Find the precise centroid (as an MKMapPoint) of an MKPolygon - ios

This means excluding the area(s) of any interiorPolygons.
Once one has the centroid of the outer points polygon, how does one (i.e., in the form of an Objective-C example) adjust the centroid by the subtractive interiorPolygons? Or is there a more elegant way to compute the centroid in one go?
If you help get the code working, it will be open sourced (WIP here).
Might be helpful:
http://www.ecourses.ou.edu/cgi-bin/eBook.cgi?topic=st&chap_sec=07.2&page=case_sol
https://en.wikipedia.org/wiki/Centroid#Centroid_of_polygon

Thinking about it today, it makes qualitative sense that adding each interior centroid weighted by area to the exterior centroid would arrive at something sensible. (A square with an interior polygon (hole) on the left side would displace the centroid right, directly proportional to the area of the hole.)
Not to scale:
- (MKMapPoint)calculateCentroid
{
switch (self.pointCount) {
case 0: return MKMapPointMake(0.0,
0.0);
case 1: return MKMapPointMake(self.points[0].x,
self.points[0].y);
case 2: return MKMapPointMake((self.points[0].x + self.points[1].x) / 2.0,
(self.points[0].y + self.points[1].y) / 2.0);
}
// onward implies pointCount >= 3
MKMapPoint centroid;
MKMapPoint *previousPoint = &(self.points[self.pointCount-1]); // for i=0, wrap around to the last point
for (NSUInteger i = 0; i < self.pointCount; ++i) {
MKMapPoint *point = &(self.points[i]);
double delta = (previousPoint->x * point->y) - (point->x * previousPoint->y); // x[i-1]*y[i] + x[i]*y[i-1]
centroid.x += (previousPoint->x + point->x) * delta; // (x[i-1] + x[i]) / delta
centroid.y += (previousPoint->y + point->y) * delta; // (y[i-1] + y[i]) / delta
previousPoint = point;
}
centroid.x /= 6.0 * self.area;
centroid.y /= 6.0 * self.area;
// interiorPolygons are holes (subtractive geometry model)
for (MKPolygon *interiorPoly in self.interiorPolygons) {
if (interiorPoly.area == 0.0) {
continue; // avoid div-by-zero
}
centroid.x += interiorPoly.centroid.x / interiorPoly.area;
centroid.y += interiorPoly.centroid.y / interiorPoly.area;
}
return centroid;
}

in Swift 5
private func centroidForCoordinates(_ coords: [CLLocationCoordinate2D]) -> CLLocationCoordinate2D? {
guard let firstCoordinate = coordinates.first else {
return nil
}
guard coords.count > 1 else {
return firstCoordinate
}
var minX = firstCoordinate.longitude
var maxX = firstCoordinate.longitude
var minY = firstCoordinate.latitude
var maxY = firstCoordinate.latitude
for i in 1..<coords.count {
let current = coords[i]
if minX > current.longitude {
minX = current.longitude
} else if maxX < current.longitude {
maxX = current.longitude
} else if minY > current.latitude {
minY = current.latitude
} else if maxY < current.latitude {
maxY = current.latitude
}
}
let centerX = minX + ((maxX - minX) / 2)
let centerY = minY + ((maxY - minY) / 2)
return CLLocationCoordinate2D(latitude: centerY, longitude: centerX)
}

Related

Get area of intersecting line (CGPoints)

