Is there a nice piece of code (simple if statements that are easy to understand) to test whether a coordinate (lat, lng) falls in a rectangle (if bottomLeft and topRight coordinates are known).
UPD: I did find some similar questions but they the answers look hacky.
You can simply check:
that the target latitude if between these of the bottomLeft and topRight points
same for longitude
If both conditions are true, the target point falls inside the rectangle.
PS: the Earth is (more or less) spherical, so if the bottomLeft and topRight points are far away, the rectangle will have rounded sides.
Related
For example , it assumes that the black rectangle is a square.
and The gps coordinates of the red circle , which is in the square is lat 126.993611 long 37.5727
i want to know the yellow circle and blue circle coordinates displayed on the screen.
Units of length are all M(meter)
please let me know best calculate formula!
You can use geofence for this.
https://developer.android.com/training/location/geofencing.html
or you write a method which calculates your distance by coordinates.
Then you can use a rise and this method to get the new location.
I have an application that is using bing maps.
I can get the boundary of the current show map - for example, if my map is on Canada, then I will get the boundary of Canada (A rectangle):
LocationRect bounds = map.Bounds;
bounds has - Height, Width, East(point of type double), West, North, South, Center.
How can I get the bounds * 2? (In math I think it's Area * 2).
Explanation:
I have the bounds of the map (A rectangle).
I want to enlarge this bounds to be bigger twice.
If my rectangle was 2cm, 5cm -> it will become 4cm, 10cm.
If I understand it right it is a simple question if not... :-)
You have a Black rectangle that you know the ABCD coordinates.
Calculate the distance between AD Which will be your Y and the distance between AB that will be your X.
In order to get get the EFHG coordinates of a new rectangle that you need to follow the sketch, just add or subtract from vertexes coordinates of the Black rectangle in order to get the Blue rectangle.
Of course, you need to check all the time that the coordinates of the blue rectangle do not exceed the maximum coordinates of the map.
How do I make a infinite/repeating world that handles rotation, just like in this game:
http://bloodfromastone.co.uk/retaliation.html
I have coded my rotating moving world by having a hierarchy like this:
Scene
- mainLayer (CCLayer)
- rotationLayer(CCNode)
- positionLayer(CCNode)
The rotationLayer and positionLayer have the same size (4000x4000 px right now).
I rotate the whole world by rotating the rotationLayer, and I move the whole world by moving the positionLayer, so that the player always stays centered on the device screen and it is the world that moves and rotates.
Now I would like to make it so that if the player reaches the bounds of the world (the world is moved so that the worlds bounds gets in to contact with the device screen bounds), then the world is "wrapped" to the opposite bounds so that the world is infinite. If the world did not rotate that would be easy, but now that it does I have no idea how to do this. I am a fool at math and in thinking mathematically, so I need some help here.
Now I do not think I need any cocos2d-iphone related help here. What I need is some way to calculate if my player is outside the bounds of the world, and then some way to calculate what new position I must give the world to wrap the world.
I think I have to calculate a radius for a circle that will be my foundry inside the square world, that no matter what angle the square world is in, will ensure that the visible rectangle (the screen) will always be inside the bounds of the world square. And then I need a way to calculate if the visible rectangle bounds are outside the bounds circle, and if so I need a way to calculate the new opposite position in the bounds circle to move the world to. So to illustrate I have added 5 images.
Visible rectangle well inside bounds circle inside a rotated square world:
Top of visible rectangle hitting bounds circle inside a rotated square world:
Rotated square world moved to opposite vertical position so that bottom of visible rectangle now hitting bounds circle inside rotated world:
Another example of top of visible rectangle hitting bounds circle inside a rotated square world to illustrate a different scenario:
And again rotated square world moved to opposite vertical position so that bottom of visible rectangle now hitting bounds circle inside rotated world:
Moving the positionLayer in a non-rotated situation is the math that I did figure out, as I said I can figure this one out as long as the world does not get rotate, but it does. The world/CCNode (positionLayer) that gets moved/positioned is inside a world/CCNode (rotationLayer) that gets rotated. The anchor point for the rotationLayer that rotates is on the center of screen always, but as the positionLayer that gets moved is inside the rotating rotationLayer it gets rotated around the rotationLayer's anchor point. And then I am lost... When I e.g. move the positionLayer down enough so that its top border hits the top of the screen I need to wrap that positionLayer as JohnPS describes but not so simple, I need it to wrap in a vector based on the rotation of the rotationLayer CCNode. This I do not know how to do.
Thank you
Søren
Like John said, the easiest thing to do is to build a torus world. Imagine that your ship is a point on the surface of the donut and it can only move on the surface. Say you are located at the point where the two circles (red and purple in the picture) intersect:
.
