Position Subviews Relative to Screen Estate - objective-c

I'd like to display multiple small UIViews as Subviews relative to the screen estate. This should work across different screen sizes (iPad, iPhone)/portrait/landscape modes.
Each subview to display has two NSNumber objects with an unsigned int ranging from -100 (min) to 100 (max) which needs to be mapped to the correct x and y coordinates for positioning.
What's the best way to translate those values (-100...100) to use them for positioning UIViews on the screen?
How do I position them in a relative rather then an absolute way, so that the code works across screen rotation and screen sizes?

Ok, so if I understand correctly you want a -100 in the x direction to map to the left most point on the screen, 100 in the x to map to the right most point on the screen, -100 in the y direction to map to the lowest point on the screen, and 100 in the y to map to the highest point on screen (or maybe you want the y inverted from what I have so that it agrees with the screen coordinate system in which y becomes bigger the lower on the screen you get?).
And we also want to account for rotation.
As far as I understand it, asking UIScreen for its height and width:
CGFloat width = [UIScreen mainScreen].bounds.size.width;
CGFloat height = [UIScreen mainScreen].bounds.size.height;
but this does not account for rotation. The only other way I am aware of that is pretty straightforward would be to ask a UIView covering the screen for its width and height (most simply, you could make your viewcontroller's view cover the whole screen).
If you had a UIView that perfectly covered the whole screen (let's call it myView), you could try:
CGFloat width = myView.frame.size.width;
CGFloat height = myView.frame.size.height;
these should adjust for orientation by themselves (from my experience, it should definitely work if you get the height and width in viewDidAppear:animated: or anything after. also the UIView needs to either be the UIViewControllers view property or a subview of this view. if not, you'll have to implement didRotateFromInterfaceOrientation: or find some other way to tell your view about any rotations). Once we have the 'width' and 'height' of the screen, we can convert from your int's to screen position. Try something like:
(CGPoint)convertX:(NSNumber *)x andY:(NSNumber *)y intoPoint
{
pointX = ([x intValue] + 100.0)*width/200.0;
pointY = (-[y intValue] + 100.0)*height/200.0; // remove the - sign at the front of the expression for y to grow as you move down the screen
return CGPointMake(pointX, pointY);
}
to convert from -100 to 100 in x and y to their respective points on the screen.

If you're working with a range of +/-100, then you may want to use the underlying CALayers to position your views. The nice part about CALayers, is that their anchor points are mapped to a device-agnostic grid that ranges from 0.0 to +1.0 on a Cartesian plane.

Related

How could I make an "iPod Wheel" type control on iPhone?

I want to create a sort of "iPod Wheel" control in a Swift project that I'm doing. I've got everything drawn out, but not it's time to actually make this thing work.
What would be the best way to recognize "spinning" so to speak, or to describe that more clearly, when the user is actively pressing the wheel and spinning his/her thumb around the wheel in a clockwise or counter-clockwise direction.
I will no doubt want to use touchesBegan/touchesMoved/touchesEnded. What's the best way to figure out spinning though?
I'm thinking
a) determine in touchesMoved if the users touch is within circle, by determining the radius from the center point. Center point and radius are easily obtainable. Using these however, how can I determine the outer edge of the circle/wheel... to know whether the user is within the actually circle (their touch could still be in the view, but outside the actual wheel portion)
b) Determine the current angle and how it has changed the previous angle. By that I mean... I would use the center point of the circle as one point, and the users current touch as the second point. This gives me my vector. I would also have a baseline angle. Likely center point to 12 c'clock. I would compare the two vectors (I already have a VectorMath class for this from something else I'm doing) and see my angle is 0. If the users touch were at 3 oclock, and I compared it to our baseline angle... I would see the angle is 90 degrees. I would continually calculate the angle, and perhaps every 5 degrees of change... would warrant a change in the controls output (depending on desired sensitivity).
Does this seem like the best way to do this? I think this would be an ideal way, but am still not sure on how to calculate the circles outer edge, and determine if a users touch is within it.
You are on the right track. I think approach b) will work.
Remember the starting position of the finger at the touchesBegan
event.
Imagine a line from the finger position to the middle of the button
circle.
For the touchesMoved event, again, imagine a virtual line from the
new position to the center of the circle.
Using the formula from
http://mathworld.wolfram.com/Line-LineAngle.html (or some code) you can determine
the angle between the two lines. If it's a negative angle the user
is turning the wheel counter-clockwise, otherwise it's clockwise.
To determine if the touch event was inside the ring, calculate the distance from the center of the circle to the point of touch. It should be between the minimum and the maximum distance (inner circle and outer circle radius). Calculating the distance between to two points is explained at https://www.mathsisfun.com/algebra/distance-2-points.html
I think you're almost there, although I'd do something slightly different on your point b.
If you think about it, when you start "spinning" on your iPod, you don't need to start from a precise position, you start spinning from "where you started", therefore I wouldn't set my "baseline angle" at π/2, I'd set my baseline (or 0°) angle at the point the user taps for the first time, and starting from then, I'd count the offset angles, clockwise and counterclockwise.
I don't think there would be much difference, except maybe from some calculations you'll do on the angles, on the two approaches, practically speaking; it just makes more sense imho to start counting from the first input rater than setting a baseline to π/2 and counting the first angle.
I am answering in parts.
// Get a position based on the angle
float xPosition = center.x + (radiusX * sinf(angleInRadians)) - (CGRectGetWidth([cell frame]) / 2);
float yPosition = center.y + (radiusY * cosf(angleInRadians)) - (CGRectGetHeight([cell frame]) / 2);
float scale = 0.75f + 0.25f * (cosf(angleInRadians) + 1.0);
next
[cell setTransform:CGAffineTransformScale(CGAffineTransformMakeTranslation(xPosition, yPosition), scale, scale)];
// Tweak alpha using the same system as applied for scale, this
// time with 0.3 the minimum and a semicircle range of 0.5
[cell setAlpha:(0.3f + 0.5f * (cosf(angleInRadians) + 1.0))];
and
- (void)spin:(SpinGestureRecognizer *)recognizer
{
CGFloat angleInRadians = -[recognizer rotation];
CGFloat degrees = 180.0 * angleInRadians / M_PI; // Radians to degrees
[self setCurrentAngle:[self currentAngle] + degrees];
[self setAngle:[self currentAngle]];
}
again check the wheelview.m of photowheel in github.

