you want to able to do this?
If so, you need to consider the friction between a spinning ball and the surface.
For example in a "simple" straight-on collision between a rolling ball and a stationary one, if we assume perfectly elastic (again not quite true):
the initial collision stops the moving ball 'A'
the stationary ball 'B' starts moving at the impact speed of 'A'
'A' still has spin, it grips the surface and picks up some small velocity
'B' starts without spin and has to match it to its speed in order to roll. This results in it slowing slightly.
For the simplistic case, the calculation is much easier if you transform to the coordinates of the centre of mass. In that frame, the collision is always a straight-on collision, reversing the direction of the balls. You then just transform back to get the resultants.
Assuming indetical masses and speeds prior to the impact of v1 and w1.
V0 = centre of mass speed = (v1+w1)/2
v1_prime = v of mass_1 in transformed coords = v1 - V0
w1_prime = w1 - V0
Post collision, we have a simple reflection:
v2_prime = -v1_prime (== w1_prime)
w2_prime = -vw_prime (== v1_prime)
v2 = v2_prime + V0
w2 = w2_prime + V0
It simply reflects from the stationary ball. So compute the point of contact (the centres of the balls will be R0 + R1 apart) and the axis of reflection will be the line joining the centres.
EDIT: By which I mean the line joining the centres at the point of contact will have an angle, and you can use this angle to help compute the new angle of the moving ball.
Related
I need the simplest solution for resolving 2D elastic collision between circles, each circle has equal mass with the others.
The environment is Android canvas in which the Y axis is growing towards down. Logic representation of the circle is class PlayerBall which has successful collision detection. PlayerBall has fields:
x and y position of center of the circle
velX and velY velocity vector of the circle movement represented as two scalar values which can be positive or negative.
diameter - diameter of the circle
public static void resolveCollision(PlayerBall ballOne, PlayerBall ballTwo)
{
double collisionAngle = Math.atan2(ballTwo.y - ballOne.y, ballTwo.x - ballOne.x); // angle for ball one
// calculating new velocities between ballOne and ballTwo
...
// setting the new velocities for both balls
ballOne.setVelocity((float)ballOneVelX, (float)ballOneVelY);
ballTwo.setVelocity((float)ballTwoVelX, (float)ballwTwoVelY);
}
I am expecting that velocities of the balls change according to formula defined in this article https://en.wikipedia.org/wiki/Elastic_collision#Two-dimensional_collision_with_two_moving_objects
If you know the x and y velocity of both masses then you don't actually need the angle of collision. The resultant x and y forces on the balls can be calculated as a product of their respective masses, and velocities.
You can define this relationship by the formula:
Where V_x1 represents the velocity purely in the x plane, m1 and m2 are the masses of the balls. This will give you the resultant x velocity. You can apply the same logic to calculate the resultant forces in the y direction.
let newVelX1 = (vel1.vX * (m1 - m2) + (2 * m2 * vel2.vX)) / (m1 + m2);
let newVelY1 = (vel1.vY * (m1 - m2) + (2 * m2 * vel2.vY)) / (m1 + m2);
I'm trying to create a method which calculates the x and y in a grid and eventually the distance between that point and the middle. The problem is, I only know a few values. To explain the case a bit better an image:
(the values between '(..,..)' are lat/long combinations).
As you can see, I know the following values:
start of canvas: xy(0,0)
middle of canvas: xy(540,800) and lat/long(52.3702160, 4.8951680)
max dimension of canvas: x 1080, y 1600
point: xy(?,?) and lat/long(52.4167267, 4.8052174)
point: xy(?,?) and lat/long(52,2422306, 5.1129068)
First, I need to do something to calculate the missing x and y's from the points.
I already tried doing the following:
double mapWidth = screenWidth;
double mapHeight = screenHeight;
// get x value
x = (location.getLongitude()+180)*(mapWidth/360);
// convert from degrees to radians
double latRad = location.getLatitude()*Math.PI/180;
// get y value
double mercN = Math.log(Math.tan((Math.PI/4)+(latRad/2)));
y = (mapHeight/2)-(mapWidth*mercN/(2*Math.PI));
point = new PointF((float)x,(float)y);
This works but I'm getting the wrong x and y values.
For example if my points lat/long are further away the x and y's are getting bigger (more to the middle). But they need to be more at the side because the lat/long point is further away.
