<p>You compare the position of the moon with that of the background of stars.  If you both take a picture of the moon at the same time, and compare the angle that it makes with a distant star the moon will be in a different position, and from this you can calculate the angle.  Here is your diagram showing, in slightly simplified way what I mean.</p>
<p><a href="https://i.sstatic.net/X4WVq.png" rel="noreferrer"><img src="https://i.sstatic.net/X4WVq.png" alt="enter image description here" /></a></p>
<p>In practice the stars won't be exactly on your baseline so you'll need to do some measuring of the position from the photos as in these simulated images.</p>
<p><a href="https://i.sstatic.net/RWl9L.png" rel="noreferrer"><img src="https://i.sstatic.net/RWl9L.png" alt="enter image description here" /></a></p>

CC BY-SA 4.0
2021-05-08T09:53:47.877Z
And with the moon you want to make sure to really take photos at the same time. So make sure to synchronize your clocks reasonably well. It moves with about 30" per minute. Take that into account for your error analysis, given the accuracy of your concurrency of the photos to compare
planetmaker
CC BY-SA 4.0
2021-05-08T13:30:57.793Z
Yes, that's alright with the method but there are two problems, first, there is about a $9$ Hour difference between the two cities so it's hard to find the moon with stars. And even if it does, there is too much light pollution and I don't have that sort of camera to capture such an image for measurement.
Himanshu
CC BY-SA 4.0
2021-05-08T13:34:30.727Z
Yes. Nobody said it was easy.
James K
CC BY-SA 4.0
2021-05-08T22:25:19.790Z
I have the urge to put your example images in my stereopticon/Viewmaster...
DJohnM
CC BY-SA 4.0
2021-05-08T22:59:28.403Z
https://imaginary.org/sites/default/files/parallaxmeasurementinstructionsfinal.pdf
James K
CC BY-SA 4.0
2021-05-08T07:50:21.753Z
2021-05-08T07:50:21.753Z
James K
<p>Using user38308's approach:</p>
<p>First, find the distance between you and your friend, knowing lat/long of your positions. You need to solve a spherical triangle with vertices at the north pole and the locations of you and your friend. Piece of cake. You have 2 sides And their included angle. The (included) angle at the pole is the difference in your longitudes. The sides are 90 degrees minus the latitudes of you and your friend. Using the law of cosines for spherical triangles find the arc distance between you and your friend. Don't stop there. Find the angles at you and your friend using the law of sines for spherical triangles. You'll need them later. Note that this calculation can/should be done months ahead of time. Note also that the sum of the angles in a spherical triangle is usually more than 180 degrees.</p>
<p>Good news is that we're done with spherical trig, and moving on to plane trig for the rest of the calculation. You and your friend are trying to find the altitude of the Moon when its center is on the plane having you, your friend and the center of the Earth. That will happen, for both of you, when its azimuth is equal to the angles of the spherical triangle (prior step) and, maybe, plus or minus 180 degrees.</p>
<p>Still with me?</p>
<p>Then we move on to the isosceles triangle with vertices at you, your friend and the center of the Earth. You know the angle at the center of the Earth from the spherical triangle analysis. You can find the unequal side knowing the other 2 sides (Earth radii) and that angle (memorized formula or law of cosines for plane triangles). That leaves you with a triangle having sides the chord between you and your friend, and the distances from you and your friend to the center of the moon.</p>
<p>Home stretch.</p>
<p>The Moon will pass through the plane of you, your friend and the center of the Earth when its azimuth is equal to the angles (or angles plus or minus 180 degrees) you calculated previously. You both should see that happen at about the same time, but a few minutes difference shouldn't matter. Measure the Moon's altitude at that time. After correcting for refraction and semi-diameter, you'll both find the Moon's altitude is less than you'd see if it was a very distant object. The angles of the triangle at you and your friend will be the observed altitudes minus half the angle of the isosceles triangle included between its radii. (Needs a picture - watch this space.) You now Know the parallax angle: 180 degrees minus the two angles at you and your friend. You can find the distances from you and your friend to the center of the Moon with the law of sines for plane triangles. You could carry on to find the distance between the Earth/Moon centers, but you wouldn't get a much more accurate answer.</p>

