I have two geo points (-27, 152) and (9, -83) and an intermediate point is (?, 180). I want to calculate Latitude of the intermediate point.
Conditions:
1) On Google Map all three points are on straight line.
So please suggest a formula to calculate it.
//For iOS developers
In actual I am trying to draw polyline on Apple Map in iOS. The polyline does not work when path is crossing 180 meridian.
You need to figure out the slope of the line given the two coordinates that you have. Then use the slope and one point on the line to figure out the x coordinate (latitude).
Here is a link that will teach you the basic geometry and formula: Find X-Coordinate of a point on a line, given another point and slope
Related
Does Here map iOS sdk provides Highway direction?
Example: While driving on US Highways/Interstate, Here map is providing information like I-5 or WA-14.
Can we get which highway/interstate direction we are heading into. Like I-5 N or WA-14 S.
You can allways use course infromation taken from currentPosition(NMAGeoPositon object) of the NMAPositionManager.
Swift
NMAPositionManager.sharedInstance().currentPosition?.course
or Objective-C
[NMAPositionManager sharedPositioningManager].currentPosition.course
Valid course values are in the range [0, 360), with 0 degrees representing north and values increasing clockwise. Thus, east is 90 degrees, south is 180 degrees, and so on. Will be NMAGeoPositionUnknownValue if unknown.
Also it can be obtained from waypoint course.
calculatedRoute.waypoints[idx].course;
or aliases 'start', 'destination' can be used
calculatedRoute.start.course;
calculatedRoute.destination.course;
Another one is to use mapOrientation of maneuver
calculatedRoute.maneuvers[idx].mapOrientation
The angle (from north) at the start of the maneuver, in degrees. Zero represents true-north, with increasing values representing a clockwise progression of map orientation.
I need to calculate the minimum distance between a 3D point (latitude, longitude, elevation) and a line (defined as two points).
The elevation is not necessary on the ground, I need to consider flying objects.
I only found an article that explains how to do that on a generic space but my points are defined with lat/lon/altitude(meters).
Thank you for pointing in the right direction, in my case I need to do that in Javascript but couldn't find any library that takes into consideration the altitude.
Point-Line Distance--3-Dimensional
If you want to compare a 3d point to a 2d line, I suppose you mean a "line" on our earth, at elevation 0. Take a look at st_distance in postgis.
If I understand you correctly, that'll give you what you want.
https://postgis.net/docs/ST_Distance.html
I am trying to convert a normal panorama to stereographic projection Using Opencv .
The resulting image is the one shown shown in this link.
http://www.photographymad.com/pages/view/little-planet-photos-5-simple-steps-to-making-panorama-worlds
In the steps to create this, there is a step to convert from rectangular to polar in photoshop.Can some explain the logic behind d conversion
Yes, there is.
http://mathworld.wolfram.com/StereographicProjection.html
Check out this wiki link: https://en.wikipedia.org/wiki/Equirectangular_projection
Think about your 360 panorama as a equirectangular projection of a sphere. That is, the x corresponds to longitude which ranges from -180 on the left to 0 at center then to 180 deg on the right. The y corresponds to the latitude which ranges from -90 deg at top to 0 at middle to 90 at bottom.
Follow the inverse map formula in the first link to build a map_x, map_y for each pixel, then use the cv2.remap function to get the stereographic projection of your image, aka a little planet.
The result would be better if your image has width = 2 * height because of the longitude, latitude range mentioned above.
And I learn all that from this awesome blogpost http://www.semifluid.com/2014/04/20/equirectangular-to-stereographic-projections-little-planets-in-matlab/
In my project my users can choose to be put in a random position inside a given, circular area.
I have the latitude and longitude of the center and the radius: how can I calculate the latitude and longitude of a random point inside the given area?
(I use PHP but examples in any language will fit anyway)
You need two randomly generated numbers.
Thinking about this using rectangular (Cartesian) (x,y) coordinates is somewhat unnatural for the problem space. Given a radius, it's somewhat difficult to think about how to directly compute an (Δx,Δy) delta that falls within the circle defined by the center and radius.
Better to use polar coordinates to analyze the problem - in which the dimensions are (r1, Θ). Compute one random distance, bounded by the radius. Compute a random angle, from 0 to 360 degrees. Then convert the (r,Θ) to Cartesian (Δx,Δy), where the Cartesian quantities are simply offsets from your circle center, using the simple trigonometry relations.
