# A lawyer, an accountant and a doctor debate science with a 12 year old...

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EDIT: FYI, I've discovered that my brilliant, alcohol-fueled 'proof' of an acute angle between straight up and the sun at sunset is false!  Read on for entertainment / edification if you want.  I'm still interested to know if (despite being wrong about the angle) the method works for figuring out what star was blinking.

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OP Below:

I'm standing on a hill overlooking the ocean at sunset. I look straight up and see a star begin to blink...(not twinkle, blink)

… I know that based on this information I should be able to pinpoint what star the signal is coming from.  Sadly, I don't have the maths for this.  But can you guys help me to know whether my scratch-paper figuring is correct?

This little thought experiment came about through a conversation with a couple of friends and my son several nights ago.  The debate meandered about which direction is 'straight up' at night after sunset, and eventually what the angle was between straight up and the sun.  After drawing circles and rays on a napkin I was able to convince them that 'straight up' immediately after sunset is a pretty sharply acute angle with the sun, despite all appearances to the contrary.

So I've tried to figure this out on my own, but what I'd like to do, now, is get someone here to help me understand it mathematically  (Mind you, both of my degrees are non-science degrees, so please bear with me!)

I know from reading Patrick O'Brian's Aubrey / Maturin novels (e.g. Master and Commander, et. al.) that for an observer about 100 feet above sea level that his or her horizon is actually about 11 nautical miles.  With the earth being 21.000 NM in circumference, if we divide that by 360, we get 60 NM for each degree, and 11 NM is roughly about 1/5th of that... so if I'm correct (and ignoring atmospheric effects) -- "Straight Up just after sunset should be about 1/5th of a degree from the edge of the sun.

Am I correct so far?

So having figured that out, if I determine my distance from the equator - let's say my hill is in Santa Barbara, California (about 2069 NM) - which is about 34 degrees - shouldn't I be able to combine the two angles and the date to figure out what star just happens to be blinking?

I'm assuming at this point I'd have to find some resource that shows where the visible stars are in relation to earth's orbit on a given date.. but would my two angles theory described above be enough to narrow it down?

Edited by JoeSchmuckatelli
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I think that the angle to the horizon is pretty much a red herring. The star nearest the zentith (the point right above you) is determined by you latitude, the position of earth in it's orbit and your local time. You could calculate the local time from the fact that it is sunset, the position of earth in it's orbit and your lattitude if you had too though.

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32 minutes ago, JoeSchmuckatelli said:

… I know that based on this information I should be able to pinpoint what star the signal is coming from.  Sadly, I don't have the maths for this.  But can you guys help me to know whether my scratch-paper figuring is correct?

This little thought experiment came about through a conversation with a couple of friends and my son several nights ago.  The debate meandered about which direction is 'straight up' at night after sunset, and eventually what the angle was between straight up and the sun.  After drawing circles and rays on a napkin I was able to convince them that 'straight up' immediately after sunset is a pretty sharply acute angle with the sun, despite all appearances to the contrary.

So I've tried to figure this out on my own, but what I'd like to do, now, is get someone here to help me understand it mathematically  (Mind you, both of my degrees are non-science degrees, so please bear with me!)

I know from reading Patrick O'Brian's Aubrey / Maturin novels (e.g. Master and Commander, et. al.) that for an observer about 100 feet above sea level that his or her horizon is actually about 11 nautical miles.  With the earth being 21.000 NM in circumference, if we divide that by 360, we get 60 NM for each degree, and 11 NM is roughly about 1/5th of that... so if I'm correct (and ignoring atmospheric effects) -- "Straight Up just after sunset should be about 1/5th of a degree from the edge of the sun.

Am I correct so far?

So having figured that out, if I determine my distance from the equator - let's say my hill is in Santa Barbara, California (about 2069 NM) - which is about 34 degrees - shouldn't I be able to combine the two angles and the date to figure out what star just happens to be blinking?

I'm assuming at this point I'd have to find some resource that shows where the visible stars are in relation to earth's orbit on a given date.. but would my two angles theory described above be enough to narrow it down?

