Can you provide the equation you used ?
Be careful what you wish for.
Here is the method I used to calculate apparent magnitude (brightness) of our lighting. I had not references for this, so another method might be better. My hope was to be within one or two magnitudes (2.5x to 6.5x the actual brightness).
[The green are the input variables, yellow are the key results needed]
Step 1: Choose a reference light.
I chose a star like Vega or something initially, but then chose the Sun, which is a better reference source. [We know the luminisoty (watts) of the Sun and we know how bright it is, so we can assume a brightness level and compute the wattage needed if we make the adjustments for the differences between the Sun and a given light source, which is mainly the efficacy.]
Step 2: Convert the Sun's magnitude for the distance desired.
[This is the distance to the ISS.]
Enter Beginning mag.: -26.70
Enter Beginning dist., km: 149,476,000
Enter new dist., km:
350
Distance in a.u.: 0.0000066845
Brightness change: 5.4827E-12
Magnitude change: -28.15
Resulting mag.:
-54.85
[
Note 1: converting brightness for a moving distance, use the inverse square law. So, take the square of the distance ratio: (350 km/ 149,476,000 km)^2]
[
Note 2: Convert this brightness change to magnitude change, though you could eliminate this step throughout the procedure, but I like magnitude values. The log (base 2.512) will do the conversion to mag. change.]
Step 3: Convert solar wattage into useable (ie visible) brightness
Bolometric Luminosity, watts: 3.94E+26
Hemisphere luminosity, watts: 1.9695E+26
Visible percentage: 45%
Visible luminosity, watts: 8.86275E+25
Step 4: Set the desired brightness (in mag.) you want for your light as will be seen at the distance you set above (ie ISS distance), and compute wattage equivalent for a solar source.
Desired mag.
0
Required Mag. reduction: 54.85
Brightness reduction: 8.7E+21
Required Luminosity, watts:
10,152
[Mag. 0 is found for the star Vega, which is a bright star, and was my original target brightness that I assumed would be just adequate for Don to see if he knew about where to look. Robert Reeves had already given Don a map that Robert made of just where we would be in relation to San Antonio and Austin.]
What this says is that a 10,152 watt glowing ball with the same spectrum of the Sun would appear as a 0 magnitude light source seen from space (no atmosphere issues considered yet) at a distance of 350 km.
But this is for a projection of light that is hemispherical. Here is where the laser shines. [Ooh, that has to be a nasty pun in this forum. oh well]
Step 5: Time to calculate the gain with narrow divergence:
Angular dia. of beam:
0.086 deg.
Angular area of beam: 0.0058
No. of sq. degrees in hemisphere: 20,627
Gain in beam by area:
3,552,703
Wow! That is some gain, assuming the spec. on the laser of 1.5mRad is correct, and it is close to the measurement I got later (1.7 mRad) done after Keith made Don and Dan look blue.
Step 6: Time to adjust the perfect gain above to a more realistic gain by guessing the inefficiencies:
Percent of light in main beam:
80%
Imperfections in lenses, etc. will send some light well outside our main beam.
Internal thermal efficiency:
40%
“Thermal efficiency” likely still means overall efficiency of the device. There will be process that take place that rob power and expelled as heat, normally. This is essentially output light power/input electrical power.
Eye reception efficiency:
70%
Our blue color cones are only 2% in number compared to the other two color cones, yet the brain does wonders with these few “blue” cones so that they are about 70% as receptive as the mid spectrum colors. [This is a fudge, really, because we are still using the Sun as a reference “color”, yet blue is less effective.]
Net Gain due to light itself:
795,806x [compared to a hemispherical light emitting source]
Step 7: Time to convert from solar wattage to our light’s wattage requirement (input power):
Required Beam Luminosity: 0.0128 watts [if Sun-like light source]
This is simply the wattage we found in step 4 (10,152 watts) divided by the 795,806 gain.
Lumens produced: 1.28
The Sun produces about 100 lumens per watt, so we multiply by 100.
Our light source efficacy:
150 lumens
This is a guess, but I think this is a fair one. Is it?
Total Wattage Required:
0.0085
Step 8: Time to consider all the losses that takes place between the laser and the eye of the beholder, Don. We will first consider the loss due to the atmosphere. About 70% of blue light will make it through the mass of one atmosphere (AM1, distance from sea level to top of atmosphere).
There are several equations that can be used to calculate the amount of air mass light travels through given a known angle from zenith (directly overhead).
Enter the altitude angle for the ISS:
65 deg.
[This is specific to the 350km distance stated above. I used a vlookup() function for a table I set-up that gives the ISS distance and altitude angle, which I got from the Heavens Above web page for the ISS.]
Atmospheric transmittance: 67.5%
This is the 70% loss per atmosphere taken to the exponent of the air mass number, (0.7^1.10).
But how did we get the air mass number…
Well, first subtract 90 degrees from the altitude to zenith angle.
Zenith angle = 25 deg.
Use the following formula to get a value to use in a subsequent formula.
=1/COS(RADIANS(25 deg)) = 1.103377919
[Remember that Excel used radians, not degrees]
Square this number = 1.2174
Air Mass = 1.103377919 * (1-0.0012 * ( 1.2174-1))
Air Mass = 1.10 [used above]
Step 9: Final fudge for efficiencies…
Due to dust, pollen, etc., more loss will take place so enter what you think.
I used 65%
Multiply these last two percentages to get…
Net atmospheric transmittance:
44%
Step 10: Calculate the final input wattage required to get the desired apparent magnitude value set earlier.
Divide the wattage found earlier of : 0.0085 by the 44% efficiency value to get the final result.
Net lighting wattage required:
0.019 watts or 19 mW.
That ain’t much as you know. But what if we had a 1 watt light, what apparent magnitude would it look like. Enter -4.3 and you should get about 1 watt.
Altitude angle for the ISS does make a big difference. Use 1000 km at an altitude angle of 15 degrees and you will need about 22 watts.
The question I would ask this forum would be whether or not the 150 lumens per output watt is accurate. Is it close?
Congrats to any reader that made it this far.
