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How do I get the endpoints of a line in Java knowing angle, midpoint, and length

I need to find the endpoints of a line knowing its midpoint, angle, and length. First, I tried this:
public Point getEndpointA(Point midpoint, double angle, double length) {
Point a = new Point(0, 0);
a.x = midpoint.x + (int) length * Math.cos(angle);
a.y = midpoint.y + (int) length * Math.sin(angle);
return a;
}
public Point getEnpointB(Point midpoint, double angle, double length) {
Point b = new Point(0, 0);
b.x = midpoint.x + (int) length * Math.cos(-angle);
b.y = midpoint.y + (int) length * Math.sin(-angle);
return b;
}
and then I called these functions to get the enpoints of the lines. However, this was inaccurate, and the points weren't placed quite where I hoped. Then, I tried experimenting with the distance formula, but that didn't work because I am only just starting Algebra I and I was having trouble getting rid of exponents and radical signs, which is not covered in Algebra I. (I looked up the functions sin and cos, and I understand what they mean now, so that's how I know about them) So, can somebody write a function that would return the endpoints of a line based of its midpoint, angle, and length, and explain to me how it works?

You want to specify the angle in degrees like followings?
public Point getEndpointA(Point midpoint, double angleDegee, double length) {
Point a = new Point(0, 0);
// convert degrees=>radians
final double angleRad = Math.toRadians(angleDegee);
a.x = (int) (midpoint.x + (int) length * Math.cos(angleRad));
a.y = (int) (midpoint.y + (int) length * Math.sin(angleRad));
return a;
}
public Point getEndpointB(Point midpoint, double angleDeg, double length) {
Point b = new Point(0, 0);
final double angleRad = Math.toRadians(angleDeg + 180d);
b.x = (int) (midpoint.x + (int) length * Math.cos(angleRad));
b.y = (int) (midpoint.y + (int) length * Math.sin(angleRad));
return b;
}

How do I get the coordinates of a diagonal circle?

I am almost done of getting the coordinates of a diagonal circle.
Here is what I have so far.
// Center point
double centerX;
double centerY;
double centerZ;
for (double degree = 0D; degree < 360D; degree = degree + 8D) {
double angle = degree * Math.PI / 180D;
// Difference from the center
double x = 1.5D * Math.cos(angle);
double y;
if (degree >= 0D && degree < 90D) {
y = degree / 90D;
} else if (degree >= 90D && degree < 180D) {
y = 1D - ((degree - 90D) / 90D);
} else if (degree >= 180D && degree < 270D) {
y = -1D * ((degree - 180D) / 90D);
} else {
y = -1D * (1D - ((degree - 270D) / 90D));
}
double z = 1.5D * Math.sin(angle);
// New point
double pointX = centerX + x;
double pointY = centerY + y;
double pointZ = centerZ + z;
}
Here is the output in a game.
It is not perfect because it creates some edges and it looks inefficient to me.
How do I correct it?
Is there a better way to do this?

This should look similar to what you have already, but it's simpler and smoother:
double y = 1.0D * Math.sin(angle);
Now, with these dimensions, the result is not quite a circle, but a stretched ellipse. If you want a circle, make sure the coefficients on the cosine and sine obey the Pythagorean Theorem. For example:
double x = 1.5D * Math.cos(angle);
double y = 0.9D * Math.sin(angle);
double z = 1.2D * Math.sin(angle);
These coefficients will ensure that x^2 + y^2 + z^2 is a constant for every angle. You can verify that this is true, given the identity cos^2 + sin^2 = 1. (The coefficient representing the hypotenuse should be attached to the coordinate that uses a different trig function than the other two.)
For the most maintainable code, you might find it better to assign (x, y, z) = (cos, sin, 0) and then apply a rotation matrix, or a sequence of rotation matrices, to the vector (x, y, z). This will be easier to read and harder to mess up, if you want to fine-tune the amount of rotation later.

