I am trying to implement a custom painter that can draw an image (scaled down version) on the canvas and the drawn image can be rotated and scaled.
I get to know that to scale the image I have to scale the canvas using scale method.
Now the questions is how to rotate the scaled image on its center (or any other point). The rotate method of canvas allow only to rotate on top left corner.
Here is my implementation that can be extended
Had the same problem, Solution was simply making your own rotation method in three lines
void rotate(Canvas canvas, double cx, double cy, double angle) {
canvas.translate(cx, cy);
canvas.rotate(angle);
canvas.translate(-cx, -cy);
}
We thus first move the canvas towards the point you want to pivot around. We then rotate along the the topleft (default for Flutter) which in coordinate space is the pivot you want and then put the canvas back to the desired position, with the rotation applied. Method is very efficient, requiring only 4 additions for the translation and the rotation cost is identical to the original one.
This can achieve by shifting the coordinate space as illustrated in figure 1.
The translation is the difference in coordinates between C1 and C2, which are exactly as between A and B in figure 2.
With some geometry formulas, we can calculate the desired translation and produce the rotated image as in the method below
ui.Image rotatedImage({ui.Image image, double angle}) {
var pictureRecorder = ui.PictureRecorder();
Canvas canvas = Canvas(pictureRecorder);
final double r = sqrt(image.width * image.width + image.height * image.height) / 2;
final alpha = atan(image.height / image.width);
final beta = alpha + angle;
final shiftY = r * sin(beta);
final shiftX = r * cos(beta);
final translateX = image.width / 2 - shiftX;
final translateY = image.height / 2 - shiftY;
canvas.translate(translateX, translateY);
canvas.rotate(angle);
canvas.drawImage(image, Offset.zero, Paint());
return pictureRecorder.endRecording().toImage(image.width, image.height);
}
alpha, beta, angle are all in radian.
Here is the repo of the demo app
If you don't want to rotate the image around the center of the image you can use this way. You won't have to care about what the offset of the canvas should be in relation to the image rotation, because the canvas is moved back to its original position after the image is drawn.
void rotate(Canvas c, Image image, Offset focalPoint, Size screenSize, double angle) {
c.save();
c.translate(screenSize.width/2, screenSize.height/2);
c.rotate(angle);
// To rotate around the center of the image, focal point is the
// image width and height divided by 2
c.drawImage(image, focalPoint*-1, Paint());
c.translate(-screenSize.width/2, -screenSize.height/2);
c.restore();
}
I have an angle that I am calculating based on the positioning of a view from the centre of the screen. I need a way to move the view from it's current position, off the screen in the direction of the angle.
I'm sure there is a fairly simple way of calculating a new x and y value, but I haven't been able to figure out the maths. I want to do it using an animation, but I can figure that out myself once I have the coordinates.
Anyone have any suggestions?
If you have angle you can calculate new coordinates by getting sine and cosine values. You can try out following code
let pathLength = 50 as Double // total distance view should move
let piFactor = M_PI / 180
let angle = 90 as Double // direction in which you need to move it
let xCoord = outView.frame.origin.x + CGFloat(pathLength * sin(piFactor*angle)) //outView is name of view you want to animate
let yCoord = outView.frame.origin.y + CGFloat(pathLength * cos(piFactor*angle))
UIView.animateWithDuration(1, delay: 0, options: UIViewAnimationOptions.CurveEaseInOut, animations: { () -> Void in
self.outView.frame = CGRectMake(xCoord, yCoord, self.outView.frame.size.width, self.outView.frame.size.height)
}, completion: { (Bool) -> Void in
})
To me it sounds what you need to do is convert a vector from polar representation (angle and radius) to cartesian representation (x and y coordinates) which should be fairly easy.
You already got the angle so you only need to get the radius, which is the length of the vector. In you case (if I understand it correctly) is the distance from the current center of the view that needs to be animated to it's new position. While it may be complex to know that exactly (cause this part of what you are trying to calculate) you can go on the safe side and take a large enough value that will surely throw the view out of its super view frame. The length of the superview diagonal plus the length of the animated view diagonal should do the work, or even more simple just take the sum of the height and width of both views.
Once you have the complete polar representation of the vector (angle and radius) you can use that simple formula to convert to cartesian representation (x = r * cos(a), y = r * sin(a)) and finally add that vector coordinates to the center of the view you need to animate.
