I am using a couple of functions from Snap.SVG, mainly path2curve and the functions around it to build a SVG morph plugin.
I've setup a demo here on Codepen to better illustrate the issue. Basically morphing shapes simple to complex and the other way around is working properly as of Javascript functionality, however, the visual isn't very pleasing.
The first shape morph looks awful, the second looks a little better because I changed/rotated it's points a bit, but the last example is perfect.
So I need either a better path2curve or a function to prepare the path string before the other function builds the curves array. Snap.SVG has a function called getClosest that I think may be useful but it's not documented.
There isn't any documentation available on this topic so I would appreciate any suggestion/input from RaphaelJS / SnapSVG / d3.js / three/js developers.
I've provided a runnable code snippet below that uses Snap.svg and that I believe demonstrates one solution to your problem. With respect to trying to find the best way to morph a starting shape into an ending shape, this algorithm essentially rotates the points of the starting shape one position at a time, sums the squares of the distances between corresponding points on the (rotated) starting shape and the (unchanged) ending shape, and finds the minimum of all those sums. i.e. It's basically a least squares approach. The minimum value identifies the rotation that, as a first guess, will provide the "shortest" morph trajectories. In spite of these coordinate reassignments, however, all 'rotations' should result in visually identical starting shapes, as required.
This is, of course, a "blind" mathematical approach, but it might help provide you with a starting point before doing manual visual analysis. As a bonus, even if you don't like the rotation that the algorithm chose, it also provides the path 'd' attribute strings for all the other rotations, so some of that work has already been done for you.
You can modify the snippet to provide any starting and ending shapes you want. The limitations are as follows:
Each shape should have the same number of points (although the point types, e.g. 'lineto', 'cubic bezier curve', 'horizontal lineto', etc., can completely vary)
Each shape should be closed, i.e. end with "Z"
The morph desired should involve only translation. If scaling or rotation is desired, those should be applied after calculating the morph based only on translation.
By the way, in response to some of your comments, while I find Snap.svg intriguing, I also find its documentation to be somewhat lacking.
Update: The code snippet below works in Firefox (Mac or Windows) and Safari. However, Chrome seems to have trouble accessing the Snap.svg library from its external web site as written (<script...github...>). Opera and Internet Explorer also have problems. So, try the snippet in the working browsers, or try copying the snippet code as well as the Snap library code to your own computer. (Is this an issue of accessing third party libraries from within the code snippet? And why browser differences? Insightful comments would be appreciated.)
var
s = Snap(),
colors = ["red", "blue", "green", "orange"], // colour list can be any length
staPath = s.path("M25,35 l-15,-25 C35,20 25,0 40,0 L80,40Z"), // create the "start" shape
endPath = s.path("M10,110 h30 l30,20 C30,120 35,135 25,135Z"), // create the "end" shape
staSegs = getSegs(staPath), // convert the paths to absolute values, using only cubic bezier
endSegs = getSegs(endPath), // segments, & extract the pt coordinates & segment strings
numSegs = staSegs.length, // note: the # of pts is one less than the # of path segments
numPts = numSegs - 1, // b/c the path's initial 'moveto' pt is also the 'close' pt
linePaths = [],
minSumLensSqrd = Infinity,
rotNumOfMin,
rotNum = 0;
document.querySelector('button').addEventListener('click', function() {
if (rotNum < numPts) {
linePaths.forEach(function(linePath) {linePath.remove();}); // erase any previous coloured lines
var sumLensSqrd = 0;
for (var ptNum = 0; ptNum < numPts; ptNum += 1) { // draw new lines, point-to-point
var linePt1 = staSegs[(rotNum + ptNum) % numPts]; // the new line begins on the 'start' shape
var linePt2 = endSegs[ ptNum % numPts]; // and finished on the 'end' shape
var linePathStr = "M" + linePt1.x + "," + linePt1.y + "L" + linePt2.x + "," + linePt2.y;
var linePath = s.path(linePathStr).attr({stroke: colors[ptNum % colors.length]}); // draw it
var lineLen = Snap.path.getTotalLength(linePath); // calculate its length
sumLensSqrd += lineLen * lineLen; // square the length, and add it to the accumulating total
linePaths[ptNum] = linePath; // remember the path to facilitate erasing it later
}
if (sumLensSqrd < minSumLensSqrd) { // keep track of which rotation has the lowest value
minSumLensSqrd = sumLensSqrd; // of the sum of lengths squared (the 'lsq sum')
rotNumOfMin = rotNum; // as well as the corresponding rotation number
}
show("ROTATION OF POINTS #" + rotNum + ":"); // display info about this rotation
var rotInfo = getRotInfo(rotNum);
show(" point coordinates: " + rotInfo.ptsStr); // show point coordinates
show(" path 'd' string: " + rotInfo.dStr); // show 'd' string needed to draw it
show(" sum of (coloured line lengths squared) = " + sumLensSqrd); // the 'lsq sum'
rotNum += 1; // analyze the next rotation of points
} else { // once all the rotations have been analyzed individually...
