I'm making a canvas game where you travel in a spaceship over an endless repreating background. Right now I'm drawing four instances of the background at different positions based off of the player's x/y position, so they will move with the player.
ctx.translate(ax,ay);
ctx.drawImage(Ibg,Math.round(x/1080)*1080,Math.round(y/720)*720,1080,720);
ctx.drawImage(Ibg,(Math.round(x/1080)*1080)-1080,Math.round(y/720)*720,1080,720);
ctx.drawImage(Ibg,Math.round(x/1080)*1080,(Math.round(y/720)*720)-720,1080,720);
ctx.drawImage(Ibg,(Math.round(x/1080)*1080)-1080,(Math.round(y/720)*720)-720,1080,720);
Translating to ax and ay basically allows objects to scroll with the cameras the player moves, since ax and ay are relative to the player's position. I can create a parallax effect by doing this instead:
ctx.translate(ax*.5,ay*.5);
This makes the background scroll slower than other game objects, like I'd like it to. But I still haven't figured out how to adjust the rest of the code to compensate. As the player moves farther from (0,0) he sees less and less of the background, because it seems to go beyond him at a faster rate. How can I fix this?
As an option to markE's answer you don't need to use a second canvas at all (which is a good option to this).
You can simply use CSS for background image and adjust background position with the amount you need.
Demo here
The essential part is simply these lines:
Background X position where -1 can be replaced with the value you want to move it at.
bgx -= 1;
Then for each loop the background position is updated (Y position is fixed in this example):
canvas.style.backgroundPosition = bgx + 'px -30px'; // set X and Y position
When bgx somehow equals the max width of the image you just reset it to the beginning.
Use 2 canvases -- one placed directly on top of the other
A "background" canvas is on the bottom and animates more slowly.
A "game objects" canvas is on the top and animates more quickly.
That way you can create a parallax effect using different animation speeds for each canvas.
Related
Alright, so I have a good deal of experience with HTML and CSS, and some experience with Javascript (I can write basic functions and have coded in similar languages).
I'm looking to start some visual projects and am specifically interested in getting into particle systems. I have an idea for something similar to Codecademy's name generator here (https://www.codecademy.com/courses/animate-your-name/0/1) where particles are mapped to a word and move if hovered over. It seems as though alphabet.js is what's really behind Codecademy's demo however I can't understand exactly how they mapped the particles to a word, etc.
I've done some basic tutorials just creating rudimentary particles in a canvas but I'm not sure a canvas is the best way to go - demos that utilize one of the many libraries available (such as http://soulwire.github.io/sketch.js/examples/particles.html) don't use a canvas.
So my question is - what is the best way for a beginner/intermediate in Javascript to start with particle systems? Specifically to accomplish the Codecademy name effect or similar? Should I try to use canvas or which library would be best to start with and how would you recommend starting?
The code for this project is achievable for your intermediate JS programmer status.
How the CodeAcademy project works ...
Start by building each letter out of circles and saving each circle's centerpoint in an array. The alphabet.js script holds that array of circle centerpoints.
On mousemove events, test which circles are within a specified radius of the mouse position. Then animate each of those discovered circles radially outward from the mouse position using simple trigonometry.
When the mouse moves again, test which circles are no longer within the specified radius of the current mouse position. Then animate each of those "outside" circles back towards their original positions.
You can also use native html5 canvas without any libraries...
Another approach allowing any text to be "dissolved" and reassembled
Start by drawing the text on the canvas. BTW, this approach will "dissolve" any drawing, not just text.
Use context.getImageData to fetch the opacity value of every pixel on the canvas. Determine which pixels on the canvas contain parts of the text. You can tell if a pixel is part of the text because it will be opaque rather than transparent.
Now do the same procedure that CodeAcademy did with their circles -- but use your pixels:
On mousemove events, test which pixels are within a specified radius of the mouse position. Then animate each of those discovered pixels radially outward from the mouse position using simple trigonometry.
When the mouse moves again, test which pixels are no longer within the specified radius of the current mouse position. Then animate each of those "outside" pixels back towards their original positions.
