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Trigonometry with Java

I'm trying to do some basic trigonometry with Java and LibGDX on android.
I've spent a long time googling "How to find an angle in right triangles".
I still don't really understand :(
I want to give an Actor subclass a random direction to follow. So what is the angle - and what should I set xSpeed and ySpeed to, in order to move at the correct angle.
I started writing an app to help me see how it works.
There are two objects - An origin point and a touch point. User presses screen, touchPoint moves to where user touched. Methods fire to figure out the appropriate values. I know the XDistance and YDistance between the two points. That means I know the Opposite length and the Adjacent length. So all I need to do is tan-1 of (opposite / adjacent), am I right?
I just don't understand what to do with the numbers my program spits out.
Some code:
In create event of main class:
stage.addListener(new ClickListener() {
#Override
public void touchDragged(InputEvent event, float x, float y, int pointer) {
touchPoint.setX(x);
touchPoint.setY(y);
touchPoint.checkDistance(); // saves x and y distances from origin in private fields
atan2D = getAtan2(touchPoint.getYDistance(), touchPoint.getXDistance());
tanhD = getTanh(touchPoint.getYDistance(), touchPoint.getXDistance());
xDistanceLbl.setText("X Distance: " + touchPoint.getXDistance());
yDistanceLbl.setText("Y Distance: " + touchPoint.getYDistance());
atan2Lbl.setText("Atan2: " + atan2D);
tanhLbl.setText("Tanh: " + tanhD);
angleLbl.setText("Angle: No idea");
}
})
...
private double getAtan2(float adjacent, float opposite) {
return Math.atan2(adjacent, opposite);
}
private double getTanh(float adjacent, float opposite) {
return Math.tanh((adjacent / opposite));
}
These two functions give me numbers between (atan2: -pi to pi) and (tanh: -1.0 to 1.0)
How do I turn these values into angles from which I can then work backwards and get the opposite and adjacent again?
Doing this should allow me to create and object with a random direction, which I can use in 2D games.

atan2 gives you direction in radians. Direction from origin (0,0) to touchPoint. If you need direction from some object to touchPoint, then subtract object coordinates. Perhaps you also want to see direction in degrees (this is only for human eyes)
dx = x - o.x
dy = y - o.y
dir = atan2(dy, dx)
dir_in_degrees = 180 * dir / Pi
I you have direction and want to retrieve coordinate differences, you need to store distance
distance = sqrt(dx*dx + dy*dy)
later
dx = distance * cos(dir)
dy = distance * sin(dir)
But note that often storing dx and dy is better, because some calculations might be performed without trigonometric functions
Just noticed - using tanh is completely wrong, this is hyperbolic tangent function, it has no relation to geometry.
You can use arctan, but it gives angle in half-range only (compared with atan2)

How to calculate a point out of an angle and a distance?

I searched and implemented things from this forum, it doesn't come out right.
What I'm trying to achieve is to calculate a spawnPoint for player bullets relative to his position and rotation.
The spawnPoint should be and his X + his width (the player is set to point to the right by default) and y + height/2 (to spawn from his center on the Y axis).
This is what I got from this forum:
this.bulletSpawn.x = (float)(this.position.x + this.width/2 + this.width * Math.cos(rotation));
this.bulletSpawn.y = (float)(this.position.y + this.height/2 + this.height/2 * Math.sin(rotation));
The rotation is in Radians. The this is the Player class.
Images showing what I expect to happen:
Original Position
Expected Behaviour
The red dot is the spawnPoint I'm trying to calculate knowing the player position and rotation.
The player Sprite is what rotates, and it rotates related to his center x and y, which is done with a lib, i do not have these variables. The entire arrow would be the player , the arrow direction is where the player is pointing at, and the red dot would be the bulletSpawn point (or the expected one)
Using the code I posted, the bullets seem to be spawning from somewhere else. Even at the beggining they have an offset and when I rotate the player the spawnPoint seems to be relative to a different origin than what I'm expecting.
This is the bullet position code:
position.x = holder.bulletSpawn.x - (float)(this.width/2 * holder.rotation);
position.y = holder.bulletSpawn.y - (float)(this.height/2 * holder.rotation);
This is inside the Bullet class. The position variable is a Vector2 of bullet, and holder is the player instance. This code is merely to give an offset for the bullet to spawn at the center of its own size
I added some fixes related to the comments, but the bullets still have a tiny offset that looks wrong at certain angles.
Basically the distance i want to get is the width of the player, and his center y which is height/2.

Let's initial position is X0, Y0, rotation is about center point CX, CY, and rotation angle is Theta. So new position after rotation is:
NX = CX + (X0-CX) * Cos(Theta) - (Y0-CY) * Sin(Theta)
NY = CY + (X0-CX) * Sin(Theta) + (Y0-CY) * Cos(Theta)
This equations describe affine transformation of rotation of arbitrary point about center point, and affine matrix is combination of translation, rotation, and back translation matrices.
About center CX, CY - you wrote
it rotates related to his x and y origin at his bottom left
About initial point coordinate - for bullet it seems to be
X + Width, Y + Height/2

Swift extension:
extension CGSize {
static func offsetFrom(angle:CGFloat, distance:CGFloat) -> CGSize {
let rad = angle * CGFloat.pi / 180
return CGSize(width: sin(rad) * distance, height: cos(rad) * distance)
}
}

Calculating distance in pixels between two locations

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.

JAVA elastic collision of moving and non moving circles

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.

Angle that a moving ball will bounce off of an inert ball

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.