bin*_*inW 51 android google-maps zoom android-mapview
我想将地图视图缩放到1公里半径,但无法弄清楚如何？
该文档说缩放级别1将地球赤道映射到256像素.那么如何计算我需要设置的缩放级别,以便地图视图显示1KM半径的区域？
更新:
阅读几篇博文后,我写了以下代码:
private int calculateZoomLevel() {
double equatorLength = 6378140; // in meters
double widthInPixels = screenWidth;
double metersPerPixel = equatorLength / 256;
int zoomLevel = 1;
while ((metersPerPixel * widthInPixels) > 2000) {
metersPerPixel /= 2;
++zoomLevel;
}
Log.i("ADNAN", "zoom level = "+zoomLevel);
return zoomLevel;
}
我的想法是,首先我在缩放级别1中计算每像素的米数,根据谷歌显示使用256像素的地球赤道.现在,每个后续缩放级别放大2级,因此每个缩放级别的每个像素的一半.我这样做,直到我有一个缩放级别,每个像素的米乘以屏幕宽度给我小于2000,即2公里跨.
但我不认为我得到的缩放级别显示的是2Km半径的地图.有人可以告诉我这里我做错了什么吗？
Sye*_*hdi 46
虽然这个答案是合乎逻辑的,我觉得它有效,但结果不准确,我不知道为什么,但我厌倦了这种方法,这种技术更准确.
1)在具有所需半径的物体上做一个圆圈
Circle circle = mGoogleMap.addCircle(new CircleOptions().center(new LatLng(latitude, longitude)).radius(getRadiusInMeters()).strokeColor(Color.RED));
circle.setVisible(true);
getZoomLevel(circle);
2)将该对象传递给此功能并设置缩放级别这是一个链接
public int getZoomLevel(Circle circle) {
if (circle != null){
double radius = circle.getRadius();
double scale = radius / 500;
zoomLevel =(int) (16 - Math.log(scale) / Math.log(2));
}
return zoomLevel;
}
bin*_*inW 33
以下代码最终使用.鉴于屏幕宽度以及在缩放级别1处地球的赤道长度为256像素并且每个后续缩放级别使表示地球赤道所需的像素数量加倍的事实,以下函数返回缩放级别,其中屏幕将显示区域宽2Km.
private int calculateZoomLevel(int screenWidth) {
double equatorLength = 40075004; // in meters
double widthInPixels = screenWidth;
double metersPerPixel = equatorLength / 256;
int zoomLevel = 1;
while ((metersPerPixel * widthInPixels) > 2000) {
metersPerPixel /= 2;
++zoomLevel;
}
Log.i("ADNAN", "zoom level = "+zoomLevel);
return zoomLevel;
}
Ori*_*itm 10
我最终使用的工具来自:
https://github.com/googlemaps/android-maps-utils
我从lib中提取了类,因此您不需要整个库.您可以使用边界而不是设置缩放级别.结果是一样的.
代码显示正好1公里:
animateToMeters(1000);
private void animateToMeters(int meters){
int mapHeightInDP = 200;
Resources r = getResources();
int mapSideInPixels = (int) TypedValue.applyDimension(TypedValue.COMPLEX_UNIT_DIP, mapHeightInDP, r.getDisplayMetrics());
LatLng point = new LatLng(0, 0);
LatLngBounds latLngBounds = calculateBounds(point, meters);
if(latLngBounds != null){
cameraUpdate = CameraUpdateFactory.newLatLngBounds(latLngBounds, mapSideInPixels, mapSideInPixels, MARKER_BOUNDS);
if(mMap != null)
mMap.animateCamera(cameraUpdate);
}
}
private LatLngBounds calculateBounds(LatLng center, double radius) {
return new LatLngBounds.Builder().
include(SphericalUtil.computeOffset(center, radius, 0)).
include(SphericalUtil.computeOffset(center, radius, 90)).
include(SphericalUtil.computeOffset(center, radius, 180)).
include(SphericalUtil.computeOffset(center, radius, 270)).build();
}
从lib中提取(稍微改变)的类:
public class SphericalUtil {
static final double EARTH_RADIUS = 6371009;
/**
* Returns hav() of distance from (lat1, lng1) to (lat2, lng2) on the unit sphere.
