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About long burn time efficiency with low TWR

Question

I begin using nuclear and ion engines and I am not sure how to use them efficiently. I created maneuver node to Mun. Burn time is 4 minutes. I divided the burn in to two 2 minutes burns to make it more efficient. How much is my gain from it? What is the maximum acceptable burn time? If I'd burn 4 minutes at once how much delta-v would I lose?

Edited by CanOmer

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1 hour ago, Geschosskopf said:

Um, no...   Because math

The Oberth effect is the rocket gaining kinetic energy due to the exhaust having less kinetic energy.  Because the exhaust leaves the nozzle at the save velocity relative to the nozzle regardless, the difference in kinetic energy comes from less reaction mass leaving the nozzle.  Which means less fuel is being burned for the same increase in velocity if the rocket is moving faster, which where the increase in efficiency comes from.  That's the whole definition of Oberth.  Therefore, the less exhaust mass there is to start with, the less difference in exhaust mass there can be.  Therefore, engines that produce less exhaust mass benefit less from Oberth.

No, it's a function of how kinetic energy works for the ship.  What you just said is not how it works.  Exhaust velocity doesn't enter into it, nor does reaction mass.  Because Oberth effect is not about the ship, it's about the orbit.

I realize that saying "yes it is" / "no it isn't" is not a good way to convince anyone on either side of the argument, so let's actually do the math, shall we?

Let's consider two cases.  In fact, let's consider the two cases I already mentioned in my original post above.  We have a ship that starts in 80 km circular orbit.  It ends in 12000 km circular orbit.

• Case 1:  Two burns.  One burn to raise Ap from 80 km up to 12000 km, then a second burn at Ap to circularize.
• Case 2:  Three burns.  One burn to raise Ap from 80 km up to (for example) 1000 km, then a second burn at the 1000 km Ap that raises the next apsis to 12000 km, then a third burn that circularizes at 12000 km.

Both cases start in circular 80 km and end in circular 12000 km, the only difference being that Case 1 does more of its burn at a lower altitude.  We would expect Case 2 to require more total burn than Case 1, since it's losing Oberth benefit, right?

Okay, let's do the actual math to figure it out.

Math in spoiler section.  But no peeking yet!

Spoiler

It's a straightforward application of the Vis-viva equation.

The constants we plug in:

• GM = Kerbin standard gravitational parameter, 3.5315984e12 m3/s2
• Rlo = radius of initial 80 km circular orbit, 6.8e5 m
• Rhi = radius of final 12000 km circular orbit, 1.26e7 m
• Rmed = radius of intermediate 1000 km altitude burn, 1.6e6 m

So, with that in mind:

Orbital velocity for initial 80 km circular orbit is:  sqrt(GM / Rlo) = 2279 m/s

Orbital velocity for final 12000 km circular orbit is:  sqrt(GM / Rhi) = 529 m/s

For the Case 1 transfer ellipse:

• Pe velocity is sqrt(2GM * (1/Rlo - 1/(Rlo+Rhi)) = 3139 m/s
• Ap velocity is sqrt(2GM * (1/Rhi - 1/(Rlo+Rhi)) = 169 m/s

For the first Case 2 transfer ellipse:

• Pe velocity is sqrt(2GM * (1/Rlo - 1/(Rlo+Rmed)) = 2700 m/s
• Ap velocity is sqrt(2GM * (1/Rmed - 1/(Rlo+Rmed)) = 1147 m/s

For the second Case 2 transfer ellipse:

• Pe velocity is sqrt(2GM * (1/Rmed - 1/(Rmed+Rhi)) = 1979 m/s
• Ap velocity is sqrt(2GM * (1/Rhi - 1/(Rmed+Rhi)) = 251 m/s

So, putting it together.

For Case 1:

• Burn 1 = 3139 - 2279 = 860 m/s
• Burn 2 = 529 - 169 = 360 m/s
• Total burn for Case 1:  1220 m/s

For Case 2:

• Burn 1:  2700 - 2279 = 421 m/s
• Burn 2:  1979 - 1147 = 832 m/s
• Burn 3:  529 - 251 = 278 m/s
• Total burn for Case 2:  1531 m/s

So, how much did we lose due to losing Oberth benefit, from Case 1 to Case 2?

Answer:  1531 - 1220 = 311 m/s

Result:  Case 2 requires 311 m/s more dV than Case 1 does.  That's the answer right there.  The amount of dV you've wasted due to losing Oberth benefit is 311 m/s.  I'm reasonably confident that that answer's correct, assuming I didn't hit a wrong key on my calculator somewhere.  If you doubt my number, you can set it up in KSP yourself to check; just put a ship in 80 km circular orbit and drop some maneuver nodes.

Okay, now comes the time to peek inside the spoiler there.  Notice that the only numbers I needed to plug in were orbital parameters.  Kerbin's standard gravitational parameter, and orbital radii.  That's it.  Those were the only numbers I needed.  You won't see a single mention of ship mass, or fuel mass, or Isp, or exhaust velocity, or anything at all about the physical characteristics of the ship.  Because they don't matter to Oberth effect.

That's that the actual math looks like.  So, if I'm wrong... care to show me what your math looks like?  Provide some equations involving exhaust velocity and such to come out with a correct answer?

Might be nice to have a second opinion from perhaps the most respected physics nerd in the forums.  @OhioBob, care to weigh in?

Some other useful discussions, also involving people with physics background providing actual math to show that this is how it works:

Anyway, Geschosskopf, would love to see some math from you supporting your contention.  Take the orbital situation described (Case 1 versus Case 2), and use your numbers for exhaust velocity or whatever-else-about-the-ship to come up with a meters-per-second answer for how much dV is saved in one case versus the other.  Show us the numbers, please?

