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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

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-6357cos(49) = 378043,7km 6357sin(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.

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  • $\begingroup$ 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! :-) $\endgroup$
    – uhoh
    Commented May 21, 2024 at 14:29
  • $\begingroup$ 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. :-) $\endgroup$
    – uhoh
    Commented May 21, 2024 at 14:33
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    $\begingroup$ If I use this formula, the parallax for the latitude of 49 is just 0.014380286 degree? Is it really correct? $\endgroup$
    – Geographos
    Commented May 21, 2024 at 15:39
  • $\begingroup$ 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 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°. $\endgroup$
    – uhoh
    Commented May 21, 2024 at 16:13
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    $\begingroup$ 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. $\endgroup$
    – Geographos
    Commented May 21, 2024 at 19:52

1 Answer 1

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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? 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

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    $\begingroup$ I've used the formula for 2 latitudes and calculated the results, but I don't know what to do next. $\endgroup$
    – Geographos
    Commented May 21, 2024 at 10:25
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    $\begingroup$ @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. $\endgroup$
    – uhoh
    Commented May 21, 2024 at 10:42
  • $\begingroup$ 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. $\endgroup$
    – Geographos
    Commented May 21, 2024 at 11:36
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    $\begingroup$ Ok, I've updated my answer. I hope that it will be clear enough for you. I am stuck at this point exactly. $\endgroup$
    – Geographos
    Commented May 21, 2024 at 13:17

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