I am trying to figure out if a latitude/longitude point is contained within a polygon defined by vertexes that represent points on the earth (also lat/lon's, in clockwise order). This is trivial for polygons that can be mapped to the 2D lat/lon space.
Where this becomes increasingly difficult is circle's (now switching back to 3D) that may go from pole to pole covering half the earth. The translation to lat/lon looks like a sine wave. The 2D point in polygon test no longer applies to this case. Is there an algorithm that exists that solves this problem?
================== Clarifications on comments below: ===================
The polygon is defined as (lon, lat) pairs in degrees, i.e., (60, 90), (60, 110), (-30, 110), (-30, 90).
I do have code that implements the ray casting algorithm, and that works. however, certain polygons on the surface of the earth do not translate to closed polygons in the 2D space.
As stated by denniston.t, if you are only interested in circles, and you have a radius, you can simply check if the Great Circle Distance between the center point and the point is less than the radius. To find the great circle distance you typically use the Haversine Formula. The following is my implementation in python:
from math import radians, sin, cos, asin, sqrt
def haversine(point1, point2):
"""Gives the distance between two points on earth.
The haversine formula, given two sets of latitude and longitude,
returns the distance along the surface of the earth in miles,
ignoring potential changes in elevation. The points must be in
decimal degrees.
"""
earth_radius_miles = 3956
lat1, lon1 = (radians(coord) for coord in point1)
lat2, lon2 = (radians(coord) for coord in point2)
dlat, dlon = (lat2 - lat1, lon2 - lon1)
a = sin(dlat/2.0)**2 + cos(lat1) * cos(lat2) * sin(dlon/2.0)**2
great_circle_distance = 2 * asin(min(1,sqrt(a)))
d = earth_radius_miles * great_circle_distance
return d
If you have the center point and radius of your circle drawn on the surface of the sphere, calculate the Great-circle distance between the center point and target point. If it's less than the radius of the circle, the target point lies in the circle.
This will not generalize to arbitrary polygons drawn on your sphere, but you only asked about circles, so I don't know if it matters to you.
containsLocation(point:LatLng, polygon:Polygon)
Related
I am developing a GPS waypoint application. I have started by drawing my compass but am finding it difficult to implement degree text around the circle. Can anyone help me with a solution? The compass image am working on] 1 here shows the circle of the compass I have drawn.
This image here shows what I want to achieve, that is implementing degree text round the compass [Image of what I want to achieve] 2
Assuming you're doing this in a custom view, you need to use one of the drawText methods on the Canvas passed in to onDraw.
You'll have to do a little trigonometry to get the x, y position of the text - basically if there's a circle with radius r you're placing the text origins on (i.e. how far out from the centre they are), and you're placing one at angle θ:
x = r * cosθ
y = r * sinθ
The sin and cos functions take a value in radians, so you'll have to convert that if you're using degrees:
val radians = (degrees.toDouble() / 360.0) * (2.0 * Math.PI)
and 0 degrees is at 3 o'clock on the circle, not 12, so you'll have to subtract 90 degrees from your usual compass positions (e.g. 90 degrees on the compass is 0 degrees in the local coordinates). The negative values you get are fine, -90 is the same as 270. If you're trying to replicate the image you posted (where the numbers and everything else are rotating while the needle stays at the top) you'll have to apply an angle offset anyway!
These x and y values are distance from the centre of the circle, which probably needs to be the centre of your view (which you've probably already calculated to draw your circle). You'll also need to account for the extra space you need to draw those labels, scaling everything so it all fits in the View
Path image
I need to calculate distance from subscriber position to Position B in the image. I have the GPS co-ordinate of the subscriber and the "B" position. How can I calculate the distance?
Simple case: Express lat and long values in decimal form and use the standard geometry distance formula if subscriber is less than 100 miles from position B. distance = sqrt((lat1-lat2)^2 - (long1-long2)^2).
