I've question about units in fields Db_GpsPt_dX,Db_GpsPt_dY from table Table_GpsPtArray_Point_Array located in every *.db (sqllite format) file maintained by Landstar application for GPS-RTK measurement. Unfortunatelly (CHC) Producer's support didn't response for my request, so i hope StackOverflow will be more helpful.
I know that for base points units are in ECEF units (WGS cartesian) but for rest of measured points there are values, that aren't corrected delta values either in ECEF, local XY oraz WGS84 (decimal). It is weird, units in given column should be written in the same format, but it isn't for that file.
Anyone has ideas about meaning of mentioned fields? Thanks for help.
OK problem solved - i've just realized that 0,898 0,344 coefficients are an absolute coordinates in WGS84 datum written in radians.
Related
I have two .shp files whose .prj files are identical and whose extents are different. I'd like to set them to the same extent so they line up on the map.
In ArcGis I have tried:
Exporting both to a new coordinate system-defined feature dataset
Removing the .prj file and then re defining the projection of each file
"project"ing both to the same Coordinate system.
Setting the data frame Coordinate System, then reintroducing the shapefiles in hopes that they'll project "on the fly"
In QGIS I have tried:
Setting project CRS
Setting layer to project and project to layer
Saving the shapefiles in specific CRS.
It seems odd to me that this is an issue in the first place: why can't Arc or Q detect this asynchronicity and give the user the option to choose one over the other?
What am I missing here?
Should I look into creating a spatial reference for one of the files, that matches the other?
Any clues/ suggestions/ clarifications?
I know this is a popular issue, and I figure there must be some simple explanation for the above, but I'm not finding it anywhere despite spending hours puzzling over the situation. Perhaps I just don't have the right vocabulary to ask the question. Any help appreciated.
some information about the files:
Extents for shp1:
top: 672344.187336 ft
bottom: 629117.938976 ft
right:7660465.885171 ft
left: 7627858.786745 ft
Extents for shp2:
top: 5984.800593 ft
bottom: 4784.800593 ft
right: 4616.411043 ft
left: 3776.411043 ft
Layer Properties-Source for both shp1 and shp2:
Projected Coordinate System: NAD_1983_HARN_StatePlane_Oregon_North_FIPS_3601_Feet_Intl
Projection: Lambert_Conformal_Conic
False_Easting: 8202099.73753281
False_Northing: 0.00000000
Central_Meridian: -120.50000000
Standard_Parallel_1: 44.33333333
Standard_Parallel_2: 46.00000000
Latitude_Of_Origin: 43.66666667
Linear Unit: Foot
Geographic Coordinate System: GCS_North_American_1983_HARN
Datum: D_North_American_1983_HARN
Prime Meridian: Greenwich
Angular Unit: Degree
.prj data for both shp1 and shp2:
PROJCS["NAD_1983_HARN_StatePlane_Oregon_North_FIPS_3601_Feet_Intl",GEOGCS["GCS_North_American_1983_HARN",DATUM["D_North_American_1983_HARN",SPHEROID["GRS_1980",6378137.0,298.257222101]],PRIMEM["Greenwich",0.0],UNIT["Degree",0.0174532925199433]],PROJECTION["Lambert_Conformal_Conic"],PARAMETER["False_Easting",8202099.737532808],PARAMETER["False_Northing",0.0],PARAMETER["Central_Meridian",-120.5],PARAMETER["Standard_Parallel_1",44.33333333333334],PARAMETER["Standard_Parallel_2",46.0],PARAMETER["Latitude_Of_Origin",43.66666666666666],UNIT["Foot",0.3048]]
For those who might have the same problem, in the course of answering this question (r/gis/reddit proved the most useful) I learned:
There are local coordinate systems which might not show up on Arc or Q. In this case, the data was derived from a CAD file, a DWG that I converted for a project.
One can change the data frame projected coordinate system within Arc trying to find a fit. In this case there was none, no matter how long or hard I tried
Some Coordinate systems are simply so local they aren't useful, or as a redditor put it, "garbage", so don't waste your time trying to make them fit.
4.The Spatial Adjust tool can be used to effectively georeference a dataset with an unknown coordinate system to one which is known. So in my case, I took the .shp and then used the spatial adjust toolset to "rubbersheet" it to some clearly defined data. Easy. Took a few minutes. Made it fit, defined projection, called it good.
I've been searching extensively for a way of converting from ordanance survey coords to valid esri coordinates. I've found quite a few pages that convert to lat long (if a little off) but nothing to convert to esri (which I believe is utm.)
