I try to make a clock, in swift, but now i want to make something strange. I want make border radius settable. This is the easy part (is easy because I already did that). I drew 60 ticks around the clock. The problem is that 60 ticks are a perfect circle. If I change the border radius I obtain this clock:
All ticks are made with NSBezierPath, and code for calculate position for every tick is :
tickPath.moveToPoint(CGPoint(
x: center.x + cos(angle) * point1 ,
y: center.y + sin(angle) * point1
))
tickPath.lineToPoint(CGPoint(
x: center.x + cos(angle) * point2,
y: center.y + sin(angle) * point2
))
point1 and point2 are points for 12 clock tick.
My clock background is made with bezier path:
let bezierPath = NSBezierPath(roundedRect:self.bounds, xRadius:currentRadius, yRadius:currentRadius)
currentRadius - is a settable var , so my background cam be, from a perfect circle (when corner radius = height / 2) to a square (when corner radius = 0 ).
Is any formula to calculate position for every tick so, for any border radius , in the end all ticks to be at same distance to border ?
The maths is rather complicated to explain without recourse to graphics diagrams, but basically if you consider a polar coordinates approach with the origin at the clock centre then there are two cases:
where the spoke from the origin hits the straight side of the square - easy by trigonometry
where it hits the circle arc at the corner - we use the cosine rule to solve the triangle formed by the centre of the clock, the centre of the corner circle and the point where the spoke crosses the corner. The origin-wards angle of that triangle is 45º - angleOfSpoke, and two of the sides are of known length. Solve the cosine equation as a quadratic and you have it.
This function does it:
func radiusAtAngle(angleOfSpoke: Double, radius: Double, cornerRadius: Double) -> Double {
// radius is the half-width of the square, = the full radius of the circle
// cornerRadius is, of course, the corner radius.
// angleOfSpoke is the (maths convention) angle of the spoke
// the function returns the radius of the spoke.
let theta = atan((radius - cornerRadius) / radius) // This determines which case
let modAngle = angleOfSpoke % M_PI_2 // By symmetry we need only consider the first quadrant
if modAngle <= theta { // it's on the vertical flat
return radius / cos(modAngle)
} else if modAngle > M_PI_2 - theta { // it's on the horizontal flat
return radius / cos(M_PI_2 - modAngle)
} else { // it's on the corner arc
// We are using the cosine rule to solve the triangle formed by
// the clock centre, the curved corner's centre,
// and the point of intersection of the spoke.
// Then use quadratic solution to solve for the radius.
let diagonal = hypot(radius - cornerRadius, radius - cornerRadius)
let rcosa = diagonal * cos(M_PI_4 - modAngle)
let sqrTerm = rcosa * rcosa - diagonal * diagonal + cornerRadius * cornerRadius
if sqrTerm < 0.0 {
println("Aaargh - Negative term") // Doesn't happen - use assert in production
return 0.0
} else {
return rcosa + sqrt(sqrTerm) // larger of the two solutions
}
}
}
In the diagram OP = diagonal, OA = radius, PS = PB = cornerRadius, OS = function return, BÔX = theta, SÔX = angleOfSpoke
Related
I have a problem and although I serached everywhere I couldn't find a solution.
I have a stacked sprite and I'm rotating this sprite around the center of the screen. So I iterate over a list of sprites (stacked) and increase the y-coordinate by 2 every loop (rotation is increased step by step by 0.01f outside of the loop):
foreach(var s in stacked)
{
Vector2 origin = new Vector2(Basic.width / 2, Basic.height / 2);
Rectangle newPosition = new Rectangle(position.X, position.Y - y, position.Width, position.Height);
float angle = 0f;
Matrix transform = Matrix.CreateTranslation(-origin.X, -origin.Y, 0f) *
Matrix.CreateRotationZ(rotation) *
Matrix.CreateTranslation(origin.X, origin.Y, 0f);
Vector2 pos = new Vector2(newPosition.X, newPosition.Y);
pos = Vector2.Transform(pos, transform);
newPosition.X = (int)pos.X;
newPosition.Y = (int)pos.Y;
angle += rotation;
s.Draw(newPosition, origin, angle, Color.White);
y += 2;
}
This works fine. But now my problem. I want not only to rotate the sprite around the center of the screen but also around itself. How to achieve this? I can only set one origin and one rotation per Draw. I would like to rotate the sprite around the origin 'Basic.width / 2, Basic.height / 2' and while it rotates, around 'position.Width / 2, position.Height / 2'. With different rotation speed each. How is this possible?
