Oberth effect

[GoSlash27]

Something else worth mentioning is that the mathematics governing orbit insertions is exactly the same as orbit ejections. Although all my examples talked about ejection ÃŽâ€v, the formulas and methods work just the same for orbit insertion. The only difference is the direction of travel. Therefore, if you are doing something like placing a station in orbit around another planet (where orbit altitude doesn't matter), the information presented in this thread can be use to select the most ÃŽâ€v friendly orbit.

Ohio Bob,

I had noticed the same thing as I was manipulating the spreadsheet. I'm going to try to pad it out to predict DV budgets from any planet to any other.

I don't have hyperedit, so I'd need somebody else to get empirical results to confirm it.

Best,

-Slashy

[NecroBones]

I haven't taken the time to look at it in depth yet, but for anyone wanting to see more orbital mechanics math, here's a page that seems to go into a lot of detail:

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

[xtoro]

I'm curious now about what would happen if I parked an asteroid and an extraplanetary Launchpad shipyard out at 68,000Km and launched ships from there. Since it would take very little to escape Kerbin SOI, would it still be worth lowering the PE by 67,930Km first?

[LostOblivion]

I'm curious now about what would happen if I parked an asteroid and an extraplanetary Launchpad shipyard out at 68,000Km and launched ships from there. Since it would take very little to escape Kerbin SOI, would it still be worth lowering the PE by 67,930Km first?

As they've been saying, it depends where you are going.

[OhioBob]

xtoro said:

I'm curious now about what would happen if I parked an asteroid and an extraplanetary Launchpad shipyard out at 68,000Km and launched ships from there. Since it would take very little to escape Kerbin SOI, would it still be worth lowering the PE by 67,930Km first?

It depends on the magnitude of V_∞. The break even point at an altitude of 68,000 km is V_∞ = 321 m/s. Therefore, if V_∞ > 321 m/s, then drop your periapsis and perform two burns. If V_∞ < 321 m/s, then just burn direct from your current orbit. The thing is, you can't get to any planet in the system for under 321 m/s V_∞, so you are always going to be better off performing a two-burn ejection from 68,000 km. The only exception might be if you are traveling to another nearby asteroid, where V_∞ might be < 321 m/s.
 

Edited May 24, 2016 by OhioBob

[GoSlash27]

I've finished the transfer calculator and worked out "optimal" transfers between Kerbin and the other planets.

The altitudes lead to some interesting conclusions. Also, the optimal altitudes are identical in both directions.

Kerbin->Moho

K 680km 1,661 m/sec

M 30 km 2,416 m/sec

Kerbin ->Eve

K 11 Mm 551 m/sec

E 23 Mm 598 m/sec

Kerbin ->Duna

K 7.7Mm 649 m/sec

D 570 km 584 m/sec

Kerbin -> Dres

K 980km 1,477 m/sec

D 30 km 1,292 m/sec

Kerbin -> Jool

K 350 km 1,918 m/sec

J 180 Mm 1,242 m/sec

Kerbin -> Eeloo

K 220 km 2,089 m/sec

E 20 km 1,366 m/sec

In the case of Jool, the most efficient altitude for the return is out near Pol's orbit, but of course you can't aerobrake out there. It would therefore seem most efficient to aerobrake at Laythe as well as safer.

Best,

-Slashy

[magnemoe]

My experience with this:

- If launching from KSC it is better to use as low a parking orbit as practical, any savings from burning at higher altitude are negated by the expenditure to get there.

- If starting from orbit around Minmus it is almost always better to drop to low Kerbin Pe before ejecting.

- If using Minmus-harvested fuel for refueling it is better to bring the fuel to the ship in LKO rather than bring the ship to Minmus.

Agree here except that benefit of going to Minmus to fill the tanks and then drop down for the burn let you get away with less fuel tanks and lower trust requirements. One serious problem then burning for Moho in particular is that you need pretty decent ISP to not get large errors then you do the burn as it takes time. You also don't have to bring all the fuel needed for getting into orbit and landing down to LKO.

Downside is that this is an more time consuming process. first go to Minmus then enter high orbit around Kerbin find the drop point where your PE is in the right position at the right time and then burn.

[xtoro]

It's interesting how the further you need to go means the optimal altitude is lower. Are those based on a 2 burn using OE or just single ejection burn?

