Jump to content


CNers have asked about a donation box for Cloudy Nights over the years, so here you go. Donation is not required by any means, so please enjoy your stay.


Need Math Verification for Measuring Prominences, Please

2 replies to this topic

#1 matt_astro_tx


    Fly Me to the Moon

  • *****
  • topic starter
  • Posts: 6,911
  • Joined: 19 Jan 2021
  • Loc: Dallas, Texas

Posted 23 April 2024 - 07:00 AM

After sharing my eclipse images with friends I had someone ask if I knew how large the prominences were.  So I dusted off the trusty slide rule, remembered I have no idea how to use one, and put it back in the drawer.  What I came up with is fairly straightforward and based on simple math but I really would like a vote of confidence from some of you before I share it with a wider audience.


Would y'all mind giving me a sanity check?


Deep breath, here goes:


First, the sun's radius is a relatively stable 6.9574 x 105 km (source: https://doi.org/10.1...743921309992304).  Converting that radius to diameter in miles:

- Radius = 6.9574 x 10^5 km = 695,740 km
- Diameter = 695,740 km x 2 = 1,391,480 km
- Convert to miles = 1,391,480 km x 0.621371 = 864,625.32 miles


Ok, now we determine how many pixels wide the sun was on my camera sensor.  (I initially started down the road of using my system's resolution of 1.72"/pixel and the sun's average angular diameter but that all seemed like a waste of time when I was done.  Determining the ratio of pixels to miles was much simpler and more logical to me and I think also to a layperson.)


2024 TSE Measure 001


So then if we divide the width in miles by the width in pixels we determine the image scale.

- 864,625.32 miles / 1,115 pixels = Each pixel represents a linear distance of 775.45 miles


While we’re at it we can determine how large the Earth would appear at the same distance (in pixels).

- 7,917.5 miles / 775.45 miles per pixel = 10.21 pixels


2024 TSE Measure 002
Then I measured the length and width in pixels of two prominences using the measuring tool and arrived at the following figures.  The first prominence is 53 x 57 pixels.  Using the scaling factor of 1 = 775.45 miles, that’s 41,098.85 miles by 44,200.65 miles.
2024 TSE Measure 003
The next one measured 80 pixels in length which is 62,036 miles. Roughly 1/4 the distance from the Earth to the Moon, so I included that for scale as well.
2024 TSE Measure 004


I also created my own scale image of the solar system because I couldn't find a CC licensed one that I liked.


Solar System To Scale


Thanks y'all!

  • zjc26138, kfiscus, 2001 and 1 other like this

#2 aabusara



  • *****
  • Posts: 1,337
  • Joined: 08 Feb 2023
  • Loc: South Texas

Posted 24 April 2024 - 09:28 AM

This is so cool Matt. The Earth size perspective always gets me.

  • matt_astro_tx likes this

#3 twinion


    Lift Off

  • -----
  • Posts: 12
  • Joined: 19 Feb 2024

Posted 24 April 2024 - 02:02 PM

I did the same math on one of my images and got a max height of that big arcing prominence as ~36k miles, or 4.5x the diameter of Earth. A user on Reddit a couple weeks ago had about the same math, but I couldn't find any more authoritative sources. I did my measurements very quickly, so I won't argue my math is any closer to correct, but we're all within a similar order of magnitude, so I'd feel pretty comfortable telling people that arc is 4-5 earths tall.

  • matt_astro_tx likes this

Reply to this topic


CNers have asked about a donation box for Cloudy Nights over the years, so here you go. Donation is not required by any means, so please enjoy your stay.

Recent Topics

Cloudy Nights LLC
Cloudy Nights Sponsor: Astronomics