Jump to content

oberth effect


JtPB
 Share

Recommended Posts

Actually I meant say an AP of 150k and a PE of 80k, then setting a node to burn to encounter between PE and AP, since you would have more velocity than in a circular orbit where AP and PE velocity are the same. (or setting the maneuver node on the "bottom" half of the orbit, where velocity is higher). Although I don't really know if the velocity in that case would be higher than a circular velocity, I'm making a pretty big assumption.

Edited by xcorps
Link to comment
Share on other sites

Why wouldn't you want an elliptical orbit to transfer to a satellite in the same SOI of Kerbin? Getting an elliptical is easier than climbing to altitude then burning for circular then burning incline then burning for Hohmann transfer isn't it?

I think just about anything wold be easier than what you just described...

For starters, you're almost always better off leaving the launchpad at or close to the inclination you want. If you time it right (launch just before KSC passes under the ascending/descending node of the target) you can match orbital planes very closely right from the start, and the extra dV you miss out on by not going due east is more than made up for by not trying to change your inclination later.

You don't always need to circularize immediately after reaching orbit and then proceed to a transfer maneuver. If you time it right you can go directly to the desired orbit.

If you must change your inclination by a lot after you're in orbit, you want to be going slow... so raise your Ap way, way up, change inclination, then lower it back down. If you only need a relatively small change, you can usually combine that into your transfer burn with minimal penalty.

Ultimately, unless your timing is perfect, you DO want your orbit to be slightly more or less elliptical than the target's... with either your Ap or Pe matching. This will mean your orbital periods will be slightly different and allow the two craft to eventually meet up.

=Smidge=

Link to comment
Share on other sites

Actually I meant say an AP of 150k and a PE of 80k, then setting a node to burn to encounter between PE and AP, since you would have more velocity than in a circular orbit where AP and PE velocity are the same. (or setting the maneuver node on the "bottom" half of the orbit, where velocity is higher). Although I don't really know if the velocity in that case would be higher than a circular velocity, I'm making a pretty big assumption.

That would just be splitting the transfer burn into two separate burns, compared to burning from an 80kmx80km orbit. The burn to make your orbit 150kmx80km would be identical to the start of a transfer burn, the second burn at periapsis would be the same as the finish of a single burn, so I don't think there'd be any net gain.

Unless, I suppose, your TWR is low enough that a significant part of the transfer burn would be away from periapsis, then it might be worthwhile.

Link to comment
Share on other sites

Stationary planets aren't actually very common, you know...
In this context that doesn't seem to be true. "Rogues" are looking very common, quite a bit more common than planets that actually do orbit stars.

Not that that has anything to do with KSP...

Link to comment
Share on other sites

What's a good margin for Incline? I've been getting it to zero because I read somewhere it was important. I usually get into orbit around Kerbin with less than 1.5 difference, mostly less than 1 with respect to the Mun.

The reason I'm curious is I'm having difficulty getting Kerbonauts back from the Mun, I have no trouble at all with Minmus or probes.

Link to comment
Share on other sites

In this context that doesn't seem to be true. "Rogues" are looking very common, quite a bit more common than planets that actually do orbit stars.

Not that that has anything to do with KSP...

<pedantic>Not orbiting a star doesn't mean its "stationary". Everything in the galaxy is orbiting around the galactic centre and even extra-galactic rogues are still moving (and constantly being accelerated by gravity).</pedantic>

Link to comment
Share on other sites

The Oberth effect allows us to get more kinetic energy change from the same amount dV spent, and it is the kinetic energy change that is important. Think of dV as the currency spent to buy a change in kinetic energy. The Oberth effect allows you to buy at a discount.

Awesome explanation :D

Link to comment
Share on other sites

<pedantic>Not orbiting a star doesn't mean its "stationary". Everything in the galaxy is orbiting around the galactic centre and even extra-galactic rogues are still moving (and constantly being accelerated by gravity).</pedantic>

Nothing is stationary, and everything is stationary. As always, it depends on your point of view, or frame of reference to be pedantic.

