Building my responsive website, I would like to build my funny timeline, but I cannot come up with a solution.
It would be a sprite such as a rocket or flying saucer taking off at the bottom of middle of the page and coming out with smoke.
Smoke would remain more or less and disclose my timeline.
Sketch
Is anyone does have an idea how to make that possible?
To simulate smoke, you have to use a particle system.
As you maybe know, WebGL is able to draw triangles, lines and points.
This last one is what we need. The smoke is made of hundreds of semi-transparent white disks of slighly different sizes. Each point is defined by 7 attributes :
x, y: starting position.
vx, vy: direction.
radius: maximal radius.
life: number of milliseconds before it disappears.
delay: Number of milliseconds to wait before its birth.
One trick is to create points along a vertical centered axis. The more you go up, the more the delay increases. The other trick is to make the point more more transparent as it reaches it end of live.
Here is how you create such vertices :
function createVertices() {
var x, y, vx, vy, radius, life, delay;
var vertices = [];
for( delay=0; delay<1; delay+=0.01 ) {
for( var loops=0; loops<5; loops++ ) {
// Going left.
x = rnd(0.01);
y = (2.2 * delay - 1) + rnd(-0.01, 0.01);
vx = -rnd(0, 1.5) * 0.0001;
vy = -rnd(0.001);
radius = rnd(0.1, 0.25) / 1000;
life = rnd(2000, 5000);
vertices.push( x, y, vx, vy, radius, life, delay );
// Going right.
x = -rnd(0.01);
y = (2.2 * delay - 1) + rnd(-0.01, 0.01);
vx = rnd(0, 1.5) * 0.0001;
vy = -rnd(0.001);
radius = rnd(0.1, 0.25) / 1000;
life = rnd(2000, 5000);
vertices.push( x, y, vx, vy, radius, life, delay );
}
}
var buff = gl.createBuffer();
gl.bindBuffer( gl.ARRAY_BUFFER, buff );
gl.bufferData( gl.ARRAY_BUFFER, new Float32Array(vertices), gl.STATIC_DRAW );
return Math.floor( vertices.length / 7 );
}
As you can see, I created points going right and points going left to get a growing fuzzy triangle.
Then you need a vertex shader controling the position and size of the points.
WebGL provide the output variable gl_PointSize which is the size (in pixels) of the square to draw for the current point.
uniform float uniWidth;
uniform float uniHeight;
uniform float uniTime;
attribute vec2 attCoords;
attribute vec2 attDirection;
attribute float attRadius;
attribute float attLife;
attribute float attDelay;
varying float varAlpha;
const float PERIOD = 10000.0;
const float TRAVEL_TIME = 2000.0;
void main() {
float time = mod( uniTime, PERIOD );
time -= TRAVEL_TIME * attDelay;
if( time < 0.0 || time > attLife) return;
vec2 pos = attCoords + time * attDirection;
gl_Position = vec4( pos.xy, 0, 1 );
gl_PointSize = time * attRadius * min(uniWidth, uniHeight);
varAlpha = 1.0 - (time / attLife);
}
Finally, the fragment shader will display a point in white. but the more you go far from the center, the more transparent the fragments become.
To know where you are in the square drawn for the current point, you can read the global WebGL variable gl_PointCoord.
precision mediump float;
varying float varAlpha;
void main() {
float x = gl_PointCoord.x - 0.5;
float y = gl_PointCoord.y - 0.5;
float radius = x * x + y * y;
if( radius > 0.25 ) discard;
float alpha = varAlpha * 0.8 * (0.25 - radius);
gl_FragColor = vec4(1, 1, 1, alpha);
}
Here is a live example : https://jsfiddle.net/m1a9qry6/1/
Let's imagine I have some character that can rotate in 360 degrees, and I am tracking its position as x and y.
I am using sin and cos to determine its movement:
x += speed * cos(rotation)
y += speed * sin(rotation)
This works fine, but I figure I want an acceleration based system instead. So rather than increasing speed, I increase acceleration which then increases speed:
x_velocity += speed * cos(rotation)
y_velocity += speed * sin(rotation)
x += x_velocity
y += y_velocity
However, I don't want this character to be able to accelerate indefinitely and therefore I want to limit its top speed. Now, I could do this:
if x_velocity > maxspeed then x_velocity = maxspeed
if y_velocity > maxspeed then y_velocity = maxspeed
but this is imperfect, as it means you can travel 44% faster if you move at a perfect diagonal than if you travel in a cardinal direction.
So I use the pythagorean theorem, but I immediately run into a problem.
if sqrt(x_velocity^2 + y_velocity^2) > maxspeed then ???
