<p>I won't work the whole problem for you, but as long as you assume Earth is a sphere, try rewriting the equation using the following diagram.</p>
<p>We have a right triangle with the long side = <span class="math-container">$d - R \cos(\text{lat})$</span> and the short side = <span class="math-container">$R \sin(\text{lat})$</span></p>
<p>Calculate the angles for both latitudes, then subtract the two angles to get the change in angle for a given change in latitude.</p>
<p>The left side of that equation should really be <span class="math-container">$\tan{\theta}$</span>, so you'll calculate that ratio then take the arctangent to get the angle. See <a href="https://astronomy.stackexchange.com/a/36190/7982">this answer to <em>Do parsecs scale as an inverse proportion with parallax angle?</em></a> to understand why we shouldn't use the small angle approximation, especially when the angle isn't so small (the Moon is pretty close!)</p>
<p><a href="https://i.sstatic.net/ZMFQznmS.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/ZMFQznmS.png" alt="enter image description here" /></a></p>

CC BY-SA 4.0
2024-05-21T10:25:30.317Z
I've used the formula for 2 latitudes and calculated the results, but I don't know what to do next.
Geographos
CC BY-SA 4.0
2024-05-21T10:42:04.133Z
@Geographos The answer says "Calculate the angles for both latitudes, *then subtract the two angles to get the change in angle for a given change in latitude.* If you let me know what those two angles are I can take a look and help debug if necessary, but if you did it right then all you need to do is subtract.
uhoh
CC BY-SA 4.0
2024-05-21T11:36:04.193Z
After calculating my first latitude 49 by using formula you proposed I have received 378043.7081 and second formula 94.88267065. For latitude of 50 I have received accordingly 378043.7373 and 96.81890345 and I don't know what to do next with it.
Geographos
CC BY-SA 4.0
2024-05-21T13:17:00.447Z
Ok, I've updated my answer. I hope that it will be clear enough for you. I am stuck at this point exactly.
Geographos
CC BY-SA 4.0
2024-05-18T00:23:28.820Z
2024-05-18T00:23:28.820Z
uhoh
<p>I want to compute lunar parallax based on the distance between two points. In fact I know the formula, but I guess they're explained wrong.</p>
<p><a href="https://i.sstatic.net/OlVzLNC1.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/OlVzLNC1.png" alt="enter image description here" /></a></p>
<p>By looking at the image below we can see the polar radius, on which the parallax computations should be based.
What is the relation of the part of polar radius to the distance measured on Earth?
If I move for example from point A at latitude of 50N to point B at latitude of 48N, the distance on the surface will be approximately 222km. What this distance can be calculated with respect to Earth's polar radius? Is there any formula for it?</p>
<p>UPDATE:</p>
<p>Based on the answer below I used the following formulae:</p>
<p>d - Rcos(lat)<br />
Rsin(lat)
where I got respectively:</p>
<p>384400km-6357<em>cos(49) = 378043,7km
6357</em>sin(49) = 94,88km</p>
<p>And did the same for latitude of 50
with respect values:</p>
<p>378043,73km
96,82km</p>
<p>and I don't know what to do next.</p>

CC BY-SA 4.0
2024-05-21T14:29:21.623Z
1) Let's not mix commas and periods to indicate decimal points, please stick to using one or the other consistently. I think it would be better if you used periods, they are more internationally recognized. 2) Always keep units with values. You did a great job here but in your comment under the answer, when I saw numbers like 90 I thought they were angles. And always a single space between the number and the units 3) The sine of 49 degrees is 0.75471, so 6357 (km) sin(49°) is 4797.68 km, not "94,88km" I still can't understand how you got that number. But we're making progress! :-)
uhoh
CC BY-SA 4.0
2024-05-21T14:33:04.667Z
Finally, for each latitude, estimate the parallax for each latitude as $$\theta \approx \frac{R \sin(\text{lat})}{d - R \cos(\text{lat})}$$ and subtract those two angles for the parallax difference between your two latitudes. Let me know how it goes - and if it works you are welcome to post your results as an answer post to your question. :-)
uhoh
CC BY-SA 4.0
2024-05-21T15:39:42.547Z
If I use this formula, the parallax for the latitude of 49 is just 0.014380286 degree? Is it really correct?
Geographos
CC BY-SA 4.0
2024-05-21T16:13:38.770Z
If I use $R$ = 6357 km, $d$ = 384400 km, and lat=49°, I get $R \sin(\text{lat})$ = 4797.69 km and $d - R \cos(\text{lat})$ = 380229.43 km. For 4797.69 km / 380229.43 km I get 0.01262, not 0.01438. It's close, not exact agreement. But when you use the [small angle approximation](https://en.wikipedia.org/wiki/Small-angle_approximation) of $\sin(\theta) \approx \tan(\theta) \approx \theta$ your angle $\theta$ is in radians, not degrees. To get degrees, multiply by 180/$\pi$. So the parallax between the Equator and 49° latitude is roughly 0.75°.
uhoh
CC BY-SA 4.0
2024-05-21T19:52:19.443Z
OK got it! I followed exactly what you pointed out and received 0.722951, which seems to be quite right. Didn't realize that my result was in radians.
Geographos
CC BY-SA 4.0
2024-05-27T13:44:53.740Z
Excellent, thanks for sharing your good news with me! :-) If you have time, feel free to post another answer here, explaining what exactly you did and how you got an answer, and in a few days after that you can click "accept" for your answer. It's always better for the site when questions have conclusive answers, and it seems you have exactly what you need now.
uhoh
<earth><distances><parallax><angular-distance>
Calculating lunar parallax based on a distance between two objects at certain latitude
Geographos
# Calculating lunar parallax based on a distance between two objects at certain latitude

(Asked by: username_0 on 2024-05-17)

I want to compute lunar parallax based on the distance between two points. In fact I know the formula, but I guess they're explained wrong.

