4 hour of search for wave in html, svg or canvas but didn't find anything what would look like in a picture.
How can I create wave like this? with smooth ending and filled with color ?
This shape can be achieved using bezierCurveTo() in canvas. I'm sure the shape is possible in SVG as well, but I'm not as familiar with it, so below is a canvas demonstration.
The snippet is borrowed and adapted from the MDN article. Basically, you want a bezier curve that keeps the control points on the same y-axis as the start and end points. Make the curve more or less dramatic by moving the control points along the x-axis. The farther the first control point and second control point are to the start and end points (respectively) the more dramatic the curve will be.
I would start by building the graph in canvas using straight lines and adapting them to be bezier curves after the fact.
var canvas = document.getElementById('canvas');
var ctx = canvas.getContext('2d');
var hw = 10, ho = hw / 2;
var x1 = 50, y1 = 20,
x2 = 230, y2 = 100,
cp1x = 120, cp1y = 20,
cp2x = 160, cp2y = 100;
ctx.beginPath();
ctx.moveTo(x1,y1);
ctx.bezierCurveTo(cp1x, cp1y, cp2x, cp2y, x2, y2);
ctx.stroke();
ctx.fillStyle = 'blue';
// start point
ctx.fillRect(x1 - ho, y1 - ho, hw, hw);
// end point
ctx.fillRect(x2 - ho, y2 - ho, hw, hw);
ctx.fillStyle = 'red';
// control point one
ctx.fillRect(cp1x - ho, cp1y - ho, hw, hw);
// control point two
ctx.fillRect(cp2x - ho, cp2y - ho, hw, hw);
<canvas id="canvas"></canvas>
Related
I've seen this and this discussion about removing antialiasing in canvases, but I don't think this is the same thing.
After scaling an html5 canvas by an arbitrary value (i.e., making it responsive), I've noticed that if I draw two rectangles of the same size and in the same location, the edges of the scaled side of the first rectangle remain visible.
I've included an example snippet where I draw a grey rectangle, then draw an red rectangle on top of it. There's a one-pixel red vertical line on the left and right edges of the grey rectangle. I know it may seem trivial, but it's very noticeable in my situation.
How do I fix this? Thanks!
var example = document.getElementById("example");
var ctx = example.getContext('2d');
ctx.scale(1.13,1);
ctx.fillStyle = "LightGrey";
ctx.fillRect(10,10,50,30);
ctx.fillStyle = "Black";
ctx.font = "20px Arial";
ctx.fillText("< Looks good.",70,30);
ctx.fillStyle = "Red";
ctx.fillRect(10,50,50,30);
// This light grey rectangle should completely cover the previous red one, but it doesn't!
ctx.fillStyle = "LightGrey";
ctx.fillRect(10,50,50,30);
ctx.fillStyle = "Black";
ctx.font = "20px Arial";
ctx.fillText("< Do you see red?",70,70);
<canvas id="example"></canvas>
You are scaling the transform matrix by a factor of 1.13 on the X axis.
So your coordinate 10, will actually end up on at coordinate 11.3 on the real pixels matrix.
You can't draw on fraction of pixels, so indeed antialiasing will kick in here.
So why does the first one looks better?
Because the mix between grey and white* is more neutral than the one between red grey and white. But even your first rect is antialiased.
Just zoom in your canvas and you'll see it, there is a one pixel band on both sides that is actually semi-transparent.
* "White" here is the one of the page's background
var example = document.createElement("canvas");
var ctx = example.getContext('2d');
ctx.scale(1.13,1);
ctx.fillStyle = "LightGrey";
ctx.fillRect(10,10,50,30);
ctx.fillStyle = "Red";
ctx.fillRect(10,50,50,30);
ctx.fillStyle = "LightGrey";
ctx.fillRect(10,50,50,30);
// draw bigger with no antialiasing
var z_ctx = zoomed.getContext('2d');
zoomed.width = example.width * 10;
zoomed.height = example.height * 10;
z_ctx.imageSmoothingEnabled = false;
z_ctx.drawImage(example, 0,0, zoomed.width, zoomed.height);
<canvas id="zoomed"></canvas>
So how to avoid this?
Well simply avoid filling at non integer pixel coordinates. This means you have to be constantly aware of your context transformation matrix too, not only of the values you pass to the drawing functions.
