I'm new to PIXI.js to make a game.
How can I draw a half of the circle like this without using image?
Thank you in advance!
Make sure you read the official documentation before you resort to stackoverflow. :)
Use the PIXI.Graphics class to draw shapes programmatically. It has an arc method for drawing arcs, curves, or parts of circles.
var semicircle = new PIXI.Graphics();
semicircle.beginFill(0xff0000);
semicircle.lineStyle(2, 0xffffff);
semicircle.arc(0, 0, 100, 0, Math.PI); // cx, cy, radius, startAngle, endAngle
semicircle.position = {x: sW/2, y: sH/2};
stage.addChild(semicircle);
https://jsfiddle.net/_jered/kydtg9jq/
Related
I'm running a sketch with an array of points in 3D space (P3D). I'd like to add an interface to it by drawing text as if it were "onscreen"/2D, only using "X, Y" parameters.
When I tried just adding "text("!##$%", width/2, height/2);" it rendered in 3D space.
Is it possible? I tried "textMode(SCREEN) but doesnt exist in processing 2 anymore.
Here is what I found, I guess on the Processing Forum
You can use:
a PMatrix3D for your 3D content
and code your 2D stuff the plain old way
I wish it helps
PMatrix3D baseMat;
float alpha =0;
void setup() {
size(400, 400, P3D);
// Remember the start model view matrix values
baseMat = getMatrix(baseMat);
}
void draw() {
background(40);
pushMatrix();
camera(0, 0, 400, 0, 0, 0, 0, 1, 0);
directionalLight(255, 255, 255, -100, 150, -100);
ambientLight(40, 40, 40);
// 3D drawing stuff here
rotateY(alpha);
box(100);
alpha += 0.05;
popMatrix();
// Restore the base matrix and lighting ready for 2D
this.setMatrix(baseMat);
ambientLight(255, 255, 255);
// draw 2D stuff here
rect(10, 10, 50, 10);
textSize(25);
text("voila", mouseX, mouseY);
}
A workaround that comes to mind is to create a 2D PGraphic that has the same width/height as your sketch, give it a transparent background, draw your text where you want on it, and then draw the PGraphic onto your real sketch like you would if you were copying over image source data.
How do I draw semi-circular arc along the y-axis ( perpendicular axis ) of an svg using D3 ?
Here is an image of what I want to achieve target
Here is my code so far :-
var canvas = d3.select("body").append("svg")
.attr("width",width)
.attr("height",height);
var group = canvas.append("g")
.attr("transform","translate(0,200)");
var origin = d3.svg.arc()
.outerRadius(50)
.innerRadius(0)
.startAngle(-1.5755)
.endAngle(1.5755);
var arcs1 = group.append('g')
.attr('class','arc')
.attr("transform","translate(50,0)");
var yAxis = group.append("line")
.attr("transform","translate(50,0)")
.attr("x1", 0)
.attr("y1", -200)
.attr("x2", 0)
.attr("y2", height)
.attr("stroke-width", 2)
.attr("stroke", "black");
First, to draw the arc like your target you want your startAngle to start at 0 and endAngle to be 3.14 or Math.PI.
var origin = d3.svg.arc()
.outerRadius(50)
.innerRadius(0)
.startAngle(0)
.endAngle(Math.PI);
Then just append a path to arcs1 and use your origin function to draw it.
arcs1.append("path").attr("d", origin);
You can take a look at a live demo at http://jsfiddle.net/7EfS5/.
Edit
To change the color of your arc, you can either use .attr("fill", color) or .style("fill", color)
arcs1.append("path").attr("d", origin)
.attr("fill", "red");
Hope this help.
When you make a d3 arc, an angle of 0 represents the top of a circle. So to make an arc like the one shown in your image, you will want your start angle to be 0 and your end angle to be halfway around a circle, which is Pi radians:
var arc = d3.svg.arc()
.outerRadius(50)
.innerRadius(0)
.startAngle(0)
.endAngle(Math.PI);
As far as positioning your arc, you can use an svg transform to translate the origin of the arc to the position you want. For instance, if your y-axis has height h you can translate the origin to the center of the axis when you create the path element for your arc like this:
svg.append('path')
.attr('transform', 'translate(0,' + h/2 + ')')
.attr('d', arc);
HERE is a demo.
I've just started playing with SVG and Raphael.js and am trying to draw a filled arc but not sure how to start. I'm trying to draw a basic padlock and have the body and two parts of the bolt to fit the arc on top. It's basically a filled 180deg arc, 10 pixels wide. I'm guessing I need to use the .path() but not sure of the syntax or whether I want to be using "curveTo" or "arc" - struggling to find any good SVG or Raphael tutorial sites to be honest.
var padlockBody = paper.rect(100, 100, 100, 100, 5);
padlockBody.attr("fill", "#000000");
var leftBoltPart = paper.rect(120, 70, 10, 30);
leftBoltPart.attr("fill", "#000000");
var rightBoltPart = paper.rect(170, 70, 10, 30);
rightBoltPart.attr("fill", "#000000");
// TODO: filled arc to fit on top of left/right bolt parts
Arcs are notoriously challenging to write by hand (in pure SVG or in Raphael, it's the same).