[![enter image description here](https://i.sstatic.net/OlVzLNC1.png)](https://i.sstatic.net/OlVzLNC1.png)

By looking at the image below we can see the polar radius, on which the parallax computations should be based.
What is the relation of the part of polar radius to the distance measured on Earth?
If I move for example from point A at latitude of 50N to point B at latitude of 48N, the distance on the surface will be approximately 222km. What this distance can be calculated with respect to Earth's polar radius? Is there any formula for it?

UPDATE:

Based on the answer below I used the following formulae:

d - Rcos(lat)  
Rsin(lat)
where I got respectively:

384400km-6357*cos(49) = 378043,7km
6357*sin(49) = 94,88km

And did the same for latitude of 50
with respect values:

378043,73km
96,82km

and I don't know what to do next.

**Tags:** earth, distances, parallax, angular-distance
> **Comments:**
> **username_1**: 1) Let's not mix commas and periods to indicate decimal points, please stick to using one or the other consistently. I think it would be better if you used periods, they are more internationally recognized. 2) Always keep units with values. You did a great job here but in your comment under the answer, when I saw numbers like 90 I thought they were angles. And always a single space between the number and the units 3) The sine of 49 degrees is 0.75471, so 6357 (km) sin(49°) is 4797.68 km, not "94,88km" I still can't understand how you got that number. But we're making progress! :-)  
> **username_1**: Finally, for each latitude, estimate the parallax for each latitude as $$\theta \approx \frac{R \sin(\text{lat})}{d - R \cos(\text{lat})}$$ and subtract those two angles for the parallax difference between your two latitudes. Let me know how it goes - and if it works you are welcome to post your results as an answer post to your question. :-)  
> **username_0**: If I use this formula, the parallax for the latitude of 49 is just 0.014380286 degree? Is it really correct?  
> **username_1**: If I use $R$ = 6357 km, $d$ = 384400 km, and lat=49°, I get $R \sin(\text{lat})$ = 4797.69 km and $d - R \cos(\text{lat})$ = 380229.43 km. For 4797.69 km / 380229.43 km I get 0.01262, not 0.01438. It's close, not exact agreement. But when you use the [small angle approximation](https://en.wikipedia.org/wiki/Small-angle_approximation) of $\sin(\theta) \approx \tan(\theta) \approx \theta$ your angle $\theta$ is in radians, not degrees. To get degrees, multiply by 180/$\pi$. So the parallax between the Equator and 49° latitude is roughly 0.75°.  
> **username_0**: OK got it! I followed exactly what you pointed out and received 0.722951, which seems to be quite right. Didn't realize that my result was in radians.  
> **username_1**: Excellent, thanks for sharing your good news with me! :-) If you have time, feel free to post another answer here, explaining what exactly you did and how you got an answer, and in a few days after that you can click "accept" for your answer. It's always better for the site when questions have conclusive answers, and it seems you have exactly what you need now.  

---

## 1 answers

### Answer by username_1 on 2024-05-18. Score: 3 | Accepted answer

I won't work the whole problem for you, but as long as you assume Earth is a sphere, try rewriting the equation using the following diagram.

We have a right triangle with the long side = $d - R \cos(\text{lat})$ and the short side = $R \sin(\text{lat})$

Calculate the angles for both latitudes, then subtract the two angles to get the change in angle for a given change in latitude.

The left side of that equation should really be $\tan{\theta}$, so you'll calculate that ratio then take the arctangent to get the angle. See [this answer to *Do parsecs scale as an inverse proportion with parallax angle?*](https://astronomy.stackexchange.com/a/36190/7982) to understand why we shouldn't use the small angle approximation, especially when the angle isn't so small (the Moon is pretty close!)

[![enter image description here](https://i.sstatic.net/ZMFQznmS.png)](https://i.sstatic.net/ZMFQznmS.png)

> **Comments:**
> **username_0**: I've used the formula for 2 latitudes and calculated the results, but I don't know what to do next.  
> **username_1**: @Geographos The answer says "Calculate the angles for both latitudes, *then subtract the two angles to get the change in angle for a given change in latitude.* If you let me know what those two angles are I can take a look and help debug if necessary, but if you did it right then all you need to do is subtract.  
> **username_0**: After calculating my first latitude 49 by using formula you proposed I have received 378043.7081 and second formula 94.88267065. For latitude of 50 I have received accordingly 378043.7373 and 96.81890345 and I don't know what to do next with it.  
> **username_0**: Ok, I've updated my answer. I hope that it will be clear enough for you. I am stuck at this point exactly.  

---

