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).
Related
I don't know if this is basic math I have to compute, or my inexperience with the pitch, roll, and yaw values. At the moment I have an image object that moves based on my accelerometer values.
//Move the ball based on accelerator values
delta.x = CGFloat(acceleration.x * 10)
delta.y = CGFloat(acceleration.y * 10)
ball.center = CGPointMake(ball.center.x + delta.x, ball.center.y + delta.y)
I can calculate the pitch through the attitude and get the angle. What I want to do is line up my "ball" in the center of the screen only when the angle of the phone is a certain angle, lets say 45 degrees. How can I move my ball so that it lines up in the center based on specific angles given?
Your screen height is Η pixels.
Your screen width is W pixels.
The horizontal centre of the screen is x = W / 2
I'm assuming from your question you want the ball centre to vary between the top (x, 0) when the screen is flat and bottom (x, H) when the screen is vertical.
If the angle of your phone θ varies between 0 and π, then y = θ / π * H
ball.center = CGPoint(x: W / 2, y: θ / π * H)
All you need is the trig to work out θ based on the gyro readings
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).
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)
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 attempting to simply make objects orbit around a center point, e.g.
The green and blue objects represent objects which should keep their distance to the center point, while rotating, based on an angle which I pass into method.
I have attempted to create a function, in objective-c, but it doesn't work right without a static number. e.g. (It rotates around the center, but not from the true starting point or distance from the object.)
-(void) rotateGear: (UIImageView*) view heading:(int)heading
{
// int distanceX = 160 - view.frame.origin.x;
// int distanceY = 240 - view.frame.origin.y;
float x = 160 - view.image.size.width / 2 + (50 * cos(heading * (M_PI / 180)));
float y = 240 - view.image.size.height / 2 + (50 * sin(heading * (M_PI / 180)));
view.frame = CGRectMake(x, y, view.image.size.width, view.image.size.height);
}
My magic numbers 160, and 240 are the center of the canvas in which I'm drawing the images onto. 50 is a static number (and the problem), which allows the function to work partially correctly -- without maintaining the starting poisition of the object or correct distance. I don't know what to put here unfortunately.
heading is a parameter that passes in a degree, from 0 to 359. It is calculated by a timer and increments outside of this class.
Essentially what I would like to be able to drop any image onto my canvas, and based on the starting point of the image, it would rotate around the center of my circle. This means, if I were to drop an image at Point (10,10), the distance to the center of the circle would persist, using (10,10) as a starting point. The object would rotate 360 degrees around the center, and reach it's original starting point.
The expected result would be to pass for instance (10,10) into the method, based off of zero degrees, and get back out, (15,25) (not real) at 5 degrees.
I know this is very simple (and this problem description is entirely overkill), but I'm going cross eyed trying to figure out where I'm hosing things up. I don't care about what language examples you use, if any. I'll be able to decipher your meanings.
Failure Update
I've gotten farther, but I still cannot get the right calculation. My new code looks like the following:
heading is set to 1 degree.
-(void) rotateGear: (UIImageView*) view heading:(int)heading
{
float y1 = view.frame.origin.y + (view.frame.size.height/2); // 152
float x1 = view.frame.origin.x + (view.frame.size.width/2); // 140.5
float radius = sqrtf(powf(160 - x1 ,2.0f) + powf(240 - y1, 2.0f)); // 90.13
// I know that I need to calculate 90.13 pixels from my center, at 1 degree.
float x = 160 + radius * (cos(heading * (M_PI / 180.0f))); // 250.12
float y = 240 + radius * (sin(heading * (M_PI / 180.0f))); // 241.57
// The numbers are very skewed.
view.frame = CGRectMake(x, y, view.image.size.width, view.image.size.height);
}
I'm getting results that are no where close to where the point should be. The problem is with the assignment of x and y. Where am I going wrong?
You can find the distance of the point from the centre pretty easily:
radius = sqrt((160 - x)^2 + (240 - y)^2)
where (x, y) is the initial position of the centre of your object. Then just replace 50 by the radius.
http://en.wikipedia.org/wiki/Pythagorean_theorem
You can then figure out the initial angle using trigonometry (tan = opposite / adjacent, so draw a right-angled triangle using the centre mass and the centre of your orbiting object to visualize this):
angle = arctan((y - 240) / (x - 160))
if x > 160, or:
angle = arctan((y - 240) / (x - 160)) + 180
if x < 160
http://en.wikipedia.org/wiki/Inverse_trigonometric_functions
Edit: bear in mind I don't actually know any Objective-C but this is basically what I think you should do (you should be able to translate this to correct Obj-C pretty easily, this is just for demonstration):
// Your object gets created here somewhere
float x1 = view.frame.origin.x + (view.frame.size.width/2); // 140.5
float y1 = view.frame.origin.y + (view.frame.size.height/2); // 152
float radius = sqrtf(powf(160 - x1 ,2.0f) + powf(240 - y1, 2.0f)); // 90.13
// Calculate the initial angle here, as per the first part of my answer
float initialAngle = atan((y1 - 240) / (x1 - 160)) * 180.0f / M_PI;
if(x1 < 160)
initialAngle += 180;
// Calculate the adjustment we need to add to heading
int adjustment = (int)(initialAngle - heading);
So we only execute the code above once (when the object gets created). We need to remember radius and adjustment for later. Then we alter rotateGear to take an angle and a radius as inputs instead of heading (this is much more flexible anyway):
-(void) rotateGear: (UIImageView*) view radius:(float)radius angle:(int)angle
{
float x = 160 + radius * (cos(angle * (M_PI / 180.0f)));
float y = 240 + radius * (sin(angle * (M_PI / 180.0f)));
// The numbers are very skewed.
view.frame = CGRectMake(x, y, view.image.size.width, view.image.size.height);
}
And each time we want to update the position we make a call like this:
[objectName rotateGear radius:radius angle:(adjustment + heading)];
Btw, once you manage to get this working, I'd strongly recommend converting all your angles so you're using radians all the way through, it makes it much neater/easier to follow!
The formula for x and y coordinates of a point on a circle, based on radians, radius, and center point:
x = cos(angle) * radius + center_x
y = sin(angle) * radius + center_y
You can find the radius with HappyPixel's formula.
Once you figure out the radius and the center point, you can simply vary the angle to get all the points on the circle that you'd want.
If I understand correctly, you want to do InitObject(x,y). followed by UpdateObject(angle) where angle sweeps from 0 to 360. (But use radians instead of degrees for the math)
So you need to track the angle and radius for each object.:
InitObject(x,y)
relative_x = x-center.x
relative_y = y-center.y
object.radius = sqrt((relative_x)^2, (relative_y)^2)
object.initial_angle = atan(relative_y,relative_x);
And
UpdateObject(angle)
newangle = (object.initial_angle + angle) % (2*PI )
object.x = cos(newangle) * object.radius + center.x
object.y = sin(newangle) * object.radius + center.y
dx=dropx-centerx; //target-source
dy=-(dropy-centery); //minus = invert screen coords to cartesian coords
radius=sqrt(dy*dy+dx*dx); //faster if your compiler optimizer is bad
if dx=0 then dx=0.000001; //hackpatchfudgenudge*
angle=atan(dy/dx); //set this as start angle for the angle-incrementer
Then go with the code you have and you'll be fine. You seem to be calculating radius from current position each time though? This, like the angle, should only be done once, when the object is dropped, or else the radius might not be constant.
*instead of handling 3 special cases for dx=0, if you need < 1/100 degree precision for the start angle go with those instead, google Polar Arctan.