3

When using mods like SCANsat, I find I spend a lot of time having to wait for my spacecraft to complete a certain number of orbits in order to complete whatever task it is I launched it for. To minimize this time, I want to find out what altitude is best to orbit at, such that the orbit period divided by the maximum time warp multiplier results in the least amount of real-time spent per orbit.

I know there's plenty of charts online that provide the values for maximum time warp allowed at certain altitudes. What I need now is a reference or formula to get the orbital periods at those altitudes, so I can figure out which will have the least real-time orbit period.

Is there an easy way to do this, or is this one of the harder bits of rocket science that I'm going to have to learn some advanced maths for?

Improve this question
1
  • 3
    The joy of ScanSAT is that you can fly other missions and let the scanner tick away, it doesn't have to be the active vessel, but if you want the quickest result then see SaintWacko's answer
    – CyanAngel
    Oct 6, 2014 at 12:47

1 Answer 1

Reset to default
5

The short answer is that the shortest real-time orbit will always be at the altitude at which you can use the maximum time warp. On Kerbin, going from the lowest possible orbit (70km) to the orbit at which you can use max (100000x) time warp (600km), will increase your orbital period from 1834 seconds to 4395 seconds (240%), while the time warp available increases from 50x to 100000x (200000%). This means that an orbit at 600km will take approximately 0.1198% of the real time that an orbit at 70km will take. While the exact numbers will vary on different bodies, I feel confident in saying that the altitude for max time warp will always be the fastest real-time orbit.

In order to calculate this I used the advice in this tutorial. I used the masses of Kerbin and the Mun from that thread, as they were pre-formatted in the proper manner*. I then entered into Wolfram Alpha orbital period <mass> <equatorial radius* + orbit altitude> e.g. orbital period 5.2915793*10^22 kg 840 km. This allowed me to find the orbital period at each minimum time warp altitude, and from there I was able to divide each orbital period by the max time warp at that altitude to find the real-time orbital period.

In conclusion, yes, it's one of the harder bits of rocket science, but, like all bits of rocket science, there are equations for it, and Wolfram Alpha knows all of the equations.

*Equatorial Radius and Mass can both be obtained from the sidebar on the KSP Wiki. 

Improve this answer
2
  • 2
    So, in short: The lowest altitude at which 100kX warp is available will be the fastest, because time warp increases much faster with altitude than orbital period?
    – Iszi
    Oct 6, 2014 at 13:58
  • Correct. Sorry, just now saw your comment.
    – SaintWacko
    Oct 14, 2014 at 15:52

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .