多少距离0.001 ^ \保监会在经度或纬度美元代表什么?- 江南体育网页版- - - - -地球科学堆江南电子竞技平台栈交换 最近30从www.hoelymoley.com 2023 - 07 - 07 - t23:36:38z //www.hoelymoley.com/feeds/question/6843 https://creativecommons.org/licenses/by-sa/4.0/rdf //www.hoelymoley.com/q/6843 14 多少距离0.001 ^ \保监会在经度或纬度美元代表什么? 马克·索恩 //www.hoelymoley.com/users/4744 2015 - 11 - 04 - t05:59:50z 2022 - 06 - 18岁t06:13:37z < p >如果我站在纬度30.000 ^ \保监会和长30.000美元$ ^ \保监会$,我搬到一个纬度30.001 ^ \保监会和长30.000美元$ ^ \保监会$,我走了多远?有小数点后和纬度之间的关系和长途吗? < / p > //www.hoelymoley.com/questions/6843/how -多-距离- - 0 - 001 -中国保监会在经度纬度-或- represent/6844 # 6844 9 答案由f。索普多少距离为0.001美元美元^ \保监会在经度或纬度代表什么? f.thorpe //www.hoelymoley.com/users/543 2015 - 11 - 04 - t06:16:21z 2020 - 12 - 21 - t22:18:23z

Latitudes are "parallels" while longitudes are "meridians" that all meet at the poles. For latitude, there is a set distance traveled per degree latitude no matter where you are on a spherical globe. For longitude, it depends what latitude you are at.

Lines of Latitude and Longitude
(source: learner.org)

.......................... Latitude .............. Longitude

Image source: https://www.learner.org/jnorth/tm/LongitudeIntro.html

Image Credit: Illinois State University

Here is a nice calculator for you online: http://www.stevemorse.org/nearest/distance.php

1 degree of latitude in physical distance is 68.94 statute miles or 59.91 nautical miles (110.95 km) -- for a spherical earth assumption. So, a change of 0.001 degrees latitude is 0.06894 statute miles or 0.05991 nautical miles. Though, an ellipsoidal earth approximation does have small variance in latitudinal distances. Incidentally, longitude has the same distance between each degree if you are at the equator. At the poles, the distance between lines of longitude is zero. Note that all distances are "as the crow flies" which means the terrain is disregarded in the distance calculation.

 //www.hoelymoley.com/questions/6843/how -多-距离- - 0 - 001 -中国保监会在经度纬度-或- represent/6846 # 6846 20. 回答由Eubie画多少距离0.001 ^ \保监会在经度或纬度美元代表什么? Eubie画 //www.hoelymoley.com/users/4487 2015 - 11 - 04 - t07:27:33z 2015 - 11 - 04 - t19:01:31z < p >我不明白最后一句话小数点后,但我可以告诉你关于之间的关系< em > lat < / em >, < em > < / em >和距离。< / p > < p >在两个世纪前,< a href = " http://en.wikipedia.org/wiki/Metre子午" >仪表被定义为10000000(1/10 000 000)的一个象限沿着地球子午线的长度;也就是说,从赤道到北极距离< / >。因此,对于< em > < / em >纬度度从极点到赤道的数量是90美元^ \保监会\ !美元,和米的数量是1000万(或10000公里)。这意味着美元1 ^ \保监会\ !美元的< em > < / em >是纬度10000/90 = 111公里,美元和0.001美元^ \保监会= 0.111公里或111美元美元米,本质上是一个美式橄榄球场+两个开始。< / p > < p >赤道的总长度约等于四次,从极点到赤道的距离。(略,因为地球的图有点扁,像一个稍扁球。)想到一个大圆从北极到赤道,南极,赤道的对面,然后回到北极(360美元^ \保监会\ \ !一个圆的美元)。这将是(几乎)相当于大圆(360美元^ \保监会\ \ !美元的< em >经度赤道的< / em >)。所以,在赤道,^ \保监会\ 0.001美元!$ of longitude will also be $111$ meters. But...

The other circles of latitude are smaller than the great circle of the Equator, yet still have $360^\circ\!$, so those degrees cover less than $111$ kilometers.

enter image description here

In fact, the size of a degree of longitude as a function of latitude scales as the cosine of the latitude. So $0.001^\circ\!$ of longitude change at $30^\circ\!$ latitude would be $111 \times \cos (30^\circ\!) = 111 \times 0.866 = 96$ meters.

enter image description here

 //www.hoelymoley.com/questions/6843/how -多-距离- - 0 - 001 -中国保监会在经度纬度-或- represent/23932 # 23932 1 回答由fischertranscripts多少距离0.001 ^ \保监会在经度或纬度美元代表什么? fischertranscripts //www.hoelymoley.com/users/27036 2022 - 06 - 18岁t06:13:37z 2022 - 06 - 18岁t06:13:37z

I am using "PARI" a free open source math program for Mac or PC, for these exact (vs GRS80 Ellipsoid) equations: "longscale"gives meters per (second of Longitude) at a constant latitude, so below I multiply it by "3.6" to get vs 0.001 degrees as requested by the original poster:

re=6378137;\ flattening=298.2572221008827112431628366;\ a=(1-1/flattening)^2;\ reh= (put 6378137 meters plus your height above the ellipsoid here) longscale(u)=1/sqrt(1 + a *(1/(cos(u))^2 - 1))*reh*2*Pi/60^4*10; longscale(((00/60+00)/60+30)*Pi/180)*3.6 %452 = 96.4862802512923113

Latitude is harder because as the arc gets bigger the scale changes, whereas scale stays exactly the same along constant latitudes. So say the OP also wanted the distance from 30degrees latitude to 30.001. Then either of the following integrals give the same answer, but might have different speeds or accuracy in your computer language:

fl=flattening;\ ma1(lat)=e2=1-(1-1/fl)^2;\ (1-1/fl)^2*intnum(th=0,lat,1/(1-e2*(sin(th))^2)^(3/2))*reh ma4(lat)=e2=1-(1-1/fl)^2;\ (intnum(th=0,lat,sqrt(1-e2*(sin(th))^2));\ -e2/2*sin(2*lat)/sqrt(1-e2*(sin(lat))^2))*reh

Thus the distance is the meridian arc length of the second point minus the meridian arc length of the first:

lat1=((00/60+00)/60+30)*Pi/180;\ lat2=((3.6/60+00)/60+30)*Pi/180;\ ma4(lat2)-ma4(lat1) cpu time = 195 ms, real time = 196 ms. %456 = 110.852450919981499

If he likewise wanted the scale at 30 degrees Latitude in meters per (second of Latitude), then differentiate the equation above like this:

lat1=((00/60+00)/60+30)*Pi/180;\ derivnum(lat=lat1,ma4(lat))*Pi/60^4*20 cpu time = 648 ms, real time = 649 ms. %455 = 30.7923451370347380
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