<p>Latitudes are &quot;parallels&quot; while longitudes are &quot;meridians&quot; 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.</p>
<p><a href="https://i.sstatic.net/BBmMe.gif" rel="nofollow noreferrer"><img src="https://i.sstatic.net/BBmMe.gif" alt="Lines of Latitude and Longitude" /></a><br />
<sub>(source: <a href="https://www.learner.org/jnorth/images/graphics/mclass/Lat_Long.gif" rel="nofollow noreferrer">learner.org</a>)</sub></p>
<p>.......................... <img src="https://i.sstatic.net/Ms2xq.gif" alt="Latitude" /> .............. <a href="https://i.sstatic.net/e2uRI.gif" rel="nofollow noreferrer"><img src="https://i.sstatic.net/e2uRI.gif" alt="Longitude" /></a></p>
<p>Image source: <a href="https://www.learner.org/jnorth/tm/LongitudeIntro.html" rel="nofollow noreferrer">https://www.learner.org/jnorth/tm/LongitudeIntro.html</a></p>
<p>Image Credit: Illinois State University</p>
<p>Here is a nice calculator for you online:
<a href="http://www.stevemorse.org/nearest/distance.php" rel="nofollow noreferrer">http://www.stevemorse.org/nearest/distance.php</a></p>
<p>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 <em>if you are at the equator</em>.  At the poles, the distance between lines of longitude is zero.  Note that all distances are &quot;as the crow flies&quot; which means the terrain is disregarded in the distance calculation.</p>

CC BY-SA 3.0
2015-11-04T07:30:34.913Z
That illustration of *latitude* is misleading because at that projection, parallels of *latitude* would NOT appear evenly spaced.
Eubie Drew
CC BY-SA 3.0
2015-11-04T10:21:01.633Z
As I remember from my mathematics classes, in curved space, longitude lines are actually parallel and latitude lines aren't even straight lines.   That said, the more common definition is what farrenthrope said.   Latitude lines are called Parallel and Longitude are called meridians that meet at the poles.
userLTK
CC BY-SA 3.0
2015-11-04T10:26:00.460Z
Globe with the apparent uneven spacing if interested, per Aabaakawad's comment:   http://www.geolounge.com/wp-content/uploads/2014/08/latitude.png
userLTK
CC BY-SA 3.0
2015-11-05T05:35:53.200Z
Thank you everyone for your comments on the original image used.  I have replaced it with a better one where latitudinal lines are not "straight".
f.thorpe
CC BY-SA 3.0
2017-06-19T14:32:52.280Z
This actually brings up an interesting thought... was there a time where latitude/longitude was slightly off given the (22 km)[https://en.wikipedia.org/wiki/Spheroid] radius difference? That'd amount to almost 3 km per 10° if all error were one direction, or at least 150 m per °. However, (this)[https://en.wikipedia.org/wiki/History_of_latitude_measurements] suggests lat was initially measured by day length, and lon was probably measured by sun/star too, so they wouldn't make that error (though more recent idealized models might?) Elevation might have hurt actual measurement though.
JeopardyTempest
CC BY-SA 4.0
2023-02-22T01:44:21.467Z
@JeopardyTempest. Welcome to the world of elliptic integrals.
Mad Physicist
CC BY-SA 4.0
2020-12-22T18:40:24.903Z
@JeopardyTempest Because the Earth is not a sphere, the answer is technically wrong: The ground distance for a given change in latitude depends on the latitude you start from.
Spencer
CC BY-SA 4.0
2015-11-04T06:16:21.170Z
2020-12-21T22:18:23.150Z
2020-12-21T22:18:23.150Z
f.thorpe
<p>I don't understand the last sentence about decimal places, but I can tell you about the relationship between <em>lat</em>, <em>long</em> and distance.</p>

<p>Over two centuries ago, <a href="http://en.wikipedia.org/wiki/Metre#Meridional">the meter was defined as one ten-millionth (1/10 000 000) of the length of a quadrant along the Earth's meridian; that is, the distance from the Equator to the North Pole</a>. So, for <em>latitude</em> the number of degrees from the pole to the equator is $90^\circ\!$, and the number of meters is 10 million (or 10,000 kilometers). That means $1^\circ\!$ of <em>latitude</em> is $10,000/90 = 111$ kilometers, and $0.001^\circ = 0.111$ kilometers or $111$ meters, essentially an American football field plus both endzones.</p>

<p>The total length of the Equator is about equal to four times the distance from a Pole to the Equator. (Slightly more because the Earth's figure is a little oblate, like a slightly flattened ball.) Think of a great circle going from the North pole, to the Equator, to the South pole, to the Equator on the opposite side of the world, then back to North Pole ($360^\circ\!\!$ of a circle). That would be (almost) equivalent to the great circle ($360^\circ\!\!$ of <em>longitude</em>) of the Equator. So, at the Equator, $0.001^\circ\!$ of <em>longitude</em> will also be $111$ meters. But...</p>

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

<p><a href="https://i.sstatic.net/SRU4p.gif"><img src="https://i.sstatic.net/SRU4p.gif" alt="enter image description here"></a></p>

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

<p><a href="https://i.sstatic.net/2ipxf.png"><img src="https://i.sstatic.net/2ipxf.png" alt="enter image description here"></a></p>

CC BY-SA 4.0
2022-06-18T13:38:41.897Z
Even the ditsance of a degree of latitude changes, because Earth is not a perfect sphere.
Spencer
CC BY-SA 3.0
2015-11-04T07:27:33.450Z
2015-11-04T19:01:31.507Z
2015-11-04T19:01:31.507Z
Eubie Drew
<p>I am using &quot;PARI&quot; a free open source math program for Mac or PC,
for these exact (vs GRS80 Ellipsoid) equations:
&quot;longscale&quot;gives meters per (second of Longitude) at a constant latitude,
so below I multiply it by &quot;3.6&quot; to get
vs 0.001 degrees as requested by the original poster:</p>
<pre><code>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
</code></pre>
<p>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 <em>latitude</em> to 30.001.
Then either of the following integrals give the same answer,
but might have different speeds or accuracy in your computer language:</p>
<pre><code>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
</code></pre>
<p>Thus the distance
is the meridian arc length of the second point minus
the meridian arc length of the first:</p>
<pre><code>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
</code></pre>
<p>If he likewise wanted the scale at 30 degrees Latitude in
meters per (second of Latitude),
then differentiate the equation above like this:</p>
<pre><code>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
</code></pre>

CC BY-SA 4.0
2023-02-22T01:24:48.720Z
Thank you. I was looking for the equation with elliptical integrals and too lazy to derive it myself. You probably don't need the numerical derivative in the distance/angle at the end since the integrals are trivially differentiable.
Mad Physicist
CC BY-SA 4.0
2022-06-18T06:13:37.793Z
2022-06-18T06:13:37.793Z
fischertranscripts
<p>If I'm standing at a lat of 30.000$^\circ$ and a long of 30.000$^\circ$ and I move to a lat of 30.001$^\circ$ and a long 30.000$^\circ$, how far have I gone? Is there a relationship between decimal places and lat and long distances?</p>

CC BY-SA 4.0
2023-02-22T01:43:26.397Z
There is a relationship between angles and distances and vice-versa, which is dependent on the starting latitude for both latitude and longitude.
Mad Physicist
<measurements><geography><cartography>
How much distance does $0.001^\circ$ in latitude or longitude represent?
Mark Thornburg
# How much distance does $0.001^\circ$ in latitude or longitude represent?

(Asked by: username_0 on 2015-11-04)

If I'm standing at a lat of 30.000$^\circ$ and a long of 30.000$^\circ$ and I move to a lat of 30.001$^\circ$ and a long 30.000$^\circ$, how far have I gone? Is there a relationship between decimal places and lat and long distances?

