Given a location(lat,lng), I want to get its coordinate in an Azimuthal Equidistant Projection. The formulas are explained here.
Below is the screenshot of the web page.
at the end of that page, it states
It looks like given any location (lat [-Pi/2, +Pi/2], lng [0, +2Pi)), and the center of the projection (latCenter, lngCenter), I can calculate its coordinate (x, y) in the map, and since no map Radius is provided, the value of x and y will fall in the range of [-1, +1] or [-Pi, +Pi].
My question is, what is the c in the formulas? If it is a value calculated from (x, y), how can it be used to calculate (x, y)?
Can somebody help me understand these formulas?
Use equation 4 to compute c when projecting from lat/long to x,y. Equation 7 is for computing the inverse, ie going from x,y to lat/long. For your purposes, making a map, ignore equation 7.
c is the angle subtended at the centre of the Earth by the arc of the great circle from the centre of the projection (phi0, lambda0) to another point (phi, lambda).
As you don't state the programming language you're working with here is an implementation in F# from a recent blogpost.
open System
module AzimuthalEquidistantProjection =
let inline degToRad d = 0.0174532925199433 * d; // (1.0/180.0 * Math.PI) * d
let project centerlon centerlat lon lat =
// http://mathworld.wolfram.com/AzimuthalEquidistantProjection.html
// http://www.radicalcartography.net/?projectionref
let t:float = degToRad lat
let l:float = degToRad lon
let t1 = degToRad centerlat // latitude center of projection
let l0 = degToRad centerlon // longitude center of projection
let c = Math.Acos ((sin t1) * (sin t) + (cos t1) * (cos t) * (cos (l-l0)))
let k = c / (sin c)
let x = k * (cos t) * (sin (l-l0))
let y = k * (cos t1) * (sin t) - (sin t1) * (cos t) * (cos (l-l0))
(x, y)
Other versions (F# with units of measure, Python and Julia)
Related
I come accross a math problem about Interactive Computer Graphics.
I summarize and abstract this problem as follows:
I'm going to rotation a 3d coordinate P(x1,y1,z1) around a point O(x0,y0,z0)
and there are 2 vectors u and v which we already know.
u is the direction to O before transformation.
v is the direction to O after transformation.
I want to know how to conduct the calculation and get the coordinate of Q
Thanks a lot.
Solution:
Rotation About an Arbitrary Axis in 3 Dimensions using the following matrix:
rotation axis vector (normalized): (u,v,w)
position coordinate of the rotation center: (a,b,c)
rotation angel: theta
Reference:
https://docs.google.com/viewer?a=v&pid=sites&srcid=ZGVmYXVsdGRvbWFpbnxnbGVubm11cnJheXxneDoyMTJiZTZlNzVlMjFiZTFi
for just single point no rotations is needed ... so knowns are:
u,v,O,P
so we now the distance is not changing:
|P-O| = |Q-O|
and directions are parallel to u,v so:
Q = O + v*(|P-O|/|v|)
But I suspect you want to construct rotation (transform matrix) such that more points (mesh perhaps) are transformed. If that is true then you need at least one known to get this right. Because there is infinite possible rotations transforming P -> Q but the rest of the mesh will be different for each ... so you need to know at least 2 non trivial points pair P0,P1 -> Q0,Q1 or axis of rotation or plane parallel to rotation or any other data known ...
