Time Dilation

[Andetch]

Maybe this is just me imagining things, but on my missions to the outer planets (Jool mainly) I think I have noticed some time dilation in that the days measured by the ships systems seem to have less minutes than days at the KSC.

Anyone else noticed anything like this?

[sal_vager]

Time dilation is not modeled in KSP, so I don't know what you're seeing, can you provide more information?

[munlander1]

I think that's just your computer. On launch, with a moderatly sized rocket the launch will take about 2 seconds real time for 1 second ingame time. For my computer.

Edited March 3, 2017 by munlander1

[Loren Pechtel]

On 3/3/2017 at 1:06 AM, Andetch said:

Maybe this is just me imagining things, but on my missions to the outer planets (Jool mainly) I think I have noticed some time dilation in that the days measured by the ships systems seem to have less minutes than days at the KSC.

Anyone else noticed anything like this?

Unless you're under Kracken drive you won't be going fast enough for time dilation to show up at the precision at which the game displays time anyway.

(Now, with the Kracken it's another matter.  I've seen a chute head outsystem at about 2/3 of lightspeed and others have reported FTL velocities from such incidents.  The game is Newtonian, though.)

[Andetch]

Ahh yes, the Krakken drive..... I have a 300T mining rig that can flip itself 50 - 100 KM above Minimus using a Krakken drive.... hehe

What I was seeing was that the period of 1 day as displayed in distance to next node was actually slightly less than the standard X hour Kerbin day at KSC.

Edited March 5, 2017 by Andetch

[FullMetalMachinist]

On 3/5/2017 at 5:22 AM, Andetch said:

What I was seeing was that the period of 1 day as displayed in distance to next node was actually slightly less than the standard X hour Kerbin day at KSC.

Not sure exactly what your circumstances are when you saw that, but it was probably the difference between a  Kerbin day and a Kerbin's sidereal period. 

On the surface a 'day' is measured from one sunrise to the next. The sidereal period is how long it takes Kerbin to make one full rotation relative to the stationary distant stars. 

To exaggerate and make the difference easier to grasp, imagine that a Kerbin day on the surface is exactly 6 hours (like it is in the game now). But instead of the 400 and something days it takes now for a year, imagine that it takes only four days, so that each day Kerbin moves 90 degrees around it's orbit. That means that in order to go from one sunrise to the next, Kerbin actually has to rotate the 360 degree for a full rotation, but then it has to keep going another 90 degrees in order to present the same spot on the surface to the sun. So if 360 + 90 = 450 degrees takes 6 hours, then how long does it take to go 360 degrees compared to a stationary reference (ie the distant stars)? It takes 4 hours 48 minutes. 

Now take that extreme example and put it back in terms of the current game. A day on the surface still takes 6 hours, but instead of also moving 90 degrees around it's solar orbit it moves less than a degree. That makes the sidereal period (as reported in the wiki) 5h, 59m, 9.4s. 

So that less-than-a-minute difference is probably what you're seeing. Unless it's not, and I've wasted a bunch of time telling you something that you already knew. 

[Andetch]

On 06/03/2017 at 2:06 PM, FullMetalMachinist said:

Not sure exactly what your circumstances are when you saw that, but it was probably the difference between a  Kerbin day and a Kerbin's sidereal period. 

On the surface a 'day' is measured from one sunrise to the next. The sidereal period is how long it takes Kerbin to make one full rotation relative to the stationary distant stars. 

To exaggerate and make the difference easier to grasp, imagine that a Kerbin day on the surface is exactly 6 hours (like it is in the game now). But instead of the 400 and something days it takes now for a year, imagine that it takes only four days, so that each day Kerbin moves 90 degrees around it's orbit. That means that in order to go from one sunrise to the next, Kerbin actually has to rotate the 360 degree for a full rotation, but then it has to keep going another 90 degrees in order to present the same spot on the surface to the sun. So if 360 + 90 = 450 degrees takes 6 hours, then how long does it take to go 360 degrees compared to a stationary reference (ie the distant stars)? It takes 4 hours 48 minutes. 

Now take that extreme example and put it back in terms of the current game. A day on the surface still takes 6 hours, but instead of also moving 90 degrees around it's solar orbit it moves less than a degree. That makes the sidereal period (as reported in the wiki) 5h, 59m, 9.4s. 

So that less-than-a-minute difference is probably what you're seeing. Unless it's not, and I've wasted a bunch of time telling you something that you already knew. 

No, I think that is probably it.....