Is there a solution? I cannot simply define it equal to maxspeed like before, but I also cannot completely allocate all the velocity to one axis, since that would change the direction. How do I determine the velocity to limit it at?
Just use overall velocity
velocity = Min(max_velocity, velocity + speed) //really acceleration instead of speed
x_velocity = velocity * cos(rotation)
y_velocity = velocity * sin(rotation)
x += x_velocity
y += y_velocity
If you have to maintain x_velocity and y_velocity (by some strange reason), then you can limit overall velocity through it's components with direction preserving:
... update x_velocity, y_velocity
vel = sqrt(x_velocity^2 + y_velocity^2)
if vel > max_velocity then
x_velocity = x_velocity * max_velocity / vel
y_velocity = y_velocity * max_velocity / vel
DRAG
You want to add drag (friction) that will set a max speed (like freefall drag creates a terminal velocity)
To define movement
acceleration = 2.0; // the constant acceleration applied
vx = 0.0; // velocity x
vy = 0.0; // velocity y
x = 0.0; // current position
y = 0.0;
direction = 0.0; // direction of acceleration
drag = 0.01; // drag as a fraction of speed
To apply acceleration
vx += cos(direction) * acceleration ;
vy += sin(direction) * acceleration ;
To apply drag
vx *= 1 - drag;
vy *= 1 - drag;
To get current speed
speed = sqrt(vx * vx + vy * vy);
Calculate drag for max speed.
If you want to set the max speed in terms of drag.
maxSpeed = 100.0;
drag = accel / (maxSpeed + accel);
The just apply acceleration and drag (in that order) to the velocity and the character will never go above maxSpeed
An important note is that if the character changes direction when near maxSpeed there will be a drop in speed related to the amount of change in velocity. EG if direction changes 180 then the character will slow down at an amount equal to accel + speed * drag (drag is helping the deacceleration) until the velocity starts to increase again at which point acceleration is accel - speed * drag
The final solution to set a max speed in terms of drag,
acceleration = 2.0; // the constant acceleration applied
vx = 0.0; // velocity x
vy = 0.0; // velocity y
x = 0.0; // current position
y = 0.0;
direction = 0.0; // direction of acceleration
maxSpeed = 100.0;
// set drag to give a maxSpeed
drag = acceleration / (maxSpeed + acceleration);
The apply acceleration and drag as follows
vx += cos(direction) * acceleration ;
vy += sin(direction) * acceleration ;
vx *= 1 - drag;
vy *= 1 - drag;
x += vx;
y += vy;
Speed will never go over maxSpeed. But also speed will never reach max speed, it will just keep on getting closer, but never quite getting there, though we are talking insignificant fractions of a unit 0.00000000001 pixels per frame
For some odd reason, the code below does not produce the desired effect. The sinewave is fine and functional, and the snippet of code below works perfectly, except for the movement forward, which is jolty.
I am using millis() and dividing by 1000 to convert to seconds, but the effect still produces what I would expect form using the second() function - the sinewave moving forward every 1 second. I want the lines to move horizontally smoothly, as the sinewave line does vertically. What have I done wrong?
int cyclesBeforeStopping = 4;
int distanceBeforeStopping = 400;
float frequency = 0.2; // in Hz
float peak = 25; // Highest point of wave
float trough = 275; // Lowest point of wave
float amplitudeOffset;
float forwardOffset;
float timeUntilStop = (1 / frequency) * cyclesBeforeStopping;
void setup(){
size(600,300);
frameRate(50);
forwardOffset = ForwardOffset();
}
void draw(){
background(255);
float forwardOffsetNow = forwardOffset * (millis() / 1000);
// Guidelines
line(50 + forwardOffsetNow, peak, 100 + forwardOffsetNow, peak);
line(50 + forwardOffsetNow, trough, 100 + forwardOffsetNow, trough);
// Sine line
float newPosition = NewPosition();
line(50 + forwardOffsetNow, newPosition, 100 + forwardOffsetNow, newPosition);
}
float ForwardOffset() {
float forwardOffsetVar = (distanceBeforeStopping) / timeUntilStop;
return forwardOffsetVar;
}
float NewPosition() {
float omega = TWO_PI * frequency;
float amplitude = trough - peak;
float halfway = peak + amplitude / 2;
float newPosition = halfway - (amplitude * sin(omega * millis() / 1000) / 2);
return newPosition;
}
A one dot solution :) just do:
float forwardOffsetNow = forwardOffset * (millis() / 1000.);//mind the dot after 1000
and also:
float newPosition = halfway - (amplitude * sin(omega * millis() / 1000.) / 2);
This is called integer division, well documented in Processing's wiki troubleshooting, here.