*/
static double havDistance(double lat1, double lat2, double dLng) {
return hav(lat1 - lat2) + hav(dLng) * cos(lat1) * cos(lat2);
}
/**
* Returns haversine(angle-in-radians).
* hav(x) == (1 - cos(x)) / 2 == sin(x / 2)^2.
*/
static double hav(double x) {
double sinHalf = sin(x * 0.5);
return sinHalf * sinHalf;
}
/**
* Computes inverse haversine. Has good numerical stability around 0.
* arcHav(x) == acos(1 - 2 * x) == 2 * asin(sqrt(x)).
* The argument must be in [0, 1], and the result is positive.
*/
static double arcHav(double x) {
return 2 * asin(sqrt(x));
}
private SphericalUtil() {}
/**
* Returns the heading from one LatLng to another LatLng. Headings are
* expressed in degrees clockwise from North within the range [-180,180).
* @return The heading in degrees clockwise from north.
*/
public static double computeHeading(LatLng from, LatLng to) {
// http://williams.best.vwh.net/avform.htm#Crs
double fromLat = toRadians(from.latitude);
double fromLng = toRadians(from.longitude);
double toLat = toRadians(to.latitude);
double toLng = toRadians(to.longitude);
double dLng = toLng - fromLng;
double heading = atan2(
sin(dLng) * cos(toLat),
cos(fromLat) * sin(toLat) - sin(fromLat) * cos(toLat) * cos(dLng));
return wrap(toDegrees(heading), -180, 180);
}
/**
* Returns the LatLng resulting from moving a distance from an origin
* in the specified heading (expressed in degrees clockwise from north).
* @param from The LatLng from which to start.
* @param distance The distance to travel.
* @param heading The heading in degrees clockwise from north.
*/
public static LatLng computeOffset(LatLng from, double distance, double heading) {
distance /= EARTH_RADIUS;
heading = toRadians(heading);
// http://williams.best.vwh.net/avform.htm#LL
double fromLat = toRadians(from.latitude);
double fromLng = toRadians(from.longitude);
double cosDistance = cos(distance);
double sinDistance = sin(distance);
double sinFromLat = sin(fromLat);
double cosFromLat = cos(fromLat);
double sinLat = cosDistance * sinFromLat + sinDistance * cosFromLat * cos(heading);
double dLng = atan2(
sinDistance * cosFromLat * sin(heading),
cosDistance - sinFromLat * sinLat);
return new LatLng(toDegrees(asin(sinLat)), toDegrees(fromLng + dLng));
}
/**
* Returns the location of origin when provided with a LatLng destination,
* meters travelled and original heading. Headings are expressed in degrees
* clockwise from North. This function returns null when no solution is
* available.
* @param to The destination LatLng.
* @param distance The distance travelled, in meters.
* @param heading The heading in degrees clockwise from north.
*/
public static LatLng computeOffsetOrigin(LatLng to, double distance, double heading) {
heading = toRadians(heading);
distance /= EARTH_RADIUS;
// http://lists.maptools.org/pipermail/proj/2008-October/003939.html
double n1 = cos(distance);
double n2 = sin(distance) * cos(heading);
double n3 = sin(distance) * sin(heading);
double n4 = sin(toRadians(to.latitude));
// There are two solutions for b. b = n2 * n4 +/- sqrt(), one solution results
// in the latitude outside the [-90, 90] range. We first try one solution and
// back off to the other if we are outside that range.
double n12 = n1 * n1;
double discriminant = n2 * n2 * n12 + n12 * n12 - n12 * n4 * n4;
if (discriminant < 0) {
// No real solution which would make sense in LatLng-space.
