I have to display 5x5 degrees pie chart on a leaflet map. I can display pie chart using the great leaflet-dvf library, but I have to provide the radius in pixel, and it is static so far.
I would like to have it dynamic so that at any zoom level, the pie chart fills the 5x5 square (aka radius = 5x5 degree length).
How can I know the length in pixel of the side of a 5x5 degrees square, depending on the zoom level?
Thanks
This will work out metres per pixel
metresPerPixel = 40075016.686 * Math.abs(Math.cos(map.getCenter().lat / 180 * Math.PI)) / Math.pow(2, map.getZoom()+8);
I used the following page that provides the pixel per meter on a leaflet map depending on the zoom level:
http://wiki.openstreetmap.org/wiki/Zoom_levels
Then I computed the length of a 5x5 square at the equator: 556000 meters.
Then I store the length ratio for zoom level = 0:
$scope.lengthRatio = 556000 / 156412 // in meter / pixel
Finally, I get the radius of a pie chart depending on the zoom level ($scope.mapZoom):
var radius = $scope.lengthRatio * Math.pow(2,$scope.mapZoom)) / 2
The /2 is because I want the radius and not the diameter
Simple!
Related
I'm trying to find a way to calculate zoom level given a radius in meters.
For example, I want to show 100km radius within point x,y, how do I calculate the proper zoom level for that point?
This will depend on:
the number of tiles of your map
the latitude of your view
The resolution (meters/pixel) of a web-Mercator map at zoom 0 (one tile for the whole map) is given by:
resolution = cos(latitude_rad) * earth circumference / tile width
with the latitude in radians, the earth circumference in meters (2 * pi * radius) and the map width given by:
map width = 256 as the tiles are 256 pixel wide.
At each zoom step, the width (in pixel) is doubled, so at zoom z:
map width = 256*2^zoom
so :
resolution=cos(latitude_rad) * 2*pi* 6371008 / (256*2^zoom)
but as you are using a certain number n of 256 pixel tiles:
widthmap = cos(latitude_rad) * 2*pi* 6371008*n / (2^zoom)
Now, to get the zoom from the distance:
with latitude in degrees, radius is your circle in meters, and n the number of tiles to display:
zoom = floor(log2 ((cos(latitude *pi/180) * 2*pi* 6371008*n)/radius))
(floor as you want an integer and to make sure your whole circle is on the map)
For example, at 45 degrees N/S, 100m and 1 tile will give you a zoom of 18.
For a visualization I need an optical satellite image for a specific rectangular AOI, that is defined by two lat/long coordinates. I tried Mapbox Static Images API, which takes a lat/long bounding box and a resolution in width/height pixel for the output. The problem is that it looks like to me that if ratio of the lat/long box is not the same as the w/h pixels, it will add padding to the lat/long bounding box to fill the w/h of the pixel image.
And this would prevent me from combining the optical image with the other data, because I would not know which image pixel would (roughly) correspond to which lat/long coordinate.
I see three "solutions", but I don't know how to achive any of them.
"Make" Mapbox return the images with out padding.
Compute the ratio for the correct w/h pixel ratio using the lat/long coordinate, so there would be no padding. Maybe with https://en.wikipedia.org/wiki/Equirectangular_projection like discussed here: https://stackoverflow.com/a/16271669/380038?
Find a way to determine the lat/long coordinates of the optical satellite image so I can cut off the possible padding.
I checked How can I extract a satellite image from google maps given a Lat Long Rectangle?, but I would prefer to use my existing paid Mapbox account and I got the impression that I still wouldn't get the exact optical image or the exact corner coordinates of the optical image.
Mapbox Static Images API serves maps
You have optical image from other source
You want to overlay these data
Right?
Note the Red and Green pins: the waypoints are at opposite corners on Mapbox.
After Equirectangular correction Mapbox matches Openstreetmaps (little wonder), but Google coordinates are quite close too.
curl -g "https://api.mapbox.com/styles/v1/mapbox/streets-v11/static/[17.55490,47.10434,17.55718,47.10543]/600x419?access_token=YOUR_TOKEN_HERE" --output example-walk-600x419-nopad.png
What is your scale? 1 km - 100 km?
What is your source of optical image?
What is the required accuracy?
Just to mention, optical images have their own sources of distortions.
In practice:
You must have the extent of your non optical satellite data (let's preserve the mist around...) I'll call it ((x1, y1), (x2, y2)) We are coders, not cartographers - right!?
If you feed your extent to https://docs.mapbox.com/playground/static/ as
min longitude = x1, min lattitude = y1, max longitude = x2, max lattitude = y2
Select "Bounding box" entry! Do you see mapbox around your data!? Don't mind the exact dimensions, just check if mapbox is related to your data! May be you have to swap some values to get to the right corner of the globe.
