Test if MKCircle intersects with MKPolygon - objective-c

I'm looking for some guidance in testing if a MKPolygon intersects an MKCircle. Currently I'm using:
if ([circle intersectsMapRect:[poly boundingMapRect]]) {
//they do intersect
}
I'm finding this returns inaccurate results simply b/c it draws a rectangle around my circle, thus giving me intersections that shouldn't otherwise be.
Searching the topic has lead me to Chad Saxon's polygon-polygon intersection project. This could be useful if I could somehow convert my MKCircle to a multi-sided polygon - which could be possible but ultimately I believe this is the round-about way to solve this.
I'm ultimately wondering if there is a simple solution I've overlooked before delving into porting my own custom geometry-ray-testing-algorithm implementation.

A couple of thoughts:
If you use that polygon intersection project, be aware that it has a few leaks in it. I issued a pull request that fixes a few of them (and a few other random observations). I would be wary of adopting any of that view controller code (as it has other issues), too, but the category seems ok if you're ok with the various limitations it entails (notably, the clockwise limitation, which is not really an issue if you're only determining if they intersected).
Rather than converting the circle to a series of polygons and then using that polygon intersection class, I might consider an alternative approach leveraging that you can detect the intersection with a circle by leveraging the fact that you can look at the distance between the relevant points in the polygon and the radius of a circle. It seems that there are three aspects of the problem:
If the distance between any of the polygon's vertices and the center of the circle is less than the radius of the circle, then the polygon and circle intersect.
Does the polygon encompass the circle (this is that special case where the distance from all of the sides of the polygon would be greater than the circle's radius, but the circle and polygon still obviously intersect). This is easily achieved by checking to see if the CGPath of the polygon's view encompasses the center of the circle using CGPathContainsPoint.
The only complicated part is to check to see if any side of the polygon intersects the circle, namely that the minimum distance between the sides of the polygon and the center of the circle less than the radius of the circle;
To calculate the distance of each side from the center of the circle, I might therefore iterate through each side of the polygon and for those sides facing the circle's center (i.e. for which the circle's center is perpendicular to the segment, meaning that an imaginary line perpendicular to the polygon's side that goes through the center of the circle actually crosses the line segment), you could:
Calculate the constants a, b, and c for this side of the polygon for the equation ax + by + c = 0 for a line segment going between vertices of the polygon (x1, y1) and (x2, y2):
a = (y1 – y2)
b = (x2 – x1)
c = (x1y2 – x2y1)
Calculate the distance from a point to a line, using (x0, y0) as the center of the circle:
If that distance is less than the radius of the circle, then you know that the polygon intersects the circle.
I put a sample project employing this technique on github.

Just for those to get a bit of substance on the solution, here is a useful MKCircle extension I wrote that checks if a point (in this case a polygon point) is inside the circle or not. Enjoy!
//MKCircle+PointInCircle.h
#import <Foundation/Foundation.h>
#import <MapKit/MapKit.h>
#interface MKCircle (PointInCircle)
-(BOOL)coordInCircle:(CLLocationCoordinate2D)coord;
#end
//MKCircle+PointInCircle.m
#import "MKCircle+PointInCircle.h"
#implementation MKCircle (PointInCircle)
-(BOOL)coordInCircle:(CLLocationCoordinate2D)coord {
CLLocation *locFrom = [[CLLocation alloc] initWithLatitude:self.coordinate.latitude longitude:self.coordinate.longitude];
CLLocation *locTo = [[CLLocation alloc] initWithLatitude:coord.latitude longitude:coord.longitude];
double distance = [locFrom distanceFromLocation:locTo];
BOOL isInside = (distance <= self.radius);
return isInside;
}
#end

Related

How could I make an "iPod Wheel" type control on iPhone?

