CGPoint pointA = [self.appDelegate.points[0] CGPointValue];//first point
CGPoint pointB = [self.appDelegate.points[1] CGPointValue];// second point
CGPoint pointC = [self.appDelegate.points[2] CGPointValue];//third point
CGFloat slopeAB = (pointB.y - pointA.y)/(pointB.x - pointA.x);//slope ab
CGFloat slopeBC = (pointC.y - pointB.y)/(pointC.x - pointB.x);//slope bc
self.ang=(slopeAB-slopeBC)/(1+(slopeAB)*(slopeBC));//slope
CGFloat finalAngle = atanf(self.ang);// angle tan inverse slope
CGFloat angle = (finalAngle * (180.0/M_PI));
NSLog(#"The angle is: %.2f degrees",angle);
calculated the slope of each line
calculated angle by tan inverse
Use the atan2() function. From the manual page:
#include <math.h>
double
atan2(double y, double x);
The atan2() function computes the principal value of the arc tangent of
y/x, using the signs of both arguments to determine the quadrant of the
return value.
To do this for the three points you'll need to call atan2() twice: once to find the angle of AB, and once for the angle of BC. Take the difference between these two to find the angle between AB and BC:
double angle_ab = atan2(pointA.y - pointB.y, pointA.x - pointB.x);
double angle_cb = atan2(pointC.y - pointB.y, pointC.x - pointB.x);
double angle_abc = angle_ab - angle_cb;
Note that this is assuming that B is the "center" point of the angle you're interested in. Adjust appropriately if I've assumed wrongly.
Related
I am showing an image in an UIImageView and i'd like to convert coordinates to x/y values so i can show cities on this image.
This is what i tried based on my research:
CGFloat height = mapView.frame.size.height;
CGFloat width = mapView.frame.size.width;
int x = (int) ((width/360.0) * (180 + 8.242493)); // Mainz lon
int y = (int) ((height/180.0) * (90 - 49.993615)); // Mainz lat
NSLog(#"x: %i y: %i", x, y);
PinView *pinView = [[PinView alloc]initPinViewWithPoint:x andY:y];
[self.view addSubview:pinView];
which gives me 167 as x and y=104 but this example should have the values x=73 and y=294.
mapView is my UIImageView, just for clarification.
So my second try was to use the MKMapKit:
CLLocationCoordinate2D coord = CLLocationCoordinate2DMake(49.993615, 8.242493);
MKMapPoint point = MKMapPointForCoordinate(coord);
NSLog(#"x is %f and y is %f",point.x,point.y);
But this gives me some really strange values:
x = 140363776.241755 and y is 91045888.536491.
So do you have an idea what i have to do to get this working ?
Thanks so much!
To make this work you need to know 4 pieces of data:
Latitude and longitude of the top left corner of the image.
Latitude and longitude of the bottom right corner of the image.
Width and height of the image (in points).
Latitude and longitude of the data point.
With that info you can do the following:
// These should roughly box Germany - use the actual values appropriate to your image
double minLat = 54.8;
double minLong = 5.5;
double maxLat = 47.2;
double maxLong = 15.1;
// Map image size (in points)
CGSize mapSize = mapView.frame.size;
// Determine the map scale (points per degree)
double xScale = mapSize.width / (maxLong - minLong);
double yScale = mapSize.height / (maxLat - minLat);
// Latitude and longitude of city
double spotLat = 49.993615;
double spotLong = 8.242493;
// position of map image for point
CGFloat x = (spotLong - minLong) * xScale;
CGFloat y = (spotLat - minLat) * yScale;
If x or y are negative or greater than the image's size, then the point is off of the map.
This simple solution assumes the map image uses the basic cylindrical projection (Mercator) where all lines of latitude and longitude are straight lines.
Edit:
To convert an image point back to a coordinate, just reverse the calculation:
double pointLong = pointX / xScale + minLong;
double pointLat = pointY / yScale + minLat;
where pointX and pointY represent a point on the image in screen points. (0, 0) is the top left corner of the image.
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 know how to draw simple lines using Core Graphics. I now need to draw a Dimension line for measurements. See the image below for an example of what I need to draw (in red). The top line would be easy, but drawing the perpendicular on a diagonal line will require some math that I'm having a difficult time figuring out right now.
