I'm animating the height of a drawRoundedRect instance, however, because it starts drawing from the upper left corner, it's animating from top to bottom, and I need it to start from the bottom.
Is it possible to flip my graphics instance (I tried by setting the scale to inverse, but this doesn't render anything, perhaps it only works on sprites), or to start drawing a rounded rectangle from the bottom?
EDIT:
Okay so I found out it's possible to animate my height going in to the other direction by just multiplying my interpolated value by -1:
graphics.drawRoundedRect(
x,
y,
barsWidth,
interpolatedHeight * -1,
10
);
However, now the radius isn't working anymore, it's just drawing square rectangles..
TIA!
I am not entirely sure that I did understand the question but if a got it right, all you need to do is to change the pivot of the rectangle you are animating. In your case, pivot.y should be equal to the rectangle's height rectangle.pivot.y = rectangle.height. After you change the pivot you also need to position it /the rectangle/ accordingly, so it could retain its visual position rectangle.y += rectangle.pivot.y
a simple demo https://jsfiddle.net/4L1od09n/23/
Related
browser.actions().dragAndDrop(elem, target).perform();
I can clearly understand the above code but I cannot get how to specify this element and target.
Take this example
browser.actions().dragAndDrop(slider,{x:100, y:0}).perform();
In the website in which I'm working on, I cannot find any x, y or anything I can match with that and develop.
So it will be helpful if someone explains with some example for x and y so that I can relate to it and make I work.
The dragAndDrop() has two ways to work.
One starts with the element to drag. Here elem works as normal ElementFinder, so something like dragAndDrop(element(by.css('div.my-class')), target).perform();.
Now the target works in two ways: Either as another ElementFinder like in elem or as coordinates to move, starting from the position of elem, moving x pixels horizontally and y pixels vertically (plus to the right or top, minus to the left or bottom). So {x:100, y:0} will move your slider 100 pixels to the right from the starting position.
dragAndDrop(element(by.css('div.my-class')), {x:100, y:0}).perform(); will therefore move the element(by.css('div.my-class')) 100 pixels to the right.
I've been working on this problem for days. I am trying to implement a "free transform" tool for svgs. Similar to that of Raphael.FreeTransform or how you would move/rotate/scale images in MS Word. (Yes, I am aware there are libraries) The following jSFiddle displays my problem: https://jsfiddle.net/hLjvrep7/12/
There are 5 functions in the jsFiddle: rotate-t. shrink-t, grow-t, shrink, grow. The functions suffixed with '-t' also apply the current rotation transformation. e.g.:
grow-t
rect.attr({height : height * 1.25, width : width * 1.25}).transform('r' + degree);
grow
rect.attr({height : height * 1.25, width : width * 1.25});
Once an svg is rotated, then scaled. If you try to rotate the svg again (after scale), the svg jumps. To see this, go top the fiddle:
Hit rotate-t twice. Svg should rotate a total of 30 degrees from the rectangles origin.
Hit grow (not grow-t) twice. Note the top left position of the svg stays the same.
Hit rotate-t once. Note the svg jumps to a different position, then rotates.
Note hitting rotate-t subsequent times will continue to rotate the image around the origin (which is what I want the first time rotate-t is clicked)
One solution I had was to apply the current rotation transformation whenever changing the height and width. This fixes my previous problem, but introduces another problem. To see an example of this, go to the fiddle, and:
Hit rotate-t twice.
Hit grow-t a couple times. Notice the svg grows, but the top left position of the rectangle moves. That's a problem for me. I want the svg to grow without the top left corner to move.
Notes on using the jsFiddle:
Any combination of rotate-t, grow-t, shrink-t will exhibit the ideal rotation behavior (about the origin, no jumping). But this also demonstrates the undesired growing and shrinking (top left position moved when svg is on angle).
Any combination pf rotate-t, grow, shrink will exhibit the ideal scaling behavior (top left corner of svg doesn't move). But this also demonstrates the undesired rotation property (will jump around after different rotations and scales).
Bottom line: I want to be able to the svg rotate around the origin. Then grow the image, while the top left position remains the same. Then rotate the svg again, around the origin without any jumping.
I am aware the how the transform function impacts the local coordinate system of the svg. I'm leaning towards using rotate-t, grow, shrink combo and simply apply some x-y offsets to remove the "jumping" effect. I would imagine there must be some sort of offset I could apply to avoid jumping or shifting during rotation or scaling, but its not clear to me how to calculate such offsets. Any help would be appreciated.
Please don't hesitate to ask anymore questions. Like I said, I've been digging into this for days. Clearly, I don't understand it all, but am somewhat intimate with what's happening and happy to explain anything in more detail.
My solutions for scale, rotate, move back and front etc:
$scope.back = function () {
if($scope.currentImage !==null) {
if($scope.currentImage.prev!=undefined) {
var bot = $scope.currentImage.prev;
$scope.currentImage.insertBefore(bot);
ft.apply();
}
}
};
//Function for moving front
$scope.front = function () {
if($scope.currentImage !==null) {
if($scope.currentImage.next!=undefined) {
var top = $scope.currentImage.next;
if($scope.currentImage.next.node.localName == "image")
$scope.currentImage.insertAfter(top);
ft.apply();
}
}
};
//ZOOM
$scope.zoomIn = function () {
if ($scope.currentImage!= null) {
var ft = paper.freeTransform($scope.currentImage);
if(ft.attrs.scale.y<4) {
$scope.currentImage.toFront();
ft.attrs.scale.y = ft.attrs.scale.y *(1.1);
ft.attrs.scale.x = ft.attrs.scale.x *(1.1);
ft.apply();
ft.updateHandles();
}
}
};
Here is the graphics:
http://snag.gy/aVFGA.jpg
the big rectangle is canvas element, the small rectangle is the image object in the canvas. I want to find what is the real distance from the left.
values are such from what I see in console:
regX: 564.256
regY: 41.4
scaleX: 0.4491319444444445
scaleY: 0.4491319444444445
x: 363.3333333333333
y: 409.77777777777777
So as I see x is not real. It somehow relates with regX and scaleX. But I am not finding how it relates. From the image I think the x should be about 100 - 150 px.