Δx = r * cos(Θ)
Δy = r * sin(Θ)
Then your new point is simply
xnew = x + Δx
ynew = y + Δy
This works for small r, in which case the geometry of the earth can be approximated by Euclidean (flat plane) geometry.
As r becomes larger, the curvature of the earth means that the Euclidean approximation does not match the reality of the situation. In that case you will need to use the formulas for geodesic distance, which take into account the 3d curvature of the earth. This begins to make sense, let's say, above distances of 100 km. Of course it depends on the degree of accuracy you need, but I'm assuming you have quite a bit of wiggle room.
In 3d-geometry, you once again need to compute 2 quantities - the angle and the distance. The distance is again bound by your r radius, except in this case the distance is measured on the surface of the earth, and is known as a "great circle distance". Randomly generate a number less than or equal to your r, for the first quantity.
The great-circle geometry relation
d = R Δσ
...states that d, a great-circle distance, is proportional to the radius of the sphere and the central angle subtended by two points on the surface of the sphere. "central angle" refers to an angle described by three points, with the center of the sphere at the vertex, and the other two points on the surface of the sphere.
In your problem, this d must be a random quantity bound by your original 'r'. Calculating a d then gives you the central angle, in other words Δσ , since the R for the earth is known (about 6371.01 km).
That gives you the absolute (random) distance along the great circle, away from your original lat/long. Now you need the direction, quantified by an angle, describing the N/S/E/W direction of travel from your original point. Again, use a 0-360 degree random number, where zero represents due east, if you like.
The change in latitude can be calculated by d sin(Θ) , the change in longitude by d cos(Θ). That gives the great-circle distance in the same dimensions as r (presumably km), but you want lat/long degrees, so you'll need to convert. To get from latitudinal distance to degrees is easy: it's about 111.32 km per degree regardless of latitude. The conversion from longitudinal distance to longitudinal degrees is more complicated, because the longitudinal lines are closer to each other nearer the poles. So you need to use some more complex formulae to compute the change in longitude corresponding to the selected d (great distance) and angle. Remember you may need to hop over the +/- 180° barrier. (The designers of the F22 Raptor warplane forgot this, and their airplanes nearly crashed when attempting to cross the 180th meridian.)
Because of the error that may accumulate in successive approximations, you will want to check that the new point fits your constraints. Use the formula
Δσ = arccos( cos(Δlat) - cos(lat1)*cos(lat2)*(1 - cos(Δlong) ) .
where Δlat is the change in latitude, etc.
This gives you Δσ , the central angle between the new and old lat/long points. Verify that the central angle you've calcuated here is the same as the central angle you randomly selected previously. In other words verify that the computed d (great-circle-distance) between the calculated points is the same as the great circle distance you randomly selected. If the computed d varies from your selected d, you can use numerical approximation to improve the accuracy, altering the latitude or longitude slightly until it meets your criterion.
This can simply be done by calculating a random bearing (between 0 and 2*pi) and a random distance between 0 and your desired maximum radius. Then compute the new (random) lat/lon using the random bearing/range from your given lat/lon center point. See the section 'Destination point given distance and bearing from start point' at this web site: http://www.movable-type.co.uk/scripts/latlong.html
Note: the formula given expects all angles as radians (including lat/lon). The resulting lat/lon with be in radians also so you will need to convert to degrees.
I have a series of lat/lon which represents the center of some object. I need to draw a line through this point that is x meters on either side of the center and it needs to be perpendicular to the heading (imagine a capital T)
Ultimately I want to get the lat/lon of this line's endpoints.
Thanks!
The basic calculation is in this similar question's answer: Calculate second point knowing the starting point and distance. Calculate the points for the two headings perpendicular to the main heading the distance away you want.
Have a look at: Core Location extensions for bearing and distance
With those extensions and two points on the initial line you should be able to get the bearing, add/subtract pi/2 and find points to either side like this:
double bearing = [bottomOfT bearingInRadiansTowardsLocation:topOfT];
CLLocation *left = [topOfT newLocationAtDistance:meters
alongBearingradians:bearing+M_PI/2];
CLLocation *right = [topOfT newLocationAtDistance:meters
alongBearingradians:bearing-M_PI/2];