My problem is that when creating faces for a country, there are faces which overlap the border.
Below is the image when all coordinates are used when creating the faces
image with all values (coordinates)
values_axis1 contains all the X coordinates [63.0613691022453, 65.1611029239906, 66.0721609548093, 68.8109022381195, 71.1822098033678..]
values_axis2 contains all the Y coordinates [34.37065245791981, 35.30003249470688, 38.11207931425505, 38.228254752215044..]
values_axis3 contains all the Z coordinates
for (var i = 0; i < values_axis1.length; i ++) {
object_geometry.vertices.push(new THREE.Vector3(values_axis1[i], values_axis2[i], values_axis3[i]));
object_geometry.faces.push(new THREE.Face3(0, i + 1, i));
}
Buy changing i++ to i+=2 we get this image of skipping some values
Arrow obviously shows the point zero where all triangles are drawn from.
Below is the part of code for that.
object_geometry.faces.push(new THREE.Face3(DRAW_FROM_HERE, i + 1, i));
This isn't per say a programming problem but more like a 'is there a algorithm for this'.
I could check for faces that collide with a border from point zero and skip them but then there would be holes and that would need to be filled by drawing from some other location. That could be done manually but doing that for all the countries in the world would take ages. I'm sure this could be done by somehow calculating where the holes are but I have no idea how that would be done.
If someone comes up even with a really bad solution performance wise it would be great! I'm planning on putting all the results in a file which could then be loaded and drawn since there is really no need to do this processing every time since the borders never move.
EDIT: ShapeGeometry was suggested and has been tested. I was wrong when I said that Z coordinates are not relevant but obviously they are due to the curvature of the globe.
Image of the shapeGeometry. 2d but otherwise perfect.
Question edited.
Edit: Solution
Thanks to #Gilles-Philippe Paillé for suggesting ear clipping. Found a great library https://github.com/mapbox/earcut
Here is updated code for others who might have the same issue.
function createVertexForEachPoint(object_geometry, values_axis1, values_axis2,
values_axis3) {
var values_x_y = []; // add x and y values to the same list [x,y,x,y,x,y..]
for (var i = 0; i < values_axis1.length; i++) {
values_x_y.push(values_axis1[i], values_axis2[i]);
}
// https://github.com/mapbox/earcut
var triangles = earcut(values_x_y);
// triangles = [37, 36, 35, 35, 34, 33, 32, 31, 30, 30, …] [a,b,c,a,b,c,a,b,c..]
// triangles contain only the verticies that are needed
// previously all possible faces were added which resulted in overlapping borders
// add all points, in this case coordinates. Same as before
for (var i = 0; i < values_axis1.length; i++) {
object_geometry.vertices.push(new THREE.Vector3(values_axis1[i], values_axis2[i], values_axis3[i]));
}
// go through the list and pick the corners (a,b,c) of the triangles and add them to faces
for (var i = 0; i < triangles.length; i += 3) {
var point_1 = triangles[i];
var point_2 = triangles[i+1];
var point_3 = triangles[i+2];
object_geometry.faces.push(new THREE.Face3(point_1, point_2, point_3));
}
}
I use a geoJSON which has 'polygon' and 'multipolygon' shapes. This code works for polygons atm but shouldn't need too much tweaking to work with multipolygons since the library supports also holes earcut(vertices[, holes, dimensions = 2]).
image of the result
I have been developing a basic game engine just to learn the process and I have hit a issue with my rotation function.
It works fine except that the object shrinks and appears to invert.
Here is a jsfiddle that illustrates my point.
I think the problem would be in the rotation code its self but i'm not positive.
function Rotation(vec, rot){
if(Math.acos((vec.x + vec.y + vec.z -1)/2) === 0) { return vec; }
var qVec = new Quaternion(vec.x, vec.y, vec.z, 0);
qVec = Quaternions.multiply(qVec, rot);
qVec = Quaternions.multiply(qVec, rot.conjugate());
return new Vector3(qVec.x, qVec.y, qVec.z);
}
Couple things:
First, the rotation quaternion is not normalized, so it's inverse is not the same as its conjugate. Rotation by a quaternion is defined by:
Where q is the vector you're rotating around, p is the vector you're rotating, and p' is the final rotated vector.