I usually extend Raphael with these methods (credits belong to the55)
// http://www.w3.org/TR/SVG/paths.html#PathDataEllipticalArcCommands
Raphael.fn.arc = function(startX, startY, endX, endY, radius1, radius2, angle) {
var arcSVG = [radius1, radius2, angle, 0, 1, endX, endY].join(' ');
return this.path('M'+startX+' '+startY + " a " + arcSVG);
};
Raphael.fn.circularArc = function(centerX, centerY, radius, startAngle, endAngle) {
var startX = centerX+radius*Math.cos(startAngle*Math.PI/180);
var startY = centerY+radius*Math.sin(startAngle*Math.PI/180);
var endX = centerX+radius*Math.cos(endAngle*Math.PI/180);
var endY = centerY+radius*Math.sin(endAngle*Math.PI/180);
return this.arc(startX, startY, endX-startX, endY-startY, radius, radius, 0);
};
You can find a demo here: http://jsfiddle.net/cahT9/
I'm working on my first project using openTk. I'm creating virtual arcball for 3D model rotation. It works fine, but I need to add circle which won't rotate with model. This circle should visualize arcball.
My code to achieve rotation is:
private void SetCamera()
{
GL.MatrixMode(MatrixMode.Modelview);
Matrix4 scale = Matrix4.Scale(magnification / diameter);
Matrix4 translation1 = Matrix4.CreateTranslation(-center);
Matrix4 rotation = Matrix4.CreateFromAxisAngle(axisOfRotation, angleOfRotation*(float)numericSensitivity.Value);
Matrix4 translation2 = Matrix4.CreateTranslation(0.0f, 0.0f, -1.5f);
if (rotationChanged)
{
oldRotation *= rotation;
rotationChanged = false;
}
modelview = translation1 * scale * oldRotation * translation2;
GL.LoadMatrix(ref modelview);
}
So I would like to ask if there is some way how to draw circle, which wil be unaffected by this rotattion (will be on same position on a screen).
If I understand your question correctly, then all you need to do is set the modelview matrix back to the identity before you draw your circle. You can easily do that using the PushMatrix() and PopMatrix() functions. Something like this:
//Draw normal things
GL.MatrixMode(MatrixMode.Modelview);
GL.PushMatrix();
GL.LoadIdentity();
//Draw un-rotated circle
GL.PopMatrix();
PushMatrix() saves the current matrix onto a stack, and PopMatrix() pops the top matrix off of that stack. This means PopMatrix() will take you back to your normal rotated frame of reference after you're done with the circle.
I need a radial gradient in the shape of an oval or ellipse and it seems like it CGContextDrawRadialGradient can only draw a perfect circle. I've been drawing to a square context then copying/drawing into a rectangular context.
Any better way to do this?
Thanks!
The only way I've found to do this is as Mark F suggested, but I think the answer needs an example to be easier to understand.
Draw an elliptical gradient in a view in iOS (and using ARC):
- (void)drawRect:(CGRect)rect {
CGContextRef ctx = UIGraphicsGetCurrentContext();
// Create gradient
CGColorSpaceRef colorSpace = CGColorSpaceCreateDeviceRGB();
CGFloat locations[] = {0.0, 1.0};
UIColor *centerColor = [UIColor orangeColor];
UIColor *edgeColor = [UIColor purpleColor];
NSArray *colors = [NSArray arrayWithObjects:(__bridge id)centerColor.CGColor, (__bridge id)edgeColor.CGColor, nil];
CGGradientRef gradient = CGGradientCreateWithColors(colorSpace, (__bridge CFArrayRef)colors, locations);
// Scaling transformation and keeping track of the inverse
CGAffineTransform scaleT = CGAffineTransformMakeScale(2, 1.0);
CGAffineTransform invScaleT = CGAffineTransformInvert(scaleT);
// Extract the Sx and Sy elements from the inverse matrix
// (See the Quartz documentation for the math behind the matrices)
CGPoint invS = CGPointMake(invScaleT.a, invScaleT.d);
// Transform center and radius of gradient with the inverse
CGPoint center = CGPointMake((self.bounds.size.width / 2) * invS.x, (self.bounds.size.height / 2) * invS.y);
CGFloat radius = (self.bounds.size.width / 2) * invS.x;
// Draw the gradient with the scale transform on the context
CGContextScaleCTM(ctx, scaleT.a, scaleT.d);
CGContextDrawRadialGradient(ctx, gradient, center, 0, center, radius, kCGGradientDrawsBeforeStartLocation);
// Reset the context
CGContextScaleCTM(ctx, invS.x, invS.y);
// Continue to draw whatever else ...
// Clean up the memory used by Quartz
CGGradientRelease(gradient);
CGColorSpaceRelease(colorSpace);
}
Put in a view with a black background you get:
You can change the transform of the context to draw an ellipse (for example, apply CGContextScaleCTM(context, 2.0, 1.0) just before calling CGContextDrawRadialGradient () to draw an elliptical gradient that's twice as wide as it is high). Just remember to apply the inverse transform to your start and end points, though.