All lat/long points inside 2km diameter need to be in my grid, if the point is for example 0.9km away from the center it needs to be nearly at the side.
After that I need to calculate the distance between the two points. I already got that part using the following:
Math.sqrt((point.x - point2.x) * (point.x - point2.x) + (point.y - point2.y) * (point.y - point2.y));
My main problem is calculating the x and y from my lat/long points.
If anyone wants to help, thanks in advance!
I completely rethought my way of calculating the x and y.
I solved it by doing the following:
double distanceMeters = mCurrentLocation.distanceTo(location);
x = ((1000+distanceMeters)*mMiddleCoords.x)/1000; //1000 = radius.
y = ((1000+distanceMeters)*mMiddleCoords.y)/1000;
You can't directly use the distance formula from latitude and longitude. You'll have to take into account the curvature of the sphere to calculate the distance.
The minimum distance between two points on a sphere (and hence earth, simplifying it to a perfect sphere) is the length of the chord on what is called the Great Circle running through those points. A Great Circle is a circle with its center running through the center of the sphere).
From Wikipedia:
C = SQRT(X^2 + Y^2 + Z^2)
where:
X = cos(lat2) * cos(long2) - cos(lat1) * cos(long1)
Y = cos(lat2) * sin(long2) - cos(lat1) * sin(long1)
Z = sin(lat2) - sin(lat1)
And the distance is (2 * R * arcsin(C/2)) where R is the radius of the earth or 6371 km
The other alternative - if you know you will always have the Android libraries - is the Location.distanceTo() method.
I'm trying to write a java mobile application (J2ME) and I got stuck with a problem: in my project there are moving circles called shots, and non moving circles called orbs. When a shot hits an orb, it should bounce off by classical physical laws. However I couldn't find any algorithm of this sort.
The movement of a shot is described by velocity on axis x and y (pixels/update). all the information about the circles is known: their location, radius and the speed (on axis x and y) of the shot.
Note: the orb does not start moving after the collision, it stays at its place. The collision is an elastic collision between the two while the orb remains static
here is the collision solution method in class Shot:
public void collision(Orb o)
{
//the orb's center point
Point oc=new Point(o.getTopLeft().x+o.getWidth()/2,o.getTopLeft().y+o.getWidth()/2);
//the shot's center point
Point sc=new Point(topLeft.x+width/2,topLeft.y+width/2);
//variables vx and vy are the shot's velocity on axis x and y
if(oc.x==sc.x)
{
vy=-vy;
return ;
}
if(oc.y==sc.y)
{
vx=-vx;
return ;
}
// o.getWidth() returns the orb's width, width is the shot's width
double angle=0; //here should be some sort of calculation of the shot's angle
setAngle(angle);
}
public void setAngle(double angle)
{
double v=Math.sqrt(vx*vx+vy*vy);
vx=Math.cos(Math.toRadians(angle))*v;
vy=-Math.sin(Math.toRadians(angle))*v;
}
thanks in advance for all helpers
At the point of collision, momentum, angular momentum and energy are preserved. Set m1, m2 the masses of the disks, p1=(p1x,p1y), p2=(p2x,p2y) the positions of the centers of the disks at collition time, u1, u2 the velocities before and v1,v2 the velocities after collision. Then the conservation laws demand that
0 = m1*(u1-v1)+m2*(u2-v2)
0 = m1*cross(p1,u1-v1)+m2*cross(p2,u2-v2)
0 = m1*dot(u1-v1,u1+v1)+m2*dot(u2-v2,u2+v2)
Eliminate u2-v2 using the first equation
0 = m1*cross(p1-p2,u1-v1)
0 = m1*dot(u1-v1,u1+v1-u2-v2)
The first tells us that (u1-v1) and thus (u2-v2) is a multiple of (p1-p2), the impulse exchange is in the normal or radial direction, no tangential interaction. Conservation of impulse and energy now leads to a interaction constant a so that
u1-v1 = m2*a*(p1-p2)
u2-v2 = m1*a*(p2-p1)
0 = dot(m2*a*(p1-p2), 2*u1-m2*a*(p1-p2)-2*u2+m1*a*(p2-p1))
resulting in a condition for the non-zero interaction term a
2 * dot(p1-p2, u1-u2) = (m1+m2) * dot(p1-p2,p1-p2) * a
which can now be solved using the fraction
b = dot(p1-p2, u1-u2) / dot(p1-p2, p1-p2)
as
a = 2/(m1+m2) * b
v1 = u1 - 2 * m2/(m1+m2) * b * (p1-p2)
v2 = u2 - 2 * m1/(m1+m2) * b * (p2-p1)
To get the second disk stationary, set u2=0 and its mass m2 to be very large or infinite, then the second formula says v2=u2=0 and the first
v1 = u1 - 2 * dot(p1-p2, u1) / dot(p1-p2, p1-p2) * (p1-p2)
that is, v1 is the reflection of u1 on the plane that has (p1-p2) as its normal. Note that the point of collision is characterized by norm(p1-p2)=r1+r2 or
dot(p1-p2, p1-p2) = (r1+r2)^2
so that the denominator is already known from collision detection.
