> center = [50, 50]
> radius = 10
> db.places.find({"loc" : {"$within" : {"$center" : [center, radius]}}})
Is it 10 km, miles, feet, or meters?
Lat/long?
I must be the same units than your data. If you store location in meters, your query will operate in meters. You have to choose what units you want to use.
I you need to use lat/long, the units for the position of your objects will be in decimal degrees and the units for the distances will be in radians. You need to use the spherical model to handle the fact that earth is round.
They're in degrees - so 1 ~= 69 miles.
Related
I would like to simplify a GPS coordinate to a "TOP LEFT" position of a defined virtual grid (for example 100 kilometers).
If my GPS coordinate is in a cell of the grid, then we use the GPS position of the "TOP LEFT" of the cell.
The new coordinate is not intended to be displayed on a map but just to be communicated and manipulated.
This imaginary grid would have an editable distance (e.g. 1 kilometer or 100 kilometers).
I had imagined calculating the distance between two known points:
latitude = 0 / longitude = 0 (Null Island)
has GPS coordinate (ex: lat 48.858284648396626 lng 2.294501066207886)
we calculate distances for lat/lng (i use leaflet distanceTo function )
distance_latitude: 5432.79 km (?) : distanceTo( [0,0],[48.858284648396626,0])
distance_longitude: 255.14 km (?) : distanceTo( [0,0],[0,2.294501066207886])
var distance=100; // in kilometers
distance_latitude= Math.floor(distance_latitude/ distance) * distance;
= 5400
distance_longitude= Math.floor(distance_longitude / distance) * distance;
= 200
but after that... how to transform these kilometers (from Null Island) to a new coordinate? (so the top left of the current cell where the poi is)
(grid not in scale!)
Sometimes it's better to wait for a good night's sleep!
answer: just move nullIsland the distance between the poi and... nullIsland... simply! So all the dots (red) take the TopLeft position -blue marker- of the imaginary grid (in my screenshot: a 5Km grid).
I'm having trouble understanding what precisely the output of meshgrat means and how this relates to the lat and lon parameters of pcolorm(lat,lon,Z). I have a grid of global data, I'll call Z, at a 1.5 degree latitude x 1.5 degree longitude spatial resolution. Thus I have a matrix that's 120 x 240 (180 degrees of latitude / 1.5 = 120, 360 degrees of longitude / 1.5 = 240). Row 1 is 90 N and column 1 is 180 W (-180).
If I follow the MATLAB documentation, I can use meshgrat to produce the lat and lon arguments that I need to supply to pcolorm as follows.
latlim = [-90 90];
lonlim = [-180 180];
[lat,lon] = meshgrat(latlim,lonlim,[120 240]);
However, I don't understand why the spacing of the output is the way it is. For example, the first five values of lat are [-90.0000, -88.4874, -86.9748,-85.4622,-83.9496...]. The lon values follow the same spacing. The spacing is very close to 1.5 degrees, but it isn't. Why is there a discrepancy? The documentation claims that the paired lat and lon values are the location of the graticule vertices. In that case, these values make some sense, since there will always be one more vertex than actual grid cells. To test this, I made the following adjustment to the meshgrat code by adding one extra row and column:
latlim2 = [-90 90];
lonlim2 = [-180 180];
[lat2,lon2] = meshgrat(latlim2,lonlim2,[121 241]);
This did, indeed, produce the expected output, with the spacing now exactly at 1.5 degrees (i.e [-90.0000, -88.5000, -87.0000, -85.5000, -84.0000...]). Again, this is logical if these are viewed as vertices. But under this scenario lat and lon no longer match Z in size, which goes against how the documentation says to treat lat and lon in this case.
There seems to be a mismatch here: either the spacing in the lat lon grids are not accurate, or the girds are not the same size as the data, which would be fine in my mind as long as MATLAB knows how to interpret them accordingly, but the documentation does not seem to suggest using it this way. I have no detailed knowledge of how the MATLAB functions work at a finer level. Can someone explain to me what I'm missing?
Thus I have a matrix that's 120 x 240 (180 degrees of latitude / 1.5 = 120, 360 degrees of longitude / 1.5 = 240).
180/1.5 is indeed 120. But you also have an element at 0deg (presumably). That's 121.