How to change the anchor point from the top-left corner of a transformation matrix to the bottom-left corner?

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 :)

Visualizing the Anchor Point of a UIImageView

Is there an easy way of putting a mark (like a cross for example) on the anchor point of an UIImageView? I'm trying to line up several rotating images by their anchor point, and being able to see these points would make the job a lot easier.
Many thanks.
You are asking how to visualize the anchor point within a view but it seem to me that you are asking for it so that you can help align the anchor points. I'll try and answer both questions.
Visualizing the anchor point.
Every view on iOS have an underlying layer that has an anchor point. The anchor point is in unit coordinate space of the layer (x and y goes from 0 to 1). This means that you can multiply x by the width and y by the height to get the position of the anchor point inside the layer in the coordinate space of the view/layer. You can then place a subview/sublayer there to show the location of the anchor point.
In code you could do something like this to display a small black dot where the anchor point is.
CALayer *anchorPointLayer = [CALayer layer];
anchorPointLayer.backgroundColor = [UIColor blackColor].CGColor;
anchorPointLayer.bounds = CGRectMake(0, 0, 6, 6);
anchorPointLayer.cornerRadius = 3;
CGPoint anchor = viewWithVisibleAnchorPoint.layer.anchorPoint;
CGSize size = viewWithVisibleAnchorPoint.layer.bounds.size;
anchorPointLayer.position = CGPointMake(anchor.x * size.width,
anchor.y * size.height);
[viewWithVisibleAnchorPoint.layer addSublayer:anchorPointLayer];
You can see the result in the image below for four different rotations.
Aligning layers by their anchor point
That is cool and all but it's actually easier then that to align anchor points.
The key trick is that the position and the anchorPoint is always the same point, only in two different coordinate spaces. The position is specified in the coordinate space of the super layer. The anchor point is specified in the unit coordinate space of the layer.
The nice thing about this is that views that have their position property aligned will automatically have their anchorPoint aligned. Note that the content is drawn relative to the anchor point. Below is an example of a bunch of views that all have the same y component of their position, thus they are aligned in y.
There really isn't any special code to do this. Just make sure that the position properties are aligned.

Randomly placing UIViews

I'd like to know what the best approach is to placing multiple UIViews on screen with random coordinates without them touching each other or touching the parent view's edges.
Thanks!
Use arc4random() to generate random numbers. You'll want to constrain these to be between 0 and your parent views frame width for the x-coordinate, and between 0 and your views frame height for the y-coordinate. Subtract your random views frame width/height from the extent assuming you want it fully visible.
NSInteger randomX = arc4random() % ((int)view.frame.size.width - randomView.frame.size.width);
NSInteger randomY = arc4random() % ((int)view.frame.size.height - randomView.frame.size.height);
CGRect frameRandom = CGRectMake (randomX, randomY, randomView.frame.size.width,randomView.frame.size.height);
Then you'll need to enumerate all your existing random views and ensure that your new random view frame doesn't intersect with the frame of your existing views.
Use CGRectIntersectsRect to see if two UIView frames intersect:
How to figure out if two CGRect intersect?

Convert point coordinates to a new coordinate system

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.