Entering SOI's Clockwise or Counter-clockwise

[Smidge204]

I think the problem is that 7000Nitro is forgetting that atmosphere moves with the body, and therefore it's pushing you sideways.

I don't think the game models atmosphere like this... rather, I think it's just a simple resistance to motion that scales with velocity. If you do manage to fall STRAIGHT DOWN through an atmosphere, I do not think you will be pulled sideways at all. Don't have time to experiment this morning. :/

The surface will rotate out from underneath you, though, that's for sure.

=Smidge=

[700NitroXpress]

A picture is worth a thousand words

Your picture is physically impossible. Your orbital speed and altitude is fine. But you can't have a different velocity going one way as opposed to the other due to Newton's Third Law "To every action there is always an equal and opposite reaction: or the forces of two bodies on each other are always equal and are directed in opposite directions."

Your orbital speed of 2290 m/s at 75,000 meters is the same clockwise as it is counterclockwise, and lets not forget that gravity is constant. When you are in orbit, the force of friction isn't acting on your vessel. The rotation of the planet isn't acting on your vessel. Therefore there is no extra force to cancel. Thus, there is no additional fuel consumed to land from orbit. When you are taking off from a planet, you inherit the planets rotational speed because you are on the surface. As soon as you leave that surface, that surface no longer is causing any force that is acting on you, but you are still moving at the surface speed in the horizontal direction, but you don't notice it because both you and the surface are moving at the same speed. When you launch from a planet you are already going at the speed of the rotation. This means that your horizontal velocity while stationary on the surface of Kerban is equal to it's rotational speed of 174 m/s. However, your vertical speed is 0 and your speed in relation to a specific point on the ground is also 0.

[Dunbaratu]

I don't think the game models atmosphere like this... rather, I think it's just a simple resistance to motion that scales with velocity

Yes but it scales with SURFACE velocity, not ORBITAL, which does end up doing what I said it does. If you're falling straight down according to the orbital mode of the navball, then you've got a sideways component of 174 m/s according to the surface mode of the navball, so the drag calculation will drag you as if you are going 174 m/s sideways because it's calculating your velocity in surface mode and using that.

Edited November 11, 2013 by Steven Mading

[700NitroXpress]

Gravity isn't constant.

Newton says that you're wrong. http://en.wikipedia.org/wiki/Gravitational_constant

[700NitroXpress]

As a simplified analogy, let's say you're a bird flying in calm weather at air speed of 2m/s. Below you, there are two conveyors both moving 0.5m/s (relative to the ground), one in same direction as you and the other in the opposite direction. So your relative speed to one of the conveyors is 1.5m/s and 2.5m/s to the other. Do you need to slow your horizontal speed more or less to land on one of the conveyors? Or if you make your air speed (analogous to orbital speed, here) zero, what is your speed relative to the conveyors, can you land softly on either?

I hear linking to wiki articles makes me more right so here's one

Thank god someone here understands Newton's Third Law.

You would have to slow down 2m/s in order to land on both conveyors. Both conveyors are moving at the same rate. So it doesn't matter if you land on one or the other. If you compensate for the rate of the conveyors, then you have to either speed up or slow down 0.5m/s in order to land on both without being swept away. This means that you have to either speed up or slow down at the same speed no matter which direction it's going in order to land on a specific point in relation to you.

Edited November 11, 2013 by 700NitroXpress

[Dunbaratu]

Thus, there is no additional fuel consumed to land from orbit

Only if your definition of "land" doesn't include having the craft survive the landing. You're still ignoring that hill coming at you sideways.

[kurja]

Thank god someone here understands Newton's Third Law.

Which is funny because I just contradicted your claim.

Surface speeds are different for clockwise and counterclockwise orbits. Just accept it. Or better yet, try it yourself and see how it is. Just get two crafts on opposite orbits with same altitude and thus orbital speed, then check their surface speeds. Are they the same? No, they're not. Unless you're saying that horizontal surface speed does not matter when landing? Forget newton, forget friction, just look at the surface speeds.

[Dunbaratu]

Newton says that you're wrong. http://en.wikipedia.org/wiki/Gravitational_constant

Newton says you aren't using vocabulary correctly by pretending the word "gravity" and the words "gravitational constant" are synonyms.

