I need to make a rectangular view [ ] appear as if it's top is rotated back, while the bottom is pinned in place: / \ . The resulting image is isometric with the bottom being wider than the top.
Which CGAffineTransform do I need to accomplish this goal?
As others have pointed out, you can't do this with a CGAffineTransform.
However, it's relatively easy to do with a CATransform3D, as I describe in this answer. You'll need to adjust the m34 component of the CATransform3D to give the transform some degree of perspective, rotate the view about the X axis, and potentially scale it so that the bottom edge remains at the same width as for your original unrotated view.
Alternatively, you might be able to adjust the anchorPoint of your view's underlying layer to be at the bottom, rather than the center. Rotations will then be applied from that edge, which should keep the bottom edge length constant and give you a receding perspective effect for the view. I believe a value of (0.5, 1.0) will set the anchorPoint to the lower edge.
Brad, I found this example (by you!) on how to do a perspective transformation:
http://www.sunsetlakesoftware.com/2008/10/22/3-d-rotation-without-trackball
For some reason it does not work in my code. My buttons have the 3d transform applied, but not the scaling effect.
Related
I have the following view where I'm using a pan gesture in the upwards or downwards direction to rotate it positively or negatively:
I'm wondering, is there a mathematical equation to precisely covert the amount panned to the amount it should be rotated so the timing is correct to keep the users finger on the view while it is rotating? For instance, if the pan translation comes back as 1, what would the proper amount be to rotate it?.
There are a few details you need to provide to give a meaningful answer:
Are you rotating the view about its centre (the default) or is there an anchor point?
Since the view is rotating, while the touch is moving strictly vertically in the superview, what's the expected behaviour as the view rotates further away from the vertical line defining the pan?
Is there a reason you're using a pan gesture instead of a rotation gesture, or even just direct touch tracking? It seems like it creates more problems than it solves.
I'm going to assume the view is rotating about its centre for the sake of simplicity, and I'll use a pan starting on the right side of the view as an example, with the rotation not exceeding ±90°. Here are two options:
Movement up and down translates linearly to the angle of rotation, i.e., a pan of a given distance rotates the view the same amount, no matter where the pan starts. In that case, you need to decide what the top and bottom limits of the pan are. They might be the bounds of the superview. Regardless, you want to convert the distance travelled in the Y direction to a value between -1 and 1, where -1 represents the bottom limit and 1 represents the top limit. Something like 2 * (dy / superview.bounds.size.height - 0.5). Multiply that by π/2 (M_PI_2 in math.h) to scale from the range [-1, 1] to the range [-π/2, π/2] and you've got the angle to add/subtract from the view's rotation at the beginning of the gesture.
The view tracks the touch so that its right edge is always "pointed at" the touch. In this case, pan isn't terribly useful because you only need the location of the touch in the superview, not the distance travelled. Calculate dx and dy as the difference in x and y coordinates from the view's centre to the touch location. Then calculate atan2(dy, dx) and you've got the absolute angle of rotation for the view.
I hope this puts you on the right track.
The answer is the angle would be panAmount.y / rectangleWidth.
Here is proof: https://math.stackexchange.com/questions/322694/angle-of-rotated-line-segment
I want to ask a question about the perspective that is achieved through CATransform3D.
I know that if you have a view that is 320x480 and then apply this:
CATransform3D perspective = CATransform3DIdentity;
CGFloat zDistance = 1000;
perspective.m34 = 1.0 / -zDistance;
view.layer.sublayerTransform = perspective;
you create a perspective that makes it look like the observer is looking straight at the center of the screen and therefore the same transformation looks different, depending on where the subview that is being transformed is located on the screen. For example, tilting a view looks like this when the view is in the middle of the screen:
And it looks like this if it's in the lower left corner:
Now, my problem is that making the perspective relative to the screen only works if the view I'm transforming is a subview of another view that is 320x480px big. But what if the view I want to transform is a subview of a view that is only 100x100px? Is there a way to make the perspective relative to the whole screen if the superview isn't the size of the screen?
Thanks in advance
According to apple
"The anchorPoint property is a CGPoint that specifies a location within the bounds of a layer that corresponds with the position coordinate. The anchor point specifies how the bounds are positioned relative to the position property, as well as serving as the point that transforms are applied around."
