# Entering SOI's Clockwise or Counter-clockwise

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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.

Relative speed: Sum of two vectors, think of vector as speed in a given direction.

1000 m/s west + 50 m/s west = 1000 + 50 = 1050

1000 m/s west - 50 m/s east = 1000 - 50 = 950

To reach a relative speed of zero you would have to change your speed by either 1050 m/s or 90 m/s.

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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

YES!

So in total you changed your speed by

-0.5 - 2 = -2.5

or

0.5 - 2 = -1.5

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

No, what speed are you going to begin with?

1000 m/s

1000 - 1000 = 0

Your target is moving at 50 m/s in one direction so you need 50 m/s to catch up to it

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YES!

So in total you changed your speed by

-0.5 - 2 = -2.5

or

0.5 - 2 = -1.5

I changed my speed by 0.5 because I was moving at 2 to begin with. If I stop moving I'm now going 0

Both conveyors are going 0.5. I now go 0.5 and stand on the conveyor.

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No, what speed are you going to begin with?

1000 m/s

1000 - 1000 = 0

Your target is moving at 50 m/s in one direction so you need 50 m/s to catch up to it

Depends on the direction of you and the target.

One question first, you do understand that we are only talking about horizontal movement?

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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 ... Your orbital speed of 2290 m/s at 75,000 meters is the same clockwise as it is counterclockwise

Your diagram reading skills are severely lacking.

If you are orbiting in the same direction as the planet is rotating, then you must go from 2,290 m/s to 174 m/s for a total change of 2,116 m/s. If you are orbiting in the opposite direction, you must go from 2,290 m/s to 174 m/s in the opposite direction. That's a total change of 2,464 m/s.

Here's a fun thing to try: Put yourself into a stable, circular, equatorial orbit. Switch your velocity between "Orbit" and "Surface" and note the difference. Now do it again in a retrograde orbit and note the difference. "Surface" speed is what you need to cancel out, and it will be a higher value for retrograde orbits. The only case where this is NOT true is in a synchronous/Kerbal-stationary orbit.

=Smidge=

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I changed my speed by 0.5 because I was moving at 2 to begin with. If I stop moving I'm now going 0

Both conveyors are going 0.5. I now go 0.5 and stand on the conveyor.

If you change your speed by 0.5 from 2.0, you will be going at 1.5 or 2.5 depending if you go 0.5 faster or slower, but the conveyer will still go at 0.5 and thereby your relative speed will still be not 0 and the conveyer will rip of your feet while landing.

If you stop moving from 2.0 your horizontal speed will be 0, but not your relative speed.

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If you change your speed by 0.5 from 2.0, you will be going at 1.5 or 2.5 depending if you go 0.5 faster or slower, but the conveyer will still go at 0.5 and thereby your relative speed will still be not 0 and the conveyer will rip of your feet while landing.

If you stop moving from 2.0 your horizontal speed will be 0, but not your relative speed.

Yes, that change is to equalize my relative speed to the conveyor. If the conveyor is going faster than I am, 2.5 and I'm going 2, I need to increase my speed by 0.5. Now I'm going at 2.5 and it is going at 2.5, 2.5 - 2.5 = 0

If it's going slower than I am, then I need to slow down 0.5 so now I'm going 1.5 and it's going 1.5. 1.5 - 1.5 = 0

The change in speed for both is the same in order to make the relative speeds equal.

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Yes, that change is to equalize my relative speed to the conveyor. If the conveyor is going faster than I am, 2.5 and I'm going 2, I need to increase my speed by 0.5. Now I'm going at 2.5 and it is going at 2.5, 2.5 - 2.5 = 0

If it's going slower than I am, then I need to slow down 0.5 so now I'm going 1.5 and it's going 1.5. 1.5 - 1.5 = 0

The change in speed for both is the same in order to make the relative speeds equal.

This is mathmatically sound, but has nothing to do with the situation discussed in this thread:

Landing from a clockwise or counter-clockwise orbit - on Kerbin and Earth meaning east to west or west to east.

The rotational speed (on the equator) is constant. In your latest example you chose two random values to make your math work, but you cannot make a planet spin faster or slower as you want it to, just because to only want to change your speed by a set value, disregarding your orbital direction.

