2 replies to this topic

[matt_astro_tx]

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.)

 

 

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

 

 

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.

 

 

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.

 

 

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

 

 

Thanks y'all!

[aabusara]

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

[twinion]

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.