Develop Reference

How to use AffineTransform with very little coordinates?

I have a set of two dimensions points. Their X and Y are greater than -2 and lesser than 2. Such point could be : (-0.00012 ; 1.2334 ).
I would want to display these points on a graph, using rectangles (a rectangle illustrates a point, and has its coordinates set to its point's ones - moreover, it has a size of 10*10).
Rectangles like (... ; Y) should be displayed above any rectangles like (... ; Y-1) (positive Y direction is up). Thus, I must set the graph's origin not at the top-left hand-corner, but somewhere else.
I'm trying to use Graphics2D's AffineTransform to do that.
I get the minimal value for all the X coordinates
I get the minimal value for all the Y coordinates
I get the maximal value for all the X coordinates
I get the maximal value for all the Y coordinates
I get the distance xmax-xmin and ymax-ymin
Then, I wrote the code I give you below.
Screenshots
Some days ago, using my own method to scale, I had this graph:
(so as I explained, Y are inverted and that's not a good thing)
For the moment, i.e., with the code I give you below, I have only one point that takes all the graph's place! Not good at all.
I would want to have:
(without lines, and without graph's axis. The important here is that points are correctly displayed, according to their coordinates).
Code
To get min and max coordinates value:
x_min = Double.parseDouble((String) list_all_points.get(0).get(0));
x_max = Double.parseDouble((String) list_all_points.get(0).get(0));
y_min = Double.parseDouble((String) list_all_points.get(0).get(1));
y_max = Double.parseDouble((String) list_all_points.get(0).get(1));
for(StorableData s : list_all_points) {
if(Double.parseDouble((String) s.get(0)) < x_min) {
x_min = Double.parseDouble((String) s.get(0));
}
if(Double.parseDouble((String) s.get(0)) > x_max) {
x_max = Double.parseDouble((String) s.get(0));
}
if(Double.parseDouble((String) s.get(1)) < y_min) {
y_min = Double.parseDouble((String) s.get(1));
}
if(Double.parseDouble((String) s.get(1)) > y_max) {
y_max = Double.parseDouble((String) s.get(1));
}
}
To draw a point:
int x, y;
private void drawPoint(Cupple storable_data) {
//x = (int) (storable_data.getNumber(0) * scaling_coef + move_x);
//y = (int) (storable_data.getNumber(1) * scaling_coef + move_y);
x = storable_data.getNumber(0).intValue();
y = storable_data.getNumber(1).intValue();
graphics.fillRect(x, y, 10, 10);
graphics.drawString(storable_data.toString(), x - 5, y - 5);
}
To paint the graph:
#Override
public void paint(Graphics graphics) {
this.graphics = graphics;
Graphics2D graphics_2d = ((Graphics2D) this.graphics);
AffineTransform affine_transform = graphics_2d.getTransform();
affine_transform.scale(getWidth()/(x_max - x_min), getHeight()/(y_max - y_min));
affine_transform.translate(x_min, y_min);
graphics_2d.transform(affine_transform);
for(StorableData storable_data : list_all_points) {
graphics_2d.setColor(Color.WHITE);
this.drawPoint((Cupple) storable_data);
}

