UT clock confusion...

[GoSlash27]

I decided to try messing around with time- based orbits to simplify astronavigation and ran into a weird problem.

According to my math, an orbit at 147,319m of should have a period of 36 minutes on the dot. I orbited a ship at exactly this altitude and timed my passage over KSC. It actually takes precisely 40 minutes according to the UTC clock.

This has to be confusion on my part about sidereal vs. solar time, but I'm having a hard time wrapping my head around it. Why am I missing 4 minutes per orbit?? *scratches head*...

Anyone run into this problem or have any ideas?

Best,

-Slashy

Edited March 18, 2015 by GoSlash27

[AlexinTokyo]

Your passage over KSC appears to be delayed (compared to your orbital period) because Kerbin is rotating underneath you.

After your 36-minute orbit, you return to the exact same point with regard to the fixed stars. However, Kerbin has rotated 36 x 360° / 6 x 60 = 36°. To catch up to KSC you have to orbit that additional 36°, plus the amount Kerbin rotates while you're completing that additional orbit.

[GoSlash27]

Your passage over KSC appears to be delayed (compared to your orbital period) because Kerbin is rotating underneath you.

After your 36-minute orbit, you return to the exact same point with regard to the fixed stars. However, Kerbin has rotated 36 x 360° / 6 x 60 = 36°. To catch up to KSC you have to orbit that additional 36°, plus the amount Kerbin rotates while you're completing that additional orbit.

I think this might be it.

If I want to orbit 10 times a day WRT the ground, I have to select an orbit that cycles 11 times a day to make up for Kerbin's 1 orbit. That would explain why the discrepancy is 4 minutes per orbit exactly...

I'll try that. If it works, I'll switch it to "answered".

Thanks!

-Slashy

Edited March 17, 2015 by GoSlash27

[Brainlord Mesomorph]

4 minutes per orbit sounds about right.

Whenever I try to do a burn relative to a precise geographic location at Kerbin, I always have to advance the nav point 2 to 3+ minutes when I finally get there, to compensate for planetary rotation.

[AlexinTokyo]

Having had time now to jot down some notes about this, I think the easy way to think about it is in terms of rotational velocities.

Your 36-minute orbit has a rotational velocity of 10°/minute (I'm using minutes because it makes the calculations easy in this case)

Kerbin has a rotational velocity of 1°/minute.

So your relative rotational velocity is 10-1 = 9°/minute.

Therefore to complete a 360° orbit relative to a fixed point on the ground requires 360/9 = 40 minutes.

This is, of course, the simplified case not accounting for orbital inclination.

[GoSlash27]

Yup, that's what it was. I ginned up a new orbit and it worked like a charm. Thanks!

-Slashy