(Ps: also remember that stroke is an even eviler beast since it start drawing from the middle of the line, so in this case, you even have to take into considerations the lineWidth, see this Q/A on the matter).
If I have this,
var canvas = document.getElementById("my-canvas");
var ctx = canvas.getContext("2d");
ctx.fillStyle = '#f00';
ctx.beginPath();
ctx.moveTo(25, 25);
ctx.lineTo(150, 25);
ctx.rotate(Math.PI*7/4); //315 degrees
ctx.lineTo(150,90);
ctx.stroke();
It does not draw the line (the ctx.lineTo(150,90);) at the angle I thought it would which is from the end of the first lineTo at 25,25 to a 150,90 at a 45 degree angle. If I use -Math.PI*7/4 I get what looks like a 45 degree angle, but it points the wrong way. Math.PI*5/4 goes the wrong way or rotation.
My question asks about the Unit Circle. I don't really need them all at once, just the ability to know how to draw them if I need them.
The canvas is rotating this way:
Consider the following example, without the y offset:
Given by the code:
var canvas = document.getElementById("my-canvas");
var ctx = canvas.getContext("2d");
ctx.fillStyle = '#f00';
ctx.beginPath();
ctx.moveTo(25, 0);
ctx.lineTo(250, 0);
ctx.rotate(45*Math.PI/180); //45 degrees
ctx.lineTo(150,0);
ctx.stroke();
ctx.beginPath();
ctx.moveTo(0,0);
ctx.strokeStyle = '#ff0000';
ctx.lineTo(350,0);
ctx.font = "30px Arial";
ctx.fillText("New x axis",100,50);
ctx.stroke();
The red line is the new x axis, and the black line touches it exactly at a distance of 150 from the origin.
Now overlaying the same red line with your coordinates (and rotating by 45 degrees, so that it doesn't go off screen), we get
The black line ends at a point which is indeed (150,90), however at the new coordinate system. JSFiddle here
If you want the angle between the two line segments to be exactly what you pass in rotate(...), then keep drawing "horizontal" lines, but translate the origin to the end of each line segment, so you rotate about this point, not the top-left corner, using ctx.translate(<current point>). For example: https://jsfiddle.net/Douma37/65z1p38z/ (this goes off canvas, but has minimal interferance with your code).
Hope that helps!
Im try to drew custom shape, but since I use moveTo ..I cant be filled it, so That my question is there's any way can may be determined points on screen to fill shape? or to do that I most use or drew another real shape in the same block as layer ...
Look at my example here to drew a simple shape:
to I can fill image with blue color Im drew a Fill rectangle, so That is is a true way?
Code for shape for before fill:
var canvas3 = document.getElementById('canvas3');
var c3 = canvas3.getContext('2d');
c3.fillStyle = 'green';
c3.beginPath();
c3.moveTo(10,30);
c3.lineTo(200,30);
c3.moveTo(10,80);
c3.lineTo(200,80);
c3.moveTo(10,30);
c3.lineTo(10,180);
c3.moveTo(200,30);
c3.lineTo(200,180);
//c3.closePath();
c3.fill();
c3.lineWidth = 5;
c3.strokeStyle = 'orange';
c3.stroke();
Code for shape after fill:
var canvas3 = document.getElementById('canvas3');
var c3 = canvas3.getContext('2d');
c3.fillStyle = 'blue';
c3.beginPath();
c3.moveTo(10,30);
c3.fillRect(10,30,190,60);
c3.moveTo(10,30);
c3.lineTo(10,180);
c3.moveTo(10,90);
c3.lineTo(200,90);
c3.moveTo(200,30);
c3.lineTo(200,180);
c3.moveTo(10,30);
c3.lineTo(200,30);
//c3.closePath();
c3.fill();
c3.lineWidth = 5;
c3.strokeStyle = 'orange';
c3.stroke();
and finally which is a best way to I can drew shapes like this?
Note: Im new on html5 canvas and I read from this online book.
is there any way can may be determined points on screen to fill
shape? or to do that I most use or drew another real shape in the same
block as layer
Just draw a shape in the same place. Fill first then stroke afterwards. A little planning may be required with canvas as to in which order to draw things.
You can define objects to hold the geometrical data if you plan to redraw often or move them around. This will certainly simplify the objective later on.
which is a best way to I can drew shapes like this?