Each main line will have (x,y) as a starting point and (x1,y1) as an ending point. I then need to draw the perpendicular lines that intersect at each of the points (x,y) and (x1,y1).
What is the math required to calculate the points for these perpendicular lines?
The following code computes a vector of length 1 that is perpendicular to
the line from p = (x, y) to p1 = (x1, y1):
CGPoint p = CGPointMake(x, y);
CGPoint p1 = CGPointMake(x1, y1);
// Vector from p to p1;
CGPoint diff = CGPointMake(p1.x - p.x, p1.y - p.y);
// Distance from p to p1:
CGFloat length = hypotf(diff.x, diff.y);
// Normalize difference vector to length 1:
diff.x /= length;
diff.y /= length;
// Compute perpendicular vector:
CGPoint perp = CGPointMake(-diff.y, diff.x);
Now you add and subtract a multiple of that perpendicular vector to the first point
to get the endpoints of the first marker line at p:
CGFloat markLength = 3.0; // Whatever you need ...
CGPoint a = CGPointMake(p.x + perp.x * markLength/2, p.y + perp.y * markLength/2);
CGPoint b = CGPointMake(p.x - perp.x * markLength/2, p.y - perp.y * markLength/2);
For the second marker line, just repeat the last calculation with p1 instead of p.
I calculate angle between two CGPoints :
//calculate radian and degree
CGPoint diff = ccpSub(center, location);//return ccp(v1.x - v2.x, v1.y - v2.y);
float rads = atan2f( diff.y, diff.x);
float degs = -CC_RADIANS_TO_DEGREES(rads);
NSLog(#"Rad %.2f Degs %.2f",rads,degs);
Now In another function where I have a pre known CGPoint and the degree of above function, I want to calculate closest point that satisfies the degree.
I was thinking about maybe below code would help me but in below code start point and rotation point is known, in my situation I only know start point.
-(void) rotateAroundPoint:(CGPoint)rotationPoint angle:(CGFloat)angle {
CGFloat x = cos(CC_DEGREES_TO_RADIANS(-angle)) * (self.position.x-rotationPoint.x) - sin(CC_DEGREES_TO_RADIANS(-angle)) * (self.position.y-rotationPoint.y) + rotationPoint.x;
CGFloat y = sin(CC_DEGREES_TO_RADIANS(-angle)) * (self.position.x-rotationPoint.x) + cos(CC_DEGREES_TO_RADIANS(-angle)) * (self.position.y-rotationPoint.y) + rotationPoint.y;
Lets say I have a point 800,600 and I have a degree of 70, how can I calculate closest point with that point and that degree?
EDIT:::
Normally in my game sprites are moved with a button therefore all rotation,movement,speed etc are handled when button pressed [sprite moveToPreGivenPostion:CGPoint]
But now a compass is added and when user choose an angle on the compass I need to move the sprite in the direction of degree on compass, since [sprite moveToPreGivenPostion:CGPoint] already handles rotation and other stuff I just want to determine that what CGPoint should I send to moveToPreGivenPostion function.
As #trumpetlicks said you cant find the closest point like that, but I guess I understood what you want and that function -(void) rotateAroundPoint:(CGPoint)rotationPoint angle:(CGFloat)angle you are trying to use is perfectly fine to achieve what you want.
all you need to do is choose float radius.
you know your current point and lets say your radius is 1, basically you can calculate your previous point without a degree, assuming 0 degrees is left of your point and lets say your point is 200,200 with 1 radius 0 degree your previous point automatically becomes 199,200.
So now you have a reference point so now calculate the point you want to move your sprite:
//choose a feasable radius
float radius = 0.5;
//position_ is your preknown position as you said
//find a the point to roate
//position_.x-radius is always 0 degrees of your current point
CGFloat x = cos(rads) * ((position_.x-radius)-position_.x) - sin(rads) * ((position_.y)-position_.y) + position_.x;
CGFloat y = sin(rads) * ((position_.x-radius)-position_.x) + cos(rads) * ((position_.y)-position_.y) + position_.y;
//get the new point
CGPoint newLocation = ccp(x, y);
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 ;-)