I want to create a sort of "iPod Wheel" control in a Swift project that I'm doing. I've got everything drawn out, but not it's time to actually make this thing work.
What would be the best way to recognize "spinning" so to speak, or to describe that more clearly, when the user is actively pressing the wheel and spinning his/her thumb around the wheel in a clockwise or counter-clockwise direction.
I will no doubt want to use touchesBegan/touchesMoved/touchesEnded. What's the best way to figure out spinning though?
I'm thinking
a) determine in touchesMoved if the users touch is within circle, by determining the radius from the center point. Center point and radius are easily obtainable. Using these however, how can I determine the outer edge of the circle/wheel... to know whether the user is within the actually circle (their touch could still be in the view, but outside the actual wheel portion)
b) Determine the current angle and how it has changed the previous angle. By that I mean... I would use the center point of the circle as one point, and the users current touch as the second point. This gives me my vector. I would also have a baseline angle. Likely center point to 12 c'clock. I would compare the two vectors (I already have a VectorMath class for this from something else I'm doing) and see my angle is 0. If the users touch were at 3 oclock, and I compared it to our baseline angle... I would see the angle is 90 degrees. I would continually calculate the angle, and perhaps every 5 degrees of change... would warrant a change in the controls output (depending on desired sensitivity).
Does this seem like the best way to do this? I think this would be an ideal way, but am still not sure on how to calculate the circles outer edge, and determine if a users touch is within it.
You are on the right track. I think approach b) will work.
Remember the starting position of the finger at the touchesBegan
event.
Imagine a line from the finger position to the middle of the button
circle.
For the touchesMoved event, again, imagine a virtual line from the
new position to the center of the circle.
Using the formula from
http://mathworld.wolfram.com/Line-LineAngle.html (or some code) you can determine
the angle between the two lines. If it's a negative angle the user
is turning the wheel counter-clockwise, otherwise it's clockwise.
To determine if the touch event was inside the ring, calculate the distance from the center of the circle to the point of touch. It should be between the minimum and the maximum distance (inner circle and outer circle radius). Calculating the distance between to two points is explained at https://www.mathsisfun.com/algebra/distance-2-points.html
I think you're almost there, although I'd do something slightly different on your point b.
If you think about it, when you start "spinning" on your iPod, you don't need to start from a precise position, you start spinning from "where you started", therefore I wouldn't set my "baseline angle" at π/2, I'd set my baseline (or 0°) angle at the point the user taps for the first time, and starting from then, I'd count the offset angles, clockwise and counterclockwise.
I don't think there would be much difference, except maybe from some calculations you'll do on the angles, on the two approaches, practically speaking; it just makes more sense imho to start counting from the first input rater than setting a baseline to π/2 and counting the first angle.
I am answering in parts.
// Get a position based on the angle
float xPosition = center.x + (radiusX * sinf(angleInRadians)) - (CGRectGetWidth([cell frame]) / 2);
float yPosition = center.y + (radiusY * cosf(angleInRadians)) - (CGRectGetHeight([cell frame]) / 2);
float scale = 0.75f + 0.25f * (cosf(angleInRadians) + 1.0);
next
[cell setTransform:CGAffineTransformScale(CGAffineTransformMakeTranslation(xPosition, yPosition), scale, scale)];
// Tweak alpha using the same system as applied for scale, this
// time with 0.3 the minimum and a semicircle range of 0.5
[cell setAlpha:(0.3f + 0.5f * (cosf(angleInRadians) + 1.0))];
and
- (void)spin:(SpinGestureRecognizer *)recognizer
{
CGFloat angleInRadians = -[recognizer rotation];
CGFloat degrees = 180.0 * angleInRadians / M_PI; // Radians to degrees
[self setCurrentAngle:[self currentAngle] + degrees];
[self setAngle:[self currentAngle]];
}
again check the wheelview.m of photowheel in github.

Computing Minkowski Difference For Circles and Convex Polygons

I'm needing to implement a Minkowski sum function that can return the Minkowski sum of either 2 circles, 2 convex polygons or a circle and a convex polygon. I found this thread that explained how to do this for convex polygons, but I'm not sure how to do this for a circle and polygon. Also, how would I even represent the answer?! I'd like the algorithm to run in O(n) time but beggars can't be choosers.
Circle is trivial -- just add the center points, and add the radii. Circle + ConvexPoly is nearly as simple: move each segment perpendicularly outward by the circle radius, and connect adjacent segments with circular arcs centered at the original poly vertices. Translate the whole by the circle center point.
As for how you represent the answer: Well, it depends on what you want to do with it. You could convert it to a NURBS if you just want to draw it with a vector drawing library. You could approximate the circular arcs with polylines if you just want a polygonal approximation. Or you might store it as is -- "this polygon, expanded by such-and-such a radius". That would be the best choice for things like raycasting, for instance. Or as a compromise, you could connect adjacent segments linearly instead of with circular arcs, and store it as the union of the (new) convex polygon and a list of circles at the vertices.
Oh, about ConvexPoly + ConvexPoly. That's the trickiest one, but still straightforward. The basic idea is that you take the list of segment vectors for each polygon (starting from some particular extremal point, like the point on each poly with the lowest X coordinate), then merge the two lists together, keeping it sorted by angle. Sum the two points you started with, then apply each vector from the merged vector list to produce the other points.