Thank you in advance!
Just to be clear:
When using SpriteBatch.Draw() with origin and angle, there is only one rotation: the final angle of the sprite.
The other rotations are positional offsets.
The origin in the Draw() call is a translation, rotation, translate back. Your transform matrix shows this quite well:
Matrix transform = Matrix.CreateTranslation(-origin.X, -origin.Y, 0f) *
Matrix.CreateRotationZ(rotation) *
Matrix.CreateTranslation(origin.X, origin.Y, 0f);
//Class level variables:
float ScreenRotation, ScreenRotationSpeed;
float ObjectRotation, ObjectRotationSpeed;
Vector2 ScreenOrigin, SpriteOrigin;
// ...
// In constructor and resize events:
ScreenOrigin = new Vector2(Basic.width <<1, Basic.height <<1);
// shifts are faster for `int` type. If "Basic.width" is `float`:
//ScreenOrigin = new Vector2(Basic.width, Basic.height) * 0.5f;
// In Update():
ScreenRotation += ScreenRotationSpeed; // * gameTime.ElapsedGameTime.Seconds; // for FPS invariant speed where speed = 60 * single frame speed
ObjectRotation+= ObjectRotationSpeed;
//Calculate the screen center rotation once per step
Matrix baseTransform = Matrix.CreateTranslation(-ScreenOrigin.X, -ScreenOrigin.Y, 0f) *
Matrix.CreateRotationZ(ScreenRotation) *
Matrix.CreateTranslation(ScreenOrigin.X, ScreenOrigin.Y, 0f);
// In Draw() at the start of your code snippet posted:
// moved outside of the loop for a translationally invariant vertical y interpretation
// or move it inside the loop and apply -y to position.Y for an elliptical effect
Vector2 ObjectOrigin = new Vector2(position.X, position.Y);
Matrix transform = baseTransform *
Matrix.CreateTranslation(-ObjectOrigin.X, -ObjectOrigin.Y, 0f) *
Matrix.CreateRotationZ(ObjectRotation) *
Matrix.CreateTranslation(ObjectOrigin.X, ObjectOrigin.Y, 0f);
foreach(var s in stacked)
{
Vector2 pos = new Vector2(ObjectOrigin.X, ObjectOrigin.Y - y);
pos = Vector2.Transform(pos, transform);
float DrawAngle = ObjectRotation;
// or float DrawAngle = ScreenRotation;
// or float DrawAngle = ScreenRotation + ObjectRotation;
// or float DrawAngle = 0;
s.Draw(pos, SpriteOrigin, DrawAngle, Color.White);
}
I suggest moving the Draw() parameter away from destinationRectangle and use the Vector2 position directly with scaling. Rotations within square rectangles can differ up to SQRT(2) in aspect ratio, i.e. stretching/squashing. Using Vector2 incurs a cost of higher collision complexity.
I am sorry for the ors, but without complete knowledge of the problem...YMMV
In my 2D projects, I use the vector form of polar coordinates.
The Matrix class requires more calculations than the polar equivalents in 2D. Matrix operates in 3D, wasting cycles calculating Z components.
With normalized direction vectors (cos t,sin t) and a radius(vector length),in many cases I use Vector2.LengthSquared() to avoid the square root when possible.