I have an Array of CGPoints and I would like to find those points, which build a shape. Please see the attached image:
The red circles just mark the points I have.
How can the area with the question mark be found?
Thanks.
You are going to have to start with your first line segment and check for intersections. Obviously if the first two line segments intersect then they are the same line and your shape is just a line, so ignore that case. As you continue down your line segments once you find a segment pair that intersect then you have your shape.
Check line segment 2 against line segment 1. Then check line segment 3 against line segment 2, then against line segment 1. Then check 4 against 3, then 2, then 1, etc... If you find that line segment 7 intersects with line segment 3, delete the first point of line segment 3 and se it to the intersection point you found. Then delete the last point of line segment 7 and set it to the intersection point you found. There you have your shape.
Here is an example method to find the intersection of 2 line segments (written in C#, but it's straight math so it should be very easy to convert to any language you would like). Taken from here:
// Determines if the lines AB and CD intersect.
static bool LinesIntersect(PointF A, PointF B, PointF C, PointF D)
{
PointF CmP = new PointF(C.X - A.X, C.Y - A.Y);
PointF r = new PointF(B.X - A.X, B.Y - A.Y);
PointF s = new PointF(D.X - C.X, D.Y - C.Y);
float CmPxr = CmP.X * r.Y - CmP.Y * r.X;
float CmPxs = CmP.X * s.Y - CmP.Y * s.X;
float rxs = r.X * s.Y - r.Y * s.X;
if (CmPxr == 0f)
{
// Lines are collinear, and so intersect if they have any overlap
return ((C.X - A.X < 0f) != (C.X - B.X < 0f))
|| ((C.Y - A.Y < 0f) != (C.Y - B.Y < 0f));
}
if (rxs == 0f)
return false; // Lines are parallel.
float rxsr = 1f / rxs;
float t = CmPxs * rxsr;
float u = CmPxr * rxsr;
return (t >= 0f) && (t <= 1f) && (u >= 0f) && (u <= 1f);
}
I've figured out the solution.
This function returns a polygon for each area that gets closed by intersecting lines.
func intersectionOfLineFrom(p1: CGPoint, to p2: CGPoint, withLineFrom p3: CGPoint, to p4: CGPoint) -> NSValue? {
let d: CGFloat = (p2.x - p1.x) * (p4.y - p3.y) - (p2.y - p1.y) * (p4.x - p3.x)
if d == 0 {
return nil
}
// parallel lines
let u: CGFloat = ((p3.x - p1.x) * (p4.y - p3.y) - (p3.y - p1.y) * (p4.x - p3.x)) / d
let v: CGFloat = ((p3.x - p1.x) * (p2.y - p1.y) - (p3.y - p1.y) * (p2.x - p1.x)) / d
if u < 0.0 || u > 1.0 {
return nil
}
// intersection point not between p1 and p2
if v < 0.0 || v > 1.0 {
return nil
}
// intersection point not between p3 and p4
var intersection: CGPoint = CGPointZero
intersection.x = p1.x + u * (p2.x - p1.x)
intersection.y = p1.y + u * (p2.y - p1.y)
return NSValue(CGPoint: intersection)
}
func intersectedPolygons(points: [CGPoint]) -> [[CGPoint]] {
var removeIndexBelow : Int = 0
var removeIndexAbove : Int = 0
var resultArrays : [[CGPoint]] = [[CGPoint]]()
for i in 1..<points.count {
let firstLineStart = points[i-1] as CGPoint
let firstLineEnd = points[i] as CGPoint
for var j = points.count-1; j > i+1; j-- {
let lastLineStart = points[j-1] as CGPoint
let lastLineEnd = points[j] as CGPoint
if let intersect: NSValue = self.intersectionOfLineFrom(firstLineStart, to: firstLineEnd, withLineFrom: lastLineStart, to: lastLineEnd){
var pointsCopy = points
let intersection = intersect.CGPointValue()
pointsCopy[i-1] = intersection
pointsCopy[j] = intersection
removeIndexBelow = i
removeIndexAbove = j
let fullPoly = Array(pointsCopy[removeIndexBelow-1..<removeIndexAbove])
resultArrays.append(fullPoly)
break;
}
}
}
return resultArrays
}

