What exactly is the length of a Kerbin day?

[CaelumEtAstra]

I'm setting up a RemoteTech network in a keostationary orbit. Before 1.0, the day was 59h, 59m, 59s. Now I'm hearing that it is exactly 6h. I'm asking so I can get the orbital period right to prevent the satellites from drifting. Thanks in advance! 

[Chaos_Klaus]

The wiki is your friend:

http://wiki.kerbalspaceprogram.com/wiki/Kerbin

Look for sidereal rotation period. It's not exactly 6h.

[FancyMouse]

It was a recent change (post 1.0) that makes a solar day a whole day, so sidereal rotation period is a few seconds shorter.

Found the thread: 

 

[GoSlash27]

Well... that depends on what you call a "day".

A "solar" day (sunrise/ sunset/ clocks) is precisely 6 hours.

 A "sidereal" day (Kerbin's rotational period in inertial space; WRT distant stars) is 5 hours, 59 minutes, 9.4 seconds

Best,
-Slashy

 

Edited March 1, 2016 by GoSlash27

[Kerbart]

3 minutes ago, GoSlash27 said:

Well... that depends on what you call a "day".

Best,
-Slashy

 

Let's call it "a day!"

[OhioBob]

On ‎2‎/‎28‎/‎2016 at 6:48 PM, FancyMouse said:

It was a recent change (post 1.0) that makes a solar day a whole day, so sidereal rotation period is a few seconds shorter.

Found the thread: 

 

The thread to which you link is from before the change to the 6 hour solar day.  The measurements presented in that thread are no longer applicable.

For a keostationary orbit, it is the sidereal rotation period that you want to match.  Starting with version 1.05, Kerbin's sidereal rotation period is 21549.4251830898 seconds.  This period is attained with an orbit having a semi-major axis of 3463334.0595 meters.

 

Edited March 1, 2016 by OhioBob