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Please explain the purpose of the bitwise or in this javascript function

I am converting a javascript function to java, and don't understand the purpose of the bitwise ors in the code below:
(Math.tan(PHId)) ^ 2) - is this ensuring the number always ends in 2?
(Et ^ 6)
The code is part of a library to convert Irish Grid References to/from Latitude and Longitude: http://www.nearby.org.uk/tests/geotools2.js
GT_Math.E_N_to_Lat = function(East, North, a, b, e0, n0, f0, PHI0, LAM0)
{
//Un-project Transverse Mercator eastings and northings back to latitude.
//Input: - _
//eastings (East) and northings (North) in meters; _
//ellipsoid axis dimensions (a & b) in meters; _
//eastings (e0) and northings (n0) of false origin in meters; _
//central meridian scale factor (f0) and _
//latitude (PHI0) and longitude (LAM0) of false origin in decimal degrees.
//'REQUIRES THE "Marc" AND "InitialLat" FUNCTIONS
//Convert angle measures to radians
var Pi = 3.14159265358979;
var RadPHI0 = PHI0 * (Pi / 180);
var RadLAM0 = LAM0 * (Pi / 180);
//Compute af0, bf0, e squared (e2), n and Et
var af0 = a * f0;
var bf0 = b * f0;
var e2 = (Math.pow(af0,2) - Math.pow(bf0,2)) / Math.pow(af0,2);
var n = (af0 - bf0) / (af0 + bf0);
var Et = East - e0;
//Compute initial value for latitude (PHI) in radians
var PHId = GT_Math.InitialLat(North, n0, af0, RadPHI0, n, bf0);
//Compute nu, rho and eta2 using value for PHId
var nu = af0 / (Math.sqrt(1 - (e2 * ( Math.pow(Math.sin(PHId),2)))));
var rho = (nu * (1 - e2)) / (1 - (e2 * Math.pow(Math.sin(PHId),2)));
var eta2 = (nu / rho) - 1;
//Compute Latitude
var VII = (Math.tan(PHId)) / (2 * rho * nu);
var VIII = ((Math.tan(PHId)) / (24 * rho * Math.pow(nu,3))) * (5 + (3 * (Math.pow(Math.tan(PHId),2))) + eta2 - (9 * eta2 * (Math.pow(Math.tan(PHId),2))));
var IX = ((Math.tan(PHId)) / (720 * rho * Math.pow(nu,5))) * (61 + (90 * ((Math.tan(PHId)) ^ 2)) + (45 * (Math.pow(Math.tan(PHId),4))));
var E_N_to_Lat = (180 / Pi) * (PHId - (Math.pow(Et,2) * VII) + (Math.pow(Et,4) * VIII) - ((Et ^ 6) * IX));
return (E_N_to_Lat);
}

I recommend to ask the author of the script.
However, I am reasonably certain that this is simply a mistake, and what was meant is Math.tan(PHId) ** 2 / Math.pow(Math.tan(PHId), 2) and Et ** 6/ Math.pow(Et, 6), i.e. exponentiation instead of bitwise OR. I believe this because
bitwise OR just doesn't make sense in numeric code
this looks very much like a series expansion - the preceeding terms also use exponentiation, and the mistake likely wasn't noticed because it introduces only a small error
All the other methods in the script (E_N_to_Long, Lat_Long_to_East, Lat_Long_to_North) use Math.pow everywhere, E_N_to_Lat is the only one to use ^
I am converting a javascript function to java
Notice the comment at the top of the script:
* Credits
* The algorithm used by the script for WGS84-OSGB36 conversions is derived
* from an OSGB spreadsheet (www.gps.gov.uk) with permission. This has been
* adapted into PHP by Ian Harris, and Irish added by Barry Hunter
I would advise to start from these primary sources, instead of translating the JavaScript translation of a PHP translation of a spreadsheet formula translation of mathematics into Java.

Find intersection coordinates of two circles on earth?

I'm trying to find a second intersection point of two circles. One of the points that I already know was used to calculate a distance and then used as the circle radius (exemple). The problem is that im not getting the know point, im getting two new coordinates, even thou they are similar. The problem is probably related to the earth curvature but I have searched for some solution and found nothing.
The circles radius are calculated with the earth curvature. And this is the code I have:
function GET_coordinates_of_circles(position1,r1, position2,r2) {
var deg2rad = function (deg) { return deg * (Math.PI / 180); };
x1=position1.lng;
y1=position1.lat;
x2=position2.lng;
y2=position2.lat;
var centerdx = deg2rad(x1 - x2);
var centerdy = deg2rad(y1 - y2);
var R = Math.sqrt(centerdx * centerdx + centerdy * centerdy);
if (!(Math.abs(r1 - r2) <= R && R <= r1 + r2)) { // no intersection
console.log("nope");
return []; // empty list of results
}
// intersection(s) should exist
var R2 = R*R;
var R4 = R2*R2;
var a = (r1*r1 - r2*r2) / (2 * R2);
var r2r2 = (r1*r1 - r2*r2);
var c = Math.sqrt(2 * (r1*r1 + r2*r2) / R2 - (r2r2 * r2r2) / R4 - 1);
var fx = (x1+x2) / 2 + a * (x2 - x1);
var gx = c * (y2 - y1) / 2;
var ix1 = fx + gx;
var ix2 = fx - gx;
var fy = (y1+y2) / 2 + a * (y2 - y1);
var gy = c * (x1 - x2) / 2;
var iy1 = fy + gy;
var iy2 = fy - gy;
// note if gy == 0 and gx == 0 then the circles are tangent and there is only one solution
// but that one solution will just be duplicated as the code is currently written
return [[iy1, ix1], [iy2, ix2]];
}
The deg2rad variable it is suppose to adjust the other calculations with the earth curvature.
Thank you for any help.

Your calculations for R and so on are wrong because plane Pythagorean formula does not work for spherical trigonometry (for example - we can have triangle with all three right angles on the sphere!). Instead we should use special formulas. Some of them are taken from this page.
At first find big circle arcs in radians for both radii using R = Earth radius = 6,371km
a1 = r1 / R
a2 = r2 / R
And distance (again arc in radians) between circle center using haversine formula
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 ad = 2 * Math.atan2(Math.sqrt(a), Math.sqrt(1-a));
And bearing from position1 to position 2:
//where φ1,λ1 is the start point, φ2,λ2 the end point
//(Δλ is the difference in longitude)
var y = Math.sin(λ2-λ1) * Math.cos(φ2);
var x = Math.cos(φ1)*Math.sin(φ2) -
Math.sin(φ1)*Math.cos(φ2)*Math.cos(λ2-λ1);
var brng = Math.atan2(y, x);
Now look at the picture from my answer considering equal radii case.
(Here circle radii might be distinct and we should use another approach to find needed arcs)
We have spherical right-angle triangles ACB and FCB (similar to plane case BD is perpendicular to AF in point C and BCA angle is right).
Spherical Pythagorean theorem (from the book on sph. trig) says that
cos(AB) = cos(BC) * cos(AC)
cos(FB) = cos(BC) * cos(FC)
or (using x for AC, y for BC and (ad-x) for FC)
cos(a1) = cos(y) * cos(x)
cos(a2) = cos(y) * cos(ad-x)
divide equations to eliminate cos(y)
cos(a1)*cos(ad-x) = cos(a2) * cos(x)
cos(a1)*(cos(ad)*cos(x) + sin(ad)*sin(x)) = cos(a2) * cos(x)
cos(ad)*cos(x) + sin(ad)*sin(x) = cos(a2) * cos(x) / cos(a1)
sin(ad)*sin(x) = cos(a2) * cos(x) / cos(a1) - cos(ad)*cos(x)
sin(ad)*sin(x) = cos(x) * (cos(a2) / cos(a1) - cos(ad))
TAC = tg(x) = (cos(a2) / cos(a1) - cos(ad)) / sin(ad)
Having hypotenuse and cathetus of ACB triangle we can find angle between AC and AB directions (Napier's rules for right spherical triangles) - note we already know TAC = tg(AC) and a1 = AB
cos(CAB)= tg(AC) * ctg(AB)
CAB = Math.acos(TAC * ctg(a1))
Now we can calculate intersection points - they lie at arc distance a1 from position1 along bearings brng-CAB and brng+CAB
B_bearing = brng - CAB
D_bearing = brng + CAB
Intersection points' coordinates:
var latB = Math.asin( Math.sin(lat1)*Math.cos(a1) +
Math.cos(lat1)*Math.sin(a1)*Math.cos(B_bearing) );
var lonB = lon1.toRad() + Math.atan2(Math.sin(B_bearing)*Math.sin(a1)*Math.cos(lat1),
Math.cos(a1)-Math.sin(lat1)*Math.sin(lat2));
and the same for D_bearing
latB, lonB are in radians

I had a similar need ( Intersection coordinates (lat/lon) of two circles (given the coordinates of the center and the radius) on earth ) and hereby I share the solution in python in case it might help someone:
'''
FINDING THE INTERSECTION COORDINATES (LAT/LON) OF TWO CIRCLES (GIVEN THE COORDINATES OF THE CENTER AND THE RADII)
Many thanks to Ture Pålsson who directed me to the right source, the code below is based on whuber's brilliant work here:
https://gis.stackexchange.com/questions/48937/calculating-intersection-of-two-circles
The idea is that;
1. The points in question are the mutual intersections of three spheres: a sphere centered beneath location x1 (on the
earth's surface) of a given radius, a sphere centered beneath location x2 (on the earth's surface) of a given radius, and
the earth itself, which is a sphere centered at O = (0,0,0) of a given radius.
2. The intersection of each of the first two spheres with the earth's surface is a circle, which defines two planes.
The mutual intersections of all three spheres therefore lies on the intersection of those two planes: a line.
Consequently, the problem is reduced to intersecting a line with a sphere.
Note that "Decimal" is used to have higher precision which is important if the distance between two points are a few
meters.
'''
from decimal import Decimal
from math import cos, sin, sqrt
import math
import numpy as np
def intersection(p1, r1_meter, p2, r2_meter):
# p1 = Coordinates of Point 1: latitude, longitude. This serves as the center of circle 1. Ex: (36.110174, -90.953524)
# r1_meter = Radius of circle 1 in meters
# p2 = Coordinates of Point 2: latitude, longitude. This serves as the center of circle 1. Ex: (36.110174, -90.953524)
# r2_meter = Radius of circle 2 in meters
'''
1. Convert (lat, lon) to (x,y,z) geocentric coordinates.
As usual, because we may choose units of measurement in which the earth has a unit radius
'''
x_p1 = Decimal(cos(math.radians(p1[1]))*cos(math.radians(p1[0]))) # x = cos(lon)*cos(lat)
y_p1 = Decimal(sin(math.radians(p1[1]))*cos(math.radians(p1[0]))) # y = sin(lon)*cos(lat)
z_p1 = Decimal(sin(math.radians(p1[0]))) # z = sin(lat)
x1 = (x_p1, y_p1, z_p1)
x_p2 = Decimal(cos(math.radians(p2[1]))*cos(math.radians(p2[0]))) # x = cos(lon)*cos(lat)
y_p2 = Decimal(sin(math.radians(p2[1]))*cos(math.radians(p2[0]))) # y = sin(lon)*cos(lat)
z_p2 = Decimal(sin(math.radians(p2[0]))) # z = sin(lat)
x2 = (x_p2, y_p2, z_p2)
'''
2. Convert the radii r1 and r2 (which are measured along the sphere) to angles along the sphere.
By definition, one nautical mile (NM) is 1/60 degree of arc (which is pi/180 * 1/60 = 0.0002908888 radians).
'''
r1 = Decimal(math.radians((r1_meter/1852) / 60)) # r1_meter/1852 converts meter to Nautical mile.
r2 = Decimal(math.radians((r2_meter/1852) / 60))
'''
3. The geodesic circle of radius r1 around x1 is the intersection of the earth's surface with an Euclidean sphere
of radius sin(r1) centered at cos(r1)*x1.
4. The plane determined by the intersection of the sphere of radius sin(r1) around cos(r1)*x1 and the earth's surface
is perpendicular to x1 and passes through the point cos(r1)x1, whence its equation is x.x1 = cos(r1)
