From my experiments, the angle returned by RotatedRect's angle variable goes from -90 to 0 degrees, which is not sufficient to determine if the object is leaned to the left or right.
For example, if the angle is -45 degrees, we cannot say if we need to rotate +45 or -45 degrees to deskew it.
An excerpt of the code I'm using:
RotatedRect rotated_rect = minAreaRect(contour);
float blob_angle_deg = rotated_rect.angle;
Mat mapMatrix = getRotationMatrix2D(center, blob_angle_deg, 1.0);
Leaning the object in one direction I get angles from 0 to -90 degrees, while leaning the object to the other direction I get angles from -90 to 0 degrees.
How can I find the angle by which I should rotate my image to deskew it?
After learning from Sebastian Schmitz and Michael Burdinov answers this is how I solved it:
RotatedRect rotated_rect = minAreaRect(contour);
float blob_angle_deg = rotated_rect.angle;
if (rotated_rect.size.width < rotated_rect.size.height) {
blob_angle_deg = 90 + blob_angle_deg;
}
Mat mapMatrix = getRotationMatrix2D(center, blob_angle_deg, 1.0);
So, in fact, RotatedRect's angle does not provide enough information for knowing an object's angle, you must also use RotatedRect's size.width and size.height.
I explained how you can convert the angle of the rectangle into [0-180] in this thread.
The Angle is always calculated along the longer side.
Switching values of width and height of rectangle is the same as rotating it by 90 degrees. So if the range of angles was 180 degrees instead of 90 than same rectangle would have 2 representations (width, height, angle) and (height, width, angle+90). Having range of 90 degrees you can represent every rectangle and you can do that in only one way.
This is what I use (c is my contour). Basically, I get the longest line's du and dv, and then use atan2()
rect = cv2.minAreaRect(c)
box = cv2.boxPoints(rect)
origin = box[0]
rect_width, rect_height = rect[1]
if rect_width > rect_height:
target = box[3]
else:
target = box[1]
dv = target[1] - origin[1]
du = target[0] - origin[0]
angle_rads = math.atan2(dv, du)
Related
I need to define a rotated rectangle from its 4 corners. The rotated rectangle is defined by a center point, a size couple (width, height), and an angle.
How is it decided which size is the height, and which one is the width?
The width is not the length of the most horizontal edge, is it? E.g. if the angle is bigger than 90°, does it swap?
height should be the largest side, width is the other one, and angle is the rotation angle (in degrees) in a clockwise direction.
Otherwise, you can get an equivalent rectangle with height and width swapped, rotated by 90 degrees.
You can use minAreaRect to find a RotatedRect:
vector<Point> pts = {pt1, pt2, pt3, pt4}
RotatedRect box = minAreaRect(pts);
// Be sure that largest side is the height
if (box.size.width > box.size.height)
{
swap(box.size.width, box.size.height);
box.angle += 90.f;
}
Ok, with Miki's help, and with some tests, I got it clearer...
It seems that the rotated rectangle is an upright rectangle (width and height are clearly defined, then)... that is rotated!
In image coords, y is directed to the bottom, the angle is given clockwise. In usual math coords (y to the top), the angle is given counter-clockwise. Then, it fits with c++ <math.h> included atan2(y,x) function for example (except that it returns radians).
Then, to summarize, if we consider one given edge of the rectangle (two corners), its length can be considered as the width if we retrieve the angle with atan2 on its y difference and x difference. Something like:
Point pt1, pt2, pt3, pt4;
RotatedRect rect;
rect.center = (pt1 + pt2 + pt3 + pt4)/4;
// assuming the points are already sorted
rect.size.width = distance(pt1, pt2); // sqrt(...)
rect.size.height = distance(pt2, pt3);
rect.angle = atan2(pt2.y-pt1.y, pt2.x-pt1.x);
and this can be improved with width being the mean value of dist(pt1,pt2) and dist(pt3,pt4) for example. The same for height.
angle can also be calculated as being the mean value of atan for (pt1, pt2) and atan for (pt3, pt4).
I have 5 subviews(White) added to the superview(Gray), when I rotate the superview I want to know the angle(like 1 and 2) of each of the subview with the red circle.(the center of the subviews and the red circle are ON the same circle)
Start Position:
Rotated Position:
From your comment you appear to want to determine the coordinates of the centres of your five circles for a given rotation. The centres will all lie on a circle. So your question boils down to what are the coordinates of a point on a circle of radius r for an angle θ. The parametric equations for a circle give you that:
x = r cos θ
y = r sin θ
The angle, θ, in these equations is measured in radians from the positive x-axis in an anti-clockwise direction. If your angle are in degrees you will find the M_PI constant for π useful as:
360 degrees = 2 π radians
The rest is simple math, take your angle of rotation to give you the angle for A (remembering to adjust for 0 being the x-axis and measuring anti-clockwise if needed), the other centres are multiples of 72 degrees (0.4 π radians) from this.
