Math problem, RotationOffset

[rthorvald]

Well, i have struggled with this problem all day, and simply cannot solve
it. Probably because i really don??t understand the problem; i have no
math background at all. Hope someone here can show me how to solve
it - not using math symbols, which is greek to me (sorry, couldn??t help
that one):-)


Here goes:
-------------------
- I have a comet with a rotationperiod of 12 hours

- On the surface, i have a balloon. It, too, has a rotationperiod of 12 hours
to make it stand still on the surface. The balloon rotates along it??s poles, not it??s equator.

Next, we have the Earth. As the comet passes the Earth, the balloon
travels along an XYZ path from the comet surface to the Earth surface.
The trip takes 86 minutes.

Now, before takeoff, the comet stands upright on the Comet surface.
This means it arrives on Earth??s surface upside down. So, right before
takeoff, i switch the ballon model to a different SSC entry that can have it??s own RotationPeriod.

This new rotationperiod should let the balloon rotate around itself exactly one half turn
before arriving on Earth. That means it should have a rotationperiod of 2.86 hours,
but since the XYZ is attached to the comet, one has to take into account
the rotation of the comet during the flight (1.43 hours, as the XYZ
starting point will move.

So, what is the rotationperiod of the comet, to make it stand upright on
both the comet surface AND the Earth?

If anyone is interested, here??s the zipped up XYZ and SSC for the
objects - together with CelUrl??s for the start and end points of the XYZ:
http://runar.thorvaldsen.net/celestia/temp/balloondata.zip

Any help at all would be greatly appreciated; i have tried everything, and has now
resorted to experimenting wildly, which, it seems, i can do for weeks
without getting any usable result...

PS: if you wonder what this is all about, see this thread:
http://shatters.net/forum/viewtopic.php?t=7683


Thanks,
- rthorvald

[rthorvald]

Well, i have struggled with this problem all day

Nevermind.
Since nobody bothered, i solved it by using the placeholder trick i used in Jestr??s Huygens addon instead. Not as elegant, but the same end result. Forget it.

-rthorvald

[ElChristou]

Sorry Runar, I couldn't help on this.

[Spaceman Spiff]

Runar,

it may be that no one helped because no one could help. I think there might not be enough control over how the ballon is orineted, so no one could provide a mathematical solution.

I'm downloading the .zip now, hope I can take a look, check if I'm right.

And by the way, your Add Ons are always great!

Spiff.

[rthorvald]

Runar, it may be that no one helped because no one could help.

I see that. Thanks for the replies, both.
The solution i came up with works very well, it just made the ssc file more cluttered. But it works, so don??t put too much work into it...

- rthorvald

[Spaceman Spiff]

Oh good. 'cos it can't be done otherwise ...

Spiff.

[cpotting]

I know you have already solved this otherwise, but I thought it would be a good excersise to go through anyway...

If I understand your problem correctly...
The balloon begins the trip with a rotation of 12hrs which will remain during the journey to Earth because it is defined as an object orbiting the comet every 12hrs.

Therefore:
- the balloon rotates 360d in 720min = 2d/min
- after the 86min trip, the balloon will have rotated 86min * 2d/min = 172d
- you want it to rotate 180d, so an additional rotation of 180d - 172d = 6d is needed.
- to rotate 6d in 86min is to rotate 360d in 5160min or 86hrs.

If the rotation of the ballon is changed to 86hrs when it leaves the comet, then it should arrive at Earth 86m later, having rotated 180d relative to the Earth.

[cpotting]

- you want it to rotate 180d, so an additional rotation of 180d - 172d = 6d is needed.
- to rotate 6d in 86min is to rotate 360d in 5160min or 86hrs.

Oh maaaaaaaannnnnn!!!!!
It's a good thing I'm not in a line of work where people's lives depend on my math skills. Let's try this again....

- you want it to rotate 180d, so an additional rotation of 180d - 172d = 8d is needed.
- to rotate 8d in 86min is to rotate 360d in 3870min or 64.5hrs.

My father was the pilot of the Gimli Glider, my brother was a programmer for Mariner 18, and I'm decended from the man who helped Columbus figure how far it was to Cathay.
"Well, Captain Smith, I went down to see for myself. Seems like the iceberg popped a few rivets, but I figure the bilge pumps will handle it..."

[rthorvald]

Thanks for the effort, guys! I really do appreciate it, though i solved it with a hack.

-rthorvald