So this is defined using the inverse of q, which is defined as conjugate(q) / magnitude(q)^2. In the case where q is normalized, magnitude(q)^2 == 1, so it's the same as just multiplying by the conjugate.
Also note the order of operations here. Quat multiplications are non-commutative so their order matters.
You can fix this by normalizing your rotation quaternion, and fixing the order of operations:
var qVec = new Quaternion(vec.x, vec.y, vec.z, 0);
qVec = Quaternions.multiply(rot.normalize(), qVec);
qVec = Quaternions.multiply(qVec, rot.conjugate());
return new Vector3(qVec.x, qVec.y, qVec.z);
Second, you want to define your rotation quat as normal to the plane you want to rotate around. In this case, you want to rotate around the x-y plane. The z-axis is normal to this plane, so we want to define the rotation vector along the z-axis:
function update(){
for(var i = 0; i < gameObjects.length; i++){
gameObjects[i].rotation = euler(new Vector3(0, 0, frames/100));
}
}
With these changes, I was able to get the box rotating correctly.
(In terms of why it was scaling up/down, I'm not 100% sure. Still trying to figure that out.)
Struggeling translating the position of the mouse to the location of the tiles in my grid. When it's all flat, the math looks like this:
this.position.x = Math.floor(((pos.y - 240) / 24) + ((pos.x - 320) / 48));
this.position.y = Math.floor(((pos.y - 240) / 24) - ((pos.x - 320) / 48));
where pos.x and pos.y are the position of the mouse, 240 and 320 are the offset, 24 and 48 the size of the tile. Position then contains the grid coordinate of the tile I'm hovering over. This works reasonably well on a flat surface.
Now I'm adding height, which the math does not take into account.
This grid is a 2D grid containing noise, that's being translated to height and tile type. Height is really just an adjustment to the 'Y' position of the tile, so it's possible for two tiles to be drawn in the same spot.
I don't know how to determine which tile I'm hovering over.
edit:
Made some headway... Before, I was depending on the mouseover event to calculate grid position. I just changed this to do the calculation in the draw loop itself, and check if the coordinates are within the limits of the tile currently being drawn. creates some overhead tho, not sure if I'm super happy with it but I'll confirm if it works.
edit 2018:
I have no answer, but since this ha[sd] an open bounty, help yourself to some code and a demo
The grid itself is, simplified;
let grid = [[10,15],[12,23]];
which leads to a drawing like:
for (var i = 0; i < grid.length; i++) {
for (var j = 0; j < grid[0].length; j++) {
let x = (j - i) * resourceWidth;
let y = ((i + j) * resourceHeight) + (grid[i][j] * -resourceHeight);
// the "+" bit is the adjustment for height according to perlin noise values
}
}
edit post-bounty:
See GIF. The accepted answer works. The delay is my fault, the screen doesn't update on mousemove (yet) and the frame rate is low-ish. It's clearly bringing back the right tile.
Source
Intresting task.
Lets try to simplify it - lets resolve this concrete case
Solution
Working version is here: https://github.com/amuzalevskiy/perlin-landscape (changes https://github.com/jorgt/perlin-landscape/pull/1 )
Explanation
First what came into mind is:
Just two steps:
find an vertical column, which matches some set of tiles
iterate tiles in set from bottom to top, checking if cursor is placed lower than top line
Step 1
We need two functions here:
Detects column:
function getColumn(mouseX, firstTileXShiftAtScreen, columnWidth) {
return (mouseX - firstTileXShiftAtScreen) / columnWidth;
}
Function which extracts an array of tiles which correspond to this column.