Making minecraft player look at point

im trying to make the player look at at specific point. What i have looks at the point but only if its in its field of vision. and i would like for it to look up and down as well
public static void lookAt(double x, double y, double z) {
double l;
double w;
double c;
l = x - playerX;
w = z - playerZ;
c = Math.sqrt(l * l + w + w);
double alpha1 = -Math.asin(l / c) / Math.PI * 180;
double alpha2 = Math.acos(w / c) / Math.PI * 180;
if (alpha2 > 180) {
playerYaw = 180 - (float) alpha1;
} else {
playerYaw = (float) alpha1;
}
}
i looked for a pitch algorithm but none seem to work.

Late answer but oh well, you can calculate the pitch and yaw with simple trigonometry if you can get a unit vector that represents the direction from your position to the point you want to look at.
To start you need to subtract the your position by the point and normalize the result, that will give us the directional vector(unit vector) that we need.
Now we can calculate the angles, the arc sine of Y will give us the the pitch angle, and the arc tan of X and Z will give us the yaw angle.
To understand why: https://en.wikipedia.org/wiki/Inverse_trigonometric_functions
and now you should have the pitch and yaw, don't forget to convert them to degrees!
here's some code:
public static void LookAt(double px, double py, double pz , EntityPlayer me)
{
double dirx = me.getPosition(0f).X - px;
double diry = me.getPosition(0f).Y - py;
double dirz = me.getPosition(0f).Z - pz;
double len = Math.sqrt(dirx*dirx + diry*diry + dirz*dirz);
dirx /= len
diry /= len;
dirz /= len;
double pitch = Math.asin(dir.y);
double yaw = Math.atan2(dir.z, dir.x);
//to degree
pitch = pitch * 180.0 / Math.PI;
yaw = yaw * 180.0 / Math.PI;
yaw += 90f;
me.rotationPitch = (float)pitch;
me.rotationYaw = (float)yaw;
}

Convert latitude/longitude point to a pixels (x,y) on mercator projection

I'm trying to convert a lat/long point into a 2d point so that I can display it on an image of the world-which is a mercator projection.
I've seen various ways of doing this and a few questions on stack overflow-I've tried out the different code snippets and although I get the correct longitude to pixel, the latitude is always off-seems to be getting more reasonable though.
I need the formula to take into account the image size, width etc.
I've tried this piece of code:
double minLat = -85.05112878;
double minLong = -180;
double maxLat = 85.05112878;
double maxLong = 180;
// Map image size (in points)
double mapHeight = 768.0;
double mapWidth = 991.0;
// Determine the map scale (points per degree)
double xScale = mapWidth/ (maxLong - minLong);
double yScale = mapHeight / (maxLat - minLat);
// position of map image for point
double x = (lon - minLong) * xScale;
double y = - (lat + minLat) * yScale;
System.out.println("final coords: " + x + " " + y);
The latitude seems to be off by about 30px in the example I'm trying. Any help or advice?
Update
Based on this question:Lat/lon to xy
I've tried to use the code provided but I'm still having some problems with latitude conversion, longitude is fine.
int mapWidth = 991;
int mapHeight = 768;
double mapLonLeft = -180;
double mapLonRight = 180;
double mapLonDelta = mapLonRight - mapLonLeft;
double mapLatBottom = -85.05112878;
double mapLatBottomDegree = mapLatBottom * Math.PI / 180;
double worldMapWidth = ((mapWidth / mapLonDelta) * 360) / (2 * Math.PI);
double mapOffsetY = (worldMapWidth / 2 * Math.log((1 + Math.sin(mapLatBottomDegree)) / (1 - Math.sin(mapLatBottomDegree))));
double x = (lon - mapLonLeft) * (mapWidth / mapLonDelta);
double y = 0.1;
if (lat < 0) {
lat = lat * Math.PI / 180;
y = mapHeight - ((worldMapWidth / 2 * Math.log((1 + Math.sin(lat)) / (1 - Math.sin(lat)))) - mapOffsetY);
} else if (lat > 0) {
lat = lat * Math.PI / 180;
lat = lat * -1;
y = mapHeight - ((worldMapWidth / 2 * Math.log((1 + Math.sin(lat)) / (1 - Math.sin(lat)))) - mapOffsetY);
System.out.println("y before minus: " + y);
y = mapHeight - y;
} else {
y = mapHeight / 2;
}
System.out.println(x);
System.out.println(y);
When using the original code if the latitude value is positive it returned a negative point, so I modified it slightly and tested with the extreme latitudes-which should be point 0 and point 766, it works fine. However when I try a different latitude value ex: 58.07 (just north of the UK) it displays as north of Spain.

