<p>45E is directly opposite 135W.
It's therefore obvious (hopefully) that the shortest path is straight over the North pole (which is at 90 degrees North). The path takes us round 120 degrees of the Earth's circumference, or 1/3rd of it.</p>

CC BY-SA 3.0
2016-02-11T02:13:28.773Z
Nice. Not a general solution but probably why it was asked using those locations.
haresfur
CC BY-SA 3.0
2016-02-10T15:25:28.570Z
2016-02-10T15:25:28.570Z
Martin
<p>This is not about the Math involved but, the Question of what really is the Shortest distance between two points on the surface of a Sphere. But,what seems to be missing when explaining this topic is. There are &quot;two&quot; Geodesic arcs between any two points on the surface of a Sphere both equidistant and,a mirror image of each other unless, the points are 180 degrees apart  then it would be &quot;0ne&quot; path only over the curve. Thank You .</p>

CC BY-SA 4.0
2023-07-18T16:18:34.920Z
Welcome Mark. Can you clarify what is meant by "two geodesic arcs between 2 points on a surface both equidistant"? Are these two arcs the same length and the shortest distance between the points? (The shortest distance is along a great circle, of which I thought there was only 1 on the surface of a sphere.)
JohnHoltz
CC BY-SA 4.0
2023-07-18T15:06:07.583Z
2023-07-18T15:08:13.537Z
2023-07-18T15:08:13.537Z
Mark Halpin
<p>If you don't want to do the calculation yourself, you can use an online calculator like the one provided by <a href="http://www.nhc.noaa.gov/gccalc.shtml" rel="nofollow">NOAA</a>.</p>

<p>Alternatively, if you do want to do the calculation yourself you can use the <a href="http://www.movable-type.co.uk/scripts/latlong.html" rel="nofollow">haversine</a> formula. </p>

<blockquote>
  <p>This uses the ‘haversine’ formula to calculate the great-circle distance between two points – that is, the shortest distance over the earth’s surface – giving an ‘as-the-crow-flies’ distance between the points (ignoring any hills they fly over, of course!).</p>
  
  <p>$$a = \sin^2\left(\frac{\Delta\phi}{2}\right) + \cos \phi_1 ⋅ \cos \phi_2 ⋅ \sin^2\left(\frac{\Delta\lambda}{2}\right)$$</p>
  
  <p>$$c = 2 ⋅ \text{atan2}( \sqrt a, \sqrt {1−a} )$$</p>
  
  <p>$$d = R ⋅ c$$ </p>
  
  <p>where $\phi$ is latitude, $\lambda$ is longitude, $R$ is earth’s radius (mean radius = 6,371km)</p>
</blockquote>

CC BY-SA 3.0
2016-02-11T00:24:27.177Z
This also ignores the earth's obliquity, right? For example, if you wanted to get from 0°E to 180°E on the equator, it would be shorter to go via the poles than via the equator, but this algorithm wouldn't tell you that.
naught101
CC BY-SA 3.0
2016-02-11T18:50:18.887Z
@naught101 Yes, you're correct. The haversine formula treats the earth as a sphere, not an ellipsoid.
mkennedy
CC BY-SA 3.0
2016-02-10T07:44:04.150Z
2016-02-22T19:32:41.733Z
2016-02-22T19:32:41.733Z
Fred
<p>I am a Math student and don't know about longitude and latitude. This question was asked in CSIR-NET Mathematics. Please help me to solve it.</p>

<p>Two points $A$ and $B$ on the surface of the Earth have the following latitude and longitude co-ordinates:</p>

<p><strong>A:</strong> $30^\circ $ N , $ 45^\circ$ E</p>

<p><strong>B:</strong> $30^\circ $ N , $ 135^\circ$ W</p>

<p>If $R$ is the radius of the Earth, what is the length of the shortest path from $A$ and $B$?</p>

CC BY-SA 3.0
2016-02-10T14:14:15.503Z
This question was in "general aptitude" section. Is there is a simple way to calculate? I am wondering how a "maths student" can answer this question?
Learner
CC BY-SA 3.0
2016-02-11T02:12:57.827Z
You could transform the locations to spherical coordinates, then solve for the distance.
haresfur
CC BY-SA 4.0
2023-07-19T08:20:22.480Z
Its called the [great-circle distance](https://en.wikipedia.org/wiki/Great-circle_distance), of which the haversine distance or the cosine formula are approximations.
hschoell
CC BY-SA 4.0
2023-07-21T23:42:31.157Z
The Great Circle is used by the airlines and the military
TheOldRanger
<geography><homework><geometry>
What is the shortest path between two points on Earth's surface?
Learner
# What is the shortest path between two points on Earth's surface?