(the "." represents the usual dot product); likewise for the other plane. There will be a unique point x0 on the
intersection of those two planes that is a linear combination of x1 and x2. Writing x0 = ax1 + b*x2 the two planar
equations are;
cos(r1) = x.x1 = (a*x1 + b*x2).x1 = a + b*(x2.x1)
cos(r2) = x.x2 = (a*x1 + b*x2).x2 = a*(x1.x2) + b
Using the fact that x2.x1 = x1.x2, which I shall write as q, the solution (if it exists) is given by
a = (cos(r1) - cos(r2)*q) / (1 - q^2),
b = (cos(r2) - cos(r1)*q) / (1 - q^2).
'''
q = Decimal(np.dot(x1, x2))
if q**2 != 1 :
a = (Decimal(cos(r1)) - Decimal(cos(r2))*q) / (1 - q**2)
b = (Decimal(cos(r2)) - Decimal(cos(r1))*q) / (1 - q**2)
'''
5. Now all other points on the line of intersection of the two planes differ from x0 by some multiple of a vector
n which is mutually perpendicular to both planes. The cross product n = x1~Cross~x2 does the job provided n is
nonzero: once again, this means that x1 and x2 are neither coincident nor diametrically opposite. (We need to
take care to compute the cross product with high precision, because it involves subtractions with a lot of
cancellation when x1 and x2 are close to each other.)
'''
n = np.cross(x1, x2)
'''
6. Therefore, we seek up to two points of the form x0 + t*n which lie on the earth's surface: that is, their length
equals 1. Equivalently, their squared length is 1:
1 = squared length = (x0 + t*n).(x0 + t*n) = x0.x0 + 2t*x0.n + t^2*n.n = x0.x0 + t^2*n.n
'''
x0_1 = [a*f for f in x1]
x0_2 = [b*f for f in x2]
x0 = [sum(f) for f in zip(x0_1, x0_2)]
'''
The term with x0.n disappears because x0 (being a linear combination of x1 and x2) is perpendicular to n.
The two solutions easily are t = sqrt((1 - x0.x0)/n.n) and its negative. Once again high precision
is called for, because when x1 and x2 are close, x0.x0 is very close to 1, leading to some loss of
floating point precision.
'''
if (np.dot(x0, x0) <= 1) & (np.dot(n,n) != 0): # This is to secure that (1 - np.dot(x0, x0)) / np.dot(n,n) > 0
t = Decimal(sqrt((1 - np.dot(x0, x0)) / np.dot(n,n)))
t1 = t
t2 = -t
i1 = x0 + t1*n
i2 = x0 + t2*n
'''
7. Finally, we may convert these solutions back to (lat, lon) by converting geocentric (x,y,z) to geographic
coordinates. For the longitude, use the generalized arctangent returning values in the range -180 to 180
degrees (in computing applications, this function takes both x and y as arguments rather than just the
ratio y/x; it is sometimes called "ATan2").
'''
i1_lat = math.degrees( math.asin(i1[2]))
i1_lon = math.degrees( math.atan2(i1[1], i1[0] ) )
ip1 = (i1_lat, i1_lon)
i2_lat = math.degrees( math.asin(i2[2]))
i2_lon = math.degrees( math.atan2(i2[1], i2[0] ) )
ip2 = (i2_lat, i2_lon)
return [ip1, ip2]
elif (np.dot(n,n) == 0):
return("The centers of the circles can be neither the same point nor antipodal points.")
else:
return("The circles do not intersect")
else:
return("The centers of the circles can be neither the same point nor antipodal points.")
'''
Example: The output of below is [(36.989311051533505, -88.15142628069133), (38.2383796094578, -92.39048549120287)]
intersection_points = intersection((37.673442, -90.234036), 107.5*1852, (36.109997, -90.953669), 145*1852)
print(intersection_points)
'''
Any feedback is appreciated.

