Orbital darkness time for polar & highly eccentric orbits?

[Cooper42]

The Wiki has the equations for orbital darkness time: https://wiki.kerbalspaceprogram.com/wiki/Orbit_darkness_time

And there are a number of calculators, spreadsheets & mods out there that calculate this.

However, all are based upon the presumption of an equatorial orbit (or low inclination orbit).

Are there any thoughts on how to calculate this for highly inclined & eccentric orbits?

 

I'm planning a number of polar orbit, high Ap, high eccentricity communications relays for RemoteTech.

I realise KSP doesn't calculate ec use for on rails craft, but I'd like to have some idea for when building to aim for some kind of plausibility.

However, an equatorial orbit with a high Ap would need >70k batteries!...

 

There's a post from 2016 by 'Hamsterjam' regarding the general case on the Wiki talk pages, but the link is long dead: : https://wiki.kerbalspaceprogram.com/wiki/Talk:Orbit_darkness_time#What_about_eccentric_polar_orbits.3F

I can find no more information on how to begin calculating this, and all the tools, spreadsheets and mods use the formula that presumes an equatorial orbit...

 

[overkill13]

Non equatorial orbits are going to have differing amounts of daylight depending on when in the year you are orbiting.  As the planet orbits around the sun, your orbit angle relative to the sun will differ and as a result how long the planet blocks your line of sight to the sun will differ over time.  At some points you may have a continuous view of the sun, other times you may not see the sun for days and any possibility in between.  It may be possible to calculate for a particular day, but you are going to be all over the place over a planetary year.

[Cooper42]

12 hours ago, overkill13 said:

Non equatorial orbits are going to have differing amounts of daylight depending on when in the year you are orbiting.  As the planet orbits around the sun, your orbit angle relative to the sun will differ and as a result how long the planet blocks your line of sight to the sun will differ over time.  At some points you may have a continuous view of the sun, other times you may not see the sun for days and any possibility in between.  It may be possible to calculate for a particular day, but you are going to be all over the place over a planetary year.

Yeah, I realise that, I'm just looking for a way to calculate the worst-case scenario to allow for maximum up-time.

 

But the worst case scenario for a highly eccentric polar orbit will still be much less than the same eccentricity but equatorial.

[AHHans]

55 minutes ago, Cooper42 said:

Yeah, I realise that, I'm just looking for a way to calculate the worst-case scenario to allow for maximum up-time.

Well, if you are O.K. with something that is not be 100% right but is probably overestimating the darkness time, then you for an orbit with its Ap above one of the poles you could use the general result for a circular orbit with the radius of the Pe. If the Ap is exactly above one of the poles then this value will be an overestimation. But if the Ap is not directly above a pole but somewhat offset then the actual maximal darkness time will become larger with a lower latitude of the Ap. And I don't know at what latitude this simple estimation stops working. But I guess(!) for latitudes > 70 deg it is probably good enough.

[Corax]

On 4/1/2020 at 4:18 PM, Cooper42 said:

the worst case scenario for a highly eccentric polar orbit will still be much less than the same eccentricity but equatorial

No it won't.

Worst case is when your AP (the slowest point in your orbit) is directly opposite the sun, viewing through the planet's centre; for that it does not matter whether it's polar or equatorial, you're still occluded by the same planetary disc, which takes exactly the same amount for otherwise identical polar or equatorial orbits (maybe unless you take the planet's precession along its orbit into account too, but that would be minuscule compared to your orbital period).

On average (or in total, whichever way you want to look at it), insolation will be better with polar orbits since you'll spend much of the time in a plane along the planet's path above/below the planet's limb, with a clear view of the sun, and much fewer short periods where it is occluded by the planet as opposed to an equatorial orbit, where the occlusion happens on every orbit.

TL;DR:

The maximally possible time, i.e. worst case, your satellite will be in darkness does not depend on the orbit's inclination for otherwise identical orbits (same AP/PE). You'll need the same battery capacity for either case.

[Rhomphaia]

16 hours ago, Corax said:

No it won't.

Worst case is when your AP (the slowest point in your orbit) is directly opposite the sun, viewing through the planet's centre; for that it does not matter whether it's polar or equatorial, you're still occluded by the same planetary disc, which takes exactly the same amount for otherwise identical polar or equatorial orbits (maybe unless you take the planet's precession along its orbit into account too, but that would be minuscule compared to your orbital period).

On average (or in total, whichever way you want to look at it), insolation will be better with polar orbits since you'll spend much of the time in a plane along the planet's path above/below the planet's limb, with a clear view of the sun, and much fewer short periods where it is occluded by the planet as opposed to an equatorial orbit, where the occlusion happens on every orbit.

TL;DR:

The maximally possible time, i.e. worst case, your satellite will be in darkness does not depend on the orbit's inclination for otherwise identical orbits (same AP/PE). You'll need the same battery capacity for either case.

Unless your AP is over a pole, or at at least at enough latitude and altitude to stay out of preumbra. that will cut the maximal tine considerably.

[Corax]

3 hours ago, Rhomphaia said:

Unless your AP is over a pole, or at at least at enough latitude and altitude to stay out of preumbra. that will cut the maximal tine considerably.

