I am using a linear gradient with d3 to make a bunch of lines go from transparent to black.
The issue is that they're all rotated at random angles and I've so far used this:
.style('stroke', function rotateGrad (d)
{
linearGradient.attr('gradientTransform', `rotate(${d.angle.toFixed()})`)
return 'url(#linear-gradient)'
})
.attr('rerotate', function rotateGrad (d)
{
console.log(d.angle.toFixed());
linearGradient.attr('gradientTransform', `rotate(${-(d.angle.toFixed())})`)
})
Which in itself is probably a horrible way to go about.
It works for most, but I think the gradient which gets reused and continuously rotated back and forth goes a bit off and doesn't start at angle 0 again. So for some lines that are rendered later I can tell that the rotation is off.
Is there a way to reset the rotation back to 0?
I mainly followed Mike's example here.
Many thanks in advance!
Related
Heyho. I just came up with a tricky question.
I did something like this: http://jsfiddle.net/LspdF/1/
As you cann see there are more than 4 sites. Thats why I did the following:
Whenevery the pic-numer is greater than the last pic-number I turn the cube left, otherwise rigt. Thats why it can happen, that the css is somewhat like this:
transform: rotateY(450deg)
Since a circle only has 360 deg it turns and turns but always in the right direction.
Now I wanted to add a nice effect. Something like this: http://jsfiddle.net/p8a2t/
For this effect I need the 14th value of the 3d matrix created by the browser (z-value of translation). Since this value is poorly not the same as translateZ(), I have to use the matrix3d()-attribute.
Thats why I calculate the rotateY myself. But as you know the sin and cos are periodic and wont work with my 450deg. They reset the cube to 90deg which makes the cube spin back very fast.
My Question: How to denie that? Is there any possibility to change the matrix for spinning more than 360deg?
PS: The effect is realized using transition. There may be code in the fiddles which isnt used since I had to create both examples at once.
PPS: Sometimes the calculation has fatal numeric errors (near zero but not rly zero). I tried to avoid that using toFixed but for some reason that wont work sometimes. Same with Math.round. Note that you can break the second example clicking many links while the animation is still not done. But thats no point here :)
Any help is appreciated!
First let me say that I'm not very good with math. I have a canvas with multiple text "boxes" that are rotated to 300°, which basically makes them parallelograms. They are very similar to this:
I'm trying to detect if the mouse is over one of them, but I don't know how to do that. Please help. Thank you!
The simplest method is to use the inverse transform on the mouse point and then do simple rectangle testing on the transformed point. As long as the affine transform you're using doesn't map everything to a line, it will have a well-defined inverse.
Each parallelogram can first of all be contained in a rectangular bounding box like the one illustrated above. If the mouse is not within that rectangle, then it is definitely not a hit. You have many easy tests for that already. The rest of the space can be decomposed into the parallelogram of interest in green, and the areas you don't want. So we just need to test if the mouse is in the red areas with the following tests:
Left: x < a - (a/h)*y
Right: x > (a+b) - (a/h)*y
If either of those conditions is true, then the mouse is outside the parallelogram.
Note, in this case I am assuming y is 0 at the top and increases as you move down, and x is zero at the left and increases as you move right.
For more information about the value of a, we can turn to trig.
If we know the angle theta and h, then
a = h tan(Ɵ)
If I have an object performing this CAKeyframe Animation path (it is just an oval shape in the upper region of an iphone..
UIBezierPath *trackPath = [UIBezierPath bezierPathWithOvalInRect:CGRectMake(20, 100, 280, 150)];
Now imagine that at anytime during this paths travels I want to create a smooth path from its current position at a point in time and some point at the bottom of the screen.
By the way I am assuming that once I have a path I can stop the current CAKeyframeAnimation and add this path to a new CAKeyframe animation, but maybe if this is incorrect you can give me pointers here as well please.
So I said to myself "Self.....it looks like you are going to have to answer this one yourself as nobody even wants to edit it or say that it is a duplicate......"
So I was hoping for some easier or ready made way to do this. From what I have found there is no easy way. It seems the steps are going to be;
get the current position of the animated layer
calculate yourself with all your own code a nice smooth curve
add this path to an animation and animate it.
As for the calculation of a Bezier Curve I am still looking to find some class or code where points can be plugged into it and the two control points are produced for the UIBezier class to produce a curve.
In my case I am going to only animate "out" of my above questioned shape a determined points and thus have ready made smooth curves which I have prepared.
Animating a smooth exit from an oval at any point is very complicated and In my case just not worth it. So I have not done that.
I'm doing some image processing, and I need to find some information on line growing algorithms - not sure if I'm using the right terminology here, so please call me out on this is needs be.
Imagine my input image is simply a circle on a black background. I'd basically like extract the coordinates, so that I may draw this circle elsewhere based on the coordinates.
Note: I am already using edge detection image filters, but I thought it best to explain with a simple example.
Basically what I'm looking to do is detect lines in an image, and store the result in a data type where by I have say a class called Line, and various different Point objects (containing X/Y coordinates).
class Line
{
Point points[];
}
class Point
{
int X, Y;
}
And this is how I'd like to use it...
Line line;
for each pixel in image
{
if pixel should be added to line
{
add pixel coordinates to line;
}
}
I have no idea how to approach this as you can probably establish, so pointers to any subject matter would be greatly appreciated.
