I have a map of a mountainous landscape, http://skimap.org/data/989/60/1218033025.jpg. It contains a number of known points, the lat-longs of which can be easily found out using Google maps. I wish to be able to pin any latitude longitude coordinate on the map, of course within the bounds of the landscape.
For this, I tried an approach that seems to be largely failing. I assumed the map to be equivalent to an aerial photograph of the Swiss landscape, without any info about the altitude or other coordinates of the camera. So, I assumed the plane perpendicular to the camera lens normal to be Ax+By+Cz-d=0.
I attempt to find the plane constants, using the known points. I fix my origin at a point, with z=0 at the sea level. I take two known points in the landscape, and using the equation for a line in 3D, I find the length of the projection of this line segment joining the two known points, on the plane. I multiply it by another constant K to account for the resizing of this length on a static 2d representation of this 3D image. The length between the two points on a 2d static representation of this image on this screen can be easily found in pixels, and the actual length of the line joining the two points, can be easily found, since I can calculate the distance between the two points with their lat-longs, and their heights above sea level.
So, I end up with an equation directly relating the distance between the two points on the screen 2d representation, lets call it Ls, and the actual length in the landscape, L. I have many other known points, so plugging them into the equation should give me values of the 4 constants. For this, I needed 8 known points (known parameters being their name, lat-long, and heights above sea level), one being my orogin, and the second being a fixed reference point. The rest 6 points generate a system of 6 linear equations in A^2, B^2, C^2, AB, BC and CA. Solving the system using a online tool, I get the result that the system has a unique solution with all 6 constants being 0.
So, it seems that the assumption that the map is equivalent to an aerial photograph taken from an aircraft, is faulty. Can someone please give me some pointers or any other ideas to get this to work? Do open street maps have a Mercator projection?
I would say that this impossible to do in an automatic way. The skimap should be considered as an image rather than a map, a map is an projection of the real world into one plane, since this doesn't fit skimaps very well they are drawn instead.
The best way is probably to manually define a lot of points in the skimap with known or estimated coordinates and use them to estimate the points betwween. To get an acceptable result you probably have to assign coordinates to each pixel in the skimap.
You could do something like the following: http://magazin.unic.com/en/2012/02/16/making-of-interactive-mobile-piste-map-by-laax/
I am solving the exact same issue. It is pretty hard and lots of maths. Taking me a few weeks to solve it. Interpolation is the key as well with lots of manual mapping. I would say that for a ski mountain it will take at least 1000/1500 points to be able to get the very basic. So, not a trivial task unless you can automate the collection of these points (what I am doing!) ;)
Related
I'm neither a geometry student or a native speaker, so apologies if my question isn't clear enough.
As part of my master's thesis, I have to plot bounded regions of the night sky onto a 2D plane. My current solution consists of a rectangular mapping where (ra, dec) values are plotted to (x,y) coordinates. While this approach works well enough for small regions in relatively low ascension values, the resulting plots get progressively distorted for higher ||dec|| values, as expected.
At some point I'll have to change this to a more versatile approach. Thing is, I'm not exactly clear on what to search for. I guess I have to be able to map angular coordinates to a square (or hexagon) subgrid, but most search results I get are concerned with full-surface mapping.
I know I won't be able to achieve a perfect, distortion-free plotting, but I don't require perfect solutions; only a more general projection that will work well near the poles. Something like this, where I put my Photoshop skills to work and try to simulate a 20ยบ region under my current approach and the one I'm looking for:
What I want:
What I have:
TL;DR: how do I convert between coordinates on a sphere (ra/dec) to cartesian coordinates on a locally-defined grid?
I'm writing a data analysis program and part of it requires finding the volume of a shape. The shape information comes in the form of a lost of points, giving the radius and the angular coordinates of the point.
If the data points were uniformly distributed in coordinate space I would be able to perform the integral, but unfortunately the data points are basically randomly distributed.
My inefficient approach would be to find the nearest neighbours to each point and stitch the shape together like that, finding the volume of the stitched together parts.
Does anyone have a better approach to take?
Thanks.
IF those are surface points, one good way to do it would be to discretize the surface as triangles and convert the volume integral to a surface integral using Green's Theorem. Then you can use simple Gauss quadrature over the triangles.
Ok, here it is, along duffymo's lines I think.
First, triangulate the surface, and make sure you have consistent orientation of the triangles. Meaning that orientation of neighbouring triangle is such that the common edge is traversed in opposite directions.
Second, for each triangle ABC compute this expression: H*cross2D(B-A,C-A), where cross2D computes cross product using coordinates X and Y only, ignoring the Z coordinates, and H is the Z-coordinate of any convenient point in the triangle (although the barycentre would improve precision).
Third, sum up all the above expressions. The result would be the signed volume inside the surface (plus or minus depending on the choice of orientation).
Sounds like you want the convex hull of a point cloud. Fortunately, there are efficient ways of getting you there. Check out scipy.spatial.ConvexHull.
I have a city square with people, cars, trees and buildings in pcl format. I need to automatically determine the ground plane and project this objects on that ground plane to get a 2D map with occupied places.
Any idea?
I think the best thing to do here would be to familiarise yourself with the following two PCL tutorials:
http://pointclouds.org/documentation/tutorials/planar_segmentation.php
http://pointclouds.org/documentation/tutorials/project_inliers.php
The first tutorial makes use of the RANSAC algorithm to find a dominant plane in a scene. I use it to find tables and floors in robotics scenarios. You would use it to find your dominant ground plane.
The second tutorial shows how to project points directly onto a plane. This is what you would use to make your 3D point cloud into a 2D one. Note that, despite the "inlier" keyword, you can pass your whole point cloud to be projected onto the plane.
Actually, if you are after "occupied" places, you might want to project all of the points that aren't in the ground plane (i.e. the outliers), and that are above it (you can use a PCL filter, such as PlaneClipper3D, for example, or just the complement of the outliers from the plane-segmentation operation.
If the plane that you end up with (containing all your projected points) is not in the coordinate frame you want, you may wish to rotate the whole lot, for example, to align with the coordinate axes so that all z-coordinates are zero. See pcl::transformPointCloud for this (the transform will be obtainable from the plane coefficients returned from the plane segmentation).
I hope this is helpful and not at too basic a level, though the question was rather general so I suppose it should be okay.
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.