I am kinda new on this. I am about to convert math equation to code lately. So I started practicing on simple formulas such as Fibonacci numbers and haversine formula.
But I am kinda confused with the haversine code on the following references;
Using the Haversine Formula in Javascript
Calculate distance between two latitude-longitude points? (Haversine formula)
Generally, what they do is,
var R = 6371e3; // metres
var φ1 = lat1.toRadians();
var φ2 = lat2.toRadians();
var Δφ = (lat2-lat1).toRadians();
var Δλ = (lon2-lon1).toRadians();
var a = Math.sin(Δφ/2) * Math.sin(Δφ/2) +
Math.cos(φ1) * Math.cos(φ2) *
Math.sin(Δλ/2) * Math.sin(Δλ/2);
var c = 2 * Math.atan2(Math.sqrt(a), Math.sqrt(1-a));
var d = R * c;
And I have the reference of original haversine formula from here.
Then, notice of this line,
var c = 2 * Math.atan2(Math.sqrt(a), Math.sqrt(1-a));
Which part of it that comes from original haversine formula that is in wikipedia? I haven't found any Tangent at all on those graphs. Am I missing something here? Any suggestion is appreciated.
By definition
hav(θ)=sin2(θ/2)
If hav(θ)=a, then
θ = 2 arcsin(sqrt(a))
After expressing arcsine with arctangent, this becomes:
θ = 2 arctan(sqrt(a)/(sqrt(1-a)))
UPDATE Clarification in response to OP's comment:
The haversine function hav(θ) for some angle θ is a shorthand for sin2(θ/2). Given two points on a sphere and θ being the flat angle between radii connecting those points with the center of the sphere, the haversine formula expresses the haversine function with the lattitude (φ) and longitude (λ) values of those points. Thus, you can calculate the value of the haversine function (which is represented in your code by the variable a) through known values φ and λ. Then you can find from a the value of θ (which is represented in your code by the variable c). My explanation above focuses exactly on that part. Then with a known value of θ (and the radius R) you can compute the great-circle distance d.
It's just another formulation of the Haversine formula.
From https://en.wikipedia.org/wiki/Inverse_trigonometric_functions#Relationships_among_the_inverse_trigonometric_functions:
arctan(x) = arcsin(x / sqrt(x^2 + 1))
With X = x / sqrt(x^2 + 1), you end up with:
x = X / sqrt(1 - X^2)
Hence:
arcsin(X) = arctan(X / sqrt(1 - X^2))
Then in the Haversine formula from Wikipedia:
d = 2R * arcsin(sqrt(a))
= 2R * arctan(sqrt(a) / sqrt(1 - a))
Related
I was reading chapter 7 of "Competitive Programming 3" by Steven and Felix Halim. I then found this problem, given in picture:
Here two intersecting points of two circles and corresponding radius are also given. I have to find centres of two circles. They have given the solution code. But I didn't understand the technique behind it. The given code is:
I have searched many times, but I didn't find. Can anyone explain the technique?
Anyway, thanks in advance :)
Let's first visualize variables d2 and h defined in the code:
As we can see, d2 stands for the square of the distance between p1 and p2. Let's put d = sqrt(d2). Then d^2 = d2.
Using Pythagoras: r^2 = h^2 + d^2/4. Hence, h^2 = r^2 - d^2/4.
The unitary (norm = 1) vector in the direction of the line joining p1 and p2 is:
v := (p2 - p1)/d = (p2.x - p1.x, p2.y - p1.y)/d.
its perpendicular is
w := (p2.y - p1.y, p1.x - p2.x)/d
Now we can express c1 as a point in the perpendicular direction:
c1 = q + w*h = (p1 + p2)/2 + w*h,
because w has norm 1 and h is precisely the distance between c1 and q. Therefore,
c1 = (p1.x + p2.x, p1.y + p2.y)/2 + (p2.y - p1.y, p1.x - p2.x)*h/d
where
h/d = sqrt(r^2 - d^2/4)/d = sqrt(r^2/d2 - 1/4)
which explains the code.
Notes
From the picture that r is always ge than d/2. Thus, r^2 ≥ d^2/4 or (r/d)^2 ≥ 1/4 and there is no need to check whether det < 0, because it never is (provided the circles intersect).