I just started learning metal and can best show you my frustration with the following series of screenshots. From top to bottom we have
(1) My model where the model matrix is the identity matrix
(2) My model rotated 60 deg about the x axis with orthogonal projection
(3) My model rotated 60 deg about the y axis with orthogonal projection
(4) My model rotated 60 deg about the z axis
So I use the following function for conversion into normalized device coordinates:
- (CGPoint)normalizedDevicePointForViewPoint:(CGPoint)point
{
CGPoint p = [self convertPoint:point toCoordinateSpace:self.window.screen.fixedCoordinateSpace];
CGFloat halfWidth = CGRectGetMidX(self.window.screen.bounds);
CGFloat halfHeight = CGRectGetMidY(self.window.screen.bounds);
CGFloat px = ( p.x - halfWidth ) / halfWidth;
CGFloat py = ( p.y - halfHeight ) / halfHeight;
return CGPointMake(px, -py);
}
The following rotates and orthogonally projects the model:
- (matrix_float4x4)zRotation
{
self.rotationZ = M_PI / 3;
const vector_float3 zAxis = { 0, 0, 1 };
const matrix_float4x4 zRot = matrix_float4x4_rotation(zAxis, self.rotationZ);
const matrix_float4x4 modelMatrix = zRot;
return matrix_multiply( matrix_float4x4_orthogonal_projection_on_z_plane(), modelMatrix );
}
As you can see when I use the exact same method for rotating about the other two axes, it looks fine-not distorted. What am I doing wrong? Is there some sort of scaling/aspect ratio thing I should be setting somewhere? What things could it be? I've been staring at this for an embarrassingly long period of time so any help/ideas that can lead me in the right direction are much appreciated. Thank you in advance.
There's nothing wrong with your rotation or projection matrices. The visual oddity arises from the fact that you move your vertices into NDC space prior to rotation. A rectangle doesn't preserve its aspect ratio when rotating in NDC space, because the mapping from NDC back to screen coordinates is not 1:1.
I would recommend not working in NDC until the very end of the vertex pipeline (i.e., pass vertices into your vertex function in "world" space, and out to the rasterizer as NDC). You can do this with a classic construction of the orthographic projection matrix that scales and biases the vertices, correctly accounting for the non-square aspect ratio of window coordinates.
i'm trying to build a game in xna, i got a circle which i want the player to move around it, as you can see in the following picture, its working great except the drawing part which i'm not pleased with
here's a link to an image http://s12.postimage.org/poiip0gtp/circle.png
i want to center the player object to the edge of the circle so it won't look like the player is standing on air
this is how i calculate the position of the player
rad = (degree * Math.PI / 180);
rotationDegree = (float)((Math.PI * degree) / 180);
currentPosition.X = (float)(Math.Cos(rad) * Earth.radius + (GraphicsDevice.Viewport.Width / 2));
currentPosition.Y = (float)(Math.Sin(rad) * Earth.radius + (GraphicsDevice.Viewport.Height / 2));
and this is how i draw the player
spriteBatch.Draw(texture,currentPosition, null, Color.White,rotationDegree, Vector2.Zero,1f,SpriteEffects.None, 1f);
thank you.
Use the origin overload for spritebatch. Which is where the sprite is drawn according to the position.
Spritebatch.Draw(texture,Position, null,Color.White,0f,new Vector2(texture.Width / 2,texture.Height /2),1f,SpriteEffects.None, 0);
Using texture.Width / 2,texture.Height /2 for origin will center it.
It looks like what you want to do here is adjust the sprite's origin, which is the vector that you're passing into SpriteBatch.Draw(). This is used to determine the "center point" of your sprite; {0, 0} represents the sprite's upper-left corner, while {spriteWidth, spriteHeight} represents the bottom-right corner. Your sprite will be positioned and rotated relative to this origin.
Given two rectangles with x, y, width, height in pixels and a rotation value in degrees -- how do I calculate the closest distance of their outlines toward each other?
Background: In a game written in Lua I'm randomly generating maps, but want to ensure certain rectangles aren't too close to each other -- this is needed because maps become unsolvable if the rectangles get into certain close-distance position, as a ball needs to pass between them. Speed isn't a huge issue as I don't have many rectangles and the map is just generated once per level. Previous links I found on StackOverflow are this and this
Many thanks in advance!