Anyway in current state you can use as rotation axis vector perpendicular to u,v and angle obtained from dot product:
axis = cross (u,v)
ang = +/-acos(dot(u,v))
You just need to find out the sign of angle so try both and use the one for which the resultinq Q is where it should be so dot(Q-O,v) is max. To rotate around arbitrary axis and point use:
Rodrigues_rotation_formula
Also this might be helpfull:
Understanding 4x4 homogenous transform matrices
By computing dot product between v and u get the angle l between the vectors. Do a cross product of v and u (normalized) to produce axis of rotation vector a. Let w be a vector along vector u from O to P. To rotate point P into Q apply the following actions (in pseudo code) having axis a and angle l computed above:
float4 Rotate(float4 w, float l, float4 a)
{
float4x4 Mr = IDENTITY;
quat_t quat = IDENTITY;
float4 t = ZERO;
float xx, yy, zz, xy, xz, yz, wx, wy, wz;
quat[X] = a[X] * sin((-l / 2.0f));
quat[Y] = a[Y] * sin((-l / 2.0f));
quat[Z] = a[Z] * sin((-l / 2.0f));
quat[W] = cos((-l / 2.0f));
xx = quat[X] * quat[X];
yy = quat[Y] * quat[Y];
zz = quat[Z] * quat[Z];
xy = quat[X] * quat[Y];
xz = quat[X] * quat[Z];
yz = quat[Y] * quat[Z];
wx = quat[W] * quat[X];
wy = quat[W] * quat[Y];
wz = quat[W] * quat[Z];
Mr[0][0] = 1.0f - 2.0f * (yy + zz);
Mr[0][1] = 2.0f * (xy + wz);
Mr[0][2] = 2.0f * (xz - wy);
Mr[0][3] = 0.0f;
Mr[1][0] = 2.0f * (xy - wz);
Mr[1][1] = 1.0f - 2.0f * (xx + zz);
Mr[1][2] = 2.0f * (yz + wx);
Mr[1][3] = 0.0f;
Mr[2][0] = 2.0f * (xz + wy);
Mr[2][1] = 2.0f * (yz - wx);
Mr[2][2] = 1.0f - 2.0f * (xx + yy);
Mr[2][3] = 0.0f;
Mr[3][0] = 0.0f;
Mr[3][1] = 0.0f;
Mr[3][2] = 0.0f;
Mr[3][3] = 1.0f;
w = Mr * w;
return w;
}
Point Q is at the end of the rotated vector w. Algorithm used in the pseudo code is quaternion rotation.
If you know u, v, P, and O then I would suggest that you compute |OP| which should be preserved under rotations. Then multiply this length by the unit vector -v (I assumed u, v are unit vectors: if not - normalize them) and translate the origin by this -|OP|v vector. The negative sign in front of v comes from the description given in your question:"v is the direction to O after transformation".
P and Q are at the same distance R to O
R = sqrt( (x1-x0)^2 + (y1-y0)^2 + (z1-z0)^2 )
and OQ is collinear to v, so OQ = v * R / ||v|| where ||v|| is the norm of v
||v|| = sqrt( xv^2 + yv^2 + zv^2 )
So the coordinates of Q(xq,yq,zq) are:
xq= xo + xv * R / ||v||
yq= yo + yv * R / ||v||
zq= zo + zv * R / ||v||
I'm using some stuff I've put together from various methods around the internet to make this work yet my bullets fly off in seemingly random directions. I've tried throwing negative signs in front of stuff and switching up the trig but to no avail. I've tried using the rotation of the player's arm because that accurately points to the user's mouse but that failed to give me any more accuracy.
I've tried to determine if the bullets follow a pattern like how I needed to invert the Y variable for my arm, but I cannot find a pattern here.
local x, y = objects.PlayerArm:GetPos()
local bullet = createBullet( x + objects.Player:GetWide(), y )
local mX, mY = gui.MouseX(), gui.MouseY()
local shootAngle = math.atan2((mY - y), (mX - x))
shootAngle = math.deg( math.Clamp(shootAngle, -90, 90) )
--shootAngle = objects.PlayerArm.Rotation
bullet.VelocityX = math.cos(shootAngle) * 5
bullet.VelocityY = math.sin(shootAngle) * 5
--bullet.Rotation = shootAngle
print("Angle", shootAngle, "Velocity X and Y", bullet.VelocityX, bullet.VelocityY)
Here is some of what was printed in console each time I shot a bullet.