I suggest you map each data point to a point on the screen, thus avoiding the following coordinate system pitfalls. Take your list of points and create from them a list of points to draw. Take into account that:
The drawing is pixel-based, so you will want to scale your points (or you would have rectangles 1 to 4 pixels wide...).
You will need to translate all your points because negative values will be outside the boundaries of the component on which you draw.
The direction of the y axis is reversed in the drawing coordinates.
Once that is done, use the new list of points for the drawing and the initial one for calculations. Here is an example:
public class Graph extends JPanel {
private static int gridSize = 6;
private static int scale = 100;
private static int size = gridSize * scale;
private static int translate = size / 2;
private static int pointSize = 10;
List<Point> dataPoints, scaledPoints;
Graph() {
setBackground(Color.WHITE);
// points taken from your example
Point p1 = new Point(-1, -2);
Point p2 = new Point(-1, 0);
Point p3 = new Point(1, 0);
Point p4 = new Point(1, -2);
dataPoints = Arrays.asList(p1, p2, p3, p4);
scaledPoints = dataPoints.stream()
.map(p -> new Point(p.x * scale + translate, -p.y * scale + translate))
.collect(Collectors.toList());
}
#Override
public Dimension getPreferredSize() {
return new Dimension(size, size);
}
#Override
protected void paintComponent(Graphics g) {
super.paintComponent(g);
Graphics2D g2d = (Graphics2D) g;
// draw a grid
for (int i = 0; i < gridSize; i++) {
g2d.drawLine(i * scale, 0, i * scale, size);
g2d.drawLine(0, i * scale, size, i * scale);
}
// draw the rectangle
g2d.setPaint(Color.RED);
g2d.drawPolygon(scaledPoints.stream().mapToInt(p -> p.x).toArray(),
scaledPoints.stream().mapToInt(p -> p.y).toArray(),
scaledPoints.size());
// draw the points
g2d.setPaint(Color.BLUE);
// origin
g2d.fillRect(translate, translate, pointSize, pointSize);
g2d.drawString("(0, 0)", translate, translate);
// data
for (int i = 0; i < dataPoints.size(); i++) {
Point sp = scaledPoints.get(i);
Point dp = dataPoints.get(i);
g2d.fillRect(sp.x, sp.y, pointSize, pointSize);
g2d.drawString("(" + dp.x + ", " + dp.y + ")", sp.x, sp.y);
}
}
public static void main(String[] args) {
JFrame frame = new JFrame();
frame.setContentPane(new Graph());
frame.setDefaultCloseOperation(JFrame.EXIT_ON_CLOSE);
frame.pack();
frame.setLocationRelativeTo(null);
frame.setVisible(true);
}
}
And another:
You might want to have the points aligned on the grid intersections and not below and to the right of them. I trust you will figure this one out.
Also, I ordered the points so that drawPolygon will paint the lines in the correct order. If your points are arbitrarily arranged, look for ways to find the outline. If you want lines between all points like in your example, iterate over all combinations of them with drawLine.

Java: Drawing a Star from a Polygon Circle w/Point Connections

So, I have been working on this program for drawing a star from a circle created using g.fillPolygon(int, int, int). I was intially having issues with creating an entire circle, but changed double angle = (i * 360) to (i * 720) to fix that (may be a band-aid fix, not sure yet). Now I'm in the process of attempting to connect all the points together (as shown in the "Target Output" section).
Note: I believe that the labeling of the points shown in the modification section was not done wih Java.
My Code: (Where I'm at right now)
import java.awt.*;
public class StarSampler {
public static void main(String[] args)
{
DrawingPanel panel = new DrawingPanel(500, 500);
Graphics2D g = panel.getGraphics();
g.setColor(Color.YELLOW);
fillStar(g, 250, 250, 150, 50, .7);
}
public static void fillStar(Graphics2D g, int ctrX, int ctrY, int radius, int nPoints, double spikiness)
{
double xDouble[] = new double[2*nPoints];
double yDouble[] = new double[2*nPoints];
int xPoint[] = new int[100];
int yPoint[] = new int[100];
for (int i = 0; i < 2*nPoints; i++)
{
double iRadius = (i % 2 == 0) ? radius : (radius * spikiness);
double angle = (i * 720.0) / (2*nPoints);
xDouble[i] = ctrX + iRadius * Math.cos(Math.toRadians(angle));
yDouble[i] = ctrY + iRadius * Math.sin(Math.toRadians(angle));
for (int j = 0; j < nPoints; j++) // Casts for ints and doubles
{
xPoint[j] = (int) xDouble[j];
yPoint[j] = (int) yDouble[j];
}
}
g.fillPolygon(xPoint, yPoint, nPoints); // Creates polygon but
}
}
My Code's Output:
Target Output (What I'm generally aiming for, not both together):