[Dunbaratu]

This has made me think that having a planet or moon with big patches of no-friction smooth surfaces could be a fun addition to the Kerbin system, especially for doing EVA's (you'd have to ice-skate using your jetpack). It would be the one and only place where 700Nitro would be correct about being able to land safely while ignoring surface velocity.

[KerbMav]

atmosphere moves with the body, and therefore it's pushing you sideways. With atmosphere present, you end up eventually being pushed sideways until you asymptotically approach matching speeds with the ground anyway even when you don't try to

Is this correctly simulated in KSP?

Nitro, let go of the vertical component of the landing, here you are mostly correct.

Imagine driving a bike at 6m/s (21.6 km/h) in a circle around a carousel going at 2m/s (7.2 km/h).

If you were driving in the same direction as horsy, piggy and the helicopter are, you would be driving at 4m/s relative speed, else at 8m/s.

Now, if you stop your bike and start walking towards the carousel - as if gravity would pull you towards it, thinking of the center of it as the center of a planet - horsy would still hit you at 2/ms.

So, rotating a bit faster, planet piggy would hit your lander at 100+ m/s (360+ km/s) from the side, ripping of part by part from bottom to top, rolling it over and smashing it to pieces.

Edited November 11, 2013 by KerbMav

[700NitroXpress]

Which is funny because I just contradicted your claim.

Surface speeds are different for clockwise and counterclockwise orbits. Just accept it. Or better yet, try it yourself and see how it is. Just get two crafts on opposite orbits with same altitude and thus orbital speed, then check their surface speeds. Are they the same? No, they're not. Unless you're saying that horizontal surface speed does not matter when landing? Forget newton, forget friction, just look at the surface speeds.

OMG, no, the surface speed is the same. The planet is not accelerating which is the only thing that would cause the speed to be different. Are the speeds of the conveyors different? No, they are both going at 0.5 m/s are they not? You are going at 2m/s are you not? If you stop going 2m/s, how fast are the conveyors going? I think they're still going at 0.5 m/s. In order to go the same speed that they are going, I have to increase or decrease my speed 0.5 m/s for both, do I not?

Therefore lets look at this:

Conveyor A is going 0.5 m/s and Conveyor B is going at 0.5 m/s in the opposite direction. I'm going at 2 m/s in the same direction as conveyor A. Therefor, my speed in relation to Conveyor A is 1.5 m/s and my speed in relation to conveyor B is 2.5 m/s

2 + 0.5 = 2.5

2 - 0.5 = 1.5

In order to compensate for the speed of both conveyors, I need to make my speed in relation to the conveyor 0.

I have 0.5 units of fuel to do this.

0 = (2 + 0.5) - 0.5

0 = (2 - 0.5) + 0.5

In order to go at the same speed and land on go on both conveyors I need to compensate for the equal opposite speed due to Newton's third law. I used the same compensation speed of 0.5 to equalize both. Any questions?

[KerbMav]

OMG, no, the surface speed is the same.

By the amount, but not the direction (vector).

The planet is not accelerating which is the only thing that would cause the speed to be different.

The conveyors are standing in for the planets surface here, as is the bird for the lander!

Are the speeds of the conveyors different? No, they are both going at 0.5 m/s are they not? You are going at 2m/s are you not? If you stop going 2m/s, how fast are the conveyors going? I think they're still going at 0.5 m/s. In order to go the same speed that they are going, I have to increase or decrease my speed 0.5 m/s for both, do I not?

Almost.

Therefore lets look at this:

Conveyor A is going 0.5 m/s and Conveyor B is going at 0.5 m/s in the opposite direction. I'm going at 2 m/s in the same direction as conveyor A. Therefor, my speed in relation to Conveyor A is 1.5 m/s and my speed in relation to conveyor B is 2.5 m/s

2 + 0.5 = 2.5

2 - 0.5 = 1.5

Keep thinking about exactly this point a bit more and you will get it right!

In order to compensate for the speed of both conveyors, I need to make my speed in relation to the conveyor 0.

Yes?!

I have 0.5 units of fuel to do this.

