When I have a specific area of the operation (like city or town) but would like to travel from point A in that area to point B outside of that area how can I check the exact point of the intersection of a route and border of area on google map?
ts like A -> [ C ] -> B
where A and B are start and finish point of route
and C is exact point for route / border - so i need distance (A;C) and then distance (C;B) with information where C is (intersection of route and border)
its urgent for me,
cheers, Marcin
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I am having trouble mapping Nebraska school districts in D3 (v4). (See bl.ock here.) I can map Nebraska counties no problem, but the same code modified for school districts--and pointing to a school district TopoJSON file--gives me a blank page.
Here's how I created the JSON, based on Mike Bostock's excellent instructions :
curl "https://www2.census.gov/geo/tiger/GENZ2017/shp/cb_2017_31_unsd_500k.zip" -o cb_2017_31_unsd_500k.zip
unzip -o cb_2017_31_unsd_500k.zip
shp2json cb_2017_31_unsd_500k.shp -o ne_district.json
ndjson-split "d.features" < ne_district.json > ne_district.ndjson
ndjson-map "d.id = d.properties.GEOID, d" < ne_district.ndjson > ne_district-id.ndjson
geo2topo -n districts=ne_district-id.ndjson > ne_district-id-topo.json
And here's my projection:
var projection = d3.geoConicConformal()
.parallels([40, 43])
.rotate([100, 0])
.scale(8000);
Thanks for your help and apologies in advance for anything important I left out!
The issue is you haven't finished setting your projection parameters. You have rotate the map, which is how you should center a conic projection along the x axis. But you haven't centered the map on the y axis, it is centered on the equator. You
For a conical projection, you can do this one of three ways:
Center the map on a central latitude : projection.center([0,y])
You don't need to use .center with an x value because the map is already centered on the x by rotation, rotation and centering are cumulative
Rotate the map to a central latitude and longitude: projection.rotate([-x,-y])
On a conical projection the rotation on the meridian does not warp the map (generally), we rotate by the negative as we move the earth under us. This option does slightly distort the map relative to the other options - this may be preferrable.
Use the projection translation to center the map
The easiest way is to translate the result while automatically scaling (though you can do this manually too) with projection.fitSize or projection.fitExtent. These methods modify projection.scale and projection.translate. As with centering with .center, you need to keep your rotation - otherwise you'll get an odd tilt to the map.
These methods set translate and scale to appropriate values so that your map area contains the desired features:
var featureCollection = topojson.feature(ne, ne.objects.districts);
projection.fitSize([width,height],featureCollection);
These methods must take objects, not arrays, so we use the featureCollection, not the features as an array
Both methods take an array specifying the size to stretch a provided geojson object over:
projection.fitSize([mapwidth,mapheight],geojsonObject)
projection.fitExtent([[left,top],[right,bottom]],geojsonObject)
Here's an updated gist using fitSize.
When using ol3-cesium and the map is in 3d mode, calling map.getView().getZoom() returns undefined. This might affect setZoom as well.
I understand we are in a 3d world, so there are no z-levels as in the tiled maps. On the other hand, Google Maps calculates a z-equivalent when coming back grom 3d to 2d.
How can I convert from height to a z-equivalent? Any formula, taking into account the latitude and altitude, to get the z equivalent?
There's no easy formula to get a 2D "Z" value from 3D, because the 3D camera can be tilted, can see different levels of tiles in the foreground vs the background, etc.
For individual tiles however, there are specific known "Level" values from the imagery quadtree. You can see these in Cesium Inspector by clicking the little + next to the word Terrain on the right side, and then put a checkmark on Show tile coordinates. The coordinates shown include L, X, and Y, where L is the tile's level (0 being the most zoomed-out, higher numbers more zoomed in), and X and Y are 2D positions within the imagery layer.
I posted an answer on GIS SE that shows how to reach in and grab these tiles, the same way Cesium Inspector does, along with a number of caveats involved. You could potentially look for the highest-level visible tile, and use that as your "Z" value.
I know this is not accurate, but sharing in case this is of use to anyone.
I have moved to several altitudes in Google Maps, switching between the 2D and 3D maps, writing down the z or altitude shown in the address bar:
z altitude (metres)
----- -----------------
3 10311040
4 5932713
5 2966357
6 1483178
7 741589
8.6 243624
11.35 36310
13.85 6410
15.26 2411
17.01 717
18.27 214
19.6 119
20.77 50
21 44
With the above correspondences, I have approximated the following function:
function altitudeToZoom(altitude) {
var A = 40487.57;
var B = 0.00007096758;
var C = 91610.74;
var D = -40467.74;
return D+(A-D)/(1+Math.pow(altitude/C, B));
}
Based on your formula, the reverse conversion should be:
altitude = C * Math.pow((A-D)/(zoomLevel-D) -1, 1/B);
I'm playing with body animation in AS3. I did a body with all parts (excluding fingers) and make a XML with the "skeleton". The XML got the instances of each part and the place of the articulation of the next part. I make it work with cardinal coordinates (x,y) and the body moves when I rotate a part and recalculate all the links again (each part in each articulation).
However, this will demand some calculation each little modification of the body, so now I'm optimizing it. As for de design x,y is easier, so when the body instance is created, the class re-build the XML converting coordinates to Polar system (r,t), like this ("Quadro" is the node with coordinates):
dx = Quadro.#x;
dy = Quadro.#y;
Quadro.#r = Math.sqrt(Math.pow(dx,2) + Math.pow(dy,2));
Quadro.#t = (dy>0)? Math.asin(dx/Quadro.#r) : Math.acos(dy/Quadro.#r);
I did some changes to make it work but at list one quadrant is always wrong! In this case, the upper left is wrong. The neck and the head should be in this place and they are in upper right (mirrored).
Any tips for a right conversion in AS3?
Try to use this:
Quadro.#t=Math.atan2(dy,dx);
From Wikipedia:
The Cartesian coordinates x and y can be converted to polar
coordinates r and φ with r ≥ 0 and φ in the interval (−π, π] by:
I have the next routes:
The red one is goes from A to B, the blue one goes from A' to B' and the green one goes from A'' to B''. I have the lat/long for A,B, A', B', A'' and B'' (I'm using Google Maps API) and I want to know if there is an algorithm to determinate if two routes have the same direction.
For example, in the picture, the red and blue route have the same direction.
Find the bearing of B from A. Movable-type's page is the standard reference.
However you can also use the geometry library in Google Maps API v3:
var a = new google.maps.LatLng(...);
var b = new google.maps.LatLng(...);
var bearing = google.maps.geometry.spherical.computeHeading(a,b)
Is it possible to plot two routes at same time with Google maps api.
For Example
Route from city A to B
and Route from city C to D
In this case A,B and C,D are separated.
Thanks.
Here are two polylines on the same map; there doesn't appear to be a limit.
http://jsfiddle.net/Lqcde/3/
I didn't add the div panels (becomes too cluttered) but you can also show two sets of text directions. It's a bit confusing because both start points are shown as "A" and endpoints, "B". Add the following lines:
directionsDisplay.setPanel(document.getElementById("dir-1"));
directionsDisplay2.setPanel(document.getElementById("dir-2"));
<div id="dir-1"></div>
<div id="dir-2"></div>