How to determine the coordinates of the next rectangle on JPanel

i have an arraylist RecArray of objects with each object containing two int values, one for the width and for the height of a rectangle. Each rectangle's height and width are a multiple of ten. the rectangles have to be passed on to the surface as in the given order in RecArray from left to right and from top to bottom. my problem is i can not find the x,y coordinates of the next rectangle. what im trying to do is, starting at the coordinate (0,0) i generate the first rectangle, add it to an arraylist RecList. Then i set the x and y coordinates. x becomes x = x+RecArray.get(0).getLength1() + 1. if x is greater than the width of the jpanel surface then it becomes 0 and y becomes y = y + 10 . starting from the second object in the RecArray i try to generate rectangles with the given coordinates and width&height. Then i try to compare them with all the previous rectangles to see if there is any overlapping. if there is no overlapping, the rectangle will be drawn, if there is overlapping, the x coordinate of the rec becomes x = RecList.get(j).width+1 and if that exceeds the width x becomes 0 and y is y=y+10. Then i regenate the current rectangle with the new coordinates and compare with the other rectangles in RecList again till i find the right spot for the current rectangle.ive been dealing with that issue for the last 5 days and am really fed up now. i would greatly appreciate any tipps. and Please be patient with me. im still learning programming.
super.paintComponent(g);
Graphics2D g2 = (Graphics2D) g;
Rectangle rec = new Rectangle(x, y, RecArray.get(0).getWidth(),
RecArray.get(0).getHeight());
RecList.add(rec);
recPaint(g2,RecArray.get(0));
x = x + RecArray.get(0).getWidth() + 1;
int i;
for (i = 1; i < RecArray.size(); i++) {
if (x >= this.getArea().getWidth()) {
x = 0;
y = y + 10;
}
Rectangle rec1 = new Rectangle(x, y, RecArray.get(i)
.getWidth(), RecArray.get(i).getheight());
for (int j= 0; j < RecList.size(); j++) {
if (!recIntersect(rec1, RecList.get(j))) {
RecList.add(rec1);
recPaint(g2,RecArray.get(i));
break;
}
else {
x = RecList.get(j).width;
if (x >= this.getFlaeche().getLength1()) {
x = 0;
y = y + 10;
}
rec1 = new Rectangle(x, y,RecArray.get(i). .getWidth(),
RecArray.get(i).getHeight());
}
x = x + RecArray.get(i).getWidth();
}
//With this method using the given rec parameter a rectangle will be drawn on the g2 and filled in blue colour
private void recPaint (Graphics2D g2, RecType rec){
g2.setColor(Color.BLUE);
g2.fillRect(x, y, rec.getWidth(),
rec.getLength2());
g2.setColor(Color.BLACK);
g2.drawRect(x, y, rec.getHeight(),
rec.getLength2());
}
// returns true, if two rectangles overlap
private boolean recIntersect(Rectangle rec1, Rectangle rec2) {
if( rec1.intersects(rec2)){
return true;
}
return false;
}
Edit: apparently i haven't stated clearly what my problem is. my problem is, that the way i generate (x,y) coordinates of the rectangles is obviously wrong. the way my algorithm works doesnt get the results i want. i want my rectangles to be placed neatly next to/above/below each other WITHOUT overlapping, which is not the case.