Rotate image 45 deg in mind. The red numbers are columnNo. 3 column is highlighted. X axis is horizontal
function tileExists(x, y, width, height) {
return x >= 0 & y >= 0 & x < width & y < height;
}
function getTilesInColumn(columnNo, width, height) {
let startTileX = 0, startTileY = 0;
let xShift = true;
for (let i = 0; i < columnNo; i++) {
if (tileExists(startTileX + 1, startTileY, width, height)) {
startTileX++;
} else {
if (xShift) {
xShift = false;
} else {
startTileY++;
}
}
}
let tilesInColumn = [];
while(tileExists(startTileX, startTileY, width, height)) {
tilesInColumn.push({x: startTileX, y: startTileY, isLeft: xShift});
if (xShift) {
startTileX--;
} else {
startTileY++;
}
xShift = !xShift;
}
return tilesInColumn;
}
Step 2
A list of tiles to check is ready. Now for each tile we need to find a top line. Also we have two types of tiles: left and right. We already stored this info during building matching tiles set.
function getTileYIncrementByTileZ(tileZ) {
// implement here
return 0;
}
function findExactTile(mouseX, mouseY, tilesInColumn, tiles2d,
firstTileXShiftAtScreen, firstTileYShiftAtScreenAt0Height,
tileWidth, tileHeight) {
// we built a set of tiles where bottom ones come first
// iterate tiles from bottom to top
for(var i = 0; i < tilesInColumn; i++) {
let tileInfo = tilesInColumn[i];
let lineAB = findABForTopLineOfTile(tileInfo.x, tileInfo.y, tiles2d[tileInfo.x][tileInfo.y],
tileInfo.isLeft, tileWidth, tileHeight);
if ((mouseY - firstTileYShiftAtScreenAt0Height) >
(mouseX - firstTileXShiftAtScreen)*lineAB.a + lineAB.b) {
// WOHOO !!!
return tileInfo;
}
}
}
function findABForTopLineOfTile(tileX, tileY, tileZ, isLeftTopLine, tileWidth, tileHeight) {
// find a top line ~~~ a,b
// y = a * x + b;
let a = tileWidth / tileHeight;
if (isLeftTopLine) {
a = -a;
}
let b = isLeftTopLine ?
tileY * 2 * tileHeight :
- (tileX + 1) * 2 * tileHeight;
b -= getTileYIncrementByTileZ(tileZ);
return {a: a, b: b};
}
Please don't judge me as I am not posting any code. I am just suggesting an algorithm that can solve it without high memory usage.
The Algorithm:
Actually to determine which tile is on mouse hover we don't need to check all the tiles. At first we think the surface is 2D and find which tile the mouse pointer goes over with the formula OP posted. This is the farthest probable tile mouse cursor can point at this cursor position.
This tile can receive mouse pointer if it's at 0 height, by checking it's current height we can verify if this is really at the height to receive pointer, we mark it and move forward.
Then we find the next probable tile which is closer to the screen by incrementing or decrementing x,y grid values depending on the cursor position.
Then we keep on moving forward in a zigzag fashion until we reach a tile which cannot receive pointer even if it is at it's maximum height.
When we reach this point the last tile found that were at a height to receive pointer is the tile that we are looking for.
In this case we only checked 8 tiles to determine which tile is currently receiving pointer. This is very memory efficient in comparison to checking all the tiles present in the grid and yields faster result.
One way to solve this would be to follow the ray that goes from the clicked pixel on the screen into the map. For that, just determine the camera position in relation to the map and the direction it is looking at:
const camPos = {x: -5, y: -5, z: -5}
const camDirection = { x: 1, y:1, z:1}
The next step is to get the touch Position in the 3D world. In this certain perspective that is quite simple:
const touchPos = {
x: camPos.x + touch.x / Math.sqrt(2),
y: camPos.y - touch.x / Math.sqrt(2),
z: camPos.z - touch.y / Math.sqrt(2)
};
Now you just need to follow the ray into the layer (scale the directions so that they are smaller than one of your tiles dimensions):
for(let delta = 0; delta < 100; delta++){
const x = touchPos.x + camDirection.x * delta;
const y = touchPos.y + camDirection.y * delta;
const z = touchPos.z + camDirection.z * delta;
Now just take the tile at xz and check if y is smaller than its height;
const absX = ~~( x / 24 );
const absZ = ~~( z / 24 );
if(tiles[absX][absZ].height >= y){
// hanfle the over event
}
I had same situation on a game. first I tried with mathematics, but when I found that the clients wants to change the map type every day, I changed the solution with some graphical solution and pass it to the designer of the team. I captured the mouse position by listening the SVG elements click.
the main graphic directly used to capture and translate the mouse position to my required pixel.