This is for use in python or JavaScript / actionscript etc - I'm not too worried about syntax more an understanding of the maths involved.
Thanks
Ian
This type of conversion is called a "geodetic transformation". OS and UTM are both "transverse mercator" projections, wherein the ellipsoid of the earth is unwrapped into a cylinder, which is then unrolled into a flat sheet and sub-divided into grid sections. OS coordinates are specific to regions (eg: OSGB for Great Britain), whereas UTM is a "universal" system and specifies a system of grids for the whole earth. Regional grids are used in order to reduce the side-effects of distortion introduced by the mercator projection. It follows that converting between such systems is possible, but can also be quite complex depending in the accuracy desired.
It seems there are only indirect methods, as you have already referred to, the most common being to convert from OSGB36 to WGS84 (lat/long) and then to UTM.
Here are some resources which might be helpful:
Convert WGS84 lat/long to UTM: http://www.uwgb.edu/dutchs/usefuldata/utmformulas.htm. Note the inclusion of specific parameters for each region. For example, if you were converting coordinates for Britain, the parameters for "Airy 1830" would be used. (also links to a spreadsheet and webpage with conversions).
Similar information as above on Wikipedia.
JavaScript to convert OSGB36 to WGS84 (7 metre accuracy): http://www.nearby.org.uk/tests/GeoTools.html
A more accurate JavaScript conversion using a Helmert transformation (5 metre accuracy): http://www.movable-type.co.uk/scripts/latlong-convert-coords.html and http://www.movable-type.co.uk/scripts/latlong-gridref.html
Comprehensive coverage of the OSGB36 coordinate system, including transformations to and from other coordinate systems: http://www.ordnancesurvey.co.uk/oswebsite/gps/docs/A_Guide_to_Coordinate_Systems_in_Great_Britain.pdf
Miscellaneous links and resources: http://www.ordnancesurvey.co.uk/oswebsite/gps/information/resourceinfolinks/gpslinks.html
As for accuracy, it is summed up in this excerpt from ordnancesurvey.co.uk:
... OSGB36 contains randomly variable scale errors, mainly due to it being
computed in blocks and the fact that scale and azimuth were
controlled entirely by the 11 stations from the
Principle Triangulation. These scale variations
mean that OSGB36 can be described as inhomogeneous ...
The inhomogenity of OSGB36 does not affect its
adequacy as a mapping datum but it does make a
simple transformation between ETRS89 and OSGB36 too inaccurate for national use.
For example, the accuracy of a national 7 parameter (3
shifts, 3 rotations and a scale change) transformation is approximately 5 metres
Here is a link to more comprehensive information regarding the ARC/INFO file format.
Quick google search: http://google-maps-utility-library-v3.googlecode.com/svn/trunk/arcgislink/docs/examples.html
I am currently working on a simple and small Kalman-Filter for GPS-Navigation. I am getting from my GPS-Sensor the current location, course angle and speed. So the Kalman-Filter should fuse the current measurement and the linear movement beginning from the previous location assuming constant speed and course-angle.
My problem is to select a useful space where the Kalman-Filter is able to perform in a good way.
Local coordinate system approach:
If I choose a local coordinate system (north [meter], east [meter]) with the previous location at origin I will be able to predict the new location easily but how to convert the new measurement (latitude/longitude) into my local coordinate system using the wgs-84 ellipsoid? and how to convert my new predicted in my local coordinate system to latitude/longitude also using the wgs-84 ellipsoid?
So I need two functions:
f:=(lat_ref, lng_ref, lat, lng) -> (x,y)
g:=(lat_ref, lng_ref, x, y) -> (lat, lng) (this could by also done using Vincenty)
Global coordinate system approach:
I found the Vincenty-Algorithm which calculates the new location from a reference location, distance and course-angle on any ellipsoid. This algorithm works fine but I dont see how to use this algorithm inside a kalman-filter which works in a global coordinate system.
Are there any ideas or suggestions how to solve one of my problems?
I've worked through many scientific papers and found the answer on wikipedia
The main trick is to convert latitude/longitude/height (llh) to earth-centered-earth-fixed (ecef) and then to my local coordinate system earth/north/up (enu) and vice versa.
I found some other interesting links to this topic.
direct and faster conversion from llh to enu in matlab
ecef2llh and llh2ecef in matlab
Check out FWTOOLS and in particular cs2cs and proj...
They can handle the coord system transformations from projected to geographic coordinate systems. It's important to remember to consider any datum shifts.
