I'm currently going through this tutorial on rendering shapes WebGL (specifically a sphere in this case) and I understand the math behind the generation of each point on the sphere. In the tutorial though, the author defines one method to find all of the vertices and another to generate all of the squares that will comprise the sphere.
A couple of things are unclear from what is done in the tutorial. First, how exactly are the vertices generated by the parametric equation being connected to the squares (triangle strips) being generated? I've made a bare bones program in plain javascript and HTML5 before doing the same thing just using the vertices generated so I'm not seeing how and why they have to be used in conjunction with the triangle strips. The other point of confusion is specifically regarding the function that generates the squares:
var indexData = [];
for (var latNumber = 0; latNumber < latitudeBands; latNumber++) {
for (var longNumber = 0; longNumber < longitudeBands; longNumber++) {
var first = (latNumber * (longitudeBands + 1)) + longNumber;
var second = first + longitudeBands + 1;
indexData.push(first);
indexData.push(second);
indexData.push(first + 1);
indexData.push(second);
indexData.push(second + 1);
indexData.push(first + 1);
}
}
To generate the first point of each square (point on the top left corner) the following is done: var first = (latNumber * (longitudeBands + 1)) + longNumber;
I'm not sure why the number of the lattitude line needs to be multiplied by the total number of longitude lines (plus 1 to fully wrap around) at each step.
The code for both functions is toward the bottom of the tutorial. A general explanation of the use of triangle strips in a case like this could also be helpful, thanks.
how exactly are the vertices generated by the parametric equation being connected to the squares (triangle strips) being generated?
A: Vertices are basically points. So its basically generating points using math. Quote from tutorial :
"for a sphere of radius r, with m latitude bands and n longitude bands, we can generate values for x, y, and z by taking a range of values for θ by splitting the range 0 to π up into m parts, and taking a range of values for φ by splitting the range 0 to 2π into n parts, and then just calculating:
x = r sinθ cosφ
y = r cosθ
z = r sinθ sinφ"
how and why they have to be used in conjunction with the triangle strips
A: they are not triangle STRIPS as in the primitive type gl.TRIANGLE_STRIP, but merely regular triangles defined with 3 points.
regarding the function that generates the squares
A: They are not generation squares per se, but using the points generated from the parametric equation to create triangles for the GPU to render. The code you shown in the OP basically divides a square into 2 triangles.
Given a series of JSON co-ordinates typically in the format:
{from: {x:0, y:0}, to: {x:0, y:10}, ...}
I would like to draw a series of straight dotted paths which are connected with simple, fixed radius rounded corners. I have been looking at Slope Intercept Form to calculate the points along the straight line but I am a little perplexed as to the approach for calcualting the points along the (Bezier?) curves.
e.g. I want to draw curves between p1 and p2 and p3 and p4. Despite what the poor mockup might imply I am happy for the corners to be a fixed radius e.g. 10px
I would like to abstract out the drawing logic and therefore am seeking a generalised approach to returning a JavaScript point array which I can then render in a number of ways (hence I am avoiding using any inbuilt functions provided by SVG, Canvas etc).
What you want is a cubic bezier curve.
http://www.blackpawn.com/texts/splines/
Look at the first applet on this page. If A is p1, D is p2, the direction A-B is line 1's angle and the direction C-D is line 2's angle you can see how this gives you the properties you need - it starts at angle 1 and ends at angle 2 and is flush with the points.
So, to get your points C and D, one way to do this would be to take the line segment 1, copy it, place it starting at p1 - and say where the new line ends is B, and similar with line segment 2 and p2 for D. (And you could do things like have a factor that multiplies into the copied line segments' distance to make the curves stick out more or less... etc)
Then just do the math :)
http://en.wikipedia.org/wiki/B%C3%A9zier_curve#Cubic_B.C3.A9zier_curves
And once you have your equation for the curve, step through it with a delta t of the desired precision (e.g. every 0.1 of t, every 0.01...) and spit out every pair of points on the curve as a line segment.
I've been working in JavaScript to code a line drawing system. I'd like the lines drawn to be selectable, so I've been attempting to implement line-highlighting. As you can see in the image below, I have a line (in black) with known coordinates and an equation in slope-intercept (y=mx+b). How can I calculate the corners' (circled in green) coordinates, knowing the box's radius?
This is easiest to think of in terms of vectors.
Start off by defining the point at the end of the line as A, and the other end as B
var A = new Vector(1, 1)
var B = new Vector(5, 3)
Now find the unit direction vector of the line (a vector of length 1 pointing from A to B), and its perpendicular:
var dir = B.minus(A).normalize();
var dir_perp = new Vector(dir.y, -dir.x)
And extend them to be of length thickness:
dir = dir.times(thickness);
dir_perp = dir_perp.times(thickness)
The four corners are then:
[
A.minus(dir).plus(dir_perp),
A.minus(dir).minus(dir_perp),
B.plus(dir).minus(dir_perp),
B.plus(dir).plus(dir_perp)
]
This obviously assumes you have some sort of vector math library. Here's one I made earlier
I would like draw 3D points represented in image to 3D rectangle. Any idea how could I represent these in x,y and z axis
Here projection type is orthographic.