https://blog.lavrton.com/hit-region-detection-for-html5-canvas-and-how-to-listen-to-click-events-on-canvas-shapes-815034d7e9f8
https://code.sololearn.com/Wq2bwzSxSnjl/#html
Here is the grid input I would define for the sake of this discussion. The output should be some tile (coordinate_1, coordinate_2) based on visibility on the users screen of the mouse:
I can offer two solutions from different perspectives, but you will need to convert this back into your problem domain. The first methodology is based on coloring tiles and can be more useful if the map is changing dynamically. The second solution is based on drawing coordinate bounding boxes based on the fact that tiles closer to the viewer like (0, 0) can never be occluded by tiles behind it (1,1).
Approach 1: Transparently Colored Tiles
The first approach is based on drawing and elaborated on here. I must give the credit to #haldagan for a particularly beautiful solution. In summary it relies on drawing a perfectly opaque layer on top of the original canvas and coloring every tile with a different color. This top layer should be subject to the same height transformations as the underlying layer. When the mouse hovers over a particular layer you can detect the color through canvas and thus the tile itself. This is the solution I would probably go with and this seems to be a not so rare issue in computer visualization and graphics (finding positions in a 3d isometric world).
Approach 2: Finding the Bounding Tile
This is based on the conjecture that the "front" row can never be occluded by "back" rows behind it. Furthermore, "closer to the screen" tiles cannot be occluded by tiles "farther from the screen". To make precise the meaning of "front", "back", "closer to the screen" and "farther from the screen", take a look at the following:
.
Based on this principle the approach is to build a set of polygons for each tile. So firstly we determine the coordinates on the canvas of just box (0, 0) after height scaling. Note that the height scale operation is simply a trapezoid stretched vertically based on height.
Then we determine the coordinates on the canvas of boxes (1, 0), (0, 1), (1, 1) after height scaling (we would need to subtract anything from those polygons which overlap with the polygon (0, 0)).
Proceed to build each boxes bounding coordinates by subtracting any occlusions from polygons closer to the screen, to eventually get coordinates of polygons for all boxes.
With these coordinates and some care you can ultimately determine which tile is pointed to by a binary search style through overlapping polygons by searching through bottom rows up.
It also matters what else is on the screen. Maths attempts work if your tiles are pretty much uniform. However if you are displaying various objects and want the user to pick them, it is far easier to have a canvas-sized map of identifiers.
function poly(ctx){var a=arguments;ctx.beginPath();ctx.moveTo(a[1],a[2]);
for(var i=3;i<a.length;i+=2)ctx.lineTo(a[i],a[i+1]);ctx.closePath();ctx.fill();ctx.stroke();}
function circle(ctx,x,y,r){ctx.beginPath();ctx.arc(x,y,r,0,2*Math.PI);ctx.fill();ctx.stroke();}
function Tile(h,c,f){
var cnv=document.createElement("canvas");cnv.width=100;cnv.height=h;
var ctx=cnv.getContext("2d");ctx.lineWidth=3;ctx.lineStyle="black";
ctx.fillStyle=c;poly(ctx,2,h-50,50,h-75,98,h-50,50,h-25);
poly(ctx,50,h-25,2,h-50,2,h-25,50,h-2);
poly(ctx,50,h-25,98,h-50,98,h-25,50,h-2);
f(ctx);return ctx.getImageData(0,0,100,h);
}
function put(x,y,tile,image,id,map){
var iw=image.width,tw=tile.width,th=tile.height,bdat=image.data,fdat=tile.data;
for(var i=0;i<tw;i++)
for(var j=0;j<th;j++){
var ijtw4=(i+j*tw)*4,a=fdat[ijtw4+3];
if(a!==0){
var xiyjiw=x+i+(y+j)*iw;