Your orbital speed - when speaking of the speed you have to maintain to remain at a given altitude in orbit - is the same in both situations.

But your surface speed changes with the direction you are flying around the planet - by the rotational speed of said planet, which is either added or subtracted from your orbital speed to calculate your surface speed -> your speed relative to the surface.

And this surface speed is what you have to reduce to zero for a safe landing.

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Ok, everyone who likes physics, here is your proof, this is a clockwise orbit and a counter clockwise orbit with the same craft. I slowed the craft accurately down to 20 m/s from the orbital speed of 2288.5 and 2285.8

So, if you give or take for the error of speed and the orbital altitude, and piloting for the deorbit burn and what time the vessel entered the rim on the atmosphere during the burn. The fuel consumption for both orbits is exactly the same. And the craft accelerates towards the surface at the same rate in both instances because we have a gravitational constant.

Craft on the launch pad:

Clockwise orbit:

Clockwise orbit fuel consumption to kill orbital velocity to about 20 m/s vertical.

Counter Clockwise orbit:

Counter Clockwise Orbit Fuel consumption to get about 20 m/s vertical:

Edited by 700NitroXpress
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Yes, that change is to equalize my relative speed to the conveyor. If the conveyor is going faster than I am, 2.5 and I'm going 2, I need to increase my speed by 0.5. Now I'm going at 2.5 and it is going at 2.5, 2.5 - 2.5 = 0

If it's going slower than I am, then I need to slow down 0.5 so now I'm going 1.5 and it's going 1.5. 1.5 - 1.5 = 0

The change in speed for both is the same in order to make the relative speeds equal.

Now you're just changing the facts... if you're going 2m/s and the conveyor is going 0.5 in the same direction you are moving faster than the conveyor, not the other way round. Your relative velocity is 1.5 m/s, and that is how much you need to change your speed. When the conveyor moves to the opposite direction, it's 2 + 0.5 = 2.5 m/s, which is again how much you need to change your speed. 1.5 does not equal 2.5.

As for your second version here, if conveyor moves at 2.5 and you're moving at 2, then your relative speed is .5 (and that's how much you need to change it). BUT if it's moving at the same speed in the other direction your relative speed is 4.5 which obviously is not the same.

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Counter Clockwise Orbit Fuel consumption to get about 20 m/s vertical:

your horizontal surface speed is not zero. Try landing that way...

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If you would actually read what others are writing, then you would have understood that surface speed is not the speed at which you are getting closer to the surface.

Just for the fun of it I would like to read your definition of air speed.

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your horizontal surface speed is not zero. Try landing that way...

You're horizontal surface speed changes once your gauge switches from orbit to surface. In orbit the retrograde marker was right on top of the nav ball. The orbital speed was canceled out. Now if I were going in for a landing, the atmosphere deceleration would take care of the horizontal surface speed for me and I would hit the ground at terminal velocity.

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Here is orbital speed canceled from a 100,000 meter orbit. Altitude accurate to within 10 meters, orbital speed is 4.6m/s on engine shutdown. Clockwise Orbit.

Here is the maximum speed achieved while falling to the surface under the force of gravity alone before the air resistance kicks in.

Here is right before it hits the ground. You can see by the prograde marker on the nav ball that the horizontal velocity is 0

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You're horizontal surface speed changes once your gauge switches from orbit to surface. In orbit the retrograde marker was right on top of the nav ball. The orbital speed was canceled out. Now if I were going in for a landing, the atmosphere deceleration would take care of the horizontal surface speed for me and I would hit the ground at terminal velocity.

Now that you've admitted that the only reason it works is because the atmosphere takes care of the problem for you, and you finally admitted that the horizontal velocity of the surface of the planet is a problem that needs to be dealt with in order to land, will you finally admit that we were correct to claim that there's a difference between landing retrograde or prograde and the only reason it works out the same on Kerbin is because of the atmosphere doing the work for you? Now do what I mentioned earlier and do it somewhere without atmosphere that has rotation. If you come down from prograde rotation you don't HAVE to kill all the orbital velocity because coming down to kiss the ground at the ground's speed is what you want. It's stupid and pointless to kill all the orbital velocity first, fall straight down, and then speed up again to match up with the ground when you didn't have to slow down enough to fall straight. Falling down in an arc was good enough, and in fact preferred. And that's why we have been 100% correct to point out that landing prograde uses less fuel than landing retrograde, and if you already knew that then you've been trolling and should be ashamed of yourself.