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There's no really simple answer, because it depends on a case-by-case basis what your trajectory is, what your TWR is, etc.

Basically, what you lose from doing a really long burn is Oberth benefit.  Maximum dV efficiency is when you do as much of your burn as possible as low as possible.  The ideal burn is right-at-periapsis.  If you have a really low TWR and have to burn until you've climbed from a really low Pe to some much higher altitude, then you're doing a lot of your burn at a higher altitude, which is less efficient from an Oberth perspective.

Breaking your burn into multiple segments (i.e. giving it a "kick" each time you swing 'round to periapsis again) isn't more efficient because it's multiple burns per se, but rather simply because it allows you to do more of your burn at the low altitude.

One way to get a rough idea of how much dV you save:  Try this little experiment (you can do it with any ship, because we're only adding up maneuver nodes here):

Case 1:  Put a ship in circular orbit at, say, 80 km.  Drop a maneuver node on it and give it enough to raise your projected Ap to, say, 12000 km. Note the amount of dV on the node.

Case 2:  Put a ship in circular orbit at 80 km, again.  Drop a maneuver node but this time only give it enough burn to raise the Ap to, say, 1000 km.  Then, drop a second maneuver node right at your projected Ap, and give that one enough to boost the Ap up to 12000 km.  Now calculate the sum of the dV for the two nodes.

Case 1 is the dV for doing a single high-TWR burn right at Pe.  It's also the dV for doing multiple low-TWR burns (doesn't matter how many) when you're right at the 80 km Pe.

Case 2 is, roughly, what happens when you have a really long, low-TWR burn.  You'll note that it's a bigger total amount of burn.  That's the efficiency loss.  You can play with case 2 to see what happens to the total burn amount if the initial burn only raises you to 200 km, or 500 km, or whatever.  That'll give you a rough idea of the kinds of numbers we're dealing with.

None of these give you the exact number for a single really long burn, because that's a hard number to calculate (you basically need to do it numerically, I don't think there's a simple analytical solution).  But it'll get you in the ballpark.

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10 minutes ago, Geschosskopf said:

Oberth is a very slippery concept.  There have been many discussons/arguments/head-beating-against-the-wall episodes about what exactly it means.  I've been in a few myself before, so I"m not surprised we're having another one here    Thus, I toast all participants and hope you all understand that I mean no disrespect when I disagree with anybody.  Now, on with the show....

This is not correct..  A given change in velocity produces the same change in Ek due to Ek = 1/2 * m * v^2.

Rather, the Oberth effect is stating that the faster a ship is moving, the more Ek it gains from burning the same amount of fuel.  Or, put another way, the same amount of fuel results in more dV the faster the rocket is moving at the time of the burn, which means that ship can burn less fuel to achieve the same amount of dV at higher speeds than lower ones.

So that's the net result, which is why Oberth is beneficial.  But for that to happen, the following things must all be true:

#1:  Conservation of Energy must be obeyed.  When looking at the ship in isolation, it appears to be getting free energy.  This cannot happen.  The increase in the ship's Ek per unit of fuel burned comes at the expense of the Ek of the exhaust stream.

#2:  The engine is functioning the same throughout, regardless of the ship's velocity.  This means the same mass of fuel is being burned per second, at the same temperature, producing the same amount of thrust, which creates the same mass of exhaust leaving the nozzle per second with the same velocity relative to the nozzle.

#3:  Despite all this, the ship still burns less fuel mass to get the same amount of dV as it would if the same burn was performed at a lower initial velocity.

The only way this works is if the dV (IOW, the burn) happens in less time.  The fuel is entering the engine at the same rate in l/sec, the engine's thrust is constant, and the exhaust velocity relative to the engine is also constant, so the exhaust mass per second is also constant.  Thus, the only way to burn less fuel is to run the engine for less time.  Because the exhaust velocity is constant, less burn time means less fuel mass burned, which means less exhaust Ek during that time.  Less fuel mass burned to achieve the same dV is the definition of increased efficiency.  If it were not for this, there would not be an Oberth effect.

Therefore, the greater the difference in fuel mass burned to produce the same dV at different velocities, the more the ship benefits from Oberth effect because this allows the greatest difference in exhaust Ek.  This means that ships with high mass flow rates reap the greatest Oberth benefits because they're throwing out the most reaction mass to start with.  You get a greater Ek change from reducing exhaust mass by several tons of LFO than you get from reducing it by several ounces of Xenon.

On the first misconception, that velocity produces the same change in kinetic energy:

We are agreed that kinetic energy obeys this equation: KE = 1/2 m*v^2. Thus, for a 2 kg object:

0 m/sec: 0.5 * 2 kg * (0 m/sec)^2 = 0

1 m/sec: 0.5 * 2 kg * (1 m/sec)^2 = 1 kg * 1 m^2/s^2 = 1 J

Thus, by accelerating your vessel from 0 to 1 m/sec, your vessel has gained 1 J of energy.

1000 m/sec: 0.5 * 2 kg * (1000 m/sec)^2 = 1 kg * 1000000 m^2/s^2 = 1000000 J

1001 m/sec: 0.5 * 2 kg * (1001 m/sec)^2 = 1 kg * 1002001 m^2/s^2 = 1002001 J

Thus, by accelerating your vessel from 1000 m/sec to 1001 m/sec, your vessel has gained 2001 J of energy.

Thus, the same change in velocity, the same delta-V, has produced massively different changes in kinetic energy dependent only on your starting velocity. The reason rockets don't horribly break conservation of energy is that the exhaust loses a commensurate amount of kinetic energy at the same time as the rocket gains energy.