for(var k=0;k<3;k++)bdat[xiyjiw*4+k]=(bdat[xiyjiw*4+k]*(255-a)+fdat[ijtw4+k]*a)/255;
bdat[xiyjiw*4+3]=255;
map[xiyjiw]=id;
}
}
}
var cleanimage;
var pickmap;
function startup(){
var water=Tile(77,"blue",function(){});
var field=Tile(77,"lime",function(){});
var tree=Tile(200,"lime",function(ctx){
ctx.fillStyle="brown";poly(ctx,50,50,70,150,30,150);
ctx.fillStyle="forestgreen";circle(ctx,60,40,30);circle(ctx,68,70,30);circle(ctx,32,60,30);
});
var sheep=Tile(200,"lime",function(ctx){
ctx.fillStyle="white";poly(ctx,25,155,25,100);poly(ctx,75,155,75,100);
circle(ctx,50,100,45);circle(ctx,50,80,30);
poly(ctx,40,70,35,80);poly(ctx,60,70,65,80);
});
var cnv=document.getElementById("scape");
cnv.width=500;cnv.height=400;
var ctx=cnv.getContext("2d");
cleanimage=ctx.getImageData(0,0,500,400);
pickmap=new Uint8Array(500*400);
var tiles=[water,field,tree,sheep];
var map=[[[0,0],[1,1],[1,1],[1,1],[1,1]],
[[0,0],[1,1],[1,2],[3,2],[1,1]],
[[0,0],[1,1],[2,2],[3,2],[1,1]],
[[0,0],[1,1],[1,1],[1,1],[1,1]],
[[0,0],[0,0],[0,0],[0,0],[0,0]]];
for(var x=0;x<5;x++)
for(var y=0;y<5;y++){
var desc=map[y][x],tile=tiles[desc[0]];
put(200+x*50-y*50,200+x*25+y*25-tile.height-desc[1]*20,
tile,cleanimage,x+1+(y+1)*10,pickmap);
}
ctx.putImageData(cleanimage,0,0);
}
var mx,my,pick;
function mmove(event){
mx=Math.round(event.offsetX);
my=Math.round(event.offsetY);
if(mx>=0 && my>=0 && mx<cleanimage.width && my<cleanimage.height && pick!==pickmap[mx+my*cleanimage.width])
requestAnimationFrame(redraw);
}
function redraw(){
pick=pickmap[mx+my*cleanimage.width];
document.getElementById("pick").innerHTML=pick;
var ctx=document.getElementById("scape").getContext("2d");
ctx.putImageData(cleanimage,0,0);
if(pick!==0){
var temp=ctx.getImageData(0,0,cleanimage.width,cleanimage.height);
for(var i=0;i<pickmap.length;i++)
if(pickmap[i]===pick)
temp.data[i*4]=255;
ctx.putImageData(temp,0,0);
}
}
startup(); // in place of body.onload
<div id="pick">Move around</div>
<canvas id="scape" onmousemove="mmove(event)"></canvas>
Here the "id" is a simple x+1+(y+1)*10 (so it is nice when displayed) and fits into a byte (Uint8Array), which could go up to 15x15 display grid already, and there are wider types available too.
(Tried to draw it small, and it looked ok on the snippet editor screen but apparently it is still too large here)
Computer graphics is fun, right?
This is a special case of the more standard computational geometry "point location problem". You could also express it as a nearest neighbour search.
To make this look like a point location problem you just need to express your tiles as non-overlapping polygons in a 2D plane. If you want to keep your shapes in a 3D space (e.g. with a z buffer) this becomes the related "ray casting problem".
One source of good geometry algorithms is W. Randolf Franklin's website and turf.js contains an implementation of his PNPOLY algorithm.
For this special case we can be even faster than the general algorithms by treating our prior knowledge about the shape of the tiles as a coarse R-tree (a type of spatial index).
I am an amateur pilot working on some airspace modeling experiments. What I am trying to achieve is a tool for easily creating cross-sections of airspace, i.e. 3D airspace to 2D. So in the end I would like to have an image similar to this created: airspace example. Accuracy is not important as these cross-sections would not be used for navigational purposes, but for training/visualization only. Hence coordinate geometry is enough, and no geodesic calculations are needed.
Currently I am storing GeoJSON 2D geometries (all polygons) in my database with additional metadata containing lower and upper altitude limits of each airspace element. Therefore I am effectively only showing 2D data on my OpenLayers and Leaflet.js maps.