- - - Updated - - -

Agree here except that benefit of going to Minmus to fill the tanks and then drop down for the burn let you get away with less fuel tanks and lower trust requirements. One serious problem then burning for Moho in particular is that you need pretty decent ISP to not get large errors then you do the burn as it takes time. You also don't have to bring all the fuel needed for getting into orbit and landing down to LKO.

Downside is that this is an more time consuming process. first go to Minmus then enter high orbit around Kerbin find the drop point where your PE is in the right position at the right time and then burn.

I think you mean TWR not ISP?

[GarrisonChisholm]

I haven't taken the time to look at it in depth yet, but for anyone wanting to see more orbital mechanics math, here's a page that seems to go into a lot of detail:

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

Ironically, this is the page that OhioBob maintains. :]

[Volix]

Why do I keep coming in under the estimated value??

I'm guessing it might be due to the game's patched conics. We don't have to actually reach escape velocity to escape, we just have to reach the SOI. That should mean that the prediction overestimates the amount of DV required.

I took a qualitative look at this problem and think the patched conics can explain the difference, but there are a couple issues. In N-body (real world) physics the gravity from any body is felt out to infinity. In patched conics however, there is some "cutoff" altitude where one body's gravity is "turned off", and the second's is "turned on". Leaving Kerbin then, once you reached this altitude any extra altitude from Kerbin does not come with a potential energy increase, letting you keep more kinetic energy than you otherwise could. This savings wouldn't be much, but Slashy's numbers aren't that much different.

I mentioned there are some issues. I spent some time manipulating the OP's equations and let V_esc be a little lower than predicted to account for patched conics. I found the effect increased V_{∞}, but when the two are added later to find dV (rather their squares are added) the effect cancels. This is likely because these equations are N-body equations and need more modification to properly handle patched conics.

Of course this also happens when arriving at a body. Energy you would have gained "falling" from infinity to the cutoff altitude is now not gained and your arrival velocity is a little lower than it would have been. This should show a small savings in the capture dV as well.

Edited October 8, 2015 by Volix
typos

[xtoro]

You could try installing the nbody mod http://forum.kerbalspaceprogram.com/threads/68502-WIP-1-0-4-Principia-Bourbaki-(2015-08-15)-N-Body-and-Extended-Body-Gravitation

I'm going to be trying it soon. Hopefully they get maneuver nodes working properly soon!

[NecroBones]

Ironically, this is the page that OhioBob maintains. :]

Yeah, Red Iron Crown told me. That was a forehead-slapping moment for me, for sure.

[OhioBob]

I just discovered something I had not previously noticed. In an earlier post I gave the equation,

r = 2μ/v_∞²

which is the radius at which the ejection/insertion Δv is minimum for a given value of v_∞. The interesting part is if we rearrange this equation we get,

v_∞ = (2μ/r)^0.5

Some of you may recognize that as the equation for escape velocity. Therefore, what we have found is that Δv is minimum when v_∞ = v_esc.

The equation for v_∞ is,

v_∞² = v_bo² - v_esc²

If we substitute v_esc for v_∞ and rearrange, we obtain

v_bo² = 2v_esc²

Recognizing that v_esc² = 2v_orb², we substitute and obtain

v_bo² = 4v_orb²

Taking the square root of each side, we have

v_bo = 2v_orb

or, expressed another way

Δv = v_orb

We therefore see that ejection Δv is minimum when we are at such altitude that our ejection velocity is exactly double our initial orbital velocity. I am not entirely certain what the implications of that is, but I find it an interesting oddity.
 

Edited April 9, 2016 by OhioBob

[Volix]

I wasn't satisfied with my qualitative look so I tried something more quantitative. I calculated the potential energy difference between the edge of a SOI in patched conics and infinity to get an idea of the size of numbers we're looking at. I'll put the math in a spoiler for those who are curious or want to check my work and post the relevant results below that. None of the math is beyond introductory physics or calculus 2 (there's a simple integral).