return null;
}
double b = n2 * n4 + sqrt(discriminant);
b /= n1 * n1 + n2 * n2;
double a = (n4 - n2 * b) / n1;
double fromLatRadians = atan2(a, b);
if (fromLatRadians < -PI / 2 || fromLatRadians > PI / 2) {
b = n2 * n4 - sqrt(discriminant);
b /= n1 * n1 + n2 * n2;
fromLatRadians = atan2(a, b);
}
if (fromLatRadians < -PI / 2 || fromLatRadians > PI / 2) {
// No solution which would make sense in LatLng-space.
return null;
}
double fromLngRadians = toRadians(to.longitude) -
atan2(n3, n1 * cos(fromLatRadians) - n2 * sin(fromLatRadians));
return new LatLng(toDegrees(fromLatRadians), toDegrees(fromLngRadians));
}
/**
* Returns the LatLng which lies the given fraction of the way between the
* origin LatLng and the destination LatLng.
* @param from The LatLng from which to start.
* @param to The LatLng toward which to travel.
* @param fraction A fraction of the distance to travel.
* @return The interpolated LatLng.
*/
public static LatLng interpolate(LatLng from, LatLng to, double fraction) {
// http://en.wikipedia.org/wiki/Slerp
double fromLat = toRadians(from.latitude);
double fromLng = toRadians(from.longitude);
double toLat = toRadians(to.latitude);
double toLng = toRadians(to.longitude);
double cosFromLat = cos(fromLat);
double cosToLat = cos(toLat);
// Computes Spherical interpolation coefficients.
double angle = computeAngleBetween(from, to);
double sinAngle = sin(angle);
if (sinAngle < 1E-6) {
return from;
}
double a = sin((1 - fraction) * angle) / sinAngle;
double b = sin(fraction * angle) / sinAngle;
// Converts from polar to vector and interpolate.
double x = a * cosFromLat * cos(fromLng) + b * cosToLat * cos(toLng);
double y = a * cosFromLat * sin(fromLng) + b * cosToLat * sin(toLng);
double z = a * sin(fromLat) + b * sin(toLat);
// Converts interpolated vector back to polar.
double lat = atan2(z, sqrt(x * x + y * y));
double lng = atan2(y, x);
return new LatLng(toDegrees(lat), toDegrees(lng));
}
/**
* Returns distance on the unit sphere; the arguments are in radians.
*/
private static double distanceRadians(double lat1, double lng1, double lat2, double lng2) {
return arcHav(havDistance(lat1, lat2, lng1 - lng2));
}
/**
* Returns the angle between two LatLngs, in radians. This is the same as the distance
* on the unit sphere.
*/
static double computeAngleBetween(LatLng from, LatLng to) {
return distanceRadians(toRadians(from.latitude), toRadians(from.longitude),
toRadians(to.latitude), toRadians(to.longitude));
}
/**
* Returns the distance between two LatLngs, in meters.
*/
public static double computeDistanceBetween(LatLng from, LatLng to) {
return computeAngleBetween(from, to) * EARTH_RADIUS;
}
/**
* Returns the length of the given path, in meters, on Earth.
*/
public static double computeLength(List<LatLng> path) {
if (path.size() < 2) {
return 0;
}
double length = 0;
LatLng prev = path.get(0);
double prevLat = toRadians(prev.latitude);
double prevLng = toRadians(prev.longitude);
for (LatLng point : path) {
double lat = toRadians(point.latitude);
double lng = toRadians(point.longitude);
length += distanceRadians(prevLat, prevLng, lat, lng);
prevLat = lat;
prevLng = lng;
}
return length * EARTH_RADIUS;
}
/**
* Returns the area of a closed path on Earth.
* @param path A closed path.
* @return The path's area in square meters.
*/
public static double computeArea(List<LatLng> path) {
return abs(computeSignedArea(path));
}
/**
* Returns the signed area of a closed path on Earth. The sign of the area may be used to
* determine the orientation of the path.
* "inside" is the surface that does not contain the South Pole.
* @param path A closed path.
* @return The loop's area in square meters.