Angle 47.920721521 Velocity X and Y -3.4948799788437 -3.5757256513158
Angle 24.928474135461 Velocity X and Y 4.8960495864893 -1.0142477244922
Angle 16.837472625676 Velocity X and Y -2.1355174970471 -4.5210137159497
Angle 10.684912400003 Velocity X and Y -1.5284445365972 -4.7606572338855
Angle -1.029154804306 Velocity X and Y 2.5777162320797 -4.2843178018061
Angle -11.63363399894 Velocity X and Y 2.978190104641 4.0162648942293
Angle -22.671343621981 Velocity X and Y -3.8872502993046 3.1447233758403
http://i.gyazo.com/e8ed605098a91bd450b10fda7d484975.png
As #iamnotmaynard suspected, Lua uses C's math library and so all the trig functions use radians instead of degrees. It is best to store all angles in radians and just print them in degrees for a more friendly format. Otherwise you have a lot of conversions to and from radians every time an angle is used. Below is the code updated to only use radians and print in degrees.
local mX, mY = gui.MouseX(), gui.MouseY()
local shootAngle = math.atan2((mY - y), (mX - x))
shootAngle = math.max(-math.pi/2, math.min(shootAngle, math.pi/2))
bullet.VelocityX = math.cos(shootAngle) * 5
bullet.VelocityY = math.sin(shootAngle) * 5
print("Angle (radians)", shootAngle, "(degrees)", math.deg(shootAngle),
"Velocity X and Y", bullet.VelocityX, bullet.VelocityY)
However to compute velocity in the x and y directions angles are not necessary at all. The function below computes VelocityX and VelocityY using only the displacements and makes sure that the velocity is only in the lower right and upper right quadrants as well.
function shoot(x, y, dirx, diry, vel)
local dx = math.max(dirx - x, 0)
local dy = diry - y
local sdist = dx * dx + dy * dy
if sdist > 0 then
local m = vel / math.sqrt(sdist)
return dx * m, dy * m
end
end
bullet.VelocityX, bullet.VeclocityY = shoot(x, y, gui.MouseX(), gui.MouseY(), 5)
I'm currently working on a game in LBP2 that has modify the way a controller gives input. This question:
How can I convert coordinates on a square to coordinates on a circle?
Has helped me quite a lot with what I am doing, but I do have one problem. I need the inverse function of the one they give. They go from square -> circle, and I've tried searching all over for how to map a circle to a square.
The function given in the previous question is:
xCircle = xSquare * sqrt(1 - 0.5*ySquare^2)
yCircle = ySquare * sqrt(1 - 0.5*xSquare^2)
From Mapping a Square to a Circle
My question is given xCircle and yCircle... how do I find xSquare and ySquare?
I've tried all of the algebra I know, filled up two pages of notes, tried to get wolfram alpha to get the inverse functions, but this problem is beyond my abilities.
Thank you for taking a look.
x = ½ √( 2 + u² - v² + 2u√2 ) - ½ √( 2 + u² - v² - 2u√2 )
y = ½ √( 2 - u² + v² + 2v√2 ) - ½ √( 2 - u² + v² - 2v√2 )
Note on notation: I'm using x = xSquare , y = ySquare, u = xCircle and v = yCircle;
i.e. (u,v) are circular disc coordinates and (x,y) are square coordinates.
For a C++ implementation of the equations, go to
http://squircular.blogspot.com/2015/09/mapping-circle-to-square.html
See http://squircular.blogspot.com
for more example images.
Also, see http://arxiv.org/abs/1509.06344 for the proof/derivation
This mapping is the inverse of
u = x √( 1 - ½ y² )
v = y √( 1 - ½ x² )
P.S. The mapping is not unique. There are other mappings out there. The picture below illustrates the non-uniqueness of the mapping.
if you have xCircle and yCircle that means that you're on a circle with radius R = sqrt(xCircle^2 + yCircle^2). Now you need to extend that circle to a square with half-side = R,
if (xCircle < yCircle)
ySquare = R, xSquare = xCircle * R/yCircle
else
xSquare = R, ySquare = yCircle * R/xCircle
this is for the first quadrant, for others you need some trivial tweaking with the signs
There are many ways you could do this; here's one simple way.
Imagine a circle of radius R centred on the origin, and a square of side 2R centred on the origin, we want to map all of the points within and on the boundary of the circle (with coordinates (x,y)) to points within and on the boundary of the square. Note that we can also describe points within the circle using polar coordinates (r, ø) (that's supposed to be a phi), where
x = r cos ø,
y = r sin ø
(ie r^2 = x^2 + y^2 and r <= 1). Then imagine other coordinates x' = a(ø) x = a(ø) r cos ø, and y' = a(ø) y (ie, we decide that a won't depend on r).