Triangle Gradient With Core Graphics

I'm trying to draw a triangle like this one in a view (one UIView, one NSView):
My first thought was CoreGraphics, but I couldn't find any information that would help me draw a gradient between three points of arbitrary color.
Any help?
Thanks!
Actually it's pretty simple with CoreGraphics. Below you can find code that renders given triangle, but first let's think how we can solve this problem.
Theory
Imagine equilateral triangle with side length w. All three angles are equal to 60 degrees:
Each angle will represent component of a pixel: red, green or blue.
Lets analyze intensity of a green component in a pixel near top angle:
The more closer pixel to the angle, the more component intense it'll have and vice versa. Here we can decompose our main goal to smaller ones:
Draw triangle pixel by pixel.
For each pixel calculate value for each component based on distance from corresponding angle.
To solve first task we will use CoreGraphics bitmap context. It will have four components per pixel each 8 bits long. This means that component value may vary from 0 to 255. Fourth component is alpha channel and will be always equal to max value - 255. Here is example of how values will be interpolated for the top angle:
Now we need to think how we can calculate value for component.
First, let's define main color for each angle:
Now let's choose an arbitrary point A with coordinates (x,y) on the triangle:
Next, we draw a line from an angle associated with red component and it passes through the A till it intersects with opposite side of a triangle:
If we could find d and c their quotient will equal to normalized value of component, so value can be calculated easily:
(source: sciweavers.org)
Formula for finding distance between two points is simple:
(source: sciweavers.org)
We can easily find distance for d, but not for c, because we don't have coordinates of intersection. Actually it's not that hard. We just need to build line equations for line that passes through A and line that describes opposite side of a triangle and find their intersection:
Having intersection point we can apply distance formula to find c and finally calculate component value for current point.
Same flow applies for another components.
Code
Here is the code that implements concepts above:
+ (UIImage *)triangleWithSideLength:(CGFloat)sideLength {
return [self triangleWithSideLength:sideLength scale:[UIScreen mainScreen].scale];
}
+ (UIImage *)triangleWithSideLength:(CGFloat)sideLength
scale:(CGFloat)scale {
UIImage *image = nil;
CGSize size = CGSizeApplyAffineTransform((CGSize){sideLength, sideLength * sin(M_PI / 3)}, CGAffineTransformMakeScale(scale, scale));
size_t const numberOfComponents = 4;
size_t width = ceilf(size.width);
size_t height = ceilf(size.height);
size_t realBytesPerRow = width * numberOfComponents;
size_t alignedBytesPerRow = (realBytesPerRow + 0xFF) & ~0xFF;
size_t alignedPixelsPerRow = alignedBytesPerRow / numberOfComponents;
CGContextRef ctx = CGBitmapContextCreate(NULL,
width,
height,
8,
alignedBytesPerRow,
CGColorSpaceCreateDeviceRGB(),
(CGBitmapInfo)kCGImageAlphaPremultipliedLast);
char *data = CGBitmapContextGetData(ctx);
for (int i = 0; i < height; i++) {
for (int j = 0; j < width; j++) {
int edge = ceilf((height - i) / sqrt(3));
if (j < edge || j > width - edge) {
continue;
}
CGFloat redNormalized = 0;
CGFloat greenNormalized = 0;
CGFloat blueNormalized = 0;
CGPoint currentTrianglePoint = (CGPoint){j / scale, (height - i) / scale};
[self calculateCurrentValuesAtGiventPoint:currentTrianglePoint
sideLength:sideLength
sideOne:&redNormalized
sideTwo:&greenNormalized
sideThree:&blueNormalized];
int32_t red = redNormalized * 0xFF;
int32_t green = greenNormalized * 0xFF;
int32_t blue = blueNormalized * 0xFF;
char *pixel = data + (j + i * alignedPixelsPerRow) * numberOfComponents;
*pixel = red;
*(pixel + 1) = green;
*(pixel + 2) = blue;
*(pixel + 3) = 0xFF;
}
}
CGImageRef cgImage = CGBitmapContextCreateImage(ctx);
image = [[UIImage alloc] initWithCGImage:cgImage];
CGContextRelease(ctx);
CGImageRelease(cgImage);
return image;
}
+ (void)calculateCurrentValuesAtGiventPoint:(CGPoint)point
sideLength:(CGFloat)length
sideOne:(out CGFloat *)sideOne
sideTwo:(out CGFloat *)sideTwo
sideThree:(out CGFloat *)sideThree {
CGFloat height = sin(M_PI / 3) * length;
if (sideOne != NULL) {
// Side one is at 0, 0
CGFloat currentDistance = sqrt(point.x * point.x + point.y * point.y);
if (currentDistance != 0) {
CGFloat a = point.y / point.x;
CGFloat b = 0;
CGFloat c = -height / (length / 2);
CGFloat d = 2 * height;
CGPoint intersection = (CGPoint){(d - b) / (a - c), (a * d - c * b) / (a - c)};
CGFloat currentH = sqrt(intersection.x * intersection.x + intersection.y * intersection.y);
*sideOne = 1 - currentDistance / currentH;
} else {
*sideOne = 1;
}
}
if (sideTwo != NULL) {
// Side two is at w, 0
CGFloat currentDistance = sqrt(pow((point.x - length), 2) + point.y * point.y);
if (currentDistance != 0) {
CGFloat a = point.y / (point.x - length);
CGFloat b = height / (length / 2);
CGFloat c = a * -point.x + point.y;
CGFloat d = b * -length / 2 + height;
CGPoint intersection = (CGPoint){(d - c) / (a - b), (a * d - b * c) / (a - b)};
CGFloat currentH = sqrt(pow(length - intersection.x, 2) + intersection.y * intersection.y);
*sideTwo = 1 - currentDistance / currentH;
} else {
*sideTwo = 1;
}
}
if (sideThree != NULL) {
// Side three is at w / 2, w * sin60 degrees
CGFloat currentDistance = sqrt(pow((point.x - length / 2), 2) + pow(point.y - height, 2));
if (currentDistance != 0) {
float dy = point.y - height;
float dx = (point.x - length / 2);
if (fabs(dx) > FLT_EPSILON) {
CGFloat a = dy / dx;
CGFloat b = 0;
CGFloat c = a * -point.x + point.y;
CGFloat d = 0;
CGPoint intersection = (CGPoint){(d - c) / (a - b), (a * d - b * c) / (a - b)};
CGFloat currentH = sqrt(pow(length / 2 - intersection.x, 2) + pow(height - intersection.y, 2));
*sideThree = 1 - currentDistance / currentH;
} else {
*sideThree = 1 - currentDistance / height;
}
} else {
*sideThree = 1;
}
}
}
Here is a triangle image produced by this code:

Get rotated rectangle UIView corners' coordinates iOS

I'm trying to find a rotated rectangle UIView's four corners' coordinates.
I think one way I can do is to use recognizer.rotation, find the rotated angle then calculate the origins. But that requires some geometry calculation.
- (IBAction)handlePan:(UIRotationGestureRecognizer*)recognizer {
NSLog(#"Rotation in degrees since last change: %f", [recognizer rotation] * (180 / M_PI));
recognizer.view.transform = CGAffineTransformRotate(recognizer.view.transform, recognizer.rotation);
NSLog(#"%#",recognizer);
recognizer.rotation = 0;
NSLog(#"bound is %f and %f, frame is %f and %f, %f and %f.",recognizer.view.bounds.size.width,recognizer.view.bounds.size.height, recognizer.view.frame.size.width,recognizer.view.frame.size.height, recognizer.view.frame.origin.x, recognizer.view.frame.origin.y);
}
I'm just wondering if there are any other easier ways to get the coordinates?
Thanks!
EDIT:
Looks like we have a great answer here(see answer below). I have managed to calculate the corners through a stupid way -- using rotation angle and geometry. It works but not easy and light. I'm sharing my code here just in case some one may want to use it(Even though I doubt it.)
float r = 100;
NSLog(#"radius is %f.",r);
float AAngle = M_PI/3+self.rotatedAngle;
float AY = recognizer.view.center.y - sin(AAngle)*r;
float AX = recognizer.view.center.x - cos(AAngle)*r;
self.pointPADA = CGPointMake(AX, AY);
NSLog(#"View Center is (%f,%f)",recognizer.view.center.x,recognizer.view.center.y);
NSLog(#"Point A has coordinate (%f,%f)",self.pointPADA.x,self.pointPADA.y);
float BAngle = M_PI/3-self.rotatedAngle;
float BY = recognizer.view.center.y - sin(BAngle)*r;
float BX = recognizer.view.center.x + cos(BAngle)*r;
self.pointPADB = CGPointMake(BX, BY);
NSLog(#"Point B has coordinate (%f,%f)",BX,BY);
float CY = recognizer.view.center.y + sin(AAngle)*r;
float CX = recognizer.view.center.x + cos(AAngle)*r;
self.pointPADC = CGPointMake(CX, CY);
NSLog(#"Point C has coordinate (%f,%f)",CX,CY);
float DY = recognizer.view.center.y + sin(BAngle)*r;
float DX = recognizer.view.center.x - cos(BAngle)*r;
self.pointPADD = CGPointMake(DX, DY);
NSLog(#"Point D has coordinate (%f,%f)",DX,DY);
Here's my solution though I wonder if there's a more succinct way:
CGPoint originalCenter = CGPointApplyAffineTransform(theView.center,
CGAffineTransformInvert(theView.transform));
CGPoint topLeft = originalCenter;
topLeft.x -= theView.bounds.size.width / 2;
topLeft.y -= theView.bounds.size.height / 2;
topLeft = CGPointApplyAffineTransform(topLeft, theView.transform);
CGPoint topRight = originalCenter;
topRight.x += theView.bounds.size.width / 2;
topRight.y -= theView.bounds.size.height / 2;
topRight = CGPointApplyAffineTransform(topRight, theView.transform);
CGPoint bottomLeft = originalCenter;
bottomLeft.x -= theView.bounds.size.width / 2;
bottomLeft.y += theView.bounds.size.height / 2;
bottomLeft = CGPointApplyAffineTransform(bottomLeft, theView.transform);
CGPoint bottomRight = originalCenter;
bottomRight.x += theView.bounds.size.width / 2;
bottomRight.y += theView.bounds.size.height / 2;
bottomRight = CGPointApplyAffineTransform(bottomRight, theView.transform);
Checked answer in Swift
struct ViewCorners {
private(set) var topLeft: CGPoint!
private(set) var topRight: CGPoint!
private(set) var bottomLeft: CGPoint!
private(set) var bottomRight: CGPoint!
private let originalCenter: CGPoint
private let transformedView: UIView
private func pointWith(multipliedWidth: CGFloat, multipliedHeight: CGFloat) -> CGPoint {
var x = originalCenter.x
x += transformedView.bounds.width / 2 * multipliedWidth
var y = originalCenter.y
y += transformedView.bounds.height / 2 * multipliedHeight
var result = CGPoint(x: x, y: y).applying(transformedView.transform)
result.x += transformedView.transform.tx
result.y += transformedView.transform.ty
return result
}
init(view: UIView) {
transformedView = view
originalCenter = view.center.applying(view.transform.inverted())
topLeft = pointWith(multipliedWidth:-1, multipliedHeight:-1)
topRight = pointWith(multipliedWidth: 1, multipliedHeight:-1)
bottomLeft = pointWith(multipliedWidth:-1, multipliedHeight: 1)
bottomRight = pointWith(multipliedWidth: 1, multipliedHeight: 1)
}
}
Then create struct instance and take transformed rect corners new points.
let view = UIView(frame: CGRect(x: 40, y: 20, width: 100, height: 50))
view.transform = .init(rotationAngle: .pi / 4)
let corners = ViewCorners(view: view)
print(corners.topLeft,
corners.topRight,
corners.bottomLeft,
corners.bottomRight,
separator: "\n")
Using Swift 4, We can get the bounds of rotated view by simple way.
let transformedBounds = view.bounds.applying(view.transform)

