Calculating Speed relative to Kerbin

[Ydoow]

I know this is provided in-game, but I, for some reason, tried doing it myself as a means of double checking.

KSP Tells me my craft at 200km is orbiting at 2.101km/s and is travelling at 1.868km/s relative to Kerbin, the orbit has an eccentricity of the orbit is aproximately .000012 or so.

It\'s circular enough to assume it\'s a perfect circle for this situation.

With a craft sitting on the launch pad, I find Kerbin rotates at .1746 km/s

Here\'s where I\'m baffled.

2.101 km/s - .1746 km/s = 1.926 km/s

I would expect this equation to show me the crafts velocity relative to kerbin\'s rotation, but clearly it does not as 1.926 != 1.868

So what gives? Why does Kerbin\'s rotational speed seemingly change?

I don\'t understand or know what could cause this difference.

[SubOrbitalGuy]

edit: sorry misread what you wrote.

[ferram4]

It has to do with the fact that it\'s calculating your speed in a rotating reference frame, not with respect to Kerbin\'s surface.

Let\'s back up a bit. Take your velocity on the surface of 0.1746km/s and divide by Kerbin\'s radius of 600km. This will give you a rotation rate of 2.91*10^-4 1/s. Now multiply your orbital radius (not altitude!) by this and you get 0.2328km/s. If you subtract your orbital velocity of 2.101km/s from your 'surface' velocity of 1.868km/s, you get approximately that number. The surface velocity is your velocity with respect to the surface, but instead your velocity in a reference frame rotating with Kerbin.

[Ydoow]

Soo...I\'m a little lost. I think I don\'t know what you mean by \'rotating reference frame\'

I understood the math bit, in taking the radius instead of altitude, but I don\'t what it means compared to the \'rotating reference frame\'

[semininja]

Soo...I\'m a little lost. I think I don\'t know what you mean by \'rotating reference frame\'

I understood the math bit, in taking the radius instead of altitude, but I don\'t what it means compared to the \'rotating reference frame\'.

I\'m not sure I understand what the difference is, myself. It doesn\'t seem, to me, that there would be any.

[ferram4]

Ydoow:

Here\'s what a rotating reference frame is:

This is a coordinate system located at the center of Earth:

Wish I\'d found an animated pic.

In the fixed reference frame (the one used for orbital mechanics) the X and Y axes always point in the same direction relative to the entire universe, but not the planet. In a rotating reference frame the coordinate system rotates along with the planet, so that the X and Y directions rotate about the Z axis at the same rate as the Earth spins so that the X direction always points to the same longitude. The former is good for orbital mechanics, the latter for problems on the planet.

If you want to have a good idea of what a rotating reference frame is like, imagine standing on a carousel as it spins. You would measure everything off the carousel as moving because you are rotating with the carousel. When the game uses surface velocity, it\'s like you\'re measuring the velocity of things while you\'re on the carousel; when the game uses orbital velocity, it\'s like you\'re measuring the velocity of things while you\'re off the carousel.

Hope that helps.

[semininja]

How does velocity in a rotating reference frame differ from velocity w.r.t. surface observer?

[ferram4]

semininja:

It doesn\'t. Your understanding of it is correct, as far as I understand.

[Ydoow]

Ah so a fixed reference frame uses a universal coordinate system, whereas Kerbin (or being on any other planet) uses a global coordinate system meaning that the coordinate system spins relative to the universe, and the inverse when in space.

I still don\'t exactly see how this would create the velocity difference I\'m seeing though.

Unless you mean to say I need to calculate how fast Kerbin is spinning relative to the Universe, not to my craft.

[maltesh]

The Surface that your speed is relative to is the nearest point on surface of a sphere, rotating about the same axis as Kerbin, with same six-hour period as Kerbin, but with a radius equivalent to your current distance from Kerbin\'s center.

It\'s that point on that hypothetical surface that you\'d have to match velocities with to remain motionless above a particular point above Kerbin.

(It should also be noted that KSC isn\'t /precisely/ on the equator of Kerbin, but that difference is a much smaller effect than the previous one.)

[Ydoow]

Sorry I\'m still lost >.<

I understand the rotating reference frame now, but I don\'t see why the numbers are different.

I\'m moving relative to the center of Kerbin, is that what you\'re saying? That\'s why we take the orbital radius?

Completely lost :l

[Ydoow]

Perhaps this is beyond what I need though.

Basically what I\'m trying to do is be able to predict the MET when my orbit craft will be at the same longitude as KSC is.

Should I just use the two speed values KSP gives me? orbital/surface

[Maraz]

Please think about a pole 200km high infixed in Kerbin\'s ground.

The pole is rigid, so it moves as a whole.

But the top of the pole moves faster than the base. Because both points are running along a circumference, but the circumference of the point at the top of the pole is larger.

It\'s the top of the pole the point used as the reference to measure your speed.

Hope this helps.

Maraz

[Ydoow]

Okay that makes sense, and I guess I kinda always knew that. I was wondering why the numbers between subtracting surface velocity from my orbital velocity are different though. I don\'t exactly see the explanation there.

You\'re analogy makes sense.

