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Mercury and the AU: Measuring the AU with a Transit of Mercury

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#1 astro151

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Posted 24 October 2019 - 04:05 PM

For anyone interested in trying to measure the Astronomical Unit (AU) for yourself using Sir Edmund Halley’s technique…

 

The principles are simple.  You need two observers spaced far apart on the globe (the further the better) to take simultaneous images of the transit.  Technically you can team-up with anyone.  However, finding someone with the same equipment as you can make things easier.

 

Basic Data Collection Principles

 

Ultimately, you need to take images from the two locations and align them as precisely as possible to see how the silhouette of Mercury’s disc shifts due to parallax.

 

There are several factors involved:

 

1.) Image scale – You can always re-size images so that the Sun’s disc is the same diameter.  Telescopes with the same focal length and accessories (flatteners, filters, reducers, etc.) should produce images where the sun is the same size.  To potentially avoid the need to re-size images, find someone with the same focal length and accessories as you.

 

2.) Image alignment – centering the Sun’s disc in the final picture is fairly easy and can be done in any number of ways.

 

3.) Image rotation – the chances of having cameras at different locations oriented at the same angle is very small.  At some point you will probably need use software to rotate one of the images to match the other so that the same features (except for Mercury) overlap.

 

NOTE: Mercury’s disc will only appear to shift by a very small amount!

 

Finding an Imaging Partner

 

If you do not already know someone at great distance from you, you can use this very basic Mercury and the AU Observers List (Google Sheet).  Enter your Cloudy Nights username, approximate location, the telescope and camera you plan to use.  Then look for others to partner-up with.  The further apart you are, the better (as long as you can still take simultaneous images).  Contact each other using Cloudy Nights

Personal Messages (PMs).

 

Google Sheet Rules:

 

1.) This is just intended to help people self-organize.  I will not be managing or maintaining the list or answering questions.

 

 

2.) Share only the info you feel comfortable with.

 

 

3.) The Google Sheet will be deleted on Monday, Dec. 2nd (three weeks after the transit) so your info is not permanently “out there”.

 

Timing

 

You and your imaging partner(s) need to take simultaneous images.  Depending on where you are on the globe, you may not both be able to see the entire transit.  Choose times which you can both see and at which the Sun is as high in the sky as possible.  Coordinate this amongst yourselves.

 

Image Processing

 

Everyone likes to post-process images in their own way.  The recommendation is that you share your unprocessed images with each other.  Re-size images so that the Sun has the exact same diameter in all images.

 

Aligning images will be the difficult part.  The best is if you know how to use software to align the images so you can measure Mercury’s shift by counting the pixels rather than printing images and measuring by hand.  Physical measurements done very carefully can still yield good results.

 

Data Analysis

 

Now that you have a pair of images, how do you analyze them and calculate the AU?  Here is a slightly stripped-down version of the AU Lab I designed for my astronomy students.


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#2 SantiagusDelSerif

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Posted 14 November 2019 - 12:29 PM

Hi everyone! I was trying to calculate the AU following this guide but I've encountered with some problems, I thought that perhaps somebody here can give me a hand and help me figure what is it that I am doing wrong. I'll try to keep it brief.

 

I'm using images from the GONG Observatory. I tried using the last transit of Mercury but it's hard to measure the displacement in the positions of the planet, so I decided I'd use images from the transit of Venus from 2012. Since Venus is bigger and closer, the shift in position is easier to measure for me.

 

My problem is that with my calculations I end up getting a figure for the AU that's almost twice as big as the actual value (296,148,304 km). I think I have narrowed it down and found where the problem is, but I can't tell where my mistake is. I end up getting a distance between Venus and the Earth that's almost twice as big (86,535,228 km) as it should really be.

 

My two locations are Uadipur, India and Mauna Loa, Hawaii. The distance between the two observatories is 12761 km. The date and time I'm using for the images is 06/06/2012 - 02:41 UTC.

 

I open both images in Photoshop, align them and measure the planet displacement between the two images (28.89 pixels) and the diameter of the Sun (1796.36 pixels).

 

Venus' maximum elongation in 2012 was in August 15th, and it was 45.8°, so the distance between Venus and the Sun is Sin(45.8) = 0.71691060765048276706127664049078 AU. So far so good.

 

I checked the Sun apparent size on that date in Stellarium and it was 0° 31' 31.31" (or 0.5253639°). So, I calculate the parallax angle of Venus:

 

(28.89px / 1796.36px) * 0.5253639° = 0.00844917670789819412589904028146°

 

And the tangent: 1.4746595370458040135944999175482e-4

 

Now, here comes the crux of the biscuit. When I try to calculate the distance between Venus and the Earth:

12761 km / 1.4746595370458040135944999175482e-4 = 86535228.501381427147220605120532 km

 

I get 86,535,228 kilometers. That's about twice the real distance between Venus and Earth (40,000,000 or 0,27 AU). Why? I can't seem to find out.

 

Then, of course, everything goes downhill from here. With that value, I calculate the Astronomical Unit:

 

86535228.501381427147220605120532 km / (1 AU - 0.71691060765048276706127664049078 AU) = 305681635.69527334206442387328539 km

 

And I end up with 305,681,635 kilometers for the AU, about twice the real value; and as expected since I started with an Earth-Venus distance that was twice as big. But why? Something must be wrong, obviously, but I can't grasp what. The logic in the calculations is flawless. Dividing the baseline distance (the opposite side in a right triangle) by the tangent of the parallax angle in order to get the distance to Venus (the adjacent side in the right triangle) makes perfect sense. So perhaps my measurement of the displacement of the planet over the Sun is wrong? I tried measuring different sets of images (at 03:00 UTC and 03:50 UTC) but I got very similar values.

 

Guess I'm going to try to rest a bit now and perhaps tomorrow with a clearer head I'll give it another shot, maybe trying different observatories to see what happens.

 

Thank you in advance and clear skies for everyone!



#3 James Bates

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Posted 22 November 2019 - 12:05 PM

 

My two locations are Uadipur, India and Mauna Loa, Hawaii. The distance between the two observatories is 12761 km.

I suspect that is the distance along the Earth's surface, which is of course curved. That isn't the actual straight-line distance between those two points in 3D space.

What's more, you don't want the distance between the two points, but the projection of the straight-line distance onto the plane perpendicular to the Earth-Sun line at the time of measurement.
 


Edited by James Bates, 22 November 2019 - 12:07 PM.



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