纬度和经度的最大长度是多少？ [英] What is the maximum length of latitude and longitude?
问题描述
我的Windows Phone设备发送的长度很长：
纬度= -63572375290155
经度= 106744840359415
这超出了我的表列大小，我收到错误。
纬度的有效范围是以度为单位，分别为南半球和北半球，强度为-90和+90 。经度位于范围 -180和+180 ，分别指定主子午线西和东的坐标。
为了参考， Equator 的纬度为0°，北极北纬90度（写90°或+ 90°），而南极的纬度为-90°。
Prime Meridian 经度为0°，通过英格兰格林威治。 国际日期行（IDL）大致遵循180°经度。东半球的经度为正值，西半球的负值下降。
小数精度
使用十进制度数符号的坐标中的六（6）位小数位精度为10厘米（或0.1米）分辨率。坐标十进制度的每个.000001差距约为10厘米。例如，Google地球和Google地图的图像通常为1米分辨率，某些地方的分辨率为每像素1英寸。可以使用5位小数表示一米分辨率，因此该分辨率超过6位小数位是无关紧要的。赤道经度之间的距离与纬度相同，经线之间的距离在两极之间达到零，因为经线在该点收敛。
如果纬度报告为-6.3572375290155或-63.572375290155，那么您可以舍入并存储最多6位小数位，为10厘米（或0.1米）精度。
对于毫米（mm）精度，则以十进制格式表示具有8位小数位的lat / lon。由于大多数应用程序不需要精确度，所以在大多数情况下，6位小数就足够了。
另一方面，整数十进制度代表约111公里的距离（或60海里），0.1十进制差表示约11平方公里的面积。
这里是一个＃度数和以米为单位的估计距离，以0,0为起点。
十进制小数距离
放置度（以米计）
------- --------- -----------
1 0.1000000 11,057.43 11 km
2 0.0100000 1,105.74 1公里
3 0.0010000 110.57
4 0.0001000 11.06
5 0.0000100 1.11
6 0.0000010 0.11 11厘米
7 0.0000001 0.01 1厘米
度数分秒秒（DMS）表示
对于DMS符号1弧秒= 1/60/60度=〜30米长，0.1弧秒增量为〜3米。
示例：
- 0°0'0W，0°0'0N - > 0°0'0W，0°0' 1N => 30.715米
- 0°0'0W，0°0'0N - > 0°0'0W，0°0'0.1N => 3.0715米
1弧分= 1/60度=〜2000米（2公里）
How long can latitude and longitude be?
I am getting very long lengths sent by a Windows Phone device:
Latitude=-63572375290155
Longitude=106744840359415
This is exceeding my table column size and I am getting errors.
The valid range of latitude in degrees is -90 and +90 for the southern and northern hemisphere respectively. Longitude is in the range -180 and +180 specifying coordinates west and east of the Prime Meridian, respectively.
For reference, the Equator has a latitude of 0°, the North pole has a latitude of 90° north (written 90° N or +90°), and the South pole has a latitude of -90°.
The Prime Meridian has a longitude of 0° that goes through Greenwich, England. The International Date Line (IDL) roughly follows the 180° longitude. A longitude with a positive value falls in the eastern hemisphere and negative value falls in the western hemisphere.
Decimal degrees precision
Six (6) decimal places precision in coordinates using decimal degrees notation is at a 10 cm (or 0.1 meter) resolution. Each .000001 difference in coordinate decimal degree is approximately 10 cm in length. For example, the imagery of Google Earth and Google Maps is typically at the 1 meter resolution, and some places have the highest resolution of 1 inch per pixel. One meter resolution can be represented using 5 decimal places so more than 6 decimal places is extraneous for that resolution. The distance between longitudes at the equator is the same as latitude, but the distance between longitudes reaches zero at the poles as the lines of meridian converge at that point.
If Latitude value is reported as -6.3572375290155 or -63.572375290155 then you could round-off and store up to 6 decimal places for 10 cm (or 0.1 meter) precision.
For millimeter (mm) precision then represent lat/lon with 8 decimal places in decimal degrees format. Since most applications don't need that level of precision 6 decimal places is sufficient for most cases.
In the other direction, whole decimal degrees represents a distance of ~111 km (or 60 nautical miles) and a 0.1 decimal degree difference represents a ~11 sq. Km area.
Here is a table of # decimal places difference in latitude with the delta degrees and the estimated distance in meters using 0,0 as the starting point.
decimal decimal distance
places degrees (in meters)
------- --------- -----------
1 0.1000000 11,057.43 11 km
2 0.0100000 1,105.74 1 km
3 0.0010000 110.57
4 0.0001000 11.06
5 0.0000100 1.11
6 0.0000010 0.11 11 cm
7 0.0000001 0.01 1 cm
Degrees-minute-second (DMS) representation
For DMS notation 1 arc second = 1/60/60 degree = ~30 meter length and 0.1 arc sec delta is ~3 meters.
Example:
- 0° 0' 0" W, 0° 0' 0" N -> 0° 0' 0" W, 0° 0' 1" N => 30.715 meters
- 0° 0' 0" W, 0° 0' 0" N -> 0° 0' 0" W, 0° 0' 0.1" N => 3.0715 meters
1 arc minute = 1/60 degree = ~2000m (2km)
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