How can I track a point on a texture in OpenGL ES1? - ios

In my iOS application I have a texture applied to a sphere rendered in OpenGLES1. The sphere can be rotated by the user. How can I track where a given point on the texture is in 2D space at any given time?
For example, given point (200, 200) on a texture that's 1000px x 1000px, I'd like to place a UIButton on top of my OpenGL view that tracks the point as the sphere is manipulated.
What's the best way to do this?
On my first attempt, I tried to use a color-picking technique where I have a separate sphere in an off-screen framebuffer that uses a black texture with a red square at point (200, 200). Then, I used glReadPixels() to track the position of the red square and I moved my button accordingly. Unfortunately, grabbing all the pixel data and iterating it 60 times a second just isn't possible for obvious performance reasons. I tried a number of ways to optimize this hack (eg: iterating only the red pixels, iterating every 4th red pixel, etc), but it just didn't prove to be reliable.
I'm an OpenGL noob, so I'd appreciate any guidance. Is there a better solution? Thanks!

I think it's easier to keep track of where your ball is instead of searching for it with pixels. Then just have a couple of functions to translate your ball's coordinates to your view's coordinates (and back), then set your subview's center to the translated coordinates.
CGPoint translatePointFromGLCoordinatesToUIView(CGPoint coordinates, UIView *myGLView){
//if your drawing coordinates were between (horizontal {-1.0 -> 1.0} vertical {-1 -> 1})
CGFloat leftMostGLCoord = -1;
CGFloat rightMostGLCoord = 1;
CGFloat bottomMostGLCoord = -1;
CGFloat topMostGLCoord = 1;
CGPoint scale;
scale.x = (rightMostGLCoord - leftMostGLCoord) / myGLView.bounds.size.width;
scale.y = (topMostGLCoord - bottomMostGLCoord) / myGLView.bounds.size.height;
coordinates.x -= leftMostGLCoord;
coordinates.y -= bottomMostGLCoord;
CGPoint translatedPoint;
translatedPoint.x = coordinates.x / scale.x;
translatedPoint.y =coordinates.y / scale.y;
//flip y for iOS coordinates
translatedPoint.y = myGLView.bounds.size.height - translatedPoint.y;
return translatedPoint;
}
CGPoint translatePointFromUIViewToGLCoordinates(CGPoint pointInView, UIView *myGLView){
//if your drawing coordinates were between (horizontal {-1.0 -> 1.0} vertical {-1 -> 1})
CGFloat leftMostGLCoord = -1;
CGFloat rightMostGLCoord = 1;
CGFloat bottomMostGLCoord = -1;
CGFloat topMostGLCoord = 1;
CGPoint scale;
scale.x = (rightMostGLCoord - leftMostGLCoord) / myGLView.bounds.size.width;
scale.y = (topMostGLCoord - bottomMostGLCoord) / myGLView.bounds.size.height;
//flip y for iOS coordinates
pointInView.y = myGLView.bounds.size.height - pointInView.y;
CGPoint translatedPoint;
translatedPoint.x = leftMostGLCoord + (pointInView.x * scale.x);
translatedPoint.y = bottomMostGLCoord + (pointInView.y * scale.y);
return translatedPoint;
}
In my app I choose to use the iOS coordinate system for my drawing too. I just apply a projection matrix to my whole glkView the reconciles the coordinate system.
static GLKMatrix4 GLKMatrix4MakeIOSCoordsWithSize(CGSize screenSize){
GLKMatrix4 matrix4 = GLKMatrix4MakeScale(
2.0 / screenSize.width,
-2.0 / screenSize.height,
1.0);
matrix4 = GLKMatrix4Translate(matrix4,-screenSize.width / 2.0, -screenSize.height / 2.0, 0);
return matrix4;
}
This way you don't have to translate anything.