CC BY-SA 4.0
2021-07-06T03:28:53.220Z
2021-07-06T03:28:53.220Z
stretch
<p>While the right answer is measuring the angles of the two telescopes with respect to each other this is not trivial.</p>
<p>The technique <a href="https://www.youtube.com/watch?v=o0_SlQP49Yc" rel="nofollow noreferrer">used here</a> by Scott Manley side steps that by picking a time when two objects with a known angular distance between them will be in the same photo, and using that distance as a calibration to measure the moon parallax. He used planets which made the process easier but narrows the usable windows to achieve it. An earlier video did <a href="https://www.youtube.com/watch?v=GTtce3rJxhc" rel="nofollow noreferrer">it with stars</a>.</p>
<p>Second video also discusses a method using two photos taken from the same location at moonrise and 'moon noon' and determining the change in apparent size due to being closer due to Earth curvature.</p>

CC BY-SA 4.0
2021-05-09T03:57:46.933Z
2021-05-09T04:05:29.677Z
2021-05-09T04:05:29.677Z
GremlinWranger
<p>I want to use the parallax method (from scratch) to find the distance between the earth and the moon. To get a significant parallax, I need to choose a large baseline. I live in India and my friend in the US and so the baseline will be about <span class="math-container">$13,000$</span> Km. That's quite a large so I suppose I would get a significant parallax.</p>
<p>What I don't think of, is taking of angle measurement?</p>
<p><a href="https://i.sstatic.net/Vacos.jpg" rel="noreferrer"><img src="https://i.sstatic.net/Vacos.jpg" alt="enter image description here" /></a></p>
<p>How can I measure the angle <span class="math-container">$\alpha$</span>?</p>

CC BY-SA 4.0
2021-05-08T05:59:40.180Z
Don’t forget that while *ground* distance is around 13,000 km, *physical* distance between you and your friend is less, as you need to account for Earth’s curvature, as your drawing makes clear. The Earth’s diameter is 12,756 km. Also, you can’t directly measure α, as this would imply you’re on the Moon, but you can measure the other two angles…
Pierre Paquette
CC BY-SA 4.0
2021-05-08T06:41:34.117Z
How can I, Can you show through figure and little calculation? And method through which I can measure those angle.
Himanshu
CC BY-SA 4.0
2021-05-08T06:51:30.507Z
You basically need to compare where the moon is on the background field of fixed stars, for you and you friend (at the same time). Since the fixed stars are effectively infinitely far away, they appear at the same angles for both you and your friend. Take a picture each with the moon and some bright stars, find out the angular distance between the stars from a catalogue or software, and compare the moon's parallax to that in your pictures.
Kristoffer Sjöö
CC BY-SA 4.0
2021-05-08T07:46:54.097Z
Isn't any other way, where I don't use the catalogs?
Himanshu
CC BY-SA 4.0
2021-05-08T20:42:51.170Z
@YoungKindaichi: You don't need calatogues. You just need to compare the two photos with each other. (In fact, you don't even need a friend; you can take two pictures at different times.)
TonyK
CC BY-SA 4.0
2021-05-09T03:05:36.293Z
@TonyK And How I can do this? Can you tell me more about this?
Himanshu
<the-moon><parallax>
Using parallax to find distance to the moon
Himanshu
# Using parallax to find distance to the moon

(Asked by: username_0 on 2021-05-08)

I want to use the parallax method (from scratch) to find the distance between the earth and the moon. To get a significant parallax, I need to choose a large baseline. I live in India and my friend in the US and so the baseline will be about $13,000$ Km. That's quite a large so I suppose I would get a significant parallax.

What I don't think of, is taking of angle measurement?

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

How can I measure the angle $\alpha$?

**Tags:** the-moon, parallax
> **Comments:**
> **username_1**: Don’t forget that while *ground* distance is around 13,000 km, *physical* distance between you and your friend is less, as you need to account for Earth’s curvature, as your drawing makes clear. The Earth’s diameter is 12,756 km. Also, you can’t directly measure α, as this would imply you’re on the Moon, but you can measure the other two angles…  
> **username_0**: How can I, Can you show through figure and little calculation? And method through which I can measure those angle.  
> **username_2**: You basically need to compare where the moon is on the background field of fixed stars, for you and you friend (at the same time). Since the fixed stars are effectively infinitely far away, they appear at the same angles for both you and your friend. Take a picture each with the moon and some bright stars, find out the angular distance between the stars from a catalogue or software, and compare the moon's parallax to that in your pictures.  
> **username_0**: Isn't any other way, where I don't use the catalogs?  
> **username_3**: @YoungKindaichi: You don't need calatogues. You just need to compare the two photos with each other. (In fact, you don't even need a friend; you can take two pictures at different times.)  
> **username_0**: @TonyK And How I can do this? Can you tell me more about this?  