It's all relative :D

Link to comment
Share on other sites

could you explain to me what exactly happens in oberth effect, & why it happens? (graphic illustration may very help)

i tried to read about it on wikipedia but i understood nothing :(

Put as basically as possible:

If your rocket is already in motion, acceleration along the direction of that motion is more efficient. Prograde burns at periapsis use less energy (because you are already in motion in that direction). Low altitude burns use less energy (because you are moving faster at low altitude than at a higher altitude). If you are standing still and fire your engines, you burn with the minimum efficiency. The same amount of force from a standstill gives you less delta-v than that equal amount of force if you are already in motion.

Link to comment
Share on other sites

*Sigh* Oberth works for both prograde and retrograde, and Oberth does not change delta-V but instead increases kinetic energy change from expending delta-V.

This happens because the relationship between speed and kinetic energy is not linear, being "in motion in that direction" has nothing to do with it.

Edited by Red Iron Crown
Link to comment
Share on other sites

So you can see, that even though the same delta V was expended, the change in kinetic energy is an order of magnitude larger when burning at periapsis due to the Oberth effect.

First off, I think we're talking past each other a bit here, which I'll get into later. But for the moment, I'm afraid I have to disagree with your math above. Those calculations have nothing to do with the Oberth Effect. What you've done is simple vector addition, demonstrating the physical "leverage" that causes you to make Ap changes at Pe anyway, Oberth or not. Also, you only added vectors at the point of the burn (Ap or Pe), without taking into account the resulting velocity change at the other end of the of the burn (Pe or Ap) caused by the burn.

Anyway, back to Oberth being merely a fuel-efficiency thing we need not worry about.....

Energy isn't something you can measure directly, but only indirectly by its effects. That's why we have so many different types of energy: kinetic energy, thermal energy, gravitational potential energy, chemical potential energy, etc., each of which indirectly measures energy by looking changes in speed, temperature, altitude, etc.

Physically, spaceflight boils down simply to just moving from 1 orbit to another. You move from high orbits to low orbits, or parking to transfer orbits, etc., but you're always in an orbit (even if you're on the ground). All orbits have velocities directly associated with them, meaning spaceflight is all about changing the ship's velocity when and as needed to move from 1 orbit into the next. Because of this, kintetic energy is what's important to rockets; they convert the chemical potential energy in their fuel into kinetic energy to give them the velocity they need for the desired orbit. Thus, delta-V is the currency of spaceflight.

However, it's essential to understand that the delta-V required to move from an arbitrary point along Orbit A into desired Orbit B is exclusively a function of vector math. The specifics of the ship involved have no effect at all on the delta-V required; all ships of whatever size and TWR require the same delta-V to make the same trip between the same 2 end points. You can easily see this is true in the game by switching to a new ship and plotting a maneuver for it before doing anything else to it. You will see the amount of delta-V required but the burn time will say "N.A." because the game doesn't know the TWR of the ship until you burn its engines. Thus, you don't need to know anything about the ship to determine the amount of delta-V needed for a maneuver.

But anyway, the delta-V required to do any given burn is determined solely by vector math. When creating, moving, and tweaking thrust direction gizmos on maneuver nodes, the game is doing vector math under the hood and you're just seeing the sums of the vectors.

Now, given this, there's Oberth. Oberth has nothing to do with the amount of delta-V required by a burn; that's solely a vector math thing. Obeth therefore can only come into play during the burn. Oberth can't change the fixed amount of delta-V required by the burn, so the only benefit he can provide is making the burn more efficient. The ONLY way to make a burn more efficient is by allowing the ship to consume less fuel to reach its required delta-V total than the stats of its engine/fuel system would dictate on their own.