How to get area for MKPolygon in iOS

How to you get the area of a MKPolygon or MKOverlay in iOS?
I have been able to breakup the Polygon into triangles and do some math to get the area. But, doesn't work well with irregular polygons.
I was thinking about doing something like the "A more complex case" here: http://www.mathopenref.com/coordpolygonarea2.html
I was hoping there is a simpler solution with MapKit.
Thanks,
Tim
Here's the implementation I'm using.
#define kEarthRadius 6378137
#implementation MKPolygon (AreaCalculation)
- (double) area {
double area = 0;
NSMutableArray *coords = [[self coordinates] mutableCopy];
[coords addObject:[coords firstObject]];
if (coords.count > 2) {
CLLocationCoordinate2D p1, p2;
for (int i = 0; i < coords.count - 1; i++) {
p1 = [coords[i] MKCoordinateValue];
p2 = [coords[i + 1] MKCoordinateValue];
area += degreesToRadians(p2.longitude - p1.longitude) * (2 + sinf(degreesToRadians(p1.latitude)) + sinf(degreesToRadians(p2.latitude)));
}
area = - (area * kEarthRadius * kEarthRadius / 2);
}
return area;
}
- (NSArray *)coordinates {
NSMutableArray *points = [NSMutableArray arrayWithCapacity:self.pointCount];
for (int i = 0; i < self.pointCount; i++) {
MKMapPoint *point = &self.points[i];
[points addObject:[NSValue valueWithMKCoordinate:MKCoordinateForMapPoint(* point)]];
}
return points.copy;
}
double degreesToRadians(double radius) {
return radius * M_PI / 180;
}
In Swift 3:
let kEarthRadius = 6378137.0
extension MKPolygon {
func degreesToRadians(_ radius: Double) -> Double {
return radius * .pi / 180.0
}
func area() -> Double {
var area: Double = 0
var coords = self.coordinates()
coords.append(coords.first!)
if (coords.count > 2) {
var p1: CLLocationCoordinate2D, p2: CLLocationCoordinate2D
for i in 0..<coords.count-1 {
p1 = coords[i]
p2 = coords[i+1]
area += degreesToRadians(p2.longitude - p1.longitude) * (2 + sin(degreesToRadians(p1.latitude)) + sin(degreesToRadians(p2.latitude)))
}
area = abs(area * kEarthRadius * kEarthRadius / 2)
}
return area
}
func coordinates() -> [CLLocationCoordinate2D] {
var points: [CLLocationCoordinate2D] = []
for i in 0..<self.pointCount {
let point = self.points()[i]
points.append(MKCoordinateForMapPoint(point))
}
return Array(points)
}
}
I figured this out by doing a little loop through the points in the polygon. For every 3 points, I check if the center of that triangle is in the polygon. If it is continue, if not, connect the polygon so that there are no dips in the polygon. Once done, get the triangles in the polygon and do the math to get the area. Then subtract the triangles that were removed.
Hope this helps someone.

