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I wanna to produce a Pie Chart on a Hexagon. There are probably several solutions for this. In the picture are my Hexagon and two Ideas:
My Hexagon (6 vertices, 4 faces)
How it should look at the end (without the gray lines)
Math: Can I get some informations from the object to dynamically calculate new vertices (from the center to each point) to add colored faces?
Clipping: On a sphere a Pie-Chart is easy, maybe I can clip the THREE Object (WITHOUT SVG.js!) so I just see the Hexagon with the clipped Chart?
Well the whole clipping thing in three.js is already solved here : Object Overflow Clipping Three JS, with a fiddle that shows it works and all.
So I'll go for the "vertices" option, or rather, a function that, given a list of values gives back a list of polygons, one for each value, that are portions of the hexagon, such that
they all have the centre point as a vertex
the angle they have at that point is proportional to the value
they form a partition the hexagon
Let us suppose the hexagon is inscribed in a circle of radius R, and defined by the vertices :
{(R sqrt(3)/2, R/2), (0,R), (-R sqrt(3)/2, R/2), (-R sqrt(3)/2, -R/2), (0,-R), (R sqrt(3)/2, -R/2)}
This comes easily from the values cos(Pi/6), sin(Pi/6) and various symmetries.
Getting the angles at the centre for each polygon is pretty simple, since it is the same as for a circle. Now we need to know the position of the points that are on the hexagon.
Note that if you use the symmetries of the coordinate axes, there are only two cases : [0,Pi/6] and [Pi/6,Pi/2], and you then get your result by mirroring. If you use the rotational symmetry by Pi/3, you only have one case : [-Pi/6,Pi/6], and you get the result by rotation.
Using rotational symmetry
Thus for every point, you can consider it's angle to be between [-Pi/6,Pi/6]. Any point on the hexagon in that part has x=R sqrt(3)/2, which simplifies the problem a lot : we only have to find it's y value.
Now we assumed that we know the polar coordinate angle for our point, since it is the same as for a circle. Let us call it beta, and alpha its value in [-Pi/6,Pi/6] (modulo Pi/3). We don't know at what distance d it is from the centre, and thus we have the following system :
Which is trivially solved since cos is never 0 in the range [-Pi/6,Pi/6].
Thus d=R sqrt(3)/( 2 cos(alpha) ), and y=d sin(alpha)
So now we know
the angle from the centre beta
it's distance d from the centre, thanks to rotational symmetry
So our point is (d cos(beta), d sin(beta))
Code
Yeah, I got curious, so I ended up coding it. Sorry if you wanted to play with it yourself. It's working, and pretty ugly in the end (at least with this dataset), see the jsfiddle : http://jsfiddle.net/vb7on8vo/5/
var R = 100;
var hexagon = [{x:R*Math.sqrt(3)/2, y:R/2}, {x:0, y:R}, {x:-R*Math.sqrt(3)/2, y:R/2}, {x:-R*Math.sqrt(3)/2, y:-R/2}, {x:0, y:-R}, {x:R*Math.sqrt(3)/2, y:-R/2}];
var hex_angles = [Math.PI / 6, Math.PI / 2, 5*Math.PI / 6, 7*Math.PI / 6, 3*Math.PI / 2, 11*Math.PI / 6];
function regions(values)
{
var i, total = 0, regions = [];
for(i=0; i<values.length; i++)
total += values[i];
// first (0 rad) and last (2Pi rad) points are always at x=R Math.sqrt(3)/2, y=0
var prev_point = {x:hexagon[0].x, y:0}, last_angle = 0;
for(i=0; i<values.length; i++)
{
var j, theta, p = [{x:0,y:0}, prev_point], beta = last_angle + values[i] * 2 * Math.PI / total;
for( j=0; j<hexagon.length; j++)
{
theta = hex_angles[j];
if( theta <= last_angle )
continue;
else if( theta >= beta )
break;
else
p.push( hexagon[j] );
}
var alpha = beta - (Math.PI * (j % 6) / 3); // segment 6 is segment 0
var d = hexagon[0].x / Math.cos(alpha);
var point = {x:d*Math.cos(beta), y:d*Math.sin(beta)};
p.push( point );
regions.push(p.slice(0));
last_angle = beta;
prev_point = {x:point.x, y:point.y};
}
return regions;
}
I would like to draw a textured circle in Direct3D which looks like a real 3D sphere. For this purpose, I took a texture of a billard ball and tried to write a pixel shader in HLSL, which maps it onto a simple pre-transformed quad in such a way that it looks like a 3-dimensional sphere (apart from the lighting, of course).