Per your code, oc{x,y} contains the center of the fixed disk or orb, sc{x,y} the center and {vx,vy} the velocity of the moving disk.
Compute dc={sc.x-oc.x, sc.y-oc.y} and dist2=dc.x*dc.x+dc.y*dc.y
1.a Check that sqrt(dist2) is sufficiently close to sc.radius+oc.radius. Common lore says that comparing the squares is more efficient. Fine-tune the location of the intersection point if dist2 is too small.
Compute dot = dc.x*vx+dcy*vy and dot = dot/dist2
Update vx = vx - 2*dot*dc.x, vy = vy - 2*dot*dc.y
The special cases are contained inside these formulas, e.g., for dc.y==0, that is, oc.y==sc.y one gets dot=vx/dc.x, so that vx=-vx, vy=vy results.
Considering that one circle is static I would say that including energy and momentum is redundant. The system's momentum will be preserved as long as the moving ball contains the same speed before and after the collision. Thus the only thing you need to change is the angle at which the ball is moving.
I know there's a lot of opinions against using trigonometric functions if you can solve the issue using vector math. However, once you know the contact point between the two circles, the trigonometric way of dealing with the issue is this simple:
dx = -dx; //Reverse direction
dy = -dy;
double speed = Math.sqrt(dx*dx + dy*dy);
double currentAngle = Math.atan2(dy, dx);
//The angle between the ball's center and the orbs center
double reflectionAngle = Math.atan2(oc.y - sc.y, oc.x - sc.x);
//The outcome of this "static" collision is just a angular reflection with preserved speed
double newAngle = 2*reflectionAngle - currentAngle;
dx = speed * Math.cos(newAngle); //Setting new velocity
dy = speed * Math.sin(newAngle);
Using the orb's coordinates in the calculation is an approximation that gains accuracy the closer your shot is to the actual impact point in time when this method is executed. Thus you might want to do one of the following:
Replace the orb's coordinates by the actual point of impact (a tad more accurate)
Replace the shot's coordinates by the position it has exactly when the impact will/did occur. This is the best scenario in respect to the outcome angle, however may lead to slight positional displacements compared to a fully realistic scenario.
I'm trying to create rope physics for a 2D game, so as a starting point I have a small rotating image and I need to add another piece of rope to the end of it. Unfortunately I'm having trouble trying to track the bottom part of the image as the rotation occurs at the top of it. I've managed to track the (0,0) coordinate of the image using the following code but I need to be able to track point (32,57). This is what I have so far:
xr = xm + (xPos - xm) * Math.cos(a) - (yPos - ym) * Math.sin(a);
yr = ym + (xPos - xm) * Math.sin(a) + (yPos - ym) * Math.cos(a);
Any help is appreciated!
EDIT:
So hey, I got it working =D Using polar coordinates turned out to be a lot easier then whatever I had going on before.
The top 2 variables are constant and stay the same:
theta0 = Math.atan2(y, x);
r = 25;
theta = theta0 + a;
xr = (r * Math.cos(theta)) + xm;
yr = (r * Math.sin(theta)) + ym;
xm and ym are the positions of my image.
Use polar coordinates. Set your origin at the point of rotation of your image, and pick your favorite angular reference (say 0 degrees is directly to the right, and positive rotations go counterclockwise from there).
Compute the polar coordinates of your desired point (32, 57) relative to this coordinate system. Say the answer is (r, theta).
Now, the only thing that's changing as you spin the image around is the value of theta. Now you can go back to x-y coordinates with your new value of theta.
Hope this helps.
What is the best way to go about moving a Point2D.Double x distance closer to another Point2D.Double?
