I've got two LatLon (latitude-longitude) objects which represent two locations on the surface of the globe. I want to find the angle (in radians) between the center of the earth and these two LatLon objects.
I'm going to use this angle and the radius of the earth to calculate the arc length between the two locations (I figure this will give better precision than using simple Pythagoras, and be faster than computing the great circle distance).
I already have code to give me the Pythagorean distance and the great circle distance.
Using something like this - how to calculate the angle between two vectors
I thought this at first (after some calc on paper) is this Pythagorean thing?
angle_between_radian = sqrt(deltaLA^2 + deltaLO^2)*PI /180
edit: delta = delta>180?360-delta:delta
We working on sphere then above must wrong ^^. But this link may help:Calculate distance, bearing and more between Latitude/Longitude points.
Related
what's the algorithm to be able locate and display places around me within a particular distance such as 100m,using easting and northing and name of the place where I'm based .
To be more clear, lets suppose I'm based in charing cross and I want to find all places within 100m using easting and northing data for example, easting =10000m and easting=20000m.
Thank you
Pythagoras is the relevant maths.
If your position is (x,y) then you can calc a distance to any other point (x2,y2) with:
distance = sqrt((x2-x)^2 + (y2-y)^2)
So you could just loop over all points, calc their distance and order the results by nearest first.
For large data sets this may become impractical, in which case you'll want to partition the points into large rectangles. The first stage then is to identify which rectangle your (x,y) is within and the adjacent rectangles, then loop through all points in those rectangles. You need the adjacent rectangles because your (x,y) might be right on the boundary of its rectangle.
More generally this partitioning approach comes under the general heading of spatial hashing. For very large areas you want a tree structure known as a quadtree, that breaks large areas down into smaller and smaller regions, but that might be overkill for what you want.
I am assuming by Cartesian coordinates you also mean linear. If you are trying to do this using actual earth coordinates this answer gets more complicated (as we aren't on a flat earth). For simple linear coordinates you could do something like:
bool contains( x, y)
{
return (x >= minx) && (x <= maxx) && (y >= miny) && (y <= maxy);
}
The min, max coordinates would be your current position + how far out you wanted to go. I think this is what you wanted to know. If you need accurate earth coordinates you might look into some geospatial libraries. If you need and estimate you can still use the algorithm above but I would use something like Rhumb lines to calculate the min, max coordinates.
Hello Programming World,
I am using the Google Maps API v3 and I have a map overlay that I need specific coordinates from.
In the image, I've managed to convert the two pixel coordinate sets from the top left and bottom right into latitude and longitude coordinates manually. I've found these equalities:
Top Left:
(1px [x-axis], 1px [y-axis]) in pixels equals...
(-109.05005006119609° [Longitude], 41.00062580014626° [Latitude]) in latitude/longitude
Bottom Right:
(575px [x-axis], 423px [y-axis]) in pixels equals...
(-102.0423399657011° [Longitude], 36.99314427042206° [Latitude]) in latitude/longitude
However, my boss later decided that he'd rather have Google Maps draw this with Google's polyline drawing class (so I need every border angle that I drew in my overlay in latitude/longitude coordinates).
My question is, given these two sets of coordinates, is there some sort of equation or formula that I can use so that I can find the pixel coordinate (which I already have in an image map file) and convert it to a latitude/longitude coordinate.
Example: Point of interest at 132px x-axis and 10px y-axis = a° Longitude and b° Latitude
Find a and b.
I appreciate the help,
Llewgnolm
I talked to a friend of mine and he answered my question, so I figured I'd post the results here for anyone who'd like to know.
What we did was create a system of linear equations:
Lat/Lng = Multiplier*Pixels+Constant (L=m*p+c)
We substituted the numbers that I found manually into two equations for the system, i.e. (for Longitude):
-109.05...=m*(1)+c and
-102.04...=m*575+c and solved both for m and c
Eventually, the numbers came out to be (for longitude):
m = 0.01220895749767
c = -109.06228374530811
We can then use our values for m and c to plug in any pixel value to this equation and find the Latitude/Longitude, i.e (from the example numbers listed above again, for Longitude):
L = 0.01220895749767(132) - 109.06228374530811
L (Longitude at 132px in the x-direction) = -107.45070135561566°
This finds the degree values based off of the images pixels down to the 14th decimal (more than enough). For latitude, I just made a different system of equations with the latitude-based numbers inside to find different m and c values. Hope this helps anyone who stumbles across a similar problem.
I am trying to create a Circle Geometry in MySQL using the co-ordinates of the center and a radius. I searched everywhere...all i could find in the MySQL doc on the site were for polygons. May be i am looking in the wrong place. can anybody help me with an appropriate SQL that can help me create a table that stores this Circle geometry as one of the columns in the table?
Also, i am not even sure if there is a way to do so in MySQL?..The version i am using is MySQL 5.6.
Thanks in advance.
As of MySQL v5.6.1, you can use Buffer(g, d):
Returns a geometry that represents all points whose distance from the geometry value g is less than or equal to a distance of d.
Obviously, in your case g should be the point at the centre of the circle and d should be its radius.
There are two Parts:
A.For given tested points you have to check their relation with given circle.
B.You want to generate points on circumference of given circle.