The derivation above will actually produce two solutions for c1, one to the right of the blue line from p1 to p2, which is the one in the drawing, and another on the left of said line. In fact, these correspond to the equations
c1 = q ± w*h = q + w*(±h)
In order to decide whether we should use +h or -h, we can apply one of the well-known criteria for establishing whether a point lies on the left or the right of a directed segment. For example, we could compute the sign of the determinant
| 1 p1.x p1.y |
D = | 1 p2.x p2.y | = (p2.x-p1.x)(c1.y-p1.y) - (p2.y-p1.y)(c1.x-p1.x)
| 1 c1.x c1.y |
which will have a sign for +h and the opposite for -h. The value of h that makes D < 0 is the one that leaves c1 on the right of the segment from p1 to p2.
Apply Pythagoras in the right triangles. This gives you the distances between the midpoint of p1p2 and the centers.
From unit vectors parallel then orthogonal to p1p2, the offset vectors are easily obtained.
We have a location specified by a lat & lng
We want to generate some other random lat & lng locations within say 20km of this location
Does anyone have a good formula for this
Generate two uniform random values r and Fi in range 0..1
Calculate distance as d = Radius * Sqrt(r) (description here for plane circle)
Calculate bearing as Theta=2 * Pi * Fi
Find lat/lon coordinates for given central point and calculated d and Theta as described here in section Destination point given distance and bearing from start point
JavaScript:
(all angles
in radians)
var φ2 = Math.asin( Math.sin(φ1)*Math.cos(d/R) +
Math.cos(φ1)*Math.sin(d/R)*Math.cos(θ) );
var λ2 = λ1 + Math.atan2(Math.sin(θ)*Math.sin(d/R)*Math.cos(φ1),
Math.cos(d/R)-Math.sin(φ1)*Math.sin(φ2));
where φ is latitude, λ is longitude,
θ is the bearing (clockwise from north), d being the distance travelled,
R the earth’s radius
Suppose I want to approximate a half-cosine curve in SVG using bezier paths. The half cosine should look like this:
and runs from [x0,y0] (the left-hand control point) to [x1,y1] (the right-hand one).
How can I find an acceptable set of coefficients for a good approximation of this function?
Bonus question: how is it possible to generalize the formula for, for example, a quarter of cosine?
Please note that I don't want to approximate the cosine with a series of interconnected segments, I'd like to calculate a good approximation using a Bezier curve.
I tried the solution in comments, but, with those coefficients, the curve seems to end after the second point.
Let's assume you want to keep the tangent horizontal on both ends. So naturally the solution is going to be symmetric, and boils down to finding a first control point in horizontal direction.
I wrote a program to do this:
/*
* Find the best cubic Bézier curve approximation of a sine curve.
*
* We want a cubic Bézier curve made out of points (0,0), (0,K), (1-K,1), (1,1) that approximates
* the shifted sine curve (y = a⋅sin(bx + c) + d) which has its minimum at (0,0) and maximum at (1,1).
* This is useful for CSS animation functions.
*
* ↑ P2 P3
* 1 ו••••••***×
* | ***
* | **
* | *
* | **
* | ***
* ×***•••••••×------1-→
* P0 P1
*/
const sampleSize = 10000; // number of points to compare when determining the root-mean-square deviation
const iterations = 12; // each iteration gives one more digit
// f(x) = (sin(π⋅(x - 1/2)) + 1) / 2 = (1 - cos(πx)) / 2
const f = x => (1 - Math.cos(Math.PI * x)) / 2;
const sum = function (a, b, c) {
if (Array.isArray(c)) {
return [...arguments].reduce(sum);
}
return [a[0] + b[0], a[1] + b[1]];
};
const times = (c, [x0, x1]) => [c * x0, c * x1];
// starting points for our iteration
let [left, right] = [0, 1];
for (let digits = 1; digits <= iterations; digits++) {
// left and right are always integers (digits after 0), this keeps rounding errors low
// In each iteration, we divide them by a higher power of 10
let power = Math.pow(10, digits);
let min = [null, Infinity];
for (let K = 10 * left; K <= 10 * right; K+= 1) { // note that the candidates for K have one more digit than previous `left` and `right`
const P1 = [K / power, 0];
const P2 = [1 - K / power, 1];
const P3 = [1, 1];
let bezierPoint = t => sum(
times(3 * t * (1 - t) * (1 - t), P1),
times(3 * t * t * (1 - t), P2),
times(t * t * t, P3)
);
// determine the error (root-mean-square)
let squaredErrorSum = 0;
for (let i = 0; i < sampleSize; i++) {
let t = i / sampleSize / 2;
let P = bezierPoint(t);
let delta = P[1] - f(P[0]);
squaredErrorSum += delta * delta;
}
let deviation = Math.sqrt(squaredErrorSum); // no need to divide by sampleSize, since it is constant
if (deviation < min[1]) {
// this is the best K value with ${digits + 1} digits
min = [K, deviation];
}
}
left = min[0] - 1;
right = min[0] + 1;
console.log(`.${min[0]}`);
}
To simplify calculations, I use the normalized sine curve, which passes through (0,0) and (1,1) as its minimal / maximal points. This is also useful for CSS animations.