*/
public static double computeSignedArea(List<LatLng> path) {
return computeSignedArea(path, EARTH_RADIUS);
}
/**
* Returns the signed area of a closed path on a sphere of given radius.
* The computed area uses the same units as the radius squared.
* Used by SphericalUtilTest.
*/
static double computeSignedArea(List<LatLng> path, double radius) {
int size = path.size();
if (size < 3) { return 0; }
double total = 0;
LatLng prev = path.get(size - 1);
double prevTanLat = tan((PI / 2 - toRadians(prev.latitude)) / 2);
double prevLng = toRadians(prev.longitude);
// For each edge, accumulate the signed area of the triangle formed by the North Pole
// and that edge ("polar triangle").
for (LatLng point : path) {
double tanLat = tan((PI / 2 - toRadians(point.latitude)) / 2);
double lng = toRadians(point.longitude);
total += polarTriangleArea(tanLat, lng, prevTanLat, prevLng);
prevTanLat = tanLat;
prevLng = lng;
}
return total * (radius * radius);
}
/**
* Returns the signed area of a triangle which has North Pole as a vertex.
* Formula derived from "Area of a spherical triangle given two edges and the included angle"
* as per "Spherical Trigonometry" by Todhunter, page 71, section 103, point 2.
* See http://books.google.com/books?id=3uBHAAAAIAAJ&pg=PA71
* The arguments named "tan" are tan((pi/2 - latitude)/2).
*/
private static double polarTriangleArea(double tan1, double lng1, double tan2, double lng2) {
double deltaLng = lng1 - lng2;
double t = tan1 * tan2;
return 2 * atan2(t * sin(deltaLng), 1 + t * cos(deltaLng));
}
/**
* Wraps the given value into the inclusive-exclusive interval between min and max.
* @param n The value to wrap.
* @param min The minimum.
* @param max The maximum.
*/
static double wrap(double n, double min, double max) {
return (n >= min && n < max) ? n : (mod(n - min, max - min) + min);
}
/**
* Returns the non-negative remainder of x / m.
* @param x The operand.
* @param m The modulus.
*/
static double mod(double x, double m) {
return ((x % m) + m) % m;
}
}
小智 9
Google地图似乎与英里/像素紧密配合.在zoom = 13,1英里= 100像素.2英里= 200像素.每个变焦杆增加或减少2倍.因此,在变焦14,1英里= 200像素和变焦12,1英里= 50像素.
我已经将接受的答案转换为返回双倍值,因为Android Google地图库使用浮点缩放级别,并且还考虑了远离赤道的纬度.
public static double getZoomForMetersWide (
final double desiredMeters,
final double mapWidth,
final double latitude )
{
final double latitudinalAdjustment = Math.cos( Math.PI * latitude / 180.0 );
final double arg = EQUATOR_LENGTH * mapWidth * latitudinalAdjustment / ( desiredMeters * 256.0 );
return Math.log( arg ) / Math.log( 2.0 );
}
另外,为了获得Android上的最佳效果,不要传递视图的实际像素数,而是按照设备的像素密度缩放尺寸.
DisplayMetrics metrics = getResources().getDisplayMetrics();
float mapWidth = mapView.getWidth() / metrics.scaledDensity;
希望这有助于某人.