**Tags:** measurements, geography, cartography

> **Comments:**
> **username_1**: There is a relationship between angles and distances and vice-versa, which is dependent on the starting latitude for both latitude and longitude.  

---

## 3 answers

### Answer by username_3 on 2015-11-04. Score: 20

I don't understand the last sentence about decimal places, but I can tell you about the relationship between *lat*, *long* and distance.

Over two centuries ago, [the meter was defined as one ten-millionth (1/10 000 000) of the length of a quadrant along the Earth's meridian; that is, the distance from the Equator to the North Pole](http://en.wikipedia.org/wiki/Metre#Meridional). So, for *latitude* the number of degrees from the pole to the equator is $90^\circ\!$, and the number of meters is 10 million (or 10,000 kilometers). That means $1^\circ\!$ of *latitude* is $10,000/90 = 111$ kilometers, and $0.001^\circ = 0.111$ kilometers or $111$ meters, essentially an American football field plus both endzones.

The total length of the Equator is about equal to four times the distance from a Pole to the Equator. (Slightly more because the Earth's figure is a little oblate, like a slightly flattened ball.) Think of a great circle going from the North pole, to the Equator, to the South pole, to the Equator on the opposite side of the world, then back to North Pole ($360^\circ\!\!$ of a circle). That would be (almost) equivalent to the great circle ($360^\circ\!\!$ of *longitude*) of the Equator. So, at the Equator, $0.001^\circ\!$ 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](https://i.sstatic.net/SRU4p.gif)](https://i.sstatic.net/SRU4p.gif)

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](https://i.sstatic.net/2ipxf.png)](https://i.sstatic.net/2ipxf.png)

> **Comments:**
> **username_6**: Even the ditsance of a degree of latitude changes, because Earth is not a perfect sphere.  

---

### Answer by username_2 on 2015-11-04. Score: 9

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](https://i.sstatic.net/BBmMe.gif)](https://i.sstatic.net/BBmMe.gif)  
(source: [learner.org](https://www.learner.org/jnorth/images/graphics/mclass/Lat_Long.gif))

.......................... ![Latitude](https://i.sstatic.net/Ms2xq.gif) .............. [![Longitude](https://i.sstatic.net/e2uRI.gif)](https://i.sstatic.net/e2uRI.gif)

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.

> **Comments:**
> **username_3**: That illustration of *latitude* is misleading because at that projection, parallels of *latitude* would NOT appear evenly spaced.  
> **username_4**: As I remember from my mathematics classes, in curved space, longitude lines are actually parallel and latitude lines aren't even straight lines.   That said, the more common definition is what farrenthrope said.   Latitude lines are called Parallel and Longitude are called meridians that meet at the poles.  
> **username_4**: Globe with the apparent uneven spacing if interested, per Aabaakawad's comment:   http://www.geolounge.com/wp-content/uploads/2014/08/latitude.png  
> **username_2**: Thank you everyone for your comments on the original image used.  I have replaced it with a better one where latitudinal lines are not "straight".  
> **username_5**: This actually brings up an interesting thought... was there a time where latitude/longitude was slightly off given the (22 km)[https://en.wikipedia.org/wiki/Spheroid] radius difference? That'd amount to almost 3 km per 10° if all error were one direction, or at least 150 m per °. However, (this)[https://en.wikipedia.org/wiki/History_of_latitude_measurements] suggests lat was initially measured by day length, and lon was probably measured by sun/star too, so they wouldn't make that error (though more recent idealized models might?) Elevation might have hurt actual measurement though.  
> **username_6**: @username_5 Because the Earth is not a sphere, the answer is technically wrong: The ground distance for a given change in latitude depends on the latitude you start from.  
> **username_1**: @username_5. Welcome to the world of elliptic integrals.  

---

### Answer by username_7 on 2022-06-18. Score: 1

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

```

> **Comments:**
> **username_1**: Thank you. I was looking for the equation with elliptical integrals and too lazy to derive it myself. You probably don't need the numerical derivative in the distance/angle at the end since the integrals are trivially differentiable.  

---

