Calculating Day/night Cycles

[meatballcannon]

So I've been planning a kethane base on the mun, and I was wondering how much battery power would be required to fill my reserves while my kerbals are on the dark side of the mun.

Rather than fudging it or using a generator, I decided to simply learn how to do the math.

As I understood it, if the mun was tidally locked, Then it my base would have a day of 1/2 of a munar orbit. However, since the mun, Kerbin, and Kerbol are all in the same plane, There would be a rather lengthy eclipse every day, right? And to make matters worse, the mun isn't tidally locked.

So as I understand it, the rotation of the mun is its sidereal rotation, and the time for one of these to complete is 1 day, 14 hours, 36 minutes, and 24.4 seconds. Am I doing this right?

Anyway, this looks like a related rates problem to me, but I don't know how to set it up, and I'm not sure how to handle the eclipse. Any ideas?

[TheShadow1138]

According to the wiki, the Mün is indeed tidally locked with Kerbin, and i don't believe that the eclipse that would take place every Münar day actually affects the solar panels as it would in reality. I believe you should use the Synodic Rotation, which in the case of the Mün (someone correct me if I'm wrong here), should be the same as the Synodic Orbital Period since it is tidally locked. So, it's actually, as far as the game is concerned, a fairly simple problem. You only need to have enough battery power to get you through half of the Synodic Orbital Period, or 19h 35m and 57.7 seconds.

[RenevB]

I'd say don't worry about the eclipse. The Mun is going fast enough that it'll be negligible compared to the night. If your base gets through the night, it will certainly survive the eclipse.

As said above, the Mun is tidally locked and the Munar night is half the synodic orbital period, which is indeed the number mentioned by the poster above me.

[meatballcannon]

Ok. I guess sidereal rotation wasn't what I needed. Thanks.

[Yasmy]

No need to get fancy.

Kerbin is 1.2 Mm in diameter and 12 Mm away from the moon.

Thus the angular size of Kerbin from the Munar surface is atan(0.1) ~= 0.1 radians ~= 5.7 degrees.

If your solar panels track Kerbol, then you lose 0.1/pi ~= 5.7/180 ~= 3.2% of your Munar daylight to eclipse.

If your solar panels do not track Kerbol, then you lose sin(0.1)/2 = 5% of your solar power.

Edited May 8, 2014 by Yasmy