Is there an obvious mathematical formula to convert a lat/long coordinates into a Cardinal directions (north, south, east, west)?
I think you will find this example of converting lat long to include the cardinal directions to be very interesting:
#here is a code snippet from Luciano Ramalho - Pythonic Objects: idiomatic OOP in Python - #PyCon 2019
def __str__(self): ns =”NS”[self.lat <0} #boolean expression using 0 or 1, and 1 means S we =”EW”[self.long <0] return f”{abs(self.lat):.1f}{ns},{abs(self.long):.1f}{we}”
#A link to the video is here
#https://youtu.be/mUu_4k6a5-I?t=2445
So the trick is to use booleans as they inherit from the in class, so your string "NS" will yield the 0 position or the 1st position because of the boolean. then the return takes the absolute value and puts the cardinal direction as a suffix. Very cool idea from Mr. Ramalho.
Related
I want to convert Dutch RD coordinates to longitude and latitude decimal degree coordinates. Example:
108519 438598
108518 438578
108517 438578
to
51.93391 4.71134
51.93382 4.71133
51.93373 4.71131
Which packages and what code can I use to apply this on a bigger dataset?
For coordinate conversion one usually uses the proj.4 lib.
Its available for many programming languages, like python, java, c
First you need to find out the projection number as EPSG number.
e.g https://epsg.io/28992
On that page under "export" there is a section for the proj.4 definition of that projection, which gives this string:
"+proj=sterea +lat_0=52.15616055555555 +lon_0=5.38763888888889 +k=0.9999079 +x_0=155000 +y_0=463000 +ellps=bessel +towgs84=565.417,50.3319,465.552,-0.398957,0.343988,-1.8774,4.0725 +units=m +no_defs"
Using the proj4 lib, you can then convert to WGS84 latitude and longitude, this is the format you want.
MapKit has function MKMapPointForCoordinate, It accept lat lng as argument and return point x,y.
https://developer.apple.com/library/prerelease/ios/documentation/MapKit/Reference/MapKitFunctionsReference/index.html
lat = 59.90738808515509
lng = 10.724523067474365
if we pass above lat, lng then function return
x = 142214284, y = 78089986
I check with lag lng wot UTM but it gives different result
http://www.latlong.net/lat-long-utm.html
MKMapPointForCoordinate doesn't return UTM Coordinates.
Coordinates refer to a position on the earth (a pseudo-sphere), but sometimes you need to do calculation refering to a 2D map (much simpler) and then convert again to coordinates. This is the goal of the conversion.
So, the MKMapPoint struct returned by MKMapPointForCoordinate is a 2D representation of the coordinates, but it doesn't match any standard known.
At this link: https://developer.apple.com/library/prerelease/ios/documentation/MapKit/Reference/MapKitDataTypesReference/index.html#//apple_ref/doc/c_ref/MKMapPoint
in the MKMapPoint documentation, you can read:
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.
EDIT
for Coordinates-UTM Conversion in a previous project I used this Open Source Code
How to find a few coordinates that are in the straight line, between 2 coordinates?
For example:
Start coordinate: Lat=X1 Long=Y1
End coordinate: Lat=X2 Long=Y2
Make a straight line from X1,Y1 to X2,Y2.
Then find 5 points that are located in that line, that are spread in the same distance.
Anyone can help to find the algorithm and calculation?
The coordinate is in decimal format, e.g. 50.123456, 6.123456
Thanks.
There are no "straight lines" on a sphere (or ellipsoid).
Anyway, you'll need to:
Calculate the distance and initial azimuth from (x1,y1) to (x2,y2).
You can use Vincenty's inverse method.
Calculate the coordinate of points with distance (0,25d, 0.5d, 0.75d) from (x1,y1) at that azimuth (plus points (x1,y1) and (x2,y2) of course).
You can use Vincenty's direct method.
Both direct and inverse methods are described on Wikipedia.
An extremely accurate implementations for both direct and inverse problems are available as part of GeographicLib.