Assuming the engine runs full throttle the whole burn, the rate of fuel consumption is constant throughout due to the engine's fixed vacuum Isp. Therefore, the only way to reduce the amount of fuel consumed in the burn is to make the burn time shorter. But the amount of delta-V required is still the same regardless of burn time. Therefore, the only way to decrease burn time for the fixed amount of delta-V is to accelerate faster, so you "delta your V" faster. Which essentially means Oberth, if he exists in the game, is providing you with extra thrust.

The Oberth Effect wiki page says this right off the top:

The Oberth effect occurs because the propellant has more usable energy due to its kinetic energy on top of its chemical potential energy. The vehicle is able to employ this kinetic energy to generate more mechanical power

IOW, a given ship with a given engine burns X amount of fuel and gets Y amount of change in Ek purely from the combustion. Oberth then adds an extra amount Z of Ek change on top of this. Thus, the total Ek change is Y (from the fuel alone) + Z (from Oberth). The mass change in burning the fuel is the same for both so both Y and Z can only be measured as changes in velocity AKA delta-V. So there it is: Oberth gives you more delta-V per unit of fuel burned than you get from the fuel alone. Therefore, Oberth is a fuel-efficiency thing.

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

In practical KSP gameplay terms, Oberth makes no difference to how we do things. If the game takes Herr Oberth into account, it does so totally "under the hood" and there's nothing we can do about it. We create and manipulate a maneuver node, doing vector math under the hood, and this determines the delta-V required by the burn. We place our Pe's as low as possible and do our Ap-affecting burns there because the vector math says this minimizes the delta-V required. That this is also the best way to benefit from Herr Oberth, assuming he's in the game, is just icing on the cake. The presence or absence of Oberth doesn't change the way we do business. Hence, there's no point worrying about him. If he's there, we save some fuel. If he's not, then we don't. But we're still going to put our nodes where we do either way.

But despite it being immaterial, some folks might still lie awake at night stewing over Oberth is really in the game. Determining this will require some math I'm afraid. Put a ship in LKO and create for it a prograde burn node of say 2000m/s or more. The bigger the better. Note the burn time the game calculates. Now you can calculate the burn time yourself, ignoring Oberth, and if yours is longer, then Oberth is in the game. You can also calculate the final mass of the rocket after the burn, then have MJ do the burn and see how the final mass comes out in the game. If your mass is lower, then Oberth is in the game.

Link to comment
Share on other sites

A lot of people have mentioned that the Oberth effect gives you more energy for the same dV, but not many people have explained where that extra energy comes from. When you're accelerating a rocket, you're accelerating your remaining propellant as well as your payload. Your fuel is moving as fast as the rest of the rocket is, and as a result is has kinetic energy. Additionally, as your lift your propellant up into space and away from the planet, it's gaining potential energy. Typically we think of the energy in fuel as purely chemical energy that is released when the fuel is burned, however all three components must be added together to find the total available energy of the propellant.

When we make a burn at periapsis (prograde for simplicity), several things are happening:

1: We're combusting our propellant, extracting its latent chemical energy and using it to accelerate the exhaust gases out the nozzle and consequently accelerating our spaceship.

2: We're at the lowest point in our orbit, meaning our spaceship has the lowest potential energy. Potential energy isn't really something that we can use to accelerate our spacecraft in a helpful way, so we want to trade it off for increased kinetic energy and more speed, which we can use.

3: Since we're moving fastest at periapse, our propellant has the highest possible kinetic energy before it's burned (oftentimes this kinetic energy dwarfs its chemical energy!). When it exits the rocket nozzle, its velocity is equal to our current orbital velocity minus the rocket's exhaust velocity. When the rocket is traveling near the exhaust velocity, this number gets small, meaning our exhaust is left with a relatively small amount of kinetic energy. The best case scenario is when we're traveling exactly at the exhaust velocity - we're essentially leaving our exhaust motionless in space. Since energy is conserved in the rocket-exhaust system, our ship gains all the kinetic energy that our exhaust loses, and we fly away with the most possible energy extracted from our propellant.

lots of things!

