Confusion about deflection of ships leaving the surface

[ExpHP]

In my experience in KSP, even if all thrust is applied straight downwards, a ship's trajectory will naturally be towards the east. I assume this is why eastward gravity turns are recommended, as chances are you will naturally be moving east at your first apoapsis.

However, I can't figure out why it is exactly that they point east. I assumed this was due to the Coriolis effect, but here's the catch:

By my calculations, the Coriolis force points west. Fcor points in the direction of (- omega x vref), where omega is the absolute angular velocity (straight out of Kerbin's north pole), and vref is the velocity in the rotating reference frame (straight out of the equator).

Have I messed up somewhere, or is there another physical effect at work here?

EDIT: With help from the community, figured it out. The ship isn't deflecting eastward relative to Kerbin. The trajectory doesn't actually show where you will land, because it doesn't take Kerbin's rotation into account (as it well shouldn't, or else a geosynchronous orbit would look like a point ).

Edited December 19, 2013 by ExpHP

[GROOV3ST3R]

I simply assume that it is because Kerbin is rotating eastward. This is what gives you a slight nudge towards eastward.

My guess would be that you overestimated KSP physics ^^ They are simpler in many aspects than the real thing - though to be honest, I know too little about physics and math in general for my opinion to matter anyway

[tavert]

It's just simple surface rotation. KSC is very close to the equator, rotating 2 pi times 600 km (plus a few meters above sea level) per 6 hours, giving you about a 175 m/s eastward orbital velocity when you're stationary relative to the surface.

[ExpHP]

It's just simple surface rotation. KSC is very close to the equator, rotating 2 pi times 600 km (plus a few meters above sea level) per 6 hours, giving you about a 175 m/s eastward orbital velocity when you're stationary relative to the surface.

Aye, but therein lies my confusion.

My apologies if I'm reading your post too literally, but to my understanding, the rocket is not stationary relative to the surface. If it was, then it would be landing right where it was launched from, instead of a ways into the ocean to the east.

[Mr.Rocket]

Its sort-of like standing on top of a train, and jumping straight up. you're going to stay with the trains motion, even if you are not touching it. You wouldn't fly backwards relative to the train. I'm not sure why this is, it just is. So, if KSC is moving to the east, and you lift off from it straight up, you will also move to the east, unless you turn a direction other than east.

[ExpHP]

Its sort-of like standing on top of a train, and jumping straight up. you're going to stay with the trains motion, even if you are not touching it. You wouldn't fly backwards relative to the train

Right, absolutely. You also wouldn't fly forwards relative to the train.

Yet rockets fly forwards relative to Kerbin.

[phunk]

Are you sure about that? Keep in mind that the path drawn on the map doesn't show where the planet will have rotated to by the time you land.

[UmbralRaptor]

ExpHP: Your calculations are correct. The reason to launch east would (IIRC) be centrifugal force, not coriolis force.

[annallia]

Its sort-of like standing on top of a train, and jumping straight up. you're going to stay with the trains motion, even if you are not touching it. You wouldn't fly backwards relative to the train. I'm not sure why this is, it just is. So, if KSC is moving to the east, and you lift off from it straight up, you will also move to the east, unless you turn a direction other than east.

Once you become detached from the train drag would slow your forward momentum and you would in fact move several (insert measurement based on velocity of train you are imagining here) back. If however say you picked something on the side of the tracks like a light pole and you jumped exactly as it passed. Even though you did move several lets say inches for safeteys sake back on the train you would have been forward of the pole.

I see what the OP is saying. If he manages to drop his orbital velocity to 0 (as in a dead stop) and fall back to Kerbin he should by all counts be moving west since Kerbin is rotating to the east.

As to why this isn't happening in game for him my guess would be limitations on KSP physics or because despite appearances he is actually listing off to the east.

[Mr.Rocket]

Thats because once you get high enough your east-ward speed is higher/lower than kerbins rotational speed. Pretty much its gravity is pushing you. If you get into a Keostationary orbit, you will be orbiting the same speed as Kerbin is rotating, while moving east-ward. On a train, you are not jumping high enough to actually move noticably, although you may move forward a bit. Its the same as an orbit sort-of. The train is on the earth, so it is also moving somewhat eastward, but you are higher up.

(p.s., don't jump while on top of a train, it's dangerous. Unless you're in a movie, then it looks epic.)

Edited December 19, 2013 by Mr.Rocket

[phunk]

One thing that comes to mind is that the atmosphere in ksp moves along with the surface, so as you go up, it is probably adding some additional eastward velocity due to drag.

[ExpHP]

ExpHP: Your calculations are correct. The reason to launch east would (IIRC) be centrifugal force, not coriolis force.

Actually, I believe the centrifugal force (-omega x (omega x r)) points up, working ever so slightly against gravity.

Are you sure about that? Keep in mind that the path drawn on the map doesn't show where the planet will have rotated to by the time you land.

Ah, I believe you've nailed it.

I just tested it, and sure enough, the ship does actually land where it launches from, rather than where the blue curve initially ends. As the ship rises and falls, the blue curve of the trajectory appears to "move west" along Kerbin as Kerbin rotates.

[tavert]

Aye, but therein lies my confusion.

My apologies if I'm reading your post too literally, but to my understanding, the rocket is not stationary relative to the surface. If it was, then it would be landing right where it was launched from, instead of a ways into the ocean to the east.

It starts out stationary relative to the surface. Then if you burn straight up, there are several things going on. You increase in altitude, and the Coriolis acceleration of the rotating reference frame (see equation (20) here https://dl.dropboxusercontent.com/u/8244638/Constant%20Altitude%20Landing%20and%20Takeoff%20Derivation.pdf) will tend to decrease your horizontal orbital speed. However atmospheric drag will tend to damp out any nonzero horizontal surface-relative speed.

The elliptical arc drawn in the map screen is showing your orbital path and doesn't take into account the effect of drag. That will generally show a prograde path, but the surface of Kerbin rotates below you as you traverse that path so by the time you reach the surface again you should end up to the west of where you started.

[Apexazimuth]

I believe alot of the confusion comes from the fact that motion is measured relative to the absolute center of the reference body (I.E. the core of Kerbin), as opposed to the surface.

[ExpHP]

I believe alot of the confusion comes from the fact that motion is measured relative to the absolute center of the reference body (I.E. the core of Kerbin), as opposed to the surface.

Indeed. In fact, it just occurred to me now that it is likely this same difference which sets apart the Surface and Orbit modes on the Navball. One is in the rotating reference frame of Kerbin's surface (which makes it relevant for tasks like "point the end that isn't on fire towards space" or "land in one piece"), while the other has a fixed axis (and is more relevant for everything else).

By this analogy, one could say that the trajectory is always drawn in "orbit" mode.