linePaths.forEach(function(linePath) {linePath.remove();}); // erase any coloured lines
show(" ");
show("BEST ROTATION, i.e. rotation with lowest sum of (lengths squared): #" + rotNumOfMin);
// show which rotation to use
show("Use the shape based on this rotation of points for morphing");
$("button").off("click");
}
});
function getSegs(path) {
var absCubDStr = Snap.path.toCubic(Snap.path.toAbsolute(path.attr("d")));
return Snap.parsePathString(absCubDStr).map(function(seg, segNum) {
return {x: seg[segNum ? 5 : 1], y: seg[segNum ? 6 : 2], seg: seg.toString()};
});
}
function getRotInfo(rotNum) {
var ptsStr = "";
for (var segNum = 0; segNum < numSegs; segNum += 1) {
var oldSegNum = rotNum + segNum;
if (segNum === 0) {
var dStr = "M" + staSegs[oldSegNum].x + "," + staSegs[oldSegNum].y;
} else {
if (oldSegNum >= numSegs) oldSegNum -= numPts;
dStr += staSegs[oldSegNum].seg;
}
if (segNum !== (numSegs - 1)) {
ptsStr += "(" + staSegs[oldSegNum].x + "," + staSegs[oldSegNum].y + "), ";
}
}
ptsStr = ptsStr.slice(0, ptsStr.length - 2);
return {ptsStr: ptsStr, dStr: dStr};
}
function show(msg) {
var m = document.createElement('pre');
m.innerHTML = msg;
document.body.appendChild(m);
}
pre {
margin: 0;
padding: 0;
}
<script src="//cdn.jsdelivr.net/snap.svg/0.4.1/snap.svg-min.js"></script>
<p>Best viewed on full page</p>
<p>Coloured lines show morph trajectories for the points for that particular rotation of points. The algorithm seeks to optimize those trajectories, essentially trying to find the "shortest" cumulative routes.</p>
<p>The order of points can be seen by following the colour of the lines: red, blue, green, orange (at least when this was originally written), repeating if there are more than 4 points.</p>
<p><button>Click to show rotation of points on top shape</button></p>
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 have a canvas on which i'm drawing 4 lines.
I need to then compare the accuracy of those lines drawn against the coordinates of 4 predefined lines.
Can i do this with canvas?
thanks.
As JJPA says, give it a try!
One route you could take would be to store the coordinates of the drawn line and the expected line. Each line forms a vector. You can take the Euclidean distance between the lines using a simple formula. The closer the distance, the more accurate the lines.
I just wrote this.
var expected = {from: {x:10, y:10}, to:{x:20, y:20}}
var identical = {from: {x:10, y:10}, to:{x:20, y:20}}
var close = {from: {x:9, y:9}, to:{x:21, y:21}}
var lessClose = {from: {x:8, y:13}, to:{x:25, y:15}}
var distance = function(a, b) {
return Math.sqrt(Math.pow(a.from.x - b.from.y, 2) +
Math.pow(a.to.x - b.to.y, 2))
}
console.log("identical", distance(expected, identical))
console.log("close", distance(expected, close))
console.log("lessclose", distance(expected, lessClose))
gives
identical 0
close 1.4142135623730951
lessclose 5.830951894845301
If you don't know which way round the lines are, you can swap start and end points and take the minimum distance.