The only time I have used Matrices in 2D is display projection matrix(entire SpriteBatch) and Mouse and TouchScreen input deprojection(times the inverse of the projection matrix)
I have a program that creates pixel-based gradients (meaning it calculates the step in the gradient for each pixel, then calculates the colour at that step, then gives the pixel that colour).
I'd like to implement spiral gradients (such as below).
My program can create conic gradients (as below), where each pixel is assigned a step in the gradient according to the angle between it and the midpoint (effectively mapping the midpoint-pixel angle [0...2PI] to [0...1]).
It would seem to me that a spiral gradient is a conic gradient with some additional function applied to it, where the gradient step for a given pixel depends not only on the angle, but on some additional non-linear function applied to the euclidean distance between the midpoint and pixel.
I envisage that a solution would take the original (x, y) pixel coordinate and displace it by some amounts in the x and y axes resulting in a new coordinate (x2, y2). Then, for each pixel, I'd simply calculate the angle between the midPoint and its new displaced coordinate (x2, y2) and use this angle as the gradient step for that pixel. But it's this displacement function that I need help with... of course, there may be other, better ways.
Below is a simple white-to-black conic gradient. I show how I imagine the displacement would work, but its the specifics about this function (the non-linearity), that I'm unable to implement.
My code for conic gradient:
public void conicGradient(Gradient gradient, PVector midPoint, float angle) {
float rise, run;
double t = 0;
for (int y = 0, x; y < imageHeight; ++y) {
rise = midPoint.y - y;
run = midPoint.x;
for (x = 0; x < imageWidth; ++x) {
t = Functions.fastAtan2(rise, run) + Math.PI - angle;
// Ensure a positive value if angle is negative.
t = Functions.floorMod(t, PConstants.TWO_PI);
// Divide by TWO_PI to get value in range 0...1
step = t *= INV_TWO_PI;
pixels[imageWidth * y + x] = gradient.ColorAt(step); // pixels is 1D pixel array
run -= 1;
}
}
}
By eye, it looks like after t = ... fastAtan2..., you just need:
t += PConstants.TWO_PI * Math.sqrt( (rise*rise + run*run) / (imageWidth * imageWidth + imageHeight * imageHeight) )
This just adds the distance from the center to the angle, with appropriate scaling.
I'm using processing, and I'm trying to create a circle from the pixels i have on my display.
I managed to pull the pixels on screen and create a growing circle from them.
However i'm looking for something much more sophisticated, I want to make it seem as if the pixels on the display are moving from their current location and forming a turning circle or something like this.
This is what i have for now:
int c = 0;
int radius = 30;
allPixels = removeBlackP();
void draw {
loadPixels();
for (int alpha = 0; alpha < 360; alpha++)
{
float xf = 350 + radius*cos(alpha);
float yf = 350 + radius*sin(alpha);
int x = (int) xf;
int y = (int) yf;
if (radius > 200) {radius =30;break;}
if (c> allPixels.length) {c= 0;}
pixels[y*700 +x] = allPixels[c];
updatePixels();
}
radius++;
c++;
}
the function removeBlackP return an array with all the pixels except for the black ones.
This code works for me. There is an issue that the circle only has the numbers as int so it seems like some pixels inside the circle won't fill, i can live with that. I'm looking for something a bit more complex like I explained.
Thanks!
Fill all pixels of scanlines belonging to the circle. Using this approach, you will paint all places inside the circle. For every line calculate start coordinate (end one is symmetric). Pseudocode:
for y = center_y - radius; y <= center_y + radius; y++
dx = Sqrt(radius * radius - y * y)
for x = center_x - dx; x <= center_x + dx; x++
fill a[y, x]
When you find places for all pixels, you can make correlation between initial pixels places and calculated ones and move them step-by-step.