HTH
I'm not sure I completely understand your question, but if you just need to take a known point and rotate it a certain number of degrees, check out the docs for CGAffineTransform.
For example:
CGAffineTransform rotation = CGAffineTransformMakeRotation (angle);
CGPoint rotatedPoint = CGPointApplyAffineTransform (startingPoint, rotation);
This rotation matrix is around (0, 0) and the angle is in radians, so you will need to subtract the center of your superview's bounds to get an offset relative to the center, do the rotation, and add back in the center. Or you can build an affine transform made up of that translation, rotation, and inverse translation, and then apply that to your starting point as above.
Given that you already seem to know the main rotation angle, this will give you the angles in the range -180 .. +180 and positions of each of the white discs:
GCFloat toRads = M_PI / 180.0;
CGFloat angleA = self.rotationInDegrees;
if (angleA > 180) angleA -= 360;
CGFloat xA = self.radius * sinf(angleA * toRads);
CGFloat yA = self.radius * cosf(angleA * toRads);
CGFloat angleB = angleA + 72;
if (angleB > 180) angleB -= 360;
CGFloat xB = self.radius * sinf(angleB * toRads);
CGFloat yB = self.radius * cosf(angleB * toRads);
etc...
(This assumes your zero degrees is from the vertical. If it's from the horizontal swap cos and sin over).
I want to give Force to my SKSpriteNode at specific angle.
So, How to treat my CGVector for give force at specific angle?
I had searched for it but unfortunately not getting any good way.
What i wants to achieve :
My SKSpriteNode moving towards the screen. There are buttons on top like 30,45,60.
So if user press button(i.e. that Button contain "30") then i had to move my SKSpriteNode to 30 degree with same speed.
Please help me towards it if any of you can help me regarding this.
First, you will need to convert the angle in degrees to radians by multiplying it by pi / 180
CGFloat angleInRadians = angleInDegrees * M_PI / 180;
You can then determine the vector components in that direction by
CGFloat dx = cosf(angleInRadians);
CGFloat dy = sinf(angleInRadians);
and finally apply a force to the sprite with
[sprite.physicsBody applyForce:CGVectorMake(dx*scale, dy*scale)];
where scale determines how much force is applied.
Optionally, you can rotate the sprite to face in the same direction as its motion by
sprite.zRotation = angleInRadians + offset;
where offset is the difference in angle, in radians, between your sprite's image and zero degrees. For example, if your sprite is facing up when zRotation is zero, offset should be -M_PI_2.
This is so much an iOS question as it is my current inability to do coordinate geometry. Given a CGPoint to act as a point that the line will pass through and an angle in radians. How do I draw a line that extends across to the bounds of the screen (infinite line)?
I am using Quartz2d to do this and the API for creating a line is limited to two points as input. So how do I convert a point and angle to two points on the bounds of the iOS device?
This begins with simple trigonometry. You need to calculate the x and y coordinate of the 2nd point. With an origin of 0,0 and treating a line that goes straight to the right as 0 degrees, and going counterclockwise (anti-clockwise for some of you), you do:
double angle = ... // angle in radians
double newX = cos(angle);
double newY = sin(angle);
This assumes a radius of 1. Multiply each times a desired radius. Pick a number that will be bigger than the screen such as 480 for an iPhone or 1024 for an iPad (assuming you want points and not pixels).
Then add the original point to get the final point.
Assuming you have CGPoint start, double angle, and a length, your final point is:
double endX = cos(angle) * length + start.x;
double endY = sin(angle) * length + start.y;
CGPoint end = CGPointMake(endX, endY);
It's OK if the end point is off the screen.
I am using GLKit quaternion to rotate a unit vector by 90 degrees around z-axis.
Here is my code:
let q = GLKQuaternionMakeWithAngleAndVector3Axis(Float(M_PI_2), GLKVector3Make(0,0,-1))
let result = GLKQuaternionRotateVector3(q, GLKVector3Make(1,0,0))
But the result shows the vector (0, -1, 0) where I expected (0, 1, 0).
According to the GLKit Framework Reference, the input angle argument for GLKQuaternionMakeWithAngleAndVector3Axis is explained :
radians : The angle of the rotation in radians (a positive angle is counterclockwise).
So I expected my code rotates x-y plane by 90 degrees counterclockwise where I am looking down the x-y plane from the position where z > 0. But actually the result showed the plane was rotated clockwise.
Can someone point out what I understand wrongly?