Thanks
Okay. Let's look at a simple example of what you are trying to accomplish it, and why this is such a complicated problem.
First, lets look a some projection functions. You need a way to mathematically describe how to transform a 3D (or higher dimensional) point into a 2D space (your monitor), or a projection.
The simpiest to understand is a very simple dimetric projection. Something like:
x' = x + z/2;
y' = y + z/4;
What does this mean? Well, x' is you x coordinate 2D projection: for every unit you move backwards in space, the projection will move that point half that many units to the right. And y' represents that same projection for your y coordinate: for every unit you move backwards in space, the projection will move that point a quarter unit up.
So a point at [0,0,0] will get projected to a 2d point of [0,0]. A point at [0,0,4] will get projected to a 2d point of [2,1].
Implemented in JavaScript, it would look something like this:
// Dimetric projection functions
var dimetricTx = function(x,y,z) { return x + z/2; };
var dimetricTy = function(x,y,z) { return y + z/4; };
Once you have these projection functions -- or ways to translate from 3D space into 2D space -- you can use them to start draw your image. A simple example of that using js canvas. First, some context stuff:
var c = document.getElementById("cnvs");
var ctx = c.getContext("2d");
Now, lets make a little helper to draw a 3D point:
var drawPoint = (function(ctx,tx,ty, size) {
return function(p) {
size = size || 3;
// Draw "point"
ctx.save();
ctx.fillStyle="#f00";
ctx.translate(tx.apply(undefined, p), ty.apply(undefined,p));
ctx.beginPath();
ctx.arc(0,0,size,0,Math.PI*2);
ctx.fill();
ctx.restore();
};
})(ctx,dimetricTx,dimetricTy);
This is pretty simple function, we are injecting the canvas context as ctx, as well as our tx and ty functions, which in this case our the dimetric functions we saw earlier.
And now a polygon drawer:
var drawPoly = (function(ctx,tx,ty) {
return function() {
var args = Array.prototype.slice.call(arguments, 0);
// Begin the path
ctx.beginPath();
// Move to the first point
var p = args.pop();
if(p) {
ctx.moveTo(tx.apply(undefined, p), ty.apply(undefined, p));
}
// Draw to the next point
while((p = args.pop()) !== undefined) {
ctx.lineTo(tx.apply(undefined, p), ty.apply(undefined, p));
}
ctx.closePath();
ctx.stroke();
};
})(ctx, dimetricTx, dimetricTy);
With those two functions, you could effectively draw the kind of graph you are looking for. For example:
// The array of points
var points = [
// [x,y,z]
[20,30,40],
[100,70,110],
[30,30,75]
];
(function(width, height, depth, points) {
var c = document.getElementById("cnvs");
var ctx = c.getContext("2d");
// Set some context
ctx.save();
ctx.scale(1,-1);
ctx.translate(0,-c.height);
ctx.save();
// Move our graph
ctx.translate(100,20);
// Draw the "container"
ctx.strokeStyle="#999";
drawPoly([0,0,depth],[0,height,depth],[width,height,depth],[width,0,depth]);
drawPoly([0,0,0],[0,0,depth],[0,height,depth],[0,height,0]);
drawPoly([width,0,0],[width,0,depth],[width,height,depth],[width,height,0]);
drawPoly([0,0,0],[0,height,0],[width,height,0],[width,0,0]);
ctx.stroke();
// Draw the points
for(var i=0;i<points.length;i++) {
drawPoint(points[i]);
}
})(150,100,150,points);
However, you should now be able to start to see some of the complexity of your actual question emerge. Namely, you asked about rotation, in this example we are using an extremely simple projection (our dimetric projection) which doesn't take much other than an oversimplified relationship between depth and its influences on x,y position. As the projections become more complex, you need to know more about your relationship/orientation in 3D space in order to create a reasonable 2D projection.
A working example of the above code can be found here. The example also includes isometric projection functions that can be swapped out for the dimetric ones to see how that changes the way the graph looks. It also does some different visualization stuff that I didn't include here, like drawing "shadows" to help "visualize" the actual orientation -- the limitations of 3D to 2D projections.
It's complicated, and even a superficial discussion is kind of beyond the scope of this stackoverflow. I recommend you read more into the mathematics behind 3D, there are plenty of resources, both online and in print form. Once you have a more solid understanding of the basics of how the math works then return here if you have a specific implementation question about it.
What you want to do is impossible to do using the method you've stated - this is because a box - when rotated in 3 dimensions won't look anything like that diagram of yours. It will also vary based on the type of projection you need. You can, however get started using three.js which is a 3D drawing library for Javascript.
Hope this helps.
How to Draw 3D Rectangle?
posted in: Parallelogram | updated on: 14 Sep, 2012
To sketch 3 - Dimensional Rectangle means we are dealing with the figures which are different from 2 – D figures, which would need 3 axes to represent them. So, how to draw 3D rectangle?