I am in charge of a program that is used to create a set of nodes and paths for consumption by an autonomous ground vehicle. The program keeps track of the locations of all items in its map by indicating the item's position as being x meters north and y meters east of an origin point of 0,0. In the real world, the vehicle knows the location of the origin's lat and long, as it is determined by a dgps system and is accurate down to a couple centimeters. My program is ignorant of any lat long coordinates.
It is one of my goals to modify the program to keep track of lat long coords of items in addition to an origin point and items' x,y position in relation to that origin. At first blush, it seems that I am going to modify the program to allow the lat long coords of the origin to be passed in, and after that I desire that the program will automatically calculate the lat long of every item currently in a map. From what I've researched so far, I believe that I will need to figure out the math behind converting to lat long coords from a UTM like projection where I specify the origin points and meridians etc as opposed to whatever is defined already for UTM.
I've come to ask of you GIS programmers, am I on the right track? It seems to me like there is so much to wrap ones head around, and I'm not sure if the answer isn't something as simple as, "oh yea theres a conversion from meters to lat long, here"
Currently, due to the nature of DGPS, the system really doesn't need to care about locations more than oh, what... 40 km? radius away from the origin. Given this, and the fact that I need to make sure that the error on my coordinates is not greater than .5 meters, do I need anything more complex than a simple lat/long to meters conversion constant?
I'm knee deep in materials here. I could use some pointers about what concepts to research.
Thanks much!
Given a start point in lat/long and a distance and bearing, finding the end point is a geodesic calculation. There's a great summary of geodesic calculations and errors on the proj.4 website. They come to the conclusion that using a spherical model can get results for distance between points with at most 0.51% error. That, combined with a formula to translate between WGS-84 and ECEF (see the "LLA to ECEF" and "ECEF to LLA" sections, seems like it gets you what you need.
If you want to really get the errors nailed down by inverse projecting your flat map to WGS-84, proj.4 is a projection software package. It has source code, and comes with three command line utilities - proj, which converts to/from cartographic projection and cartesian data; cs2cs, which converts between different cartographic projections; and geod, which calculates geodesic relationships.
The USGS publishes a very comprehensive treatment of map projections.
I'd do a full-up calculation if you can. That way you'll always be as accurate as you can be.
If you happen to be using C++ the GDAL is a very good library.
For a range of 40km, you may find that approximating the world to a 2D flat surface may work, although a UTM transform would be the ideal way to go - in any case, I'd advocate using the actual WGS84 co-ordinates & ellipsoid for calculations such as great circle distance, or calculating bearings.
If you get bored, you could go down a similar line to something I've been working on, that can be used as a base class for differing datums such as OSGB36 or WGS84...
I'm writing a mapping application that I am writing in python and I need to get the lat/lon centroid of N points.
Say I have two locations
a.lat = 101
a.lon = 230
b.lat = 146
b.lon = 200
Getting the center of two points is fairly easy using a euclidean formula. I would like
to be able to do it for more then two points.
Fundamentally I'm looking to do something like http://a.placebetween.us/ where one can enter multiple addresses and find a the spot that is equidistant for everyone.
Have a look at the pdf document linked below. It explains how to apply the plane figure algorithm that Bill the Lizard mentions, but on the surface of a sphere.
poster thumbnail and some details http://img51.imageshack.us/img51/4093/centroidspostersummary.jpg
Source: http://www.jennessent.com/arcgis/shapes_poster.htm
There is also a 25 MB full-size PDF available for download.
Credit goes to mixdev for finding the link to the original source, and of course to Jenness Enterprises for making the information available. Note: I am in no way affiliated with the author of this material.
Adding to Andrew Rollings' answer.
You will also need to make sure that if you have points on either side of the 0/360 longitude line that you are measuring in the "right direction"
Is the center of (0,359) and (0, 1) at (0,0) or (0,180)?
If you are averaging angles and have to deal with them crossing the 0/360 then it is safer to sum the sin and cos of each value and then Average = atan2(sum of sines,sum of cosines)
(be careful of the argument order in your atan2 function)
The math is pretty simple if the points form a plane figure. There's no guarantee, however, that a set of latitudes and longitudes are that simple, so it may first be necessary to find the convex hull of the points.
EDIT: As eJames points out, you have to make corrections for the surface of a sphere. My fault for assuming (without thinking) that this was understood. +1 to him.
The below PDF has a bit more detail than the poster from Jenness Enterprises. It also handles conversion in both directions and for a spheroid (such as the Earth) rather than a perfect sphere.
Converting between 3-D Cartesian and ellipsoidal latitude, longitude and height coordinates
Separately average the latitudes and longitudes.