给定一定的小数位数的经纬度，如何判断覆盖区域（以米为单位）？ [英] Given a latitude / longitude to a certain number of decimal places, how to tell the area covered (in metres)?

问题描述

如果我有一个经纬度位置指定的位置，那么它可以覆盖一个方框，具体取决于位置的准确位数。

例如，

例如，给定位置为55.0°N，3.0°W（即小数点后1位），假设截断（与舍入相反），这可以覆盖55.01°至55.09°的任何角度。这将覆盖此框中的区域： http://www.openstreetmap.org/?minlat=55.0&maxlat=55.1&maxlon=-3.0&minlon=-3.1&box=yes

无论如何计算那个盒子的面积？我只想这样做一次，所以提供此计算的简单网站就足够了。

我想这样做的主要原因是因为我有一个位置非常高的小数位数，我想看看它有多精确。

虽然地球并不完全是球形您可以将其视为这些计算。

南北计算相对简单，因为从极点到极点有180°，距离为20,014公里（< a href =http://wiki.answers.com/Q/What_is_the_distance_between_the_North_Pole_and_the_South_Pole =nofollow noreferrer>来源）所以一度== 20014/180 = 111.19公里。

东/西计算更加困难，因为它取决于纬度。赤道距离为40,076公里（来源），所以一度= 40076/360 = 111.32公里。两极的周长（根据定义）为0公里。所以你可以用三角函数（ circumference * sin（latitude））计算任意纬度的圆周。

If i have a position specified in latitude and longitude, then it can cover a box, depending on how many digits of accuracy are in the position.

For example, given a position of 55.0° N, 3.0° W (i.e. to 1 decimal place), and assuming a truncation (as opposed to rounding), this could cover anything that's 55.01° to 55.09°. This would cover the area in this box: http://www.openstreetmap.org/?minlat=55.0&maxlat=55.1&maxlon=-3.0&minlon=-3.1&box=yes

Is there anyway to calculate the area of that box? I only want to do this once, so a simple website that provides this calculation would suffice.

The main reason I want to do this is because I have a position to a very high number of decimal places, and I want to see how precise it is.

解决方案

While the Earth isn't exactly spherical you can treat it as such for these calculations.

The North/South calculation is relatively simple as there are 180° from pole to pole and the distance is 20,014 km (Source) so one degree == 20014/180 = 111.19 km.

The East/West calculation is more difficult as it depends on the latitude. The Equatorial distance is 40,076 km (Source) so one degree = 40076/360 = 111.32 km. The circumference at the poles is (by definition) 0 km. So you can calculate the circumference at any latitude by trigonometry (circumference * sin(latitude)).

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