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You're horizontal surface speed changes once your gauge switches from orbit to surface. In orbit the retrograde marker was right on top of the nav ball. The orbital speed was canceled out. Now if I were going in for a landing, the atmosphere deceleration would take care of the horizontal surface speed for me and I would hit the ground at terminal velocity.

What Steven said. What did you think that deceleration due to aerodynamic drag is doing to your speed? Not changing it?

Also, you didn't mean to seriously suggest that your speed actually changes when the navball changes modes?!?

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I'm done with the troll. If he's pretending he didn't already know this then he's just lying.

Edited by Steven Mading
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Lets grab some graph paper and draw out what's happening. You have a vector going in the positive x direction at 10m/s. You have a vector 3m above that vector in positive y moving 15m/s in the positive x direction. You also have a vector 3 meters above the axis moving left in the negative x direction at 15 m/s. There are no accelerations acting on these vectors (they are straight lines, two of which are 3 "ticks" above the axis, one on the axis).

How can we get both of these vectors to equal our "surface" vector? Simple math.

Let's make the negative x direction negative (for simplicity's sake, and it will also represent retrograde as it is going the opposite direction of our "surface") and positive x positive (as well as prograde, going in the same direction as our surface). Our 10m/s vector is the "surface" and moving 10m/s positive direction. Our next vector (one we want to match speeds with our surface) is moving 15m/s in the positive direction. How do we get to 10m/s? We add -5m/s (10 - 15 = -5). That's our change in velocity (delta v).

Let's do our other one now. We have 15m/s again but in the negative direction. So how do we go from -15 to +10? Math again. 10 - (-15) = 25. Our change in velocity needed to match the surface speed is 25m/s in the positive direction.

25 > absolute value of 5 so you will use more delta v (and more fuel as well) matching surface speed from a "negative" orbit. This is irrelevant though on planets with an atmosphere or very low surface speeds, as the atmosphere will change your velocity to the surface speed (it moves at the same speed as the surface) and low surface speeds won't matter too much (a surface moving at 1m/s won't make a big difference in your change in velocity).

If you don't understand this you might want to take a physics class. This is basic vector addition.

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What's the difference? Also, AFAIK, all the planets rotate clockwise, no? Why do you always see people saying they are entering the SOI the wrong direction & flip it so that it is rotating w/ the planet?

The difference has been beaten to death by the other commenters. Basically, if you enter counterclockwise, your speed relative to the planetary surface and relative to the moons is going to be higher than if you enter clockwise. It doesn't matter for a fly-by, or if you're just intending to orbit the planet. It does matter if you want to land on the planet, or rendezvous with a moon. Still, if there's an atmosphere, you can aerobrake to do the work, so it only is a problem if you want to land on an airless body.

People end up going the wrong way often because in the default draw mode for your spacecraft trajectory, it's not intuitive to figure out which way you'll end up going when you enter the SOI. With experience you can learn how to read that map. Alternately, if you change the settings.cfg file to have "CONIC_PATCH_DRAW_MODE=0" instead of the default 3, it'll be pretty obvious.

Edited by numerobis
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Let's go to Mun.

Counter Clockwise Mun orbit.

Counter Clockwise Mun Landing.

Clockwise Mun orbit.

Clockwise Mun landing.

Notice that there is an altitude difference in the landing zones by about 2000 meters. This accounts for the small difference in fuel consumption. Both craft killed all orbital velocity and came straight down on top of the ground. No counter burns were needed. Fuel difference 1.5% which is attributed to altitude difference in both the orbit and the landing zone.

Edited by 700NitroXpress
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Mun rotates at whopping 9 m/s, your altitude difference alone contributes for a much larger difference in fuel consumption.

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Therefor, it's the same no matter what direction you come in from.

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On a body that hardly rotates at all, the difference is insignificant. That does not mean that there is no difference.

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So where in the solar system would it actually be noticed then?

Tylo?

Edited by 700NitroXpress
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