A.Yes, First of all take the distance between your given point(test Point) and the centre of circle point. Both of these points are defined in Latitude and longitude. Distance formula between two points(x1,y1) and (x2,y2) is distance d= sqrt[ (x2-x1)^2 + (y2-y1)^2 ].
Now,
If this distance is less than radius of circle then your tested point is inside your circle.
If this distance is Greater than radius then tested point is outside the circle.
If this calculated distance is equal to radius of circle then this tested point is on your circle i.e. on the circumference of your circle.
B. In a circle the total angle theta is 360 degree or 2*Pi in radians.
For given Circle whose centre is (x1, y1) and radius is r.
x = x1 + r * cos(theta)
y = y1 + r * sin(theta)
where, theta is running from Zero to 2*Pi and Pi is 3.1415.
Depending upon how you do it. Example: if you wants 10 points on circle, then increment=(2*Pi-Zero)/10.
fist theta is zero, then theta is Zero+increment, then theta is Zero +increment+increment i.e. 2* increment and then zero + 3*increment and then so on. unless you get theta equal to 2*Pi.
For all above thetas calculate x and y. These all x and y coordinate points are on the circumference of the circle.
I'm receiving data from GPS and store them in MySQL database. I have following columns:
User ID (int)
Latitude (double)
Longitude (double)
Horizontal Accuracy (double)
Horizontal accuracy is radius around Lat/Long, so my user with equivalent probability can be in any point of this area.
I need to find out probability that two users was intersecting. But I also have vision area, which is 30 meters. If horizontal accuracy would be 0 I could just measure area of intersection of two circles that have radius of 30 meters around lat/long. But in my case that's not possible because horizontal accuracy could be in range from 5 to 3000. Usually it's more than my vision area.
I think I can measure area of intersection of two cones where inner circle of this cone will have radius of horizontal accuracy + 30 meters and outer circle will have radius of horizontal accuracy. But this solution seems to be little bit complicated.
I want to hear some thoughts about that and other possible solution.
I've checked MySQL Spatial extension and as far I can see it can't do such calculations for me.
Thanks.
I worked on just such a problem as you are describing. How I approached it was to convert the Lat/Long (world coordinates) into X/Y (Cartesian coordinates) then I applied the Pythagorean Theorem a^2 + b^2 = c^2 to solve the problem.
First you need to convert the Lat/Long Coordinates.
To get X you Multiply the Radius by the cosine (cos) of the angle (NOTE: this angle has to be expressed as radians).
To get Y you do the same as above but use the sine function (sin).
To convert degrees to radials Multiply the angle by the quantity of PI (Approx. 3.14159...) / 180.
Radians = Angle * (PI / 180);
To solve for the c^2 "C Squared" c = SQRT (a*a + b*b);
For more information on Degrees to Radians: http://www.mathwarehouse.com/trigonometry/radians/convert-degee-to-radians.php
For more information on: Converting Lat/Long to X/Y coordinates: http://www.mathsisfun.com/polar-cartesian-coordinates.html
I usually find the information that I need for this kind of problem by asking a question on ask.com.
All the best.
Allan
I'm using OpenMap and I'm reading a ShapeFile using com.bbn.openmap.layer.shape.ShapeFile. The bounding box is read in as lat/long points, for example 39.583642,-104.895486. The bounding box is a lower-left point and an upper-right point which represents where the points are contained. The "points," which are named "radians" in OpenMap, are in a different format, which looks like this: [0.69086486, -1.8307719, 0.6908546, -1.8307716, 0.6908518, -1.8307717, 0.69085056, -1.8307722, 0.69084936, -1.8307728, 0.6908477, -1.8307738, 0.69084626, -1.8307749, 0.69084185, -1.8307792].
How do I convert the points like "0.69086486, -1.8307719" into x,y coordinates that are usable in normal graphics?
I believe all that's needed here is some kind of conversion, because bringing the points into Excel and graphing them creates a line whose curve matches the curve of the road at the given location (lat/long). However, the axises need to be adjusted manually and I have no reference as how to adjust the axises, since the given bounding box appears to be in a different format than the given points.
The ESRI Shapefile technical description doesn't seem to mention this (http://www.esri.com/library/whitepapers/pdfs/shapefile.pdf).
0.69086486, -1.8307719 is a latitude and a longitude in radians.
First, convert to degrees (multiply by (180/pi)), then you will have common units between your bounding box and your coordinates.
Then you can plot all of it in a local frame with the following :
x = (longitude-longitude0)*(6378137*pi/180)*cos(latitude0*pi/180)
y = (latitude-latitude0)*(6378137*pi/180)
(latitude0, longitude0) are the coordinates of a reference point (e.g. the lower-left corner of the bounding box)
units are degrees for angles and meters for distances
Edit -- explanation :
This is an orthographic projection of the Earth considered as a sphere whose radius is 6378137.0 m (semi-major axis of the WGS84 ellipsoid), centered on the point (lat0, lon0)
In OpenMap, there are a number of ways to convert from radians to decimal degrees:
Length.DECIMAL_DEGREE.fromRadians(radVal);
Math.toDegrees(radVal) // Standard java library
For an array, you can use ProjMath.arrayDegToRad(double[] radvals);
Be careful with that last one, it does the conversion in place. So if you grab a lat/lon array from an OMPoly, make a copy of it first before converting it. Otherwise, you'll mess up the internal coordinates of the OMPoly, which it expects to be in radians.