Hot answers tagged longitude
68
Accuracy is the tendency of your measurements to agree with the true values. Precision is the degree to which your measurements pin down an actual value. The question is about an interplay of accuracy and precision.
As a general principle, you don't need much more precision in recording your measurements than there is accuracy built into them. Using too ...
31
This Wikipedia page (Decimal Degrees) has a table on "Accuracy versus decimal places". Also the accuracy of your coordinates depends on the instrument used to collect the coordinates - A-GPS used in cell phones, DGPS etc.
decimal places degrees distance
0 1 111 km
1 0.1 11.1 km
2 ...
29
The terms are medieval:
latitude (n.) late 14c., "breadth," from Old French latitude (13c.) and directly from Latin latitudo "breadth, width, extent, size," from latus "wide," ... . Geographical sense also is from late 14c., literally "breadth" of a map of the known world...
In the fourteenth century, most known-world maps made in the Christian and ...
23
I'm not an expert in this area, but I have done some reading on the subject, particularly on its history. I think the reason is: Accurate measurement of latitude came first as it was based on astronomical measurements. Longitude was not accurately measurable until a highly accurate time measuring device was developed.
8
Although geodesics do look a little like sine waves in some projections, the formula is incorrect.
Here is one geodesic in an Equirectangular projection. Clearly it is not a sine wave:
(The background image is taken from http://upload.wikimedia.org/wikipedia/commons/thumb/e/ea/Equirectangular-projection.jpg/800px-Equirectangular-projection.jpg.)
...
7
Same, equivalent terms. lat/long is probably the most common, but of course "long" is a reserved word / data type in many programming languages, so you'll find other ways to shorten longitude.
In some contexts, GIS people will use long/lat (i.e. reverse order), since this is the normal X/Y coordinate reference system for a Cartesian plane.
6
Manually reversing the rotation should do the trick; there should be a formula for rotating spherical coordinate systems somewhere, but since I can't find it, here's the derivation ( ' marks the rotated coordinate system; normal geographic coordinates use plain symbols):
First convert the data in the second dataset from spherical (lon', lat') to (x',y',z') ...
6
For a simple mean, then no, you do not want to average the longitude and latitude ordinates. This might work pretty well at lower latitudes, but at higher latitudes it will beging to give poor results, and completely break down near the poles.
The method I've used for this type of thing, is to convert the longitude,latitude coordinates to 3d cartesian ...
6
Your coordinates are Ordnance Survey GB (OSGB36 Airy 1830 ellipsoid)
431833,336684 (base on a false origin with the Transverse Mercator projection.
http://en.wikipedia.org/wiki/Ordnance_Survey_National_Grid
To convert them correctly to Lat/Lng
Use this site:http://www.nearby.org.uk
example:
http://www.nearby.org.uk/coord.cgi?p=431833%2C336684
look for the ...
5
This isn't so old-fashioned: I remember having to solve exactly this problem back in the 80's when we didn't have scanners readily available and had to lift coordinates and elevations off large-format printed maps for geostatistical analysis.
In effect you can already read the longitude accurately along any line of longitude on the map. You want to ...
5
Right, a bit of trig, some simple algebra, and a ruler should get you there... assuming it is a conic projection with the north pole at the centre.
First you need to determine the location of the north pole. To do that, you need to measure the distance along the bottom of your map of two points, A and B. To keep things positive, you can add a horizontal ...
5
It's encoded here: http://ditu.google.cn/maps?daddr=%e9%bb%84%e6%b5%a6%e5%8c%ba%e5%b9%bf%e4%b8%9c%e8%b7%af20%e5%8f%b7%e5%a4%96%e6%bb%a95%e5%8f%b7
I'd guess they are using an address instead of coordinates.
5
It is used in marine navigation to make the maths simpler, as well as instrumentation like sextants.
Take the example of a navigator who wants to find the angle of a celestial body with the horizon, he'll first make coarse adjustments on the sextant to get to the nearest degree, then turn the vernier to make fine adjustments. The vernier is marked off in ...
5
You can use Python shapefile library pyshp
>>> import shapefile
>>> sf = shapefile.Reader("shapefiles/blockgroups")
>>> shapes = sf.shapes()
>>> # Read the bounding box from the 4th shape
>>> shapes[3].bbox
[-122.485792, 37.786931000000003, -122.446285, 37.811019000000002]
>>># Read the 8th point in the ...