For example, if initial coordinates relative to center point for k-th pixel are (x0, y0) and final coordinates are (x1,y1), and you want to make M steps, moving pixel by spiral, calculate intermediate coordinates:
calc values once:
r0 = Sqrt(x0*x0 + y0*y0) //Math.Hypot if available
r1 = Sqrt(x1*x1 + y1*y1)
fi0 = Math.Atan2(y0, x0)
fi1 = Math.Atan2(y1, x1)
if fi1 < fi0 then
fi1 = fi1 + 2 * Pi;
for i = 1; i <=M ; i++
x = (r0 + i / M * (r1 - r0)) * Cos(fi0 + i / M * (fi1 - fi0))
y = (r0 + i / M * (r1 - r0)) * Sin(fi0 + i / M * (fi1 - fi0))
shift by center coordinates
The way you go about drawing circles in Processing looks a little convoluted.
The simplest way is to use the ellipse() function, no pixels involved though:
If you do need to draw an ellipse and use pixels, you can make use of PGraphics which is similar to using a separate buffer/"layer" to draw into using Processing drawing commands but it also has pixels[] you can access.
Let's say you want to draw a low-res pixel circle circle, you can create a small PGraphics, disable smoothing, draw the circle, then render the circle at a higher resolution. The only catch is these drawing commands must be placed within beginDraw()/endDraw() calls:
PGraphics buffer;
void setup(){
//disable sketch's aliasing
noSmooth();
buffer = createGraphics(25,25);
buffer.beginDraw();
//disable buffer's aliasing
buffer.noSmooth();
buffer.noFill();
buffer.stroke(255);
buffer.endDraw();
}
void draw(){
background(255);
//draw small circle
float circleSize = map(sin(frameCount * .01),-1.0,1.0,0.0,20.0);
buffer.beginDraw();
buffer.background(0);
buffer.ellipse(buffer.width / 2,buffer.height / 2, circleSize,circleSize);
buffer.endDraw();
//render small circle at higher resolution (blocky - no aliasing)
image(buffer,0,0,width,height);
}
If you want to manually draw a circle using pixels[] you are on the right using the polar to cartesian conversion formula (x = cos(angle) * radius, y = sin(angle) * radius).Even though it's focusing on drawing a radial gradient, you can find an example of drawing a circle(a lot actually) using pixels in this answer
I'm trying to draw a (svg) arc connecting two rectangles. The catch is, that the arc should start at the border of the rectangles, not at the center.
To illustrate:
I have the center points, width & height of the rectangles C1 w1 h1 C2 w2 h2 and the center and x and y radius of the arc D rx ry. So basically, for drawing the purple arc, i'm missing P1 and P2.
All values are dynamic and can change, so the algorithm needs be agnostic of rx and ry, width and height of the rectangles, how the rectangles are positioned relatively to each other, etc.
Taking the rounded corners into account would be the cherry on top. But that's not really necessary..
Let's center of ellipse be coordinate origin (if not, just shift rectangle coordinates by -D.X and -D.Y).
In this system ellipse equation is
x^2/rx^2 + y^2/ry^2 = 1
Substitute rectangle edge coordinates in this equation and check if result actually belongs to rectangle.
For example, right edge of top rectangle is X = C1'.X + w1. Find Y from ellipse equation and check it is in range C1'Y - h1 .. C1'Y + h1. If yes, P1 = (C1'.X + w1, CalculatedY)
Alright, just for people who might stumble upon this in the future..
Here is what i came up with in javascript (ES6). But it should be easy to port this to other languages..
/**
* Calculate the intersection of the border of the given rectangle
* and a circular arc.
* #param {Object} rectangle Rectangle object with dimensions and position properties.
* #param {Object} arc Arc object with center and radius properties.
* #return {Object} Resulting intersection point, with x- and y-coordinates.
