Develop Reference

Calculate how far to translate an arc to avoid intersection

I have a Canvas on which I've drawn a circle / 360 degree arc. I have the arc start drawing from -90 (the top) rather than the right (0) as is default.
I want to place a rectangle at the top of the same canvas and to reduce the sweep of the arc so that the two do not intersect. I've attached an image to illustrate
So what I need to do is work out what angle is represented by half of the rectangle so that I can adjust where to start drawing my arc. The centre of the circle is at the centre of my Canvas:
[canvas.width /2, canvas.height /2]
I've read some resources like this question but they haven't helped to do much more than make me feel like I know nothing. I tried a few failing formulas ending with this
double adjustment = Math.atan2(rectangleY - circleY, rectangleX - circleX) - Math.atan2(rectangleY - circleY, (rectangleX + rectangleWidth) - circleX);
Can somebody tell me what is the right way to calculate this in Java? I'd also like to know how to find where the rectangle intersects if that's possible (I.e. the width of the shaded orange part on the image) although this is of lesser importance to me right now

Let W be the width of the rectangle, R be radius of the circle, and A be the angle you're looking for.
There's a right-angle triangle with angle A at the center, R as the hypotenuse and W/2 as the side opposite angle A, so
W/2R = sin(A)
so
A = Math.asin(0.5*W/R);
Of course you can't use asin(0.5W/R) when W > 2R
EDIT
To answer the second problem (finding the intersection)
Let H be the distance from the center of the circle to the rectangle. When the rectangle is wide enough that the circle intersects the lower side, there's a right angle triangle with A at the center, R as the hypotenuse, and H as the side adjacent to A
H/R = cos(A) and A = Math.acos(H/R). Calculate both angles and use the smaller one.

Rotating and drawing an image at a point in Java

I tried to post a different topic for this but people didn't really seem to understand what I was trying to do, so, now I've closed that one and opened this one to give more detail and rephrase the question as a whole.
Ok.
So basically, I have an application which draws an Ellipse. Now, I have a certain number of points (that can be random) in which I have to rotate an image and draw at.
Using Maths I know that to get a point on an Ellipse based by using an angle I use the following equation;
final int radiusW = (width / 2);
final int radiusH = (height / 2);
final int angle = 120;
int pointX = (int) (radiusW + (radiusW * Math.cos(Math.toRadians(angle))));
int pointY = (int) (radiusH + (radiusH * Math.sin(Math.toRadians(angle))));
And that works fine, I can locate an absolute point around the perimeter of the Ellipse.
However, now I'm trying to draw an image on this point so that the image is rotated facing the center of ellipse and is centered on the point.
So, to get the image rotated to the center of the point I do the following;
final AffineTransform at = new AffineTransform();
at.rotate(Math.toRadians(angle - 90), image.getWidth() / 2, image.getHeight() / 2);
final AffineTransformOp ato = new AffineTransformOp(at, AffineTransformOp.TYPE_BILINEAR);
And then I get my new rotated BufferedImage using;
BufferedImage rotated = ato.filter(image, null);
However, I can't seem to be able to get the central point of the image anymore.
If the angle was 0 so that the Image was the original direction then I would simply do;
graphics.drawImage(rotated, pointX - rotated.getWidth() / 2, pointY - rotated.getHeight() / 2, this);
However I'm not sure how to find the central point and draw it based upon that on a rotated image.
I know it involves using cos and sin to multiply the original pointX and pointY by the rotation matrix but everytime I try and work out a solution it always draws completely wrong.
Any help would be very much appreciated as I've spent the best part of a day trying to resolve this.
Thank you.

The thing is that if you just use sin and cos to rotate the corner of the image you will end up with the new rotated position of that corner - when actually what you want to find is the new width and height.
The center is width/2, height/2
Use this to calculate your new width and height:
Calculate Bounding box coordinates from a rotated rectangle

I have done this with OpenCV, and there the image have been rotated but the resulted image was having the same width and height as the initial on. The image was cropped if it was getting out of the initial dimentions and there were black pixels if no information (the rotated image has no pixels in that place).
If you think that rotated has more pixels than the initial image you can verify with size() or length(). Or you can play with the diagonals of the initial rectangle (image size): compute the projection of the diagonals and thake the greatest or what you think. But I am sure that it is similar to the OpenCV case.
I don't know if this can help you, but I hope so.