In order to map the boundary of the circle (r = 1) to the boundary of the square (x' = R), we must have, for ø < 45deg, x' = a(ø) R cos ø = R, so we must have a(ø) = 1/cos ø. Similarly, for 45 < ø < 90 we must have the boundary of the circle map to y' = R, giving a(ø) = 1/sin ø in that region. Continuing round the circle, we see that a(ø) must always be positive, so the final mapping from the circle to the square is
x' = a(ø) x,
y' = a(ø) y
where
ø = |arctan y/x| = arctan |y/x|
and
a(ø) = 1/cos ø, when ø <= 45 deg (ie, when x < y), and
a(ø) = 1/sin ø, when ø > 45 deg.
That immediately gives you the mapping in the other direction. If you have coordinates (x', y') on the square (where x' <= R and y' <= R), then
x = x'/a(ø)
y = y'/a(ø)
with a(ø) as above.
A much simpler mapping, though, is to calculate the (r, ø) for the desired position on the circle, and map that to x' = r and y' = ø. That also maps every point in the circle into a rectangle, and vice versa, and might have better properties, depending on what you want to do.
So that's the real question: what is it you're actually aiming to do here?
I was implementing the solution above but the results are not satisfiying.
The square coordinates are not exact.
Here is a simple counter-example:
Consider the point (x,y)=(0.75, 1) on the square.
We map it to the circle with (u,v)=(0.53, 0.85) on the circle.
Applying the expression above we get the new square coordinates
(x',y')=(u/v,r)=(0.625543242, 1) with r=(u^2+v^2)^(1/2).
This point is close but not the expected precise solution.
I solved a root finding problem in order to get the inverse expression of the mapping from square to circle like above.
you need to solve the system equations like above:
I) u = x*(1-y^2/2)^(1/2)
II) v = y*(1-x^2/2)^(1/2)
One ends up with 8 root points as solution. One of the roots I implemented into Excel-VBA which I present here below and it works very fine.
' given the circle coordinates (u,v) caluclates the x coordinate on the square
Function circ2sqrX(u As Double, v As Double) As Double
Dim r As Double, signX As Double, u2 As Double, v2 As Double, uuvv As Double, temp1 As Double
u2 = u * u
v2 = v * v
r = Sqr(u2 + v2)
signX = 1
If v = 0 Or u = 0 Then
circ2sqrX = u
Exit Function
End If
If u < 0 Then
signX = -1
End If
If Abs(u) = Abs(v) And r = 1 Then
circ2sqrX = signX
Exit Function
End If
uuvv = (u2 - v2) * (u2 - v2) / 4
temp1 = 2 * Sqr(uuvv - u2 - v2 + 1)
circ2sqrX = -((temp1 - u2 + v2 - 2) * Sqr(temp1 + u2 - v2 + 2)) / (4 * u)
End Function
' given the circle coordinates (u,v) caluclates the y coordinate on the square
' make use of symetrie property
Function circ2sqrY(u As Double, v As Double) As Double
circ2sqrY=circ2sqrX(v,u)
End Function
What is the formula for calculating the distance between 2 geocodes? I have seen some of the answers on this site but they basically say to rely on SQL Server 08 functions, I'm not on 08 yet. Any help would be appreciated.
Use an approximation of the earth and the Haversine formula. You can get a javascript version on the following url, which you can translate to your language of choice:
http://www.movable-type.co.uk/scripts/latlong.html
Here is another way: http://escience.anu.edu.au/project/04S2/SE/3DVOT/3DVOT/pHatTrack_Application/Source_code/pHatTrack/Converter.java
Take a look here for a SQL server 2000 version SQL Server Zipcode Latitude/Longitude proximity distance search
This will do it for you in c#.