The Mercator map projection is a special limiting case of the Lambert Conic Conformal map projection with
the equator as the single standard parallel. All other parallels of latitude are straight lines and the meridians
are also straight lines at right angles to the equator, equally spaced. It is the basis for the transverse and
oblique forms of the projection. It is little used for land mapping purposes but is in almost universal use for
navigation charts. As well as being conformal, it has the particular property that straight lines drawn on it are
lines of constant bearing. Thus navigators may derive their course from the angle the straight course line
makes with the meridians. [1.]
The formulas to derive projected Easting and Northing coordinates from spherical latitude φ and longitude λ
are:
E = FE + R (λ – λₒ)
N = FN + R ln[tan(π/4 + φ/2)]
where λO is the longitude of natural origin and FE and FN are false easting and false northing.
In spherical Mercator those values are actually not used, so you can simplify the formula to
Pseudo code example, so this can be adapted to every programming language.
latitude = 41.145556; // (φ)
longitude = -73.995; // (λ)
mapWidth = 200;
mapHeight = 100;
// get x value
x = (longitude+180)*(mapWidth/360)
// convert from degrees to radians
latRad = latitude*PI/180;
// get y value
mercN = ln(tan((PI/4)+(latRad/2)));
y = (mapHeight/2)-(mapWidth*mercN/(2*PI));
Sources:
OGP Geomatics Committee, Guidance Note Number 7, part 2: Coordinate Conversions and Transformation
Derivation of the Mercator projection
National Atlas: Map Projections
Mercator Map projection
EDIT
Created a working example in PHP (because I suck at Java)
https://github.com/mfeldheim/mapStuff.git
EDIT2
Nice animation of the Mercator projection
https://amp-reddit-com.cdn.ampproject.org/v/s/amp.reddit.com/r/educationalgifs/comments/5lhk8y/how_the_mercator_projection_distorts_the_poles/?usqp=mq331AQJCAEoAVgBgAEB&amp_js_v=0.1