On the second misconception, that the Oberth effect lets you get more delta-V:

Rockets are fundamentally based on Newton's third law of motion: for every force, there is an equal and opposite reaction force. This produces conservation of momentum, where momentum p = mv. This is, for our purposes, more important than conservation of energy; while total energy is conserved, kinetic energy only describes a portion of your system's total energy.

While the concept holds for rockets, it is simpler to consider a vessel propelled by a gun firing discrete chunks of mass. A 1 kg object propelled out the back of a spacecraft at 3500 m/sec will have (relative to the original position/velocity) -3500 kg*m/sec of impulse. Total impulse of the system remains constant: +3500 kg*m/sec of impulse had to be imparted to the spaceship. If the spacecraft (sans the propelled object) masses 100 kg, its velocity will be changed as (3500 kg*m/sec / 100 kg) = 35 m/sec of delta-V.

The famous Tsiolkovsky rocket equation is essentially stating "if you propel mass out the back in infinitesimally small quantities, the change in velocity is described by this equation":

delta-V = V(exhaust) * ln(full mass / empty mass) = Isp * Gm * ln(full/empty)

Exhaust velocity is determined by how fast the propellant leaves the rocket, and is fixed by the engine and what it's using as propellant. Your full/empty mass is determined by the spacecraft design. Both are wholly independent of how fast you happen to be traveling right now and in what reference frame. If you're going 100 m/sec relative to Earth or 100,000 m/sec, you still have the same delta-V.

The amount of change kinetic energy that delta-V affords you, however, is hugely different based on the Oberth effect... and it is energy that one needs to escape a gravity well.

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2 hours ago, Snark said:

@OhioBob, care to weigh in?

I think you and @Starman4308 have got it covered.  My opinion has been registered using the like button.

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On ‎1‎/‎12‎/‎2018 at 6:26 PM, CanOmer said:

I begin using nuclear and ion engines and I am not sure how to use them efficiently. I created maneuver node to Mun. Burn time is 4 minutes. I divided the burn in to two 2 minutes burns to make it more efficient. How much is my gain from it? What is the maximum acceptable burn time? If I'd burn 4 minutes at once how much delta-v would I lose?

I don't think the issue is so much about how long a burn takes so much as what fraction of the orbital arc the burn consumes. If you are burning for 4 minutes when your orbital period is 24 hours, there's probably not much difference over a single impulse burn.

@OhioBob has a site that states that the dV for a *really* slow burn -- a burn closer to the entire orbital period, is about the difference in the old and final orbital velocity. You could compare that to a Hohmann transfer to estimate the upper bound of loss.

http://www.braeunig.us/space/orbmech.htm#maneuver (see eqn 4.72)

I have done a number of high dV burns from Kerbin (say 3000+ ejection). While I would benefit from low orbit oberth effects, I use a higher orbit so that I get a predictable outcome. If I have a 10 minute burn from a 30 minute orbital period, I have a really hard time making that end up like I wanted. I find doing the same burn from an orbital period of say... 3 hours to have a more predictable outcome.

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51 minutes ago, CanOmer said:

I begin using nuclear and ion engines and I am not sure how to use them efficiently. I created maneuver node to Mun. Burn time is 4 minutes. I divided the burn in to two 2 minutes burns to make it more efficient. How much is my gain from it? What is the maximum acceptable burn time? If I'd burn 4 minutes at once how much delta-v would I lose?

The amount of efficiency lost by a long burn is a function of altitude.  The lower your altitude, the tighter the turn radius of your orbit and the faster you're moving around it.  Thus, when doing a long burn at low altitude, your ship swings through a greater arc centered on either side of the node.  The bigger this angle, the greater the cosine loss due to off-angle thrust.

When orbiting Kerbin at about 100km, your orbital period is about 28 minutes.  Thus, during a 4-minute burn, your ship will travel 1/7 of a circle, with 1/14 on either side of the node.  1/14 of 360^ is 25.7^.  The cosine of 25.7^ is 0.9010.  This means that AT WORST, at the very beginning and very end of a 4-minute burn, you're only 90% as efficient as you are when passing through the node.  Between the end points and the node, the efficiency is better than 90%.

So that's perfectly acceptable.  Even a 5-minute burn is fine.  But that's where I draw the line, FOR BURNS IN LKO, for the several different reasons:

• First and foremost, that's all the time I want to sit twiddling my thumbs in real time while nothing happens but the engine running.
• Second, burns any longer than that can push your Pe down into the atmosphere if you start at less than 100km orbit.
• Third, the cosine loss starts getting bigger than I like.

So, if the burn time is going to exceed 5 minutes, I either break it up into 2 parts or I add more TWR to the design.  To get an ejection burn from LKO <= 5 minutes, you need a TWR >= 1.75 for that stage.

NOTE, however, that the efficiency loss only really matters for chemical rockets, not nukes and ions.  They've got dV to spare.  So for them, the 1st 2 reasons above are more important than the 3rd.  Besides, Oberth effect is proportional to reaction mass, so chemical rockets benefit from it the most, nukes get only meh, and ions essentially zero benefit.   Which means, you for them, don't need to be in LKO (see the below).

Also note that the longer your orbital period (so the higher your altitude), the less cosine loss you have for the same amount of time.  This is why I only care about burn time for ejection burns.  When in interplanetary space, or when doing capture burns, you've got a huge turn radius so there's much less cosine loss.

Edited by Geschosskopf
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4 minutes ago, Geschosskopf said:

Thus, when doing a long burn at low altitude, your ship swings through a greater arc centered on either side of the node.  The bigger this angle, the greater the cosine loss due to off-angle thrust.