[Addition: mousemove event to test if circles are within mouse distance]
Note: You probably want to keep an animation frame running that moves circles closer or further from their original positions based on a flag (isInside) for each circle.
function handleMouseMove(e){
// tell the browser we're handling this event
e.preventDefault();
e.stopPropagation();
// calc the current mouse position
mouseX=parseInt(e.clientX-offsetX);
mouseY=parseInt(e.clientY-offsetY);
// test each circle to see if it's inside or outside
// radius of 40px to current mouse position
// circles[] is an array of circle objects shaped like this
// {x:,y:,r:,originalX:,originalY:,isInside:}
var radius=40;
for(var i=0;i<circles.length;i++){
var c=circles[i];
var dx=c.x-mouseX;
var dy=c.y-mouseY;
if(dx*dx+dy*dy<radius*radius){
c.isInside=true;
// move c.x & c.y away from its originalX & originalY
}else{
c.isInside=false;
// if the circle is not already back at it's originalX, originalY
// then move c.x & c.y back towards its originalX, originalY
}
}
}
How does one get and set the following values using Javascript in Adobe Photoshop CS6+:
Canvas Rotation
Canvas Zoom (https://forums.adobe.com/thread/1016213)
Horizontal Window Offset (Panning)
Vertical Window Offset (Panning)
I see that there is an app.activeDocument.rotateCanvas() function, but beyond that...
I'm not entirely sure what you mean by offset.
As for rotation. It's quite simple. The amount of rotation desired is measured in degrees & is placed between the brackets. So if you wanted to rotate counter clockwise that would be -90.
e.g.
app.activeDocument.rotateCanvas(-90.0);
I am working on a page where I can view images. I want to create a rotation tool. I've done that, but, it's not working consistently. When I set up the centre point to rotate by, the image jumps slightly, and it gets worse each time. I was experimenting, and, I have code to add a wedge to the top left corner of my top level group ( so, at 0,0 ). If I rotate the image by 45 degrees and drag it so that half of it is off the left edge of my canvas, then I call getAbsolutePosition on the wedge and on the group, I get these values:
layer.getAbsolutePosition()
Object {x: 104.66479545850302, y: 279.2748571151325}
wedge.getAbsolutePosition()
Object {x: 180.2684127179338, y: -73.48773356791764}
I think this means my y position is actually the bottom of the image, which is off screen.
What I want to do, is calculate the absolute position of the middle of my image, when the mouse moves over it, regardless of it's rotation. I have some code that works out points with rotation, which seems like it works at first, almost, but it just gets more and more broken the more I use the tool. I feel like there's something about how Kinetic is tracking these things and what it's reporting, that I am missing. Any hints would be most appreciated. Tutorials I can read are even better ( yes, I've read everything linked from the KineticJS site and searched the web ).
In a nutshell, the question is, if I have an image inside a group, and it's rotated, how do I work out the centre point of the image, taking the rotation in to account, and how do I set the offset so it will rotate from that point, and stay in the same place ?
Thanks
As you've discovered about KinetiJS:
rotation is easy
dragging is easy
dragging+rotation is difficult
After you drag your image you must reset its rotation point (offsetX/offsetY).
KineticJS makes dragging+rotation more difficult than it has to be.
Resetting the offset points of your image will cause KineticJS to automatically move your image (Noooo!!).
That's what's causing your jumping.
The solution to the "jumping" problem:
When you reset the image's rotation point (offsetX/OffsetY) you must also reset the image's X/Y position.
This code resets both XY and Offsets for an image after dragging:
A Demo: http://jsfiddle.net/m1erickson/m9Nw7/
// calc new position and offset
var pos=rect.getPosition();
var size=rect.getSize();
var offset=rect.getOffset();
var newX=pos.x-offset.x+size.width/2;
var newY=pos.y-offset.y+size.height/2;
// reset both position and offset
rect.setPosition([newX,newY]);
rect.setOffset(size.width/2,size.height/2);
I am currently experimenting with parallax effect that i am planning to implement to my HTML5-canvas game engine.
The effect itself is fairly easy to achieve, but when you add zooming and rotating, things get a little more complicated, at least for me. My goal is to achieve something like this:Youtube video.
As you can see, you can zoom in and out "to the center", and also rotate around it and get the parallax effect.
In my engine i want to have multiple canvases that are going to be my parallax layers, and i am going to translate them.
