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[Tutorial] Interplanetary How-To Guide


Kosmo-not

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If the orbits are slightly elliptical, the information in the post will get you close to the destination object.

If the orbits are more than slightly elliptical, the mathematics is going to get pretty hairy, and likely beyond the scope of something you'd post in a thread like this one. You'd really want a tool built into the game, or an app that could read the persistence file, for something like that.

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It may sound stupid, but what does the weird U mean in step 1? Do I have to fill in a specific number?

I'm just 15 years old :P

That U, stands for the gravitational parameter of the sun. You can find gravitational parameters of all celestial bodies in KSP on the KSP wiki.

Kerbol: 1.167922e9 km^3/s^2

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That U, stands for the gravitational parameter of the sun.

Kerbol: 1.167922e9 km^3/s^2

About that, what do we put into the equation? 1.167922e7 or 11,679,220 doesn't seem to get the answer that Kosmo not got. Also, what does the km^3/s^2 mean? lol D:

EDIT: Well, it turns out I might not have been doing it wrong. I guess the pi button only goes to 3.14 because my original answer was only a little off... So, finally, I got 128,969,970.14415253188733116798055, or about 1.2897e7 like he got lol

EDIT2: Anyway, thanks so much for this. I made myself a spreadsheet to do the math so I hope when the update comes out I will be able to get to the planets a little easier, now I just need a protractor :P

Edited by 2008dragon
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Also, what does the km^3/s^2 mean? lol D:

Its basically what the number means. In this case kilometers cubed divided by seconds squared. Its a complex unit for gravity's effect.

Velocity is kilometers divided by seconds (how fast you travel a second, in the direction you are headed).

Edited by air805ronin
Fixing quote
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Awesome guide. When the new planets come out I am going to calculate all of this myself and write it down and make my self a little table of interplanetary transfers. Cant wait.

How do you plan to actually measure the phase and ejection angles when piloting? Hold a protractor up to the screen?

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Awesome guide. When the new planets come out I am going to calculate all of this myself and write it down and make my self a little table of interplanetary transfers. Cant wait.

How do you plan to actually measure the phase and ejection angles when piloting? Hold a protractor up to the screen?

I'm going to be using an on-screen protractor.

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  • 3 weeks later...

But what if we dont want to wait until planets will rich calculated angle? Lets say, that on we send mission to another planet, and plan to get back after about 6 months - but waiting untill best angle would be 3 years...

Is there way to make phase angle an argument here?

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  • 2 weeks later...
i just tried out the calculator multiple times (around 4) and even used a protractor for the angles,

im missing the planets by millions of miles, not even close, are you sure it works?

I have used it 3 times and it has never failed me.

HOWEVER. All you need to do is make sure you get the angle right. Dont follow the velocity readings, they are inaccurate (for me atleast). Just follow the angles, and burn until the trajactory says that you intercepted the planet.

Works like a charm everytime.

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Dont follow the velocity readings, they are inaccurate (for me atleast).

The ejection velocity is accurate. My understanding is that it is based upon an instantanious burn (delta-V change) at the right spot. As a craft moves along the escape trajectory, it slows down.

Because it's not possible do the burn instantly, a craft starts moving along the new trajectory while still doing the burn. So if you end the burn at the target velocity, you are actually going faster than you should be at that point.

I'm sure it's possible to work out what the right velicity is to stop the burn, by factoring in the PWR. But.... just use the escape velocity as a guide and use map view get the Ap correct.

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r = 670 m

v = 3361 m/s

μ = 3530.5

equationsk.png

€ = 33612/2 - 3530.5/670 = 5648155

h = 670*3361 = 2251870

e = sqrt(1+(2*5648155*22518702)/3530.52) = 2143756

θ = cos-1(1/2143756) = 89.9999

Ejection Angle = 180° - θ = 90.0001

I've done this calculation several time and I keep getting the same answer, 90 degrees, (Which is WRONG, according to this.)

This is for a transfer from Kerbin to Duna. The answer should be close to 150 degrees and I have the right ejection velocity and μ.

So what's going on here? And yes, I am using degree mode on my calculator.

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Use radian mode

Then the answer is À/2, which as I'm sure you know is 90 degrees.

Also,

by Kosmo-not

(have this value in degrees, not radians)

Ejection Angle = 180° - θ

Edited by mossman
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Mossman, you're mixing units. You're using a velocity of 3361m/s and an altitude of 670km. Further, μ is in km^3/s^2. So you need to run the calculations with v = 3.361 km/s to make it work.

E = 0.37876 km^2/s^2

h = 2251.87 km^2/s

e = 1.14375

θ = 29.0365 degrees

Ejection Angle = 150.963 degrees.

Hope that clears it up. Try carrying the units through the calculation next time and canceling them as you go. It will help you pick up common errors like this. ;)

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