THe bigger the x - the more it is to the right.
But the bigger regX - the more it makes rectangle go to the left.
So if I would just take the difference 564.256 - 363.333 = ~200 - left corner of the rectangle should be in them middle of canvas because canvas is 400px widh. But it is not, so substraction does not help. So how do I get how many pixels are in real from the left?
You can do this by using the localToGlobal method (see here).
It depends to which object the given attributes belong.
If they belong to the shape and your rectangle inside the image / shape starts at (0,0):
var point = shape.localToGlobal(0, 0);
// this will calculate the global point of the shape's local point (0,0)
If they belong to the stage:
var point = stage.localToGlobal(yourRectObject.x, yourRectObject.y);
// point.x should contain the position on the canvas
You should use these methods in general because your method might work for the current situation but will probably break as soon as you scale the stage itself or put the shape in a scaled / positioned container.
I guess I found what by experimenting with values:
distanceFromLeft = x - scaleX * regX;
so getting 109.90793888888885 px
If someone has worked more with this library, they could confirm that its not accidental.
Manipulating the slider until the end, the circle that represents the star disappears or does a different motion. See: jsfiddle.net/NxNXJ/13 Unlike this: astro.unl.edu/naap/hr/animations/hrExplorer.html
Can you help me?? Thanks
When you supply a big luminosity, You're rendering a circle which is millions of pixels tall. The broswer might not render it because it's so big.
However, you are really only interested in a small slice of that big circle - namely, the bit that fits in your tiny window.
At some point, it doesn't make sense to increase the size of the circle, since you can't observe a change in the curvature of the circle - it just looks like a straight vertical line.
This apparent verticality occurs around when x^2 + y^2 = R^2, where R is the radius of the star, Y is half the height of your window, and x is R-1. Solve for R in terms of Y, and you get
function maximumNecessaryRadius(windowHeight){
y = windowHeight / 2;
maxRadius = (y*y - 1)/2;
return Math.round(maxRadius);
}
When resizing the star, check to make sure that its radius doesn't exceed the maximum necessary radius. Rendering it any larger than that is overkill.
Example Implementation
I'm trying to figure out how I can get the correct "active" tile under the mouse when I have "ramp" and +1 height tiles (see picture below).
When my world is flat, everything works no problem. Once I add a tile with a height of say +1, along with a ramp going back to +0, my screen -> map routine is still looking as if everything is "flat".
In the picture above, the green "ramp" is the real tile I want to render and calculate mouse -> map, however the blue tile you see "below" it is the area which gets calculated. So if you move your mouse into any of the dark green areas, it thinks you're on another tile.
Here is my map render (very simple)
canvas.width = canvas.width; // cheap clear in firefox 3.6, does not work in other browsers
for(i=0;i<map_y;i++){
for(j=0;j<map_x;j++){
var xpos = (i-j)*tile_h + current_x;
var ypos = (i+j)*tile_h/2+ current_y;
context.beginPath();
context.moveTo(xpos, ypos+(tile_h/2));
context.lineTo(xpos+(tile_w/2), ypos);
context.lineTo(xpos+(tile_w), ypos+(tile_h/2));
context.lineTo(xpos+(tile_w/2), ypos+(tile_h));
context.fill();
}
}
And here is my mouse -> map routine:
ymouse=( (2*(ev.pageY-canvas.offsetTop-current_y)-ev.pageX+canvas.offsetLeft+current_x)/2 );
xmouse=( ev.pageX+ymouse-current_x-(tile_w/2)-canvas.offsetLeft );
ymouse=Math.round(ymouse/tile_h);
xmouse=Math.round(xmouse/(tile_w/2));
current_tile=[xmouse,ymouse];
I have a feeling I'll have to start over and implement a world based map system rather than a simple screen -> map routine.
Thanks.
Your assumption is correct. In order to "pick" against world geometry, your routine needs to be aware of the world (and not just the base-level tile configuration). That is, without any concept of the height of the tiles near the one that is currently picked (by your current algorithm), there's no way to determine whether a neighboring tile (or one even further away, depending on the permitted height) should be intercepted by picking ray.
You've got the final possible point of your picking ray, already. What remains is to define the remainder of the ray, in world-space, and to check that ray for intersections with world geometry.
If, like the picture, your view angle is always 45 degrees and always from the same direction, your mouse -> map routine could use an algorithm something like:
calculate i,j of tile as you're doing currently (your final value of xmouse, ymouse)
look up height and angle of tile at i,j
given the height and angle, does this tile intersect the picking ray? If so, set lasti, lastj = i, j
increment/decrement i,j one step diagonally toward viewer
have we fallen off the edge of the map? If so, return lasti, lastj. Otherwise go back to 2.
Depending on the maximum height of a tile, you might have to check only 2 tiles, rather than going all the way to the edge of the map.
3 is the tricky part, and depends on your world geometry. Draw some triangles and you should be able to figure it out. Or you might try looking at the function intersect_quadrilateral_ray() here.