This is what I've got so far:
struct PS_INPUT
{
float2 Texture : TEXCOORD0;
};
struct PS_OUTPUT
{
float4 Color : COLOR0;
};
sampler2D Tex0;
// main function
PS_OUTPUT ps_main( PS_INPUT In )
{
// default color for points outside the sphere (alpha=0, i.e. invisible)
PS_OUTPUT Out;
Out.Color = float4(0, 0, 0, 0);
float pi = acos(-1);
// map texel coordinates to [-1, 1]
float x = 2.0 * (In.Texture.x - 0.5);
float y = 2.0 * (In.Texture.y - 0.5);
float r = sqrt(x * x + y * y);
// if the texel is not inside the sphere
if(r > 1.0f)
return Out;
// 3D position on the front half of the sphere
float p[3] = {x, y, sqrt(1 - x*x + y*y)};
// calculate UV mapping
float u = 0.5 + atan2(p[2], p[0]) / (2.0*pi);
float v = 0.5 - asin(p[1]) / pi;
// do some simple antialiasing
float alpha = saturate((1-r) * 32); // scale by half quad width
Out.Color = tex2D(Tex0, float2(u, v));
Out.Color.a = alpha;
return Out;
}
The texture coordinates of my quad range from 0 to 1, so I first map them to [-1, 1]. After that I followed the formula in this article to calculate the correct texture coordinates for the current point.
At first, the outcome looked ok, but I'd like to be able to rotate this illusion of a sphere arbitrarily. So I gradually increased u in the hope of rotating the sphere around the vertical axis. This is the result:
As you can see, the imprint of the ball looks unnaturally deformed when it reaches the edge. Can anyone see any reason for this? And additionally, how could I implement rotations around an arbitrary axis?
Thanks in advance!
I finally found the mistake by myself: The calculation of the z value which corresponds to the current point (x, y) on the front half of the sphere was wrong. It must of course be:
That's all, it works as exspected now. Furthermore, I figured out how to rotate the sphere. You just have to rotate the point p before calculating u and v by multiplying it with a 3D rotation matrix like this one for example.
The result looks like the following:
If anyone has any advice as to how I could smooth the texture a litte bit, please leave a comment.
I'm trying to come up with a proper algorithm to create a matrix which gets the quad to face the camer straight up but I'm having difficulty.
The quad I'm drawing is facing downwards, here's the code I have so far:
D3DXVECTOR3 pos;
pos = D3DXVECTOR3(-2.0f, 6.0f, 0.1f);
D3DXMATRIX transform;
D3DXMatrixTranslation(&transform, pos.x, pos.y, pos.z);
D3DXVECTOR3 axis = D3DXVECTOR3(0, -1, 0);
D3DXVECTOR3 quadtocam = pos - EmCameraManager::Get().GetPosition();
D3DXVec3Normalize(&quadtocam,&quadtocam);
D3DXVECTOR3 ortho;
D3DXVec3Cross(&ortho, &axis, &quadtocam);
float rotAngle = acos(D3DXVec3Dot(&axis,&quadtocam));
D3DXMATRIX rot;
D3DXMatrixRotationAxis(&rot,&ortho,rotAngle);
transform = rot*transform;
This works when it comes to making the quad face the camera however it doesn't stay up right when facing it from all angles.
In this screen shot: http://imgur.com/hFmzc.png On the left the quad is being viewed straight on (vector is 0,0,1), in the other its being viewed from an arbitrary angle.
Both times it's facing the camera, however when from an arbitrary angle its tilting along its local z axis. I'm not sure how to fix it and was wondering what the next step here would be?
Rotating around an arbitrary axis will always make that. You should rotate your model around y axis first to make it point(up vector) to your camera and after rotate it around your ortho axis which would be aligned with the model's right vector.
