I have a cylinder that is split into many height segments (the amount depends on the data). For each height segment I have a value for which I want the entire circle at that height to be extruded.
So essentially I end up with a cylinder that has very spiky edges.
I was intending to do this by manually moving the vertices or faces but I cannot seem to access the vertices/faces for a given segment.
So basically I need to scale the segment at N height.
Any suggestions on which direction I take? Have had a few failed attempts now and am running out of ideas.
Have a look at the source code for CylinderGeometry.js on GitHub.
You could copy this entire method and call it something different, e.g.
THREE.CylinderGeometry2 = function (...
Then change the generation of the vertices based on the number of height segments you have.
Related
I have a tile-based isometric world and I can calculate which tile is underneath specific (mouse) coordinates by using the following calculations:
function isoTo2D(pt:Point):Point{
var tempPt:Point = new Point(0, 0);
tempPt.x = (2 * pt.y + pt.x) / 2;
tempPt.y = (2 * pt.y - pt.x) / 2;
return(tempPt);
}
function getTileCoordinates(pt:Point, tileHeight:Number):Point{
var tempPt:Point = new Point(0, 0);
tempPt.x = Math.floor(pt.x / tileHeight);
tempPt.y = Math.floor(pt.y / tileHeight);
return(tempPt);
}
(Taken from http://gamedevelopment.tutsplus.com/tutorials/creating-isometric-worlds-a-primer-for-game-developers--gamedev-6511, this is a flash implementation but the maths is the same)
However, my problem comes in when I have tiles that have different elevation levels:
In these scenarios, due to the height of some tiles which have a higher elevation, the tiles (or portions of tiles) behind are covered up and shouldn't be able to be selected by the mouse, instead selecting the tile which is in front of it.
How can I calculate the tile by mouse coordinates taking into account the tiles' elevation?
I'm using a javascript and canvas implementation.
There is a technique of capturing object under the mouse on a canvas without needing to recalculate mouse coordinates into your "world" coordinates. This is not perfect, has some drawbacks and restrictions, yet it does it's job in some simple cases.
1) Position another canvas atop of your main canvas and set it's opacity to 0. Make sure your second canvas has the same size and overlaps your main one.
2) Whenever you draw your interactive objects to the main canvas, draw and fill the same objects on the second canvas, but using one unique color per object (from #000000 to #ffffff)
3) Set mouse event handling to the second canvas.
4) Use getPixel on the second canvas at mouse position to get the "id" of the object clicked/hovered over.
Main advantage is WYSIWYG principle, so (if everything is done properly) you can be sure, that objects on the main canvas are in the same place as on the second canvas, so you don't need to worry about canvas resizing or object depth (like in your case) calculations to get the right object.
Main drawback is need to "double-render" the whole scene, yet it can be optimized by not drawing on the second canvas when it's not necessary, like:
in "idling" scene state, when interactive objects are staying on their places and wait for user action.
in "locked" scene state, when some stuff is animated or smth. and user is not allowed to interact with objects.
Main restriction is a maximum number of interactive objects on the scene (up to #ffffff or 16777215 objects).
So... Not reccomended for:
Games with big amount of interactive objects on a scene. (bad performance)
Fast-paced games, where interactive objects are constantly moved/created/destroyed.(bad performance, issues with re-using id's)
Good for:
GUI's handling
Turn-based games / slow-paced puzzle games.
Your hit test function will need to have access to all your tiles in order to determine which one is hit. It will then perform test hits starting with the tallest elevation.
Assuming that you only have discreet (integer) tile heights, the general algorithm would be like this (pseudo code, assuming that tiles is a two-dimensional array of object with an elevation property):
function getTile(mousePt, tiles) {
var maxElevation = getMaxElevation(tiles);
var minElevation = getMinElevation(tiles);
var elevation;
for (elevation = maxElevation; elevation >= minElevation; elevation--) {
var pt = getTileCoordinates(mousePt, elevation);
if (tiles[pt.x][pt.y].elevation === elevation) {
return pt;
}
}
return null; // not tile hit
}
This code would need to be adjusted for arbitrary elevations and could be optimized to skip elevation that don't contain any tiles.
