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22

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 ...


19

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 ...


16

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 --> ...


10

This answer is divided into multiple sections: Analysis and Reduction of the Problem, showing how to find the desired point with "canned" routines. Illustration: a Working Prototype, giving working code. Example, showing examples of the solutions. Pitfalls, discussing potential problems and how to cope with them. ArcGIS implementation, comments about ...


10

WGS84, according to: http://support.google.com/earth/bin/answer.py?hl=en&answer=148110 Cheers, Ryan


8

The question asks how to convert local coordinates into geocentric coordinates. As an example, What would be the geocentric coordinates of a point displaced 250 meters west and 250 meters south of -108.619987 degrees longitude, 36.234600 degrees latitude (at sea level)? (For the reason why only six decimal places are used here, please see ...


7

One thing to keep in mind is that lat/long is geodetic and not geocentric: If we were to calculate elevation as a radius from the center of the ellipse, our elevation lat/long would be different than our horizontal lat/long! This is why there are two different datums. The horizontal datum is just a smooth ellipse, because it's easier to do trig ...


7

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. ...


7

I admire your enthusiasm but have to say you're not defining anything new. The simplest geographic/geodetic models of the earth are perfect spheres/globes. The only "difference" between that simple old system and your "new" system being that heights, instead of measured relative to the surface of the fixed-radius (R) sphere/globe, are now relative to the ...


6

One could use either kind of latitude to locate points on the WGS 84 ellipsoid (used by the NED) or any other ellipsoid, but "everybody knows" that the values will always be given as geodetic latitudes. However, it is surprisingly hard to find an authoritative statement to that effect! Before we go on, it helps to understand that although a datum like 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

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.


4

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


3

The difference is defined by the datum shift between the two reference ellipsoids. The shift itself is defined with a transformation (7 param. Helmert, 4 param,....) using 3D cartesian coordinates (X, Y, Z) which are calculated from ellipsoid coordinates (lat, lng, elevation). The transformation (parameters) is usually derived from a set of points which ...


3

I was curious to see how quickly @cffk's approach converges on a solution, so I wrote a test using arcobjects, which produced this output. Distances are in meters: 0 longitude: 0 latitude: 90 Distances: 3134.05443974188 2844.67237777542 3234.33025754997 Diffs: 289.382061966458 -389.657879774548 -100.27581780809 1 longitude: 106.906152157596 ...


3

As you note, this problem arises in determining maritime boundaries; it's often referred to as the "tri-point" problem and you can Google this and find several papers addressing it. One of these papers is by me(!) and I offer an accurate and rapidly convergent solution. See Section 14 of http://arxiv.org/abs/1102.1215 The method consists of the following ...


3

The pe.dll available with the free download of ArcGIS Explorer can be used to do this. See Exploiting the ESRI Projection Engine (second edition) for discussion.


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 ...


2

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

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.


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

A third option and the one I ended up using is simple and works for anywhere but with the possible exception of the poles with accuracy for the NM example at less than 2 feet (58 cm). Given a single input of the geodetic coordinate (GDC) The first two steps have a lot done "in the background" by geotrans which is available for at least C/C++ and Java. It ...


2

The question is pretty complicated. What you are asking about is calculations on the surface of the Earth, which is called spherical trigonometry. To get even more precise you need to use an ellipsoidal model of the Earth. I'd suggest you use a program that can already do this for you, but if you want to do it yourself, here's a link to start on. The ...


1

Answered by user base9geodesy at http://surveyorconnect.com/index.php?mode=thread&id=236530#p236567 You are correct that the determination of the origin and orientation of a horizontal/geometric datum are independent of the reference ellipsoid. That being said I have never seen this solution used with respect to the determination of the ellipsoid. As ...


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

This raster normalizer is capable of splitting images into tiles. For the empty areas of your tiles, I believe you'll have to render certain pixel values as transparent. http://rasternormalizer.codeplex.com/releases/view/50769


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

Yes, you must use an ellipsoid (or other mathematical surfaces). the reason is that the Geoid is a Physical surface (defined as the equipotential surface of gravity strength field). Simple meaning - it has no mathematical formula (another simple meaning - it is a surface at the height of the mean sea level that if you put a drop of water on it it wont ...



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