5
Just like any other function, you will get out the same unit of measurement you put in. In this example you are using lat/long and so will therefore get the distance between your two points (the hypotenuse) in degrees. The number of kilometers/miles/inches/whatever that the output corresponds to depends entirely on where on the surface of the earth the ...
5
When someone says that data points are tiny compared to other data layers, the usual problem is that the new points have been assigned a projected coordinate system--probably the one that the other layers are using--but the new points are latitude-longitude and should be assigned a geographic coordinate system.
The Add XY tool will default to the data ...
4
You can experiment with different transformations on a Danish Ministry of the Environment site. You may have to make multiple guesses (narrowed down perhaps by this experimentation), map out the points, overlay them on related layers, and check for reasonableness.
The second set is likely in meters in UTM 32N. "Lat" and "lon" are mixed up: they correspond ...
4
I don't think there is any way to georeference a ski resort map - which are often hand-drawn - in a meaningful way.
It's an artistic interpretation of the landscape with high degrees of freedom and a focus on being easy to read.
Maybe if you describe your final goal - e.g. what do you want to do with the georeferenced image? - we can propose a different ...
3
You need to populate your spatial_ref_sys table, e.g.:
psql -d your_db -f spatial_ref_sys.sql
This will give you SRID 4326, but not 94283, which doesn't appear to be any publicly recognized SRID.
To add 900913 (i.e. leetspeek for GOOGLE), see the PostGIS instructions at http://trac.osgeo.org/openlayers/wiki/SphericalMercator
3
The referenced article is thoughtful. However, I believe there is a "simple and elegant" solution: for geographic datasets, there are two kinds of bounding boxes. Those that do not straddle the +-180 meridian can be stored and searched as always. Those that do straddle the +-180 meridian can be stored in a semi-complementary form: namely, store the range ...
3
'Yes' is the short answer. The way you do it depends on what you want to use the data for.
The simplest way is to re-project your data into a Lat/Lon coordinate system using ST_Transform() (I think it is just 'Transform()' in SpatiaLite and something weird in Oracle). Alternatively, follow one of these recipes, perhaps using Proj4. Once you have ...
3
In general, you cannot determine the zone from the coordinates: that's why the zone is needed!
In your case, though, it sounds like you are looking within a small area that straddles neighboring zones. In that case, locations in zone 39 will have large positive x-values and locations in zone 40 will have negative or small positive x-values: that's how you ...
3
You have to distinguish between the project CRS and the layer CRS.
Your data is in lat/lon degrees, so the layer CRS should be EPSG:4326, WGS84.
Your project CRS can be anything else. In most cases a projected CRS, where units are metres or feet. On-the-fly projection should always be activated.
Just to be sure: the coordinates should have decimal points ...
3
If the image is in the least realistic--and it appears to be--then it represents a projection from a portion of Euclidean three-space onto a portion of two-space. We cannot expect this projection to have a nice mathematical formula, because it is likely the artist has introduced irrregular small distortions, but we can expect its formula to somewhat ...
3
In the Specification for the Wide Area Augmentation System (WAAS) on page 99 there are two formulae for calculating the "pierce point" of a GPS signal through the ionosphere. This may not be your application area, but may be able to adapt these to your application and they happen to work with longitude, latitude, azimuth, and elevation angle. Here's a ...
3
This is not WGS84 System, WGS system is an angular system so the latitude value cannot exceed 90, and the longitude can not exceed 360. You need to figure out the coordinate system of your data, which i guess might be in WGS84 projected system which measures distance in meters. Any GIS softwares can convert it e.g. Quantum GIS.
Roughly: 3 sec in WGS84 ...
3
Euclidean distances don't make much sense when calculated from lat/long. Haversine formula is used to calculate distance between lat/long on the surface of the earth surface. If the distance between the two places in question is large please consider using Haversine formula. Some detail and a calculator for Haversine formula can be found at:
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2
Clustering Options:
I think the conceptual buzz word that covers this type of operation is "clustering". Averaging is by far the simplest to implement and works well for most purposes. The only time I would use something else is if you are worried about outliers [Edit]-> or the poles or the international dateline. [Edit]--> also averaging, while it will ...
2
Using the example from underdark
if you - click the 分享链接 (link) button it exposes the Latitude and Longitude
Long URL
...
2
To create URLs for use by OnPoint, you can use AdvancedGenerateURL tool in the OnPoint Administration console. This tool will generate a URL that OnPoint can use, based on what parameters you want to use to set the map extent. You can then copy and paste the URL to tweak it for you needs.
Here is an example of the structure of an OnPoint URL.
To pass ...
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