Within the namespace put these:
public enum DistanceType { Miles, Kilometers };
public struct Position
{
public double Latitude;
public double Longitude;
}
class Haversine
{
public double Distance(Position pos1, Position pos2, DistanceType type)
{
double preDlat = pos2.Latitude - pos1.Latitude;
double preDlon = pos2.Longitude - pos1.Longitude;
double R = (type == DistanceType.Miles) ? 3960 : 6371;
double dLat = this.toRadian(preDlat);
double dLon = this.toRadian(preDlon);
double a = Math.Sin(dLat / 2) * Math.Sin(dLat / 2) +
Math.Cos(this.toRadian(pos1.Latitude)) * Math.Cos(this.toRadian(pos2.Latitude)) *
Math.Sin(dLon / 2) * Math.Sin(dLon / 2);
double c = 2 * Math.Asin(Math.Min(1, Math.Sqrt(a)));
double d = R * c;
return d;
}
private double toRadian(double val)
{
return (Math.PI / 180) * val;
}
Then to utilise these in the main code:
Position pos1 = new Position();
pos1.Latitude = Convert.ToDouble(hotelx.latitude);
pos1.Longitude = Convert.ToDouble(hotelx.longitude);
Position pos2 = new Position();
pos2.Latitude = Convert.ToDouble(lat);
pos2.Longitude = Convert.ToDouble(lng);
Haversine calc = new Haversine();
double result = calc.Distance(pos1, pos2, DistanceType.Miles);
If
you know that the 2 points are "not too far from each other"
and you tolerate a "reasonably small" error.
Then, consider that the earth is flat between the 2 points :
The distance difference in the latitude direction is EarthRadius * latitude difference
The distance difference in the longitude direction is EarthRadius * longitude difference * cos(latitude).
We multiply by cos(lat) because the longitude degrees don't make the same km distance if the latitude changes. As P1 and P2 are close, cos(latP1) is close from cos(latP2)
Then Pythagore
In JavaScript :
function ClosePointsDistance(latP1, lngP1, latP2, lngP2) {
var d2r = Math.PI / 180,
R=6371; // Earth Radius in km
latP1 *= d2r; lngP1 *= d2r; latP2 *= d2r; lngP2 *= d2r; // convert to radians
dlat = latP2 - latP1,
dlng = (lngP2 - lngP1) * Math.cos(latP1);
return R * Math.sqrt( dlat*dlat + dlng*dlng );
}
I tested it between Paris and Orleans (France) : the formula finds 110.9 km whereas the (exact) Haversine formula finds 111.0 km.
!!! Beware of situations around the meridian 0 (you may shift it) : if P1 is at Lng 359 and P2 is at Lng 0, the function will consider them abnormally far !!!
The pythagorean theorem as offered up by others here doesn't work so well.
The best, simple answer is to approximate the earth as a sphere (its actually a slightly flattened sphere, but this is very close). In Haskell, for instance you might use the following, but the math can be transcribed into pretty much anything:
distRadians (lat1,lon1) (lat2,lon2) =
radius_of_earth *
acos (cos lat1 * cos lon1 * cos lat2 * cos lon2 +
cos lat1 * sin lon1 * cos lat2 * sin lon2 +
sin lat1 * sin lat2) where
radius_of_earth = 6378 -- kilometers
distDegrees a b = distRadians (coord2rad a) (coord2rad b) where
deg2rad d = d * pi / 180
coord2rad (lat,lon) = (deg2rad lat, deg2rad lon)
distRadians requires your angles to be given in radians.
distDegrees is a helper function that can take lattitudes and longitudes in degrees.
See this series of posts for more information on the derivation of this formula.
If you really need the extra precision granted by assuming the earth is ellipsoidal, see this FAQ: http://www.movable-type.co.uk/scripts/gis-faq-5.1.html
Here is a way to do it if you are using sql server.
CREATE function dist (#Lat1 varchar(50), #Lng1 varchar(50),#Lat2 varchar(50), #Lng2 varchar(50))
returns float
as
begin
declare #p1 geography
declare #p2 geography
set #p1 = geography::STGeomFromText('POINT('+ #Lng1+' '+ #Lat1 +')', 4326)
set #p2 = geography::STGeomFromText('POINT('+ #Lng2+' '+ #Lat2 +')', 4326)
return #p1.STDistance(#p2)
end
You're looking for the length of the Great Circle Path between two points on a sphere. Try looking up "Great Circle Path" or "Great Circle Distance" on Google.