It can be avoided by using a dot or an f after the number, like:
millis()/1000.;
or
millis()/1000.0;
or
millis()/1000f;
All will work.
I have a set of latitudes and longitudes of locations.
How to find distance from one location in the set to another?
Is there a formula ?
The Haversine formula assumes a spherical earth. However, the shape of the earh is more complex. An oblate spheroid model will give better results.
If such accuracy is needed, you should better use Vincenty inverse formula.
See http://en.wikipedia.org/wiki/Vincenty's_formulae for details. Using it, you can get a 0.5mm accuracy for the spheroid model.
There is no perfect formula, since the real shape of the earth is too complex to be expressed by a formula. Moreover, the shape of earth changes due to climate events (see http://www.nasa.gov/centers/goddard/earthandsun/earthshape.html), and also changes over time due to the rotation of the earth.
You should also note that the method above does not take altitudes into account, and assumes a sea-level oblate spheroid.
Edit 10-Jul-2010: I found out that there are rare situations for which Vincenty inverse formula does not converge to the declared accuracy. A better idea is to use GeographicLib (see http://sourceforge.net/projects/geographiclib/) which is also more accurate.
Here's one: http://www.movable-type.co.uk/scripts/latlong.html
Using Haversine formula:
R = earth’s radius (mean radius = 6,371km)
Δlat = lat2− lat1
Δlong = long2− long1
a = sin²(Δlat/2) + cos(lat1).cos(lat2).sin²(Δlong/2)
c = 2.atan2(√a, √(1−a))
d = R.c
Apply the Haversine formula to find the distance. See the C# code below to find the distance between 2 coordinates. Better still if you want to say find a list of stores within a certain radius, you could apply a WHERE clause in SQL or a LINQ filter in C# to it.
The formula here is in kilometres, you will have to change the relevant numbers and it will work for miles.
E.g: Convert 6371.392896 to miles.
DECLARE #radiusInKm AS FLOAT
DECLARE #lat2Compare AS FLOAT
DECLARE #long2Compare AS FLOAT
SET #radiusInKm = 5.000
SET #lat2Compare = insert_your_lat_to_compare_here
SET #long2Compare = insert_you_long_to_compare_here
SELECT * FROM insert_your_table_here WITH(NOLOCK)
WHERE (6371.392896*2*ATN2(SQRT((sin((radians(GeoLatitude - #lat2Compare)) / 2) * sin((radians(GeoLatitude - #lat2Compare)) / 2)) + (cos(radians(GeoLatitude)) * cos(radians(#lat2Compare)) * sin(radians(GeoLongitude - #long2Compare)/2) * sin(radians(GeoLongitude - #long2Compare)/2)))
, SQRT(1-((sin((radians(GeoLatitude - #lat2Compare)) / 2) * sin((radians(GeoLatitude - #lat2Compare)) / 2)) + (cos(radians(GeoLatitude)) * cos(radians(#lat2Compare)) * sin(radians(GeoLongitude - #long2Compare)/2) * sin(radians(GeoLongitude - #long2Compare)/2)))
))) <= #radiusInKm
If you would like to perform the Haversine formula in C#,
double resultDistance = 0.0;
double avgRadiusOfEarth = 6371.392896; //Radius of the earth differ, I'm taking the average.
//Haversine formula
//distance = R * 2 * aTan2 ( square root of A, square root of 1 - A )
// where A = sinus squared (difference in latitude / 2) + (cosine of latitude 1 * cosine of latitude 2 * sinus squared (difference in longitude / 2))
// and R = the circumference of the earth
double differenceInLat = DegreeToRadian(currentLatitude - latitudeToCompare);
double differenceInLong = DegreeToRadian(currentLongitude - longtitudeToCompare);
double aInnerFormula = Math.Cos(DegreeToRadian(currentLatitude)) * Math.Cos(DegreeToRadian(latitudeToCompare)) * Math.Sin(differenceInLong / 2) * Math.Sin(differenceInLong / 2);
double aFormula = (Math.Sin((differenceInLat) / 2) * Math.Sin((differenceInLat) / 2)) + (aInnerFormula);
resultDistance = avgRadiusOfEarth * 2 * Math.Atan2(Math.Sqrt(aFormula), Math.Sqrt(1 - aFormula));
DegreesToRadian is a function I custom created, its is a simple 1 liner of"Math.PI * angle / 180.0
My blog entry - SQL Haversine
Are you looking for
Haversine formula
The haversine formula is an equation
important in navigation, giving
great-circle distances between two
points on a sphere from their
longitudes and latitudes. It is a
special case of a more general formula
in spherical trigonometry, the law of
haversines, relating the sides and
angles of spherical "triangles".
Have a look at this.. has a javascript example as well.
