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).
Related
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.
There are a lot of similar questions but I can't get a clear answer out of them. So, I want to represent latitude and longitude in a 2D space such that I can calculate the distances if necessary.
There is the equirectangular approach which can calculate the distances but this is not exactly what I want.
There is the UTM but it seems there are many zones and letters. So the distance should take into consideration the changing of zone which is not trivial.
I want to have a representation such that i can deal with x,y as numbers in Euclidean space and perform the standard distance formula on them without multiplying with the diameter of Earth every time I need to calculate the distance between two points.
Is there anything in Matlab that can change lat/long to x,y in Euclidean space?
I am not a matlab speciallist but the answer is not limited to matlab. Generally in GIS when you want to perform calculations in Euclidean space you have to apply 'projection' to the data. There are various types of projections, one of the most popular being Transverse Mercator
The common feature of such projections is the fact you can't precisely represent whole world with it. I mean the projection is based on chosen meridian and is precise enough up to some distance from it (e.g. Gauss Krueger projection is quite accurate around +-500km from the meridian.
You will always have to choose some kind of 'zone' or 'meridian', regardless of what projection you choose, because it is impossible to transform a sphere into plane without any deformations (be it distance, angle or area).
So if you are working on a set of data located around some geographical area you can simply transform (project) the data and treat it as normal Enclidean 2d space.
But if you think of processing data located around the whole world you will have to properly cluster and project it using proper zone.
I have problems understanding the depth (Z) value in 3D point cloud resulted from 3d sparse reconstruction like this example in MATLAB: http://www.mathworks.com/help/vision/ug/sparse-3-d-reconstruction-from-multiple-views.html
I have attached a picture showing the reconstructed 3D point cloud in the above example. I have put some datatips on the figure so we know the (x,y,z) coordinates of the points. here are my questions:
1- what does the Z value in point cloud represent? is it the distance in millimeters from the camera? if that's the case then it does not make sense based on the picture I attached since I am sure the distance of the sphere and checkerboard from the camera must be greater than 200 mm.
Or maybe it is from some reference point in space? then what is this reference point? and how can I make a 3D point cloud that the Z values indicate the distance from the camera?
2- why is there negative values for Z? what does that mean in terms of distance to the camera?
I appreciate if someone can explain.
In this example the world coordinates are defined by the checkerboard. The checkerboard defines the X-Y plane, and the Z-axis points into the checkerboard, as explained in the documentation:
Since your 3D points are above the checkerboard, they have negative Z-coordinates.
Your (x,y,z) coordinates are in world units, which are completely disconnected from metric values (unless you build a scale between world and metric, there are various methods to do it). So the z value tells you about the depth of each point in world coordinates.
If you have the pose of each camera, and you multiply each point by the camera projection matrix, you will get the (x',y',z') points in camera coordinates. At that point, if z' is negative, it means it's behind the camera.
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
I have an array of CGPoints (basic struct with two floats: x and y). I want to use OpenGL ES to draw a textured curve using these points. I can do this fine with just two points, but it gets harder when I need to make a line from several points.
Currently I draw a line horizontally, calculate its angle from the points given, and then rotate it. I don't think doing this for all lines in a curve is a good idea. There's probably a faster way.
I'm thinking that I can "enlarge" or "constrict" all the points at once to make a curve with some sort of width.
I'm not positive what you want to accomplish, but consider this:
Based on a ordered list of points, you can draw a polyline using those points. If you want to have a polyline with a 2D texture on it, you can draw a series of quadrilaterals (using two triangles each, of course). You can generate these quadrilaterals using an idea similar to catmul-rom spline generation.
Consider a series of points p[i-1], p[i], p[i+1]. Now, for each i, you can find two points each an epsilon distance away from p[i] along the line perpendicular to the line connecting p[i-1] and p[i+1]. You can determine the two points generated for the endpoints in various ways, like using the perpendicular to the line from p[0] to p[1].
I'm not sure if this will be faster than your method, but you should be caching the results. If you are planning on doing this every frame, another type of solution to your problem may be needed.