You cannot merely transpose from longitude/latitude to x/y like that because the world isn't flat. Have you look at this post? Converting longitude/latitude to X/Y coordinate
UPDATE - 1/18/13
I decided to give this a stab, and here's how I do it:-
public class MapService {
// CHANGE THIS: the output path of the image to be created
private static final String IMAGE_FILE_PATH = "/some/user/path/map.png";
// CHANGE THIS: image width in pixel
private static final int IMAGE_WIDTH_IN_PX = 300;
// CHANGE THIS: image height in pixel
private static final int IMAGE_HEIGHT_IN_PX = 500;
// CHANGE THIS: minimum padding in pixel
private static final int MINIMUM_IMAGE_PADDING_IN_PX = 50;
// formula for quarter PI
private final static double QUARTERPI = Math.PI / 4.0;
// some service that provides the county boundaries data in longitude and latitude
private CountyService countyService;
public void run() throws Exception {
// configuring the buffered image and graphics to draw the map
BufferedImage bufferedImage = new BufferedImage(IMAGE_WIDTH_IN_PX,
IMAGE_HEIGHT_IN_PX,
BufferedImage.TYPE_INT_RGB);
Graphics2D g = bufferedImage.createGraphics();
Map<RenderingHints.Key, Object> map = new HashMap<RenderingHints.Key, Object>();
map.put(RenderingHints.KEY_INTERPOLATION, RenderingHints.VALUE_INTERPOLATION_BICUBIC);
map.put(RenderingHints.KEY_RENDERING, RenderingHints.VALUE_RENDER_QUALITY);
map.put(RenderingHints.KEY_ANTIALIASING, RenderingHints.VALUE_ANTIALIAS_ON);
RenderingHints renderHints = new RenderingHints(map);
g.setRenderingHints(renderHints);
// min and max coordinates, used in the computation below
Point2D.Double minXY = new Point2D.Double(-1, -1);
Point2D.Double maxXY = new Point2D.Double(-1, -1);
// a list of counties where each county contains a list of coordinates that form the county boundary
Collection<Collection<Point2D.Double>> countyBoundaries = new ArrayList<Collection<Point2D.Double>>();
// for every county, convert the longitude/latitude to X/Y using Mercator projection formula
for (County county : countyService.getAllCounties()) {
Collection<Point2D.Double> lonLat = new ArrayList<Point2D.Double>();
for (CountyBoundary countyBoundary : county.getCountyBoundaries()) {
// convert to radian
double longitude = countyBoundary.getLongitude() * Math.PI / 180;
double latitude = countyBoundary.getLatitude() * Math.PI / 180;
Point2D.Double xy = new Point2D.Double();
xy.x = longitude;
xy.y = Math.log(Math.tan(QUARTERPI + 0.5 * latitude));
// The reason we need to determine the min X and Y values is because in order to draw the map,
// we need to offset the position so that there will be no negative X and Y values
minXY.x = (minXY.x == -1) ? xy.x : Math.min(minXY.x, xy.x);
minXY.y = (minXY.y == -1) ? xy.y : Math.min(minXY.y, xy.y);
lonLat.add(xy);
}
countyBoundaries.add(lonLat);
}
// readjust coordinate to ensure there are no negative values
for (Collection<Point2D.Double> points : countyBoundaries) {
for (Point2D.Double point : points) {
point.x = point.x - minXY.x;
point.y = point.y - minXY.y;
// now, we need to keep track the max X and Y values
maxXY.x = (maxXY.x == -1) ? point.x : Math.max(maxXY.x, point.x);
maxXY.y = (maxXY.y == -1) ? point.y : Math.max(maxXY.y, point.y);
}
}
int paddingBothSides = MINIMUM_IMAGE_PADDING_IN_PX * 2;
// the actual drawing space for the map on the image
int mapWidth = IMAGE_WIDTH_IN_PX - paddingBothSides;
int mapHeight = IMAGE_HEIGHT_IN_PX - paddingBothSides;
// determine the width and height ratio because we need to magnify the map to fit into the given image dimension
double mapWidthRatio = mapWidth / maxXY.x;
double mapHeightRatio = mapHeight / maxXY.y;
// using different ratios for width and height will cause the map to be stretched. So, we have to determine
// the global ratio that will perfectly fit into the given image dimension
double globalRatio = Math.min(mapWidthRatio, mapHeightRatio);
// now we need to readjust the padding to ensure the map is always drawn on the center of the given image dimension
double heightPadding = (IMAGE_HEIGHT_IN_PX - (globalRatio * maxXY.y)) / 2;
double widthPadding = (IMAGE_WIDTH_IN_PX - (globalRatio * maxXY.x)) / 2;
// for each country, draw the boundary using polygon
for (Collection<Point2D.Double> points : countyBoundaries) {
Polygon polygon = new Polygon();
for (Point2D.Double point : points) {
int adjustedX = (int) (widthPadding + (point.getX() * globalRatio));
// need to invert the Y since 0,0 starts at top left
int adjustedY = (int) (IMAGE_HEIGHT_IN_PX - heightPadding - (point.getY() * globalRatio));
polygon.addPoint(adjustedX, adjustedY);
}
g.drawPolygon(polygon);
}
// create the image file
ImageIO.write(bufferedImage, "PNG", new File(IMAGE_FILE_PATH));
}
}
RESULT: Image width = 600px, Image height = 600px, Image padding = 50px
RESULT: Image width = 300px, Image height = 500px, Image padding = 50px