Your question is not entirely clear but it seems to me that you have a misunderstanding of the nature of the angle used to parametrise your ellipse. The angle as you have it is merely used to parametrise the form of an ellipse equation. It is not the same as the polar angle (except at particular angles). That is to say if you evaluate a point on your ellipse using an angle of (pi/4) radians (45 degrees), then measured the angle that the line from the ellipse centre to your point makes with the axis, it will not measure 45 degrees (except for the case where the ellipse is actually a circle).
That is to say that
int pointX = (int) (radiusW + (radiusW * Math.cos(Math.toRadians(angle))));
int pointY = (int) (radiusH + (radiusH * Math.sin(Math.toRadians(angle))));
is just a parametrisation of an ellipse and that angle is not a polar angle and treating this angle as a rotation angle will not give accurate results (except at integer multiples of (pi/2) radians)
It seems to me that you require the polar form of an ellipse relative to its centre in order for your code to make sense in the context of using this angle for rotation.
It is also possible that I have misunderstood your question though, in which case this answer will be downvoted on a grand scale and I will delete it.

Making a rectangle go around in a circle - Java/Swing

My maths isn't that good so I'm having a bit of trouble in one of my applications that I'm trying to do where I want a rectangle to represent a vehicle and I want that vehicle/rectangle to "drive" around in a circle. Imagine a roundabout with only 1 vehicle in it, just circling around forever.
If I can get some help how to do that then I'll be able to build on the example and most importantly learn.
If someone could write up a simple example for me I'd be grateful. No background no images, just a rectangle "driving" around in a circle. I'm using java and Swing.

Sorry, I am not sure if could understand clear you exactly need. If you need to draw rectangle which is moving around inside of circle, you can use sin/cos functions.
Something like that:
double r = 50.0; // radius (it might radius of your circle, but consider dimensions of rectangle to make sure you are drawing inside of circle, e.g. circleRadius - rectangeDimesion / 2.0)
for (int f = 0; f < 360; f++) {
double x = Math.sin(Math.toRadians((double)f)) * r;
double y = Math.cos(Math.toRadians((double)f)) * r;
// draw rectangle on [x, y] coordinates
}

If you know the radius of the round about, all the you would need would be a trigonometric function and the angle which the vehicle makes to the round about. You could take a look at this simple introduction which should get you started in the right direction.
On another hand, another approach would be to use a Transformation Matrix where you start with a matrix containing two points (your X and Y co-ordinates) and you transform them to become the new co-ordinates.
You can then rotate the rectangle to mimic a vehicle turning.
If you have a limited background in Mathematics, the first option might be easier for you to grasp.

This is more an extended comment than an answer.
I would divide the problem up into several easier problems, and work on each of them separately:
Draw your rectangle with a specified center location and long axis orientation.
Determine the center point and long axis orientation for an object orbiting around the origin. Note that to get make the long axis a tangent it needs to be perpendicular to the radius through the center.
Translate the whole system so that it orbits the desired point, rather than the origin.

Rotating quaternions based on mouse movement (OpenGL and Java)