Certainly that can be an issue if you just point and ignore as it changes.  But long, low-TWR burns are less efficient even if you've got SAS set to hold (in which case there wouldn't be any cosine losses).

The real (unavoidable) limit is Oberth loss.

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1 minute ago, Snark said:

The real (unavoidable) limit is Oberth loss.

The amount of Oberth effect you get is proportional to reaction mass.  Therefore, it really helps LFO rockets, is meh for nukes, and practically non-existent for ions.  The OP is using nukes and ions, so Oberth is not of great concern to him, but cosine loss is.

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2 minutes ago, Geschosskopf said:

The amount of Oberth effect you get is proportional to reaction mass.  Therefore, it really helps LFO rockets, is meh for nukes, and practically non-existent for ions.

No, it's not, even slightly.  Oberth benefit is simply a question of m/s, reaction mass doesn't enter into it.  It's in the math.

It's true that an ion-powered ship would care less, but it's because it has more dV to spare (and therefore the lost Oberth benefit would probably be of less concern), not because it has less reaction mass.  The dV lost to Oberth is the same, either way.

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6 minutes ago, Snark said:

No, it's not, even slightly.  Oberth benefit is simply a question of m/s, reaction mass doesn't enter into it.  It's in the math.

It's true that an ion-powered ship would care less, but it's because it has more dV to spare (and therefore the lost Oberth benefit would probably be of less concern), not because it has less reaction mass.  The dV lost to Oberth is the same, either way.

Um, no...   Because math

The Oberth effect is the rocket gaining kinetic energy due to the exhaust having less kinetic energy.  Because the exhaust leaves the nozzle at the save velocity relative to the nozzle regardless, the difference in kinetic energy comes from less reaction mass leaving the nozzle.  Which means less fuel is being burned for the same increase in velocity if the rocket is moving faster, which where the increase in efficiency comes from.  That's the whole definition of Oberth.  Therefore, the less exhaust mass there is to start with, the less difference in exhaust mass there can be.  Therefore, engines that produce less exhaust mass benefit less from Oberth.

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@Snark is correct on the Oberth Effect. It's a consequence of kinetic energy (in the classical limit) being proportional to the square of current velocity.

For a 2 kg test object:

0 to 1 m/sec is a gain of 1 J.

1000 to 1001 m/sec is a gain of 2001 J.

The Oberth effect is essentially stating that a change to your velocity produces greater kinetic energy changes when you are going fast. This benefits maneuvers at periapses, where velocity is highest, or otherwise deep in gravity wells. It's why high TWR burns are more dV-efficient, because you can complete the requisite velocity change while giving the gravity well less time to reduce your velocity.

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Oberth is a very slippery concept.  There have been many discussons/arguments/head-beating-against-the-wall episodes about what exactly it means.  I've been in a few myself before, so I"m not surprised we're having another one here    Thus, I toast all participants and hope you all understand that I mean no disrespect when I disagree with anybody.  Now, on with the show....

59 minutes ago, Starman4308 said:

The Oberth effect is essentially stating that a change to your velocity produces greater kinetic energy changes when you are going fast.

This is not correct..  A given change in velocity produces the same change in Ek due to Ek = 1/2 * m * v^2.

Rather, the Oberth effect is stating that the faster a ship is moving, the more Ek it gains from burning the same amount of fuel.  Or, put another way, the same amount of fuel results in more dV the faster the rocket is moving at the time of the burn, which means that ship can burn less fuel to achieve the same amount of dV at higher speeds than lower ones.

So that's the net result, which is why Oberth is beneficial.  But for that to happen, the following things must all be true:

#1:  Conservation of Energy must be obeyed.  When looking at the ship in isolation, it appears to be getting free energy.  This cannot happen.  The increase in the ship's Ek per unit of fuel burned comes at the expense of the Ek of the exhaust stream.

#2:  The engine is functioning the same throughout, regardless of the ship's velocity.  This means the same mass of fuel is being burned per second, at the same temperature, producing the same amount of thrust, which creates the same mass of exhaust leaving the nozzle per second with the same velocity relative to the nozzle.

#3:  Despite all this, the ship still burns less fuel mass to get the same amount of dV as it would if the same burn was performed at a lower initial velocity.

The only way this works is if the dV (IOW, the burn) happens in less time.  The fuel is entering the engine at the same rate in l/sec, the engine's thrust is constant, and the exhaust velocity relative to the engine is also constant, so the exhaust mass per second is also constant.  Thus, the only way to burn less fuel is to run the engine for less time.  Because the exhaust velocity is constant, less burn time means less fuel mass burned, which means less exhaust Ek during that time.  Less fuel mass burned to achieve the same dV is the definition of increased efficiency.  If it were not for this, there would not be an Oberth effect.

Therefore, the greater the difference in fuel mass burned to produce the same dV at different velocities, the more the ship benefits from Oberth effect because this allows the greatest difference in exhaust Ek.  This means that ships with high mass flow rates reap the greatest Oberth benefits because they're throwing out the most reaction mass to start with.  You get a greater Ek change from reducing exhaust mass by several tons of LFO than you get from reducing it by several ounces of Xenon.

Edited by Geschosskopf
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2 minutes ago, Snark said:

That's that the actual math looks like.  So, if I'm wrong... care to show me what your math looks like?  Provide some equations involving exhaust velocity and such to come out with a correct answer?

Sure thing.  Your math, while it adds up, is focusing on the wrong values, so is not an illustration of the Oberth effect. The Oberth effect is not measured in the difference in amounts of m/s used when getting to the same place via different sets of maneuvers.  It is instead measured in terms of fuel mass remaining after changing velocity by the same amount when starting at different speeds.

