Equations needed

[Pretorian28715]

I have been looking at the following, to try to understand what I am/can do in KSP more.

However some of the Equations are missing/not showing up.

Does anyone know what they should be?

[Ohm is Futile]

Not a bad guide, although it's pretty technical and the pictures are broken. I would think some googling should actually yield the answers. I'm sure some of my more technical fellows can answer in more detail, as for me, I can tell you you don't need those to go anywhere in the stock game. In fact, you may very well manage to do pretty much anything in the game without being aware of all the math involved. Blasphemy, I know.

[WanderingKid]

Have you gone interplanetary?  If not, I recommend you actually start here:

http://ksp.olex.biz/

This lets you eyeball your departure times to play with maneuver nodes and get a rough idea of the dV required for intercepts.  Then, for more accurate build outs, you go here:

http://alexmoon.github.io/ksp/

To get that kind of precision on your builds, you want a mod such as KER or MechJeb.  I personally prefer KER (Kerbal Engineer Redux) for a number of reasons, but pick your poison.

Once you understand those two sites and their links and instructions, the rest of the tutorials and whatnot out there will probably make a lot more sense.  Here's the gist of it: The angle between the planets when you leave dictates where the planet will be when you get to its orbit, and the angle you're departing the home system from your orbit around it will dictate if you're 'slowing down' or 'speeding up' relative to other orbits in the solar system.

This is one of those cases where doing, then learning the math behind it, can simplify the learning by having visual reference points.

[GoSlash27]

pretorian28715,

 It must be pretty screwed up, 'cuz I can't tell which equations are missing

 Try this link:

http://www.braeunig.us/space/orbmech.htm

It covers pretty much everything you need for interplanetary flight. If you find it useful, drop a quarter in @OhioBob's tip jar.

Best,
-Slashy

[WanderingKid]

Self removing poorly written, unhelpful comment.

Edited June 7, 2017 by WanderingKid
Chagrin

[Pretorian28715]

8 hours ago, GoSlash27 said:

'cuz I can't tell which equations are missing

I think it is the Velocity on Escape and Ejection Angle equations

[Pretorian28715]

Thanks @GoSlash27, I have been looking at that link already, but don't really have the time to absorb it [at the moment, but I will keep going back to it], all I really need are the Velocity on Escape and Ejection Angle equations.

@OhioBob can you provide these please? as I can't figure which one it is for the moment.

[OhioBob]

9 hours ago, Pretorian28715 said:

Thanks @GoSlash27, I have been looking at that link already, but don't really have the time to absorb it [at the moment, but I will keep going back to it], all I really need are the Velocity on Escape and Ejection Angle equations.

@OhioBob can you provide these please? as I can't figure which one it is for the moment.

 

By “velocity on escape” I presume you’re referring to what we call hyperbolic excess velocity, which is given by equation (4.88) in my web page.

It’s a little different in KSP, however, because of the way patched conics are used.  I’ve derived the following equation:

V_soi² = 2μ * (1 / R_soi – 1 / R_bo) + V_bo²

where V_soi is the velocity at the sphere of influence, V_bo is the burnout velocity, R_soi is the radius at the sphere of influence, R_bo is the radius at burnout, and μ is the gravitational parameter.

This is the KSP version of V_∞² = V_bo² - V_esc², except it takes into account that patched conics are used by giving the velocity at the sphere of influence rather than at infinity.  Note that the equations are the same if we set R_soi = ∞.  In the KSP universe, V_soi is functionally equivalent to V_∞ because it represents the velocity remaining after escaping the planet’s gravity.

There is no single or easy equation used to compute ejection angles, it can get rather complex.  This section of my web site explains it.

 

Edited June 7, 2017 by OhioBob

[Pretorian28715]

1 hour ago, OhioBob said:

V_soi is the velocity at the sphere of influence, V_bo is the burnout velocity, R_soi is the radius at the sphere of influence, R_bo is the radius at burnout

Just to clarify, it's been a while since I've done this level of math.

I thought it might be 4.88 but could not find the V_bo Equation [thinking about it its probably on a different page].

R_bo = the Periapsis I guess? but would I straddle the burn [half either side of Pe] or burn to Periapsis [all before]?