I want the user to be able to draw a linestring over the map (see the green linestring in the picture below, stroke-width dramatically increased for demonstration purposes). This I can do with OpenLayers or Leaflet. The outcome should be a 1-dimensional cross-section of the 2D elements intersecting with the user-drawn line, as in this very artistic illustration by me:
Clarification: if the length of the output 1-dimensional cross-section is for example 1L, then the contents of the cross-section in the example should be a set of the following geometries: 1) black line between 0.1L and 0.5L and 2) a red line between 0.7L and 0.825L.
The user interface part is doable, and it would be running on top of OpenLayers or Leaflet. I have also found several algorithms in various languages for determining if two lines intersect and even to find out the intersection point. I can use Raphael.js then to draw the cross-section.
I should be able to do all this in a day or two... But I was wondering if there was an easier path to take? For example, does anyone know of a software library that would enable calculation of such cross-sections that I am trying to achieve? Oh, please don't mention those $10,000 GIS packages :-).
As this will be a web application, I am mostly looking into Javascript, Perl or PHP solutions. GeoJSON Utilities for JavaScript looks quite promising for calculating intersections, but I wonder if there are others?
Posting answer to my own question for future reference. I was able to come up with a solution for creating the airspace cross-sections. It is based on coordinate geometry.
1) Input data is the start and end coordinates of the cross-section.
2) Loop through all areas and check if the start and/or end coordinates fall inside any of the airspace areas (polygons). PHP code slightly modified from another SO answer to check if a point is inside a polygon:
// Point class, storage of lat/long-pairs
class Point {
var $lat;
var $long;
function Point($lat, $long) {
$this->lat = $lat;
$this->long = $long;
}
}
function pointInPolygon($p, $polygon) {
$c = 0;
$p1 = $polygon[0];
$n = count($polygon);
for ($i=1; $i<=$n; $i++) {
$p2 = $polygon[$i % $n];
if ($p->long > min($p1->long, $p2->long)
&& $p->long <= max($p1->long, $p2->long)
&& $p->lat <= max($p1->lat, $p2->lat)
&& $p1->long != $p2->long) {
$xinters = ($p->long - $p1->long) * ($p2->lat - $p1->lat) / ($p2->long - $p1->long)
if ($p1->lat == $p2->lat || $p->lat <= $xinters) {
$c++;
}
}
$p1 = $p2;
}
// if the number of edges we passed through is even, then it's not in the poly.
return $c%2!=0;
}
3) Loop through all the line segments of each and every area (polygon) containing the airspace data. PHP code slightly modified from another SO answer to return the intersection point ofe two line segments:
function Det2($x1, $x2, $y1, $y2) {
return ($x1 * $y2 - $y1 * $x2);
}
function lineIntersection ($v1Y, $v1X, $v2Y, $v2X, $v3Y, $v3X, $v4Y, $v4X) {
$tolerance = 0.000001;
$a = Det2($v1X - $v2X, $v1Y - $v2Y, $v3X - $v4X, $v3Y - $v4Y);
if (abs($a) < $tolerance) return null; // Lines are parallel
$d1 = Det2($v1X, $v1Y, $v2X, $v2Y);
$d2 = Det2($v3X, $v3Y, $v4X, $v4Y);
$x = Det2($d1, $v1X - $v2X, $d2, $v3X - $v4X) / $a;
$y = Det2($d1, $v1Y - $v2Y, $d2, $v3Y - $v4Y) / $a;
if ($x < min($v1X, $v2X) - $tolerance || $x > max($v1X, $v2X) + $tolerance) return null;
if ($y < min($v1Y, $v2Y) - $tolerance || $y > max($v1Y, $v2Y) + $tolerance) return null;
if ($x < min($v3X, $v4X) - $tolerance || $x > max($v3X, $v4X) + $tolerance) return null;
if ($y < min($v3Y, $v4Y) - $tolerance || $y > max($v3Y, $v4Y) + $tolerance) return null;
return array($x, $y);
}
4) If there are intersecting segments, figure out the distance of the intersection (as result of the previous step) from the start coordinates of the cross-section. Divide this with the overall length of the cross-section to get the relative location of the airspace area boundary on the cross-section line.
5) Based on results of points 2 and 4, draw SVG polygons. Relative intersection locations are translated to X coordinates of polygons and altitude data (lower and upper limit) becomes the Y coordinates of the polygon.