Well, you can get the pixels of your canvas in an rgb-array in javascript:
var pixelarray= canvas.getImageData(x, y, width, height).data;
then pixelarray[0] as integer is the red-value of the first pixel, pixelarray[3] the red-value of the second and so on.
In a canvas of 100px width (pixel-index 0 to 299 in first line), pixelarray[299] would be the blue-value of the last pixel in the first line and pixelarray[300] the red-value of the first in second line.
Then you could get either the functions representing the 4 lines you want to compare to and do something like
function isblack(x,y,width,height){
return pixelarray[(y*width+x)*3]==255&&pixelarray[(y*width+x)*3+1]==255&&pixelarray[(y*width+x)*3+2]==255||x<0||x>width||y<0||y>height;
}
function f(x){return a*x+b;}
var lineblackness=0;
for(var x=0;x<picturewidth;x++)if(isblack(x,f(x),picturewidth,pictureheight))lineblackness++;
if(lineblackness==picturewidth)line_is_drawn();
Where fx(x) would represent the line ...
well, if you use a picture or coordinates for the line, it might get easier:
getcoord(data){
var ar=new array();
for(var i=0;i<data.length/3;i++)if(data[i*3]==255)ar.push(i);
return ar;
}
var comparecoords=getcoord(comparecanvas.getImageData(x, y, width, height).data);
var thisdata=canvas.getImageData(x, y, width, height).data;
var linehits=0;
for(var i=0;i<comparecoords.length;i++){if(thisdata[comparecoords[i]*3]==255)linehits++;}
if(comparecoords.length==linehits)all_coords_are_hit();
Here I only compared the red-value, assuming you have a black&White-image.
If you want to compare lines that aren't exactly fits, you can change the ==255 with a few additional ifs e.g.
function isblack(x,y,width,height,blurr,treshhold){
for(var i=-blurr;i<blurr;i++)if(x-i>0&&x+i<width&&pixelarray[(y*width+(x+i))*3]>=treshhold)return true;
for(var i=-blurr;i<blurr;i++)if(y-i>0&&y+i<height&&pixelarray[((y+i)*width+(x))*3]>=treshhold)return true;
return false;
}
This function uses also only red, but uses a treshhold to let also gray be valid. it also accepts a blurr-value for checking in x-direction before/after the pixel and then again in y-direction.
You could also combine the two loops for checking in an rectangle around the pixel.
Changing the return true into return i will also give you the distance from the pixel you want to check. if you combine the two loops into an i and a j-value, return sqrt(i^2+j^2) and sum all checks up, you can get an indicator how accurate your line is.
To implement sqareroot-error, only comparing the y-distance and summing up the squared distance i^2 should be enough.
Please note that these programs are suggestions, better implementations may exists and my codes could still contain errors.
I've a 3d model of a tube geometry. There are 18000 co-ordinates on production side. I am taking every 9th co-ordinate so that actually plotting 9000 co-ordinates to build a tube geometry. I've to use CanvasRenderer only.
Now when I use vertexColors: THREE.VertexColors in WebGLRenderer, the model displays different color on each face. When I change it to CanvasRenderer, the model turns into white color only. Even I change vertexColors: THREE.FaceColors, the result is same.
Please find below the link of jsfiddle and link of my previous where mrdoob added support for material.vertexColors = THREE.FaceColors to CanvasRenderer.
support for vertex color in canvas rendering
tube in canvas rendering
Please find below the image to apply colors based on values.
As shown in the image there are 12 values at 12 different degrees for every co-ordinate. So I've created a tube with radius segment of 12. Then I've stored these values into JSON file but as there 18000 points, the file becomes to heavy. Even though I am plotting 2000 points it takes too much time. For 2000 segments and each segment has 12 faces, there are 24000 faces on a tube.
Please find below the programming logic to apply color based on value of a parameter.