Find Distance
Use the Great Circle Distance Formula.
here is a fiddle with finding locations / near locations to long/lat by given IP:
http://jsfiddle.net/bassta/zrgd9qc3/2/
And here is the function I use to calculate the distance in straight line:
function distance(lat1, lng1, lat2, lng2) {
var radlat1 = Math.PI * lat1 / 180;
var radlat2 = Math.PI * lat2 / 180;
var radlon1 = Math.PI * lng1 / 180;
var radlon2 = Math.PI * lng2 / 180;
var theta = lng1 - lng2;
var radtheta = Math.PI * theta / 180;
var dist = Math.sin(radlat1) * Math.sin(radlat2) + Math.cos(radlat1) * Math.cos(radlat2) * Math.cos(radtheta);
dist = Math.acos(dist);
dist = dist * 180 / Math.PI;
dist = dist * 60 * 1.1515;
//Get in in kilometers
dist = dist * 1.609344;
return dist;
}
It returns the distance in Kilometers
If you are measuring distances less than (perhaps) 1 degree lat/long change, are looking for a very high performance approximation, and are willing to accept more inaccuracy than Haversine formula, consider these two alternatives:
(1) "Polar Coordinate Flat-Earth Formula" from Computing Distances:
a = pi/2 - lat1
b = pi/2 - lat2
c = sqrt( a^2 + b^2 - 2 * a * b * cos(lon2 - lon1) )
d = R * c
(2) Pythagorean theorem adjusted for latitude, as seen in Ewan Todd's SO post:
d_ew = (long1 - long0) * cos(average(lat0, lat1))
d_ns = (lat1 - lat0)
d = sqrt(d_ew * d_ew + d_ns * d_ns)
NOTES:
Compared to Ewan's post, I've substituted average(lat0, lat1) for lat0 inside of cos( lat0 ).
#2 is vague on whether values are degrees, radians, or kilometers; you will need some conversion code as well. See my complete code at bottom of this post.
#1 is designed to work well even near the poles, though if you are measuring a distance whose endpoints are on "opposite" sides of the pole (longitudes differ by more than 90 degrees?), Haversine is recommended instead, even for small distances.
I haven't thoroughly measured errors of these approaches, so you should take representative points for your application, and compare results to some high-quality library, to decide if the accuracies are acceptable. For distances less than a few kilometers my gut sense is that these are within 1% of correct measurement.
An alternative way to gain high performance (when applicable):
If you have a large set of static points, within one or two degrees of longitude/latitude, that you will then be calculating distances from a small number of dynamic (moving) points, consider converting your static points ONCE to the containing UTM zone (or to any other local Cartesian coordinate system), and then doing all your math in that Cartesian coordinate system.
Cartesian = flat earth = Pythagorean theorem applies, so distance = sqrt(dx^2 + dy^2).
Then the cost of accurately converting the few moving points to UTM is easily afforded.
CAVEAT for #1 (Polar): May be very wrong for distances less than 0.1 (?) meter. Even with double precision math, the following coordinates, whose true distance is about 0.005 meters, was given as "zero" by my implementation of Polar algorithm:
inputs:
lon1Xdeg 16.6564465477996 double
lat1Ydeg 57.7760262271983 double
lon2Xdeg 16.6564466358281 double
lat2Ydeg 57.776026248554 double
results:
Oblate spheroid formula:
0.00575254911118364 double
Haversine:
0.00573422966122257 double
Polar:
0
this was due to the two factors u and v exactly canceling each other:
u 0.632619944868587 double
v -0.632619944868587 double
In another case, it gave a distance of 0.067129 m when the oblate spheroid answer was 0.002887 m. The problem was that cos(lon2 - lon1) was too close to 1, so cos function returned exactly 1.
Other than measuring sub-meter distances, the max errors (compared to an oblate spheroid formula) I found for the limited small-distance data I've fed in so far:
maxHaversineErrorRatio 0.00350976281908381 double
maxPolarErrorRatio 0.0510789996931342 double
where "1" would represent a 100% error in the answer; e.g. when it returned "0", that was an error of "1" (excluded from above "maxPolar"). So "0.01" would be an error of "1 part in 100" or 1%.
Comparing Polar error with Haversine error over distances less than 2000 meters to see how much worse this simpler formula is. So far, the worst I've seen is 51 parts per 1000 for Polar vs 4 parts per 1000 for Haversine. At about 58 degrees latitude.
Now implemented "Pythagorean with Latitude Adjustment".
It is MUCH more consistent than Polar for distances < 2000 m.
I originally thought the Polar problems were only when < 1 m,
but the result shown immediately below is quite troubling.
As distances approach zero, pythagorean/latitude approaches haversine.