Java version of original Google Maps JavaScript API v3 java script code is as following, it works with no problem
public final class GoogleMapsProjection2
{
private final int TILE_SIZE = 256;
private PointF _pixelOrigin;
private double _pixelsPerLonDegree;
private double _pixelsPerLonRadian;
public GoogleMapsProjection2()
{
this._pixelOrigin = new PointF(TILE_SIZE / 2.0,TILE_SIZE / 2.0);
this._pixelsPerLonDegree = TILE_SIZE / 360.0;
this._pixelsPerLonRadian = TILE_SIZE / (2 * Math.PI);
}
double bound(double val, double valMin, double valMax)
{
double res;
res = Math.max(val, valMin);
res = Math.min(res, valMax);
return res;
}
double degreesToRadians(double deg)
{
return deg * (Math.PI / 180);
}
double radiansToDegrees(double rad)
{
return rad / (Math.PI / 180);
}
PointF fromLatLngToPoint(double lat, double lng, int zoom)
{
PointF point = new PointF(0, 0);
point.x = _pixelOrigin.x + lng * _pixelsPerLonDegree;
// Truncating to 0.9999 effectively limits latitude to 89.189. This is
// about a third of a tile past the edge of the world tile.
double siny = bound(Math.sin(degreesToRadians(lat)), -0.9999,0.9999);
point.y = _pixelOrigin.y + 0.5 * Math.log((1 + siny) / (1 - siny)) *- _pixelsPerLonRadian;
int numTiles = 1 << zoom;
point.x = point.x * numTiles;
point.y = point.y * numTiles;
return point;
}
PointF fromPointToLatLng(PointF point, int zoom)
{
int numTiles = 1 << zoom;
point.x = point.x / numTiles;
point.y = point.y / numTiles;
double lng = (point.x - _pixelOrigin.x) / _pixelsPerLonDegree;
double latRadians = (point.y - _pixelOrigin.y) / - _pixelsPerLonRadian;
double lat = radiansToDegrees(2 * Math.atan(Math.exp(latRadians)) - Math.PI / 2);
return new PointF(lat, lng);
}
public static void main(String []args)
{
GoogleMapsProjection2 gmap2 = new GoogleMapsProjection2();
PointF point1 = gmap2.fromLatLngToPoint(41.850033, -87.6500523, 15);
System.out.println(point1.x+" "+point1.y);
PointF point2 = gmap2.fromPointToLatLng(point1,15);
System.out.println(point2.x+" "+point2.y);
}
}
public final class PointF
{
public double x;
public double y;
public PointF(double x, double y)
{
this.x = x;
this.y = y;
}
}

JAVA only?
Python code here! Refer to Convert latitude/longitude point to a pixels (x,y) on mercator projection
import math
from numpy import log as ln
# Define the size of map
mapWidth = 200
mapHeight = 100
def convert(latitude, longitude):
# get x value
x = (longitude + 180) * (mapWidth / 360)
# convert from degrees to radians
latRad = (latitude * math.pi) / 180
# get y value
mercN = ln(math.tan((math.pi / 4) + (latRad / 2)))
y = (mapHeight / 2) - (mapWidth * mercN / (2 * math.pi))
return x, y
print(convert(41.145556, 121.2322))
Answer:
(167.35122222222225, 24.877939817552335)

public static String getTileNumber(final double lat, final double lon, final int zoom) {
int xtile = (int)Math.floor( (lon + 180) / 360 * (1<<zoom) ) ;
int ytile = (int)Math.floor( (1 - Math.log(Math.tan(Math.toRadians(lat)) + 1 / Math.cos(Math.toRadians(lat))) / Math.PI) / 2 * (1<<zoom) ) ;
if (xtile < 0)
xtile=0;
if (xtile >= (1<<zoom))
xtile=((1<<zoom)-1);
if (ytile < 0)
ytile=0;
if (ytile >= (1<<zoom))
ytile=((1<<zoom)-1);
return("" + zoom + "/" + xtile + "/" + ytile);
}
}

I'm new here, just to write, as I've been following the community for some years. I'm happy to be able to contribute.
Well, it took me practically a day in search of that and your question encouraged me to continue the search.
I arrived at the following function, which works! Credits for this article: https://towardsdatascience.com/geotiff-coordinate-querying-with-javascript-5e6caaaf88cf
var bbox = [minLong, minLat, maxLong, maxLat];
var pixelWidth = mapWidth;
var pixelHeight = mapHeight;
var bboxWidth = bbox[2] - bbox[0];
var bboxHeight = bbox[3] - bbox[1];
var convertToXY = function(latitude, longitude) {
var widthPct = ( longitude - bbox[0] ) / bboxWidth;
var heightPct = ( latitude - bbox[1] ) / bboxHeight;
var x = Math.floor( pixelWidth * widthPct );
var y = Math.floor( pixelHeight * ( 1 - heightPct ) );
return { x, y };
}

