I have a problem with JSlider in Java I have drawn a circle A, and I want to put ANOTHER circle B inside the first circle A. I want to place the CENTRE of the second circle B at the same coordinate with the centre of the first circle A, and then I want to use JSlider to INCREASE or DECREASE the radius of circle B. The trouble is, when you increase or decrease the slider, the CENTRE of circle B does not stay aligned with the centre of A. Basically, I want two circles with the SAME centre. Can someone point out my mistake, please?
slider1 = new JSlider(JSlider.HORIZONTAL,10,100,10);
window.add(slider1);
slider1.addChangeListener(this);
Graphics paper = panel.getGraphics();
int slider1Value = slider1.getValue();
paper.setColor(Color.white);
paper.fillRect(0, 0, 500, 500);
paper.setColor(Color.pink);
paper.fillOval(20,20,100,100); // this is circle A
paper.drawOval(60,60,slider1Value,slider1Value); // this is circle B slider
Because you have to change position of top-left "corner" of circle. If you change radius, the circle is bigger/smaller so it's obvious if you don't change position of top-left cornet, centers of 2 circles won't be aligned
Theory
From Graphics.drawOval() javadoc. When you draw a circle their (x,y) coordinates are not its center but its top-left corner. In order to make your B circle aligned with A circle, you need to calculate its (x,y) coordinates relatives to the last one:
Circle A: (x,y) = (20,20); diameter = 100 ==> radius = 50; center = (x + radius, y + radius) = (70,70)
Well, you have your center now: (70,70). Now, slider1Value is your new radius so you need to calculate circle B (x,y) coordinates:
Circle B: center = (70,70); (x,y) = (centerX - radius, centerY - radius) = (70 - slider1Value, 70 - slider1Value)
Finally, circle B width and height are equals to its diameter:radius * 2 = slider1Value * 2.
Facts
Make this change and it will work like a charm:
paper.drawOval(70 - slider1Value, 70 - slider1Value, 2*slider1Value, 2*slider1Value);
Related
I have the following code which makes a basic Pane with a Rectangle inside it:
public class Canvas extends Pane{
public Canvas() {
Rectangle rect = new Rectangle(50,50, Color.GOLD);
this.setBackground(new Background(new BackgroundFill(Color.INDIANRED,new CornerRadii(0),new Insets(0))));
this.setPrefSize(200, 200);
rect.setX((this.getMaxWidth()/2) + (rect.getWidth()));
rect.setY((this.getMaxHeight()/2) + (rect.getHeight()));
this.getChildren().add(rect);
}
I already tried a few things to get the center of the triangle at the center of the Pane (one example in the above code), but because in JavaFX the X and Y coordinates of a Rectangle represent the top left corner it has made it more difficult to center it within my pane. I am very bad at maths and dont know the correct formula to achieve this.
How can i get the center of my Rectangle to align with the center of my Pane?
Without using Stackpane etc...
You want to align the midpoint of the Rectangle with the midpoint of the Pane.
You have to find the midpoint of the pane (x and y), and the rectangle (x and y).
// Find the midpoint of the Pane.
// x midpoint
float xPaneMidpoint = pane.width / 2
// y midpoint
float yPaneMidpoint = pane.height/ 2
You then move the Rectangle to the position of the midpoint of the Pane, then offset it by half of its size
// Take the midpoint of the pane, and move the rectangle to that point.
// Then offset it by half its width/height
rectangle.setX( xPaneMidpoint - (rectangle.width / 2));
rectangle.setX( yPaneMidpoint - (rectangle.height/ 2));
I'm trying to draw a circle using Pixmap. To make the problem clearer, I'm filling the entire Pixmap area in white, then drawing the circle in a different color. Here is the code that I feel should work.
I'm setting the width/height of the Pixmap to twice the size of the radius of the circle.
Then I'm drawing a circle in the middle of the Pixmap at (radius, radius).
public static Texture circle(int radius, Color color) {
Pixmap pixmap = new Pixmap(radius * 2, radius * 2, Pixmap.Format.RGBA4444);
pixmap.setColor(Color.WHITE);
pixmap.fill();
pixmap.setColor(color);
pixmap.fillCircle(radius, radius, radius);
Texture texture = new Texture(pixmap);
pixmap.dispose();
}
Unfortunately, the Pixmap cuts off the circle on the right and bottom sides. For example:
If I increase the size of the Pixmap by 1 in both the width and height, then it looks fine:
I can just arbitrarily add an extra pixel but I'd like to understand why this is necessary. Why does setting the radius of the circle to X result in a diameter that is actually X + 1?
To get the result you want, the location of the circle's center would have to fall between two pixels, so that there are a similar number of whole pixels on either side of that location. My guess is that the Pixmap code defines a pixel's location to mean the center of a pixel. So the point (radius, radius) is closer to the right edge than the left, and (radius-1, radius-1) is closer to the left edge than the right. With this definition of location, the center of your circle should be at location (radius-.5, radius-.5).
If you have to put the center of the circle in the middle of a pixel, then it makes sense that you'd use the location (radius, radius) for the circle and that you'd need the width and height of the Pixmap to be (2*radius + 1, 2*radius+1). This way, there are the same number of pixels, radius+.5 pixels, on either side of the center of the circle. You might at that point want to draw a circle of radius radius + .5 if the library will take that.
Because it draws a circle centered on a pixel, not between pixels.
