I have :
+37.785834 (lat)
-122.406417 (long)
The LAT positive means North of the equator
The LONG negative means to the West of 0
So far so good.
But what does the number represent exactly ? For example suppose I added 0.000001 to the LAT what would this actually represent ? That I have moved in some Northerly direction by 1 metre, for example or....
If I understand what these numbers actually represent then I can use them intelligently.
Longitude and latitude are measured in degrees. The latitude ranges from -90° (south pole) to +90° (north pole). The longitude ranges from -180° to +180°.
Related
I have a set of Lat-long values of various locations on the map.
Eg.
(18.605906724686033, 73.14103329678608)
(18.606069075261992, 73.13979277511669)
(18.606066194308486, 73.13996845980716)
(18.60624633887108, 73.1400100340469)
I need to plot them in a graph (in meters) with the first position as the origin.
Please help me with the formulae for the conversion.
Thank you for answering.
x0= 18.605906724686033 //origin longitude
y0= 73.14103329678608 //origin latitude
x1= 18.606069075261992 //longidute
y1= 73.13979277511669 //latitude
dlong=x1-x2
dlat=y1-y2.
x=dlong*111139. //the x-cordinate for the given longitude
y=dlat*111190 //The y-cordinate for the given latitude
I want to create a group of polygons for a city that are 80km x 80km. Given a starting Lat and Long, my thought is I can add 80km to that point so that I get 4 points to create the polygon.
(x,y) -> (x+80km, y) -> (x+80km, y+80km) -> (x, y+80km) -> (x,y)
Where I'm having difficulty is finding a way to calculate the point +80km. I've found the SQL Server Spatial Tools and there is a function
SqlGeography LocateAlongGeog(SqlGeography g, double distance)
But so far I haven't been able to figure out how to use it. I will continue to play with this but if there are any other approaches I can take, or if anyone knows how to properly use this function, I'd be grateful.
Longitude is a "great circle" measure, i.e. if you draw a circle representing a particular longitude round the Earth, it's always a circle whose centre is the centre of the Earth - so to circumnavigate the Earth at a constant longitude, you always travel the same distance:
2 * PI * 6378 /* 6378 is the radius of the Earth in km */
So, moving North (i.e travelling along the same longitude) 80 km will increase your latitude by:
360 * 80 / (2 * PI * 6378)
Latitude is trickier cos the distance travelled when you circumnavigate the Earth at the same latitude changes depending on the latitude at which you're travelling: however, the formula is simple and I looked it up at: http://www.newton.dep.anl.gov/askasci/env99/env086.htm
2 * PI * 6378 * COS(LAT) /* where LAT is your Latitude */
So, if you are at latitude LAT, and move 80km East, you will increase your longitude by:
360 * 80 / (2 * PI * 6378 * COS(LAT))
Couple of things to note:
a) 6378 is only accurate to the nearest km
b) The East/West between your two Northerly points will not be precisely 80km - not significantly different for Latitudes between about 80 degrees North and 80 degrees South - as long as you're not looking for high-precision pinpoint accuracy (which I'm guessing with base measurements of 80 km you're not) it'll do just nicely (and point nicelt at Bing or Google, say)
c) SQL calculates trigonometry functions using radians not degrees - so in SQL your cosine will need to be:
COS(PI * LAT / 180)
HTH and makes some sort of sense
I want to overlay great circle arcs between airports on a map using longitudes and latitudes.
I can already get the distance and bearing from the initial and final coordinates but now I need to produce the points on the curve to plot through.
What I would like is a formula which takes the origin, destination, and a distance, and returns the latitude/longitude of the point that lies at that distance on the path between the origin and the destination.
I'm currently approximating the earth by a sphere and using radians -- eventually I'll add in spheroid corrections.
currlat = oldlat + d * sin (angle)/ (radius);
currlon = oldlon + d * cos (angle)/ (radius * cos(oldlat));
where d is distance travelled and angle is in radians. This is assuming circumference of earth at 40000km both at equator and through the poles. You can convert in radians...
