I'm using UIBezierPath to draw lines(in multiple angles) with two points, but I want to draw lines a little shorter than the distance between the two points.
I tried the following codes to find a point between the two points:
let x3 = x2 + 0.9 * (x1 - x2);
let y3 = y2 + 0.9 * (y1 - y2);
It works in 1 or 2 angles but fails in others. How can I get the correct point? Thanks.
=== Edited ===
By now I got some idea from the search, but I still cannot get it works
Get the distance between the two points, and then minus 15, since I want it shorter
let distance = sqrt(pow((p2.x - p1.x), 2) + pow((p2.y - p1.y), 2)) - 15
Get the line angle:
let angle = (p2.y - p1.y) / (p2.x - p1.x)
Get point 3 with distance and angle:
let x = p1.x + (distance * cos(angle))
let y = p1.y - (distance * sin(angle))
It's a problem of wrong angle, function atan2 gives a correct angle value. Now the whole code work perfect.
let angle = atan2((p2.y - p1.y), (p2.x - p1.x))
Related
I'm trying to better understand the calibrateCamera and SolvePnP functions in OpenCV, specifically the rotation vectors returned by these functions which I believe is an axis-angle rotation vector (NOT as I had thought initially the yaw,pitch,roll angles). I would like to know the rotation around the x,y and z axis of my checkerboard image. The OpenCV functions return a rotation vector in the form rot = [a,b,c]
Using this answer
as a guide I calculate the angle theta with theta = sqrt(a^2,b^2,c^2) and the rotation axis v = [a/theta, b/theta, c/theta];
Then I take these values and use the Axis-Angle To Euler conversion on euclideanspace.com. shown here:
heading = atan2(y * sin(angle)- x * z * (1 - cos(angle)) , 1 - (y^2 + z^2 ) * (1 - cos(angle)))
attitude = asin(x * y * (1 - cos(angle)) + z * sin(angle))
bank = atan2(x * sin(angle)-y * z * (1 - cos(angle)) , 1 - (x^2 + z^2) * (1 - cos(angle)))
I'm using one of the example OpenCV checkerboard images (Left01.jpg), shown below (note the frame axes in the upper left corner with red = x, green = y, blue = z
Using this image I get a rotation vector from calibrateCamera of [0.166,0.294,0.014]
Running these values through the calculations discussed and converting to degrees I get:
heading = 16.7 deg
attitude = 1.7 deg
bank = 9.3 deg
I believe these correspond to yaw,pitch,roll? The 16.7 degree heading seems high looking at the image, but it's hard to tell. Does this make sense? What would be the correct way to figure out the euler angles (angles around each axis) given the OpenCV rotation vector? Snippets of my code are shown below.
double RMSError = calibrateCamera(
objectPointsArray,
imagePointsArray,
img.size(),
intrinsics,
distortion,
rotation,
translation,
CALIB_ZERO_TANGENT_DIST |
CALIB_FIX_K3 | CALIB_FIX_K4 | CALIB_FIX_K5 |
CALIB_FIX_ASPECT_RATIO);
Mat rvec = rotation.at(0);
//try and get the rotation angles here
//https://stackoverflow.com/questions/12933284/rodrigues-into-eulerangles-and-vice-versa
float theta = sqrt(pow(rvec.at<double>(0),2) + pow(rvec.at<double>(1),2) + pow(rvec.at<double>(2),2));
Mat axis = (Mat_<double>(1, 3) << rvec.at<double>(0) / theta, rvec.at<double>(1) / theta, rvec.at<double>(2) / theta);
float x_ = axis.at<double>(0);
float y_ = axis.at<double>(1);
float z_ = axis.at<double>(2);
//this is yaw,pitch,roll respectively...maybe
float heading = atan2(y_ * sin(theta) - x_ * z_ * (1 - cos(theta)), 1 - (pow(y_,2) + pow(z_,2)) * (1 - static_cast<double>(cos(theta))));
float attitude = asin(x_ * y_ * (1 - cos(theta) + z_ * sin(theta)));
float bank = atan2(x_ * sin(theta) - y_ * z_ * (1 - cos(theta)), 1 - (pow(x_, 2) + pow(z_, 2)) * (1 - static_cast<double>(cos(theta))));
float headingDeg = heading * (180 / 3.14);
float attitudeDeg = attitude * (180 / 3.14);
float bankDeg = bank * (180 / 3.14);
In my application, a user taps 3 times and an angle will be created by the 3 points that were tapped. It draws the angle perfectly. I am trying to calculate the angle at the second tap, but I think I am doing it wrong (probably a math error). I haven't covered this in my calculus class yet, so I am going off of a formula on wikipedia.
http://en.wikipedia.org/wiki/Law_of_cosines
Here is what I am trying:
Note: First, Second, and Third are CGPoints created at the user's tap.