Related

Choosing a random point within a circular image in a UIImageView

I have an app with a color wheel and I'm trying to pick a random color within the color wheel. However, I'm having problems verifying that the random point falls within the color wheel.
Here's the code as it currently is:
CGPoint randomPoint = CGPointMake(arc4random() % (int)colorWheel.bounds.size.width, arc4random() % (int)colorWheel.bounds.size.height);
UIColor *randomColor = [self colorOfPoint:randomPoint];
CGPoint pointInView = [colorWheel convertPoint:randomPoint fromView:colorWheel.window];
if (CGRectContainsPoint(colorWheel.bounds, pointInView)) {
NSLog(#"%#", randomColor);
}
else {
NSLog(#"out of bounds");
}
A couple of other methods of verifying the point that I've tried with no luck:
if (CGRectContainsPoint(colorWheel.frame, randomPoint)) {
NSLog(#"%#", randomColor);
}
if ([colorWheel pointInside:[self.view convertPoint:randomPoint toView: colorWheel] withEvent: nil]) {
NSLog(#"%#", randomColor);
}
Sometimes it'll output "out of bounds", and sometimes it'll just output that the color is white (the background around the color wheel is currently white but there's no white in the color wheel image).
The color wheel image is a circle, so I'm not sure if that's throwing off the test, although it seems like white pops up way too frequently for it to just be a transparent square outline around the image giving a white color.
If you want to generate a random point in a circle, you would do better to pick your point in polar coordinates and then convert it to Cartesian.
The polar coordinate space uses two dimesions, radius and angle. Radius is just the distance from the center, and angle usually starts at "due east" for 0, and goes around counter-clockwise up to 2π (that's in radians, 360˚ of course in degrees).
Presumably your wheel is divided into simple wedges, so the radius actually doesn't matter; you just need to pick a random angle.
uint32_t angle = arc4random_uniform(360);
// Radius will just be halfway from the center to the edge.
// This assumes the circle is exactly enclosed, i.e., diameter == width
CGFloat radius = colorWheel.bounds.size.width / 4;
This function will give you a Cartesian point from your polar coordinates. Wikipedia explains the simple math if you're interested.
/** Convert the polar point (radius, theta) to a Cartesian (x,y). */
CGPoint poltocar(CGFloat radius, CGFloat theta)
{
return (CGPoint){radius * cos(theta), radius * sin(theta)};
}
The function uses radians for theta, because sin() and cos() do, so change the angle to radians, and then you can convert:
CGFloat theta = (angle * M_PI) / 180.0
CGPoint randomPoint = poltocar(radius, theta);
One last step: this circle has its origin at the same place as the view, that is, in the corner, so you need to translate the point to use the center as the origin.
CGPoint addPoints(CGPoint lhs, CGPoint rhs)
{
return (CGPoint){lhs.x + rhs.x, lhs.y, rhs.y};
}
CGPoint offset = (CGPoint){colorWheel.bounds.size.width / 2,
colorWheel.bounds.size.height / 2};
randomPoint = addPoints(randomPoint, offset);
And your new randomPoint will always be within the circle.
I agree with #JoshCaswell's approach, but FYI, the reason the OP code is not working is that the test for inside a circle is incorrect.
The coordinate conversion is unnecessary, and the test against a rectangle is sure to be wrong. Instead, work out how far the random point is from the center and compare that with the radius.
CGFloat centerX = colorWheel.bounds.size.width / 2.0;
CGFloat centerY = colorWheel.bounds.size.height / 2.0;
CGFloat distanceX = centerX - randomPoint.x;
CGFloat distanceY = centerY - randomPoint.y;
CGFloat distance = distanceX*distanceX + distanceY*distanceY;
CGFloat radius = colorWheel.bounds.size.width / 2.0; // just a guess
CGFloat r2 = radius*radius;
// this compares the square of the distance with r^2, to save a sqrt operation
BOOL isInCircle = distance < r2;