Not in Lua, a Python code based on M Katz's suggestion:
def rect_distance((x1, y1, x1b, y1b), (x2, y2, x2b, y2b)):
left = x2b < x1
right = x1b < x2
bottom = y2b < y1
top = y1b < y2
if top and left:
return dist((x1, y1b), (x2b, y2))
elif left and bottom:
return dist((x1, y1), (x2b, y2b))
elif bottom and right:
return dist((x1b, y1), (x2, y2b))
elif right and top:
return dist((x1b, y1b), (x2, y2))
elif left:
return x1 - x2b
elif right:
return x2 - x1b
elif bottom:
return y1 - y2b
elif top:
return y2 - y1b
else: # rectangles intersect
return 0.
where
dist is the euclidean distance between points
rect. 1 is formed by points (x1, y1) and (x1b, y1b)
rect. 2 is formed by points (x2, y2) and (x2b, y2b)
Edit: As OK points out, this solution assumes all the rectangles are upright. To make it work for rotated rectangles as the OP asks you'd also have to compute the distance from the corners of each rectangle to the closest side of the other rectangle. But you can avoid doing that computation in most cases if the point is above or below both end points of the line segment, and to the left or right of both line segments (in telephone positions 1, 3, 7, or 9 with respect to the line segment).
Agnius's answer relies on a DistanceBetweenLineSegments() function. Here is a case analysis that does not:
(1) Check if the rects intersect. If so, the distance between them is 0.
(2) If not, think of r2 as the center of a telephone key pad, #5.
(3) r1 may be fully in one of the extreme quadrants (#1, #3, #7, or #9). If so
the distance is the distance from one rect corner to another (e.g., if r1 is
in quadrant #1, the distance is the distance from the lower-right corner of
r1 to the upper-left corner of r2).
(4) Otherwise r1 is to the left, right, above, or below r2 and the distance is
the distance between the relevant sides (e.g., if r1 is above, the distance
is the distance between r1's low y and r2's high y).
Actually there is a fast mathematical solution.
Length(Max((0, 0), Abs(Center - otherCenter) - (Extent + otherExtent)))
Where Center = ((Maximum - Minimum) / 2) + Minimum and Extent = (Maximum - Minimum) / 2.
Basically the code above zero's axis which are overlapping and therefore the distance is always correct.
It's preferable to keep the rectangle in this format as it's preferable in many situations ( a.e. rotations are much easier ).
Pseudo-code:
distance_between_rectangles = some_scary_big_number;
For each edge1 in Rectangle1:
For each edge2 in Rectangle2:
distance = calculate shortest distance between edge1 and edge2
if (distance < distance_between_rectangles)
distance_between_rectangles = distance
There are many algorithms to solve this and Agnius algorithm works fine. However I prefer the below since it seems more intuitive (you can do it on a piece of paper) and they don't rely on finding the smallest distance between lines but rather the distance between a point and a line.
The hard part is implementing the mathematical functions to find the distance between a line and a point, and to find if a point is facing a line. You can solve all this with simple trigonometry though. I have below the methodologies to do this.
For polygons (triangles, rectangles, hexagons, etc.) in arbitrary angles
If polygons overlap, return 0
Draw a line between the centres of the two polygons.
Choose the intersecting edge from each polygon. (Here we reduce the problem)
Find the smallest distance from these two edges. (You could just loop through each 4 points and look for the smallest distance to the edge of the other shape).
These algorithms work as long as any two edges of the shape don't create angles more than 180 degrees. The reason is that if something is above 180 degrees then it means that the some corners are inflated inside, like in a star.
Smallest distance between an edge and a point
If point is not facing the face, then return the smallest of the two distances between the point and the edge cornerns.
Draw a triangle from the three points (edge's points plus the solo point).
We can easily get the distances between the three drawn lines with Pythagorean Theorem.
Get the area of the triangle with Heron's formula.
Calculate the height now with Area = 12⋅base⋅height with base being the edge's length.
Check to see if a point faces an edge
As before you make a triangle from an edge and a point. Now using the Cosine law you can find all the angles with just knowing the edge distances. As long as each angle from the edge to the point is below 90 degrees, the point is facing the edge.
I have an implementation in Python for all this here if you are interested.
This question depends on what kind of distance. Do you want, distance of centers, distance of edges or distance of closest corners?
I assume you mean the last one. If the X and Y values indicate the center of the rectangle then you can find each the corners by applying this trick
//Pseudo code
Vector2 BottomLeftCorner = new Vector2(width / 2, heigth / 2);
BottomLeftCorner = BottomLeftCorner * Matrix.CreateRotation(MathHelper.ToRadians(degrees));
//If LUA has no built in Vector/Matrix calculus search for "rotate Vector" on the web.
//this helps: http://www.kirupa.com/forum/archive/index.php/t-12181.html
BottomLeftCorner += new Vector2(X, Y); //add the origin so that we have to world position.
