# A planetary atmospheric timing question



## Gryphos (Dec 3, 2015)

So I have a world in which their are giant cosmic eyeballs hovering all over the sky at geostationary orbit. Every night 'Eyesight' occurs, and all the Eyes Above light up. The thing is, they only light up once the sun has set from their perspective, so obviously the way it works is that form ground level, the sun sets, followed by a brief period of darkness as the eyes all light up in canon, like a wave going from east to west.

I know that the moment the sun sets on the western horizon, the Eyes on level with the eastern horizon would light up. My question is, from the moment of sunset, how long would it take for the wave of Eyelight to reach the western horizon?

I've decided that the Eyes are about 10km above the earth, on the lower boundary of the stratosphere.


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## NerdyCavegirl (Dec 3, 2015)

I'm physically incapable of that math, but I just gotta say that's an awesome freakish idea. Oh and I replied to a character Q&A with eyes in the sky the other night, I think it might've been yours.


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## Seraphim (Dec 3, 2015)

So this is a very difficult problem for two reasons. The distance to the horizon is different based on you latitude (distance north or south from equator), and two, because the earth spins at a different relative rate based on latitude so the answer is different based on your location. It spins at a constant radian/sec everywhere but at different km/hr at different latitudes because the distance to the Earth's axis of rotation is different( at the poles the Earth spin is approximately 0).

However, at the equator the Earth spins approx 15degrees/hr(pi/12 rad/hr) and the altitude of a single eye is 10 km while a person's eyes altitude is approx 1.2m. Using trigonometry to solve for the radians between the twos distances, it takes the Earth then approximately 12 mins and 41 secs between a human's sunset and the eyeball's sunset.

You may use this approximate for most locations, but the farther away from the Equator, the value decreases or increases depending on the season. (in summer, the farther north you go the time increases, and winter it decreases. The opposite is the case for moving south.) 

This value doesn't account for the atmosphere. Light actual bends when it hits the atmosphere so the sun is setting, it is actually lower than it appears. The affect of the bend varies based on altitude.


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## Gryphos (Dec 4, 2015)

NerdyCavegirl said:


> I'm physically incapable of that math, but I just gotta say that's an awesome freakish idea. Oh and I replied to a character Q&A with eyes in the sky the other night, I think it might've been yours.



Thanks, I'm glad you like the idea. I did see that you responded to the Q&A, but I don't think I can continue it because since posting it, this world and that character have gone through total revision.


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## Gryphos (Dec 4, 2015)

Seraphim said:


> So this is a very difficult problem for two reasons. The distance to the horizon is different based on you latitude (distance north or south from equator), and two, because the earth spins at a different relative rate based on latitude so the answer is different based on your location. It spins at a constant radian/sec everywhere but at different km/hr at different latitudes because the distance to the Earth's axis of rotation is different( at the poles the Earth spin is approximately 0).
> 
> However, at the equator the Earth spins approx 15degrees/hr(pi/12 rad/hr) and the altitude of a single eye is 10 km while a person's eyes altitude is approx 1.2m. Using trigonometry to solve for the radians between the twos distances, it takes the Earth then approximately 12 mins and 41 secs between a human's sunset and the eyeball's sunset.
> 
> ...



Tis is SO helpful, thank you so much. The exact timing doesn't really matter that much, I just needed a ballpark figure in my head. About 10-15 minutes it is.


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