Separate out your List of Rectangles. Calculate the X, Y coordinates once.
Since I didn't have your object class, I used the Dimension class, which holds a width and a length. I used the Rectangle class to hold the objects that will eventually be drawn in your Swing GUI.
Divide and conquer. Separate out your GUI model, view, and controller(s). This way, you can focus on one piece of the puzzle at a time.
Here are the results of my test code when I ran it with a drawing area of 500, 400.
java.awt.Rectangle[x=0,y=0,width=100,height=100]
java.awt.Rectangle[x=100,y=0,width=20,height=10]
java.awt.Rectangle[x=120,y=0,width=40,height=20]
java.awt.Rectangle[x=160,y=0,width=60,height=40]
java.awt.Rectangle[x=220,y=0,width=80,height=60]
java.awt.Rectangle[x=300,y=0,width=20,height=10]
java.awt.Rectangle[x=320,y=0,width=120,height=110]
Here are the results of my test code when I ran it with a drawing area of 200, 200.
java.awt.Rectangle[x=0,y=0,width=100,height=100]
java.awt.Rectangle[x=100,y=0,width=20,height=10]
java.awt.Rectangle[x=120,y=0,width=40,height=20]
java.awt.Rectangle[x=0,y=100,width=60,height=40]
java.awt.Rectangle[x=60,y=100,width=80,height=60]
java.awt.Rectangle[x=140,y=100,width=20,height=10]
And here's the code. I fit rectangles on the X axis until I can't fit another rectangle. Then I add the maximum height to Y, reset the X to zero, reset the maximum height and fit the next row of rectangles.
Create test applications like I did here and make sure that you can create the GUI model long before you create the GUI view and GUI controller.
package com.ggl.testing;
import java.awt.Dimension;
import java.awt.Rectangle;
import java.util.ArrayList;
import java.util.List;
public class CalculatingRectangles {
public static void main(String[] args) {
CalculatingRectangles calculatingRectangles = new CalculatingRectangles();
Dimension drawingArea = new Dimension(200, 200);
List<Dimension> dimensions = new ArrayList<>();
dimensions.add(new Dimension(100, 100));
dimensions.add(new Dimension(20, 10));
dimensions.add(new Dimension(40, 20));
dimensions.add(new Dimension(60, 40));
dimensions.add(new Dimension(80, 60));
dimensions.add(new Dimension(20, 10));
dimensions.add(new Dimension(120, 110));
List<Rectangle> rectangles = calculatingRectangles
.calculatingRectangles(drawingArea, dimensions);
System.out.println(displayRectangles(rectangles));
}
private static String displayRectangles(List<Rectangle> rectangles) {
StringBuilder builder = new StringBuilder();
for (Rectangle r : rectangles) {
builder.append(r);
builder.append(System.getProperty("line.separator"));
}
return builder.toString();
}
public List<Rectangle> calculatingRectangles(Dimension drawingArea,
List<Dimension> dimensions) {
int width = drawingArea.width;
int height = drawingArea.height;
int x = 0;
int y = 0;
int index = 0;
int maxHeight = 0;
boolean hasRoom = dimensions.size() > index;
List<Rectangle> rectangles = new ArrayList<>();
while (hasRoom) {
Dimension d = dimensions.get(index);
maxHeight = Math.max(maxHeight, d.height);
if ((x + d.width) <= width && (y + maxHeight) <= height) {
Rectangle r = new Rectangle(x, y, d.width, d.height);
x += d.width;
rectangles.add(r);
index++;
if (index >= dimensions.size()) {
hasRoom = false;
}
} else {
y += maxHeight;
if (y > height) {
hasRoom = false;
}
x = 0;
}
}
return rectangles;
}
}

Is there a method that computes a smaller rectangle out of two given rectangles?

One of the exercises in my Java textbook says "Consult the API documentation to find methods for:
Computing the smallest rectangle that contains two given rectangles. • Returning a random floating-point number."
I've looked at the Java API for class Rectangle, but I can't find one that computes the smaller rectangle. The closest methods I've found are union and bounds, but I don't think that's correct.
I found min from the Java Math class and wrote a test program to see if it would work, but min cannot have arguments of rectangles.
Here's the code I wrote:
import java.awt.Rectangle;
public class RectangleSize {
public static void main(String[] args)
{
Rectangle a = new Rectangle(5, 5, 10, 10);
Rectangle b = new Rectangle(5, 5, 20, 20);
int min = Math.min(a, b); //In Eclipse, I get an error.
System.out.println(min);
}
}

I am studying java recently using the textbook of Big Java Early Object. I found the same question in chapter 2. Eventually I found the answer by looking at Java SE8 API list.
First use method .add() to combine 2 rectangles.
Note: actually, this combined one is already the one you need. But you can get a new rectangle (same size and location) at step 2.
Them use .getBounds() to get the smallest rectangle containing the combined one.
.
import java.awt.*;
public class RectangleTester01 {
public static void main(String[] args) {
Rectangle box1 = new Rectangle(10, 20, 40, 40);
Rectangle box2 = new Rectangle(20, 30, 60, 60);
box1.add(box2);
System.out.println(box1);
Rectangle box3 = box1.getBounds();
System.out.println(box3);
}
}
output is:
java.awt.Rectangle[x=10,y=20,width=70,height=70]
java.awt.Rectangle[x=10,y=20,width=70,height=70]

You want to use Rectangle.contains . You would be given many rectangles. You'd need to loop through all the rectangles and see if it contains the two rectangles given. If it does you should calculate the size of that rectangle. In the end you take the rectangle with the smallest size.
public Rectangle getSmallest(Rectangle one, Rectangle two, Rectangle[] rectangles) {
Rectangle smallest = null;
double area = Double.MAX_VALUE;
for (Rectangle r: rectangles) {
if (r.contains(one) && r.contains(two)) {
calculatedArea = r.getWidth() * r.getHeight();
if (calculatedArea < area) {
area = calculatedArea;
smallest = r;
}
}
}
return r;
}

I think the method you're looking for is Rectangle2D.createUnion. It combines two rectangles to make a bigger one that contains both with a minimum of extra space.