If you have the right ((x1, y1), (x2, y2)) coordinates, do the equirectangular transformation to get the right pixel size.
You've called it Solution #2.
Let's say the with of your non optical satellite data is Wd, the height is Hd.
The mapbox image will fit your data, if you ask for Wm widht, and Hm height of mapbox data where
Wm = Wd
Hm = Wd * (y2 - y1) * cos(x1) / (x2 - x1)
Now you can pull the mapbox by
curl -g "https://api.mapbox.com/styles/v1/mapbox/streets-v11/static/[<x1>,<y1>,<x2>,<y2>]/<Wm>x<Hm>?access_token=<YOUR_TOKEN>" --output overlay.png
If (Hd == Hm)
then {you are lucky :) the two images just fit each other}
else { the two images are for the same area, but you have to scale the height of one of the images to make match }
Well... almost. You have not revealed what size of area you want to cover. The equation above is just an approximation which works up to the size of a smaller country (~100 km or so). For continent scale you probably have to apply more accurate formulas.
In my opinion, your #2 idea is the way to go. You do have the LLng bbox, so all that remains is calculate its "real" size in pixels.
Let us say that you want (or can allow, or can afford) a resolution of 50m per pixel, and the area is small enough not to have distortions (i.e., a rectangle of say 1 arcsecond of latitude and 1 arcsecond of longitude has top and bottom sides of the same length, with an error less than your chosen resolution). These are, I believe, very loose requisites and easy to fulfill.
Then, you just need to calculate the distance between the (Lat1, Lon1) and (Lat1, Lon2) points, and betwen (Lat1, Lon1) and (Lat2, Lon1). Divide that distance in meters by 50, and you'll get the exact number of pixels:
Lon1 Lon2
Lat1 +---------------+
| |
| |
Lat2 +---------------+
And you have a formula for that - the haversine formula.
If you need a higher precision, you could recourse to the Vincenty oblate spheroid (here a Javascript library). On the MT site (first link) there is a live calculator that you can use to plug data from your calls, and verify whether the approach is indeed working. I.e. you plug in your bounding box, get the distance in meters, divide and get the pixel size of the image (if the image is good, chances are that you can go with the simpler haversine. If it isn't, then there has to be some further quirk in the maps API - its projection, perhaps - that doesn't return the expected bounding box. But it seems unlikely).
I've had this exact problem when using a satellite image on an apple watch. I overlay some markers and a path. I convert everything from coordinates to pixels. Below is my code to determine the exact bbox result
var maxHoleLat = 52.5738902
var maxHoleLon = 4.9577606
var minHoleLat = 52.563994
var minHoleLon = 4.922364
var mapMaxLat = 0.0
var mapMaxLon = 0.0
var mapMinLat = 0.0
var mapMinLon = 0.0
let token = "your token"
var resX = 1000.0
var resY = 1000.0
let screenX = 184.0
let screenY = 224.0 // 448/2 = 224 - navbarHeight
let navbarHeight = 0.0
var latDist = 111000.0
var lonDist = 111000.0
var dx = 0.0
var dy = 0.0
func latLonDist(){
//calgary.rasc.ca/latlong.htm
let latRad = maxHoleLat * .pi / 180
//distance between 1 degree of longitude at given latitude
self.lonDist = 111412.88 * cos(latRad) - 0.09350*cos(3 * latRad) + 0.00012 * cos(5 * latRad)
print("lonDist = \(self.lonDist)")
//distance between 1 degree of latitude at a given longitude
self.latDist = 111132.95 - 0.55982 * cos(2 * latRad) + 0.00117 * cos(4 * latRad)
print("latDist = \(self.latDist)")
}
func getMapUrl(){
self.dx = (maxHoleLon - minHoleLon) * lonDist
self.dy = (maxHoleLat - minHoleLat) * latDist
//the map is square, but the hole not
//check if the hole has less x than y
if dx < dy {
mapMaxLat = maxHoleLat
mapMinLat = minHoleLat
let midLon = (maxHoleLon + minHoleLon ) / 2
mapMaxLon = midLon + dy / 2 / lonDist
mapMinLon = midLon - dy / 2 / lonDist
} else {
mapMaxLon = maxHoleLon
mapMinLon = minHoleLon
let midLat = (maxHoleLat + minHoleLat ) / 2
mapMaxLat = midLat + dx / 2 / latDist
mapMinLat = midLat - dx / 2 / latDist
}
self.imageUrl = URL(string:"https://api.mapbox.com/styles/v1/mapbox/satellite-v9/static/[\(mapMinLon),\(mapMinLat),\(mapMaxLon),\(mapMaxLat)]/1000x1000?logo=false&access_token=\(token)")
print("\(imageUrl)")
}
I'm trying to resize and reposition a ROI (region of interest) correctly from a low resolution image (256x256) to a higher resolution image (512x512). It should also be mentioned that the two images cover different field of view - the low and high resolution image have 330mm x 330mm and 180mm x 180mm FoV, respectively.