SpriteKit move rotated physicsBody with applyImpulse

I want to move a physicsBody with the applyImpulse method in a direction based on the physicsBody rotation.
Foe example, the physicsBody is a square in shape, I call a "move" which will apply an impulse to make it move up vertically. I then call a method to rotate the physicsBody 45 degrees right. If I call the "move" method again, the physicsBody will move diagonally right and up.
I suggest that you follow Sprite Kit’s coordinate and rotation conventions. Specifically, your sprite image should be facing right at zero degrees (the default value), and a positive value is a counter-clockwise rotation. That said, here's one way to apply an impulse in the direction a sprite is facing:
// Specify the force to apply to the SKPhysicsBody
CGFloat r = 5;
// Create a vector in the direction the sprite is facing
CGFloat dx = r * cos (sprite.zRotation);
CGFloat dy = r * sin (sprite.zRotation);
// Apply impulse to physics body
[sprite.physicsBody applyImpulse:CGVectorMake(dx,dy)];
UPDATED:
Fixed with the below thanks to #0x141E
-(void)characterJump {
CGFloat radianFactor = 0.0174532925;
CGFloat rotationInDegrees = _body.zRotation / radianFactor;
CGFloat newRotationDegrees = rotationInDegrees + 90;
CGFloat newRotationRadians = newRotationDegrees * radianFactor;
CGFloat r = 500;
CGFloat dx = r * cos(newRotationRadians);
CGFloat dy = r * sin(newRotationRadians);
[_body.physicsBody applyImpulse:CGVectorMake(dx, dy)];
}

iOS: transforming a view into cylindrical shape

With Quartz 2D we can transform our views on the x, yand z axis.
In some cases we could even make them look 3D by changing the values of the matrixes.
I was wondering if it could be possible to transform a view into a cylinder shape like in the following picture?
Please ignore the top part of the cylinder. I am more curious to know whether it would be possible warping an UIView around like the side of the cylinder as in the image.
Is that possible only making use of Quartz 2D, layers and transformations (not OpenGL)? If not, is it possible to at least draw it in CGContext to make a view appear like so?
You definitely can't do this with a transform. What you could do is create your UIView off-screen, get the context for the view, get an image from that, and then map the image to a new image, using a non-linear mapping.
So:
Create an image context with UIGraphicsBeginImageContext()
Render the view there, with view.layer.renderInContext()
Get an image of the result with CGBitmapContextCreateImage()
Write a mapping function that takes the x/y screen coordinates and maps them to coordinates on the cylinder.
Create a new image the size of the screen view, and call the mapping
function to copy pixels from the source to the destination.
Draw the destination bitmap to the screen.
None of these steps is particularly-difficult, and you might come up with various ways to simplify. For example, you can just render strips of the original view, offsetting the Y coordinate based on the coordinates of a circle, if you are okay with not doing perspective transformations.
If you want the view to actually be interactive, then you'd need to do the transform in the opposite direction when handling touch events.
No you can't bend a view using a transform.
The transform can only manipulate the four corners of the view so no matter what you do it will still be a plane.
I realize this goes beyond Quartz2D... You could try adding SceneKit.
Obtain the view's image via UIGraphicsBeginImageContext(), view.layer.renderInContext(), CGBitmapContextCreateImage().
Create a SCNMaterial with the diffuse property set to the image of your view
Create an SCNCylinder and apply the material to it.
Add the cylinder to an SCNScene.
Create an SCNView and set its scene.
Add the SCNView to your view hierarchy.
Reference : Using OpenGL ES 2.0 with iOS, how do I draw a cylinder between two points?
I have also used the same code for one of my project:
Check this one where it is mentioned to draw cone shape; it's dated but after adapting the algorithm, it works.
See code below for solution. Self represents the mesh and contains the vertices, indices, and such.
- (instancetype)initWithOriginRadius:(CGFloat)originRadius
atOriginPoint:(GLKVector3)originPoint
andEndRadius:(CGFloat)endRadius
atEndPoint:(GLKVector3)endPoint
withPrecision:(NSInteger)precision
andColor:(GLKVector4)color
{
self = [super init];
if (self) {
// normal pointing from origin point to end point
GLKVector3 normal = GLKVector3Make(originPoint.x - endPoint.x,
originPoint.y - endPoint.y,
originPoint.z - endPoint.z);
// create two perpendicular vectors - perp and q
GLKVector3 perp = normal;
if (normal.x == 0 && normal.z == 0) {
perp.x += 1;
} else {
perp.y += 1;
}
// cross product
GLKVector3 q = GLKVector3CrossProduct(perp, normal);
perp = GLKVector3CrossProduct(normal, q);
// normalize vectors
perp = GLKVector3Normalize(perp);
q = GLKVector3Normalize(q);
// calculate vertices
CGFloat twoPi = 2 * PI;
NSInteger index = 0;
for (NSInteger i = 0; i < precision + 1; i++) {
CGFloat theta = ((CGFloat) i) / precision * twoPi; // go around circle and get points
// normals
normal.x = cosf(theta) * perp.x + sinf(theta) * q.x;
normal.y = cosf(theta) * perp.y + sinf(theta) * q.y;
normal.z = cosf(theta) * perp.z + sinf(theta) * q.z;
AGLKMeshVertex meshVertex;
AGLKMeshVertexDynamic colorVertex;
// top vertex
meshVertex.position.x = endPoint.x + endRadius * normal.x;
meshVertex.position.y = endPoint.y + endRadius * normal.y;
meshVertex.position.z = endPoint.z + endRadius * normal.z;
meshVertex.normal = normal;
meshVertex.originalColor = color;
// append vertex
[self appendVertex:meshVertex];
// append color vertex
colorVertex.colors = color;
[self appendColorVertex:colorVertex];
// append index
[self appendIndex:index++];
// bottom vertex
meshVertex.position.x = originPoint.x + originRadius * normal.x;
meshVertex.position.y = originPoint.y + originRadius * normal.y;
meshVertex.position.z = originPoint.z + originRadius * normal.z;
meshVertex.normal = normal;
meshVertex.originalColor = color;
// append vertex
[self appendVertex:meshVertex];
// append color vertex
[self appendColorVertex:colorVertex];
// append index
[self appendIndex:index++];
}
// draw command
[self appendCommand:GL_TRIANGLE_STRIP firstIndex:0 numberOfIndices:self.numberOfIndices materialName:#""];
}
return self;
}