We'll start with the general work equation

W = F*d

where W is the work done (potential energy change), F is the force exerted on the craft, and d is the distance traveled along the direction of force (in this case radial from the planet, d is the altitude in this case). The force due to gravity is given by

F_g = G*M*m/r²

where G is the gravitational constant 6.674×10^{−11} Nâ‹…m²/kg, (capital) M is the planet mass, (lower-case) m is the mass of the craft, and r is the distance between them. We put this into the work equation and turn it into a differential so we can integrate. This integral will be

int(G*M*m/r^2 dr)

with limits corresponding to the altitudes we're traveling between. In this case I chose "x" (edge of the SOI) and infinity. (sidenote: Is there a good way to get an integral symbol on these forums?) The result is the energy that isn't being accounted for by using the patched conics model

W = G*M*m/x

Notice that as x gets larger this energy gets small, so the larger your SOI the more accurate the approximation. Of course, what we're really after is the dV, which can be found by setting the kinetic energy equal to the work done

KE = (1/2)*m*v² = G*M*m/x = W

which yields

v = (2*G*M/x)^1/2

This is the difference between the velocity you'd have at infinity for patched conics vs N-body. G*M and x are both given by the wiki for Kerbin as

G*M = 3.5316000×10¹² m³/s²

x = 84 159 286 m

Putting these numbers in the equation yields v = 289.57 m/s.

So, the "free" velocity gained by turning off gravity at the SOI for any planet of mass M and SOI at altitude x is (G is the gravitational constant 6.674×10^{−11} Nâ‹…m²/kg)

v = (2*G*M/x)^1/2

For Kerbin this works out to 289.57 m/s! Another way to think of this is if Kerbin was the only other thing in the universe, and you were very very far away. If you started "falling" toward Kerbin due to gravity you'd be going 290 m/s when you crossed it's "SOI". This is pretty significant, and due to the inverse square law most of this would be gained within the Kerbol system. I did the integral again but used 1.1*10¹¹ m instead of infinity (around the distance from Kerbol to Eeloo) and got 288.57 m/s, the rest of the way to infinity would only slow you by another 1 m/s from the edge of the system (again, if Kerbin is the only other thing in the universe, I'm neglecting every other planet and the sun).

Back to the issue at hand, how does this compare to the difference in Slashy's testing and the theoretical numbers? Slashy was finding a difference of 10-20 m/s. Compared to the 290 m/s difference from the patched conics that I've calculated his numbers are pretty much identical to theory. So, did I make a mistake in my math? Is Kerbin to Eve just not that far? Does the game somehow throw in some constant offset when you cross SOI to account for this? Something else?

[regex]

We therefore see that ejection ÃŽâ€v is minimum when we are at such altitude that our ejection velocity is exactly double our initial orbital velocity. I am not entirely certain what the implications of that is, but I find it an interesting oddity.

We aim to place our interplanetary craft into a parking orbit where the velocity is exactly half of what the ejection velocity would be from the altitude? Is there a quick and dirty algorithm to figure that out?

[GarrisonChisholm]

I just love reading intelligent conversation even though I am ill equipped to participate. Dialogue; the world should use it!

[OhioBob]

regex said:

We aim to place our interplanetary craft into a parking orbit where the velocity is exactly half of what the ejection velocity would be from the altitude? Is there a quick and dirty algorithm to figure that out?

All the math is described in post #40, however I'm not suggesting that you maneuver your spacecraft into that orbit just because the Δv is lower. The Δv you spend to get into the orbit can be more than what you save. If you are launching from the ground, you should seek the lowest practical orbit. However, there may be other applications (perhaps as yet undiscovered) where knowing the optimum altitude could be useful.
 

Edited April 9, 2016 by OhioBob

[GoSlash27]

regex,

We wouldn't necessarily aim to place our craft into that parking orbit. It's only when we're planning on fueling it from a source that's at that altitude or higher. Actually placing the craft there from low orbit costs more DV than launching from there saves.

Plus there's a whole boatload of practical considerations like hitting transfer windows, possible interference from moons, etc.

But to answer your question,

r = 2μ/v_{∞}² would be the simplest algorithm.

You would calculate V_{∞} by a Hohmann transfer about Kerbol from your source planet to the destination.

That's solved in the usual way; know your source and destination planets' SMA and Kerbol's gravity well μ. That's in the wiki.

from that, you have the SMA of the transfer orbit; (r1+r2)/2

You also need the source planet's orbit velocity, which is sqrt(μ/r1)

So now all you need is your orbital velocity leaving the source planet on the transfer v_tx= sqrt(μ(2/r1-1/SMA_tx))

Subtract the source planet's orbital velocity from the transfer velocity and that leaves V_{∞}.