最终工作解决方案：
public static void getZoomForMetersWide(GoogleMap googleMap, int mapViewWidth, LatLng latLngPoint, int desiredMeters) {
DisplayMetrics metrics = App.getAppCtx().getResources().getDisplayMetrics();
float mapWidth = mapViewWidth / metrics.density;
final int EQUATOR_LENGTH = 40075004;
final int TIME_ANIMATION_MILIS = 1500;
final double latitudinalAdjustment = Math.cos(Math.PI * latLngPoint.latitude / 180.0);
final double arg = EQUATOR_LENGTH * mapWidth * latitudinalAdjustment / (desiredMeters * 256.0);
double valToZoom = Math.log(arg) / Math.log(2.0);
googleMap.animateCamera(CameraUpdateFactory.newLatLngZoom(latLngPoint, Float.valueOf(String.valueOf(valToZoom))), TIME_ANIMATION_MILIS , null);
}
ps 使用@sho 回答和@Lionel Briand 评论
使用循环来计算缩放级别非常幼稚。使用数学要好得多。
这里的函数（返回类型：float）
public static double calcZoom(int visible_distance, int img_width)
{
// visible_distance -> in meters
// img_width -> in pixels
visible_distance = Math.abs(visible_distance);
double equator_length = 40075016; // in meters
// for an immage of 256 pixel pixel
double zoom256 = Math.log(equator_length/visible_distance)/Math.log(2);
// adapt the zoom to the image size
int x = (int) (Math.log(img_width/256)/Math.log(2));
double zoom = zoom256 + x;
return zoom;
}
示例调用：
public static void main(String[] args)
{
// computes the zoom for 1km=1000m for an image having 256 width
double zoom = MainClass.calcZoom(1000, 256);
System.out.println("zoom: " + String.valueOf(zoom));
return;
}
计算缩放级别的数学公式为：
equator_length = 40075016
zoom_level = logE(equator_length/distance)/logE(2) + logE(img_width/256)/logE(2)
// The zoom_level computed here is a float number.
那是所有人！:-)
注意：上面的解决方案作为可接受的答案仅适用于赤道旁的缩放级别。如果您想要一个适用于所有纬度的解决方案，则需要与要计算的缩放级别相同纬度的平行线长度。该calcZoom方法的变化
private double calcZoom(int visible_distance, int img_width, double atLatitude) {
// visible_distance -> in meters
// img_width -> in pixels
double parallel_length = this.calcParallelLegth(atLatitude); // in meters
// for an immage of 256 pixel pixel
zoom256 = Math.log(parallel_length/visible_distance))/Math.log(2)
// adapt the zoom to the image size
x = (int) Math.log(img_width/256)/Math.log(2)
zoom = zoom256 + x
return zoom;
}
此处this.calcParallelLegth(atLatitude)表示atLatitude纬度上平行线的长度。
您可以使用一些库自己计算长度（最好使用Vincenty公式）。
或者
如果你没有这样的库（或者您不搜索库，或者你只是想要一个完整的代码作品）在这个答案的底部，你可以找到整个工作代码与一个实现double calcParallelLegth(double atLatitude)了使用表格（使用Vincenty公式计算），在所有纬度上具有平行长度，且倾斜度为3％。
注意：如果
您需要阅读下面的内容，请理解以下内容（或检查该公式是否正确）
公式说明如下：
简而言之！
让我们将问题分为两部分。
第1部分
计算256x256尺寸图像的缩放
第2部分
适用于不同尺寸图像的缩放
解决第1部分的
图像大小为256x256。缩放级别0显示整个赤道。
每个后续的缩放级别让我看到之前的一半。
赤道长40,075,016米（根据WGS-84 （* 1）和Vincenty公式（* 2））
zoom=0 -> 40,075,016 / 1 = 40,075,016 meters visible Note: 2^0=1
zoom=1 -> 40,075,016 / 2 = 20,037,508 meters visible Note: 2^1=2
zoom=2 -> 40,075,016 / 4 = 10,018,754 meters visible Note: 2^2=4
zoom=3 -> 40,075,016 / 8 = 5,009,377 meters visible Note: 2^3=8
zoom=4 -> 40,075,016 / 16 = 2,504,688.5 meters visible Note: 2^4=16
zoom=5 -> 40,075,016 / 2^5 = 1,252,344.25 meters visible Note= 2^5=32
zoom=6 -> 40,075,016 / 2^6 = 636,172.125 meters visible Note= 2^6=64
...
