Size of Mun eclipse shadow on Kerbin (derivation found & tested)

[Drew Kerman]

I was looking for a way to calculate the size of the shadow that Mun would cast on Kerbin when it eclipses the sun as seen from the equator, where it would be total. After some searching I came across this blog entry with a derivation for calculating the shadow size of the Moon on Earth. I went in and plugged the numbers to see if I was doing it right and ended up with the proper result (273.14km). So I then went and tried it for Mun and Kerbin. Interestingly, it gave me a negative number:

rSun = 261600000m

rMun = 200000m

dSun = 13599840256m

dMun = 12000000m

rKerbin = 600000m

angle = 0.019238 = arcsin((rSun - rMun)/(dSun - dMun))

dUmbra = 10396700 = rMun/sin(angle)

-38607.73 = 2*(dUmbra - (dMun - rKerbin))*tan(angle)

So! My first question - is this equation even valid for Mun/Kerbin? If so, what does the negative result mean?

[*Aqua*]

The calculation appears to be correct. A negative value at the end means there's no umbra.

It becomes clear when calculating the length of the umbra:

dUmbra = 10396700 = rMun/sin(angle)

That's 10.4 million meters or Megameters.

The Mun orbits Kerbin at a distance of 11 Megameters. That means it can never cause a total solar eclipse and that's the reason why there'll no umbra on the surface of Kerbin.

[Drew Kerman]

It becomes clear when calculating the length of the umbra

Well, that's obvious I knew I was missing something teach me not to just look at the result heh.

Very cool, I always suspected the eclipses were annular but it's really hard to tell in the game whether you're looking at just the corona or still some part of the surface

Further question - does anything think 38607.73m is the size of the antumbra?

Edited April 16, 2015 by Gaiiden

[Fel]

*Being really lazy and making stuff up*

Is the general idea just to figure out what happens when two circles of different sizes overlap?

Okay, so you're making a cone, basically.

dUmbra = dMun = rMun/sin(angle_special)

Okay, basically, you need to solve the equation for angle_special.

angle_special = arcsin((rSun_special - rMun)/(dSun - dMun))

Then solve for rSun_special.

The result will be just how much sun is visible at the "best attempt" location. The antumbra I think.

Edited April 16, 2015 by Fel