Need a function to limit a line (known by its coordinates) in its length

I need a function which takes a line (known by its coordinates)
and return a line with same angle, but limited to certain length.
My code gives correct values only when the line is turned 'right'
(proven only empirically, sorry).
Am I missing something?
public static double getAngleOfLine(int x1, int y1, int x2, int y2) {
double opposite = y2 - y1;
double adjacent = x2 - x1;
if (adjacent == Double.NaN) {
return 0;
}
return Math.atan(opposite / adjacent);
}
// returns newly calculated destX and destY values as int array
public static int[] getLengthLimitedLine(int startX, int startY,
int destX, int destY, int lengthLimit) {
double angle = getAngleOfLine(startX, startY, destX, destY);
return new int[]{
(int) (Math.cos(angle) * lengthLimit) + startX,
(int) (Math.sin(angle) * lengthLimit) + startY
};
}
BTW: I know that returning arrays in Java is stupid,
but it's just for the example.

It would be easier to just treat it as a vector. Normalize it by dividing my its magnitude then multiply by a factor of the desired length.
In your example, however, try Math.atan2.

In Python because I don't have a Java compiler handy:
import math
def getLengthLimitedLine(x1, y1, x2, y2, lengthLimit):
length = math.sqrt((x2-x1)**2 + (y2-y1)**2)
if length > lengthLimit:
shrink_factor = lengthLimit / length
x2 = x1 + (x2-x1) * shrink_factor
y2 = y1 + (y2-y1) * shrink_factor
return x2, y2
print getLengthLimitedLine(10, 20, 25, -5, 12)
# Prints (16.17, 9.71) which looks right to me 8-)

It's an easy problem if you understand something about vectors.
Given two points (x1, y1) and (x2, y2), you can calculate the vector from point 1 to 2:
v12 = (x2-x1)i + (y2-y2)j
where i and j are unit vectors in the x and y directions.
You can calculate the magnitude of v by taking the square root of the sum of squares of the components:
v = sqrt((x2-x2)^2 + (y2-y1)^2)
The unit vector from point 1 to point 2 equals v12 divided by its magnitude.
Given that, you can calculate the point along the unit vector that's the desired distance away by multiply the unit vector times the length and adding that to point 1.

Encapsulate Line in a class, add a unit method and a scale method.
public class Line {
private float x;
private float y;
public Line(float x1, float x2, float y1, float y2) {
this(x2 - x1, y2 - y1);
}
public Line(float x, float y) {
this.x = x;
this.y = y;
}
public float getLength() {
return (float) Math.sqrt((x * x) + (y * y));
}
public Line unit() {
return scale(1 / getLength());
}
public Line scale(float scale) {
return new Line(x * scale, y * scale);
}
}
Now you can get a line of arbitrary length l by calling
Line result = new Line(x1, x2, y1, y2).unit().scale(l);

No need to use trig, which can have some nasty edge cases. Just use similar triangles:
public static int[] getLengthLimitedLine(int startX, int startY,
int destX, int destY, int lengthLimit)
{
int deltaX = destX - startX;
int deltaY = destY - startY;
int lengthSquared = deltaX * deltaX + deltaY * deltaY;
// already short enough
if(lengthSquared <= lengthLimit * lengthLimit)
return new int[]{destX, destY};
double length = Math.sqrt(lengthSquared);
double newDeltaX = deltaX * lengthLimit / length;
double newDeltaY = deltaY * lengthLimit / length;
return new int[]{(int)(startX + newDeltaX), (int)(startY + newDeltaY)};
}

Just use the Pythagorean theorem, like so:
public static int[] getLengthLimitedLine(int start[], int dest[], int lengthLimit) {
int xlen = dest[0] - start[0]
int ylen = dest[1] - start[1]
double length = Math.sqrt(xlen * xlen + ylen * ylen)
if (length > lengthLimit) {
return new int[] {start[0], start[1],
start[0] + xlen / lengthLimit,
start[1] + ylen / lengthLimit}
} else {
return new int[] {start[0], start[1], dest[0], dest[1];}
}
}