Tile to CGPoint conversion with Retina display

I have a project that uses a tilemap. I have a separate tilemap for low-res (29x29 Tilesize) and high-res (58x58). I have these methods to calculate tileCoord to position and back again.
- (CGPoint)tileCoordForPosition:(CGPoint)position {
int x = position.x / _tileMap.tileSize.width;
int y = ((_tileMap.mapSize.height * _tileMap.tileSize.height) - position.y) / _tileMap.tileSize.height;
return ccp(x, y);
}
- (CGPoint)positionForTileCoord:(CGPoint)tileCoord {
int x = (tileCoord.x * _tileMap.tileSize.width) + _tileMap.tileSize.width/2;
int y = (_tileMap.mapSize.height * _tileMap.tileSize.height) - (tileCoord.y * _tileMap.tileSize.height) - _tileMap.tileSize.height/2;
return ccp(x, y);
}
I got this from RayWenderLich and I do honeslty not understand how it works, and why it has to be so complicated. But this doesn't work when I use retina tilemaps, only on 480x320. Can someone clever come up with a way to make this work for HD? Does not have to work on low-res either, I do not plan on supporting sub-iOS 7.
I want the output to be in the low-res coordinate scale tho, as you might know, cocos2d does the resizing to HD for you. (By multiplying by two)
i think this will work
- (CGPoint)tileCoordForPosition:(CGPoint)position {
    int x = position.x/29;
    int y = ((11*29)-position.y) / 29;
    
