I want to flip image view on axis which is 45 degree left or right from Y axis. How can I obtain this new axis of rotation and how can I apply it to image view. Here is the link
of what I want to achieve.
https://www.youtube.com/watch?v=hDlp-aWq-uM&feature=youtu.be
Thanks!
Use Camera to achieve 3D rotation like View.setRotationY.
camera.save();
camera.rotateY(rotate);
camera.getMatrix(matrix);
matrix.preTranslate(-left, 0);
matrix.postTranslate(left, 0);
camera.restore();
canvas.save();
canvas.concat(matrix);
canvas.drawBitmap(...);
canvas.restore();
The result for my personal project can be viewed below.
You can achieve this with Matrix class
Matrix mtrx=new Matrix();
imageView.setScaleType(ScaleType.MATRIX);
mtrx.postRotate((float) angle, pivX, pivY);
imageView.setImageMatrix(mtrx);
you should override draw method or create your own image view class by extending ImageView, then to rotate the axis rotate the canvas in draw(), and for the flipping you can user objectanimator which is available for API 11+, and for pre-11 you can user nineoldandroids.com library located at below url
http://nineoldandroids.com/
Related
Ok, so I rotated an image by 25 degrees like this :
MyImage.layer.affineTransform = CGAffineTransformMakeRotation(25);
Now I want to rotate the image again by 25 more degrees (50 degrees in total)
So my question is, How do I rotate X more degrees an image ?
If you have any other way to rotate an image or better way please include your code.
There is no need to know the current angle! Instead of setting the affineTransform, which is only a shortcut, apply an actual 3D transform. Now you can call CATransform3DRotate, which rotates an existing transform — and thus is additive.
Say, I have an image on an HTML page.
I apply an affine transformation to the image using CSS3 matrix function.
It looks like:
img#myimage {
transform: matrix(a, b, c, d, tx, ty);
/* use -webkit-transform, -moz-transform etc. */
}
The origin of an HTML page is the top-left corner and the y-axis is inverted.
I'm trying to put the same image in an environment (cocos2d) where the origin is the bottom-left corner and the y-axis is upright.
To get the same result in the other environment, I need to transform the origin somehow and reflect that in the resulting CGAffineTransform.
It would be great if I can get some help with the matrix math that goes here. (I'm not so good with matrices.)
The following formula would work,
for converting the position from CSS3 to Cocos2d:
(screen Size - "y" position in CSS3 - height of object)
Explanation:
To make the origin for the Cocos environment same as for the CSS3 environment we would only have to add the screen size to the cocos2d's bodies y co-ordinate.
Eg. The screen size is (100,100) and the body is a point object if you place it at (0,0) in CSS3 it would be at the top left corner. If we add the screen size to the y co-ordinates for cocos2d the object would be placed at (0,100) which is the top-left corner for cocos2d as well
To make the co-ordinates same, since the Y axis is inverted, we have to subtract the "Y" co-ordinate given in CSS3 from the Screen Size for Cocos2d. Suppose we place the same point object in the previous example at (0,10) in CSS3 we would place it at (0, 100 - 10) in cocos2d which would be the same positions on the screen
Since our body would NOT always be a point object we have to take care of its anchor point as well. If suppose the body's height is 20 and we place it at (0,10) in CSS3 then it would be placed at the top-left position and would be coming down because the Y axis is inverted
Hence we would also have to subtract the body's total height from the screen size and "y" co-ordinate to place it at the same position which would be (0, 100 - 10 - 20) putting the body at the same place in cocos2d environment
I hope I am correct and clear :)
I am building an augmented reality application and I have the yaw, pitch, and roll for the camera. I want to start placing objects in the 3D environment. I want to make it so that when the user clicks, a 3D point pops up right where the camera is pointed (center of the 2D screen) and when the user moves, the point moves accordingly in 3D space. The camera does not change position, only orientation. Is there a proper way to recover the 3D location of this point? We can assume that all points are equidistant from the camera location.
I am able to accomplish this independently for two axes (OpenGL default orientation). This works for changes in the vertical axis:
x = -sin(pitch)
y = cos(pitch)
z = 0
This also works for changes in the horizontal axis:
x = 0
y = -sin(yaw)
z = cos(yaw)
I was thinking that I should just make combine into:
x = -sin(pitch)
y = sin(yaw) * cos(pitch)
z = cos(yaw)
and that seems to be close, but not exactly correct. Any suggestions would be greatly appreciated!
It sounds like you just want to convert from a rotation vector (pitch,yaw,roll) to a rotation matrix. The conversion can bee seen on the Wikipedia article on rotation matrices. The idea is that once you have constructed your matrix, to transform any point simply.
final_pos = rot_mat*initial_pose
where final and initial pose are 3x1 vectors and rot_mat is a 3x3 matrix.
Let's say I have point which has the coordinates (50,100) where (0,0) is in the upper left corner of a view.
