I want to draw a oval in html5 canvas,and i found a good method for it in stackoverflow.but I have another quesition.
function drawEllipse(ctx, x, y, w, h) {
var kappa = 0.5522848;
ox = (w / 2) * kappa, // control point offset horizontal
oy = (h / 2) * kappa, // control point offset vertical
xe = x + w, // x-end
ye = y + h, // y-end
xm = x + w / 2, // x-middle
ym = y + h / 2; // y-middle
ctx.beginPath();
ctx.moveTo(x, ym);
ctx.bezierCurveTo(x, ym - oy, xm - ox, y, xm, y);
ctx.bezierCurveTo(xm + ox, y, xe, ym - oy, xe, ym);
ctx.bezierCurveTo(xe, ym + oy, xm + ox, ye, xm, ye);
ctx.bezierCurveTo(xm - ox, ye, x, ym + oy, x, ym);
ctx.closePath();
ctx.stroke();
}
the method in the above link has use bezierCurveTo to draw a ellipse,but it has draw bezierCurveTo 4 times. but I think just 2 bezierCurveTo can draw a ellipse.like this:
but i'm weak in Mathematics,could someone tell me the relationship of "the control point" and "the oval point" please? or we must draw four bezier Curve to draw a oval?
thanks everybody
My background isn't in mathematics so if I'm wrong I'm sure someone will correct me, but from my understanding we can draw a pretty good approximation of an ellipse with just two cubic bezier curves but the coordinates will be a little tricky.
To answer your question about the relation between the oval point and the control points I think it best be answered by watching this video from the point I've selected if you're familiar with interpolation or from the beginning if you are not. Don't worry it is short.
One problem we're probably going to run into is that when we start from the top and do a bezierCurveTo the bottom of the ellipse with the corners of the rectangle (of the same width and height) as the control points, the ellipses width is going to be smaller than the rectangle. .75 times the size we want. So we can just scale the control points accordingly.
Our control point's x would be adjusted like so (assuming width is the width of the ellipse and we're dividing by two to get its offset from the origin)
var cpx = (width / .75) / 2;
Put together a visualization where you can play with the width and height and see the drawn ellipse.
The red ellipse is how we wanted it to be drawn, with the inner one how it would be drawn if we didnt reposition the control points. The lines illustrate De Casteljau's algorithm that was shown in the video.
Here's a screenshot of the visualization
You only need two cubic bezier curves to draw an ellipse. Here's a simplified version of DerekR's code that uses the original function arguments that you provided--assuming you want x and y to be the center of the ellipse:
jsFiddle
function drawEllipse(ctx, x, y, w, h) {
var width_over_2 = w / 2;
var width_two_thirds = w * 2 / 3;
var height_over_2 = h / 2;
ctx.beginPath();
ctx.moveTo(x, y - height_over_2);
ctx.bezierCurveTo(x + width_two_thirds, y - height_over_2, x + width_two_thirds, y + height_over_2, x, y + height_over_2);
ctx.bezierCurveTo(x - width_two_thirds, y + height_over_2, x - width_two_thirds, y - height_over_2, x, y -height_over_2);
ctx.closePath();
ctx.stroke();
}
Big thanks to BKH.
I used his code with two bezier curves to complete my ellipse drawing with any rotation angle. Also, I created an comparison demo between ellipses drawn by bezier curves and native ellipse() function (for now implemented only in Chrome).
function drawEllipseByBezierCurves(ctx, x, y, radiusX, radiusY, rotationAngle) {
var width_two_thirds = radiusX * 4 / 3;
var dx1 = Math.sin(rotationAngle) * radiusY;
var dy1 = Math.cos(rotationAngle) * radiusY;
var dx2 = Math.cos(rotationAngle) * width_two_thirds;
var dy2 = Math.sin(rotationAngle) * width_two_thirds;
var topCenterX = x - dx1;
var topCenterY = y + dy1;
var topRightX = topCenterX + dx2;
var topRightY = topCenterY + dy2;
var topLeftX = topCenterX - dx2;
var topLeftY = topCenterY - dy2;
var bottomCenterX = x + dx1;
var bottomCenterY = y - dy1;
var bottomRightX = bottomCenterX + dx2;
var bottomRightY = bottomCenterY + dy2;
var bottomLeftX = bottomCenterX - dx2;
var bottomLeftY = bottomCenterY - dy2;
ctx.beginPath();
ctx.moveTo(bottomCenterX, bottomCenterY);
ctx.bezierCurveTo(bottomRightX, bottomRightY, topRightX, topRightY, topCenterX, topCenterY);
ctx.bezierCurveTo(topLeftX, topLeftY, bottomLeftX, bottomLeftY, bottomCenterX, bottomCenterY);
ctx.closePath();
ctx.stroke();
}
You will find this explained slightly more math-based in http://pomax.github.io/bezierinfo/#circles_cubic, but the gist is that using a cubic bezier curve for more than a quarter turn is usually not a good idea. Thankfully, using four curves makes finding the required control points rather easy. Start off with a circle, in which case each quarter circle is (1,0)--(1,0.55228)--(0.55228,1)--(0,1) with scaled coordinates for your ellipse. Draw that four times with +/- signs swapped to effect a full circle, scale the dimensions to get your ellipse, and done.
