Algorithm for creating a circular path around a center mass? - ios

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.

Related

Get the actual angle after the superview is rotated

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).

How to draw line given a center point and angle in iOS?

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.

How to determine closest CGPoint with an angle and another CGPoint

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);

How to rotate a triangle?

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 ;-)

Absolute angle between lines using inverse cosine

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)).

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