In order to cache tiles for off-line use, I tried to calculate tiles coordinates according to a certain zoom level. Calculated x coordinates were correct but the y coordinates Were not.
This Old example compares actually received coordinates with that calculated. (click in the map to display results)
I was using map.project(latlng,zoom) to get the projected coordinates and then divide by tileSize which is 256. is this approach even correct ?
EDIT :
Thanks to Ivan Sanchez for the orientation about y inversion in TMS. Actually after projecting the point with map.project(latlng,zoom) you need to inverse the y coordinate as follow :
You calculate _globalTileRange(zoom) for the corresponding zoom level, then
InvertedY = _globalTileRange(zoom).max.y - y ;
Here is another Link that shows the correct calculation of y coordinates for the current zoom of the map, for other zoom levels the globalTileRange need to be recalculated accordingly.
Regards,
Your approach is correct. However:
In order to get the tile coordinates loaded by Leaflet, you are looping through all the loaded images and outputting the min/max of those values.
The problem with this approach is that Leaflet doesn't immediately unload off-screen tiles. See the keepBuffer option, bug #4039 and PR #4650.
In order to fetch the bounds of tiles visible within the map bounds, see the private methods used internally by L.GridLayer around this line of code.
In TMS, the y coordinate goes up, and in non-TMS tiles it does down. This is because TMS was done by geographers, where the y coordinate is the northing, and non-TMS tiles were initially done by computer programmers, who interpret the y coordinate as downward pixels.
For more background, read https://wiki.openstreetmap.org/wiki/TMS#The_Y_coordinate and https://wiki.osgeo.org/wiki/Tile_Map_Service_Specification#TileMap_Diagram and https://wiki.openstreetmap.org/wiki/Slippy_map_tilenames#X_and_Y
Related
I'm combining ARKit with a CNN to constantly update ARKit nodes when they drift. So:
Get estimate of node position with ARKit and place a virtual object in the world
Use CNN to get its estimated 2D location of the object
Update node position accordingly (to refine it's location in 3D space)
The problem is that #2 takes 0,3s or so. Therefore I can't use sceneView.unprojectPoint because the point will correspond to a 3D point from the device's world position from #1.
How do I calculate the 3D vector from my old location to the CNN's 2D point?
unprojectPoint is just a matrix-math convenience function similar to those found in many graphics-oriented libraries (like DirectX, old-style OpenGL, Three.js, etc). In SceneKit, it's provided as a method on the view, which means it operates using the model/view/projection matrices and viewport the view currently uses for rendering. However, if you know how that function works, you can implement it yourself.
An Unproject function typically does two things:
Convert viewport coordinates (pixels) to the clip-space coordinate system (-1.0 to 1.0 in all directions).
Reverse the projection transform (assuming some arbitrary Z value in clip space) and the view (camera) transform to get to 3D world-space coordinates.
Given that knowledge, we can build our own function. (Warning: untested.)
func unproject(screenPoint: float3, // see below for Z depth hint discussion
modelView: float4x4,
projection: float4x4,
viewport: CGRect) -> float3 {
// viewport to clip: subtract viewport origin, divide by size,
// scale/offset from 0...1 to -1...1 coordinate space
let clip = (screenPoint - float3(viewport.x, viewport.y, 1.0))
/ float3(viewport.width, viewport.height, 1.0)
* float3(2) - float3(1)
// apply the reverse of the model-view-projection transform
let inversePM = (projection * modelView).inverse
let result = inversePM * float4(clip.x, clip.y, clip.z, 1.0)
return float3(result.x, result.y, result.z) / result.w // perspective divide
}
Now, to use it... The modelView matrix you pass to this function is the inverse of ARCamera.transform, and you can also get projectionMatrix directly from ARCamera. So, if you're grabbing a 2D position at one point in time, grab the camera matrices then, too, so that you can work backward to 3D as of that time.
There's still the issue of that "Z depth hint" I mentioned: when the renderer projects 3D to 2D it loses information (one of those D's, actually). So you have to recover or guess that information when you convert back to 3D — the screenPoint you pass in to the above function is the x and y pixel coordinates, plus a depth value between 0 and 1. Zero is closer to the camera, 1 is farther away. How you make use of that sort of depends on how the rest of your algorithm is designed. (At the very least, you can unproject both Z=0 and Z=1, and you'll get the endpoints of line segment in 3D, with your original point somewhere along that line.)
Of course, whether this can actually be put together with your novel CNN-based approach is another question entirely. But at least you learned some useful 3D graphics math!
I programatically create requests to dev.virtualearth.net (Bing static maps).