Best,

-Slashy

Edited October 8, 2015 by GoSlash27

[FishInferno]

Relevant XKCD:

[regex]

All the math is described in post #40, however I'm not suggesting that you maneuver your spacecraft into that orbit just because the ÃŽâ€v is lower. The ÃŽâ€v you spend to get into the orbit can be more than what you save. If you are launching from the ground, you should seek the lowest practical orbit. However, there may be other applications (perhaps as yet undiscovered) where knowing the optimum altitude could be useful.

Awesome, that makes sense.

r = 2μ/v_{∞}² would be the simplest algorithm.

You would calculate V_{∞} by a Hohmann transfer about Kerbol the Sun from your source planet to the destination.

Nice, thanks.

[Zoidos]

It's posts like this when i wish i wouldn't suck at math.

[nohelmet]

OhioBob,

The Oberth effect exists because kinetic energy is what actually gets you from place to place and DV is a nonlinear measure of it.

OMG, slashy, I *finally* understand the Oberth effect, and it's all thanks to you!

This makes perfect sense: a given burn will increase your speed by the same amount, regardless of the initial speed. And as you pointed out, it's kinetic energy that gets you places!

Why didn't anyone say this earlier?

[OhioBob]

Volix said:

Back to the issue at hand, how does this compare to the difference in Slashy's testing and the theoretical numbers? Slashy was finding a difference of 10-20 m/s. Compared to the 290 m/s difference from the patched conics that I've calculated his numbers are pretty much identical to theory. So, did I make a mistake in my math? Is Kerbin to Eve just not that far? Does the game somehow throw in some constant offset when you cross SOI to account for this? Something else?

We had a similar discussion to this some time ago in which I made the following post.

 

OhioBob said:

I don't know if anybody will find it useful or not, but I just 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 of the sphere of influence, R_bo is the radius at burnout, and μ is the gravitational parameter.

This equation 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.

Let's first consider real life n-body physics. Suppose we are launching from an altitude of 75 km, where V_esc = 3235 m/s, and we have V_∞ = 900 m/s. Our required burnout velocity is,

V_bo = (V_∞² + V_esc²)^0.5 = (900² + 3235²)^0.5 = 3358 m/s

Now, with KSP's patched conics. We no longer think in terms of V_∞, instead we just need to make sure that our velocity is 900 m/s relative to Kerbin when we reach the edge of the SOI. We set V_soi = 900 m/s and solve for V_bo.

V_bo = (V_soi² - 2μ * (1/R_soi - 1/R_bo))^0.5 = (900² - 2*3.5316E+12*(1/84159287 - 1/675000))^0.5 = 3345 m/s

Therefore, in KSP we require 13 m/s fewer Δv than we would in real life. This number is comparable to what Slashy found in his experiments. I feel pretty confident that the difference between Slashy's numbers and the predicted numbers is due to KSP's patched conics.
 

Edited April 9, 2016 by OhioBob

[GoSlash27]

It's posts like this when i wish i wouldn't suck at math.

Zoidos,

KSP is actually an evil scheme to encourage you to not suck at math

Best,

-Slashy

- - - Updated - - -

OMG, slashy, I *finally* understand the Oberth effect, and it's all thanks to you!

This makes perfect sense: a given burn will increase your speed by the same amount, regardless of the initial speed. And as you pointed out, it's kinetic energy that gets you places!

Why didn't anyone say this earlier?

nohelmet,

Probably because they're way smarter than me

And I have actually said this before (as have others), but I don't believe it's ever been put into an easy- to- find tutorial in such plain language.

Best,

-Slashy

- - - Updated - - -

OhioBob,

This seems to explain the disparity nicely. I think I'll just keep the error in my spreadsheet and err on the side of caution (no kerbonaut ever complains about a little extra DV).

And thanks about a gazillion for the http://www.braeunig.us/space/orbmech.htm reference. It's been absolutely indispensable!

Best,

-Slashy

[OhioBob]

OhioBob,

This seems to explain the disparity nicely. I think I'll just keep the error in my spreadsheet and err on the side of caution (no kerbonaut ever complains about a little extra DV).

Yeah, I usually do all my computations using real life concepts, even if they don't exactly match KSP's way of doing things. In most cases the math is actually easier that way. And, as you say, it is often a little conservative, which is not a bad thing.

And thanks about a gazillion for the http://www.braeunig.us/space/orbmech.htm reference. It's been absolutely indispensable!

I'm glad you've found it useful. I think my web site audience has gotten much bigger since the introduction of KSP.