Edit: Tried to edit, but so went down for maintenance. No this is not homework
I need to move a plane (A) towards the end of a runway (C) and point it in the correct direction (angle a).
alt text http://img246.imageshack.us/img246/9707/planec.png
Here is what I have so far, but it seems messy, what is the usual way to go about doing something like this?
//coordinate = plane coordinate (Point2D.Double)
//Distance = max distance the plane can travel in this frame
Triangle triangle = new Triangle(coordinate, new Coordinate(coordinate.x, landingCoordinate.y), landingCoordinate);
double angle = 0;
//Above to the left
if (coordinate.x <= landingCoordinate.x && coordinate.y <= landingCoordinate.y)
{
angle = triangle.getAngleC();
coordinate.rotate(angle, distance);
angle = (Math.PI-angle);
}
//Above to the right
else if (coordinate.x >= landingCoordinate.x && coordinate.y <= landingCoordinate.y)
{
angle = triangle.getAngleC();
coordinate.rotate(Math.PI-angle, distance);
angle = (Math.PI*1.5-angle);
}
plane.setAngle(angle);
The triangle class can be found at http://pastebin.com/RtCB2kSZ
Bearing in mind the plane can be in in any position around the runway point
You can minimize the difference along both axis by a percent (that depends on how much you want to move the points).
For example:
Point2D.Double p1, p2;
//p1 and p2 inits
// you don't use abs value and use the still point as the first one of the subtraction
double deltaX = p2.getX() - p1.getX();
double deltaY = p2.getY() - p1.getY();
// now you know how much far they are
double coeff = 0.5; //this coefficient can be tweaked to decice how much near the two points will be after the update.. 0.5 = 50% of the previous distance
p1.setLocation(p1.getX() + coeff*deltaX, p1.getY() + coeff*deltaY);
So you moved p1 halfway toward p2. The good thing avoid abs is that, if you choose which point will be moved and which one will stand still you can avoid if tests and just use the raw coefficient.
The shortest distance between two points is a line, so simply move that point x units along the line that connects the two points.
Edit: I didn't want to give away the specifics of the answer if this is homework, but this is simple enough that it can be illustrated without being too spoiler-y.
Let us assume you have two points A = (x1, y1) and B = (x2, y2). The line that includes these two points has the equation
(x1, y1) + t ยท (x2 - x1, y2 - y1)
where t is some parameter. Notice that when t = 1, the point specified by the line is B, and when t = 0, the point specified by the line is A.
Now, you would like to move B to B', a point which is a new distance d away from A:
A B' B
(+)---------------------(+)-----------(+)
<========={ d }=========>
The point B', like any other point on the line, is also governed by the equation we showed earlier. But what value of t do we use? Well, when t is 1, the equation points to B, which is |AB| units away from A. So the value of t that specifies B' is t = d/|AB|.
Solving for |AB| and plugging this into the above equation is left as an exercise to the reader.
Vectors to the rescue!
Given points A and B. Create a vector V from A to B (by doing B-A). Normalize vector V into a unit vector and then just multiply it with the distance, d, you want and finally add the resulting vector to point A. ie:
A_moved = A + |(B-A)|*d
Java(ish)
Vector2D a_moved = a.add(b.subtract(a).norm().multiply(d));
No angles, no nasty trig needed.
double angle = Math.atan2(landingCoordinate.y-coordinate.y, landingCoordinate.x-coordinate.x);
coordinate.x += Math.cos(angle)*distance;
coordinate.y += Math.sin(angle)*distance;
//Add 90 degress to the plane angle
plane.setAngle(angle + 1.57079633);
(point A is to be moved closer to B)
if (A.x > B.x)
//decrement A.x
if (A.x < B.x)
//increment A.x
that might be the basic idea in pseudocode, but there's a lot more to it than that. How do you define 'best' way?
Certain things need to be considered:
Is there a specific measure/ratio you want the points to be apart from each other, or do you just want them closer?
Should both coordinates always be modified? (e.g. if you want to move (1,50) closer to (0,0), do you make it (.5, 25) or just (1,25)?
if the point is to be 1 unit away, should it be 1 unit horizontally? directly diagonally? 1 unit away on the line between the 2 points?
Give us a little more detail and we'll see what we've got. :D
In this game, integral coordinates are used to represent squares in a grid. The method move(int row, int col) moves toward the specified row and column by advancing one square in one of eight semi-cardinal directions, as seen here.