Here is another question to find a rectangle which contains a list of rectangles.find-smallest-area-that-contains-all-the-rectangles.
Here is my brute answer which is not accuracy and did not try the union function of Rectengle and also the createUnion(Rectengle2D r) as well.
Math.min can help find the minimum number in a number array. You may write a similar one for Rectangle. This is something from 1D to 2D in my mind. BTW, it would be much interesting if you extends it to 3D or nD objects calculation.
package com.stackoverflow.q26311076;
import java.awt.Rectangle;
public class Test {
public static void main(String[] args) {
Rectangle a = new Rectangle(5, 5, 10 , 10);
Rectangle b = new Rectangle(5, 5, 20 , 20);
Rectangle min = getMin(a, b) ;
// ...
}
public static Rectangle getMin(Rectangle a, Rectangle b) {
//find the min range in X.
{
double x1 = a.getX();
double x2 = a.getX() + a.getWidth();
double x3 = b.getX();
double x4 = b.getX() + b.getWidth();
double minX1 =Math.min( Math.min(x1, x2), Math.min(x3, x4)) ;
double maxX1 =Math.max( Math.max(x1, x2), Math.max(x3, x4)) ;
}
//find the min range in Y.
{
double y1 = a.getY();
double y2 = a.getY() + a.getHeight();
double y3 = b.getY();
double y4 = b.getY() + b.getHeight();
double minY1 =Math.min( Math.min(y1, y2), Math.min(y3, y4)) ;
double maxY1 =Math.max( Math.max(y1, y2), Math.max(y3, y4)) ;
}
//build new rectangle with X & Y
Rectangle r = new Rectangle();
r.setRect(minX1, minY1, maxX1 - minX1, maxY1 - minY1);
return r;
}
}

Space ship simulator guidance computer targeting with concentric indicator squares