That is, after all, what makes the Oberth effect a buff to efficiency.  As we all know, the same mass of fuel produces a different amount of m/s dV under different circumstances, which is why you're looking at the wrong numbers.  For instance, the amount of dV produced by 1 unit of fuel increases as the ship's mass decreases.  Oberth is another such situation.  Increased efficiency means the ship has more fuel left to make future maneuvers after a burn producing the same amount of dV.  The Oberth effect therefore actually applies in cases like these:

• Case 3:  I'm flying Ship X.  I want to change my velocity from 1000m/s to 2000m/s, which is a dV of 1000m/s
• Case 4:  I'm flying an identical Ship X.  I want to change my velocity from 4000m/s to 5000m/s, which is also a dV of 1000m/s

The Oberth effect says that Case 4 will consume less fuel (IOW, be more efficient) than Case 3 to get the same 1000m/s dV.  It does this by noting the effect of initial velocity on Work:

• W = F * s which, for a prograde burn, means dEk = F * s
• Over time, dEk/dt = F * v

Therefore, with constant force (which is from the same engine running the same way), a higher initial velocity results in a greater change in Ek over the same time.  That statement right there, and nothing else, is the definition of the Oberth effect, and is why Case 4 burns less fuel than Case 3.  The Oberth effect is not a thing unto itself, it's nothing but a simple observation, which Herr Oberth was the 1st to really draw attention to, of this natural relationship between initial velocity and the change over time in kinetic energy.  Go see the oft-cited wiki page if you don't believe me

Assuming you're with me so far, where we're talking past each other is in the implications of this statement.  The only energy input into the system is the same engine burning the same fuel at the same rate to produce the same thrust in both cases.  IOW, the engine adding the same amount of energy to the system per unit time in both cases.

So where does the extra Ek of the ship in Case 4 come from?  It doesn't just miraculously appear because that would violate conservation of energy.  It instead comes from the Ek of the exhaust stream.  The total Ek of ship + exhaust is the same, it's just that the ship in Case 4 gets more of it than the ship in Case 3.  This is easily seen (again, see the wiki page) when comparing a static engine test with a ship moving at ANY non-zero velocity.  When the ship isn't moving at all, all the Ek goes to the exhaust.  When the ship is moving, some of the total Ek is from its motion, so the exhaust has less.  And the faster the ship is moving, the more Ek it has due to Ek = 1/2 * m * v^2.

HOWEVER, the ships in both Case 3 and Case 4 are burning fuel at the same rate.  This means that until one of them achieves its desired final velocity and stops its engine, both will have the same mass.  Therefore, because faster Case 4 is getting more Ek over the same amount of time than slower Case 3, it's velocity is increasing faster, again due to EK = 1/2 * m * v^2.  Therefore, Case 4 will achieve the 1000m/s in less time than Case 3.  This means that Case 4 will run its engine for less time, and thus burn less fuel, than Case 3.  This is where the efficiency buff comes from.  And that's a mathematical demonstration of the actual Oberth effect, not your calculations.

If you're still with me, we can now get back to what started this whole discussion:  my statement that the amount of Oberth fuel savings is also proportional to reaction mass.  This is a case of comparing apples to oranges because none of the variables noted as equal above are so any longer.  The ships have different starting masses, they burn fuel at different rates, and they have different thrusts.  Still, we can apply the above to each type of ship individually and then compare the results.

• Case 5:  A ship with high thrust and low Isp.
• Case 6:  A ship with low thrust and high Isp

The exhaust in Case 6 has less Ek than the exhaust in Case 5 because it contains a smaller mass of fuel and/or this fuel is moving at lower velocity.  This means that if both ships start at the same velocity, the Ek balance between ship and exhaust is already more in favor of the ship in Case 6 than in Case 5.  This in turn means that the Case 6 ship can get less extra Ek from its exhaust stream than is possible with the Case 5 ship.

2 hours ago, Starman4308 said:

On the first misconception, that velocity produces the same change in kinetic energy:

You completely missed my point.  I was explaining how the Ek is transferred from the exhaust stream to the rocket in an effort to prove my assertion that the benefits of Oberth are proportional to reaction mass.  The velocity I was talking about in what you took exception to was that of the exhaust stream compared to the nozzle, which is CONSTANT regardless of ship velocity through space.  Therefore, to decrease the Ek of the exhaust stream, it has to have less mass.  Because the mass flow rate is also constant, this can only be achieved by running the engine for less time.  Which I just demonstrated again above, coming at it from a different direction, so hopefully we're on the same page now.

2 hours ago, Starman4308 said:

On the second misconception, that the Oberth effect lets you get more delta-V:

I'm sorry, but if you believe this is a misconception, then you have no idea at all what the Oberth effect really is.  Go back and read the wiki page again.  Especially these sentences up near the top:

Quote

The gain in efficiency is explained by the Oberth effect, wherein the use of an engine at higher speeds generates greater mechanical energy than use at lower speeds....

The Oberth effect occurs because the propellant has more usable energy due to its kinetic energy in addition to its chemical potential energy.[3]:204 The vehicle is able to employ this kinetic energy to generate more mechanical power.

Efficiency means you burn less fuel to get the same result, or you can get a greater result from the same fuel.  IOW, Oberth does indeed give you more dV for the same fuel mass.  This happens at higher initial speeds as a result of the work equation noted in my reply to Snark:

• dEk/dt = F * v

This basically means that the faster you're going, the less fuel you burn to increase your velocity by the same amount.  This happens because the faster the ship is going, the more of the total EK of ship + exhaust it gets, so the less time it has to run its engine to change its velocity by the same amount.

Or do you have some other definition of "efficiency" ?

