I have a floor map that is relayed on top of a grid and have the length and width of the floor along with 3 reference points that has x,y, lat, long.
Now I have some x,y that lies within the floor map and need to derive the lat and long for those points.
Is there a way to achieve that? Preferably Java but even a formula would do or some documentation.
Related
I have a question about if something is possible using Tableau.
I already have a coastline plotted on one map using custom LatLon coordinates and I would like to take a user inputted Lat and Lon and plot a circle around it with let's say radius 10 and display it on the same map.
I was using this tutorial before to plot a circle:
https://www.crowdanalytix.com/communityBlog/customers-within-n-miles-radius-analysis-using-tableau
But I don't think the same approach can work with user-inputted fields because then it would require restructuring the data..
Okay, a (much smarter LOL) coworker helped me figure this out....
So my goal was to graph distance band (like a distance of 5 miles around a coast) . In order to do this we can use the distance between two coastline points since they are connected by a line, not a curve...From there we can find the perpendicular point a certain distance away and connect those points. Much easier than my circle idea...
Leaflet consists of multiple self worlds. General limit of latitude n longitude is -90 to +90 and -180 to +180 respectively. So for a different number of world in map area i receive {lat: 76.12621315046384, lng: 370.70826412409673}, which i normalize to the limit format and send as a param to server in order to receive points based on an algorithm. However the points that i receive are in normalized format already which will plot on the initial first map only, however i would like to plot them on that number of world on map area from which i retrieved longitude as 370.70826412409673.
I tried getting getting pane, scaleZoom, zoomScale, scale, zoom but nothing seem to work in order to get me the world number or anything that helps me de-normalize the geopoints.
You can use the function getBounds() that gives you the coordinates of the two corners of the current view and/or getCenter(). You can then change your longitude to fit this view.
An other solution is to keep the offset between the normalized and "true" coordinates, and add it back to the answer of your algorithm.
Can anyone point me in the right direction when it comes to understanding MKMapPoint?
I understand that it has to do with laying suface of the globe on a 2D surface. But I don't understand how each "point" is measured?
Can anyone give me an example in code?
There is nothing much to it.. just a structure to display point on a 2D map... here is what the documentation says..
MKMapPoint A point on a two-dimensional map projection.
typedef struct {
double x;
double y; } MKMapPoint;
Fields x The location of the point along the x-axis of the
map. y The location of the point along the y-axis of
the map. Discussion If you project the curved surface of the globe
onto a flat surface, what you get is a two-dimensional version of a
map where longitude lines appear to be parallel. Such maps are often
used to show the entire surface of the globe all at once. An
MKMapPoint data structure represents a point on this two-dimensional
map.
The actual units of a map point are tied to the underlying units used
to draw the contents of an MKMapView, but you should never need to
worry about these units directly. You use map points primarily to
simplify computations that would be complex to do using coordinate
values on a curved surface. By converting to map points, you can
perform those calculations on a flat surface, which is generally much
simpler, and then convert back as needed. You can map between
coordinate values and map points using the MKMapPointForCoordinate and
MKCoordinateForMapPoint functions.
When saving map-related data to a file, you should always save
coordinate values (latitude and longitude) and not map points.
Availability Available in iOS 4.0 and later. Declared In MKGeometry.h
If you want to draw something over a map that has a certain real-world size, for example, a scale, then in your calculations you would multiply any given length in meters with the result of MKMapPointsPerMeterAtLatitude(...) to get the size in MKMapPoint units.
If your map displays only of small portion of the globe, MKMapPointsPerMeterAtLatitude(...) will deliver roughly the same constant value for all the latitudes involved in your map. In this case, you can simply use the latitude in the middle of your map as the argument of this function.
Also read in the documentation that a Mercator projection is used for the transformation. Here is an additional observation: northing (y) axis direction seems to be in reverse direction of standard mercator projections (positive direction is down instead of up).