More general case: Use the haversine formulas using a great circle to calculate distances from points on a sphere for more accurate measurements if position B might be a continent or two away from the subscriber. Let's call the subscriber position A and say and say he is at lat[a], long[a] and the fixed point B is at lat[b], long[b]. Let r represent the radius of the earth (about 3961 miles).
distance = 2*r*arcsin(sqrt(sin^2((lat[b]-lat[a])/2) + cos(lat[a])*cos(lat[b])*sin^2((long[b]-long[a])/2)))
If you specify r in miles, your answer will come out in miles. If you use kilometers use 6373 for a good number for the earth's radius, and of course the answer will come out in kilometers.
Exact case: The haversine formula will not provide a perfect answer because the earth is not a perfect sphere. Even apart from the mountains and the canyons, the earth has a larger radius at the equator than it does at the poles. The radius at the equator is the equator is about 3963 miles, and at the poles it is about 3950 miles. So you really need to devise your own lookup table (or borrow one from google maps) if you are measuring distances halfway around the globe and you have to be exact.
The haversine formula will be accurate to less than half of a percentage point. In 1000 miles your answer will be accurate to within 5 miles.
Haversine formula: https://en.wikipedia.org/wiki/Haversine_formula
Radius of the earth: https://en.wikipedia.org/wiki/Earth_radius
I am currently trying to figure out a way to calcute the size of a given object with kinect
since I have the following data
angular field of view of the lens
distance
and width in pixels from a 800*600 resolution
I believe this can be possible to calculate. Does anyone has math skills to give me a little help?
With some trigonometry, it should be possible to approximate.
If you draw a right trangle ABC, with the camera at one of the legs (A), and the object at the far end (edge BC), where the right angle is (C), then the height of the object is going to be the height of leg BC. the distance to the pixel might be the distance of leg AC or AB. The Kinect sensor specifications are going to regulate that. If you get distance to the center of a pixel, then it will be AC. if you have distances to pixel corners then the distance will be AB.
With A representing the angle at the camera that the pixel takes up, d is the distance of the hypotenuse of a right angle and y is the distance of the far leg (edge BC):
sin(A) = y / d
y = d sin(A)
y is the length of the pixel projected into the object plane. You calculate it by multiplying the sin of the angel by the distance to the object.
Here I confess I do not know the API of the kinect, and what level of detail it provides. You say you have the angle of the field of vision. You might assume each pixel of your 800x600 pixel grid takes up an equal angle of your camera's field of vision. If you do, then you can break up that field of vision into equal pieces to measure the linear size of your object in each pixel.
You also mentioned that you have the distance to the object. I was assuming that you have a distance map for each pixel of the 800x600 grid. If this is incorrect, some calculations can be done to approximate a distance grid for the pixels involving the object of interest if you make some assumptions about the object being measured.
I'm following an example in a Processing book describing how to calculate nonorthogonal collisions (a ball bouncing on a non-horizontal plane), however, I don't really understand the logic behind these four expressions.
float groundXTemp = cosine * deltaX + sine * deltaY;
float groundYTemp = cosine * deltaY - sine * deltaX;
float velocityXTemp = cosine * velocity.vx + sine * velocity.vy;
float velocityYTemp = cosine * velocity.vy - sine * velocity.vx;
They're supposed to be calculating temporary values for the ground coordinates and velocity of the ball to calculate the collision as if it were orthogonal. Cosine and sine are the values for the rotation of the ground, and the velocity variables are the velocity of the ball. I can't grok what the expressions are actually doing to make the ground horizontal, and the book doesn't explain it very well. Any help would be appreciated.
These expressions are traditional expressions of a rotation. If you take a point (x,y), and you rotate it by an angle theta, you will obtain a point (x',y') with coordinates :
x' = cos(theta)*x - sin(theta)*y
y' = sin(theta)*x + cos(theta)*y
In your case, let's say that theta is the angle of the ground, you want to do the inverse rotation (so with angle -theta) to make the ground horizontal, this is why the sign is different from the formula above (cos(-theta) = cos(theta) and sin(-theta) = -sin(theta)).
If you want to go into details, check : http://en.wikipedia.org/wiki/Rotation_matrix
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.