Hot answers tagged geodesy
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This summarizes my understanding of some of the basic ideas. Because it is hard to find all of them clearly described and summarized in one place, I could be wrong or misleading about some of them: comments and corrections are welcome.
"Geoids" are approximations to a surface of gravitational equipotential.
The geoid is a hypothetical Earth surface ...
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After some looking around at Wikipedia and the same question/answer at StackOverflow, I figured I would take a stab at it, and try to fill in the gaps.
First off, Not sure where you got the output, but it appears to be wrong. I plotted the points in ArcMap, buffered them to the distances specified, ran intersect to on the buffers, and then captured the ...
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This question assumes an ellipsoidal model of the earth. Its reference surface is obtained by rotating an ellipse around its minor axis (plotted vertically by convention). Such an ellipse is just a circle that has been stretched horizontally by a factor of a and vertically by a factor of b. Using the standard parameterization of the unit circle,
t --> ...
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I am not a Geodesy expert, but far as I understand it, the geoid, is the shape that the surface of the oceans would take under the influence of gravity alone. It is the surface at which the intensity of gravity is the same.
The Problem isn't that it is difficult to describe mathematically, but it might be impossible to predict correctly and accurately.
...
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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
The IERS reference meridian is a weighted average of ground-based monitoring stations. Thus, tectonic monitoring must involve the motion of one's own plate relative to this global average motion.
3
Some (slightly) theoretical pointers:
Instead of focusing on attributes, one approach to the problem might focus on exploring characteristics of movement patterns. Those could be explored by calculating aggregated characteristics of movement or dividing your data into logical 'chunks' (for instance, daily trajectories of certain objects). At next stage you ...
3
I'm not sure if I'm being naive, but, if you buffer each point by size, and then intersect all three circles that would get you the correct location?
EDIT:
You can compute the intersection using spatial APIs. Examples:
GeoScript
Java Topology Suite
NET Topology Suite
GEOS
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This might work. Quickly again in python, you could put this in the body of a function xN,yN = coordinates of points, r1 & r2 = radius values
dX = x2 - x1
dY = y2 - y1
centroidDistance = math.sqrt(math.pow(e,2) + math.pow(dY,2)) #distance from centroids
distancePL = (math.pow(centroidDistance,2) + (math.pow(r1,2) - math.pow(r2,2))) / (2 * ...
2
The following notes use planarithmic geometry (i.e. you would have to project your coordinates into an appropriate local coordinate system).
My reasoning, with a worked example in Python, follows:
Take 2 of the data-points (call them a and b). Call our target point x. We already know the distances ax and bx. We can calculate the distance ab using ...
2
I worked out the trilateration problem. Here is a spreadsheet that calculates the 2 points of intersection of 3 intersecting spheres:
http://www.datafilehost.com/download-2e1bffb2.html
Here is the complementary math:
http://jons-math.bravehost.com/triangulation.html
2
I'm not an expert on the subject, so feedback from others is welcome, but I think you should store in global geographic CRS - i.e. lat / lon in decimal degrees. Here's why (and this is the bit that could be wrong / inaccurate):
When converting between projected CRSs you have to first convert to geographic and then to the target CRS. Each CRS must, as part ...
2
At least that's the formula I found at the US Data Analysis and Assessment Center (DAAC) for the Department of Defense (DoD) High Performance Computing Modernization Program (HPCMP) wiki. It does say that they borrowed heavily from Wikipedia's entry. Still, the fact that they retained that formula should count for something.
1
1- Here is a link to several courses (undergraduate and graduate) of GIS, Cartography and Remote Sensing which are offered by the Oregon State University (OSU) in U.S.A.
2- Another option are the approximately 90 free online courses offered by ESRI to manage their software ArcGis. Some of them include exercises to do. Here is a link that shows how to attend ...
1
The question sounds like homework ;-), and the answer, unfortunately, requires spherical trigonometry.
http://en.wikipedia.org/wiki/Spherical_trigonometry
http://mathworld.wolfram.com/SphericalTrigonometry.html
It is like normal trig, just more complex. You will need to know some things like eccentricities (maybe, but perhaps that's only if you have ...
1
A few resources for different languages (except C#) are listed here: http://trac.osgeo.org/proj/wiki/GeodesicCalculations
One of them is a PROJ.4 utility geod, which probably has some underlying API. Check out Proj4Net, but I've never used it, and I'm not sure how mature or complete it is.
1
Yes, I agree that I don't see how to get exact location without a distance. You would be in this situation:
So you would only be able to tell within a certain area where you were. And that's with only a 1000m buffer (radius). Your 6000m buffer has such an overlap that the potential location would be huge.
The solutions would be to:
Get some sort of ...
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