I'm writing a game in Java using OpenGL (the LWJGL binding, to be specific). Each entity, including the camera, has a quaternion that represents it's rotation. I've figured out how to apply the quaternion to the current OpenGL matrix and everything rotates just fine. The issue I'm having is getting the camera to rotate with the mouse.
Right now, every frame, the game grabs the amount that the mouse has moved on one axis, then it applies that amount onto the quaternion for the camera's rotation. Here is the code that rotates the quaternion, I'll post it since I think it's where the problem lies (although I'm always wrong about this sort of stuff):
public void rotateX(float amount){
Quaternion rot = new Quaternion(1.0f, 0.0f, 0.0f, (float)Math.toRadians(amount));
Quaternion.mul(rot, rotation, rotation);
rotation.normalise();
}
This method is supposed to rotate the quaternion around the X axis. 'rotation' is the quaternion representing the entity's rotation. 'amount' is the amount that I want to rotate the quaternion (aka the amount that the mouse was moved). 'rot' is a normalized vector along the X axis with a w value of the amount converted to radians (I guess the goal here is to give it an angle- say, 10 degrees- and have it rotate the quaternion along the given axis by that angle). Using Quaternion.mul takes the new quaternion, multiplies it by the rotation quaternion, and then stores the result as the rotation quaternion. I don't know if the normalization is necessary, since 'rot' is normal and 'rotation' should already by normalized.
The rotateY and rotateZ methods do the same thing, except for changing the vector for 'rot' (0.0, 1.0, 0.0 for y and 0.0, 0.0, 1.0 for z).
The code appears to work fine when the game starts and the camera is looking down the negative Z axis. You can spin all the way around on the Y axis OR all the way around the X axis. But as soon as you try to rotate the camera while not looking down the Z axis, everything gets really screwy (I can't even describe it, it rotates very oddly).
My end goal here is to have something to use for controlling a ship in a space with no up vector. So when you move the mouse on the Y axis, no matter what angle the ship is at, it changes the pitch of the ship (rotation along the X axis). Similarly, when you move the mouse on the X axis, it changes the yaw (rotation along the Y axis). I might be going about this the wrong way and I probably just need a push (or shove) in the right direction.
If you need more details on anything (how my rendering is done, any other maths that I'm trying to do) just ask and I'll put it up. I understood everything when I was using euler angles (which apparently are a big no-no for 3D application development... wish somebody would have told me that before I sunk a lot of time into getting them to work) but as soon as I switched over to quaternions, I got in over my head really fast. I've spent the past few months just playing with this code and reading about quaternions trying to get it to work, but I haven't really gotten anywhere at all :'(
Very, very frustrating... starting to regret trying to make something in 3D >_<

Quaternion rot = new Quaternion(1.0f, 0.0f, 0.0f, (float)Math.toRadians(amount));
OK, this is flat-out wrong.
The constructor that takes four floats assumes that they represent an actual quaternion. What you give that constructor is not a quaternion; it's a vec3 axis and an angle that you expect to rotate around.
You can't shove those into a quaternion class and expect to get a legitimate quaternion out of it.
Your quaternion class should have a constructor or some other means of creating a quaternion from an angle and an axis of rotation. But according to the documentation you linked to, it does not. So you have to do it yourself.
A quaternion is not a vec3 axis with a fourth value that is an angle. A unit quaternion representation a change in orientation is a vec3 that is the axis of rotation * the sine of half of the angle of rotation, and a scalar component that is the cosine of half the angle of rotation. This assumes that the angle of rotation is clamped on the range [-pi/2, pi/2].
Therefore, what you want is this:
float radHalfAngle = ... / 2.0; //See below
float sinVal = Math.Sin(radHalfAngle);
float cosVal = Math.Cos(radHalfAngle);
float xVal = 1.0f * sinVal;
float yVal = 0.0f * sinVal; //Here for completeness.
float zVal = 0.0f * sinVal; //Here for completeness.
Quaternion rot = new Quaternion(xVal, yVal, zVal, cosVal);
Also, converting amount to radians directly doesn't make sense, particularly so if amount is just a pixel-coordinate delta that the mouse moved. You need some kind of conversion scale between the distance the mouse moves and how much you want to rotate. And toRadians is not the kind of scale you want.
One more thing. Left-multiplying rot, as you do here, will perform a rotation about the camera space X axis. If you want a rotation about the world-space X axis, you need to right-multiply it.

Move bitmap in certain direction Android

Ok so I am working on a game in Android and right now I have a bitmap that I have drawn at the center of the screen. I can rotate the bitmap left and right by certain degrees using the Matrix class. The bitmap is a picture of a ship so when the user wants to move forward, I want the ship to move at the current angle that the ship is rotated at. Any ideas about how I should go about doing this?

Ok well with a bunch of trial and error and some reading up on trig I have managed to solve my own question. The vector that holds the current location of the ship has an X and a Y. What I then need to do, was based on the current rotation of the ship calculate a speed vector and then add that speed vector to the position vector.
speedX = (float) Math.sin(rotation*(Math.PI/180)) * speed;
speedY = (float) -Math.cos(rotation*(Math.PI/180)) * speed;
x += speedX;
y += speedY;
The rotation is in degrees so they needed to be converted to radians. Also speed is the actual speed of the ship and is applied to each speed vector. Hope that will help someone having the same problem.