I am trying to calculate distance between two geographical coordinates and I want to convert geographical coordinates to the km. Therefore I used deg2km function. However, I realise that it is not convert points properly.
For instance, I used these two points.
p_x=[5; 10]; %degree
p_y=[8; 16]; %degree
pos_y=deg2km(p_y,6378);
pos_x=deg2km(p_x,6378);
It returns as:
pos_x= [556.58549846099 1113.17099692198]
pos_y= [890.536797537587 1781.07359507517]
When I calculate distance ( sqrt((556.5-1113.2)^2+(890.5368-1781.1)^2) ) between these points I obtained distance as : 1050.2464
However I checked it google map and also other websites it should be 1042 km.
Do you have any suggestion to calculate distance and also points as kilometers properly?
Thanks in advance!
edited as :
I've points(deg)and I need to convert them km and calculate distance between points.
LAT=[41.000173;41.010134]*pi/180;
LON=[28.995882;28.995584]*pi/180;
I used this code to calculate distance. It calculates properly.
But I can not convert my points to kilometers.
LAT=[41.000173;41.010134]*pi/180;
LON=[28.995882;28.995584]*pi/180;
R=6378; %km
for i=1:length(LAT)-1
psi(i,1) = atan2( sin (LON(i+1)-LON(i)) * cos (LAT(i+1)) , cos (LAT(i)) *sin (LAT(i+1)) - sin (LAT(i)) * cos (LAT(i+1)) * cos (LON(i+1)-LON(i)) );
a=(sin((LAT(i+1)-LAT(i))/2))^2+cos(LAT(i))*cos(LAT(i+1))*(sin((LON(i+1)-LON(i))/2))^2;
c=2*atan2(sqrt(a),sqrt(1-a));
d(i,1)=R*c;
end
Reading here
http://docs.mongodb.org/manual/tutorial/query-a-2dsphere-index/
I find the following:
The following example queries grid coordinates and returns all documents
within a 10 mile radius of longitude 88 W and latitude 30 N. The example
converts the distance, 10 miles, to radians by dividing by the approximate
radius of the earth, 3959 miles:
db.places.find( { loc :
{ $geoWithin :
{ $centerSphere :
[ [ 88 , 30 ] , 10 / 3959 ]
} } } )
I think the "standard" notation is:
East is + (plus) and West is - (minus),
North is + (plus) and South is - (minus).
So why is West + (plus) in this example
on the MongoDB documentation site?
Is it really that way in MongoDB?
In fact, is there any standard which defines if West
maps to + or to - and the same for East, North, South?
See also:
Wikipedia - Latitude and longitude of the Earth
Wikipedia - Geographic coordinate system
I wanted to test the mapKit and wanted to make my own overlay to display the accuracy of my position.
If i have a zoom factor of for example .005 which radius does my circle around me has to have(If my accuracy is for example 500m)?
Would be great to get some help :)
Thanks a lot.
Look at the documentation for MKCoordinateSpan, which is part of the map's region property. One degree of latitude is always approx. 111 km, so converting the latitudeDelta to meters and then getting to the meters per pixel should be easy. For longitudinal values it is not quite so easy as the distance covered by one degree of longitude varies between 111 km (at the equator) and 0 km (at the poles).
My way to get meters per pixel:
MKMapView *mapView = ...;
CLLocationCoordinate2D coordinate = ...;
MKMapRect mapRect = mapView.visibleMapRect;
CLLocationDistance metersPerMapPoint = MKMetersPerMapPointAtLatitude(coordinate.latitude);
CGFloat metersPerPixel = metersPerMapPoint * mapRect.size.width / mapView.bounds.size.width;
To add to another answer, a difference of one minute of latitude corresponds to one nautical mile: that's how the nautical mile was defined. So, converting to statute miles, 1 nautical mile = 1.1508 statue miles, or 6076.1 ft. or 1852 meters.
When you go to longitude, the size of the longitude circles around the Earth shrink as latitude increases, as was noted on the previous answer. The correct factor is that
1 minute of longitude = (1852 meters)*cos(theta),
where theta is the latitude.
Of course, the Earth is not a perfect sphere, but the simple calculation above would never be off by more than 1%.