Scale of UIPinchGestureRecognizer in horizontal and vertical directions separately

When using UIPinchGestureRecognizer what is the best way to detect/read the pinch scale in horizontal and vertical directions individually? I saw this post
UIPinchGestureRecognizer Scale view in different x and y directions
but I noticed there were so many going back and forth for such a seemingly routine task that I am not sure that is the best answer/way.
If not using UIPinchGestureRecognizer altogether for this purpose is the answer, what's the best way to detect pinch scale in two different directions?
Basically do this,
func _mode(_ sender: UIPinchGestureRecognizer)->String {
// very important:
if sender.numberOfTouches < 2 {
print("avoided an obscure crash!!")
return ""
}
let A = sender.location(ofTouch: 0, in: self.view)
let B = sender.location(ofTouch: 1, in: self.view)
let xD = fabs( A.x - B.x )
let yD = fabs( A.y - B.y )
if (xD == 0) { return "V" }
if (yD == 0) { return "H" }
let ratio = xD / yD
// print(ratio)
if (ratio > 2) { return "H" }
if (ratio < 0.5) { return "V" }
return "D"
}
That function will return H, V, D for you .. horizontal, vertical, diagonal.
You would use it something like this ...
func yourSelector(_ sender: UIPinchGestureRecognizer) {
// your usual code such as ..
// if sender.state == .ended { return } .. etc
let mode = _mode(sender)
print("the mode is \(mode) !!!")
// in this example, we only care about horizontal pinches...
if mode != "H" { return }
let vel = sender.velocity
if vel < 0 {
print("you're squeezing the screen!")
}
}
In my C# I do the following
private double _firstDistance = 0;
private int _firstScaling = 0;
private void PinchHandler(UIPinchGestureRecognizer pinchRecognizer)
{
nfloat x1, y1, x2, y2 = 0;
var t1 = pinchRecognizer.LocationOfTouch(0, _previewView);
x1 = t1.X;
y1 = t1.Y;
var t2 = pinchRecognizer.LocationOfTouch(1, _previewView);
x2 = t2.X;
y2 = t2.Y;
if (pinchRecognizer.State == UIGestureRecognizerState.Began)
{
_firstDistance = Math.Sqrt(Math.Pow((x2 - x1), 2) + Math.Pow((y2 - y1), 2));
_firstScaling = _task.TextTemplates[_selectedTextTemplate].FontScaling;
}
if (pinchRecognizer.State == UIGestureRecognizerState.Changed)
{
var distance = Math.Sqrt(Math.Pow((x2 - x1), 2) + Math.Pow((y2 - y1), 2));
var fontScaling = Convert.ToInt32((distance - _firstDistance) / _previewView.Frame.Height * 100);
fontScaling += _firstScaling;
_task.TextTemplates[_selectedTextTemplate].FontScaling = fontScaling;
UpdateBitmapPreview();
}
}
I calculate the distance between the two points when pinch "began" and hold that value in two privates. Then I calculate a scaling (fontScaling) based on the first measured distance and the second one (in "changed").
I use my own view (_previewView) to set as base (100%), but you can use View.Bounds.height instead or the width for that matter. in my case, I always have a square view, so height == width in my app.

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