Huh?  I'm sure someone in your group argued that directly overhead was about 90 degrees from the sun.  And your argument for a very acute angle is???

Many years ago, on a vacation with my family, I derived the general case for how far you can see over the ocean, based on lines tangent to a circle.  I won't derive it again right now, but I can tell you several variables are squared or square-rooted so the approximation you made is not accurate.  IIRC standing 6' above sea level lets you see about 4 miles.   Hotel balconies let you see 10-20 miles.  At long ranges you really see mostly things above the horizon like the crest of a wave and the crest of the next wave beyond it, without seeing any surface in between.

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28 minutes ago, JoeSchmuckatelli said:

"Straight Up just after sunset should be about 1/5th of a degree from the edge of the sun.

"Straight up" is a point called the zenith. It's normal to the surface of Earth. You're suggesting that the star that is at the zenith should be less than one sun-radius above the horizon.

I think the correct statement would be "I'm standing 1/5 of a degree away on Earth's surface from the terminator." The angle the star makes with the Sun would be 90.2 degrees, then.

If you know where you are and what day and time it is, you should be able to find the star that is 90.2 degrees above the sun by figuring out the sun's azimuth at sunset (or in the case of your measly hill, about thirty seconds after sunset) and going to 90 degrees above that. Actually, you don't need to do that at all, since the zenith is a single defined point. Just figure out what time the sun sets for that day, and go to the star that's at the zenith for your location at that time. Given that you're seeing it at sunset, it's got to be a pretty bright one. For such a small hill, you won't even need to worry about the altitude difference. Even if you were on Mount Everest or cruising in an intercontinental airliner, the difference in the sky would only be 3 degrees, hardly distinguishable on a handheld starmap.

So the bottom line is, you can find out what star it is if you also know the date and time and your location, or just the date and location if you have a table of sunset times (you probably can derive the table too if you know Earth's axial tilt). Otherwise, if you don't know what day it is or where you are you're better off waiting an hour so you can see the rest of the constellation it's in and figure it out that way.

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45 minutes ago, farmerben said:

Huh?  I'm sure someone in your group argued that directly overhead was about 90 degrees from the sun.  And your argument for a very acute angle is???

Many years ago, on a vacation with my family, I derived the general case for how far you can see over the ocean, based on lines tangent to a circle.  I won't derive it again right now, but I can tell you several variables are squared or square-rooted so the approximation you made is not accurate.  IIRC standing 6' above sea level lets you see about 4 miles.   Hotel balconies let you see 10-20 miles.  At long ranges you really see mostly things above the horizon like the crest of a wave and the crest of the next wave beyond it, without seeing any surface in between.

This was the exact crux of the discussion.

… and, damnit, having revisited my napkin... I see now what I did wrong.

I drew a ray from the earth toward the sun, as if it were a finite point.  Then, on the circle I'd drawn for the earth, I drew a tiny circle to show the local horizon, with its circumference just touching the ray I'd drawn to the sun.  From the center of that circle, my ray for "Straight Up" formed an acute angle.

However - I now see what I did wrong: I shouldn't have used the sun as a finite point.  I should have recognized that the sun actually lights up 1/2 the sphere (if not more, given its true size compared to the earth), and then drawn my tiny horizon circle on the edge of that... which would indeed have made my 'straight up' ray perpendicular to the sun.

Thanks to @cubinator for helping with this: I completely forgot about the terminator

Laughing b/c this is what happens when a lawyer, an accountant and a doctor debate science with 12 year old.

FYI:

Quote

Assuming no atmospheric refraction and a spherical Earth with radius R=6,371 kilometres (3,959 mi):

• For an observer standing on the ground with h = 1.70 metres (5 ft 7 in), the horizon is at a distance of 4.7 kilometres (2.9 mi).
• For an observer standing on the ground with h = 2 metres (6 ft 7 in), the horizon is at a distance of 5 kilometres (3.1 mi).
• For an observer standing on a hill or tower 30 metres (98 ft) above sea level, the horizon is at a distance of 19.6 kilometres (12.2 mi).

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Edited by JoeSchmuckatelli

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