For example this measurement ~ 217 meters:
lon1Xdeg 16.6531667510102 double
lat1Ydeg 57.7751705615804 double
lon2Xdeg 16.6564468739869 double
lat2Ydeg 57.7760263007586 double
oblate 217.201200413731
haversine 216.518428601051
polar 226.128616011973
pythag-cos 216.518428631907
havErrRatio 0.00314349925958048
polErrRatio 0.041102054598393
pycErrRatio 0.00314349911751603
Polar has a much worse error with these inputs; either there is some mistake in my code, or in Cos function I am running on, or I have to recommend not using Polar, even though most Polar measurements were much closer than this.
OTOH, Pythagorean, even with * cos(latitude) adjustment, has error that increases more rapidly than distance (ratio of max_error/distance increases for larger distances), so you need to carefully consider the maximum distance you will measure, and the acceptable error. In addition, it is not advisable to COMPARE two nearly-equal distances using Pythagorean, to decide which is shorter, as the error is different in different DIRECTIONS (evidence not shown).
Worst case measurements, errorRatio = Abs(error) / distance (Sweden; up to 2000 m):
t_maxHaversineErrorRatio 0.00351012021578681 double
t_maxPolarErrorRatio 66.0825360597085 double
t_maxPythagoreanErrorRatio 0.00350976281416454 double
As mentioned before, the extreme polar errors are for sub-meter distances, where it could report zero instead of 6 cm, or report over 0.5 m for a distance of 1 cm (hence the "66 x" worst case shown in t_maxPolarErrorRatio), but there are also some poor results at larger distances. [Needs to be tested again with a Cosine function that is known to be highly accurate.]
Measurements taken in C# code in Xamarin.Android running on a Moto E4.
C# code:
// x=longitude, y= latitude. oblate spheroid formula. TODO: From where?
public static double calculateDistanceDD_AED( double lon1Xdeg, double lat1Ydeg, double lon2Xdeg, double lat2Ydeg )
{
double c_dblEarthRadius = 6378.135; // km
double c_dblFlattening = 1.0 / 298.257223563; // WGS84 inverse
// flattening
// Q: Why "-" for longitudes??
double p1x = -degreesToRadians( lon1Xdeg );
double p1y = degreesToRadians( lat1Ydeg );
double p2x = -degreesToRadians( lon2Xdeg );
double p2y = degreesToRadians( lat2Ydeg );
double F = (p1y + p2y) / 2;
double G = (p1y - p2y) / 2;
double L = (p1x - p2x) / 2;
double sing = Math.Sin( G );
double cosl = Math.Cos( L );
double cosf = Math.Cos( F );
double sinl = Math.Sin( L );
double sinf = Math.Sin( F );
double cosg = Math.Cos( G );
double S = sing * sing * cosl * cosl + cosf * cosf * sinl * sinl;
double C = cosg * cosg * cosl * cosl + sinf * sinf * sinl * sinl;
double W = Math.Atan2( Math.Sqrt( S ), Math.Sqrt( C ) );
if (W == 0.0)
return 0.0;
double R = Math.Sqrt( (S * C) ) / W;
double H1 = (3 * R - 1.0) / (2.0 * C);
double H2 = (3 * R + 1.0) / (2.0 * S);
double D = 2 * W * c_dblEarthRadius;
// Apply flattening factor
D = D * (1.0 + c_dblFlattening * H1 * sinf * sinf * cosg * cosg - c_dblFlattening * H2 * cosf * cosf * sing * sing);
// Transform to meters
D = D * 1000.0;
// tmstest
if (true)
{
// Compare Haversine.
double haversine = HaversineApproxDistanceGeo( lon1Xdeg, lat1Ydeg, lon2Xdeg, lat2Ydeg );
double error = haversine - D;
double absError = Math.Abs( error );
double errorRatio = absError / D;
if (errorRatio > t_maxHaversineErrorRatio)
{
if (errorRatio > t_maxHaversineErrorRatio * 1.1)
Helper.test();
t_maxHaversineErrorRatio = errorRatio;
}
// Compare Polar Coordinate Flat Earth.
double polarDistanceGeo = ApproxDistanceGeo_Polar( lon1Xdeg, lat1Ydeg, lon2Xdeg, lat2Ydeg, D );
double error2 = polarDistanceGeo - D;
double absError2 = Math.Abs( error2 );
double errorRatio2 = absError2 / D;
if (errorRatio2 > t_maxPolarErrorRatio)
{
if (polarDistanceGeo > 0)
{
if (errorRatio2 > t_maxPolarErrorRatio * 1.1)
Helper.test();
t_maxPolarErrorRatio = errorRatio2;
}
else
Helper.dubious();
}
// Compare Pythagorean Theorem with Latitude Adjustment.