If you follow those circles you'll end up where you started. Also, notice that, no matter how you move on the surface, there is no way you're going to reach an "edge". The surface of the torus has no such thing, which is why it's useful to use as an infinite 2D world. The other reason it's useful is because the equations are quite simple. You specify where on the torus you are by two angles: the angle you travel from the "origin" on the purple circle to find the red circle and the angle you travel on the red circle to find the point you are interested in. Both those angles wrap at 360 degrees. Let's call the two angles theta and phi. They are your ship's coordinates in the world, and what you change when you change velocities, etc. You basically use them as your x and y, except you have to make sure to always use the modulus when you change them (your world will only be 360 degrees in each direction, it will then wrap around).
Suppose now that your ship is at coordinates (theta_ship,phi_ship) and has orientation gamma_ship. You want to draw a square window with the ship at its center and length/width equal to some percentage n of the whole world (say you only want to see a quarter of the world at a time, then you'd set n = sqrt(1/4) = 1/2 and have the length and width of the window set to n*2*pi = pi). To do this you need a function that takes a point represented in the screen coordinates (x and y) and spits out a point in the world coordinates (theta and phi). For example, if you asked it what part of the world corresponds to (0,0) it should return back the coordinates of the ship (theta_ship,phi_ship). If the orientation of the ship is zero (x and y will be aligned with theta and phi) then some coordinate (x_0,y_0) will correspond to (theta_ship+k*x_0, phi_ship+k*y_0), where k is some scaling factor related to how much of the world one can see in a screen and the boundaries on x and y. The rotation by gamma_ship introduces a little bit of trig, detailed in the function below. See the picture for exact definitions of the quantities.
!Blue is the screen coordinate system, red is the world coordinate system and the configuration variables (the things that describe where in the world the ship is). The object
represented in world coordinates is green.
The coordinate transformation function might look something like this:
# takes a screen coordinate and returns a world coordinate
function screen2world(x,y)
# this is the angle between the (x,y) vector and the center of the screen
alpha = atan2(x,y);
radius = sqrt(x^2 + y^2); # and the distance to the center of the screen
# this takes into account the rotation of the ship with respect to the torus coords
beta = alpha - pi/2 + gamma_ship;
# find the coordinates
theta = theta_ship + n*radius*cos(beta)/(2*pi);
phi = phi_ship + n*radius*sin(beta)/(2*pi));
# return the answer, making sure it is between 0 and 2pi
return (theta%(2*pi),phi%(2*pi))
and that's pretty much it, I think. The math is just some relatively easy trig, you should make a little drawing to convince yourself that it's right. Alternatively you can get the same answer in a somewhat more automated fashion by using rotations matrices and their bigger brother, rigid body transformations (the special Euclidian group SE(2)). For the latter, I suggest reading the first few chapters of Murray, Li, Sastry, which is free online.
If you want to do the opposite (go from world coordinates to screen coordinates) you'd have to do more or less the same thing, but in reverse:
beta = atan2(phi-phi_ship, theta-theta_ship);
radius = 2*pi*(theta-theta_ship)/(n*cos(beta));
alpha = beta + pi/2 - gamma_ship;
x = radius*cos(alpha);
y = radius*sin(alpha);
You need to define what you want "opposite bounds" to mean. For 2-dimensional examples see Fundamental polygon. There are 4 ways that you can map the sides of a square to the other sides, and you get a sphere, real projective plane, Klein bottle, or torus. The classic arcade game Asteroids actually has a torus playing surface.
The idea is you need glue each of your boundary points to some other boundary point that will make sense and be consistent.
If your world is truly 3-dimensional (not just 3-D on a 2-D surface map), then I think your task becomes considerably more difficult to determine how you want to glue your edges together--your edges are now surfaces embedded in the 3-D world.
Edit:
Say you have a 2-D map and want to wrap around like in Asteroids.
If the map is 1000x1000 units, x=0 is the left border of the map, x=999 the right border, and you are looking to the right and see 20 units ahead. Then at x=995 you want to see up to 1015, but this is off the right side of the map, so 1015 should become 15.
If you are at x=5 and look to the left 20 units, then you see x=-15 which you really want to be 985.
To get these numbers (always between 0 and 999) when you are looking past the border of your map you need to use the modulo operator.
new_x = x % 1000; // in many programming languages
When x is negative each programming language handles the result of x % 1000 differently. It can even be implementation defined. i.e. it will not always be positive (between 0 and 999), so using this would be safer:
new_x = (x + 1000) % 1000; // result 0 to 999, when x >= -1000
So every time you move or change view you need to recompute the coordinates of your position and coordinates of anything in your view. You apply this operation to get back a coordinate on the map for both x and y coordinates.