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1 hour ago, Geschosskopf said:

The Oberth effect therefore actually applies in cases like these:

• Case 3:  I'm flying Ship X.  I want to change my velocity from 1000m/s to 2000m/s, which is a dV of 1000m/s
• Case 4:  I'm flying an identical Ship X.  I want to change my velocity from 4000m/s to 5000m/s, which is also a dV of 1000m/s

The Oberth effect says that Case 4 will consume less fuel (IOW, be more efficient) than Case 3 to get the same 1000m/s dV.

No, Case 3 and Case 4 will burn the same amount of propellant.  What the Oberth effect says is that Case 4 will increase the mechanical energy of the spacecraft more than in Case 3.

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39 minutes ago, OhioBob said:

No, Case 3 and Case 4 will burn the same amount of propellant.  What the Oberth effect says is that Case 4 will increase the mechanical energy of the spacecraft more than in Case 3.

The fact that it burns the same amount of propellant is the even the whole reason the Oberth effect works.

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Case 1

A 10-ton rocket is traveling 2000 m/s.  Its momentum is the product of its mass and velocity,

10 * 2000 = 20,000 t*m/s

The rocket ejects 1 ton of propellant at a velocity of -3000 m/s relative to the rocket, which is -1000 m/s in our fixed frame of reference.  The momentum of the ejected propellant is,

1 * -1000 = -1000

Since momentum must be conserved, the momentum of what is left of the rocket is the initial momentum minus the momentum of the ejected propellant,

20,000 - (-1000) = 21,000

The remaining rocket has a mass of 9 tons.  Its velocity is its final momentum divided my its final mass,

21,000 / 9 = 2333 m/s

Which is an increase of 333 m/s.

Case 2

A 10-ton rocket is traveling 4000 m/s.  Its momentum is,

10 * 4000 = 40,000 t*m/s

The rocket ejects 1 ton of propellant at -3000 m/s, or +1000 m/s in the fixed frame of reference..  The momentum of the ejected propellant is,

1 * 1000 = 1000

The momentum of what is left of the rocket is,

40,000 - 1000 = 39,000

The final velocity of the rocket is,

39,000 / 9 = 4333 m/s

Which is an increase of 333 m/s.

-------------------------------------------------------------------

In both case 1 and case 2 we have the same rocket, traveling at different initial speeds, ejecting the same amount of propellant, resulting in the same increase in velocity.  In other words, it takes the same amount of propellant to produce the same change in velocity regardless of the initial speed of the rocket.

Edited by OhioBob
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20 hours ago, CanOmer said:

I begin using nuclear and ion engines and I am not sure how to use them efficiently. I created maneuver node to Mun. Burn time is 4 minutes. I divided the burn in to two 2 minutes burns to make it more efficient. How much is my gain from it?

Very little, in the case you give.   You often want to split a four-minute burn, though, for the lack of battery capacity for the ion engines.

The benefit of burning while moving fast is, as they say, that thrust increases orbital energy at a rate thrust·velocity
and it is true that for longer burns the math is not simple because, in addition to thrust, gravity is changing your velocity,
but in your case the burn can be completed before your orbit gets noticeably high and slow.

I suggest you make a quick-save and try both ways, and post back with numbers.

Spoiler

Using an ion probe from 72km orbit, I had to cheat infinite electricity to make a balanced test, because my battery was not enough for the long burn.

The maneuver node says one burn of 4m 49s=289s, 860m/s, gets to the Mun.
It actually took 305s, 146units Xe, 893m/s delta-v used per KER
4% more than the simple math for an instantaneous burn.

Splitting into two 430m/s burns, each at periapsis but on successive orbits,
the first took 147s, 70u Xe, 434m/s, and
the second took 148s, 71u Xe, 436m/s,
for a total of 142u Xe, 870m/s,
just 1% more than an instantaneous burn.

(The math giving a 311-m/s difference from Snark was for a different case: two instantaneous burns with the second burn at apoapsis where the craft is very slow.  The low-TWR burn-splitting asked about in the top post, though, saves only 23m/s.)

Edited by OHara
I got 23m/s savings out of 860m/s
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@Geschosskopf, is it possible you could be getting confused with hyperbolic excess velocity (V∞)?  V∞ is the velocity left over after a spacecraft has escaped a celestial body's influence.  The lower a spacecraft is in the body's gravity well when it performs the escape burn (i.e. the faster it is moving), the less ΔV is takes to produce a given amount of V∞.  And since it requires less ΔV, it requires less propellant.  Therefore producing X amount of V∞ requires less propellant if the burn is performed at a lower periapsis when the vehicle is moving faster.  But V∞ is not ΔV, you could be conflating the two.

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Fundamentally, what Oberth does is not change how much dV you get from your propellant, but how much dV is needed to accomplish something. Look at it through energy methods: If you want to transition between two different energy states (in this case, trajectories, which may or may not be orbits), and since gravity is a conserving force and will neither add nor remove energy, you need to supply or remove the difference.

However, because kinetic energy has a squared function in it, the same amount of change in velocity can produce different amounts of change in energy, which depends on what the original velocity was. The higher it is, the larger the change in energy achieved by changing the velocity by a fixed amount becomes. Because rockets in vacuum always produce the same change in velocity for the same amount of propellant regardless of their velocity, they can get more energy change for the same dV and propellant use if they're going faster. This 'extra' energy comes from the stored kinetic energy of the propellant itself, for anyone worrying about CoE.

Edited by foamyesque
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The Oberth Effect simply states that the power a rocket produces is higher at higher velocities.

power = energy / time

energy = force * distance     (also referred to as work)

For example, assume a rocket engine burn for some arbitrary amount of time. At a higher velocity, the rocket travels through more distance for the same amount of time. Same force applied over a larger distance equates to more work energy done. More work energy done over the same amount of time equates to more power produced.