In my opinion this code can be drawn much simpler and in fewer lines of codes. There is no need to break up a shape in several parts as in that code if you can draw a shape using a simple method for it. In this case four lines can be replaced with one rectangle.
Knowing how these shapes are drawn internally also helps so we can take advantage of the path a rect() leaves, i.e. closing in upper-left corner so we can continue from there.
For example:
var c3 = c.getContext("2d");
// fill blue box first as it will be at the bottom
c3.rect(10, 30, 190, 50); // x, y, width, height
c3.fillStyle = 'blue';
c3.fill();
// orange stroke
// we know rect() will close at upper-left corner so continue from there with left leg
c3.lineTo(10, 180);
c3.moveTo(200, 80); // create second leg at right separately
c3.lineTo(200, 180);
c3.strokeStyle = "orange";
c3.lineWidth = 5;
c3.lineJoin = c3.lineCap = "round";
c3.stroke();
<canvas id=c height=180></canvas>
An alternative approach would be to fill then build the line path:
var c3 = c.getContext("2d");
c3.fillStyle = 'blue';
c3.fillRect(10, 30, 190, 50); // fillRect does not add to path
// orange stroke
c3.moveTo(10, 180); // create "housing" starting at bottom-left corner
c3.lineTo(10, 30); // upper-left
c3.lineTo(200, 30); // upper-right
c3.lineTo(200, 180); // bottom-right
c3.moveTo(10, 80); // add cross-bar
c3.lineTo(200, 80);
c3.strokeStyle = "orange";
c3.lineWidth = 5;
c3.lineJoin = c3.lineCap = "round";
c3.stroke();
<canvas id=c height=180></canvas>
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.
As the title states. Is this possible?
Edit: When i say doughnut I mean a top, 2D view
Is the only option to draw a segment of a circle, then draw a segment of a smaller circle with the same origin and smaller radius over the top, with the colour of the background? That would be crap if so :(
You do it by making a single path with two arcs.
You draw one circle clockwise, then draw a second circle going counter-clockwise. I won't go into the detail of it, but the way paths are constructed knows to take this as a reason to un-fill that part of the path. For more detail of what its doing you can this wiki article.
The same would work if you were drawing a "framed" rectangle. You draw a box one way (clockwise), then draw the inner box the other way (counter-clockwise) to get the effect.
Here's the code for a doughnut:
var can = document.getElementById('canvas1');
var ctx = can.getContext('2d');
// Pay attention to my last argument!
//ctx.arc(x,y,radius,startAngle,endAngle, anticlockwise);
ctx.beginPath()
ctx.arc(100,100,100,0,Math.PI*2, false); // outer (filled)
ctx.arc(100,100,55,0,Math.PI*2, true); // inner (unfills it)
ctx.fill();
Example:
http://jsfiddle.net/Hnw6a/
Drawing only a "segment" of it can be done by making the path smaller (you might need to use beziers instead of arc), or by using a clipping region. It really depends on how exactly you want a "segment"
Here's one example: http://jsfiddle.net/Hnw6a/8/
// half doughnut
ctx.beginPath()
ctx.arc(100,100,100,0,Math.PI, false); // outer (filled)
ctx.arc(100,100,55,Math.PI,Math.PI*2, true); // outer (unfills it)
ctx.fill();
You can make a 'top view doughnut' (circle with hollow center) by stroking an arc. You can see an example of this here: http://phrogz.net/tmp/connections.html
The circles (with nib) are drawn by lines 239-245:
ctx.lineWidth = half*0.2; // set a nice fat line width
var r = half*0.65; // calculate the radius
ctx.arc(0,0,r,0,Math.PI*2,false); // create the circle part of the path
// ... some commands for the nib
ctx.stroke(); // actually draw the path
Yes, I understand how old this question is :)
Here are my two cents:
(function(){
var annulus = function(centerX, centerY,
innerRadius, outerRadius,
startAngle, endAngle,
anticlockwise) {
var th1 = startAngle*Math.PI/180;
var th2 = endAngle*Math.PI/180;
var startOfOuterArcX = outerRadius*Math.cos(th2) + centerX;
var startOfOuterArcY = outerRadius*Math.sin(th2) + centerY;
this.beginPath();
this.arc(centerX, centerY, innerRadius, th1, th2, anticlockwise);
this.lineTo(startOfOuterArcX, startOfOuterArcY);
this.arc(centerX, centerY, outerRadius, th2, th1, !anticlockwise);
this.closePath();
}
CanvasRenderingContext2D.prototype.annulus = annulus;
})();
Which will add a function "annulus()" similar to "arc()" in the CanvasRenderingContext2D prototype. Making the closed path comes in handy if you want to check for point inclusion.