    return ccp(x, y);
}
- (CGPoint)positionForTileCoord:(CGPoint)tileCoord {
    double x = tileCoord.x * 29 + 14.5;
    double y = (11*29) - (tileCoord.y * 29) - 14.5;
    return ccp(x, y);
}
Here you're trying to compute your map X coordinate:
int x = position.x / _tileMap.tileSize.width;
The problem here is that (as of v0.99.5-rc0, cocos2d generally uses points for positions, but CCTMXTiledMap always uses pixels for tileSize. On a low-res device, 1 point = 1 pixel, but on a Retina device, 1 point = 2 pixels. Thus on a Retina device, you need to multiply by 2.
You can use the CC_CONTENT_SCALE_FACTOR() macro to fix this:
int x = CC_CONTENT_SCALE_FACTOR() * position.x / _tileMap.tileSize.width;
Here you're trying to compute yoru map Y coordinate:
int y = ((_tileMap.mapSize.height * _tileMap.tileSize.height) - position.y) / _tileMap.tileSize.height;
The extra math here is trying to account for the difference between Cocos2D's normal coordinate system and your map's flipped coordinate system. In standard Cartesian coordinates, the origin is at the lower left and Y coordinates increase as you move up. In a flipped coordinate system, the origin is at the upper left and Y coordinates increase as you move down. Thus you must subtract your position's Y coordinate from the height of the map (in scene units, which are points) to flip it to map coordinates.
The problem again is that _tileMap.tileSize is in pixels, not points. You can again fix that by using CC_CONTENT_SCALE_FACTOR():
CGFloat tileHeight = _tileMap.tileSize.height / CC_CONTENT_SCALE_FACTOR();
int y = ((_tileMap.mapSize.height * tileHeight) - position.y) / tileHeight;