Applied to orbital mechanics, energy is most efficiently added/subtracted from your total orbital energy (kinetic energy + potential energy) at the lowest point in your orbit, where you are travelling the fastest.

That's all there is to it.

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On 1/12/2018 at 8:24 PM, Snark said:

No, it's not, even slightly.  Oberth benefit is simply a question of m/s, reaction mass doesn't enter into it.  It's in the math.

It's true that an ion-powered ship would care less, but it's because it has more dV to spare (and therefore the lost Oberth benefit would probably be of less concern), not because it has less reaction mass.  The dV lost to Oberth is the same, either way.

PB666 wrote up an interesting article on why ion (more specifically electrical powered spacecraft) engines shouldn't use "spiral out" navigation but instead a "mangalyaan maneuver (Pe kicking)".  He works out the math and shows that doing so can give twice the delta-v (at a cost of 100 times the time to reach escape velocity).  I took the data to mean that I should bring enough reaction mass to spiral out (possibly switching from Xenon to Argon if I had to) as I don't expect to see ion engines increasing in thrust by a factor of 100 any time soon (and consider spiraling out already "slow enough").  As everywhere, it is a tradeoff and you can choose either way.

Note that the problem is pretty moot if you aren't willing to stick around in the Van Allan belts (NASA uses big rockets to send ion engines out well past them, as the belts are harsh on everything, especially solar panels).

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On 1/13/2018 at 3:25 PM, OhioBob said:

@Geschosskopf, is it possible you could be getting confused with hyperbolic excess velocity (V∞)?  V∞ is the velocity left over after a spacecraft has escaped a celestial body's influence.  The lower a spacecraft is in the body's gravity well when it performs the escape burn (i.e. the faster it is moving), the less ΔV is takes to produce a given amount of V∞.  And since it requires less ΔV, it requires less propellant.  Therefore producing X amount of V∞ requires less propellant if the burn is performed at a lower periapsis when the vehicle is moving faster.  But V∞ is not ΔV, you could be conflating the two.

The ability to get more V∞ for less fuel due to moving faster when doing the burn is EXACTLY what the Oberth effect is.  In a non-powered fly-by, the gravity provides a free change of direction, but the magnitude of the velocity when leaving the SOI is the same as when entering it  The ship accelerates due to gravity while falling to Pe, then decelerates due to gravity while climbing away from it, so gravity's net effect on the magnitude of the velocity is (effectively) zero.

V∞ after the encounter is the value of interest here.  The ship is doing a fly-by so its final destination is elsewhere, and it gets there using V∞.  The object of the game, therefore, is to increase V∞ to the required amount using as little fuel as possible.  Because you get a greater change in V∞ for a given amount of fuel if you're moving faster when you burn, a powered fly-by is more fuel-efficient than making the same increase in V∞ from a lower starting speed.

ΔV, OTOH, depends on the frame of reference.  And the amount of ΔV you get from a given amount of fuel is not constant.  These 2 things confuse a lot of people.  ΔV means "change in velocity" (aka acceleration), but you always have to qualify this with respect to what that velocity is relative to.  The amount of ΔV available shown by things like MJ and KER has a fixed frame of reference, the ship itself, based on the thrust and time-dependent mass of the ship, ΔVs = T/mt.  But what really matters for spaceflight is ΔV∞, the ship's velocity relative to the parent body.  By doing a powered fly-by, so that you burn when the magnitude of your velocity is higher than normal, you get a bigger ΔV∞ for a given amount of fuel.  Hence, in the frame of reference of the parent body, you get more ΔV out of your fuel with the Oberth effect than without it.

On 1/13/2018 at 4:12 PM, foamyesque said:

Because rockets in vacuum always produce the same change in velocity for the same amount of propellant regardless of their velocity

This is not correct.  Burning a given amount of fuel always produces the same amount of thrust, but the amount of  ΔV this results in depends on the mass of the ship at the time.   ΔV = T/m.  Thus, the 1st unit of fuel burned, when the tanks are full and the ship is at its max mass, produces far less  ΔV than the last unit of fuel, when the tanks are empty and the ship is light.  You can easily see this by comparing the  remaining  ΔV shown by MJ or KER with the remain green bars in the fuel tanks.  When the remaining  ΔV displayed is 1/2 the starting value with full tanks, the tanks will have only about 1/4 fuel left in them.  IOW, you get about 1/2 your total  ΔV (relative to the ship) from the last 1/4 of your fuel.

==============

BTW. I saw some show on the Science Channel the other day talking about the Voyager missions.  It actually had a lot of the real-life boffins who designed the things explaining how they did it.  The Voyager missions required 2 main things:  an accurate computational method for doing N-body gravity, to plot the trajectories through all the fly-bys, and the Oberth effect, to get from 1 fly-by to the next.  Apparently, nobody had really paid any attention to Herr Oberth's original paper from way back in the 1920s or so, which was long before anybody had a need of it.  Thus, when the Voyager team began planning the flightpaths, they weren't taking Oberth into account and were dismayed because it didn't look like they could do the missions.  With the fuel tanks constrained by the size of the lifter, it didn't look like they could get enough  ΔV(∞) to get everywhere.  Then somebody dug up Oberth's paper, adjusted the calculations accordingly, and it turned out they really could do it after all.

So, as I said, the Oberth effect results in more ΔV, but this is the ΔV∞ kind, not the ship-based ΔVs = T/mt kind of ΔV.

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4 hours ago, Geschosskopf said:

And the amount of ΔV you get from a given amount of fuel is not constant.

