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Solar apparent magnitude vs sky surface brightness

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#1 Krukarius

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Posted 19 February 2024 - 05:08 AM

I found the useful website, which is here:

https://stjarnhimlen.../radfaq.html#11

and we can find there the values like this:

s7KOo.png

and this:

Pg9EU.png

most of them like "Overcast sky" or "Twilight" and especially the lower table refers to the situation, when the solar apparent magnitude is -26.77.

What I am after is the same computation for the situation if our primary light source would have i.e. -25Mag, -24,5Mag, -22Mag and so forth.

I found some hint here

What's the conversion between apparent magnitude and lux?

 

and did some computations, however they refer only to the source of light itself. I am still unable to estimate the illumination for the twilight sky when the Sun is i.e. 6 degrees below the horizon, etc.

Moreover I would be interested in calculations of the sky surface brightness in various directions based on the stellar magnitude of the light source and its position against the horizon. Is there any (quite simple) formula? Which would calculate this stuff? In Stellarium when we click on the Sun, we have its apparent magnitude and the magnitude "reduced by airmasses". I would need the formula for it either if possible.

Any hint will be cordially appreciated!



 



#2 TOMDEY

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Posted 19 February 2024 - 08:42 AM

Excellent and useful data! --- Just keep in mind... the color of the sky is "sky blue" and highly-polarized, especially 90o from the sun. This is how bees navigate and also causes the Haidinger's brushes that we see in that part of the sky.   Tom

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  • 213 Haidinger's Brushes appearance Tom Dey.jpg


#3 bob71741

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Posted 19 February 2024 - 10:01 AM

Try RCA's Electro-Optics Handbook EOH-10  Section 6. You can find a PDF copy online  here: https://www.ok1rr.co...cs-Handbook.pdf

 

You can buy a used copy at thriftbooks for about $11



#4 Krukarius

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Posted 21 February 2024 - 11:40 AM

This book is promising. I've checked it, although still don't have the most crucial information - the relationship of these values with thew apparent magnitude of thew Sun. I am after the formula, which would calculate it.



#5 mrm6656

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Posted 22 February 2024 - 12:28 PM

Krukarius:

 

My understanding is that you want to calculate some or all of the values listed in the two tables that you posted but for different "suns" that each have a different magnitude from the true sun. Assuming that's what you want the solution is pretty straightforward and is based on equation 6.3 in the Electro-Optics Handbook. The handbook explains the simple derivation of the necessary equation which is

                  E1 / E2 = 2.512^(M2 - M1)

and solving for E1 gives

                     E1 = E2 * 2.512^(M2-M1)

where M1 and M2 are two stars' magnitudes and E1 and E1 are their illuminances in lux.

 

To use this equation substitute M2 = -26.7, the sun's magnitude, and E2 = 130000 lux from the table. Then enter M1 as the "sun's" magnitude and calculate. For example

                     E1 = 130000 * 2.512^(-26.7 -(-24.0))
                     E1 = 130000 * 0.083166
                     E1 = 1081 lux

This result means that under the same circumstances in which the sun produces illuminance of 130000 lux, a -24th magnitude "sun" will yield 1081 lux. The factor 0.083166 is always the ratio between the illuminance that the sun produces and the illuminance that a -24th magnitude "sun" would produce under identical circumstances.

 

Substituting E2 for the other circumstances and solving gives the following table for a -24th magnitude "sun".
                    =================================
                     Circumstance         sun           "sun"
                    ---------------------------------------------------------
                    Sun overhead         130000      1081
                    Full daylight      10000-25000    831-2079
                    Overcast day           1000           83.2
                    Very dark overcast     100            8.3
                    Twilight                          10            0.83
                    Deep Twilight                 1           0.083 
                    =================================

 

The ratio also applies to the second table. That yields, in part, the table

                   ===========================
                   Solar altitude        sun           "sun"
                   ------------------------------------------------
                        90             129000          10728
                        45               77800             6470
                        20               27400             2279
                        10               11100                923
                         5                 4540                 378
                         2                 1920                 160
                         1                 1270                 106
                         and from the Handbook, pg 63
                        -6                  3.4                 0.28
                       -12            8.31x10^-4       0.013
                       -18            6.51x10^-4       0.0099
                   ===========================
The last three table rows are for the end of civil twilight, of nautical twilight and of astronomical twilight.