I came up with something like this:
var parallax = {
target: {
x: Mouse.x,
y: Mouse.y
},
offset: {
x: -ctx.width / 2,
y: -ctx.height / 2
},
factor: {
x: 1,
y: 1
}
}
var angle = 0;
var zoomX = 1;
var zoomY = 1;
var loop = function(){
ctx.canvas.width = ctx.canvas.width; //Clear the canvas.
ctx.translate(parallax.target.x * parallax.factor.x, parallax.target.y * parallax.factor.y);
ctx.rotate(angle);
ctx.scale(zoomX, zoomY);
ctx.translate((-parallax.target.x - parallax.offset.x) * parallax.factor.x, (-parallax.target.y - parallax.offset.y) * parallax.factor.y);
Draw(); //Function that draws all the objects on the screen.
}
This is a very small and simplified part of my script, but i hope that's enough to get what i am doing. The object "parallax" contains the target position, the offset(the distance from the target), and the factor that is determining how fast the canvas is moving away relatively to the target. ctx is the canvas that is moving in the opposite direction of the target.(In this example i am using only one layer.) I am using the mouse as the "target", but i could also use the player, or some other object with x and y property. The target is also the point around which i rotate and scale the canvas.
This method works completely fine as long as the factor is equal to 1. If it is something else, the whole thing suddenly stops working correctly, and when i try to zoom, it zooms to the top-left corner, not the target. I also noticed that if i zoom out too much, the canvas is not moving in the opposite way of the target, but in the same direction.
So my question is: What is the correct way of implementing parallax with zooming and rotating?
P.S. It is important to me that i am using canvases as the layers.
To prepare for the next animation frame, you must undo any previous transforms in the reverse order they were executed:
context.translate(x,y);
context.scale(sx,sy);
context.rotate(r);
// draw stuff
context.rotate(-r);
context.scale(-sx,-sy);
context.translate(-x,-y);
Alternatively, you can use context.save / context.restore to undo the previous transforms.
Adjust your parallax values for the current frame,
Save the un-transformed context state using context.save(),
Do your transforms (translate, scale, rotate, etc),
Draw you objects as if they were in non-transformed space with [0,0] at your translate point,
Restore your context to it's untransformed state using context.restore()/
Either way will correctly give you a default-oriented canvas to use for your next animation frame.
The exact parallax effects you apply are up to your own design, but using these methods will make the canvas return to a normal default state for you to design with.
I'm using canvas for a project and I have a number of elements that I'm skewing. I'm only skewing on the y value and just want to know what the new width of the image is after skewing (so I can align it with another canvas element). Check out the code below to see what I mean
ctx.save();
//skew the context
ctx.transform(1,0,1.3,0,0,0);
//draw two images with different heights/widths
ctx.drawImage(image,0,0,42,60);
ctx.drawImage(image,0,0,32,25);
The goal would be to know that the 42 by 60 image was now a X by 60 image so I could do some translating before drawing it at 0,0. It's easy enough to measure each image individually, but I have different skew values and heights/widths throughout the project that need to be align. Currently I use this code (works decently for images between 25 and 42 widths):
var skewModifier = imageWidth*(8/6)+(19/3);
var skewAmount = 1.3; //this is dynamic in my app
var width = (skewModifier*skewAmount)+imageWidth;
As images get wider though this formula quickly falls apart (I think it's a sloping formula not a straight value like this one). Any ideas on what canvas does for skews?
You should be able to derive it mathematically. I believe:
Math.atan(skewAmount) is the angle, in radians, that something is skewed with respect to the origin.
So 1.3 would skew the object by 0.915 radians or 52 degrees.
So here's a red unskewed object next to the same object skewed (painted green). So you have a right triangle:
We know the origin angle (0.915 rads) and we know the adjacent side length, which is 60 and 25 for your two images. (red's height).
The hypotenuse is the long side thats being skewed.
And the opposite side is the triangle bottom - how much its been skewed!
Tangent gets us opposite / adjacent if I recall, so for the first one:
tan(0.915) = opposite / 60, solving for the opposite in JavaScript code we have:
opposite = Math.tan(0.915)*60
So the bottom side of the skewed object starts about 77 pixels away from the origin. Lets check our work in the canvas:
http://jsfiddle.net/LBzUt/
Looks good to me!
The triangle in question of course is the canvas origin, that black dot I painted, and the bottom-left of the red rectangle, which is the original position that we're searching for before skewing.
That was a bit of a haphazard explanation. Any questions?
Taking Simon's fiddle example one step further, so you can simply enter the degrees:
Here's the fiddle
http://jsfiddle.net/LBzUt/33/