Lat/Long Javascript 中的等式

Question

我正在尝试找到一个方程来找到 Javascript 中两个 lat/long 点之间的 lat/long 点。它会像这样工作。

getMiddle(lat1, lng1, lat2, lng2) <= 将 return [lat3, lat3] 中途距离明智（显然绕地球转）。

我找到了这个：

http://mathforum.org/library/drmath/view/51822.html

Date: 10/11/2001 at 11:41:08 From: Doctor Rick Subject: Re: Determining lat and long of a point between two given points

Hi, Chuck.

This will probably be a little more complicated than you think. The easiest way for me to do it is to think in terms of vectors. Some of the items in our Archives on the topic of latitude and longitude use the vector approach, so you can see them for background. For example:

Distance Between Two Points on the Earth
http://mathforum.org/library/drmath/view/51722.html

Let's call the two points A and B, and choose a rectangular coordinate system in which the equator is in the x-y plane and the longitude of point A is in the x-z plane. Let lat1 be the latitude of A, let lat2 be the latitude of B, and let dlat be the longitude of B minus the longitude of A. Finally, use distance units such that the radius of the earth is 1. Then the vectors from the center of the earth to A and B are

A = (cos(lat1), 0, sin(lat1)) B = (cos(lat2)*cos(dlon), cos(lat2)*sin(dlon), sin(lat2))

Point C, the midpoint of the shortest line between A and B, lies along the sum of vectors A and B. (This works because A and B have the same length, so the sum of the vectors is the diagonal of a rhombus, and this diagonal bisects the angle.)

A+B = (cos(lat1)+cos(lat2)*cos(dlon), cos(lat2)*sin(dlon), sin(lat1)+sin(lat2))

To get the actual vector C, we need to scale this vector to length R so it ends at the surface of the earth. Thus we have to divide it by |A+B|, that is, the length of vector A+B. That would get pretty messy. But we can find the latitude lat3 and longitude difference (lon3-lon1) by looking at ratios of the coordinates of A+B, and these ratios are the same whether we scale the vector or not. To see this, look back at the formula for vector B. Knowing that vector, we can recover lat2 and dlon:

dlon = tan^-1(B_y/B_x) lat2 = tan^-1(B_z/sqrt(B_x^2+B_y^2))

Here, B_x, B_y, and B_z are the x, y, and z coordinates of vector B.

We can do the same thing with vector A+B to find the latitude and longitude of point C:

dlon3 = tan^-1(cos(lat2)*sin(dlon)/ (cos(lat1)+cos(lat2)*cos(dlon))) lat3 = tan^-1((sin(lat1)+sin(lat2))/ sqrt((cos(lat1)+cos(lat2)*cos(dlon))^2+ (cos(lat2)*sin(dlon))^2))

That's the formula you seek. Since both formulas involve division, we must consider the special cases. The longitude calculation fails when C_x = 0, that is, when C is 90 degrees away from A in longitude, so dlon3 will be +90 or -90; the sign depends on the sign of dlon. The latitude calculation fails when C_x and C_y are both zero, thus we know that in this case, lat3 = 0. A complete algorithm will check for these cases, but they won't occur if you're interested only in the continental US.

When I plug in the latitudes and longitudes for LA and NYC, I get:

LA 34.122222 118.4111111 NYC 40.66972222 73.94388889 midpt 39.54707861 97.201534

I hope this is helpful to you.

Doctor Rick, The Math Forum http://mathforum.org/dr.math/

我认为有答案但不是 Javascript。

这就是我正在使用的：

我花了一些时间创建地图点并确定它们之间的距离，但要找到它们之间的中间点对我来说很难。到目前为止，我发现的转换方程式一直没有成功。

找到中点后，我想使用递归 +（根据距离在两点之间需要多少点）来构建适当间隔的虚线。

如能对本文提供任何帮助，我们将不胜感激。

Answer 1

我认为这就是您要查找的公式：

这是两点之间沿大圆路径的中点。

Formula:    Bx = cos φ2 ⋅ cos Δλ
By = cos φ2 ⋅ sin Δλ
φm = atan2( sin φ1 + sin φ2, √(cos φ1 + Bx)² + By² )
λm = λ1 + atan2(By, cos(φ1)+Bx)

JavaScript:
(all angles 
in radians)
var Bx = Math.cos(φ2) * Math.cos(λ2-λ1);
var By = Math.cos(φ2) * Math.sin(λ2-λ1);
var φ3 = Math.atan2(Math.sin(φ1) + Math.sin(φ2),
                    Math.sqrt( (Math.cos(φ1)+Bx)*(Math.cos(φ1)+Bx) + By*By ) );
var λ3 = λ1 + Math.atan2(By, Math.cos(φ1) + Bx);

来源：http://www.movable-type.co.uk/scripts/latlong.html

Answer 2

python中的答案：http://code.activestate.com/recipes/577713-midpoint-of-two-gps-points/

转换为 js
利润

另外：对于那些想直接获利的人（是的，我知道我修补了数学。无耻地从这里借来的：https://nickthecoder.wordpress.com/2012/04/15/radian-and-degree-conversion-in-javascript/）

getMidpoint: function(lat1, lng1, lat2, lng2) {
    Math.degrees = function(rad) {
        return rad * (180 / Math.PI);
    }
    Math.radians = function(deg) {
        return deg * (Math.PI / 180);
    }
    lat1 = Math.radians(lat1);
    lng1 = Math.radians(lng1);
    lat2 = Math.radians(lat2);
    lng = Math.radians(lng2);
    bx = Math.cos(lat2) * Math.cos(lng - lng1)
    by = Math.cos(lat2) * Math.sin(lng - lng1)
    lat3 = Math.atan2(Math.sin(lat1) + Math.sin(lat2), Math.sqrt((Math.cos(lat1) + bx) * (Math.cos(lat1) + bx) + Math.pow(by, 2)));
    lon3 = lng1 + Math.atan2(by, Math.cos(lat1) + bx);
    return [Math.round(Math.degrees(lat3), 5), Math.round(Math.degrees(lon3), 5)]
}

Lat/Long Javascript 中的等式

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