Question about the Trench Run

[Hardy]

Word of caution: that display is obviously not to scale.

True, but it is worthless if it isn't positionally correct.

Um-it can't be positionally correct if it isn't to scale.

By "not to scale", you mean that the planets are not at the correct relative sizes, right?

Um-that sort of was my point. No reason to show an altitude change that isn't there.

1. How do you know that the DS maintained a costant altitude?
2. Why would there be no reason to show that it wasn't there?

Oops.

No worries.

Derived from where? Given that the rebel display shows us exactly three positions for the DS

Yes. That's all I need to know to see if it conforms to Kepler's Second Law. Four positions would be preferrable and more accurate, though.

and is HIGHLY simplistic, that's a pretty shaky source.

It's simple enough to calculate the area that the radius-vector of the planet and DS sweeps. A 3D display would be a bitch to work with.
Simplicity doesn't nescesarily detract from accuracy

Err-why not? If wouldn't require them and no sane person would assume they're on without further evidence but they're certainly not forbidden.

Correct. Wording error on my part.

[Batman]

True, but it is worthless if it isn't positionally correct.

Um-it can't be positionally correct if it isn't to scale.

By "not to scale", you mean that the planets are not at the correct relative sizes, right?

That, and the distances depicted were way off.

Um-that sort of was my point. No reason to show an altitude change that isn't there.

1. How do you know that the DS maintained a costant altitude?

I DON'T. That's the whole point. Display shows DS goes lower, we do not know that DS didn't do that therefore assumption that DS DID do that valid.

2. Why would there be no reason to show that it wasn't there?

There would have been no reason to show the dip IF it wasn't there. They SHOWED the dip, therefore we can assume that it WAS there.
Just in case you haven't noticed, I AGREE with you.

Derived from where? Given that the rebel display shows us exactly three positions for the DS

Yes. That's all I need to know to see if it conforms to Kepler's Second Law. Four positions would be preferrable and more accurate, though.

Enough to show if it doesn't? Propably. But not enough to show if it does, as the DS is not required to observe orbital mechanics.
If all you want to know it if that course is compatible with orbital mechanics, that's something else.

and is HIGHLY simplistic, that's a pretty shaky source.

It's simple enough to calculate the area that the radius-vector of the planet and DS sweeps. A 3D display would be a bitch to work with.
Simplicity doesn't nescesarily detract from accuracy

Problem not being it's 2D (WAY easier to work with I agree) but that it didn't show scale, nor did it show a course, merely three data points that are by no means required to be part of a curve, orbital or otherwise.
As I said, if all you want to know is if it could have been a proper orbit, or to show that it could positively NOT, that should work.

Err-why not? If wouldn't require them and no sane person would assume they're on without further evidence but they're certainly not forbidden.

Correct. Wording error on my part.

Yeah well, I'm a nitpicky bastard. So there.

[Hardy]

That, and the distances depicted were way off.

I'll give you that. But again, the Death Star's position relative to the planet would be critical since that is the only thing that separates the Death Star's line of sight from the planet. But I can't explain the moon's position (it may be correct, however).

However, the relative sizes must've been changed to make the moon and DS visible compared to the planet.

Just in case you haven't noticed, I AGREE with you.

I'm going to go bang my head against a wall when I hit the submit button. Sorry about that.

Enough to show if it doesn't? Propably.

It goes either way. The default assumption is that it isn't ballistically orbting since it's going at "maximum velocity"

But not enough to show if it does, as the DS is not required to observe orbital mechanics.

Agreed

If all you want to know it if that course is compatible with orbital mechanics, that's something else.

Agreed

Problem not being it's 2D (WAY easier to work with I agree) but that it didn't show scale, nor did it show a course, merely three data points that are by no means required to be part of a curve, orbital or otherwise.

See above for scale.
As for the positions of the Death Star, the most basic assumption about them is that they fit into a curve. It is a valid assumption, but like you said, not completely substantiated.

As I said, if all you want to know is if it could have been a proper orbit, or to show that it could positively NOT, that should work.

Which is what I'm trying to do, of course.

Err-why not? If wouldn't require them and no sane person would assume they're on without further evidence but they're certainly not forbidden.

Correct. Wording error on my part.

Yeah well, I'm a nitpicky bastard. So there.

So I've noticed

[Batman]

That, and the distances depicted were way off.

I'll give you that. But again, the Death Star's position relative to the planet would be critical since that is the only thing that separates the Death Star's line of sight from the planet. But I can't explain the moon's position (it may be correct, however).

Absolutely. However, I don't see how you can use that information whithout knowing the distances involved.

However, the relative sizes must've been changed to make the moon and DS visible compared to the planet.

Oh, the display was completely adequate for what it was supposed to do.
I'm just saying I don't expect you to get much usefulsmeasurements out of it.

Enough to show if it doesn't? Propably.

It goes either way. The default assumption is that it isn't ballistically orbting since it's going at "maximum velocity"

Don't remind me of 'maximum velocity' in space. Yeargh.

But not enough to show if it does, as the DS is not required to observe orbital mechanics.

Agreed

Damn. You're no fun.

As for the positions of the Death Star, the most basic assumption about them is that they fit into a curve. It is a valid assumption, but like you said, not completely substantiated.

Would you stop being reasonable already ?

As I said, if all you want to know is if it could have been a proper orbit, or to show that it could positively NOT, that should work.

Which is what I'm trying to do, of course.

Yeah, well, so I'm a slow learner. Sue me

[Hardy]

Here's an easier calculation based on Saxton's derivations from Galaxy Guide 2.

The orbital period at the surface of Yavin would be 9,976 seconds. The Death Star, (which was not orbiting at the the surface of the planet and therefore should have been travelling slower) made about 1/3 of a circle around Yavin.

The Death Star covered that distance in 15 minutes, whereas the GG2 figures would say that it could do that in no less than 55 minutes. Therefore the DS was using its thrusters to accelerate around the planet.

Then again, teh evil WEG figures could be flawed.

[Hardy]

However, I don't see how you can use that information whithout knowing the distances involved.

Pixel counts and Saxton figures are helpful.

Oh, the display was completely adequate for what it was supposed to do.
I'm just saying I don't expect you to get much usefulsmeasurements out of it.

Well, the Death Star's position relative to its former position should be to scale and I can resort to pixel counts.

Enough to show if it doesn't? Propably.

It goes either way. The default assumption is that it isn't ballistically orbting since it's going at "maximum velocity"

Don't remind me of 'maximum velocity' in space. Yeargh.

Maybe it was bordering on escape velocity and didn't want to accelerate until the last moment?

But not enough to show if it does, as the DS is not required to observe orbital mechanics.

Agreed

Damn. You're no fun.

When you're right, you're right.

[drachefly]

You know, a simple solution to the issue of why the DS didn't just have a little burst of acceleration is that the DS propulsion system, though capable of high accelerations, i.e. ((d/dt)^2)x, is not capable of rapid changes in acceleration, i.e. ((d/dt)^3)x.