Using a Bezier Curve to draw a spiral

This is for an iPad application, but it is essentially a math question.
I need to draw a circular arc of varying (monotonically increasing) line width. At the beginning of the curve, it would have a starting thickness (let's say 2pts) and then the thickness would smoothly increase until the end of the arc where it would be at its greatest thickness (let's say 12pts).
I figure the best way to make this is by creating a UIBezierPath and filling the shape. My first attempt was to use two circular arcs (with offset centers), and that worked fine up to 90°, but the arc will often be between 90° and 180°, so that approach won't cut it.
My current approach is to make a slight spiral (one slightly growing from the circular arc and one slightly shrinking) using bezier quad or cubic curves. The question is where do I put the control points so that the deviation from the circular arc (aka the shape "thickness") is the value I want.
Constraints:
The shape must be able to start and end at an arbitrary angle (within 180° of each other)
The "thickness" of the shape (deviation from the circle) must start and end with the given values
The "thickness" must increase monotonically (it can't get bigger and then smaller again)
It has to look smooth to the eye, there can't be any sharp bends
I am open to other solutions as well.
My approach just constructs 2 circular arcs and fills the region in between. The tricky bit is figuring out the centers and radii of these arcs. Looks quite good provided the thicknesses are not too large. (Cut and paste and decide for yourself if it meet your needs.) Could possibly be improved by use of a clipping path.
- (void)drawRect:(CGRect)rect
{
CGContextRef context = UIGraphicsGetCurrentContext();
CGMutablePathRef path = CGPathCreateMutable();
// As appropriate for iOS, the code below assumes a coordinate system with
// the x-axis pointing to the right and the y-axis pointing down (flipped from the standard Cartesian convention).
// Therefore, 0 degrees = East, 90 degrees = South, 180 degrees = West,
// -90 degrees = 270 degrees = North (once again, flipped from the standard Cartesian convention).
CGFloat startingAngle = 90.0; // South
CGFloat endingAngle = -45.0; // North-East
BOOL weGoFromTheStartingAngleToTheEndingAngleInACounterClockwiseDirection = YES; // change this to NO if necessary
CGFloat startingThickness = 2.0;
CGFloat endingThickness = 12.0;
CGPoint center = CGPointMake(CGRectGetMidX(self.bounds), CGRectGetMidY(self.bounds));
CGFloat meanRadius = 0.9 * fminf(self.bounds.size.width / 2.0, self.bounds.size.height / 2.0);
// the parameters above should be supplied by the user
// the parameters below are derived from the parameters supplied above
CGFloat deltaAngle = fabsf(endingAngle - startingAngle);
// projectedEndingThickness is the ending thickness we would have if the two arcs
// subtended an angle of 180 degrees at their respective centers instead of deltaAngle
CGFloat projectedEndingThickness = startingThickness + (endingThickness - startingThickness) * (180.0 / deltaAngle);
CGFloat centerOffset = (projectedEndingThickness - startingThickness) / 4.0;
CGPoint centerForInnerArc = CGPointMake(center.x + centerOffset * cos(startingAngle * M_PI / 180.0),
center.y + centerOffset * sin(startingAngle * M_PI / 180.0));
CGPoint centerForOuterArc = CGPointMake(center.x - centerOffset * cos(startingAngle * M_PI / 180.0),
center.y - centerOffset * sin(startingAngle * M_PI / 180.0));
CGFloat radiusForInnerArc = meanRadius - (startingThickness + projectedEndingThickness) / 4.0;
CGFloat radiusForOuterArc = meanRadius + (startingThickness + projectedEndingThickness) / 4.0;
CGPathAddArc(path,
NULL,
centerForInnerArc.x,
centerForInnerArc.y,
radiusForInnerArc,
endingAngle * (M_PI / 180.0),
startingAngle * (M_PI / 180.0),
!weGoFromTheStartingAngleToTheEndingAngleInACounterClockwiseDirection
);
CGPathAddArc(path,
NULL,
centerForOuterArc.x,
centerForOuterArc.y,
radiusForOuterArc,
startingAngle * (M_PI / 180.0),
endingAngle * (M_PI / 180.0),
weGoFromTheStartingAngleToTheEndingAngleInACounterClockwiseDirection
);
CGContextAddPath(context, path);
CGContextSetFillColorWithColor(context, [UIColor redColor].CGColor);
CGContextFillPath(context);
CGPathRelease(path);
}
One solution could be to generate a polyline manually. This is simple but it has the disadvantage that you'd have to scale up the amount of points you generate if the control is displayed at high resolution. I don't know enough about iOS to give you iOS/ObjC sample code, but here's some python-ish pseudocode:
# lower: the starting angle
# upper: the ending angle
# radius: the radius of the circle
# we'll fill these with polar coordinates and transform later
innerSidePoints = []
outerSidePoints = []
widthStep = maxWidth / (upper - lower)
width = 0
# could use a finer step if needed
for angle in range(lower, upper):
innerSidePoints.append(angle, radius - (width / 2))
outerSidePoints.append(angle, radius + (width / 2))
width += widthStep
# now we have to flip one of the arrays and join them to make
# a continuous path. We could have built one of the arrays backwards
# from the beginning to avoid this.
outerSidePoints.reverse()
allPoints = innerSidePoints + outerSidePoints # array concatenation
xyPoints = polarToRectangular(allPoints) # if needed
A view with a spiral .. 2023
It's very easy to draw a spiral mathematically and there are plenty of examples around.
https://github.com/mabdulsubhan/UIBezierPath-Spiral/blob/master/UIBezierPath%2BSpiral.swift
Put it in a view in the obvious way:
class Example: UIView {
private lazy var spiral: CAShapeLayer = {
let s = CAShapeLayer()
s.strokeColor = UIColor.systemPurple.cgColor
s.fillColor = UIColor.clear.cgColor
s.lineWidth = 12.0
s.lineCap = .round
layer.addSublayer(s)
return s
}()
private lazy var sp: CGPath = {
let s = UIBezierPath.getSpiralPath(
center: bounds.centerOfCGRect(),
startRadius: 0,
spacePerLoop: 4,
startTheta: 0,
endTheta: CGFloat.pi * 2 * 5,
thetaStep: 10.radians)
return s.cgPath
}()
override func layoutSubviews() {
super.layoutSubviews()
clipsToBounds = true
spiral.path = sp
}
}

Resources