double pythagoreanDistanceGeo = ApproxDistanceGeo_PythagoreanCosLatitude( lon1Xdeg, lat1Ydeg, lon2Xdeg, lat2Ydeg, D );
double error3 = pythagoreanDistanceGeo - D;
double absError3 = Math.Abs( error3 );
double errorRatio3 = absError3 / D;
if (errorRatio3 > t_maxPythagoreanErrorRatio)
{
if (D < 2000)
{
if (errorRatio3 > t_maxPythagoreanErrorRatio * 1.05)
Helper.test();
t_maxPythagoreanErrorRatio = errorRatio3;
}
}
}
return D;
}
// As a fraction of the distance.
private static double t_maxHaversineErrorRatio, t_maxPolarErrorRatio, t_maxPythagoreanErrorRatio;
// Average of equatorial and polar radii (meters).
public const double EarthAvgRadius = 6371000;
public const double EarthAvgCircumference = EarthAvgRadius * 2 * PI;
// CAUTION: This is an average of great circles; won't be the actual distance of any longitude or latitude degree.
public const double EarthAvgMeterPerGreatCircleDegree = EarthAvgCircumference / 360;
// Haversine formula (assumes Earth is sphere).
// "deg" = degrees.
// Perhaps based on Haversine Formula in https://cs.nyu.edu/visual/home/proj/tiger/gisfaq.html
public static double HaversineApproxDistanceGeo(double lon1Xdeg, double lat1Ydeg, double lon2Xdeg, double lat2Ydeg)
{
double lon1 = degreesToRadians( lon1Xdeg );
double lat1 = degreesToRadians( lat1Ydeg );
double lon2 = degreesToRadians( lon2Xdeg );
double lat2 = degreesToRadians( lat2Ydeg );
double dlon = lon2 - lon1;
double dlat = lat2 - lat1;
double sinDLat2 = Sin( dlat / 2 );
double sinDLon2 = Sin( dlon / 2 );
double a = sinDLat2 * sinDLat2 + Cos( lat1 ) * Cos( lat2 ) * sinDLon2 * sinDLon2;
double c = 2 * Atan2( Sqrt( a ), Sqrt( 1 - a ) );
double d = EarthAvgRadius * c;
return d;
}
// From https://stackoverflow.com/a/19772119/199364
// Based on Polar Coordinate Flat Earth in https://cs.nyu.edu/visual/home/proj/tiger/gisfaq.html
public static double ApproxDistanceGeo_Polar( double lon1deg, double lat1deg, double lon2deg, double lat2deg, double D = 0 )
{
double approxUnitDistSq = ApproxUnitDistSq_Polar(lon1deg, lat1deg, lon2deg, lat2deg, D);
double c = Sqrt( approxUnitDistSq );
return EarthAvgRadius * c;
}
// Might be useful to avoid taking Sqrt, when comparing to some threshold.
// Threshold would have to be adjusted to match: Power(threshold / EarthAvgRadius, 2)
private static double ApproxUnitDistSq_Polar(double lon1deg, double lat1deg, double lon2deg, double lat2deg, double D = 0 )
{
const double HalfPi = PI / 2; //1.5707963267949;
double lon1 = degreesToRadians(lon1deg);
double lat1 = degreesToRadians(lat1deg);
double lon2 = degreesToRadians(lon2deg);
double lat2 = degreesToRadians(lat2deg);
double a = HalfPi - lat1;
double b = HalfPi - lat2;
double u = a * a + b * b;
double dlon21 = lon2 - lon1;
double cosDeltaLon = Cos( dlon21 );
double v = -2 * a * b * cosDeltaLon;
// TBD: Is "Abs" necessary? That is, is "u + v" ever negative?
// (I think not; "v" looks like a secondary term. Though might be round-off issue near zero when a~=b.)
double approxUnitDistSq = Abs(u + v);
//if (approxUnitDistSq.nearlyEquals(0, 1E-16))
// Helper.dubious();
//else if (D > 0)
//{
// double dba = b - a;
// double unitD = D / EarthAvgRadius;
// double unitDSq = unitD * unitD;
// if (approxUnitDistSq > 2 * unitDSq)
// Helper.dubious();
// else if (approxUnitDistSq * 2 < unitDSq)
// Helper.dubious();
//}
return approxUnitDistSq;
}
// Pythagorean Theorem with Latitude Adjustment - from Ewan Todd - https://stackoverflow.com/a/1664836/199364
// Refined by ToolmakerSteve - https://stackoverflow.com/a/53468745/199364
public static double ApproxDistanceGeo_PythagoreanCosLatitude( double lon1deg, double lat1deg, double lon2deg, double lat2deg, double D = 0 )
{
double approxDegreesSq = ApproxDegreesSq_PythagoreanCosLatitude( lon1deg, lat1deg, lon2deg, lat2deg );
// approximate degrees on the great circle between the points.