I'm working on a 3D space trading game with some people, and one of the things I've been assigned to do is to make a guidance computer 'tunnel' that the ship travels through, with the tunnel made of squares that the user flies through to their destination, increasing in number as the user gets closer to the destination.
It's only necessary to render the squares for the points ahead of the ship, since that's all that's visible to the user. On their way to a destination, the ship's computer is supposed to put up squares on the HUD that represent fixed points in space between you and the destination, which are small in the distance and get larger as the points approach the craft.
I've had a go at implementing this and can't seem to figure it out, mainly using logarithms (Math.log10(x) and such). I tried to get to get the ship position in 'logarithmic space' to help find out what index to start from when drawing the squares, but then the fact that I only have distance to the destination to work with confuses the matter, especially when you consider that the number of squares has to vary dynamically to make sure they stay fixed at the right locations in space (i.e., the squares are positioned at intervals of 200 or so before being transformed logarithmically).
With regard to this, I had a working implementation with the ship between a start of 0.0d and end of 1.0d, although the implementation wasn't so nice. Anyway, the problem essentially boils down to a 1d nature. Any advice would be appreciated with this issue, including possible workarounds to achieve the same effect or solutions.
(Also, there's a Youtube video showing this effect: http://www.youtube.com/watch?v=79F9Nj7GgfM&t=3m5s)
Cheers,
Chris
Edit: rephrased the entire question.
Edit: new testbed code:
package st;
import java.awt.BorderLayout;
import java.awt.Canvas;
import java.awt.Color;
import java.awt.Dimension;
import java.awt.Font;
import java.awt.Graphics;
import java.awt.Graphics2D;
import java.awt.GraphicsDevice;
import java.awt.GraphicsEnvironment;
import java.awt.event.ActionEvent;
import java.awt.event.ActionListener;
import java.awt.image.BufferStrategy;
import java.text.DecimalFormat;
import javax.swing.JFrame;
import javax.swing.JPanel;
import javax.swing.SwingUtilities;
import javax.swing.Timer;
public class StUI2 extends JFrame {
public static final double DEG_TO_RAD = Math.PI / 180.0d;
public static final DecimalFormat decimalFormat = new DecimalFormat("0.0000");
public static final Font MONO = new Font("Monospaced", Font.PLAIN, 10);
public class StPanel extends Canvas {
protected final Object imgLock = new Object();
protected int lastWidth = 1, lastHeight = 1;
protected boolean first = true;
protected Color bgColour = Color.DARK_GRAY, gridColour = Color.GRAY;
double shipWrap = 700;
double shipFrame = 100;
double shipPos = 0;
long lastUpdateTimeMS = -1;
long currUpdateTimeMS = -1;
public StPanel() {
setFocusable(true);
setMinimumSize(new Dimension(1, 1));
setAlwaysOnTop(true);
}
public void internalPaint(Graphics2D g) {
synchronized (imgLock) {
if (lastUpdateTimeMS < 0) {
lastUpdateTimeMS = System.currentTimeMillis();
}
currUpdateTimeMS = System.currentTimeMillis();
long diffMS = currUpdateTimeMS - lastUpdateTimeMS;
g.setFont(MONO);
shipPos += (60d * ((double)diffMS / 1000));
if (shipPos > shipWrap) {
shipPos = 0d;
}
double shipPosPerc = shipPos / shipWrap;
double distToDest = shipWrap - shipPos;
double compression = 1000d / distToDest;
g.setColor(bgColour);
Dimension d = getSize();
g.fillRect(0, 0, (int)d.getWidth(), (int)d.getHeight());
//int amnt2 = (int)unlog10((1000d / distToDest));
g.setColor(Color.WHITE);
g.drawString("shipPos: " + decimalFormat.format(shipPos), 10, 10);
g.drawString("distToDest: " + decimalFormat.format(distToDest), 10, 20);
g.drawString("shipWrap: " + decimalFormat.format(shipWrap), 150, 10);
int offset = 40;
g.setFont(MONO);
double scalingFactor = 10d;
double dist = 0;
int curri = 0;
int i = 0;
do {
curri = i;
g.setColor(Color.GREEN);
dist = distToDest - getSquareDistance(distToDest, scalingFactor, i);
double sqh = getSquareHeight(dist, 100d * DEG_TO_RAD);
g.drawLine(30 + (int)dist, (offset + 50) - (int)(sqh / 2d), 30 + (int)dist, (offset + 50) + (int)(sqh / 2d));
g.setColor(Color.LIGHT_GRAY);
g.drawString("i: " + i + ", dist: " + decimalFormat.format(dist), 10, 120 + (i * 10));
i++;
} while (dist < distToDest);
g.drawLine(10, 122, 200, 122);
g.drawString("last / i: " + curri + ", dist: " + decimalFormat.format(dist), 10, 122 + (i * 10));
g.setColor(Color.MAGENTA);
g.fillOval(30 + (int)shipPos, offset + 50, 4, 4);
lastUpdateTimeMS = currUpdateTimeMS;
}
}
public double getSquareDistance(double initialDist, double scalingFactor, int num) {
return Math.pow(scalingFactor, num) * num * initialDist;
}
public double getSquareHeight(double distance, double angle) {
return distance / Math.tan(angle);
}
/* (non-Javadoc)
* #see java.awt.Canvas#paint(java.awt.Graphics)
*/
#Override
public void paint(Graphics g) {
internalPaint((Graphics2D)g);
}
public void redraw() {
synchronized (imgLock) {
Dimension d = getSize();
if (d.width == 0) d.width = 1;
if (d.height == 0) d.height = 1;
if (first || d.getWidth() != lastWidth || d.getHeight() != lastHeight) {
first = false;
// remake buf
GraphicsEnvironment ge = GraphicsEnvironment.getLocalGraphicsEnvironment();
//create an object that represents the device that outputs to screen (video card).
GraphicsDevice gd = ge.getDefaultScreenDevice();
gd.getDefaultConfiguration();
createBufferStrategy(2);
lastWidth = (int)d.getWidth();
lastHeight = (int)d.getHeight();
}
BufferStrategy strategy = getBufferStrategy();
Graphics2D g = (Graphics2D)strategy.getDrawGraphics();
internalPaint(g);
g.dispose();
if (!strategy.contentsLost()) strategy.show();
}
}
}
protected final StPanel canvas;
protected Timer viewTimer = new Timer(1000 / 60, new ActionListener() {
#Override
public void actionPerformed(ActionEvent e) {
canvas.redraw();
}
});
{
viewTimer.setRepeats(true);
viewTimer.setCoalesce(true);
}
/**
* Create the applet.
*/
public StUI2() {
JPanel panel = new JPanel(new BorderLayout());
setContentPane(panel);
panel.add(canvas = new StPanel(), BorderLayout.CENTER);
setVisible(true);
setDefaultCloseOperation(EXIT_ON_CLOSE);
setSize(800, 300);
setTitle("Targetting indicator test #2");
viewTimer.start();
}
public static double unlog10(double x) {
return Math.pow(10d, x);
}
public static void main(String[] args) {
SwingUtilities.invokeLater(new Runnable() {
#Override
public void run() {
StUI2 ui = new StUI2();
}
});
}
}