 

19 hours ago, Corax said:

when your AP (the slowest point in your orbit) is directly opposite the sun, viewing through the planet's centre

 

[Cooper42]

  

On 4/6/2020 at 2:35 AM, Corax said:

Worst case is when your AP (the slowest point in your orbit) is directly opposite the sun, viewing through the planet's centre; for that it does not matter whether it's polar or equatorial, you're still occluded by the same planetary disc

 

Could you explain more please? As I would assume the same as Rhomphaia

 

You say the worst case scenario for darkness time doesn't change based upon inclination. But, if the Ap is directly above a pole, how could that lead to a situation where the Ap is directly opposite the sun, through the planet's centre?

Edited April 7, 2020 by Cooper42

[sturmhauke]

1 hour ago, Cooper42 said:

  

 

Could you explain more please? As I would assume the same as Rhomphaia

 

You say the worst case scenario for darkness time doesn't change based upon inclination. But, if the Ap is directly above a pole, how could that lead to a situation where the Ap is directly opposite the sun, through the planet's centre?

That's not a function of inclination, but of argument of periapsis. It describes the direction of the periapsis relative to the reference plane.

[Corax]

6 hours ago, Cooper42 said:

If the Ap is directly above a pole, how could that lead to a situation where the Ap is directly opposite the sun, through the planet's centre?

I think I may have misinterpreted your original question; you want to know the maximum possible time for the period of orbit in darkness for a specific orbit, not the general worst case which would always be with the AP opposite the sun.

My pedestrian approximation would be to assume an infinitely far away star–that is, parallel light rays–which would give me the distance I'd have to pass through shadow as equivalent to the planet's diameter. Then using trigonometry and Kepler's Second Law, it should be possible to deduce the time it takes to pass through that portion of the orbit.

[Cooper42]

Okay, thanks.

After a bit more searching and a refresh on some orbital mechanics I once knew, but have since forgotten after I stopped playing KSP for some time, I've found the calculation steps.

I'm putting it here in case anyone else comes across this, trying to work out the same thing.

Step-by-step to calculate the longest possible period of time in darkness for an elliptical orbit with the Ap directly above the pole of a body.

All values in meters, seconds and radians.

 

Need to know the following:

Ap = Apoapsis altitude

Pe = Periapsis altitude

R = radius of planet (check the KSP wiki)

GM = Standard gravitational parameter (check the KSP wiki)

 

Calculate:

R_a = Ellipse radius at Ap

R_a = Ap + R

 

R_p = Ellipse radius at Pe

R_p = Pe + R

a = semi-major axis

a =  = R_p + R_a / 2

b = semi-minor axis

b =  = sqrt(R_p * R_a)

e = eccentricity

e =   = sqrt(1 - (b/a)^2)

F = distance between centre of the ellipse (C) and the focal point / the planetary body (F₁)

F =  = sqrt(a^2 - b^2)

 

Now, presuming that the penumbra of the planet is a cylinder that extends infinitely behind the planet (which is a close enough approximation) we now have enough to use some basic trigonometry to work out the eccentric anomaly, as shown on this diagram from Wikipedia:

 

E₁ = Eccentric anomaly as satellite enters the penumbra

E₂ = Eccentric anomaly as satellite leaves the penumbra

E₁ =  = arccos((F + R) / a)

E₂ =  = arccos((F - R) / a)

M = Mean anomaly

M =  = E - e * sin(E)

Calculate M₁ and M₂ for E₁ and E₂ respectively

ΔM = M₂ - M₁

n = mean motion

n =  = sqrt(GM / a^3)

 

Finally

Δt = Time to travel between E₁ and E₂ (Which is the time spent in darkness)

Δt =  = ΔM/n

 

Note that this time in darkness won't be experienced every orbit.

 

Example.

For an orbit around Kerbin, with Ap above one pole of 60,000km and a Pe above the other pole of 120km

Δt = 809.4 seconds

 

(As I type this out, I'm sure there's almost certainly a quicker / more efficient way of doing this, but my doctorate is in a social science, so cobbling this together will do for me!)

 

On 4/1/2020 at 4:26 PM, AHHans said:

Well, if you are O.K. with something that is not be 100% right but is probably overestimating the darkness time, then you for an orbit with its Ap above one of the poles you could use the general result for a circular orbit with the radius of the Pe. If the Ap is exactly above one of the poles then this value will be an overestimation. But if the Ap is not directly above a pole but somewhat offset then the actual maximal darkness time will become larger with a lower latitude of the Ap. And I don't know at what latitude this simple estimation stops working. But I guess(!) for latitudes > 70 deg it is probably good enough.

The orbital darkness period for a circular orbit of 120km Ap / Pe is 640.51s

Which is, thankfully, quite a bit different than that which I calculated: Which is good, it means I didn't waste all that time when there was a good enough easy estimation!

Although, this is a highly eccentric orbit. For less eccentric orbits, calculating the orbital darkness time for an equatorial orbit with Pe & Ap = Pe of intended orbit, and then add a bit for good measure, would be close enough.

 

Useful sources:

https://space.stackexchange.com/a/23129

https://en.wikipedia.org/wiki/Eccentric_anomaly

and, as always:

http://www.braeunig.us/space/orbmech.htm

 

Edited April 9, 2020 by Cooper42

[AHHans]

9 hours ago, Cooper42 said:

Which is, thankfully, quite a bit different than that which I calculated: Which is good, it means I didn't waste all that time when there was a good enough easy estimation!

Oh! That's not what I expected. It doesn't happen often that my gut feeling is that wrong.

Well, you live and learn.