Edit: and this must be right as it gives me the same value that I calculated for the Hohmann transfer DeltaV
 

Had already looked at the 'Hyperbolic Departure and Approach' section, but again been awhile.

I found this: -

=((PI()/2)-ACOS(µ/(µ+(SMA/1000)*Vescape)))*(180/PI())

for the Ejection Angle, but not sure if I can get it to work? is that equivalent to your 5.37?

 

[Sorry OhioBob the below was not meant for you, it is meant as an addition to the thread for general comment, but]

------------------------------------------------------------------------------------------------------Merged Posts------------------------------------------------------------------------------------------------------------------

 

ok, in the thread in the OP there is an example question plus answers. could I consider them within the 'margin of error'?

 

A spaceship needs to get from a 700km parking orbit (100km altitude) around Kerbin to a planet that is orbiting at 3x Kerbin's orbit. (SMA = 40,799,520,768m)

Answers:

t_H = 1.2897e7 seconds

Planetary Phase Angle = 82.0°

Hohmann Δv₁ = 2.0904 km/s

Ejection Velocity = 3.7909 km/s

Ejection Angle = 122.7°

 

I get the following Answers:

t_H = 13,015,777.67 seconds [1.3016e7]    - not OK

Planetary Phase Angle = 82.02°     - OK

Hohmann Δv₁ = 2,086.64 m/s    - not OK, by 4ish m/s

Ejection Velocity = 3,800.57 m/s    - not OK, by 9.6ish m/s

Ejection Angle = 76.30° based on the Equation I mentioned earlier.    - not OK, and well out, still trying to figure this out.

Edited June 7, 2017 by Pretorian28715

[OhioBob]

1 hour ago, Pretorian28715 said:

Just to clarify, it's been a while since I've done this level of math.

I thought it might be 4.88 but could not find the V_bo Equation [thinking about it its probably on a different page].

R_bo = the Periapsis I guess? but would I straddle the burn [half either side of Pe] or burn to Periapsis [all before]?

Edit: and this must be right as it gives me the same value that I calculated for the Hohmann transfer DeltaV

V_bo is the burnout velocity.  It the velocity the spacecraft is traveling immediately upon completion of the ejection burn, i.e. at engine burnout. 

R_bo is the magnitude of the spacecraft's position vector at burnout, i.e. the distance from the center of the planet to the spacecraft.

For most computations we simplify things by assuming that the Δv is applied instantaneously.  Therefore, V_bo is the initial orbital velocity plus the burn Δv, and R_bo is the radius at the ejection point, which we also assume is the periapsis of the resulting hyperbolic trajectory.  In practice it's more complex.
 

1 hour ago, Pretorian28715 said:

Had already looked at the 'Hyperbolic Departure and Approach' section, but again been awhile.

I found this: -

=((PI()/2)-ACOS(µ/(µ+(SMA/1000)*Vescape)))*(180/PI())

for the Ejection Angle, but not sure if I can get it to work? is that equivalent to your 5.37?

I don't know.  I'll have to study that and get back to you.
 

Edited June 7, 2017 by OhioBob

[OhioBob]

Let me make sure we're talking about the same thing when referring to "ejection angle".  What I'm thinking of is the "angle to prograde" or "angle to retrograde".  Is that right?

If we assume that the departure asymptote is tangent to the orbit of the planet, then the ejection angle is simply the angle η as shown in figure 4.15 and computed by equation 4.81.

 

[Pretorian28715]

30 minutes ago, OhioBob said:

Let me make sure we're talking about the same thing when referring to "ejection angle".  What I'm thinking of is the "angle to prograde" or "angle to retrograde".  Is that right?

If we assume that the departure asymptote is tangent to the orbit of the planet, then the ejection angle is simply the angle η as shown in figure 4.15 and computed by equation 4.81.

 

Yes, ejection angle to pro/retrograde.

Gonna need to read more, but looks like I have a place to start. Thanks.

[Pretorian28715]

With Thanks to @Nexus24680 for the PDF version of the original thread.

I will ask for this to be closed if there are no further posts to the thread, within the next 2-3 weeks.