I'm not sure if I'm interpreting you right, but the standard way is to use a Hough transform. It's a two step process:
From the given image, determine whether each pixel is an edge pixel (this process creates a new "binary" image). A standard way to do this is Canny edge-detection.
Using the binary image of edge pixels, apply the Hough transform. The basic idea is: for each edge pixel, compute all lines through it, and then take the lines that went through the most edge pixels.
Edit: apparently you're looking for the boundary. Here's how you do that.
Recall that the Canny edge detector actually gives you a gradient also (not just the magnitude). So if you pick an edge pixel and follow along (or against) that vector, you'll find the next edge pixel. Keep going until you don't hit an edge pixel anymore, and there's your boundary.
What you are talking about is not an easy problem! I have found that this website is very helpful in image processing: http://homepages.inf.ed.ac.uk/rbf/HIPR2/wksheets.htm
One thing to try is the Hough Transform, which detects shapes in an image. Mind you, it's not easy to figure out.
For edge detection, the best is Canny edge detection, also a non-trivial task to implement.
Assuming the following is true:
Your image contains a single shape on a background
You can determine which pixels are background and which pixels are the shape
You only want to grab the boundary of the outside of the shape (this excludes donut-like shapes where you want to trace the inside circle)
You can use a contour tracing algorithm such as the Moore-neighbour algorithm.
Steps:
Find an initial boundary pixel. To do this, start from the bottom-left corner of the image, travel all the way up and if you reach the top, start over at the bottom moving right one pixel and repeat, until you find a shape pixel. Make sure you keep track of the location of the pixel that you were at before you found the shape pixel.
Find the next boundary pixel. Travel clockwise around the last visited boundary pixel, starting from the background pixel you last visited before finding the current boundary pixel.
Repeat step 2 until you revisit first boundary pixel. Once you visit the first boundary pixel a second time, you've traced the entire boundary of the shape and can stop.
You could take a look at http://processing.org/ the project was created to teach the fundamentals of computer programming within a visual context. There is the language, based on java, and an IDE to make 'sketches' in. It is a very good package to quickly work with visual objects and has good examples of things like edge detection that would be useful to you.
Just to echo the answers above you want to do edge detection and Hough transform.
Note that a Hough transform for a circle is slightly tricky (you are solving for 3 parameters, x,y,radius) you might want to just use a library like openCV
See also: Why is my image rotation algorithm not working?
This question isn't language specific, and is a math problem. I will however use some C++ code to explain what I need as I'm not experienced with the mathematic equations needed to express the problem (but if you know about this, I’d be interested to learn).
Here's how the image is composed:
ImageMatrix image;
image[0][0][0] = 1;
image[0][1][0] = 2;
image[0][2][0] = 1;
image[1][0][0] = 0;
image[1][1][0] = 0;
image[1][2][0] = 0;
image[2][0][0] = -1;
image[2][1][0] = -2;
image[2][2][0] = -1;
Here's the prototype for the function I'm trying to create:
ImageMatrix rotateImage(ImageMatrix image, double angle);
I'd like to rotate only the first two indices (rows and columns) but not the channel.
The usual way to solve this is by doing it backwards. Instead of calculating where each pixel in the input image ends up in the output image, you calculate where each pixel in the output image is located in the input image (by rotationg the same amount in the other direction. This way you can be sure that all pixels in the output image will have a value.
output = new Image(input.size())
for each pixel in input:
{
p2 = rotate(pixel, -angle);
value = interpolate(input, p2)
output(pixel) = value
}
There are different ways to do interpolation. For the formula of rotation I think you should check https://en.wikipedia.org/wiki/Rotation_matrix#In_two_dimensions
But just to be nice, here it is (rotation of point (x,y) angle degrees/radians):
newX = cos(angle)*x - sin(angle)*y
newY = sin(angle)*x + cos(angle)*y
To rotate an image, you create 3 points:
A----B
|
|
C
and rotate that around A. To get the new rotated image you do this:
rotate ABC around A in 2D, so this is a single euler rotation
traverse in the rotated state from A to B. For every pixel you traverse also from left to right over the horizontal line in the original image. So if the image is an image of width 100, height 50, you'll traverse from A to B in 100 steps and from A to C in 50 steps, drawing 50 lines of 100 pixels in the area formed by ABC in their rotated state.
This might sound complicated but it's not. Please see this C# code I wrote some time ago:
rotoZoomer by me
When drawing, I alter the source pointers a bit to get a rubber-like effect, but if you disable that, you'll see the code rotates the image without problems. Of course, on some angles you'll get an image which looks slightly distorted. The sourcecode contains comments what's going on so you should be able to grab the math/logic behind it easily.
If you like Java better, I also have made a java version once, 14 or so years ago ;) ->
http://www.xs4all.nl/~perseus/zoom/zoom.java
Note there's another solution apart from rotation matrices, that doesn't loose image information through aliasing.
You can separate 2D image rotation into skews and scalings, which preserve the image quality.
Here's a simpler explanation
It seems like the example you've provided is some edge detection kernel. So if what you want to is detect edges of different angles you'd better choose some continuous function (which in your case might be a parametrized gaussian of x1 multiplied by x2) and then rotate it according to formulae provided by kigurai. As a result you would be able to produce a diskrete kernel more efficiently and without aliasing.