What I've got at my disposal are:
Physical reference point (in mm) in the 256x256 and 512x512 image, which are refpoint_lowres=(-164.424,-194.462) and refpoint_highres=(-94.3052,-110.923). The reference points are located in the top left pixel (1,1) in their respective images.
Pixel coordinates of the ROI in the 256x256 image (named pxX and pxY). These coordinates are positioned relative to the reference point of the lower resolution image, refpoint_lowres=(-164.424,-194.462).
Pixel spacing for the 256x256 and 512x512 image, which are 0.7757 pixel/mm and 2.8444 pixel/mm respectively.
How can I rescale and reposition the ROI (the binary mask) to correct pixel location in the 512x512 image? Many thanks in advance!!
Attempt
% This gives correctly placed and scaled binary array in the 256x256 image
mask_lowres = double(poly2mask(pxX, pxY, 256., 256.));
% Compute translational shift in pixel
mmShift = refpoint_lowres - refpoint_highres;
pxShift = abs(mmShift./pixspacing_highres)
% This produces a binary array that is only positioned correctly in the
% 512x512 image, but it is not upscaled correctly...(?)
mask_highres = double(poly2mask(pxX + pxShift(1), pxY + pxShift(2), 512.,
512.));
So you have coordinates pxX, and pxY in pixels with respect to the low-resolution image. You can transform these coordinates to real-world coordinates:
pxX_rw = pxX / 0.7757 - 164.424;
pxY_rw = pxY / 0.7757 - 194.462;
Next you can transform these coordinates to high-res coordinates:
pxX_hr = (pxX_rw - 94.3052) * 2.8444;
pxY_hr = (pxY_rw - 110.923) * 2.8444;
Since the original coordinates fit in the low-res image, but the high-res image is smaller (in physical coordinates) than the low-res one, it is possible that these new coordinates do not fit in the high-res image. If this is the case, cropping the polygon is a non-trivial exercise, it cannot be done by simply moving the vertices to be inside the field of view. MATLAB R2017b introduces the polyshape object type, which you can intersect:
bbox = polyshape([0 0 180 180] - 94.3052, [180 0 0 180] - 110.923);
poly = polyshape(pxX_rw, pxY_rw);
poly = intersect([poly bbox]);
pxX_rw = poly.Vertices(:,1);
pxY_rw = poly.Vertices(:,2);
If you have an earlier version of MATLAB, maybe the easiest solution is to make the field of view larger to draw the polygon, then crop the resulting image to the right size. But this does require some proper calculation to get it right.
I want to use ImageOverlays as markers, because I want the images to scale with zoom. Markers icons always resize to keep their size the same when you zoom.
My problem is that I can't figure out how to transform pixels to cords, so my image isn't stretched.
For instance, I decided my south-west LatLng to be [50, 50]. My image dimensions are 24px/24px.
How do I calculate the north-east LatLng based on the image pixels?
You are probably looking for map conversion methods.
In particular, you could use:
latLngToContainerPoint: Given a geographical coordinate, returns the corresponding pixel coordinate relative to the map container.
containerPointToLatLng: Given a pixel coordinate relative to the map container, returns the corresponding geographical coordinate (for the current zoom level).
// 1) Convert LatLng into container pixel position.
var originPoint = map.latLngToContainerPoint(originLatLng);
// 2) Add the image pixel dimensions.
// Positive x to go right (East).
// Negative y to go up (North).
var nextCornerPoint = originPoint.add({x: 24, y: -24});
// 3) Convert back into LatLng.
var nextCornerLatLng = map.containerPointToLatLng(nextCornerPoint);
var imageOverlay = L.imageOverlay(
'path/to/image',
[originLatLng, nextCornerLatLng]
).addTo(map);
Demo: http://playground-leaflet.rhcloud.com/tehi/1/edit?html,output
I have a .bmp image with a map. What i know:
Height an Width of bmp image
dpi
Map Scale
Image Center's coordinates in meters.
What i want:
How can i calculate some points of image (for example corners) in meters.
Or how can i change a pixel distanse to meters?
What i do before:
For sure i know image center coordinates in pixels:
CenterXpix = Widht/2;
CenterYpix = Height/2;
But what i gonna do to find another corners coordinates. Don't think that:
metersDistance = pixelDistance*Scale;
is a correct equation.
Any advises?
If you know the height or width in both meters and pixels, you can calculate the scale in meters/pixel. You equation:
metersDistance = pixelDistance*Scale;
is correct, but only if your points are on the same axis. If your two points are diagonal from each other, you have to use good old pythagoras (in pseudocode):
X = XdistancePix*scale;
Y = YdistancePix*scale;
Distance_in_m = sqrt(X*X+Y*Y);