0 = (2 + 0.5) - 0.5

0 = (2 - 0.5) + 0.5

In order to go at the same speed and land on go on both conveyors I need to compensate for the equal opposite speed

No, to reach 0 relative speed you have to change your speed by

0 = (2 + 0.5) - 2.5

0 = (2 - 0.5) - 1.5

else your equation is wrong, because you would be saying 0 = 2 ...

due to Newton's third law. I used the same compensation speed of 0.5 to equalize both. Any questions?

Third law: When one body exerts a force on a second body, the second body simultaneously exerts a force equal in magnitude and opposite in direction to that of the first body.

What does this have to do with anything (here)?

[Dunbaratu]

(In response to the claim that the air drag is calculated using your surface velocity not your orbital velocity:)

Is this correctly simulated in KSP?

I've made retrograde orbit landings on Kerbin before using only parachutes. If it wasn't correctly simulated, that wouldn't have worked.

If it used orbital velocity to calculate it, then sitting on the launchpad prior to launch you'd already be above terminal velocity, in a gale force wind going 174 m/s sideways while sitting at the launchpad before you do anything.

[Dunbaratu]

OMG, no, the surface speed is the same. The planet is not accelerating which is the only thing that would cause the speed to be different.

Fail. You didn't perform the test. The speed is not the same. Also, fail on not understanding the difference between acceleration and velocity. If I walk east at 3 mph and you walk west at 2mph our relative velocity difference is 5 mph (the difference between 3 and -2). Even if we are both walking at constant speed and neither of us is accelerating. On the other hand if we're both walking east and I'm going 3 mph and you're going 2, then our relative speed difference is 1 mph. Even if we are both walking at constant speed and neither of us is accelerating.

[KerbMav]

(In response to the claim that the air drag is calculated using your surface velocity not your orbital velocity:)

I've made retrograde orbit landings on Kerbin before using only parachutes. If it wasn't correctly simulated, that wouldn't have worked.

If it used orbital velocity to calculate it, then sitting on the launchpad prior to launch you'd already be above terminal velocity, in a gale force wind going 174 m/s sideways while sitting at the launchpad before you do anything.

Your last point may be a clue, but not evidence!

It might simply be, that KSP does not account for the movement of the atmosphere - as there are no winds/currents in the game right now, I would think that this is the case.

Possible experiment would be to drop a probe into the atmosphere and to kill its horizontal surface speed before deploying chutes.

If the horizontal speed then changes again ... but would it ... hmmm ...

[kurja]

OMG, no, the surface speed is the same. The planet is not accelerating which is the only thing that would cause the speed to be different. Are the speeds of the conveyors different? No, they are both going at 0.5 m/s are they not? You are going at 2m/s are you not? If you stop going 2m/s, how fast are the conveyors going? I think they're still going at 0.5 m/s. In order to go the same speed that they are going, I have to increase or decrease my speed 0.5 m/s for both, do I not?

Therefore lets look at this:

Conveyor A is going 0.5 m/s and Conveyor B is going at 0.5 m/s in the opposite direction. I'm going at 2 m/s in the same direction as conveyor A. Therefor, my speed in relation to Conveyor A is 1.5 m/s and my speed in relation to conveyor B is 2.5 m/s

2 + 0.5 = 2.5

2 - 0.5 = 1.5

In order to compensate for the speed of both conveyors, I need to make my speed in relation to the conveyor 0.

I have 0.5 units of fuel to do this.

0 = (2 + 0.5) - 0.5

0 = (2 - 0.5) + 0.5

In order to go at the same speed and land on go on both conveyors I need to compensate for the equal opposite speed due to Newton's third law. I used the same compensation speed of 0.5 to equalize both. Any questions?

Questions? Yes. Why won't your example actually work?

In your example, you have cherry picked unrealistic values to give the result you want - when does anyone approach a planet at orbital speed lower than the planet's rotation? Try flying at 1000 m/s with conveyors moving at 50 m/s: your relative velocity to one is 1050 and 950 to the other. Different delta vees needed to come to a halt, you see.

(edit: also, in your example, if your relative velocities are 2.5 and 1.5 like you said, those are the amounts by which you need to change your speed to come to zero, not 0.5 or -0.5..)

Edited November 11, 2013 by kurja

[kurja]

Your last point may be a clue, but not evidence!