With this function, you could do something like:
var canvas = document.getElementById("canvas1");
var ctx = canvas.getContext("2d");
ctx.annulus(0, 0, 100, 200, 15, 45);
ctx.fill();
Or check this out: https://jsfiddle.net/rj2r0k1z/10/
Thanks!
With WebGL (one of the contexts of the HTML5 canvas) that is possible. There are even some JS libraries for browsers that don't support/implement it yet - check out these links:
http://sixrevisions.com/web-development/how-to-create-an-html5-3d-engine/
http://slides.html5rocks.com/#landing-slide
http://sebleedelisle.com/2009/09/simple-3d-in-html5-canvas/
http://www.khronos.org/webgl/
http://webdesign.about.com/od/html5tutorials/f/is-there-a-3d-context-for-html5-canvas.htm
http://code.google.com/p/html-gl/
Given the requirements, what #SimonSarris says satisfies the problem. But lets say you're like me and you instead want to "clear" a part of a shape that may be partially outside the bounds of the shape you're clearing. If you have that requirement, his solution won't get you want you want. It'll look like the "xor" in the image below.
The solution is to use context.globalCompositeOperation = 'destination-out' The blue is the first shape and the red is the second shape. As you can see, destination-out removes the section from the first shape. Here's some example code:
explosionCanvasCtx.fillStyle = "red"
drawCircle(explosionCanvasCtx, projectile.radius, projectile.radius, projectile.radius)
explosionCanvasCtx.fill()
explosionCanvasCtx.globalCompositeOperation = 'destination-out' #see https://developer.mozilla.org/samples/canvas-tutorial/6_1_canvas_composite.html
drawCircle(explosionCanvasCtx, projectile.radius + 20, projectile.radius, projectile.radius)
explosionCanvasCtx.fill()
Here's the potential problem with this: The second fill() will clear everything underneath it, including the background. Sometimes you'll want to only clear the first shape but you still want to see the layers that are underneath it.
The solution to that is to draw this on a temporary canvas and then drawImage to draw the temporary canvas onto your main canvas. The code will look like this:
diameter = projectile.radius * 2
console.log "<canvas width='" + diameter + "' height='" + diameter + "'></canvas>"
explosionCanvas = $("<canvas width='" + diameter + "' height='" + diameter + "'></canvas>")
explosionCanvasCtx = explosionCanvas[0].getContext("2d")
explosionCanvasCtx.fillStyle = "red"
drawCircle(explosionCanvasCtx, projectile.radius, projectile.radius, projectile.radius)
explosionCanvasCtx.fill()
explosionCanvasCtx.globalCompositeOperation = 'destination-out' #see https://developer.mozilla.org/samples/canvas-tutorial/6_1_canvas_composite.html
durationPercent = (projectile.startDuration - projectile.duration) / projectile.startDuration
drawCircle(explosionCanvasCtx, projectile.radius + 20, projectile.radius, projectile.radius)
explosionCanvasCtx.fill()
explosionCanvasCtx.globalCompositeOperation = 'source-over' #see https://developer.mozilla.org/samples/canvas-tutorial/6_1_canvas_composite.html
ctx.drawImage(explosionCanvas[0], projectile.pos.x - projectile.radius, projectile.pos.y - projectile.radius) #center
Adapting/simplifying #Simon Sarris's answer to easily work with any angle gives the below:
To create an arc segment you draw an outer arc (of n radians) in one direction and then an opposite arc (of the same number of radians) at a smaller radius and fill in the resulting area.
var can = document.getElementById('canvas1');
var ctx = can.getContext('2d');
var angle = (Math.PI*2)/8;
var outer_arc_radius = 100;
var inner_arc_radius = 50.;
ctx.beginPath()
//ctx.arc(x,y,radius,startAngle,endAngle, anticlockwise);
ctx.arc(100,100,outer_arc_radius,0,angle, false); // outer (filled)
// the tip of the "pen is now at 0,100
ctx.arc(100,100,inner_arc_radius,angle,0, true); // outer (unfills it)
ctx.fill();
<canvas id="canvas1" width="200" height="200"></canvas>