I am trying to rotate the camera smoothly and without altering the y-vector of the camera direction, i can use look at, and it changes the camera direction in a flash, but this is not working for me, I would like a smooth transition as the direction of the camera changes. I have been reading up, and not understanding everything, but it seems to me that quaternions are the solution to this problem.
I have this.object (my camera) moving along a set path (this.spline.points). The location of the camera at any one time is (thisx,thisy, thisz)
I have cc[i] the direction vector for the direction I would like the camera to face (formerly I was using lookat(cc[i]) which changes the direction correctly, but too quickly/instantaneously)
Using info I have read, I have tried this below, and it just resulted in the screen going black at the point when the camera is due to move.
Could anyone please explain if I am on the right track, how to correct my code.
Thanks
var thisx = this.object.matrixWorld.getPosition().x.toPrecision(3);
var thisy = this.object.matrixWorld.getPosition().y.toPrecision(3);
var thisz = this.object.matrixWorld.getPosition().z.toPrecision(3);
var i = 0;
do {
var pathx = this.spline.points[i].x.toPrecision(3);
var pathz = this.spline.points[i].z.toPrecision(3);
if (thisx == pathx && thisz == pathz){
this.object.useQuaternion = true;
this.object.quaternion = new THREE.Quaternion(thisx, thisy, thisz, 1);
var newvect;
newvect.useQuaternion = true;
newvect.quaternion = new THREE.Quaternion(thisx+cc[i].x, thisy+cc[i].y, thisz+cc[i].z, 1);
var newQuaternion = new THREE.Quaternion();
THREE.Quaternion.slerp(this.object.quaternion, newvect.quaternion, newQuaternion, 0.5);
this.object.quaternion = newQuaternion;
//this.object.lookAt( cc[i]);
i = cc.length;
} else i++;
} while(i < cc.length);
There is no need to call this.object.useQuaternion = true. That is default behavior.
Also, this.object.quaternion contains the current rotation, so no need to generate that either.
You might want to try a different approach - construct the rotation matrix from the spline position, lookAt and up vectors, creating a path of quaternions as a preprocessing step:
var eye = this.spline.points[i].clone().normalize();
var center = cc[i].normalize();
var up = this.object.up.normalize();
var rotMatrix = new THREE.Matrix4().lookAt(eye, center, up);
You could then create the quaternions from the rotation matrix:
var quaternionAtSplineCoordinates = [];
quaternionAtSplineCoordinates.push(new THREE.Quaternion().setFromRotationMatrix(rotMatrix));
Once you have that path, you could apply the quaternion to the camera in your animation loop - provided you have a large enough number of samples. Otherwise, you could consider using slerp to generate the intermediate points.
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'm using Trackball controls in a scene and I want to implement a function to rotate the camera just like the way dragging the mouse in the canvas does it. How can I accomplish something like that? I've been looking the code of the Trackball control module, but I can't find something to start.
EDIT: I've been looking several pages, the THREE documentation and whatnot, but still can't reproduce the Trackball style rotation. I've been using quaternions too but they can't reproduce the behavior(or I'm missing something, most probably). Any help?
EDIT 2 : What I'm looking for is a way to do something like this:
function rotateCam(angle) { // code }
var angle = 0.01; //some value
rotateCam(angle);
$('#button').addEventListener('mousedown', function() { rotateCam(angle); } );
Where button is an HTML element representing a button.
I noticed that the Trackball controls, apart of rotate via quaternion, do a zoom to correct some distances. I tried to read the code, and got this:
function rotate(L) {
var vector = controls.target.clone();
var l = (new THREE.Vector3()).subVectors(camera.position, vector).length();
var up = camera.up.clone();
var quaternion = new THREE.Quaternion();
// Zoom correction
camera.translateZ(L - l);
quaternion.setFromAxisAngle(up, 0.015);
camera.position.applyQuaternion(quaternion);
camera.lookAt(vector);
renderer.render(scene, camera);
}
Works like a charm...hope someone find this useful too. ;)
function cameraRotate(distance, angle){
camera.position.x = distance * Math.cos( angle );
camera.position.z = distance * Math.sin( angle );
}
This would rotate your camera at the specified distance and angle around the scene. If you want a smooth rotation you could call this with a small angle increase from the animate-loop, depending on input for example.
I'm on the right way I think...I did this:
function rotate() {
if (this.showCase) {
var vector = controls.target.clone(); // controls is a TrackballControls
var up = camera.up.clone();
var quaternion = new THREE.Quaternion();
quaternion.setFromAxisAngle(up, 0.015);
camera.position.applyQuaternion(quaternion);
camera.lookAt(vector);
renderer.render(this.scene, this.camera);
}
},
Still, it doesn't look right right like the TrackballControls rotation.