 

Other fictitious "suns" can, of course, be treated by substituting their magnitudes into the equation and proceeding as done here.

 

            --- Mike


Edited by mrm6656, 22 February 2024 - 12:38 PM.


#6 KBHornblower

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Posted 29 February 2024 - 09:46 PM

Krukarius:

 

My understanding is that you want to calculate some or all of the values listed in the two tables that you posted but for different "suns" that each have a different magnitude from the true sun. Assuming that's what you want the solution is pretty straightforward and is based on equation 6.3 in the Electro-Optics Handbook. The handbook explains the simple derivation of the necessary equation which is

                  E1 / E2 = 2.512^(M2 - M1)

and solving for E1 gives

                     E1 = E2 * 2.512^(M2-M1)

where M1 and M2 are two stars' magnitudes and E1 and E1 are their illuminances in lux.

 

To use this equation substitute M2 = -26.7, the sun's magnitude, and E2 = 130000 lux from the table. Then enter M1 as the "sun's" magnitude and calculate. For example

                     E1 = 130000 * 2.512^(-26.7 -(-24.0))
                     E1 = 130000 * 0.083166
                     E1 = 1081 lux

This result means that under the same circumstances in which the sun produces illuminance of 130000 lux, a -24th magnitude "sun" will yield 1081 lux. The factor 0.083166 is always the ratio between the illuminance that the sun produces and the illuminance that a -24th magnitude "sun" would produce under identical circumstances.

 

Substituting E2 for the other circumstances and solving gives the following table for a -24th magnitude "sun".
                    =================================
                     Circumstance         sun           "sun"
                    ---------------------------------------------------------
                    Sun overhead         130000      1081
                    Full daylight      10000-25000    831-2079
                    Overcast day           1000           83.2
                    Very dark overcast     100            8.3
                    Twilight                          10            0.83
                    Deep Twilight                 1           0.083 
                    =================================

 

The ratio also applies to the second table. That yields, in part, the table

                   ===========================
                   Solar altitude        sun           "sun"
                   ------------------------------------------------
                        90             129000          10728
                        45               77800             6470
                        20               27400             2279
                        10               11100                923
                         5                 4540                 378
                         2                 1920                 160
                         1                 1270                 106
                         and from the Handbook, pg 63
                        -6                  3.4                 0.28
                       -12            8.31x10^-4       0.013
                       -18            6.51x10^-4       0.0099
                   ===========================
The last three table rows are for the end of civil twilight, of nautical twilight and of astronomical twilight.

 

Other fictitious "suns" can, of course, be treated by substituting their magnitudes into the equation and proceeding as done here.

 

            --- Mike

My bold.  Your bottom line here is low by a factor of 10.  It should be about 10,810 lux.



#7 mrm6656

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Posted 01 March 2024 - 11:04 AM

KBHornblower:

 

Thanks for catching the 10x error and for pointing it out. I don't find a way to edit my original post now to correct that error. The values for twilight are also in error. Corrected values for them are in the following edited table.

 

                   ===========================
                   Solar altitude        sun           "sun"
                   ------------------------------------------------
                        90             129000          10728
                        45               77800             6470
                        20               27400             2279
                        10               11100                923
                         5                 4540                 378
                         2                 1920                 160
                         1                 1270                 106
                         and from the Handbook, pg 63
                        -6                  3.4                 0.28                 end of civil twilight
                       -12      8.31x10^-3       6.911x10^-4          end of nautical twilight
                       -18      6.51x10^-4       5.414x10^-5         end of astronomical twilight
                   ===========================

 

I can only speculate how I made these three dumb errors.

 

      ---- Mike



#8 Krukarius

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Posted 03 May 2024 - 03:17 AM

Fantastic! Thank you very much for this answer. It's very valuable!
 




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