(Asked by: username_0 on 2016-02-10)

I am a Math student and don't know about longitude and latitude. This question was asked in CSIR-NET Mathematics. Please help me to solve it.

Two points $A$ and $B$ on the surface of the Earth have the following latitude and longitude co-ordinates:

**A:** $30^\circ $ N , $ 45^\circ$ E

**B:** $30^\circ $ N , $ 135^\circ$ W

If $R$ is the radius of the Earth, what is the length of the shortest path from $A$ and $B$?

**Tags:** geography, homework, geometry

> **Comments:**
> **username_0**: This question was in "general aptitude" section. Is there is a simple way to calculate? I am wondering how a "maths student" can answer this question?  
> **username_1**: You could transform the locations to spherical coordinates, then solve for the distance.  
> **username_2**: Its called the [great-circle distance](https://en.wikipedia.org/wiki/Great-circle_distance), of which the haversine distance or the cosine formula are approximations.  
> **username_3**: The Great Circle is used by the airlines and the military  

---

## 3 answers

### Answer by username_7 on 2016-02-10. Score: 15 | Accepted answer

If you don't want to do the calculation yourself, you can use an online calculator like the one provided by [NOAA](http://www.nhc.noaa.gov/gccalc.shtml).

Alternatively, if you do want to do the calculation yourself you can use the [haversine](http://www.movable-type.co.uk/scripts/latlong.html) formula.

> This uses the ‘haversine’ formula to calculate the great-circle distance between two points – that is, the shortest distance over the earth’s surface – giving an ‘as-the-crow-flies’ distance between the points (ignoring any hills they fly over, of course!).
>
> $$a = \sin^2\left(\frac{\Delta\phi}{2}\right) + \cos \phi\_1 ⋅ \cos \phi\_2 ⋅ \sin^2\left(\frac{\Delta\lambda}{2}\right)$$
>
> $$c = 2 ⋅ \text{atan2}( \sqrt a, \sqrt {1−a} )$$
>
> $$d = R ⋅ c$$
>
> where $\phi$ is latitude, $\lambda$ is longitude, $R$ is earth’s radius (mean radius = 6,371km)

> **Comments:**
> **username_8**: This also ignores the earth's obliquity, right? For example, if you wanted to get from 0°E to 180°E on the equator, it would be shorter to go via the poles than via the equator, but this algorithm wouldn't tell you that.  
> **username_9**: @username_8 Yes, you're correct. The haversine formula treats the earth as a sphere, not an ellipsoid.  

---

### Answer by username_4 on 2016-02-10. Score: 11

45E is directly opposite 135W.
It's therefore obvious (hopefully) that the shortest path is straight over the North pole (which is at 90 degrees North). The path takes us round 120 degrees of the Earth's circumference, or 1/3rd of it.

> **Comments:**
> **username_1**: Nice. Not a general solution but probably why it was asked using those locations.  

---

### Answer by username_5 on 2023-07-18. Score: 0

This is not about the Math involved but, the Question of what really is the Shortest distance between two points on the surface of a Sphere. But,what seems to be missing when explaining this topic is. There are "two" Geodesic arcs between any two points on the surface of a Sphere both equidistant and,a mirror image of each other unless, the points are 180 degrees apart then it would be "0ne" path only over the curve. Thank You .

> **Comments:**
> **username_6**: Welcome Mark. Can you clarify what is meant by "two geodesic arcs between 2 points on a surface both equidistant"? Are these two arcs the same length and the shortest distance between the points? (The shortest distance is along a great circle, of which I thought there was only 1 on the surface of a sphere.)  

---