I am trying to calculate the distance between two points (Pins) on the iPhone map.
Everything works fine however if you put one of the pin close to the left edge and the other one to the right edge of the map, distanceFromLocation method always returned the shortest distance between points (of course earth is not flat). I tried switching the CLLocations but it still shows me the shortest distance.
For my app I will need to calculate both, shortest and the longest distance between two pins.
The problem seems to be trivial but I can't come out with anything to solve it.
Any help or clues appreciated. Thanks!
There is a function MKCoordinateRegionForMapRect that returns a MKCoordinateRegion. Without having tried it, seems like all you would need to do is accurately specify the rect as the region bounded at one diagonal by one pin and the other diagonal by the other pin. You could use CGRectUnion twice giving the pin origins as rect origin and an arbitrarily small width and height.
Then the distance between the pins will be the square root of the distances represented by the latitudeDelta and longitudeDelta returned in the span squared and added. (i.e. Pythagoras theorem, the two side lengths are the horizontal and vertical sides of a triangle and the diagonal is a straight line between your points.)
Finding latitudeDelta in metres is easy, each degree of latitude is 111km.
To find meters for the longitudeDelta you need a little bit more code.
#define EARTH_EQUATORIAL_RADIUS (6378137.0)
#define WGS84_CONSTANT (0.99664719)
#define degreesToRadians(x) (M_PI * (x) / 180.0)
// accepts decimal degrees. Convert from HMS first if that's what you have
double lengthOfDegreeLongitude(double degrees) {
double tanDegrees = tanf(degreesToRadians(degrees));
double beta = tanDegrees * WGS84_CONSTANT;
double lengthOfDegree = cos(atan(beta)) * EARTH_EQUATORIAL_RADIUS * M_PI / 180.0;
return lengthOfDegree;
}
Give that function the longitude of the location you are interested in and multiply the result by the longitudeDelta in degrees to get a distance in meters. Then you have a horizontal distance and a vertical and you can find the diagonal which is what you were looking for.
Inaccuracy will get bad if you are looking for a distance which is very large in N-S as the length of a degree longitude will be varying over the calculation. And you can refine the 111km for a degree of latitude a bit too, that's just from memory.
I have data describing a rotated ellipse (the center of the ellipse in latitude longitude coordinates, the lengths of the major and minor axes in kilometers, and the angle that the ellipse is oriented). I do not know the location of the foci, but assume there is a way to figure them out somehow. I would like to determine if a specific latitude longitude point is within this ellipse. I have found a good way to determine if a point is within an ellipse on a Cartesian grid, but don't know how to deal with latitude longitude points.
Any help would be appreciated.
-Cody O.
The standard way of doing this on a Cartesian plane would be with a ray-casting algorithm. Since you're on a sphere, you will need to use great circle distances to accurately represent the ellipse.
EDIT: The standard ray-casting algorithm will work on your ellipse, but its accuracy depends on a) how small your ellipse is, and b) how close to the equator it is. Keep in mind, you'd have to be aware of special cases like the date line, where it goes from 179 -> 180/-180 -> -179.
Since you already have a way to solve the problem on a cartesian grid, I would just convert your points to UTM coordinates. The points and lengths will all be in meters then and the check should be easy. Lots of matlab code is available to do this conversion from LL to UTM. Like this.
You don't mention how long the axes of the ellipse are in the description. If they are very long (say hundreds of km), this approach may not work for you and you will have to resort to thinking about great circles and so on. You will have to make sure to specify the UTM zone to which you are converting. You want all your points to end up in the same UTM zone or you won't be able to relate the points.
After some more research into my problem and posting in another forum I was able to figure out a solution. My ellipse is relatively small so I assumed it was a true (flat) ellipse. I was able to locate the lat lon of the foci of the ellipse then if the sum of the distances from the point of interest to each focus is less than 2a (the major axis radius), then it is within the ellipse. Thanks for the suggestions though.
-Cody