It returns (.3642124232,0)* as the point with the smallest root-mean-square deviation (about 0.00013).
I also created a Desmos graph that shows the accuracy:
(Click to try it out - you can drag the control point left and right)
* Note that there are rounding errors when doing math with JS, so the value is presumably accurate to no more than 5 digits or so.
Because a Bezier curve cannot exactly reconstruct a sinusoidal curve, there are many ways to create an approximation. I am going to assume that our curve starts at the point (0, 0) and ends at (1, 1).
Simple method
A simple way to approach this problem is to construct a Bezier curve B with the control points (K, 0) and ((1 - K), 1) because of the symmetry involved and the desire to keep a horizontal tangent at t=0 and t=1.
Then we just need to find a value of K such that the derivative of our Bezier curve matches that of the sinusoidal at t=0.5, i.e., .
Since the derivative of our Bezier curve is given by , this simplifies to at the point t=0.5.
Setting this equal to our desired derivative, we obtain the solution
Thus, our approximation results in:
cubic-bezier(0.3633802276324187, 0, 0.6366197723675813, 1)
and it comes very close with a root mean square deviation of about 0.000224528:
Advanced Method
For a better approximation, we may want to minimize the root mean square of their difference instead. This is more complicated to calculate, as we are now trying to find the value of K in the interval (0, 1) that minimizes the following expression:
where B is defined as follows:
cubic-bezier(0.364212423249, 0, 0.635787576751, 1)
After few tries/errors, I found that the correct ratio is K=0.37.
"M" + x1 + "," + y1
+ "C" + (x1 + K * (x2 - x1)) + "," + y1 + ","
+ (x2 - K * (x2 - x1)) + "," + y2 + ","
+ x2 + "," + y2
Look at this samples to see how Bezier matches with cosine: http://jsfiddle.net/6165Lxu6/
The green line is the real cosine, the black one is the Bezier. Scroll down to see 5 samples. Points are random at each refresh.
For the generalization, I suggest to use clipping.
I would recommend reading this article on the math of bezier curves and ellipses, as this is basicly what you want (draw a part of an ellipse):
http://www.spaceroots.org/documents/ellipse/elliptical-arc.pdf
it provides some of the insights required.
then look at this graphic:
http://www.svgopen.org/2003/papers/AnimatedMathematics/ellipse.svg
where an example is made for an ellipse
now that you get the math involved, please see this example in LUA ;)
http://commons.wikimedia.org/wiki/File:Harmonic_partials_on_strings.svg
tada...
I have triangle: a, b, c. Each vertex has a value: va, vb, vc. In my software the user drags point p around inside and outside of this triangle. I use barycentric coordinates to determine the value vp at p based on va, vb, and vc. So far, so good.
Now I want to limit p so that vp is within range min and max. If a user chooses p where vp is < min or > max, how can I find the point closest to p where vp is equal to min or max, respectively?
Edit: Here is an example where I test each point. Light gray is within min/max. How can I find the equations of the lines that make up the min/max boundary?
a = 200, 180
b = 300, 220
c = 300, 300
va = 1
vb = 1.4
vc = 3.2
min = 0.5
max = 3.5
Edit: FWIW, so far first I get the barycentric coordinates v,w for p using the triangle vertices a, b, c (standard stuff I think, but looks like this). Then to get vp:
u = 1 - w - v
vp = va * u + vb * w + vc * v
That is all fine. My trouble is that I need the line equations for min/max so I can choose a new position for p when vp is out of range. The new position for p is the point closest to p on the min or max line.
Note that p is an XY coordinate and vp is a value for that coordinate determined by the triangle and the values at each vertex. min and max are also values. The two line equations I need will give me XY coordinates for which the values determined by the triangle are min or max.
It doesn't matter if barycentric coordinates are used in the solution.
The trick is to use the ratio of value to cartesian distance to extend each triangle edge until it hits min or max. Easier to see with a pic:
The cyan lines show how the triangle edges are extended, the green Xs are points on the min or max lines. With just 2 of these points we know the slope if the line. The yellow lines show connecting the Xs aligns with the light gray.