double d_degrees = Sqrt( approxDegreesSq );
return d_degrees * EarthAvgMeterPerGreatCircleDegree;
}
public static double ApproxDegreesSq_PythagoreanCosLatitude( double lon1deg, double lat1deg, double lon2deg, double lat2deg )
{
double avgLatDeg = average( lat1deg , lat2deg );
double avgLat = degreesToRadians( avgLatDeg );
double d_ew = (lon2deg - lon1deg) * Cos( avgLat );
double d_ns = (lat2deg - lat1deg);
double approxDegreesSq = d_ew * d_ew + d_ns * d_ns;
return approxDegreesSq;
}
I am done using SQL query
select *, (acos(sin(input_lat* 0.01745329)*sin(lattitude *0.01745329) + cos(input_lat *0.01745329)*cos(lattitude *0.01745329)*cos((input_long -longitude)*0.01745329))* 57.29577951 )* 69.16 As D from table_name
Following is the module (coded in f90) containing three formulas discussed in the previous answers. You can either put this module at the top of your program
(before PROGRAM MAIN) or compile it separately and include the module directory during compilation. The following module contains three formulas. First two are great-circle distances based on the assumption that earth is spherical.
module spherical_dists
contains
subroutine great_circle_distance(lon1,lat1,lon2,lat2,dist)
!https://en.wikipedia.org/wiki/Great-circle_distance
! It takes lon, lats of two points on an assumed spherical earth and
! calculates the distance between them along the great circle connecting the two points
implicit none
real,intent(in)::lon1,lon2,lat1,lat2
real,intent(out)::dist
real,parameter::pi=3.141592,mean_earth_radius=6371.0088
real::lonr1,lonr2,latr1,latr2
real::delangl,dellon
lonr1=lon1*(pi/180.);lonr2=lon2*(pi/180.)
latr1=lat1*(pi/180.);latr2=lat2*(pi/180.)
dellon=lonr2-lonr1
delangl=acos(sin(latr1)*sin(latr2)+cos(latr1)*cos(latr2)*cos(dellon))
dist=delangl*mean_earth_radius
end subroutine
subroutine haversine_formula(lon1,lat1,lon2,lat2,dist)
! https://en.wikipedia.org/wiki/Haversine_formula
! This is similar above but numerically better conditioned for small distances
implicit none
real,intent(in)::lon1,lon2,lat1,lat2
!lon, lats of two points
real,intent(out)::dist
real,parameter::pi=3.141592,mean_earth_radius=6371.0088
real::lonr1,lonr2,latr1,latr2
real::delangl,dellon,dellat,a
! degrees are converted to radians
lonr1=lon1*(pi/180.);lonr2=lon2*(pi/180.)
latr1=lat1*(pi/180.);latr2=lat2*(pi/180.)
dellon=lonr2-lonr1 ! These dels simplify the haversine formula
dellat=latr2-latr1
! The actual haversine formula
a=(sin(dellat/2))**2+cos(latr1)*cos(latr2)*(sin(dellon/2))**2
delangl=2*asin(sqrt(a)) !2*asin(sqrt(a))
dist=delangl*mean_earth_radius
end subroutine
subroutine vincenty_formula(lon1,lat1,lon2,lat2,dist)
!https://en.wikipedia.org/wiki/Vincenty%27s_formulae
!It's a better approximation over previous two, since it considers earth to in oblate spheroid, which better approximates the shape of the earth
implicit none
real,intent(in)::lon1,lon2,lat1,lat2
real,intent(out)::dist
real,parameter::pi=3.141592,mean_earth_radius=6371.0088
real::lonr1,lonr2,latr1,latr2
real::delangl,dellon,nom,denom
lonr1=lon1*(pi/180.);lonr2=lon2*(pi/180.)
latr1=lat1*(pi/180.);latr2=lat2*(pi/180.)
dellon=lonr2-lonr1
nom=sqrt((cos(latr2)*sin(dellon))**2. + (cos(latr1)*sin(latr2)-sin(latr1)*cos(latr2)*cos(dellon))**2.)