Note that my pseudocode ignores vertical sides of a tile and clicks on them will select the (lower elevation) tile obscured by the vertical side. If vertical tiles need to be accounted for, then a more generic surface hit detection approach will be needed. You could visit every tile (from closest to farthest away) and test whether the mouse coordinates are in the "roof" or in one of the viewer facing "wall" polygons.
If map is not rotatable and exatly same as picture you posted here,
When you are drawing polygons, save each tile's polygon(s) in a polygon array. Then sort the array only once using distance of them(their tile) to you(closest first, farthest last) while keeping them grouped by tile index.
When click event happens, get x,y coordinates of mouse, and do point in polygon test starting from first element array until last element. When hit, stop at that element.
No matter how high a tile is, will not hide any tile that is closer to you(or even same distance to you).
Point in polygon test is already solved:
Point in Polygon Algorithm
How can I determine whether a 2D Point is within a Polygon?
Point in polygon
You can even check every pixel of canvas once with this function and save results into an 2d array of points, vect2[x][y] which gives i,j indexes of tiles from x,y coordinates of mouse, then use this as a very fast index finder.
Pros:
fast and parallelizable using webworkers(if there are millions of tiles)
scalable to multiple isometric maps using arrays of arrays of polygons sorted by distance to you.
Elevation doesnt decrease performance because of only 3 per tile maximum.
Doesn't need any conversion to isometric to 2d. Just the coordinates of corners of polygons on canvas and coordinates of mouse on the same canvas.
Cons:
You need coordinates of each corner if you haven't already.
Clicking a corner will pick closest tile to you while it is on four tiles at the same time.
The answer, oddly, is written up in the Wikipedia page, in the section titled "Mapping Screen to World Coordinates". Rather than try to describe the graphics, just read the section three times.
You will need to determine exactly which isomorphic projection you are using, often by measuring the tile size on the screen with a ruler.
I'm working on a visualization with pixel data and I have been able to successfully load data onto the canvas. Data is about the type of clouds on the globe. There are about 12 types. Every type has a color on the map. Right now I'm basically plotting every coordinate of the pixel where the color is of a specific cloud and then creating svg for it and loading it too the canvas. But when I plot the pixel data, it creates too many svg's.
However, some of the pixels are like next to each other and I was thinking that is there any way I can join those two pixels that are next to each other. I have a list of coordinates:
Cloud1 = [(1, 1), (1, 2), (2, 1), (2, 2), .... ]
In this case I would combine the four coordinates into one svg, since they are next to each other.
Also, one more thing, How many svg's can you load into a browser(max) at once without any lag?
Image URL: http://i.stack.imgur.com/beB8f.png
To answer the specific question of joining pixels that are adjacent to each other, use a bin method to accumulate touching/adjacent pixels.
Sort all the cloud's pixels by their pixel position (x,y). Iterate through each pixels, check against all existing bins(arrays of pixels), and if none are touching any points in any of the bins, create a new bin with that pixel as the first item. If the current pixel touches an item in the first bin found, add that pixel to that bin and go to the next pixel in the cloud.
After all the pixels in cloud have been divided into separate bins, get the outer x,y values of each bin to draw a polyline representing the subcloud. So essentially you have 1 svg representing each bin.
This should reduce your number of svg elements greatly and you have the added bonus of colouring the area of the polyline cloud with an opaque color which may overlay other cloud type very nicely.
I have a large circle with smaller ones inside made using two.js.
My problem is that these two do not rotate in their own place but in the top left axis.
I want the group of circles (circlesGroup) rotate only inside the large one in a static position. The circlesGroup and the large circle are grouped together as rotatoGroup.
two.bind('update', function(frameCount, timeDelta) {
circlesGroup.rotation = frameCount / 120;
});
two.bind('update', function(frameCount, timeDelta) {
rotatoGroup.rotation = frameCount / 60;
});
The whole code is in CodePen.