I know the following values:
Center Point (Latitude & Longitude)
Zoom Level
Map Size (X pixels, Y pixels)
After I recieved the map as a bitmap, how do I determine the Coordinates (Latitude and Longitude) of the upper left corner (basically the very first pixel) and the lower right corner (the very last pixel)?
I just need some suggestions or some pseudo code. Note, that while I know the Center Point, Zoom Level and Map Size, these aren't the same for every request.
Thank you.
You will need to do tile math: https://msdn.microsoft.com/en-us/library/bb259689.aspx
You will need to do the following:
Pass the center point into LatLongToPixelXY method to get the center global pixel value.
Knowing the pixel dimensions of the static image you created, subtract half the width from the x value of the center global pixel value. Do the same with the height and y.
This gives you a new pixel value, pass it into the PixelXYToLatLong to get the coordinate for the top left corner.
That's it :)
I have an old code sample that does this, but retrieves the static image using the old SOAP services rather than the REST services. You can find the blog post here: https://rbrundritt.wordpress.com/2008/10/25/ve-imagery-service-and-custom-icons/ See the LatLongToPixel function code that is half way down the post. That does the above three steps.
Seems like a simple question, but I have been tearing my hair out for hours now.
I have a series of files ie.
kml_image_L1_0_0.jpg
kml_image_L2_0_0.jpg
kml_image_L2_0_1.jpg
kml_image_L2_1_0.jpg
kml_image_L2_1_1.jpg
etc. However just plotting them on the leaflet map surface understandibly puts the images at 0,0 on the earths surface, and the 0 zoom level inferred by the files should really be about 15 or so.
So I want to specify the latitude and longitude where the images should originate , and what zoom level they should start at. I have tried bounds (which doesn't display anything) and I have tried playing with offsetting the zoom level.
I need this because a user needs to click on an offline map to specify where they are and I need the GPS coordinates.
I also have a KML file but it seems to be of more help for plotting vector data on the map.
Any help is much appreciated, cheers.
If I understand correctly, the "kml_image_Lz_x_y.jpg" images that you have are actually tiles, with zoom, horizontal and vertical indices in their file name?
And your issue is that they use (z,x,y) numbers as if they started from the top-most level (zoom 0, single tile for entire world), but in fact they are just a small portion of the pyramid of tiles?
And you cannot use them as is because you still want to get actual geographic coordinates (latitude, longitude), which would be totally wrong if you used the tiles as if they were showing the entire world?
In that case, you have several options as workarounds:
The most simple and reliable would probably be to simply write a small script to rename all your tiles to their true (z,x,y) numbers.
Another option would be to modify the (z,x,y) numbers before they are written in the tile src attribute, and apply the appropriate offset (constant for z, scaled by z for x and y). That should probably happen in L.TileLayer.getTileUrl() method.
Good luck! :-)
I'm trying to create a view that present a bunch of coordinates without using a map.
The user's coordinates should be at the center of the screen(in the middle of the circle),
and the rest of the coordinates will be layout relative to one another according to their real latitude and longitude.
Something like this:
I understood that I can't do this with MKMapKit because it will be a violation of Google license, so I need a way to place and manage the coordinates myself.
What is best practice to do something like this? how should I convert the coordinates to a screen points?
I'd go about it as follows:
1) I'd start by normalizing at the user's current location (translate it to 0,0 then apply the same translation to the rest of the coordinates).
2) Once you've done that, use a distance function to find out which coordinate in your list is furthest away from your current location.
3) Use the furthest away coordinate to determine the scale of your view.
4) Calculate the X & Y screen coordinates of all your locations based on the scale you come up with in #3
I'm currently working on a mapping app for iPhone. I've created some custom maps of various sizes, but I've run into an issue:
I would like to implement the ability for users' locations to be checked automatically, but since Im not using a MapView this is much more dificult. (see below)
given the different coordinate systems, I would like to receive a geolocation (green dot) and translate it into a pixel location on a custom map.
Ive got the geolocations for the 4 corners, but the rect is askew. Ive calculated the angle of rotation, but Im just generally confused.
note: the size of the maps arent big enough for the spherical nature of the earth to come into calculation.
Any help is appreciated!
To convert a geolocation to point you need to first understand the mapping. assuming you are using Mercator.
x = R*long
y = R*(1+sin(lat))/cos(lat)
where lat and long are in radians.R is radius of earth. the scale of the image would be from 0 to R*PI
so to get it within view.frame.size you may have to divide by a scale factor.
for difference between points.
x2-x1 = R* (long2-long1)
y2-y1 = R* ( (1+sin(lat2))/cos(lat2) - (1+sin(lat1))/cos(lat1) )