If you use two curves, the coordinates become (1,0)--(1,4/3)--(-1,4/3)--(-1,0), scaled for your ellipse. It may still look decent enough in your application, it depends a bit on how big your drawing ends up being.
It can be mathematically proven, that circle can not be made with Bézier curve of any degree. You can make "almost circle" by approximating it.
Say you want to draw a quarter of circle around [0,0]. Cubic bézier coordinates are:
[0 , 1 ]
[0.55, 1 ]
[1 , 0.55]
[1 , 0 ]
It is a very good approximation. Transform it linearly to get an ellpise.
I am creating a game in which I draw a series of polygons by creating points around a radius of a cricle.
Later on I rotate the shapes and I need to calculate the new location (X,Y) of the points based on the rotation. I have the Old XY of each point, the XY of the center of the shape, radius of shape and the rotation.
Have a look at my diagram of the problem.
It should be possible to use matrix transformations for this, but you can also do it manually:
http://www.siggraph.org/education/materials/HyperGraph/modeling/mod_tran/2drota.htm
So essentially;
newX = initialX * Math.cos(angle) - initialY * Math.sin(angle);
newY = initialY * Math.cos(angle) + initialX * Math.sin(angle);
//Angle is in radians btw
This assumes that the initialX/Y is relative to the center of rotation, so you would have to subtract the center point before starting, and then add it again after the calculation to place it correctly.
Hope this helps!
For each point do:
alpha = arctan2(x, y)
len = sqrt(x^2 + y^2)
newX = len * cos(alpha + rotation)
newy = len * sin(alpha + rotation)
Original [x,y] and new [newX,newY] coordinates are both relative to the center of your rotation. If your original [x,y] is absolut, you have to calculate relative first:
x = xAbs - xCenter
y = yAbs - yCenter
Make sure your arctan2 function provides a result of PI/2 or -PI/2 if x=0. Primitive arctan functions do not allow x=0.
I'm programming a flash game, I made an array of points (x and y positions) that some movieclips must follow. Those movieclips have a certain speed (they make steps of 5 pixels for now). When I want to move them horizontally or vertically, everything's fine, I have to add or remove 5 pixels of those clips' x or y. But sometimes they have to move diagonally and now that's complicated.
What I'm doing:
var angle:Number = Math.atan2(nextPoint.y - this.y, nextPoint.x - this.x) * 180 / Math.PI;
var xstep:Number = Math.cos(angle) * this.speed;
var ystep:Number = Math.sqrt(Math.pow(this.speed, 2) - Math.pow(xstep, 2));
this.x += xstep;
this.y += ystep;
It's only a fraction of the code, but I think it's all you need.
Basically, this makes my movieclip do a little step (of this.speed (currently set to 5) pixels).
If the current point and the next point have the same y position, it works fine. When they don't, it doesn't work. The angle is right at first but it slowly decreases (while it should stay the same). I don't know if it's the angle that isn't computed the right way or if it's the x and y steps, but it's one of those, I'm sure.
Try this instead:
var angle:Number = Math.atan2(nextPoint.y - this.y, nextPoint.x - this.x);
var xstep:Number = Math.cos(angle) * this.speed;
var ystep:Number = Math.sin(angle) * this.speed;
Because cos operates on angles in radians, you don't need to convert to degrees. Computing the y component of an angle uses sin, so it should be similar to x. I'm not able to test this, but it's possible that ystep will be backwards and may need to be multiplied by -1.
I'm working on a game in HTML5 canvas.
I want is draw an S-shaped cubic bezier curve between two points, but I'm looking for a way to calculate the coordinates of the control points so that the curve itself is always the same length no matter how close those points are, until it reaches the point where the curve becomes a straight line.
