perhaps this is a simple question.
I have 3 GPS coordinates (one is the current user location). What I want now is to calculate the angle between the user location and the two GPS coordinates. Imagine the user location in the center of the two other points, the three points can be seen as a triangle. And I want to calculate the angle at the user location.
I hope someone can help me because I have no idea how to do this with spherical coordinates like the GPS coordinates I have.
THX - nekro
For short distances (less than 100km, say) you can safely ignore the spherical nature of the calculation and treat the problem as a 2 cartesian coordinate problem. For large distances the spherical geometry gets pretty gnarly. I could probably figure it out, but I don't want to think that hard right now.
Edit:
All you need to do is to convert both coordinates to KM, and then treat it as a cartesian problem. (At a small scale, you can ignore the curved nature of the "lines" and treat them as normal cartesian grid lines, since the curvature is small enough to ignore at that scale)
The distance per degree of latitude is constant. The distance for a degree of longitude changes based on latitude.
Do a google search on "KM per degree of longitude" and find a link that explains it clearly. Here's one: http://www.colorado.edu/geography/gcraft/warmup/aquifer/html/distance.html
You could use thessien polygons and calculate the geometry on those from a strictly GIS perspective. If you have qgis or arcgis this should be fairly simple. These packages offer APIs which might suit your needs.
You're essentially doing two calculations (bearing to (or from) current position to two other positions) and not crosstrack (distance from a great circle line between to other points).
However, both can be found in Ed William's Aviation Formulary which has the most comprehensive collection of formulas for spherical calculations I've found.
You would be looking for "Course between points" which is listed as:
tc1=mod(atan2(sin(lon1-lon2)*cos(lat2),
cos(lat1)*sin(lat2)-sin(lat1)*cos(lat2)*cos(lon1-lon2)), 2*pi)
Related
I have an arbitrary map image, which may or may not be accurately projected to some standard geographic mapping. Probably not, though, since it's an artists rendition. Consider this map a 2D image of pixels at 0,0 onward.
I'd like to map lat/lon points in world space to this map. Since the map is not necessarily a known or accurate projection, I've got to come up with some other solution. I figure that establishing control points on the 2D image that correlate to known lat/lon values is step #1. At a minimum, 3, but maybe more, in case it's required to sort out distortion in the map image.
What algorithm or equation would I be looking for to take these control points, and identify the X,Y position on the image from any given lat/lon input?
I expect it to be inaccurate, depending on the number of control points. And I expect, for some weirder images, to have to go and add many control points in certain areas to make it line up right.
When the area depicted is small, (e.g. it fits in a square few km on the side), one thing to try is described below. I'm sorry if you find the description too terse, I wanted to keep it reasonably short.
The idea is to assume the image is in some unknown conformal projection, and to try to approximate it. Of course this may fail, if the image can not, in fact, be reasonably approximated this way.
Given your control points P[], project them into map coordinates Q[] using some conformal projection, and get hold of their image coordinates R[]. To within a metre or so -- given the assumption above -- the R[] can be obtained from the Q[] by a transformation T that is a translation, an (isotropic) scaling and a rotation. You can then find T, say by least squares, using the Q[] and R[]. You have a two stage map from the control point geographic coordinates P[] to their image coordinates R[]: first project using the chosen projection, then apply T. You could use the inverse of this map to go from arbitrary image coordinates to geographical coordinates.
If the image is larger than a few km, you may not get enough accuracy this way. All is not lost. Though a translation, scale and rotation may not suffice, any two conformal projections are related by a (complex) analytic map. So you could try to fit (an approximation to) this map using the control points as above. A suitable approximation might be a complex polynomial, or a complex rational function.
If I were doing this, I think I would first test it on artificial data. For example you could generate images of various sizes, using some projection (differing from the one used above), and see how well you fit the known points in these images.
I'm working on a SceneKit project in Objective-C, trying to rotate a sphere (representing a planet) so that a point at a given longitude/latitude coordinate is facing "forward" (i.e. that point moves to the {0,0} position). So for example, we're looking at Earth's {0,0} point, off the coast of Africa, but I need to know what rotation vector would have us looking at {37,-122}, aka San Francisco. Ideally, the planet would remain "naturally" oriented, with north remaining generally "up" from our perspective, but that's flexible.