Let\'s back up a bit. Take your velocity on the surface of 0.1746km/s and divide by Kerbin\'s radius of 600km. This will give you a rotation rate of 2.91*10^-4 1/s. Now multiply your orbital radius (not altitude!) by this and you get 0.2328km/s. If you subtract your orbital velocity of 2.101km/s from your 'surface' velocity of 1.868km/s, you get approximately that number. The surface velocity is your velocity with respect to the surface, but instead your velocity in a reference frame rotating with Kerbin.

Okay, I get that 0.000291 1/s is a ratio more or less and I was able to replicate the math to understand what\'s going on.

But I don\'t see the significance of .2328km/s, and I don\'t understand how you got the number 1.868km/s as the 'surface' velocity.

[ferram4]

Okay, I get that 0.000291 1/s is a ratio more or less and I was able to replicate the math to understand what\'s going on.

But I don\'t see the significance of .2328km/s, and I don\'t understand how you got the number 1.868km/s as the 'surface' velocity.

That is the velocity of the 'surface' of Kerbin, if it came out to the height of your orbit. Using Maraz\'s analogy, that would be the velocity of the tip of the 200km pole relative to fixed space. Your method of simply subtracting the velocity measured on the surface from the velocity in orbit would have worked if Kerbin were flat (), but we have to account for the fact that the surface of Kerbin is moving at 0.1746 km/s on a circle with a radius of 600 km but your ship is moving at 2.101 km/s on a circle with a radius of 800 km.

Basically what I\'m trying to do is be able to predict the MET when my orbit craft will be at the same longitude as KSC is.

As for this, I actually came up with a formula that should work for doing that, though I haven\'t tried it myself yet:

t = (L_KSC-L_Ship)*R_K/V_surf*180/pi

L_KSC = Longitude KSC, in degrees

L_Ship = Your Longitude, in degrees

R_K = Kerbin\'s Radius, 600km

V_surf = Your velocity, using the surface number

The 180/pi is to convert degrees into radians. The difference in longitudes multiplied by the radius is the arc length of the planet, and your velocity relative to the surface is how quickly you\'re moving relative to that point on the surface.

[Ydoow]

Okay I think I\'m mixing up the numbers your referring to.

V_Surf of Kerbin is .1746 km/s

So, dividing that by the radius, 600km gets me .000291 1/s

Then multiply that by my crafts current Orbital radius, 800km. That gets me .2838km/s

.2328 km/s should be the speed Kerbin rotates if the surface extended out to 800km, right? I felt you were trying to tell me that\'s how you got 1.868.

OOOOOOOOOHHHH!!!!!

*facepalm*

I think the decimals were throwing me off.

So 2.101 km/s is my crafts orbital speed, and IF Kerbin\'s surface was extended out to 800km (the same radius as my craft), it would rotate at .2328 km/s

2.101 km/s - .2328 km/s = 1.868 km/s

So that when I see Surface Velocity in KSP, it\'s telling me 'This is how fast your travelling relative to Kerbin\'s rotation as if it\'s surface was extended out to your orbit.'

Shoot. Now that I understand it, I wish you hadn\'t given me a formula ;P

Thanks for your patience, and the formula. It\'s been a while since I did physics, and sadly I don\'t think I could have made a formula on my own. It\'s been a long time since I\'ve done it.

You\'ve reminded me it\'s as simple as eliminating units until you\'re left with one that answers your question ;P

So thanks again! and hopefully I am properly understanding this, otherwise I\'ll feel like a bigger idiot.

[ferram4]

No problem. It also sounds like you do get it, especially if you did have one of those 'Wow, I\'m an idiot' kind of moments (Lovely feeling isn\'t it ). Sorry if I wasn\'t clear earlier, I\'m not very good at explaining math if I don\'t have a notebook to scribble on and show to people.

[Ydoow]

I\'m not always great at understanding math

The largest issue with understanding math is that numbers seem to be randomly pulled out of thin air. Of course there\'s a very good reason we use the numbers and derived them somehow, but we just overlook a step or something.

That was a chunk of my problem in this case. The other half was the conceptual matter

Edit: Oh wow. After fiddling with numbers I\'ve discovered that no matter how \'wide\' Kerbin gets, it always has the same rotational period. So the rotation rate more or less determines it\'s period.

[Ydoow]

Your equation is wrong

I did a lot of fiddling, and I finally came down to this

(Longitude_target - Longitude_target) x (pi/180) x Radius / Velocity_surface

It\'s almost exactly the same as yours, except for (pi/180)

You had (180/pi) ;P

The difference is degrees -> radians vs. radians -> degrees

lol

I did a lot more math/equations than I should have.

But I at least found an equation that can tell you how long it takes for Kerbin to rotate in any given degrees/radians.

It\'s interesting because with that rotation rate, you don\'t need to tell it at what radius you\'re interested in.

The units cancel out, and it\'s always the same. Which makes sense since the rotational period is always the same

[ferram4]

Oops. :-[ Close enough for Kerbal Math.

Edit: It technically does depend on the radius of the orbit, but that is accounted for in the V_surface term.

[Ydoow]

Actually I\'m just referring to the period at which Kerbin rotates within.

It ends up with the Radius being on the top and bottom of the division, so they cancel out and don\'t need to be added.

You do need to know the rotation rate first, which may need the V_surface though

Edit: Just wanted to say that the equation has proved to be extremely accurate (as it should). The main variation comes down to your actual orbital speed, which obviously changes slightly with elliptical orbits.