CGFloat xDistA = (second.x - third.x);
CGFloat yDistA = (second.y - third.y);
CGFloat a = sqrt((xDistA * xDistA) + (yDistA * yDistA));
CGFloat xDistB = (first.x - third.x);
CGFloat yDistB = (first.y - third.y);
CGFloat b = sqrt((xDistB * xDistB) + (yDistB * yDistB));
CGFloat xDistC = (second.x - first.x);
CGFloat yDistC = (second.y - first.y);
CGFloat c = sqrt((xDistC * xDistC) + (yDistC * yDistC));
CGFloat angle = acos(((a*a)+(b*b)-(c*c))/((2*(a)*(b))));
NSLog(#"FULL ANGLE IS: %f, ANGLE IS: %.2f",angle, angle);
Sometimes, it gives the angle as 1 which doesn't make sense to me. Can anyone explain why this is, or how to fix it please?
Not sure if this is the main problem but it is a problem
Your answer gives the angle at the wrong point:
To get the angle in green (which is probably angle you want based on your variable names "first", "second" and "third), use:
CGFloat angle = acos(((a*a)+(c*c)-(b*b))/((2*(a)*(c))));
Here's a way that circumvents the law of cosines and instead calculates the angles of the two vectors. The difference between the angles is the searched value:
CGVector vec1 = { first.x - second.x, first.y - second.y };
CGVector vec2 = { third.x - second.x, third.y - second.y };
CGFloat theta1 = atan2f(vec1.dy, vec1.dx);
CGFloat theta2 = atan2f(vec2.dy, vec2.dx);
CGFloat angle = theta1 - theta2;
NSLog(#"angle: %.1f°, ", angle / M_PI * 180);
Note the atan2 function that takes the x and y components as separate arguments and thus avoids the 0/90/180/270° ambiguity.
The cosine formula implementation looks right; did you take into account that acos() returns the angle in radians, not in degrees? In order to convert into degrees, multiply the angle by 180 and divide by Pi (3.14159...).
The way I have done it is to calculate the two angles separately using atan2(y,x) then using this function.
static inline double
AngleDiff(const double Angle1, const double Angle2)
{
double diff = 0;
diff = fabs(Angle1 - Angle2);
if (diff > <Pi>) {
diff = (<2Pi>) - diff;
}
return diff;
}
The function deals in radians, but you can change <Pi> to 180 and <2Pi> to 360
Using this answer to compute angle of the vector:
CGFloat angleForVector(CGFloat dx, CGFloat dy) {
return atan2(dx, -dy) * 180.0/M_PI;
}
// Compute angle at point Corner, that is between AC and BC:
CGFloat angle = angleForVector(A.x - Corner.x, A.y - Corner.y)
- angleForVector(B.x - Corner.x, B.y - Corner.y);
NSLog(#"FULL ANGLE IS: %f, ANGLE IS: %.2f",angle, angle);
I'm struggling with rotating a triangle resulting from a UIRotationGestureRecognizer. If you could look over my approach and offer suggestions, I'd greatly appreciate it.
I ask the gesture recognizer object for the rotation, which the documentation says is returned in radians.
My strategy had been to think of each vertex as a point on a circle that exists between the center of the triangle and the vertex, and then use the radians of rotation to find the new point on that circumference. I'm not totally sure this is a valid approach, but I wanted to at least try it. Visually I'd know whether or not it was working.
Here's the code I created in that attempt:
- (CGPoint)rotateVertex:(CGPoint)vertex byRadians:(float)radians
{
float deltaX = center.x - vertex.x;
float deltaY = center.y - vertex.y;
float currentAngle = atanf( deltaX / deltaY );
float newAngle = currentAngle + radians;
float newX = cosf(newAngle) + vertex.x;
float newY = sinf(newAngle) + vertex.y;
return CGPointMake(newX, newY);
}
When executed, there's a slight rotation at the beginning, but then as I continue rotating my fingers the vertices just start getting farther away from the center point, indicating I'm confusing something here.