So the actual radius of the circle drawn is one more than passed in, a circle with radius 1 is drawn as (numbers are coordinates in this example):
012
0 X
1XCX
2 X
This technically has a radius of 1.5, but now it's centered on a pixel (C).
I am guessing this is to allow you to place it accurately, as if it actually had a radius of 2, you wouldn't be able to place the center on a pixel.
Working on a Jbox2D program I created 2 objects a rectangle at (0,10) meters and 10 Meters Wide and 1 meter wide, and a Ball at (1,0) that has a radius of 0.5f Meters
//in RectangleObject Class
PolygonShape cs = new PolygonShape();
cs.setAsBox(width, height);
//In CircleObject Class
CircleShape cs = new CircleShape();
cs.m_radius = radius;
When my program runs the ball moves toward the platform and hits the rectangle like I expected but the numbers I get back are not what I expect
BallX[0] : 1
BallY[0] : 7.9964995
RectX[0] : 1
RectY[0] : 10
If the X and Y are calculated from the center of the ball then the ball should only be 0.5 away from the box at Y = 9.5. even if it uses the Diameter it should still be at most 1 meter away at Y = 9.
anyone know why its calculating the Y to be 2 meters away when the Radius is only 0.5?
The arguments for the polygon shape are "half-width" and "half-height". It's one of the strange carry-overs from box2d (see the manual here), and not the most intuitive behavior. Try:
cs.setAsBox(width / 2, height / 2);
I try to find a solution for drawing ellipses based on the center point, not the upper left corner as it is specified in the constructor of Ellipse2D.Double. As seen in the picture the ellipses should have the same center point and scale, is that somehow possible?
Thanks in advance for your help.
If (x,y) is the center you want to use and you can only specify the upper left corner, then use the following:
private Ellipse2D getEllipseFromCenter(double x, double y, double width, double height)
{
double newX = x - width / 2.0;
double newY = y - height / 2.0;
Ellipse2D ellipse = new Ellipse2D.Double(newX, newY, width, height);
return ellipse;
}
If called with the center point and the width and height, this will "transform" your center point to the upper left corner and create an Ellipse2D which is located just as you want it to be.
The 'Upper' coordinate is misleading , it only works assuming y >=0 ( which works fine for a screen referential , bur not if you use the primitive with y <0 , for instance calculating object collisions )
With the usual math referential , where y<0 is possible , up is at the bottom
so it lacks a general definition not to get confused
The exact definition is that x and y are the min coordinates of the bounding rectangle.
It can be 'up' or 'down' ( relatively to your screen i suppose ) depending on the y axis orientation and y coordinate sign
Im trying to get into some basic JavaFX game development and I'm getting confused with some circle maths.
I have a circle at (x:250, y:250) with a radius of 50.
My objective is to make a smaller circle to be placed on the circumference of the above circle based on the position of the mouse.
Where Im getting confused is with the coordinate space and the Trig behind it all.
My issues come from the fact that the X/Y space on the screen is not centered at 0,0. But the top left of the screen is 0,0 and the bottom right is 500,500.
My calculations are:
var xpos:Number = mouseEvent.getX();
var ypos:Number = mouseEvent.getY();
var center_pos_x:Number = 250;
var center_pos_y:Number = 250;
var length = ypos - center_pos_y;
var height = xpos - center_pos_x;
var angle_deg = Math.toDegrees(Math.atan(height / length));
var angle_rad = Math.toRadians(angle_deg);
var radius = 50;
moving_circ_xpos = (radius * Math.cos(angle_rad)) + center_pos_x;
moving_circ_ypos = (radius * Math.sin(angle_rad)) + center_pos_y;
I made the app print out the angle (angle_deg) that I have calculated when I move the mouse and my output is below:
When the mouse is (in degrees moving anti-clockwise):
directly above the circle and horizontally inline with the center, the angle is -0
to the left and vertically centered, the angle is -90
directly below the circle and horizontally inline with the center, the angle is 0
to the right and vertically centered, the angle is 90
So, what can I do to make it 0, 90, 180, 270??
I know it must be something small, but I just cant think of what it is...
Thanks for any help
(and no, this is not an assignment)
atan(height/length) is not enough to get the angle. You need to compensate for each quadrant, as well as the possibility of "division-by-zero". Most programming language libraries supply a method called atan2 which take two arguments; y and x. This method does this calculation for you.
More information on Wikipedia: atan2
You can get away without calculating the angle. Instead, use the center of your circle (250,250) and the position of the mouse (xpos,ypos) to define a line. The line intersects your circle when its length is equal to the radius of your circle:
// Calculate distance from center to mouse.
xlen = xpos - x_center_pos;
ylen = ypos - y_center_pos;
line_len = sqrt(xlen*xlen + ylen*ylen); // Pythagoras: x^2 + y^2 = distance^2
// Find the intersection with the circle.
moving_circ_xpos = x_center_pos + (xlen * radius / line_len);
moving_circ_ypos = y_center_pos + (ylen * radius / line_len);
Just verify that the mouse isn't at the center of your circle, or the line_len will be zero and the mouse will be sucked into a black hole.
There's a great book called "Graphics Gems" that can help with this kind of problem. It is a cookbook of algorithms and source code (in C I think), and allows you to quickly solve a problem using tested functionality. I would totally recommend getting your hands on it - it saved me big time when I quickly needed to add code to do fairly complex operations with normals to surfaces, and collision detections.