Assuming that z axis is coming out of the screen here's some code:
D3DXVECTOR3 Zaxis = D3DXVECTOR3(0, 1, 0);
D3DXVECTOR3 flattenedQuadtocam = quadtocam;
flattenedQuadtocam.y = 0;
float firstRotAngle = acos(D3DXVec3Dot(&Zaxis,&flattenedQuadtocam));
D3DXMATRIX firstRot;
D3DXMatrixRotationAxis(&firstRot,&Zaxis,firstRotAngle);
transform = firstRot*transform;
This should be put right after this line:
D3DXVec3Normalize(&quadtocam,&quadtocam);
Then the rest of your code should work.
Given curves of type Circle and Circular-Arc in 3D space, what is a good way to compute accurate bounding boxes (world axis aligned)?
Edit: found solution for circles, still need help with Arcs.
C# snippet for solving BoundingBoxes for Circles:
public static BoundingBox CircleBBox(Circle circle)
{
Point3d O = circle.Center;
Vector3d N = circle.Normal;
double ax = Angle(N, new Vector3d(1,0,0));
double ay = Angle(N, new Vector3d(0,1,0));
double az = Angle(N, new Vector3d(0,0,1));
Vector3d R = new Vector3d(Math.Sin(ax), Math.Sin(ay), Math.Sin(az));
R *= circle.Radius;
return new BoundingBox(O - R, O + R);
}
private static double Angle(Vector3d A, Vector3d B)
{
double dP = A * B;
if (dP <= -1.0) { return Math.PI; }
if (dP >= +1.0) { return 0.0; }
return Math.Acos(dP);
}
One thing that's not specified is how you convert that angle range to points in space. So we'll start there and assume that the angle 0 maps to O + r***X** and angle π/2 maps to O + r***Y**, where O is the center of the circle and
X = (x1,x2,x3)
and
Y = (y1,y2,y3)
are unit vectors.
So the circle is swept out by the function
P(θ) = O + rcos(θ)X + rsin(θ)Y
where θ is in the closed interval [θstart,θend].
The derivative of P is
P'(θ) = -rsin(θ)X + rcos(θ)Y
For the purpose of computing a bounding box we're interested in the points where one of the coordinates reaches an extremal value, hence points where one of the coordinates of P' is zero.
Setting -rsin(θ)xi + rcos(θ)yi = 0 we get
tan(θ) = sin(θ)/cos(θ) = yi/xi.
So we're looking for θ where θ = arctan(yi/xi) for i in {1,2,3}.
You have to watch out for the details of the range of arctan(), and avoiding divide-by-zero, and that if θ is a solution then so is θ±k*π, and I'll leave those details to you.
All you have to do is find the set of θ corresponding to extremal values in your angle range, and compute the bounding box of their corresponding points on the circle, and you're done. It's possible that there are no extremal values in the angle range, in which case you compute the bounding box of the points corresponding to θstart and θend. In fact you may as well initialize your solution set of θ's with those two values, so you don't have to special case it.
I'm writing a script where icons rotate around a given pivot (or origin). I've been able to make this work for rotating the icons around an ellipse but I also want to have them move around the perimeter of a rectangle of a certain width, height and origin.
I'm doing it this way because my current code stores all the coords in an array with each angle integer as the key, and reusing this code would be much easier to work with.
If someone could give me an example of a 100x150 rectangle, that would be great.
EDIT: to clarify, by rotating around I mean moving around the perimeter (or orbiting) of a shape.
You know the size of the rectangle and you need to split up the whole angle interval into four different, so you know if a ray from the center of the rectangle intersects right, top, left or bottom of the rectangle.
If the angle is: -atan(d/w) < alfa < atan(d/w) the ray intersects the right side of the rectangle. Then since you know that the x-displacement from the center of the rectangle to the right side is d/2, the displacement dy divided by d/2 is tan(alfa), so
dy = d/2 * tan(alfa)
You would handle this similarily with the other three angle intervals.
Ok, here goes. You have a rect with width w and depth d. In the middle you have the center point, cp. I assume you want to calculate P, for different values of the angle alfa.
I divided the rectangle in four different areas, or angle intervals (1 to 4). The interval I mentioned above is the first one to the right. I hope this makes sense to you.