Less accurate, but much simpler methods are described in Aviation Formulary.
Basically, I have a set of up to 100 co-ordinates, along with the desired tangents to the curve at the first and last point.
I have looked into various methods of curve-fitting, by which I mean an algorithm with takes the inputted data points and tangents, and outputs the equation of the cure, such as the gaussian method and interpolation, but I really struggled understanding them.
I am not asking for code (If you choose to give it, thats acceptable though :) ), I am simply looking for help into this algorithm. It will eventually be converted to Objective-C for an iPhone app, if that changes anything..
EDIT:
I know the order of all of the points. They are not too close together, so passing through all points is necessary - aka interpolation (unless anyone can suggest something else). And as far as I know, an algebraic curve is what I'm looking for. This is all being done on a 2D plane by the way
I'd recommend to consider cubic splines. There is some explanation and code to calculate them in plain C in Numerical Recipes book (chapter 3.3)
Most interpolation methods originally work with functions: given a set of x and y values, they compute a function which computes a y value for every x value, meeting the specified constraints. As a function can only ever compute a single y value for every x value, such an curve cannot loop back on itself.
To turn this into a real 2D setup, you want two functions which compute x resp. y values based on some parameter that is conventionally called t. So the first step is computing t values for your input data. You can usually get a good approximation by summing over euclidean distances: think about a polyline connecting all your points with straight segments. Then the parameter would be the distance along this line for every input pair.
So now you have two interpolation problem: one to compute x from t and the other y from t. You can formulate this as a spline interpolation, e.g. using cubic splines. That gives you a large system of linear equations which you can solve iteratively up to the desired precision.
The result of a spline interpolation will be a piecewise description of a suitable curve. If you wanted a single equation, then a lagrange interpolation would fit that bill, but the result might have odd twists and turns for many sets of input data.
How do i calculate point D for triangle 2?
I have the the following coordinates for triangle 1:
a(0,0) b(0,78) c(18,39)
point D is located at (0,39) in triangle 1.
now I change the perspective on my triangle by for example moving coordinate b and c.
the new triangle formed is called triangle 2 with coordinates:
a(0,0) b(11,72) c(37,42)
AS YOU CAN SEE POINT D IS NOT IN THE MIDDLE OF LINE a<-->b BECAUSE OF THE CHANGE IN PERSPECTIVE/SKEW.
How do i calculate point d? I have the coordinates abc of triangle 1 & 2.
Preferably answer in programcode rather than using math signs, since i am not a hero at reading math :)
You need to convert point D to barycentric coordinates using the original triangle coordinates, then convert it back to cartesian coordinates using the modified triangle coordinates.
This looks like a good introduction to triangular barycentric coordinates: http://blogs.msdn.com/b/rezanour/archive/2011/08/07/barycentric-coordinates-and-point-in-triangle-tests.aspx
Also, explicit formulae for converting a point in a triangle to barycentric coordinates are given at the end of the Converting to Barycentric Coordinates section of the Wikipedia article “Barycentric coordinate system”.
I guess there are more ways of calculating a coordinate from one perspective to another.
more on the triangle way is written by culebrón here: Transforming captured co-ordinates into screen co-ordinates
At the same link there is another way by using SVD and calculate an H-matrix which can be used to translate any coordinate from one perspective to another. I am going to use this way because i could solve this way in matlab. Next step in objective-c! i had some trouble calculating the same in objective-c. more on that here: calculate the V from A = USVt in objective-C with SVD from LAPACK in xcode
I would like to know how to solve the triangle way too! i could not figure out what a1 and a2 were in culebron's post: https://stackoverflow.com/a/1690300/1568532 neither the width and height made much sense to me.
Also i would like to know how to calculate the EYE's point of view on a triangle or quadrangle based on 3 or 4 coordinates. if you know the original size of the object.
any ideas on this?
when i search for eye or camera's point of view. there is load of result about photography.
what do i need to use in order to calculate this? maybe some example anyone?