// get res values & apply color
var lblSeg=0; var pntId; var d=0; var faceLength=tube.faces.length;
var degrees = [ '30', '60', '90', '120', '150', '180', '210', '240', '270', '300', '330' ];
var faces = tube.faces; var degreeCntr=0; var degreeProp;
//console.log(faces);
var res30=0,res60=0,res90=0,res120=0,res150=0,res180=0,res210=0,res240=0,res270=0,res300=0,res330=0;
var res; var resDegree; var pnt=0;
// fetching json data of resistivity values at different degree as //shown in the image
var result = getResValue();
for(var k=0; k<faceLength; k++){
resDegree = degrees[degreeCntr];
degreeProp = "r"+resDegree;
res = result.resistivity[pnt][degreeProp];
objects.push(result.resistivity[pnt]);
f = faces[k];
color = new THREE.Color( 0xffffff );
if(res<5){
color.setRGB( 197/255, 217/255, 241/255);
}
else if(res>=5 && res<50){
color.setRGB( 141/255, 180/255, 226/255);
}
else if(res>=50 && res<100){
color.setRGB( 83/255, 141/255, 213/255);
}
else if(res>=100 && res<200){
color.setRGB( 22, 54, 92);
}
else if(res>=200 && res<300){
color.setRGB( 15/255,36/255,62/255);
}
else if(res>=300 && res<400){
color.setRGB( 220/255, 230/255, 241/255);
}
else if(res>=400 && res<700){
color.setRGB( 184/255, 204/255, 228/255);
}
else if(res>=700 && res<1200){
color.setRGB( 149/255, 179/255, 215/255);
}
else if(res>=1200 && res<1500){
color.setRGB( 54/255, 96/255, 146/255);
}
else if(res>=1700 && res<1800){
color.setRGB( 36/255, 84/255, 98/255);
}
else if(res>1900){
color.setRGB( 128/255, 128/255, 128/255);
}
for(var j=0;j<4;j++)
{
tube.vertices.push(f.centroid);
vertexIndex = f[ faceIndices[ j ] ];
p = tube.vertices[ vertexIndex ];
f.vertexColors[ j ] = color;
}
degreeCntr++;
if(degreeCntr==10){
degreeCntr=0;
}
if(k%12==0 && k!=0){
pnt++;
}
}
This logic takes too much time to render the model and the model becomes too heavy and we can't perform other operations. The FPS on android drops at 2-3 FPS. Actually I've to render this model on iPad so have to use canvas renderer only.
So, how do I make this model lighter to load and works smoothly on iPad ? and is there any other way to apply colors on every face ? If canvas map as texture can be applied to make the model lighter, how do I build that map with all the colors based on value ?
Update:
After changing library version to r53, vertexColors: THREE.FaceColors and face.color.setRGB( Math.random(), Math.random(), Math.random()), the model displays random color for each face on canvas rendering.
So now the issue is applying colors as per requirements (either by canvas map or any feasible solution) and to make the model lighter to load it smoothly on iPad.
I believe this will give you a little bit better performance + if you could come up with some automated method of calculating colors for each angle offset, that you could set hex color directly:
for ( var i = 0; i < tube.faces.length; i ++ ) {
tube.faces[ i ].color.setHex( Math.random() * 0xffffff );
}
As I explained to you in the previous message - three.js - text next to line, using canvas textures will only increase load to you fps if you'll attempt to render so many faces.
If you really want to render 24,000 faces on canvas renderer and still hope that it gonna show up good on an iPad – you are out of your mind!))
Here is the only solution that I can think of for now:
1) Set your tube to only 1 segment.
2) Create 12 canvas elements (for every radius segment) with Width equal to your tube length (see my link above).
3) Now imagine that your 2000 segments you are going to create inside of each canvas. So, you divide your canvas length by 2000 and for every one of the portion of this division you set your calculated color!!! (Just like the Stats() FPS bar shows it’s bar, but you are going to have each bar different color).
4) Then you just apply your colored-bars-canvas-texture to each one of your 12 radius segments and you are good to go!!
This way you’ll only get initial page load (calculating 'em 24,000 colored-bars) and YOUR WHOLE TUBE ONLY GONNA BE 12 FACES!!!
Now, I know your next question is going to be: How I'll pick my faces to show my lines with tag text?
Well, very simple! Just take current face (1 of 12) pick position coordinates and translate them back to your JSON, just the same way you would do with 24,000 faces;)
Hope that helps!