I looked at what the CGContextRotateCTM could do for me, but ultimately I need to know what the vertices are after the rotation, so just rotating the graphics context doesn't appear to leave me with those changed coordinates.
I also tried the technique described here but that resulted in the triangle being flipped about the second vertex, which seems odd, but then that technique works with p and q being the x and y coordinates of the second vertex.
Thanks for taking a look!
Solved: Here is the corrected function. It assumes you have calculated the center of the triangle. I used the 1/3(x1 + x2 + x3), 1/3(y1 + y2 + y3) method described on the Wikipedia article on Centroids.
- (CGPoint)rotatePoint:(CGPoint)currentPoint byRadians:(float)radiansOfRotation
{
float deltaX = currentPoint.x - center.x;
float deltaY = currentPoint.y - center.y;
float radius = sqrtf(powf(deltaX, 2.0) + powf(deltaY, 2.0));
float currentAngle = atan2f( deltaY, deltaX );
float newAngle = currentAngle + radiansOfRotation;
float newRun = radius * cosf(newAngle);
float newX = center.x + newRun;
float newRise = radius * sinf(newAngle);
float newY = center.y + newRise;
return CGPointMake(newX, newY);
}
Of noteworthy relevance to why the first code listing did not work was that the arguments to atan2 were reversed. Also, the correct calculation of the delta values was reversed.
You're forgetting to multiply by the radius of the circle. Also, since the Y axis points down in the UIKit coordinate system, you have to subtract instead of add the radians and negate the y coordinate at the end. And you need to use atan2 only gives output in the range -pi/2 to pi/2:
float currentAngle = atan2f(deltaY, deltaX);
float newAngle = currentAngle - radians;
float radious = sqrtf(powf(deltaX, 2.0) + powf(deltaY, 2.0));
float newX = radius * cosf(newAngle) + vertex.x;
float newY = -1.0 * radius * sinf(newAngle) + vertex.y;
The answer is embedded now in the original question. Gun shy about proper decorum ;-)
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.
I want to calculate the angle between two lines formed by three points(one of the points is the point of intersection of the two lines) using inverse cosine function as follows:
CGFloat a = initialPosition.x - origin.x;
CGFloat b = initialPosition.y - origin.y;
CGFloat c = currentPosition.x - origin.x;
CGFloat d = currentPosition.y - origin.y;
CGFloat angle = (180/M_PI) * acosf(((a*c) + (b*d)) / ((sqrt(a*a + b*b)) * (sqrt(c*c + d*d))));
Unfortunately, acosf returns a value between 0 and pi only. How do I find a value between 0 and 2*pi (going, say, in the anti-clockwise manner)?
i don't know what language you are using, but typically there is an atan2 function that gives you a value from the full 360 degrees. in this case you need to use it twice and then add a little additional logic.
some pseudocode will help clear things up:
initialAngle = atan2(initialPosition.y - origin.y, initialPosition.x - origin.x)
currentAngle = atan2(currentPosition.y - origin.y, currentPosition.x - origin.x)
# angle is measured from x axis anti-clock, so lets find the value starting from
# initial and rotating anti-clock to current, as a positive number
# so we want current to be larger than initial
if (currentAngle < initialAngle) {currentAngle += 2 pi}
# and then we can subtract
return currentAngle - initialAngle
i know this isn't using acos, but that is multi-valued so to do so ends up using lots of logic about signs of differences that is bug-prone. atan2 is what you want.
found a simple solution...This comes from high school maths! First make an equation of a line made from the origin and the initialPosition of the form y = mx+c. A point lying on either side of this line will satisfy y < mx+c or y > mx+c, depending on where it is. If finding angles in the clockwise or anti-clockwise sense, make the following check:
currentPosition.y < (currentPosition.x *(initialPosition.y - origin.y) + (initialPosition .x * origin.y - initialPosition.y * origin.x)) / (initialPosition.x - origin.x)
If the above condition is true, then the line formed from origin and currentPosition makes an angle less than 180 deg (in the clockwise sense) with the line formed from origin and initialPosition. Otherwise it makes an angle more than 180 deg in the clockwise sense and less than 180 deg in the anti-clockwise sense...and so on. Depending on the requirement, the final angle is either the (angle returned by acos) or (360 - (angle returned by acos)).