Assuming you want the squares to be equal height (when you reach them), you can calculate a scaling factor based on the distance to the destination (d) and the required height of the squares upon reaching them (h).
From these two pieces of information you can calculate the inverse tangent (atan) of the angle (alpha) between the line connecting the ship to the destination (horizontal line in your image) and the line connecting the top of the squares with the destination (angled line in your image).
EDIT: corrected formula
Using the angle, you can calculate the height of the square (h') at any given distance from the destination: you know the distance to the destination (d') and the angle (alpha); The height of the square at distance d' is h'=r'*sin(alpha) -- sin(alpha)=cos(alpha)*tan(alpha) and r'=d'/cos(alpha) (the distance between the destination and the top of the square -- the "radius"). Or more easily: h'=d'*tan(alpha).
Note: adopting the algorithm to varying height (when you reach them) squares is relatively simple: when calculating the angle, just assume a (phantom) square of fixed height and scale the squares relatively to that.
If the height of the square at distance d' is calculated for you by your graphic library, all the better, you only need to figure out the distances to place the squares.
What distances to place the squares from the destination?
1) If you want a varying number of squares shown (in front of the ship), but potentially infinite number of squares to consider (based on d), you can chose the distance of the closest square to the destination (d1) and calculate the distances of other squares by the formula s^k*k*d1, where s (scaling factor) is a number > 1 for the k'th square (counting from the destination). You can stop the algorithm when the result is larger than d.
Note that if d is sufficiently large, the squares closest to the distance will block the destination (there are many of them and their heights are small due to the low angle). In this case you can introduce a minimal distance (possibly based on d), below which you do not display the squares -- you will have to experiment with the exact values to see what looks right/acceptable.
2) If you want a fixed amount of squares (sn) showing always, regardless of d, you can calculate the distances of the squares from the destination by the formula d*s^k, where s is a number < 1, k is the index of the square (counting from the ship). The consideration about small squares probably don't apply here unless sn is high.
To fix the updated code, change the relavant part to:
double dist = 0;
double d1 = 10;
int curri = 0;
int i = 1;
int maxSquareHeight = 40;
double angle = Math.atan(maxSquareHeight/distToDest);
while (true)
{
curri = i;
g.setColor(Color.GREEN);
dist = getSquareDistance(d1, scalingFactor, i);
if (dist > distToDest) {
break;
}
double sqh = getSquareHeight(dist, angle);
g.drawLine(30 + (int)(shipWrap - dist), offset+50-(int)(sqh / 2d), 30 + (int)(shipWrap - dist), offset+50+(int)(sqh / 2d));
g.setColor(Color.LIGHT_GRAY);
i++;
}
public double getSquareHeight(double distance, double angle) {
return distance * Math.tan(angle);
}
You should also reduce scalingFactor to the magnitude of ~1.5.
EDIT: If you replace the formula s^k*k*d1 with s^(k-1)*k*d1, then the first square will be exactly at distance d1.
EDIT: fixed square height calculating formula
EDIT: updated code