*/
calculateBorderPoint(rectangle, arc) {
// Clone the rectangle, because we don't want to mutate the original.
const a = Object.assign({}, rectangle.position);
// Treat center of circle as coordinate origin.
a.x -= arc.center.x;
a.y -= arc.center.y;
let points = [];
// Check east & west with possible x values
const possibleX = [
a.x - rectangle.dimensions.width / 2,
a.x + rectangle.dimensions.width / 2,
];
possibleX.forEach((x) => {
const ySquared = [
Math.sqrt(Math.pow(arc.radius, 2) - Math.pow(x, 2)),
-Math.sqrt(Math.pow(arc.radius, 2) - Math.pow(x, 2)),
];
// Check if the derived y value is in range of rectangle
ySquared.forEach((y) => {
if (y >= a.y - rectangle.dimensions.height / 2 &&
y <= a.y + rectangle.dimensions.height / 2) {
points.push({x, y});
}
});
});
// Check north & south with possible y values
const possibleY = [
a.y - rectangle.dimensions.height / 2,
a.y + rectangle.dimensions.height / 2,
];
possibleY.forEach((y) => {
const xSquared = [
Math.sqrt(Math.pow(arc.radius, 2) - Math.pow(y, 2)),
-Math.sqrt(Math.pow(arc.radius, 2) - Math.pow(y, 2)),
];
// Check if the derived x value is in range of rectangle
xSquared.forEach((x) => {
if (x >= a.x - rectangle.dimensions.width / 2 &&
x <= a.x + rectangle.dimensions.width / 2) {
points.push({x, y});
}
});
});
// At this point you will propably have multiple possible solutions,
// because the circle might intersect the rectangle at multiple points.
// You need to select the point, that makes most sense in your case.
// One solution would be to select the one closest to the other rectangle.
// Translate it back.
points[0].x += arc.center.x;
points[0].y += arc.center.y;
return points[0];
}
It's not pretty, but it works. I'm happy to hear any suggestions..
video game link
I'm trying to make a game (see link above) , and I need to have the stick rotate around himself to maintain the orientation face to center of the circle.
this is how I declare the Sprite, and how I move it around the circle:
declaration:
line = new Sprite(new Texture(Gdx.files.internal("drawable/blockLine.png")));
line.setSize(140, 20);
lineX = Gdx.graphics.getWidth()/2 - line.getWidth()/2;
lineY = (Gdx.graphics.getHeight()/2 - line.getHeight()/2) + circle.getHeight()/2;
movement:
Point point = rotatePoint(new Point(lineX, lineY), new Point(Gdx.graphics.getWidth()/2, Gdx.graphics.getHeight()/2), angle+= Gdx.graphics.getDeltaTime() * lineSpeed);
line.setPosition(point.x, point.y);
rotatePoint function:
Point rotatePoint(Point point, Point center, double angle){
angle = (angle ) * (Math.PI/180); // Convert to radians
float rotatedX = (int) (Math.cos(angle) * (point.x - center.x) - Math.sin(angle) * (point.y-center.y) + center.x);
float rotatedY = (int) (Math.sin(angle) * (point.x - center.x) + Math.cos(angle) * (point.y - center.y) + center.y);
return new Point(rotatedX,rotatedY);
}
Any sugestions ?
I can't test right now but I think the rotation of the line should simply be:
Math.atan2(rotatedPoint.getOriginX() - middlePoint.getOriginX(), rotatedPoint.getOriginY() - middlePoint.getOriginY()));
Then you'll have to adjust rad to degrees or whatever you'll use. Tell me if it doesn't work!
I would take a different approach, I just created a method that places n Buttons around a click on the screen. I am using something that looks like this:
float rotation; // in degree's
float distance; //Distance from origin (radius of circle).
vector2 originOfRotation; //Center of circle
vector2 originOfSprite; //Origin of rotation sprite we are calculating
Vector2 direction = new vector2(0, 1); //pointing up
//rotate the direction
direction.rotate(rotation);
// add distance based of the direction. Warning: originOfRotation will change because of chaining method.
// use originOfRotation.cpy() if you do not want to init each frame
originOfSprite = originOfRotation.add(direction.scl(distance));
Now you have the position of your sprite. You need to increment rotation by x each frame to have it rotate. If you want the orientation of the sprite to change you can use the direction vector, probably rotated by 180 again. Efficiency wise I'm not sure what the difference would be.