All visible shapes when invoked with two.make... ( circles, rectangles, polygons, and lines ) are oriented in the center like this Adobe Illustrator example:
When this shape's translation, rotation, or scale change those changes will be reflected as transformations about the center of the shape.
Two.Groups however do not behave this way. Think of them as display-less rectangles. They're origin, i.e group.translation vector, always begins at (0, 0). In your case you can deal with this by normalizing the translation your defining on all your circles.
Example 1: Predefined in normalized space
In this codepen example we're defining the position of all the circles around -100, 100, effectively half the radius in both positive-and-negative x-and-y directions. Once we've defined the circles within these constraints we can move the whole group with group.translation.set to place it in the center of the screen. Now when the circles rotate they are perceived as rotating around themselves.
Example 2: Normalizing after the fact
In this codepen example we're working with what we already have. A Two.Group that contains all of our shapes ( the bigger circle as well as the array of the smaller circles ). By using the method group.center(); ( line 31 ) we can normalize the children of the group to be around (0, 0). We can then change the translation of the group in order to be in the desired position.
N.B: This example is a bit complicated because it invokes underscore's defer method which forces the centering of the group after all the changes have been registered. I'm in the process of fixing this.
I'm using canvas for a project and I have a number of elements that I'm skewing. I'm only skewing on the y value and just want to know what the new width of the image is after skewing (so I can align it with another canvas element). Check out the code below to see what I mean
ctx.save();
//skew the context
ctx.transform(1,0,1.3,0,0,0);
//draw two images with different heights/widths
ctx.drawImage(image,0,0,42,60);
ctx.drawImage(image,0,0,32,25);
The goal would be to know that the 42 by 60 image was now a X by 60 image so I could do some translating before drawing it at 0,0. It's easy enough to measure each image individually, but I have different skew values and heights/widths throughout the project that need to be align. Currently I use this code (works decently for images between 25 and 42 widths):
var skewModifier = imageWidth*(8/6)+(19/3);
var skewAmount = 1.3; //this is dynamic in my app
var width = (skewModifier*skewAmount)+imageWidth;
As images get wider though this formula quickly falls apart (I think it's a sloping formula not a straight value like this one). Any ideas on what canvas does for skews?
You should be able to derive it mathematically. I believe:
Math.atan(skewAmount) is the angle, in radians, that something is skewed with respect to the origin.
So 1.3 would skew the object by 0.915 radians or 52 degrees.
So here's a red unskewed object next to the same object skewed (painted green). So you have a right triangle:
We know the origin angle (0.915 rads) and we know the adjacent side length, which is 60 and 25 for your two images. (red's height).
The hypotenuse is the long side thats being skewed.
And the opposite side is the triangle bottom - how much its been skewed!
Tangent gets us opposite / adjacent if I recall, so for the first one:
tan(0.915) = opposite / 60, solving for the opposite in JavaScript code we have:
opposite = Math.tan(0.915)*60
So the bottom side of the skewed object starts about 77 pixels away from the origin. Lets check our work in the canvas:
http://jsfiddle.net/LBzUt/
Looks good to me!
The triangle in question of course is the canvas origin, that black dot I painted, and the bottom-left of the red rectangle, which is the original position that we're searching for before skewing.
That was a bit of a haphazard explanation. Any questions?
Taking Simon's fiddle example one step further, so you can simply enter the degrees:
Here's the fiddle
http://jsfiddle.net/LBzUt/33/
Quick question involving javascript canvas... I have a set points (connected with a line) I want to graph on a 400x300 canvas element. I will constantly be adding more points. I need the line to stretch to fill the entire canvas (leaving no unnecessary space).
Example:
into this:
Thanks! C.Ruhl
You want to find the step by doing canvasWidth / (number of points - 1)
and adding X += step each time.
Example here:
http://jsfiddle.net/pDDTQ/
Distinguish between internal canvas size and visible size. 400x300 is your visible size and set by style="width:400px; height:300px". Everytime there is new point (e.g. 400,500) you set canvas.width=400; canvas.height=500; and replot the whole graph. From a certain point you might want to adjust the width of the line.