Differences in distance calculated between javascript function and sql server STD distance

DECLARE #orig geography = geography::Point(17, 78, 4326);
Select #orig.STDistance(geography::Point(17.001, 78.00001, 4326))/1000
gives
110.674385214845
function calculateDistance(lat1, lon1, lat2, lon2, units) {
var R = 6371; // Radius of the earth in km
var dLat = deg2rad(lat2 - lat1); // deg2rad below
var dLon = deg2rad(lon2 - lon1);
var a = Math.sin(dLat / 2) * Math.sin(dLat / 2) +
Math.cos(deg2rad(lat1)) * Math.cos(deg2rad(lat2)) *
Math.sin(dLon / 2) * Math.sin(dLon / 2);
var c = 2 * Math.atan2(Math.sqrt(a), Math.sqrt(1 - a));
var d = R * c; // Distance in km
return d;
}
function deg2rad(deg) {
return deg * (Math.PI / 180)
}`
gives
0.11120001099379416 km
difference of 0.6 mtrs

The reason is because the Haversine formula you have used above assumes a spherical earth, with a circumference of 6371 km whereas the Geography datatype uses the WGS 84 ellipsoid. The main difference between the two is that in WGS 84 the earth is an oblate spheroid (ie, squashed at the poles) and is assumed to have a circumference of 6378.13 km around the equator, but a circumference of 6356.75 km around the poles. The full equations for calculations using the WGS 84 ellipsoid are quite complex and Haversine is considered a good approximation in most circumstances.
If you look at the introduction to this excellent resource, http://www.movable-type.co.uk/scripts/latlong.html, the author makes this exact point and suggests that the error from ignoring the ellipsoid is about 0.3% on average (yours is around 0.4%). I repeat the author's quote here, just in case the page ever goes down.
All these formulæ are for calculations on the basis of a spherical earth (ignoring ellipsoidal effects) – which is accurate enough* for most purposes… [In fact, the earth is very slightly ellipsoidal; using a spherical model gives errors typically up to 0.3% – see notes for further details].

Calculation between “my current geolocation” and other coordinates

How can I calculate the distance between “my current geolocation” (the coordinate I get using navigator.geolocation.watchPosition) and say 10 others (fixed) coordinates simultaneously?
I want to track the distance so if “my current geolocation” get within a specific radius of one the coordinates ==> something amazing happens.
Can I solve this problem using the Haversine Distance Formula?
All the calculations will be within a smaller area (a city) so I need to get the results in meters rather than kilometers.
Thank you for your time and input !
/a.

At the equator 5 decimal places in your coordinate(0.00001) = 1.1057 meters Latidude &1.1132 meters Longitude.
As the coordinates move from the equator to the poles the Longitude value decreases.
Degree| Latidude | Longitude
------------------------------
0° | 1.1057 m |1.1132 m
15° | 1.1064 m |1.0755 m
30° | 1.1085 m |0.9648 m
45° | 1.1113 m |0.7884 m
60° | 1.1141 m |0.5580 m
75° | 1.1161 m |0.2890 m
90° | 1.1169 m |0.0000 m
If your distances are small you can either use Pythagoras with correction for latidude or use Haversine.
The following code uses Haversine
var lat1 =55.00001;//Point 1 nearest
var lng1 =-2.00001 ;
//var lat1 =55.00002;//Point 2 nearest
//var lng1 =-2.00002 ;
//Array of points
var coordArray = [[55.00000,-2.00000],[55.00003,-2.00003],[55.00006,-2.00006],[55.00009,-2.00009],[55.00012,-2.00012]];
function toRad(Value) {
/** Converts numeric degrees to radians */
return Value * Math.PI / 180;
}
function Round(Number, DecimalPlaces) {
return Math.round(parseFloat(Number) * Math.pow(10, DecimalPlaces)) / Math.pow(10, DecimalPlaces);
}
function haversine(lat1,lat2,lng1,lng2){
rad = 6371000; // meters
deltaLat = toRad(lat2-lat1);
deltaLng = toRad(lng2-lng1);
lat1 = toRad(lat1);
lat2 = toRad(lat2);
a = Math.sin(deltaLat/2) * Math.sin(deltaLat/2) + Math.sin(deltaLng/2) * Math.sin(deltaLng/2) * Math.cos(lat1) * Math.cos(lat2);
c = 2 * Math.atan2(Math.sqrt(a), Math.sqrt(1-a));
return rad * c;
}
function calculate(){
var result = haversine(lat1,coordArray [0][0],lng1,coordArray [0][1]);
var index = 0;
for (var i=1;i<coordArray.length;i++){
var ans = haversine(lat1,coordArray [i][0],lng1,coordArray [i][1]);
if (ans < result){
result = ans;
index++;
}
}
document.write("Result = " +Round(result,2)+ " Meters Point = "+ (index+1));
}
See Fiddle

