Heuristic for fastest Mun Orbit to Kerbin sub-Orbit

[Gkibarricade]

Hi,

What would be the best way to figure out a the fastest return to Kerbin from Mun Orbit? (Of course with same DV)

 

So, for example. Currently I am in a polarish Mun orbit and Have 525 DV. I can burn now and get into a high (25Mm orbit) or I can wait and burn into a lower orbit. I am using any extra DV to get more equatoral. It doesn't seem to matter how I approach it, it seems to all take about 2 days to get to Kerbin periapsis. 

Is it true that for a given dv it will always take the same time or am I missing something?

Can anyone do better than 1D: 5H?

Here is my orbit if you want to try:

Peri: 6769

Apo: 8938

Inclination: 131.56

SMA: 207.67

Arguement of Peri: 223.7

LS Longitude: 299.760

Mun Time to Apo: 5 hr : 37 min

I think that is all that is required to build the orbit

 

[king of nowhere]

1 hour ago, Gkibarricade said:

Hi,

What would be the best way to figure out a the fastest return to Kerbin from Mun Orbit? (Of course with same DV)

i don't know of any exact method besides trying. spaceflight is as much art as science.

well, of course the fastest way would be to point at kerbin and burn several km/s worth. you'll certainly be fast then. but it would be very wasteful.

Quote

So, for example. Currently I am in a polarish Mun orbit and Have 525 DV. I can burn now and get into a high (25Mm orbit) or I can wait and burn into a lower orbit. I am using any extra DV to get more equatoral. It doesn't seem to matter how I approach it, it seems to all take about 2 days to get to Kerbin periapsis. 

Is it true that for a given dv it will always take the same time or am I missing something?

yes and no.

when you have a simple system like kerbin-mun, and you always use the same optimized trajectory, then you will always get the same time, or close enough; because you are always doing the same manuever. you would get higher times if you were doing an inefficient manuever.

On the other hand, when you get to more complex realities like an interplanetary transfer, then the flight time will vary considerably, even among the same launch window. I once sent 4 probes together to jool, with similar deltaV, on what i thought were similar trajectories, but more than one year passed between the first and the last.

[Gkibarricade]

So a few trials got me 3.5 hrs for 550 DV and 1 D : 4hrs for 510 DV. The chasm between these two is what has got me scratching my head.

Anyone know of any excel based "manuever planners"?

[bewing]

Generally, an efficient trip to the Mun or back is going to take about one Kerbin day (6 hours). Yes, you can add some speed to that to make it go faster, but that's almost always a bad idea. Why? Because every m/s of dV that you add, you have to subtract back off again when you get to your destination. On most trips that means burning extra fuel and wasting dV. But going back to Kerbin suborbit means burning the hell up in the atmosphere. Kablooie!

An efficient trip has you reaching something close to zero velocity as you cross the SOI boundary.

So, usually you don't want to minimize the time of the voyage. You want to minimize the dV. Even when just going back to Kerbin, having a decent amount of fuel onboard allows you to come close to pinpointing your landing, and minimizing your reentry heating.

 

[Wobbly Av8r]

While I was unable to cheat a vessel into that orbit, what your results tell me is that the orientation of your orbit prevented you from slowing your orbital velocity around Kerbin. Using your 525 dV to minimize time to Kerbin would require that the plane of the orbit around the Mun allow the KOI (Kerbin Orbital Insertion) to leave the Mun's SOI in the most retrograde manner possible; the most efficient orbits are those that pass through or close to the Mun's retrograde 'node' in its orbit around Kerbin.

In an equatorial orbit, this position occurs once every orbit while the more inclined the orbit, the further from this point you move as the Mun orbits Kerbin resulting in a less and less efficient burn - unless you wait until the combination of your specific inclined orbit and the Mun's orbit around Kerbin RESULT in an position that passes through or close to the Mun's retrograde node.

So time, then, is relative - burn now (in some arbitrary position) and take a long time to travel to Kerbin, or wait an amount of time until your orbit relative to the Mun/Kerbin allows the 525 dV burn to produce the most efficient trip betwen the Mun and Kerbin. Either way, you're gonna need time to overcome your inclined orbit within a dV budget.

Edited September 28, 2020 by Wobbly Av8r