It might simply be, that KSP does not account for the movement of the atmosphere - as there are no winds/currents in the game right now, I would think that this is the case.

Possible experiment would be to drop a probe into the atmosphere and to kill its horizontal surface speed before deploying chutes.

If the horizontal speed then changes again ... but would it ... hmmm ...

Atmosphere would rotate at the same speed as the surface, so if you already had 0 surface speed upon re-entry, nothing would happen, right? I suppose if you made the orbital speed zero, then entered atmosphere, we should see horizontal acceleration from drag..?

Maybe.

[Dunbaratu]

Your last point may be a clue, but not evidence!

It might simply be, that KSP does not account for the movement of the atmosphere - as there are no winds/currents in the game right now, I would think that this is the case.

Possible experiment would be to drop a probe into the atmosphere and to kill its horizontal surface speed before deploying chutes.

If the horizontal speed then changes again ... but would it ... hmmm ...

In order for the drag calculations to work they have to only apply a drag force when you have a velocity, and they apply it in the opposite of the velocity direction. If you are standing still there isn't drag. So what velocity are they using to calculate it? It's got to be something. So it's either orbital velocity or it's surface velocity.

And the behavior is not consistent with it being orbital velocity. If it was based on orbital velocity it wouldn't work because standing still on the ground would give you a velocity of 174 m/s east in the calculation, and thus apply a drag force westward on you because it would THINK you're moving through the air.

The fact that when landing on parachutes you end up matching speeds with the ground *is* the proof that the drag calculations are based on surface velocity.

Edited November 11, 2013 by Steven Mading

[700NitroXpress]

[quote name=

0 = (2 + 0.5) - 2.5

0 = (2 - 0.5) - 1.5

else your equation is wrong, because you would be saying 0 = 2 ...

Third law: When one body exerts a force on a second body, the second body simultaneously exerts a force equal in magnitude and opposite in direction to that of the first body.

What does this have to do with anything (here)?

[KerbMav]

But you wanted to reach 0 relative speed ... ?

[kurja]

Right, got ahead of myself, but you'll notice that they both cancel,

so: The 0.5 m/s difference is relative to a point on the belt. So in order to get them to cancel it would be,

2 = (2 + 0.5) - 0.5

2 = (2 - 0.5) + 0.5

So to compensate for the relative speed for a point on the belt so they move at the same speed, you would either go plus or minus 0.5 m/s in order to move at the same speed as the conveyor.

No. 2 = (2 + 0.5) - 0.5, like it says there, equals two which is not zero

[700NitroXpress]

Questions? Yes. Why won't your example actually work?

In your example, you have cherry picked unrealistic values to give the result you want - when does anyone approach a planet at orbital speed lower than the planet's rotation? Try flying at 1000 m/s with conveyors moving at 50 m/s: your relative velocity to one is 1050 and 950 to the other. Different delta vees needed to come to a halt, you see.

My bad I was writing too quickly. So yes in terms of relative speed, they are different, but they are different in the same way. You have to compensate the same amount for both. If you subtract 1000 m/s from this to equal +50 and -50, you still have to compensate for 50 on both.

[KerbMav]

standing still on the ground would give you a velocity of 174 m/s east in the calculation, and thus apply a drag force westward on you because it would THINK you're moving through the air.

The fact that when landing on parachutes you end up matching speeds with the ground *is* the proof that the drag calculations are based on surface velocity.

Thank you for getting this out of my head, sure, you are right!

[kurja]

My bad I was writing too quickly. So yes in terms of relative speed, they are different, but they are different in the same way. You have to compensate the same amount for both. If you subtract 1000 m/s from this to equal +50 and -50, you still have to compensate for 50 on both.

No. I have to use 1050 m/s dv for one and 950 m/s dv for the other.

[700NitroXpress]

No. 2 = (2 + 0.5) - 0.5, like it says there, equals two which is not zero

In this equation they don't equal 0. 2 = 2.

So you are taking the relative speed which is the 2.5 and 1.5

Then you add you're compensation speed which is -0.5 and +0.5

So now your relative speed in relation to it is 2

and now you kill your relative speed so you're moving at the same speed which is -2

So now you're speed in relation to the target is 0.0