The math works like this, first get the value distance between vb and vc:
valueDistBtoC = vc - vb
Then get the cartesian distance from b to c:
cartesianDistBtoC = b.distance(c)
Then get the value distance from b to max:
valueDistBtoMax = max - vb
Now we can cross multiply to get the cartesian distance from b to max:
cartesianDistBtoMax = (valueDistBtoMax * cartesianDistBtoC) / valueDistBtoC
Do the same for min and also for a,b and c,a. The 6 points are enough to restrict the position of p.
Consider your triangle to actually be a 3D triangle, with points (ax,ay,va), (bx,by,vb), and (cx,cy,vc). These three points define a plane, containing all the possible p,vp triplets obtainable through barycentric interpolation.
Now think of your constraints as two other planes, at z>=max and z<=min. Each of these planes intersects your triangle's plane along an infinite line; the infinite beam between them, projected back down onto the xy plane, represents the area of points which satisfy the constraints. Once you have the lines (projected down), you can just find which (if either) is violated by a particular point, and move it onto that constraint (along a vector which is perpendicular to the constraint).
Now I'm not sure about your hexagon, though. That's not the shape I would expect.
Mathematically speaking the problem is simply a change of coordinates. The more difficult part is finding a good notation for the quantities involved.
You have two systems of coordinates: (x,y) are the cartesian coordinates of your display and (v,w) are the baricentric coordinates with respect to the vectors (c-a),(b-a) which determine another (non orthogonal) system.
What you need is to find the equation of the two lines in the (x,y) system, then it will be easy to project the point p on these lines.
To achieve this you could explicitly find the matrix to pass from (x,y) coordinates to (v,w) coordinates and back. The function you are using toBaryCoords makes this computation to find the coordinates (v,w) from (x,y) and we can reuse that function.
We want to find the coefficients of the transformation from world coordinates (x,y) to barycentric coordinates (v,w). It must be in the form
v = O_v + x_v * x + y_v * y
w = O_w + x_w * x + y_w * y
i.e.
(v,w) = (O_v,O_w) + (x_v,y_y) * (x,y)
and you can determine (O_v,O_w) by computing toBaryCoord(0,0), then find (x_v,x_w) by computing the coordinates of (1,0) and find (y_v,y_w)=toBaryCoord(1,0) - (O_v,O_w) and then find (y_v,y_w) by computing (y_v,y_w) = toBaryCoord(0,1)-(O_v,O_w).
This computation requires calling toBaryCoord three times, but actually the coefficients are computed inside that routine every time, so you could modify it to compute at once all six values.
The value of your function vp can be computed as follows. I will use f instead of v because we are using v for a baricenter coordinate. Hence in the following I mean f(x,y) = vp, fa = va, fb = vb, fc = vc.
You have:
f(v,w) = fa + (fb-fa)*v + (fc-fa)*w
i.e.
f(x,y) = fa + (fb-fa) (O_v + x_v * x + y_v * y) + (fc-fa) (O_w + x_w * x + y_w * y)
where (x,y) are the coordinates of your point p. You can check the validity of this equation by inserting the coordinates of the three vertices a, b, c and verify that you obtain the three values fa, fb and fc. Remember that the barycenter coordinates of a are (0,0) hence O_v + x_v * a_x + y_v * a_y = 0 and so on... (a_x and a_y are the x,y coordinates of the point a).
If you let
q = fa + (fb_fa)*O_v + (fc-fa)*O_w
fx = (fb-fa)*x_v + (fc-fa) * x_w
fy = (fb-fa)*y_v + (fc-fa) * y_w
you get
f(x,y) = q + fx*x + fy * y
Notice that q, fx and fy can be computed once from a,b,c,fa,fb,fc and you can reuse them if you only change the coordinates (x,y) of the point p.
Now if f(x,y)>max, you can easily project (x,y) on the line where max is achieved. The coordinates of the projection are:
(x',y') = (x,y) - [(x,y) * (fx,fy) - max + q]/[(fx,fy) * (fx,fy)] (fx,fy)
Now. You would like to have the code. Well here is some pseudo-code:
toBarycoord(Vector2(0,0),a,b,c,O);
toBarycoord(Vector2(1,0),a,b,c,X);
toBarycoord(Vector2(0,1),a,b,c,Y);
X.sub(O); // X = X - O
Y.sub(O); // Y = Y - O
V = Vector2(fb-fa,fc-fa);
q = fa + V.dot(O); // q = fa + V*O
N = Vector2(V.dot(X),V.dot(Y)); // N = (V*X,V*Y)
// p is the point to be considered
f = q + N.dot(p); // f = q + N*p
if (f > max) {
Vector2 tmp;
tmp.set(N);
tmp.multiply((N.dot(p) - max + q)/(N.dot(N))); // scalar multiplication
p.sub(tmp);
}
if (f < min) {
Vector2 tmp;
tmp.set(N);
tmp.multiply((N.dot(p) - min + q)/(N.dot(N))); // scalar multiplication
p.sum(tmp);
}
We think of the problem as follows: The three points are interpreted as a triangle floating in 3D space with the value being the Z-axis and the cartesian coordinates mapped to the X- and Y- axes respectively.
Then the question is to find the gradient of the plane that is defined by the three points. The lines where the plane intersects with the z = min and z = max planes are the lines you want to restrict your points to.
If you have found a point p where v(p) > max or v(p) < min we need to go in the direction of the steepest slope (the gradient) until v(p + k * g) = max or min respectively. g is the direction of the gradient and k is the factor we need to find. The coordinates you are looking for (in the cartesian coordinates) are the corresponding components of p + k * g.
In order to determine g we calculate the orthonormal vector that is perpendicular to the plane that is determined by the three points using the cross product:
// input: px, py, pz,
// output: p2x, p2y
// local variables
var v1x, v1y, v1z, v2x, v2y, v2z, nx, ny, nz, tp, k,
// two vectors pointing from b to a and c respectively
v1x = ax - bx;
v1y = ay - by;
v1z = az - bz;
v2x = cx - bx;
v2y = cy - by;
v2z = cz - bz;
// the cross poduct
nx = v2y * v1z - v2z * v1y;
ny = v2z * v1x - v2x * v1z;
nz = v2x * v1y - v2y * v1x;
// using the right triangle altitude theorem
// we can calculate the vector that is perpendicular to n
// in our triangle we are looking for q where p is nz, and h is sqrt(nx*nx+ny*ny)
// the theorem says p*q = h^2 so p = h^2 / q - we use tp to disambiguate with the point p - we need to negate the value as it points into the opposite Z direction
tp = -(nx*nx + ny*ny) / nz;
// now our vector g = (nx, ny, tp) points into the direction of the steepest slope
// and thus is perpendicular to the bounding lines
// given a point p (px, py, pz) we can now calculate the nearest point p2 (p2x, p2y, p2z) where min <= v(p2z) <= max
if (pz > max){
// find k
k = (max - pz) / tp;
p2x = px + k * nx;
p2y = py + k * ny;
// proof: p2z = v = pz + k * tp = pz + ((max - pz) / tp) * tp = pz + max - pz = max
} else if (pz < min){
// find k
k = (min - pz) / tp;
p2x = px + k * nx;
p2y = py + k * ny;
} else {
// already fits
p2x = px;
p2y = py;
}
Note that obviously if the triangle is vertically oriented (in 2D it's not a triangle anymore actually), nz becomes zero and tp cannot be calculated. That's because there are no more two lines where the value is min or max respectively. For this case you will have to choose another value on the remaining line or point.
I'm going to find the distance between two cities using Haversine formula.
below is the code in VC++.
But I could't find the distance between the points (18.567367, -68.363431) and (33.636719,-84.428067) [first value is latitude, second is longitude].
It gives something like -1.#IND.
Could you please tell me how to deal with this issue?
below is the code:
double deg2rad(double deg) {
return (deg * pi / 180);
};
double TravelDistance(double lat1d, double lon1d, double lat2d, double lon2d) {
double lat1r, lon1r, lat2r, lon2r, u, v,dblDistance;
lat1r = deg2rad(lat1d);
lon1r = deg2rad(lon1d);
lat2r = deg2rad(lat2d);
lon2r = deg2rad(lon2d);
u = sin(lat2r - lat1r);
v = sin(lon2r - lon1r);
return ( 2.0 * earthRadiusKm * asin(sqrt(u * u + cos(lat1r) * cos(lat2r) * v * v)))};
Thank in advance....!!
Looking at
http://en.wikipedia.org/wiki/Haversine_formula
it appears you have forgotten to divide by two in the arguments to sin() for u and v.
The answer you get is most likely because either the sqrt argument is < 0 or because the asin argument is > 1.