<p>The answer in a plane you linked works rather well in a sphere provided that the distance you want to measure is tiny compared with both the radius of Earth and the distance to the nearest pole - and 2.67 chains is that tiny.</p>

<p>For the remaining of the answer I'll be assuming that the Earth is an sphere, which quite accurate unless you need great accuracy - that is, enough accurate for nearly anything except for surveying.</p>

<p>The first step is to convert your distance and direction to easting and northing distances - as the linked answer tells. </p>

<p>The second step is to convert those two distances to degrees latitude and longitude, and you only need to know that one degree latitude measures <span class="math-container">$R_{Earth}\cdot\frac{\pi}{180}$</span> and a degree longitude measures <span class="math-container">$R_{Earth}\cdot\frac{\pi}{180}\cdot cos(latitude)$</span>.</p>

<p>As a final step, you just need to sum those angles to the latitude and longitude of your position.</p>

<p>With your data:</p>

<pre><code>&gt; # Earth radius in chains
&gt; (r &lt;- 3958.8*80)
[1] 316704
&gt; 
&gt; #bearing, latitude and longitude in radians
&gt; (br &lt;- -52/180*pi)
[1] -0.907571211
&gt; (latr &lt;- 37/180*pi)
[1] 0.6457718232
&gt; (lonr &lt;- -91/180*pi)
[1] -1.588249619
&gt; 
&gt; #1 Distance to North and East in chains
&gt; (east &lt;- sin(br)*2.67)
[1] -2.103988712
&gt; (north &lt;- cos(br)*2.67)
[1] 1.643816139
&gt; 
&gt; #One degree in chains
&gt; (deglat &lt;- r*pi/180)
[1] 5527.527554
&gt; (deglon &lt;- r*pi/180*cos(latr))
[1] 4414.479788
&gt; 
&gt; #2 distance in degrees
&gt; (dlat &lt;- north/deglat)
[1] 0.0002973872356
&gt; (dlon &lt;- east/deglon)
[1] -0.000476610793
&gt; 
&gt; #3 arriving point (GPS coordinates)
&gt; (lat &lt;- 37+dlat)
[1] 37.00029739
&gt; (lon &lt;- -91+dlon)
[1] -91.00047661
</code></pre>

CC BY-SA 4.0
2020-03-08T22:06:47.150Z
Awesome, thank you!
Don Branson
CC BY-SA 4.0
2020-03-08T17:35:35.707Z
2020-03-08T18:19:11.933Z
2020-03-08T18:19:11.933Z
Pere
<p>I need to calculate the gps coordinates of a point "North 52° West 2.67 chains" from a known location. What's the process to make this calculation? If possible, start with 37N-91W.</p>

<p>There's an answer to this on <a href="https://math.stackexchange.com/questions/604324/find-a-point-n-distance-away-from-a-specified-point-in-a-given-direction">Mathematics</a> on a plane. That answer, plus knowing degrees/meter around my location will provide a sufficiently accurate answer for the short distances in my case. Given that, this question is probably a good candidate to be closed. </p>

CC BY-SA 4.0
2020-03-08T17:47:03.440Z
I would like to make an example with your data, but I can't make sense of "North 52° West 2.67 chains". Does it mean a bearing of 52º measured counter-clockwise from North?
Pere
CC BY-SA 4.0
2020-03-08T17:52:09.083Z
@Pere - I think so - it's a form used in a property description from about the 1960s, perhaps older.
Don Branson
<gps>
How to calculate gps location based on direction and distance?
Don Branson
# How to calculate gps location based on direction and distance?

(Asked by: username_0 on 2020-03-07)

I need to calculate the gps coordinates of a point "North 52° West 2.67 chains" from a known location. What's the process to make this calculation? If possible, start with 37N-91W.

There's an answer to this on [Mathematics](https://math.stackexchange.com/questions/604324/find-a-point-n-distance-away-from-a-specified-point-in-a-given-direction) on a plane. That answer, plus knowing degrees/meter around my location will provide a sufficiently accurate answer for the short distances in my case. Given that, this question is probably a good candidate to be closed.

**Tags:** gps

> **Comments:**
> **username_1**: I would like to make an example with your data, but I can't make sense of "North 52° West 2.67 chains". Does it mean a bearing of 52º measured counter-clockwise from North?  
> **username_0**: @username_1 - I think so - it's a form used in a property description from about the 1960s, perhaps older.  

---

## 1 answers

### Answer by username_1 on 2020-03-08. Score: 4 | Accepted answer

The answer in a plane you linked works rather well in a sphere provided that the distance you want to measure is tiny compared with both the radius of Earth and the distance to the nearest pole - and 2.67 chains is that tiny.

For the remaining of the answer I'll be assuming that the Earth is an sphere, which quite accurate unless you need great accuracy - that is, enough accurate for nearly anything except for surveying.

The first step is to convert your distance and direction to easting and northing distances - as the linked answer tells.

The second step is to convert those two distances to degrees latitude and longitude, and you only need to know that one degree latitude measures $R\_{Earth}\cdot\frac{\pi}{180}$ and a degree longitude measures $R\_{Earth}\cdot\frac{\pi}{180}\cdot cos(latitude)$.

As a final step, you just need to sum those angles to the latitude and longitude of your position.

With your data:

```
> # Earth radius in chains
> (r <- 3958.8*80)
[1] 316704
> 
> #bearing, latitude and longitude in radians
> (br <- -52/180*pi)
[1] -0.907571211
> (latr <- 37/180*pi)
[1] 0.6457718232
> (lonr <- -91/180*pi)
[1] -1.588249619
> 
> #1 Distance to North and East in chains
> (east <- sin(br)*2.67)
[1] -2.103988712
> (north <- cos(br)*2.67)
[1] 1.643816139
> 
> #One degree in chains
> (deglat <- r*pi/180)
[1] 5527.527554
> (deglon <- r*pi/180*cos(latr))
[1] 4414.479788
> 
> #2 distance in degrees
> (dlat <- north/deglat)
[1] 0.0002973872356
> (dlon <- east/deglon)
[1] -0.000476610793
> 
> #3 arriving point (GPS coordinates)
> (lat <- 37+dlat)
[1] 37.00029739
> (lon <- -91+dlon)
[1] -91.00047661

```

> **Comments:**
> **username_0**: Awesome, thank you!  

---