Your perspective should not be relative to the center of the screen or even to the center of your layer by default, is that where you have your anchor point? Aside from that though, what you seem to be asking is how to make your perspective appear to be relative to a different point. The trick is that your perspective is created by multiplying by your perspective matrix. Setting m34 to a small number does nothing magic, you are multiplying by your projection matrix.
YourProjection = {1,0,0 ,0,
0,1,0 ,0,
0,0,1 ,0,
0,0,-1.0/zdistance,1};
Remember that you can combine successive transforms by multiplying them together. Just transform your layer to wherever you want, then apply your Projection matrix, then transform it back, presto, perspective from a different origin.
float x = your coordinates relative to the screen center;
float y = same thing
TranslationMatrix = {1,0,0,0,
0,1,0,0,
0,0,1,0,
x,y,0,1};
ReverseTranslationMatrix = {1,0,0,0,
0,1,0,0,
0,0,1,0,
-x,-y,0,1};
//now just multiply them all together
Final = ReverseTranslation*YourProjection*Translation;
You will need to do the matrix math yourself, hopefully you already have a generic 4x4 column major matrix class that can do multiplication for you, if not i suggest you make one. Also, if you are interested, you might consider reading. This for an explanation of how the matrix you are currently using works, or This for a different take on projection matrices.
Horizontal gradient is working fine. Is there any way to get a vertical color gradient from horizontal gradient?
I have seen a related question regarding this where they did this by rotating the frames.
Is there any simpler way to achieve a vertical gradient?
The default startPoint and endPoint would have the gradient display your colors from top to bottom (which in my mind is a vertical gradient). If you want the gradient to display from left to right (again in my mind this is a horizontal gradient), use this code on your CAGradientLayer:
[gradientLayer setStartPoint:CGPointMake(0.0, 0.5)];
[gradientLayer setEndPoint:CGPointMake(1.0, 0.5)];
A 3D transform is unnecessary.
How about rotating it 90º?
edit judging from your comment, it seems that you're doing this via CAGradientLayer. CAGradientLayer is a subclass of CALayer, which has a transform property. This transform property takes a CATransform3D, which is a struct that represents some sort of linear transformation to be applied to the layer (such as scaling, translation, or rotation).
So really you just need to make a rotational CATransform3D and set it as the transform property of your CAGradientLayer.
You could probably also make this work by fiddling with the startPoint and endPoint (which would actually probably be simpler).
When changing the center point of an 10 degrees rotated view, the specified coordinates for the center point will not match the coordinate system of the superview.
As an example, when you want to go straight up by y - 50, and no horizontal movement (i.e. x = theRotatedView.center.x), the shift in y-direction will be rotated. The view will go up and right, and not just up.
The rotation is done on the view, not on the layer.
I could wrap it in a superview that does the movement, while the subview does the rotation. Then I had no problems with converting. But maybe there is some simple solution. I just want to move the center of an rotated view as if the view was not rotated, i.e. when wanting y to be decremented by 10, just do so without worrying about the rotated coordinates. Are there any methods or functions for converting between the coordinate systems, so that I can just specify the center point coordinates in superview coordinates and it will just be right when applying these to the rotated view's center property?
Look at the CGAffineTransform methods, such as CGAffineTransformMake, for an easier way to generate the transforms you need. You can combine rotation and translation in one matrix.
You can multiply your coordinates by a transformation matrix. This puts your coordinates into your parent view's "coordinate space". You can also use the inverse matrix of the parent view to take coordinates that are in the parent view and return them to "world space".
http://en.wikipedia.org/wiki/Rotation_matrix
Read further down the page for pseudo-code examples.
I have an UIImageView which I want to rotate a little bit, but I need to define an special origin for rotation. How could I do that? More precisely, I have an image of a goblet, and I want it to wiggle on a table. It's socket is quadrate. So when wiggeling to left, then my rotation origin has to be the left corner of that quadrate. And when wiggeling to right, the rotation origin has to be the right corner of that quadrate.
How could I do that? I already have this:
myView.transform = CGAffineTransformMakeRotation(.25);
There was a similar question on how to achieve this here.
The best answer was to change the anchorPoint property of the underlying layer of the UIView.