Each subsequent unit of propellant produces more ΔV than the one before.  But that's because the vehicle is less massive when the second unit is ejected.  The second unit of propellant mass represents a larger fraction of the vehicle's remaining mass than the first, so the ratio mf/mi is greater for the second unit, so it produces more ΔV.  I don't think anybody here disputes that.  But that's not what you said.  You said that an "identical ship" will produce more ΔV if its initial velocity is higher.  That is false.  If two identical ships both having the same initial mass eject the same amount of propellant, each will produce the same ΔV regardless of how fast the ships were moving at the time they performed the burn.

Quote

ΔV means "change in velocity" (aka acceleration)

No, acceleration is rate of change in velocity.  It is not the same thing as ΔV.

Edited by OhioBob
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4 hours ago, Geschosskopf said:

The ability to get more V∞ for less fuel due to moving faster when doing the burn is EXACTLY what the Oberth effect is.  In a non-powered fly-by, the gravity provides a free change of direction, but the magnitude of the velocity when leaving the SOI is the same as when entering it  The ship accelerates due to gravity while falling to Pe, then decelerates due to gravity while climbing away from it, so gravity's net effect on the magnitude of the velocity is (effectively) zero.

V∞ after the encounter is the value of interest here.  The ship is doing a fly-by so its final destination is elsewhere, and it gets there using V∞.  The object of the game, therefore, is to increase V∞ to the required amount using as little fuel as possible.  Because you get a greater change in V∞ for a given amount of fuel if you're moving faster when you burn, a powered fly-by is more fuel-efficient than making the same increase in V∞ from a lower starting speed.

ΔV, OTOH, depends on the frame of reference.  And the amount of ΔV you get from a given amount of fuel is not constant.  These 2 things confuse a lot of people.  ΔV means "change in velocity" (aka acceleration), but you always have to qualify this with respect to what that velocity is relative to.  The amount of ΔV available shown by things like MJ and KER has a fixed frame of reference, the ship itself, based on the thrust and time-dependent mass of the ship, ΔVs = T/mt.  But what really matters for spaceflight is ΔV∞, the ship's velocity relative to the parent body.  By doing a powered fly-by, so that you burn when the magnitude of your velocity is higher than normal, you get a bigger ΔV∞ for a given amount of fuel.  Hence, in the frame of reference of the parent body, you get more ΔV out of your fuel with the Oberth effect than without it.

This is not correct.  Burning a given amount of fuel always produces the same amount of thrust, but the amount of  ΔV this results in depends on the mass of the ship at the time.   ΔV = T/m.  Thus, the 1st unit of fuel burned, when the tanks are full and the ship is at its max mass, produces far less  ΔV than the last unit of fuel, when the tanks are empty and the ship is light.  You can easily see this by comparing the  remaining  ΔV shown by MJ or KER with the remain green bars in the fuel tanks.  When the remaining  ΔV displayed is 1/2 the starting value with full tanks, the tanks will have only about 1/4 fuel left in them.  IOW, you get about 1/2 your total  ΔV (relative to the ship) from the last 1/4 of your fuel.

==============

BTW. I saw some show on the Science Channel the other day talking about the Voyager missions.  It actually had a lot of the real-life boffins who designed the things explaining how they did it.  The Voyager missions required 2 main things:  an accurate computational method for doing N-body gravity, to plot the trajectories through all the fly-bys, and the Oberth effect, to get from 1 fly-by to the next.  Apparently, nobody had really paid any attention to Herr Oberth's original paper from way back in the 1920s or so, which was long before anybody had a need of it.  Thus, when the Voyager team began planning the flightpaths, they weren't taking Oberth into account and were dismayed because it didn't look like they could do the missions.  With the fuel tanks constrained by the size of the lifter, it didn't look like they could get enough  ΔV(∞) to get everywhere.  Then somebody dug up Oberth's paper, adjusted the calculations accordingly, and it turned out they really could do it after all.

So, as I said, the Oberth effect results in more ΔV, but this is the ΔV∞ kind, not the ship-based ΔVs = T/mt kind of ΔV.

V∞ is only one application of the Oberth effect. The Oberth effect is the simple observation that, in whatever reference frame you choose, a change in speed produces a greater effect on kinetic energy if the speed is already high. Orbital mechanics does not play into the Oberth effect.

What almost everybody refers to with ΔV is a craft's ability to propulsively change its own velocity vector. That you can accomplish more or less with that ΔV depending on gravitational assists and the Oberth effect is immaterial; a craft's ΔV is solely a function of full mass, empty mass, and exhaust velocity. That is the definition used by almost all rocket scientists and KSP players, and adoption of this terminology will make things much easier to discuss. Otherwise, you are simply talking past everyone else, irritating those who know what they're talking about, and confusing new players who don't know what is being talked about.

As to that Science Channel show you're discussing: material presented for a general audience is often rife with oversimplifications and lack of technical correctness, and they could have been referring to just about anything.

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6 hours ago, Geschosskopf said:

This is not correct.  Burning a given amount of fuel always produces the same amount of thrust, but the amount of  ΔV this results in depends on the mass of the ship at the time.

That's trivially obvious -- so obvious, in fact, that I assumed it to be implicitly understood that the comparison involved equivalent masses: the same burn being performed by the same ship, with the only difference being the ship's original velocity when executing the burn.

Because that's the useful thing to think about; you're looking really hard for things to argue about here instead of trying to understand what people're telling you. At this point I'm not sure what to say.

EDIT:

Also, in a non-relativistic system, which KSP is, deltaV is independent of reference frame and will always be equal viewed by all observers.

Edited by foamyesque

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