denom=sin(latr1)*sin(latr2)+cos(latr1)*cos(latr2)*cos(dellon)
delangl=atan2(nom,denom)
dist=delangl*mean_earth_radius
end subroutine
end module
On this page you can see the whole code and formulas how distances of locations are calculated in Android Location class
android/location/Location.java
EDIT: According the hint from #Richard I put the code of the linked function into my answer, to avoid invalidated link:
private static void computeDistanceAndBearing(double lat1, double lon1,
double lat2, double lon2, BearingDistanceCache results) {
// Based on http://www.ngs.noaa.gov/PUBS_LIB/inverse.pdf
// using the "Inverse Formula" (section 4)
int MAXITERS = 20;
// Convert lat/long to radians
lat1 *= Math.PI / 180.0;
lat2 *= Math.PI / 180.0;
lon1 *= Math.PI / 180.0;
lon2 *= Math.PI / 180.0;
double a = 6378137.0; // WGS84 major axis
double b = 6356752.3142; // WGS84 semi-major axis
double f = (a - b) / a;
double aSqMinusBSqOverBSq = (a * a - b * b) / (b * b);
double L = lon2 - lon1;
double A = 0.0;
double U1 = Math.atan((1.0 - f) * Math.tan(lat1));
double U2 = Math.atan((1.0 - f) * Math.tan(lat2));
double cosU1 = Math.cos(U1);
double cosU2 = Math.cos(U2);
double sinU1 = Math.sin(U1);
double sinU2 = Math.sin(U2);
double cosU1cosU2 = cosU1 * cosU2;
double sinU1sinU2 = sinU1 * sinU2;
double sigma = 0.0;
double deltaSigma = 0.0;
double cosSqAlpha = 0.0;
double cos2SM = 0.0;
double cosSigma = 0.0;
double sinSigma = 0.0;
double cosLambda = 0.0;
double sinLambda = 0.0;
double lambda = L; // initial guess
for (int iter = 0; iter < MAXITERS; iter++) {
double lambdaOrig = lambda;
cosLambda = Math.cos(lambda);
sinLambda = Math.sin(lambda);
double t1 = cosU2 * sinLambda;
double t2 = cosU1 * sinU2 - sinU1 * cosU2 * cosLambda;
double sinSqSigma = t1 * t1 + t2 * t2; // (14)
sinSigma = Math.sqrt(sinSqSigma);
cosSigma = sinU1sinU2 + cosU1cosU2 * cosLambda; // (15)
sigma = Math.atan2(sinSigma, cosSigma); // (16)
double sinAlpha = (sinSigma == 0) ? 0.0 :
cosU1cosU2 * sinLambda / sinSigma; // (17)
cosSqAlpha = 1.0 - sinAlpha * sinAlpha;
cos2SM = (cosSqAlpha == 0) ? 0.0 :
cosSigma - 2.0 * sinU1sinU2 / cosSqAlpha; // (18)
double uSquared = cosSqAlpha * aSqMinusBSqOverBSq; // defn
A = 1 + (uSquared / 16384.0) * // (3)
(4096.0 + uSquared *
(-768 + uSquared * (320.0 - 175.0 * uSquared)));
double B = (uSquared / 1024.0) * // (4)
(256.0 + uSquared *
(-128.0 + uSquared * (74.0 - 47.0 * uSquared)));
double C = (f / 16.0) *
cosSqAlpha *
(4.0 + f * (4.0 - 3.0 * cosSqAlpha)); // (10)
double cos2SMSq = cos2SM * cos2SM;
deltaSigma = B * sinSigma * // (6)
(cos2SM + (B / 4.0) *
(cosSigma * (-1.0 + 2.0 * cos2SMSq) -
(B / 6.0) * cos2SM *
(-3.0 + 4.0 * sinSigma * sinSigma) *
(-3.0 + 4.0 * cos2SMSq)));
lambda = L +
(1.0 - C) * f * sinAlpha *
(sigma + C * sinSigma *
(cos2SM + C * cosSigma *
(-1.0 + 2.0 * cos2SM * cos2SM))); // (11)
double delta = (lambda - lambdaOrig) / lambda;
if (Math.abs(delta) < 1.0e-12) {
break;
}
}
float distance = (float) (b * A * (sigma - deltaSigma));
results.mDistance = distance;
float initialBearing = (float) Math.atan2(cosU2 * sinLambda,
cosU1 * sinU2 - sinU1 * cosU2 * cosLambda);
initialBearing *= 180.0 / Math.PI;
results.mInitialBearing = initialBearing;
float finalBearing = (float) Math.atan2(cosU1 * sinLambda,
-sinU1 * cosU2 + cosU1 * sinU2 * cosLambda);
finalBearing *= 180.0 / Math.PI;
results.mFinalBearing = finalBearing;
results.mLat1 = lat1;
results.mLat2 = lat2;
results.mLon1 = lon1;
results.mLon2 = lon2;
}
just use the distance formula Sqrt( (x2-x1)^2 + (y2-y1)^2 )