The harvesine formula is just a special formula. But it works for your question. If you want better result you can try the vincenty formula. It uses an ellipsoid earth. Just loop over every point and compute the distance from it.

Mercator longitude and latitude calculations to x and y on a cropped map (of the UK)

I have this image. It's a map of the UK (not including Southern Ireland):
I have successfully managed to get a latitude and longitude and plot it onto this map by taking the leftmost longitude and rightmost longitude of the UK and using them to work out where to put the point on the map.
This is the code (for use in Processing.js but could be used as js or anything):
// Size of the map
int width = 538;
int height = 811;
// X and Y boundaries
float westLong = -8.166667;
float eastLong = 1.762833;
float northLat = 58.666667;
float southLat = 49.95;
void drawPoint(float latitude, float longitude){
fill(#000000);
x = width * ((westLong-longitude)/(westLong-eastLong));
y = (height * ((northLat-latitude)/(northLat-southLat)));
console.log(x + ", " + y);
ellipseMode(RADIUS);
ellipse(x, y, 2, 2);
}
However, I haven't been able to implement a Mercator projection on these values. The plots are reasonably accurate but they are not good enough and this projection would solve it.
I can't figure out how to do it. All the examples I find are explaining how to do it for the whole world. This is a good resource of examples explaining how to implement the projection but I haven't been able to get it to work.
Another resource is the Extreme points of the United Kingdom where I got the latitude and longitude values of the bounding box around the UK. They are also here:
northLat = 58.666667;
northLong = -3.366667;
eastLat = 52.481167;
eastLong = 1.762833;
southLat = 49.95;
southLong = -5.2;
westLat = 54.45;
westLong = -8.166667;
If anyone could help me with this, I would greatly appreciate it!
Thanks

I wrote a function which does exactly what you were looking for. I know it's a bit late, but maybe there are some other people interested in.
You need a map which is a mercator projection and you need to know the lat / lon positions of your map.
You get great customized mercator maps with perfect matching lat / lon positions from TileMill which is a free software from MapBox!
I'm using this script and tested it with some google earth positions. It worked perfect on a pixel level. Actually I didnt test this on different or larger maps. I hope it helps you!
Raphael ;)
<?php
$mapWidth = 1500;
$mapHeight = 1577;
$mapLonLeft = 9.8;
$mapLonRight = 10.2;
$mapLonDelta = $mapLonRight - $mapLonLeft;
$mapLatBottom = 53.45;
$mapLatBottomDegree = $mapLatBottom * M_PI / 180;
function convertGeoToPixel($lat, $lon)
{
global $mapWidth, $mapHeight, $mapLonLeft, $mapLonDelta, $mapLatBottom, $mapLatBottomDegree;
$x = ($lon - $mapLonLeft) * ($mapWidth / $mapLonDelta);
$lat = $lat * M_PI / 180;
$worldMapWidth = (($mapWidth / $mapLonDelta) * 360) / (2 * M_PI);
$mapOffsetY = ($worldMapWidth / 2 * log((1 + sin($mapLatBottomDegree)) / (1 - sin($mapLatBottomDegree))));
$y = $mapHeight - (($worldMapWidth / 2 * log((1 + sin($lat)) / (1 - sin($lat)))) - $mapOffsetY);
return array($x, $y);
}
$position = convertGeoToPixel(53.7, 9.95);
echo "x: ".$position[0]." / ".$position[1];
?>
Here is the image I created with TileMill and which I used in this example:

In addition to what Raphael Wichmann has posted (Thanks, by the way!),
here is the reverse function, in actionscript :
function convertPixelToGeo(tx:Number, ty:Number):Point
{
/* called worldMapWidth in Raphael's Code, but I think that's the radius since it's the map width or circumference divided by 2*PI */
var worldMapRadius:Number = mapWidth / mapLonDelta * 360/(2 * Math.PI);
var mapOffsetY:Number = ( worldMapRadius / 2 * Math.log( (1 + Math.sin(mapLatBottomRadian) ) / (1 - Math.sin(mapLatBottomRadian)) ));
var equatorY:Number = mapHeight + mapOffsetY;
var a:Number = (equatorY-ty)/worldMapRadius;
var lat:Number = 180/Math.PI * (2 * Math.atan(Math.exp(a)) - Math.PI/2);
var long:Number = mapLonLeft+tx/mapWidth*mapLonDelta;
return new Point(lat,long);
}

Here's another Javascript implementation. This is a simplification of #Rob Willet's solution above. Instead of requiring computed values as parameters to the function, it only requires essential values and computes everything from them:
function convertGeoToPixel(latitude, longitude,
mapWidth, // in pixels
mapHeight, // in pixels
mapLngLeft, // in degrees. the longitude of the left side of the map (i.e. the longitude of whatever is depicted on the left-most part of the map image)
mapLngRight, // in degrees. the longitude of the right side of the map
mapLatBottom) // in degrees. the latitude of the bottom of the map
{
const mapLatBottomRad = mapLatBottom * Math.PI / 180
const latitudeRad = latitude * Math.PI / 180
const mapLngDelta = (mapLngRight - mapLngLeft)
const worldMapWidth = ((mapWidth / mapLngDelta) * 360) / (2 * Math.PI)
const mapOffsetY = (worldMapWidth / 2 * Math.log((1 + Math.sin(mapLatBottomRad)) / (1 - Math.sin(mapLatBottomRad))))
const x = (longitude - mapLngLeft) * (mapWidth / mapLngDelta)
const y = mapHeight - ((worldMapWidth / 2 * Math.log((1 + Math.sin(latitudeRad)) / (1 - Math.sin(latitudeRad)))) - mapOffsetY)
return {x, y} // the pixel x,y value of this point on the map image
}

I've converted the PHP code provided by Raphael to JavaScript and can confirm it worked and this code works myself. All credit to Raphael.
/*
var mapWidth = 1500;
var mapHeight = 1577;
var mapLonLeft = 9.8;
var mapLonRight = 10.2;
var mapLonDelta = mapLonRight - mapLonLeft;
var mapLatBottom = 53.45;
var mapLatBottomDegree = mapLatBottom * Math.PI / 180;
*/
function convertGeoToPixel(latitude, longitude ,
mapWidth , // in pixels
mapHeight , // in pixels
mapLonLeft , // in degrees
mapLonDelta , // in degrees (mapLonRight - mapLonLeft);
mapLatBottom , // in degrees
mapLatBottomDegree) // in Radians
{
var x = (longitude - mapLonLeft) * (mapWidth / mapLonDelta);
latitude = latitude * Math.PI / 180;
var worldMapWidth = ((mapWidth / mapLonDelta) * 360) / (2 * Math.PI);
var mapOffsetY = (worldMapWidth / 2 * Math.log((1 + Math.sin(mapLatBottomDegree)) / (1 - Math.sin(mapLatBottomDegree))));
var y = mapHeight - ((worldMapWidth / 2 * Math.log((1 + Math.sin(latitude)) / (1 - Math.sin(latitude)))) - mapOffsetY);
return { "x": x , "y": y};
}

I think it's worthwhile to keep in mind that not all flat maps are Mercator projections. Without knowing more about that map in particular, it's hard to be sure. You may find that most maps of a small area of the world are more likely to be a conical type projection, where the area of interest on the map is "flatter" than would be on a global Mercator projection. This is especially more important the further you get away from the equator (and the UK is far enough away for it to matter).
You may be able to get "close enough" using the calculations you're trying, but for best accuracy you may want to either use a map with a well-defined projection, or create your own map.