Do this for all corners of all rectangles, then just loop over all corners and calculate the distance (just abs(v1 - v2)).
I hope this helps you
I just wrote the code for that in n-dimensions. I couldn't find a general solution easily.
// considering a rectangle object that contains two points (min and max)
double distance(const rectangle& a, const rectangle& b) const {
// whatever type you are using for points
point_type closest_point;
for (size_t i = 0; i < b.dimensions(); ++i) {
closest_point[i] = b.min[i] > a.min[i] ? a.max[i] : a.min[i];
}
// use usual euclidian distance here
return distance(a, closest_point);
}
For calculating the distance between a rectangle and a point you can:
double distance(const rectangle& a, const point_type& p) const {
double dist = 0.0;
for (size_t i = 0; i < dimensions(); ++i) {
double di = std::max(std::max(a.min[i] - p[i], p[i] - a.max[i]), 0.0);
dist += di * di;
}
return sqrt(dist);
}
If you want to rotate one of the rectangles, you need to rotate the coordinate system.
If you want to rotate both rectangles, you can rotate the coordinate system for rectangle a. Then we have to change this line:
closest_point[i] = b.min[i] > a.min[i] ? a.max[i] : a.min[i];
because this considers there is only one candidate as the closest vertex in b. You have to change it to check the distance to all vertexes in b. It's always one of the vertexes.
See: https://i.stack.imgur.com/EKJmr.png
My approach to solving the problem:
Combine the two rectangles into one large rectangle
Subtract from the large rectangle the first rectangle and the second
rectangle
What is left after the subtraction is a rectangle between the two
rectangles, the diagonal of this rectangle is the distance between
the two rectangles.
Here is an example in C#
public static double GetRectDistance(this System.Drawing.Rectangle rect1, System.Drawing.Rectangle rect2)
{
if (rect1.IntersectsWith(rect2))
{
return 0;
}
var rectUnion = System.Drawing.Rectangle.Union(rect1, rect2);
rectUnion.Width -= rect1.Width + rect2.Width;
rectUnion.Width = Math.Max(0, rectUnion.Width);
rectUnion.Height -= rect1.Height + rect2.Height;
rectUnion.Height = Math.Max(0, rectUnion.Height);
return rectUnion.Diagonal();
}
public static double Diagonal(this System.Drawing.Rectangle rect)
{
return Math.Sqrt(rect.Height * rect.Height + rect.Width * rect.Width);
}
Please check this for Java, it has the constraint all rectangles are parallel, it returns 0 for all intersecting rectangles:
public static double findClosest(Rectangle rec1, Rectangle rec2) {
double x1, x2, y1, y2;
double w, h;
if (rec1.x > rec2.x) {
x1 = rec2.x; w = rec2.width; x2 = rec1.x;
} else {
x1 = rec1.x; w = rec1.width; x2 = rec2.x;
}
if (rec1.y > rec2.y) {
y1 = rec2.y; h = rec2.height; y2 = rec1.y;
} else {
y1 = rec1.y; h = rec1.height; y2 = rec2.y;
}
double a = Math.max(0, x2 - x1 - w);
double b = Math.max(0, y2 - y1 - h);
return Math.sqrt(a*a+b*b);
}
Another solution, which calculates a number of points on the rectangle and choses the pair with the smallest distance.
Pros: works for all polygons.
Cons: a little bit less accurate and slower.
import numpy as np
import math
POINTS_PER_LINE = 100
# get points on polygon outer lines
# format of polygons: ((x1, y1), (x2, y2), ...)
def get_points_on_polygon(poly, points_per_line=POINTS_PER_LINE):
all_res = []
for i in range(len(poly)):
a = poly[i]
if i == 0:
b = poly[-1]
else:
b = poly[i-1]
res = list(np.linspace(a, b, points_per_line))
all_res += res
return all_res
# compute minimum distance between two polygons
# format of polygons: ((x1, y1), (x2, y2), ...)
def min_poly_distance(poly1, poly2, points_per_line=POINTS_PER_LINE):
poly1_points = get_points_on_polygon(poly1, points_per_line=points_per_line)
poly2_points = get_points_on_polygon(poly2, points_per_line=points_per_line)
distance = min([math.sqrt((a[0] - b[0])**2 + (a[1] - b[1])**2) for a in poly1_points for b in poly2_points])
# slower
# distance = min([np.linalg.norm(a - b) for a in poly1_points for b in poly2_points])
return distance