First you need to calculate the angle intervals, these are determined completely by w and d. Depending on what value alfa has, calculate P accordingly, i.e. if the "ray" from CP to P intersects the upper, lower, right or left sides of the rectangle.
Cheers
This was made for and verified to work on the Pebble smartwatch, but modified to be pseudocode:
struct GPoint {
int x;
int y;
}
// Return point on rectangle edge. Rectangle is centered on (0,0) and has a width of w and height of h
GPoint getPointOnRect(int angle, int w, int h) {
var sine = sin(angle), cosine = cos(angle); // Calculate once and store, to make quicker and cleaner
var dy = sin>0 ? h/2 : h/-2; // Distance to top or bottom edge (from center)
var dx = cos>0 ? w/2 : w/-2; // Distance to left or right edge (from center)
if(abs(dx*sine) < abs(dy*cosine)) { // if (distance to vertical line) < (distance to horizontal line)
dy = (dx * sine) / cosine; // calculate distance to vertical line
} else { // else: (distance to top or bottom edge) < (distance to left or right edge)
dx = (dy * cosine) / sine; // move to top or bottom line
}
return GPoint(dx, dy); // Return point on rectangle edge
}
Use:
rectangle_width = 100;
rectangle_height = 150;
rectangle_center_x = 300;
rectangle_center_y = 300;
draw_rect(rectangle_center_x - (rectangle_width/2), rectangle_center_y - (rectangle_center_h/2), rectangle_width, rectangle_height);
GPoint point = getPointOnRect(angle, rectangle_width, rectangle_height);
point.x += rectangle_center_x;
point.y += rectangle_center_y;
draw_line(rectangle_center_x, rectangle_center_y, point.x, point.y);
One simple way to do this using an angle as a parameter is to simply clip the X and Y values using the bounds of the rectangle. In other words, calculate position as though the icon will rotate around a circular or elliptical path, then apply this:
(Assuming axis-aligned rectangle centered at (0,0), with X-axis length of XAxis and Y-axis length of YAxis):
if (X > XAxis/2)
X = XAxis/2;
if (X < 0 - XAxis/2)
X = 0 - XAxis/2;
if (Y > YAxis/2)
Y = YAxis/2;
if (Y < 0 - YAxis/2)
Y = 0 - YAxis/2;
The problem with this approach is that the angle will not be entirely accurate and the speed along the perimeter of the rectangle will not be constant. Modelling an ellipse that osculates the rectangle at its corners can minimize the effect, but if you are looking for a smooth, constant-speed "orbit," this method will not be adequate.
If think you mean rotate like the earth rotates around the sun (not the self-rotation... so your question is about how to slide along the edges of a rectangle?)
If so, you can give this a try:
# pseudo coode
for i = 0 to 499
if i < 100: x++
else if i < 250: y--
else if i < 350: x--
else y++
drawTheIcon(x, y)
Update: (please see comment below)
to use an angle, one line will be
y / x = tan(th) # th is the angle
the other lines are simple since they are just horizontal or vertical. so for example, it is x = 50 and you can put that into the line above to get the y. do that for the intersection of the horizontal line and vertical line (for example, angle is 60 degree and it shoot "NorthEast"... now you have two points. Then the point that is closest to the origin is the one that hits the rectangle first).
Use a 2D transformation matrix. Many languages (e.g. Java) support this natively (look up AffineTransformation); otherwise, write out a routine to do rotation yourself, once, debug it well, and use it forever. I must have five of them written in different languages.
Once you can do the rotation simply, find the location on the rectangle by doing line-line intersection. Find the center of the orbited icon by intersecting two lines:
A ray from your center of rotation at the angle you desire
One of the four sides, bounded by what angle you want (the four quadrants).
Draw yourself a sketch on a piece of paper with a rectangle and a centre of rotation. First translate the rectangle to centre at the origin of your coordinate system (remember the translation parameters, you'll need to reverse the translation later). Rotate the rectangle so that its sides are parallel to the coordinate axes (same reason).
Now you have a triangle with known angle at the origin, the opposite side is of known length (half of the length of one side of the rectangle), and you can now:
-- solve the triangle
-- undo the rotation
-- undo the translation