Placement of "good" control points in Bezier curves

I've been working on this problem for awhile now, and haven't been able to come up with a good solution thusfar.
The problem: I have an ordered list of three (or more) 2D points, and I want to stroke through these with a cubic Bezier curve, in such a way that it "looks good." The "looks good" part is pretty simple: I just want the wedge at the second point smoothed out (so, for example, the curve doesn't double-back on itself). So given three points, where should one place the two control points that would surround the second point in the triplet when drawing the curve.
My solution so far is as follows, but is incomplete. The idea might also help communicate the look that I'm after.
Given three points, (x1,y1), (x2,y2), (x3,y3). Take the circle inscribed by each triplet of points (if they are collinear, we just draw a straight line between them and move on). Take the line tangent to this circle at point (x2,y2) -- we will place the control points that surround (x2,y2) on this tangent line.
It's the last part that I'm stuck on. The problem I'm having is finding a way to place the two control points on this tangent line -- I have a good enough heuristic on how far from (x2,y2) on this line they should be, but of course, there are two points on this line that are that distance away. If we compute the one in the "wrong" direction, the curve loops around on itself.
To find the center of the circle described by the three points (if any of the points have the same x value, simply reorder the points in the calculation below):
double ma = (point2.y - point1.y) / (point2.x - point1.x);
double mb = (point3.y - point2.y) / (point3.x - point2.x);
CGPoint c; // Center of a circle passing through all three points.
c.x = (((ma * mb * (point1.y - point3.y)) + (mb * (point1.x + point2.x)) - (ma * (point2.x + point3.x))) / (2 * (mb - ma)));
c.y = (((-1 / ma) * (c.x - ((point1.x + point2.x) / 2))) + ((point1.y + point2.y) / 2));
Then, to find the points on the tangent line, in this case, finding the control point for the curve going from point2 to point3:
double d = ...; // distance we want the point. Based on the distance between
// point2 and point3.
// mc: Slope of the line perpendicular to the line between
// point2 and c.
double mc = - (c.x - point2.x) / (c.y - point2.y);
CGPoint tp; // point on the tangent line
double c = point2.y - mc * point2.x; // c == y intercept
tp.x = ???; // can't figure this out, the question is whether it should be
// less than point2.x, or greater than?
tp.y = mc * tp.x + c;
// then, compute a point cp that is distance d from point2 going in the direction
// of tp.
It sounds like you might need to figure out the direction the curve is going, in order to set the tangent points so that it won't double back on itself. From what I understand, it would be simply finding out the direction from (x1, y1) to (x2, y2), and then travelling on the tangent line your heuristic distance in the direction closest to the (x1, y1) -> (x2, y2) direction, and plopping the tangent point there.
If you're really confident that you have a good way of choosing how far along the tangent line your points should be, and you only need to decide which side to put each one on, then I would suggest that you look once again at that circle to which the line is tangent. You've got z1,z2,z3 in that order on the circle; imagine going around the circle from z2 towards z1, but go along the tangent line instead; that's which side the control point "before z2" should be; the control point "after z2" should be on the other side.
Note that this guarantees always to put the two control points on opposite sides of z2, which is important. (Also: you probably want them to be the same distance from z2, because otherwise you'll get a discontinuity at z2 in, er, the second derivative of your curve, which is likely to look a bit suboptimal.) I bet there will still be pathological cases.
If you don't mind a fair bit of code complexity, there's a sophisticated and very effective algorithm for exactly your problem (and more) in Don Knuth's METAFONT program (whose main purpose is drawing fonts). The algorithm is due to John Hobby. You can find a detailed explanation, and working code, in METAFONT or, perhaps better, the closely related METAPOST (which generates PostScript output instead of huge bitmaps).
Pointing you at it is a bit tricky, though, because METAFONT and METAPOST are "literate programs", which means that their source code and documentation consist of a kind of hybrid of Pascal code (for METAFONT) or C code (for METAPOST) and TeX markup. There are programs that will turn this into a beautifully typeset document, but so far as I know no one has put the result on the web anywhere. So here's a link to the source code, which you may or may not find entirely incomprehensible: http://foundry.supelec.fr/gf/project/metapost/scmsvn/?action=browse&path=%2Ftrunk%2Fsource%2Ftexk%2Fweb2c%2Fmplibdir%2Fmp.w&view=markup -- in which you should search for "Choosing control points".
(The beautifully-typeset document for METAFONT is available as a properly bound book under the title "METAFONT: the program". But it costs actual money, and the code is in Pascal.)