Sorry, I don't know what country you are in even. Are we talking about Easting and Northings (UK, Ordance Survey system) or Lat/Long or some other system?
If we are talking Easting and Northing then you can use
sqr((x1-x2)^2 + (y1-y2)^2)
This does not allow for the fact that the earth is a sphere, but for short distances you won't notice. We use it at work for distances between points within the county.
Be carful about how longer grid reference you use. I think an 8 figure reference will give you a distance in metres. I'll be able to get a definate answer at work next week if no one else has supplied it.
the pythagorean theorem?
using a Latitude and Longitude value (Point A), I am trying to calculate another Point B, X meters away bearing 0 radians from point A. Then display the point B Latitude and Longitude values.
Example (Pseudo code):
PointA_Lat = x.xxxx;
PointA_Lng = x.xxxx;
Distance = 3; //Meters
bearing = 0; //radians
new_PointB = PointA-Distance;
I was able to calculate the distance between two Points but what I want to find is the second point knowing the distance and bearing.
Preferably in PHP or Javascript.
Thank you
It seems you are measuring distance (R) in meters, and bearing (theta) counterclockwise from due east. And for your purposes (hundereds of meters), plane geometry should be accurate enough. In that case,
dx = R*cos(theta) ; theta measured counterclockwise from due east
dy = R*sin(theta) ; dx, dy same units as R
If theta is measured clockwise from due north (for example, compass bearings),
the calculation for dx and dy is slightly different:
dx = R*sin(theta) ; theta measured clockwise from due north
dy = R*cos(theta) ; dx, dy same units as R
In either case, the change in degrees longitude and latitude is:
delta_longitude = dx/(111320*cos(latitude)) ; dx, dy in meters
delta_latitude = dy/110540 ; result in degrees long/lat
The difference between the constants 110540 and 111320 is due to the earth's oblateness
(polar and equatorial circumferences are different).
Here's a worked example, using the parameters from a later question of yours:
Given a start location at longitude -87.62788 degrees, latitude 41.88592 degrees,
find the coordinates of the point 500 meters northwest from the start location.
If we're measuring angles counterclockwise from due east, "northwest" corresponds
to theta=135 degrees. R is 500 meters.
dx = R*cos(theta)
= 500 * cos(135 deg)
= -353.55 meters
dy = R*sin(theta)
= 500 * sin(135 deg)
= +353.55 meters
delta_longitude = dx/(111320*cos(latitude))
= -353.55/(111320*cos(41.88592 deg))
= -.004266 deg (approx -15.36 arcsec)
delta_latitude = dy/110540
= 353.55/110540
= .003198 deg (approx 11.51 arcsec)
Final longitude = start_longitude + delta_longitude
= -87.62788 - .004266
= -87.632146
Final latitude = start_latitude + delta_latitude
= 41.88592 + .003198
= 41.889118
It might help if you knew that 3600 seconds of arc is 1 degree (lat. or long.), that there are 1852 meters in a nautical mile, and a nautical mile is 1 second of arc. Of course you're depending on the distances being relatively short, otherwise you'd have to use spherical trigonometry.
Here is an updated version using Swift:
let location = CLLocation(latitude: 41.88592 as CLLocationDegrees, longitude: -87.62788 as CLLocationDegrees)
let distanceInMeter : Int = 500
let directionInDegrees : Int = 135
let lat = location.coordinate.latitude
let long = location.coordinate.longitude
let radDirection : CGFloat = Double(directionInDegrees).degreesToRadians
let dx = Double(distanceInMeter) * cos(Double(radDirection))
let dy = Double(distanceInMeter) * sin(Double(radDirection))
let radLat : CGFloat = Double(lat).degreesToRadians
let deltaLongitude = dx/(111320 * Double(cos(radLat)))
let deltaLatitude = dy/110540
let endLat = lat + deltaLatitude
let endLong = long + deltaLongitude
Using this extension:
extension Double {
var degreesToRadians : CGFloat {
return CGFloat(self) * CGFloat(M_PI) / 180.0
}
}
dx = sin(bearing)
dy = cos(bearing)
x = center.x + distdx;
y = center.y + distdy;