This is solvable numerically. I assume you have a cubic bezier with 4 control points.
at each step you have the first (P0) and last (P3) points, and you want to calculate P1 and P2 such that the total length is constant.
Adding this constraint removes one degree of freedom so we have 1 left (started with 4, determined the end points (-2) and the constant length is another -1). So you need to decide about that.
The bezier curve is a polynomial defined between 0 and 1, you need to integrate on the square root of the sum of elements (2d?). for a cubic bezier, this means a sqrt of a 6 degree polynomial, which wolfram doesn't know how to solve. But if you have all your other control points known (or known up to a dependency on some other constraint) you can have a save table of precalculated values for that constraint.
Is it really necessary that the curve is a bezier curve? Fitting two circular arcs whose total length is constant is much easier. And you will always get an S-shape.
Fitting of two circular arcs:
Let D be the euclidean distance between the endpoints. Let C be the constant length that we want. I got the following expression for b (drawn in the image):
b = sqrt(D*sin(C/4)/4 - (D^2)/16)
I haven't checked if it is correct so if someone gets something different, leave a comment.
EDIT: You should consider the negative solution too that I obtain when solving the equation and check which one is correct.
b = -sqrt(D*sin(C/4)/4 - (D^2)/16)
Here's a working example in SVG that's close to correct:
http://phrogz.net/svg/constant-length-bezier.xhtml
I experimentally determined that when the endpoints are on top of one another the handles should be
desiredLength × cos(30°)
away from the handles; and (of course) when the end points are at their greatest distance the handles should be on top of one another. Plotting all ideal points looks sort of like an ellipse:
The blue line is the actual ideal equation, while the red line above is an ellipse approximating the ideal. Using the equation for the ellipse (as my example above does) allows the line to get about 9% too long in the middle.
Here's the relevant JavaScript code:
// M is the MoveTo command in SVG (the first point on the path)
// C is the CurveTo command in SVG:
// C.x is the end point of the path
// C.x1 is the first control point
// C.x2 is the second control point
function makeFixedLengthSCurve(path,length){
var dx = C.x - M.x, dy = C.y - M.y;
var len = Math.sqrt(dx*dx+dy*dy);
var angle = Math.atan2(dy,dx);
if (len >= length){
C.x = M.x + 100 * Math.cos(angle);
C.y = M.y + 100 * Math.sin(angle);
C.x1 = M.x; C.y1 = M.y;
C.x2 = C.x; C.y2 = C.y;
}else{
// Ellipse of major axis length and minor axis length*cos(30°)
var a = length, b = length*Math.cos(30*Math.PI/180);
var handleDistance = Math.sqrt( b*b * ( 1 - len*len / (a*a) ) );
C.x1 = M.x + handleDistance * Math.sin(angle);
C.y1 = M.y - handleDistance * Math.cos(angle);
C.x2 = C.x - handleDistance * Math.sin(angle);
C.y2 = C.y + handleDistance * Math.cos(angle);
}
}
I am trying to write a formula in ActionScript 3 that will give me var "z" (please see image below) in degrees, which I will then convert to radians.
I will already know the value of vars "x" and "y". Using trigonometry, how can I calculate the length of the hypotenuse and therefore the variable angle of var z? A solution in either AS3 or psuedocode would be very helpful. Thanks.
What you need is this:
var h:Number = Math.sqrt(x*x + y*y);
var z:Number = Math.atan2(y, x);
That should give you the angle in radians, you might need to swap x/y and possibly add or remove 90 degrees but it should do the trick! (Note that you don't even need h to get z when you're using atan2)
I use multiplication instead of Math.pow() just because Math is pretty slow, you can do:
var h:Number = Math.sqrt(Math.pow(x, 2) + Math.pow(y, 2));
And it should be exactly the same.
z is equivalent to 180 - angle of yH. Or:
180 - arctan(x/y) //Degrees
pi - arctan(x/y) //radians
Also, if actionscript's math libraries have it, use arctan2, which takes both the x and y and deals with signs correctly.
The angle you want is the same as the angle opposed to the one wetween y and h.
Let's call a the angle between y and h, the angle you want is actually 180 - a or PI - a depending on your unit (degrees or radians).
Now geometry tells us that:
cos(a) = y/h
sin(a) = x/h
tan(a) = x/y
Using tan(), we get:
a = arctan(x/y)
As we are looking for 180 - a, you should compute:
180 - arctan(x/y)
What #Patrick said, also the hypotenuse is sqrt(x^2 + y^2).