Reflecting a circle off another circle

Working with iPhone and Objective C.
I am working on a game and I need to correctly reflect a ball off a circle object. I am trying to do it as a line and circle intersection. I have my ball position outside the circle and I have the new ball position that would be inside the circle at the next draw update. I know the intersect point of the line (ball path) and the circle. Now I want to rotate the ending point of the ball path about the intersection point to get the correct angle of reflection off the tangent.
The following are known:
ball current x,y
ball end x,y
ball radius
circle center x,y
circle radius
intersection point of ball path and circle x and y
I know I need to find the angle of incidence between the tangent line and the incoming ball path which will also equal my angle of reflection. I think once I know those two angles I can subtract them from 180 to get my rotation angle then rotate my end point about the angle of intersection by that amount. I just don't know how.
First, you should note that the center of the ball doesn't have to be inside of the circle to indicate that there's a reflection or bounce. As long as the distance between ball center and circle is less than the radius of the ball, there will be a bounce.
If the radius of the circle is R and the radius of the ball is r, things are simplified if you convert to the case where the circle has radius R+r and the ball has radius 0. For the purposes of collision detection and reflection/bouncing, this is equivalent.
If you have the point of intersection between the (enlarged) circle and the ball's path, you can easily compute the normal N to the circle at that point (it is the unit vector in the direction from the center of the circle to the collision point).
For an incoming vector V the reflected vector is V-2(N⋅V) N, where (N⋅V) is the dot product. For this problem, the incoming vector V is the vector from the intersection point to the point inside the circle.
As for the reflection formula given above, it is relatively easy to derive using vector math, but you can also Google search terms like "calculate reflection vector". The signs in the formula will vary with the assumed directions of V and N. Mathworld has a derivation although, as noted, the signs are different.
I only know the solution to the geometry part.
Let:
r1 => Radius of ball
r2 => Radius of circle
You can calculate the distance between the two circles by using Pythagoras theorem.
If the distance is less than the r1+r2 then do the collision.
For the collision,I would refer you Here. It's in python but I think it should give you an idea of what to do. Hopefully, even implement it in Objective C. The Tutorial By PeterCollingRidge.

Calculating collision for a moving circle, without overlapping the boundaries

Let's say I have circle bouncing around inside a rectangular area. At some point this circle will collide with one of the surfaces of the rectangle and reflect back. The usual way I'd do this would be to let the circle overlap that boundary and then reflect the velocity vector. The fact that the circle actually overlaps the boundary isn't usually a problem, nor really noticeable at low velocity. At high velocity it becomes quite clear that the circle is doing something it shouldn't.
What I'd like to do is to programmatically take reflection into account and place the circle at it's proper position before displaying it on the screen. This means that I have to calculate the point where it hits the boundary between it's current position and it's future position -- rather than calculating it's new position and then checking if it has hit the boundary.
This is a little bit more complicated than the usual circle/rectangle collision problem. I have a vague idea of how I should do it -- basically create a bounding rectangle between the current position and the new position, which brings up a slew of problems of it's own (Since the rectangle is rotated according to the direction of the circle's velocity). However, I'm thinking that this is a common problem, and that a common solution already exists.
Is there a common solution to this kind of problem? Perhaps some basic theories which I should look into?
Since you just have a circle and a rectangle, it's actually pretty simple. A circle of radius r bouncing around inside a rectangle of dimensions w, h can be treated the same as a point p at the circle's center, inside a rectangle (w-r), (h-r).
Now position update becomes simple. Given your point at position x, y and a per-frame velocity of dx, dy, the updated position is x+dx, y+dy - except when you cross a boundary. If, say, you end up with x+dx > W (letting W = w-r), then you do the following:
crossover = (x+dx) - W // this is how far "past" the edge your ball went
x = W - crossover // so you bring it back the same amount on the correct side
dx = -dx // and flip the velocity to the opposite direction
And similarly for y. You'll have to set up a similar (reflected) check for the opposite boundaries in each dimension.
At each step, you can calculate the projected/expected position of the circle for the next frame.
If this lies outside the rectangle, then you can then use the distance from the old circle position to the rectangle's edge and the amount "past" the rectangle's edge that the next position lies at (the interpenetration) to linearly interpolate and determine the precise time when the circle "hits" the rectangle edge.
For example, if the circle is 10 pixels away from the rectangle's edge, then is predicted to move to 5 pixels beyond it, you know that for 2/3rds of the timestep (10/15ths) it moves on its orginal path, then is reflected and continues on its new path for the remaining 1/3rd of the timestep (5/15ths). By calculating these two parts of the motion and "adding" the translations together, you can find the correct new position.
(Of course, it gets more complicated if you hit near a corner, as there may be several collisions during the timestep, off different edges. And if you have more than one circle moving, things get a lot more complex. But that's where you can start for the case you've asked about)
Reflection across a rectangular boundary is incredibly simple. Just take the amount that the object passed the boundary and subtract it from the boundary position. If the position without reflecting would be (-0.8,-0.2) for example and the upper left corner is at (0,0), the reflected position would be (0.8,0.2).