To start with, first make two lines, one vertical and another horizontal in the middle of the paper such that they represent a “t” letter of English. This is what we need to draw for temporary use and will be removed later after the construction of the 3 – D rectangle is complete. Next we draw a Square whose measure of each side is 1 inch. Square must be perfect in Geometry so that 90 degree angles that are formed at respective corners are exact in measure. Now starting from upper right corner of the square we draw a line segment that will be stretched to a measure of 2 inches in the direction at an angle of 45 degrees. Similarly, we repeat the procedure by drawing another Line Segment from the upper left corner of the square and stretching it to 2 inches length in the direction at an angle of 45 degrees. These 2 line segments are considered to be the diagonals with respect to the horizontal line that we drew temporarily in starting. Also these lines will be parallel to each other. Next we draw a line that joins the end Point of these two diagonals.
Next starting from the very right of the 2 inch diagonal end point, draw a line of measure 1 inch that is supposed to be perpendicular to the temporary horizontal line. Next we need to join the lower left corner of the square with end point of the last 1’’ line we drew in 4th step and finally we get our 3 - D rectangular. Now we can erase our initial “t”. This 3- D rectangle resembles a Cuboid.
I'm currently trying to build a kind of pie chart / voronoi diagram hybrid (in canvas/javascript) .I don't know if it's even possible. I'm very new to this, and I haven't tried any approaches yet.
Assume I have a circle, and a set of numbers 2, 3, 5, 7, 11.
I want to subdivide the circle into sections equivalent to the numbers (much like a pie chart) but forming a lattice / honeycomb like shape.
Is this even possible? Is it ridiculously difficult, especially for someone who's only done some basic pie chart rendering?
This is my view on this after a quick look.
A general solution, assuming there are to be n polygons with k vertices/edges, will depend on the solution to n equations, where each equation has no more than 2nk, (but exactly 2k non-zero) variables. The variables in each polygon's equation are the same x_1, x_2, x_3... x_nk and y_1, y_2, y_3... y_nk variables. Exactly four of x_1, x_2, x_3... x_nk have non-zero coefficients and exactly four of y_1, y_2, y_3... y_nk have non-zero coefficients for each polygon's equation. x_i and y_i are bounded differently depending on the parent shape.. For the sake of simplicity, we'll assume the shape is a circle. The boundary condition is: (x_i)^2 + (y_i)^2 <= r^2
Note: I say no more than 2nk, because I am unsure of the lowerbound, but know that it can not be more than 2nk. This is a result of polygons, as a requirement, sharing vertices.
The equations are the collection of definite, but variable-bounded, integrals representing the area of each polygon, with the area equal for the ith polygon:
A_i = pi*r^2/S_i
where r is the radius of the parent circle and S_i is the number assigned to the polygon, as in your diagram.
The four separate pairs of (x_j,y_j), both with non-zero coefficients in a polygon's equation will yield the vertices for the polygon.
This may prove to be considerably difficult.
Is the boundary fixed from the beginning, or can you deform it a bit?
If I had to solve this, I would sort the areas from large to small. Then, starting with the largest area, I would first generate a random convex polygon (vertices along a circle) with the required size. The next area would share an edge with the first area, but would be otherwise also random and convex. Each polygon after that would choose an existing edge from already-present polygons, and would also share any 'convex' edges that start from there (where 'convex edge' is one that, if used for the new polygon, would result in the new polygon still being convex).
By evaluating different prospective polygon positions for 'total boundary approaches desired boundary', you can probably generate a cheap approximation to your initial goal. This is quite similar to what word-clouds do: place things incrementally from largest to smallest while trying to fill in a more-or-less enclosed space.
Given a set of voronio centres (i.e. a list of the coordinates of the centre for each one), we can calculate the area closest to each centre:
area[i] = areaClosestTo(i,positions)
Assume these are a bit wrong, because we haven't got the centres in the right place. So we can calculate the error in our current set by comparing the areas to the ideal areas:
var areaIndexSq = 0;
var desiredAreasMagSq = 0;
for(var i = 0; i < areas.length; ++i) {
var contrib = (areas[i] - desiredAreas[i]);
areaIndexSq += contrib*contrib;
desiredAreasMagSq += desiredAreas[i]*desiredAreas[i];
}
var areaIndex = Math.sqrt(areaIndexSq/desiredAreasMagSq);
This is the vector norm of the difference vector between the areas and the desiredAreas. Think of it like a measure of how good a least squares fit line is.
We also want some kind of honeycomb pattern, so we can call that honeycombness(positions), and get an overall measure of the quality of the thing (this is just a starter, the weighting or form of this can be whatever floats your boat):
var overallMeasure = areaIndex + honeycombnessIndex;
Then we have a mechanism to know how bad a guess is, and we can combine this with a mechanism for modifying the positions; the simplest is just to add a random amount to the x and y coords of each centre. Alternatively you can try moving each point towards neighbour areas which have an area too high, and away from those with an area too low.
This is not a straight solve, but it requires minimal maths apart from calculating the area closest to each point, and it's approachable. The difficult part may be recognising local minima and dealing with them.
Incidentally, it should be fairly easy to get the start points for the process; the centroids of the pie slices shouldn't be too far from the truth.
A definite plus is that you could use the intermediate calculations to animate a transition from pie to voronoi.