I know the question was asked a while ago, but the Proj4JS library is ideal for transforming between different map projections in JavaScript.
UK maps tend to use the OSGB's National Grid which is based on a Transverse Mercator projection. Ie. like a conventional Mercator but turned 90 degrees, so that the "equator" becomes a meridian.

#Xarinko Actionscript snippet in Javascript (with some testing values)
var mapWidth = 1500;
var mapHeight = 1577;
var mapLonLeft = 9.8;
var mapLonRight = 10.2;
var mapLonDelta = mapLonRight - mapLonLeft;
var mapLatBottom = 53.45;
var mapLatBottomRadian = mapLatBottom * Math.PI / 180;
function convertPixelToGeo(tx, ty)
{
/* called worldMapWidth in Raphael's Code, but I think that's the radius since it's the map width or circumference divided by 2*PI */
var worldMapRadius = mapWidth / mapLonDelta * 360/(2 * Math.PI);
var mapOffsetY = ( worldMapRadius / 2 * Math.log( (1 + Math.sin(mapLatBottomRadian) ) / (1 - Math.sin(mapLatBottomRadian)) ));
var equatorY = mapHeight + mapOffsetY;
var a = (equatorY-ty)/worldMapRadius;
var lat = 180/Math.PI * (2 * Math.atan(Math.exp(a)) - Math.PI/2);
var long = mapLonLeft+tx/mapWidth*mapLonDelta;
return [lat,long];
}
convertPixelToGeo(241,444)

C# implementation:
private Point ConvertGeoToPixel(
double latitude, double longitude, // The coordinate to translate
int imageWidth, int imageHeight, // The dimensions of the target space (in pixels)
double mapLonLeft, double mapLonRight, double mapLatBottom // The bounds of the target space (in geo coordinates)
) {
double mapLatBottomRad = mapLatBottom * Math.PI / 180;
double latitudeRad = latitude * Math.PI / 180;
double mapLonDelta = mapLonRight - mapLonLeft;
double worldMapWidth = (imageWidth / mapLonDelta * 360) / (2 * Math.PI);
double mapOffsetY = worldMapWidth / 2 * Math.Log((1 + Math.Sin(mapLatBottomRad)) / (1 - Math.Sin(mapLatBottomRad)));
double x = (longitude - mapLonLeft) * (imageWidth / mapLonDelta);
double y = imageHeight - ((worldMapWidth / 2 * Math.Log((1 + Math.Sin(latitudeRad)) / (1 - Math.Sin(latitudeRad)))) - mapOffsetY);
return new Point()
{
X = Convert.ToInt32(x),
Y = Convert.ToInt32(y)
};
}

If you want to avoid some of the messier aspects of lat/lng projections intrinsic to Proj4JS, you can use D3, which offers many baked-in projections and renders beautifully. Here's an interactive example of several flavors of Azimuthal projections. I prefer Albers for USA maps.
If D3 is not an end-user option -- say, you need to support IE 7/8 -- you can render in D3 and then snag the xy coordinates from the resultant SVG file that D3 generates. You can then render those xy coordinates in Raphael.

This function works great for me because I want to define the mapHeight based on the map I want to plot. I'm generating PDF maps. All I need to do is pass in the map's max Lat , min Lon and it returns the pixels size for the map as [height,width].
convertGeoToPixel(maxlatitude, maxlongitude)
One note in the final step where $y is set, do not subtract the calculation from the mapHeight if your coordinate system 'xy' starts at the bottom/left , like with PDFs, this will invert the map.
$y = (($worldMapWidth / 2 * log((1 + sin($lat)) / (1 - sin($lat)))) - $mapOffsetY);