I've got the animation working and I can rotate to a given longitude OR latitude using SCNVector4, but as soon as I try to rotate to a given longitude AND latitude, it gets wildly messed up. So I've been doing some research, and I've found some resources that seem to do exactly what I need, but to be honest they're a bit beyond me. I'm a complete newbie in the world of trigonometry and matrices, and it seems there's no way around this problem without them.
Can anyone help me understand how these concepts can take Lat/Long coordinates, and calculate a SCNVector4? A method for creating a SCNVector4 based on the desired Lat/Long would be perfect, but even just help with calculating the components individually would be greatly appreciated!
I have a large data set of regions , each of which is defined by a longitude, latitude and a given radius. I have a location point with a a latitude and longitude and I need to determine which of the regions contains my point.
Currently I am using brute force : I compute the distance between the target point and each region's center longitude and latitude ; if the distance is less than the radius of the region, I include the region in in my result.
Obviously this solution is not tenable.
Can GeoHash be used to formulate a solution ?
Edit : The business problem is given a set of cell phone with known ranges and a fixed set of available locations owned by a real estate holding, where is the place for a signal repeater. There are other considerations of course besides location and distance. Otherwise someone will have drive around the country with a signal detection kit -- not optimal . Not a homework question. I have Comp Sci background but GIS is new to me and I am willing to learn.
Edit : I will continue using brute force across several ec2 instances. Not the most optimum solution but it works. Thank you all for proposed solutions but unfortunately given the time and knowledge constraints and vagueness of the methodology, I am not going to be able to try them out.
The usual approach is to use a spatial index like quad tree or kd tree.
To this index you add the rectangular bounds of all circles to build up the index.
Uisng the quad tree:
Query the index which objects overlap a quad node at given position. The result will be some circles , these you check as you described.
quad trees don't like the deletion of elements.
I'm starting a project where I am mapping a set of points on the Earth using google maps. I want to find the point on the globe which is the average (shortest total distance to all points), but I'm unsure how to handle it considering the distance may be shorter going the other way around the earth. (-178 degrees to 178 degrees longitude is only 4 degrees longitude apart, not 356). What is the best way to approach this, either via an api call or from a mathematical perspective?
I highly doubt there is a slick geometric argument giving a closed form expression for the desired point. Nonetheless here's a simple-minded algorithm which gives an answer to within any desired precision:
https://gist.github.com/amitkgupta/5019163
If you want a mathematically more satisfying solution, I recommend asking over at http://math.stackexchange.com, or they don't avail you, escalate it to http://mathoverflow.net.
I can suggest simple and fast solution (but not to exact initial task). Find the center of gravity of points, then there may be 2 situations:
it's located at center of sphere - don't know what to do (if initial points distributed close to each other - this will not happen)
in other case - consider vector with center of mass and center of sphere as finish and start points, find where such vector intersects surface of sphere, that point - is the answer.
So, you'll get point somewhat similar to 'mid-point', but only in cases when surface under consideration is very small (may be all point lay within the same city). But it is also has nothing to do with minimal average distances from result to initial points.
I am trying to find area of MKPolygonView object added to MapView. Apple documentation has method distanceFromLocation: to find distance between edges of MKPolygonView object. But I could not find anything to calculate area of the overlay.
Does Apple have any documented method for finding area?
Concerning the comments on the question post, the Earth is not a perfect sphere either. In fact, it's not a perfect anything, so "correct" answers aren't possible. What matters is how accurate of an approximation you need. Also, are you interested in a mean sea level type measurement, or do you want the actual contours of the ground (for example if your polygon is put over a mountain, then the same exact size polygon is put over some plains, should the result you calculate be the same or different)?
Depending on how big your polygon is, and which measurement you're looking for, a 2D approximation can be pretty accurate (the smaller the polygon, the closer you'll get). Something to keep in mind, if you want your area in something like square feet, the distance between two longitudinal lines is not constant (63 deg west and 62 deg west are closer (in feet) somewhere in Alaska than they are at the equator). You might have to do a unit conversion to handle this depending on how big your polygon is (or if your polygon could be placed anywhere). If you can't do the 2D approximation, I'm not even sure how you'd do that.
When I did this, I did the 2D approx, and I had to do the unit conversion. If that's the way you go, I can try to dig up some of my old notes and the links I used to get you started.