Also it assumes the angle (direction) is with reference to equator line.
Obviously this needs spheroid corrections.
if you go south sin values will turn negative and north it will go positive. If you go west cos will turn negative and east it will turn positive.
d * sin(angle) and d * cos(angle) gives you the change. and you just calculate the new lat/long on that basis scaling up against circumference of earth.
I have latitude and longitude of a point.I have to find out the latitude and longitude of another point from a relative distance from the known point.For example point A has some location with latitude and longitude.What is the latitude and longitude after moving 1000m south and 500m west from point A.Is there any direct equation to find this? Thanks in advance
Note the accepted answer is basically the flat earth projection equations:
x = δlon * EarthRadius * cos( lat )
y = δlat * EarthRadius
For better accuracy over larger distances, you should compute the final lat/lon from a typical bearing/range calculation. See the section Destination point given distance and bearing from start point at this website: http://www.movable-type.co.uk/scripts/latlong.html
Instead of looking up an equation you can calculate as follows. Let R be the radius of the Earth. Let a be the current latitude and b be the current longitude. Then if you move δx metres east (negative for west) then δy metres south, calculating the new longitude can be done as follows.
Intersecting a horizontal plane with the Earth at the current latitude will give you a circle of radius R*cos(a). So to convert δx to the change in longitude, you get something like
δlong = δx * 2π / (2π * R * cos(a)) = δx / (R * cos (a))
The change in latitude is easier, since it doesn't depend on the current position. You're always moving around a great circle through the two poles. Then δlat = δy / R. (Of course you need to mod out by 2 π at some point).
I have a table containing geographic data and I want to group rows on the proximity of the X and Y coordinates. So, given an offset, n, and a table with columns X and Y, I want to group rows where ABS(row1.X - row2.X) < n, with a count of the number of rows in the group. Is this possible with SQL or do I need a function?
How big is your spatial range ? How big is your data set ? How accurate do you need ?
This is relevant because, if they are close, you don't have to worry about the curvature of the Earth.
Degrees of latitude are parallel so the distance between each degree remains almost constant but since degrees of longitude are farthest apart at the equator and converge at the poles, their distance varies greatly.
Each degree of latitude is approximately 69 miles (111 kilometers) apart. The range varies (due to the earth's slightly ellipsoid shape) from 68.703 miles (110.567 km) at the equator to 69.407 (111.699 km) at the poles. This is convenient because each minute (1/60th of a degree) is approximately one mile.
A degree of longitude is widest at the equator at 69.172 miles (111.321) and gradually shrinks to zero at the poles. At 40° north or south the distance between a degree of longitude is 53 miles (85 km).
The other aspect is, if Fred is 4 miles from Bill and Bill is 4 miles from Tom, the distance between Fred and Tom might be 8 miles. If your proximity threshold is 5 miles, Bill and Fred are in the same group, Bill and Tom are in the same group, but Fred and Tom aren't.
The following query might give you a useful start though:
select abs(abs(a.lat) - abs(b.lat)),abs(abs(a.lon) - abs(b.lon)),
sdo_geom.sdo_distance(a.geom, b.geom, 0.005,'unit=kilometer') dist_km
from
(select sdo_geometry(2001,8314,sdo_point_type(33,151, null), null,null) geom, 33 lat, 151 lon from dual) a,
(select sdo_geometry(2001,8314,sdo_point_type(34,151, null), null,null) geom, 34 lat, 151 lon from dual) b
Do you have oracle spatial? If so there are a number of built in functions to handle this for you. If not - you'll need a function to determine catographic distances (and then group off of that). I recall seing a SO question on how to write such a function last week. (There are actually a few along these lines)
How to limit a MySQL Distance Query
Best bet. Use the spatial extensions. They'll perform much better.