How can I get the coordinates of the same point if I want the beginning of the coordinate system to be the center of the screen (ie width/2, height/2) ?
Note that I am implementing a custom View and I am drawing inside it and I just want to convert the coordinate inside that same view. I am basically implementing a graphic calculator and I need to have my coordinate system to start in the middle of the screen so the graphics could look better.
I notice you tagged it as iOS problem, so use the method Apple have built in UIView:
(CGPoint)convertPoint:(CGPoint)point fromView:(UIView *)view
Find the midpoint you will be using, so for a 100x100 screen, this would be (50,50). Then take the point you need to convert and subtract the midpoint X value from the point X value, and then subtract the point Y value from the midpoint Y value. Notice that you are not doing the same operation on both values.
So if the point is (30,25) the new point would be (-20,25) because 30 - 50 = -20 and 50 - 25 = 25.
I'm trying to draw a "conical"/"arcing" gradient (I don't know what would be the correct term for this) (Photoshop calls it an "angle" gradient —your friendly neighborhood stackoverflow editor) using Objective-C (IOS), pretty much exactly like the image shown in the following thread.
After days of googling and searching the internet to no avail, I've decided to ask for help here.
A little background on what I'm trying to do. My objective is to create a custom UIView, which is circular progress bar, a ring basicly, somewhat similar to the activity indicator as seen in the TweetBot iPhone app (displays when you drag to refresh, which can be seen in action here, around 17-18 seconds into the video, on top of the iphone screen). I want the progress indicator (the fill of the ring) to be a simple two color gradient, which can be set programmatically, and the view to be resizable.
Filling the ring shape with a gradient that "follows" the arc of the ring is where I'm stuck. The answers that I get from googling, reading Apple's Core Graphics documentation on gradients and searching on SO are either about radial gradients or linear/axial gradients, which is not what I'm trying to achieve.
The thread linked above suggests using pre-made images, but this isn't an option because the colors of the gradient should be settable, the view should be resizable and the fill of the progress bar isn't always 100% full obviously (which would be the state of the gradient as shown in the picture in the thread above).
The only solution that I've come up with is to draw the gradient "manually", so without using a CGGradientRef, clipping small slices of the gradient with single solid color fills within a circular path. I don't know exactly how well this will perform when the bar is being animated though, it shouldn't be that bad, but it might be a problem.
So my first question:
Is there an easier/different solution to draw a conical/arcing gradient in Objective-C (IOS) than the solution I've come up with?
Second question:
If I have to draw the gradient manually in my view using the solution I came up with, how can I determine or calculate (if this is even possible) the value (HEX or RGBA) of each color "slice" of the gradient that I'm trying to draw, as illustrated in the image below.
(Can't link image) gradient slice illustration
Looks to me like a job for a pixel shader. I remember seeing a Quartz Composer example that simulated a radar sweep, and that used a pixel shader to produce an effect like you're describing.
Edit:
Found it. This shader was written by Peter Graffignino:
kernel vec4 radarSweep(sampler image, __color color1,__color color2, float angle, vec4 rect)
{
vec4 val = sample(image, samplerCoord(image));
vec2 locCart = destCoord();
float theta, r, frac, angleDist;
locCart.x = (locCart.x - rect.z/2.0) / (rect.z/2.0);
locCart.y = (locCart.y - rect.w/2.0) / (rect.w/2.0);
// locCart is now normalized
theta = degrees(atan(locCart.y, locCart.x));
theta = (theta < 0.0) ? theta + 360.0 : theta;
r = length(locCart);
angleDist = theta - angle;
angleDist = (angleDist < 0.0) ? angleDist + 360.0 : angleDist;
frac = 1.0 - angleDist/360.0;
// sum up 3 decaying phosphors with different time constants
val = val*exp2(-frac/.005) + (val+.1)*exp2(-frac/.25)*color1 + val*exp2(-frac/.021)*color2;
val = r > 1.0 ? vec4(0.0, 0.0,0.0,0.0) : val; // constrain to circle
return val;
}
The thread linked above suggests using pre-made images, but this isn't an option because the colors of the gradient should be settable, the view should be resizable and the fill of the progress bar isn't always 100% full obviously (which would be the state of the gradient as shown in the picture in the thread above).
Not a problem!
Use the very black-to-white image from the other question (or a bigger version if you need one), in the following fashion:
Clip to whatever shape you want to draw the gradient in.
Fill with the color at the end of the gradient.
Use the black-to-white gradient image as a mask.
Fill with the color at the start of the gradient.
You can rotate the gradient by rotating the mask image.
This only supports the simplest case of a gradient with a color at each extreme end; it doesn't scale to three or more colors and doesn't support unusual gradient stop positioning.
FYI: here's also a good tutorial for creating a circular progress bar using Quartz drawing.
http://www.turnedondigital.com/blog/quartz-tutorial-how-to-draw-in-quartz/