---

## 3 answers

### Answer by username_4 on 2021-05-08. Score: 6

You compare the position of the moon with that of the background of stars. If you both take a picture of the moon at the same time, and compare the angle that it makes with a distant star the moon will be in a different position, and from this you can calculate the angle. Here is your diagram showing, in slightly simplified way what I mean.

[![enter image description here](https://i.sstatic.net/X4WVq.png)](https://i.sstatic.net/X4WVq.png)

In practice the stars won't be exactly on your baseline so you'll need to do some measuring of the position from the photos as in these simulated images.

[![enter image description here](https://i.sstatic.net/RWl9L.png)](https://i.sstatic.net/RWl9L.png)

> **Comments:**
> **username_5**: And with the moon you want to make sure to really take photos at the same time. So make sure to synchronize your clocks reasonably well. It moves with about 30" per minute. Take that into account for your error analysis, given the accuracy of your concurrency of the photos to compare  
> **username_0**: Yes, that's alright with the method but there are two problems, first, there is about a $9$ Hour difference between the two cities so it's hard to find the moon with stars. And even if it does, there is too much light pollution and I don't have that sort of camera to capture such an image for measurement.  
> **username_4**: Yes. Nobody said it was easy.  
> **username_6**: I have the urge to put your example images in my stereopticon/Viewmaster...  
> **username_4**: https://imaginary.org/sites/default/files/parallaxmeasurementinstructionsfinal.pdf  

---

### Answer by username_7 on 2021-07-06. Score: 1

Using user38308's approach:

First, find the distance between you and your friend, knowing lat/long of your positions. You need to solve a spherical triangle with vertices at the north pole and the locations of you and your friend. Piece of cake. You have 2 sides And their included angle. The (included) angle at the pole is the difference in your longitudes. The sides are 90 degrees minus the latitudes of you and your friend. Using the law of cosines for spherical triangles find the arc distance between you and your friend. Don't stop there. Find the angles at you and your friend using the law of sines for spherical triangles. You'll need them later. Note that this calculation can/should be done months ahead of time. Note also that the sum of the angles in a spherical triangle is usually more than 180 degrees.

Good news is that we're done with spherical trig, and moving on to plane trig for the rest of the calculation. You and your friend are trying to find the altitude of the Moon when its center is on the plane having you, your friend and the center of the Earth. That will happen, for both of you, when its azimuth is equal to the angles of the spherical triangle (prior step) and, maybe, plus or minus 180 degrees.

Still with me?

Then we move on to the isosceles triangle with vertices at you, your friend and the center of the Earth. You know the angle at the center of the Earth from the spherical triangle analysis. You can find the unequal side knowing the other 2 sides (Earth radii) and that angle (memorized formula or law of cosines for plane triangles). That leaves you with a triangle having sides the chord between you and your friend, and the distances from you and your friend to the center of the moon.

Home stretch.

The Moon will pass through the plane of you, your friend and the center of the Earth when its azimuth is equal to the angles (or angles plus or minus 180 degrees) you calculated previously. You both should see that happen at about the same time, but a few minutes difference shouldn't matter. Measure the Moon's altitude at that time. After correcting for refraction and semi-diameter, you'll both find the Moon's altitude is less than you'd see if it was a very distant object. The angles of the triangle at you and your friend will be the observed altitudes minus half the angle of the isosceles triangle included between its radii. (Needs a picture - watch this space.) You now Know the parallax angle: 180 degrees minus the two angles at you and your friend. You can find the distances from you and your friend to the center of the Moon with the law of sines for plane triangles. You could carry on to find the distance between the Earth/Moon centers, but you wouldn't get a much more accurate answer.


---

### Answer by username_8 on 2021-05-09. Score: 1

While the right answer is measuring the angles of the two telescopes with respect to each other this is not trivial.

The technique [used here](https://www.youtube.com/watch?v=o0_SlQP49Yc) by Scott Manley side steps that by picking a time when two objects with a known angular distance between them will be in the same photo, and using that distance as a calibration to measure the moon parallax. He used planets which made the process easier but narrows the usable windows to achieve it. An earlier video did [it with stars](https://www.youtube.com/watch?v=GTtce3rJxhc).

Second video also discusses a method using two photos taken from the same location at moonrise and 'moon noon' and determining the change in apparent size due to being closer due to Earth curvature.


---