I'm new to Cocos2d, but I think I can give it a try on helping you with the geometry calculation issue, since, as you said, it's not a framework question.
I'd start off by setting the anchor point of every layer you're using in the visual center of them all.
Then let's agree on the assumption that the first part to touch the edge will always be a corner.
In case you just want to check IF it's inside the circle, just check if all the four edges are inside the circle.
In case you want to know which edge is touching the circumference of the circle, just check for the one that is the furthest from point x=0 y=0, since the anchor will be at the center.
If you have a reason for not putting the anchor in the middle, you can use the same logic, just as long as you include half of the width of each object on everything.
Firstly - Z is Up in this problem.
Context: Top down 2D Game using 3D objects.
The player and all enemies are Spheres that can move in any direction on a 2D Plane (XY). They rotate as you would expect when they move. Their velocity is a 3D vector in world space and this is how I influence them. They aren't allowed to rotate on the spot.
I need to find a formula to determine the direction one of these spheres should move in order to get their Z-Axis (or any axis really) pointing a specified direction in world space.
Some examples may be in order:
X
|
Z--Y
This one is simple: The Spheres local axes matches the world so if I want the Spheres Z-Axis to point along 1,0,0 then I can move the sphere along 1,0,0.
The one that gives me trouble is this:
X
|
Y--Z
Now I know that to get the Z-Axis to point along 1,0,0 in world space I have to tell the sphere to move along 1,1,0 but I don't know/understand WHY that is the case.
I've been programming for ten years but I absolutely suck at vector maths so assume I'm an idiot when trying to explain :)
All right, I think I see what you mean.
Take a ball-- you must have one lying around. Mark a spot on it to indicate an axis of interest. Now pick a direction in which you want the axis to point. The trick is to rotate the ball in place to bring the axis to the right direction-- we'll get to the rolling in a minute.
The obvious way is to move "directly", and if you do this a few times you'll notice that the axis around which you are rotating the ball is perpendicular to the axis you're trying to move. It's as if the spot is on the equator and you're rotating around the North-South axis. Every time you pick a new direction, that direction and your marked axis determine the new equator. Also notice (this may be tricky) that you can draw a great circle (that's a circle that goes right around the sphere and divides it into equal halves) that goes between the mark and the destination, so that they're on opposite hemispheres, like mirror images. The poles are always on that circle.
Now suppose you're not free to choose the poles like that. You have a mark, you have a desired direction, so you have the great circle, and the north pole will be somewhere on the circle, but it could be anywhere. Imagine that someone else gets to choose it. The mark will still rotate to the destination, but they won't be on the equator any more, they'll be at some other latitude.
Now put the ball on the floor and roll it -- don't worry about the mark for now. Notice that it rotates around a horizontal axis, the poles, and touches the floor along a circle, the equator (which is now vertical). The poles must be somewhere on the "waist" of the sphere, halfway up from the floor (don't call it the equator). If you pick the poles on that circle, you choose the direction of rolling.
Now look at the marks, and draw the great circle that divides them. The poles must be on that circle. Look where that circle crosses the "waist"; that's where your poles must be.
Tell me if this makes sense, and we can put in the math.
If I have a given rectangle, with the width w, height h and angle r
How large does another rectangle which contains all points of the rotated rectangle need to be?
I would need this to perform fast bounding box checks for a 2D physics engine I am making
this may be what you need:
Calculate Bounding box coordinates from a rotated rectangle, answered by someone named markus.
You usually should consider rotating rectangles in a collision detection engine, since it will be quite straightforward to implement (I mean considering the rotated rectangle as it is).
In any case if you really want to simplify to have a coarse-level of collision detection the best thing is to embed the rectangle inside a circle, because it's really simple (centered over rectangle center and with a radius of the semi-diagonal of the rectangle) and compared to using a box it can be quite accurate for a coarse detection. Actually you can have an angle threshold to decide if it's better to use a circle or to consider the original rectangle (most degenerating cases are when angle is near to k*PI with k = 0,1,2,3
If you really really want to consider the rotated rectangle you can calculate it easily by choosing the topmost vertex of your rectangle (xT, yT) and the leftmost (xL, yL) (after the rotation of course) to obtain the topleft point that will be (xL, yT). Then you do the same thing for the bottomright corner taking (xR, yB) from the rightmost and lowest point of your rectangle and you have it. It will be the rectangle included in (xL, yY) (xR, yB).