zoom -> equator_length / 2^zoom = visible_distance
如上所示，每个后续缩放级别让我看到的是之前的一半。
zoom是所需的zoom_level。
visible_distance是图像水平显示多少米。
如果要比1公里，则必须使用visible_distance = 1000计算缩放
让我们找出缩放公式。
这是数学做魔术的地方（“无聊的”魔术东西）。
equator_length / 2^zoom = visible_distance ->
-> equator_length / visible_distance = 2^zoom ->
-> log2(equator_length / visible_distance) = log2(2^zoom) -> (*3)
-> log2(equator_length / visible_distance) = zoom*log2(2) -> (*4)
-> log2(equator_length / visible_distance) = zoom*1 -> (*5)
-> log2(equator_length / visible_distance) = zoom ->
-> logE(equator_length / visible_distance)/logE(2) = zoom -> (*6)
256x256图像的缩放级别公式为：
zoom256 = logE(equator_length/visible_distance) / logE(2)
第1部分完成！
解决第2部分
使缩放适应所需的图像尺寸。
每当图像宽度加倍时，看到整个赤道所需的变焦都会增加一个。
示例：
在图像512x512中，看到整个赤道所需的缩放比例是1。在图像1024x1024中，要看到整个赤道所需的缩放比例是2。在图像2048x2048中，看到整个赤道所需的缩放比例是3。
那说
width= 256 -> 256/256 = 1 -> zoom=0 (needed to see the whole equator)
width= 512 -> 512/256 = 2 -> zoom=1 (needed to see the whole equator)
width=1024 -> 1024/256 = 4 -> zoom=2 (needed to see the whole equator)
width=2048 -> 2048/256 = 8 -> zoom=3 (needed to see the whole equator)
width=4096 -> 4096/256 = 2^4 -> zoom=4 (needed to see the whole equator)
width=4096 -> 4096/256 = 2^5 -> zoom=5 (needed to see the whole equator)
... width-> width / 256 = 2 ^ x-> zoom = x（需要看到整个赤道）
这意味着（zoom_level是
- with an 512x512 image, the zoom needed is zoom256+1
- with an 1024x1024 image, the zoom needed is zoom256+2
- with an 2048x2048 image, the zoom needed is zoom256+3
...
- with an WIDTHxHEIGHT image, the zoom needed is zoom256+x
需要x来适应缩放所需的图像大小。
因此，这是从中提取x的问题
width/256 = 2^x
我们开始做吧
width/256 = 2^x ->
-> log2(width/256) = log2(2^x) -> (*3)
-> log2(width/256) = x * log2(2) -> (*4)
-> log2(width/256) = x * 1 -> (*5)
-> log2(width/256) = x ->
-> logE(width/256) / logE(2) = x -> (*6)
现在我们有了x公式。
WIDTHxHEIGHT图像的缩放级别公式为：
zoom = zoom256 + x
因此，如果您想在512x512图像中看到1公里，
zoom256 = logE(40075016/1000) / logE(2) = 15.29041547592718
x = logE(512/256) / logE(2) = 1
zoom = zoom256 + z = 15.29041547592718 + 1 = 16.29041547592718
如果必须是整数
zoom = floor(zoom) = 16
完成！
(*3) expr1=expr2 <-> log(expr1)=log(expr2)
(*4) logN(a^b) = b * logN(a)
(*5) logN(N) = 1
(*6) logN(expr) = log(expr)/log(N)
(*7) log(a/b) = log(a) - log(b)
这是计算每个纬度图像宽度下的缩放级别的完整代码。
class MainClass
{
public static int getParallelLength(double atLatitude)
{
int FR_LAT = 0; // from latitude
int TO_LAT = 1; // to latidude
int PA_LEN = 2; // parallel length in meters)
int PC_ERR = 3; // percentage error
// fr_lat| to_lat | par_len| perc_err
double tbl[][] = {
{ 0.00, 12.656250000000000, 40075016, 2.410},
{12.66, 17.402343750000000, 39107539, 2.180},
{17.40, 22.148437500000000, 38252117, 2.910},
{22.15, 25.708007812500000, 37135495, 2.700},
{25.71, 28.377685546875000, 36130924, 2.330},
{28.38, 31.047363281250000, 35285940, 2.610},
{31.05, 33.717041015625000, 34364413, 2.890},
{33.72, 35.719299316406250, 33368262, 2.380},
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