I think you're a little confused about what exactly the Oberth effect is. You're talking about how it increases fuel efficiency, and you're technically right about that, but not in the way you explained. The Oberth effect can't magically increase the chemical energy available in the fuel, or the exhaust velocity of your engine. It doesn't do that in real life, and it doesn't do that in KSP. The reason burns higher velocity are more efficient is that your rocket is extracting the maximum total energy, chemical and kinetic, from your fuel.

Additionally, saying that the Oberth effect increases the dV from a burn is incorrect, because you're wrongly equating orbital energy and dV. If my spaceship is capable of an average acceleration of 10m/s^2, and I burn my engine for 10 seconds, I will be going 100m/s faster than I was before the burn. It doesn't matter what my initial speed was, accelerating at a given rate for a given time will always result in the same change in velocity. However, the same change in velocity does not always give you the same change in orbital energy.

It's not some magic device that makes your engine better, it's just a name for a specific consequence of physics.

Edited by MockKnizzle
Link to comment
Share on other sites

I'll jump in here to just lay out the basics in a way that I figured out to help me understand why this occurs. That was actually for race car performance simulation stuff, but physics are of course the same for everything.

Power = Energy / time

Energy = Force * distance

velocity = distance / time

Force = mass * acceleration

(now some of those variables represent physically measurable entities, and some are purely imaginary mathematical values, but that's not important here)

So:

P=E/t, and E=F*s, so:

P=F*s/t, but v=s/t, so:

P=Fv, but F=ma, so:

P=mav

Power = mass * acceleration * velocity

So, if you produce some acceleration on some mass, then the faster that mass is going, the more power is used/produced/converted.

It doesn't matter if it's a rocket, a volleyball, a race car, a proton in a particle accelerator, or a bullet. It doesn't matter if this acceleration is produced by throwing some mass in some direction, by a magnetic force, by gravity, or whatever.

Rockets conveniently work by pushing on mass that is initially traveling at the same speed as the engine, so the acceleration is constant (for a given force and mass), regardless of v.

So the end result is; if you burn at double the velocity, your engine produces twice as much power! (with the same amount of fuel used, with the same Isp, naturally)

Edited by Oskar-
Link to comment
Share on other sites

But despite it being immaterial, some folks might still lie awake at night stewing over Oberth is really in the game. Determining this will require some math I'm afraid. Put a ship in LKO and create for it a prograde burn node of say 2000m/s or more. The bigger the better. Note the burn time the game calculates. Now you can calculate the burn time yourself, ignoring Oberth, and if yours is longer, then Oberth is in the game. You can also calculate the final mass of the rocket after the burn, then have MJ do the burn and see how the final mass comes out in the game. If your mass is lower, then Oberth is in the game.

Sadly the maneuver node's burn time estimation doesn't take the rising TWR as you burn fuel into account, so this won't work. You might be able to use Kerbal Engineer though, it gives you a calculated dV of your current fuel load. If you burn what KE says is 1000m/s of fuel and gain 1010m/s velocity, say Hi to Oberth. KE does not change those dV numbers based on your current speed or location, so it at least is definitely not Oberth Aware.

Personally I'd be extremely surprised if the devs put Oberth in.

You could always ask them.

Link to comment
Share on other sites

If KSP abides by normal Newtonian physics, then the Oberth effect is by default in the game. You don't have to code up anything special. It is a consequence of physics, nothing more. Doppler effect, Hall effect, Leidenfrost effect, Oberth effect... all of these are simply interesting (and useful) consequences of the underlying physical framework of the universe.

Link to comment
Share on other sites

<pedantic>Not orbiting a star doesn't mean its "stationary". Everything in the galaxy is orbiting around the galactic centre and even extra-galactic rogues are still moving (and constantly being accelerated by gravity).</pedantic>
The explicit context of the comment we were quoting was "stationary planets" vs "planet orbiting another body", not my fault if the original was a gross misuse of the term "stationary". :wink:
Link to comment
Share on other sites

Lots of people don't seem to be getting this, I will try to explain with an example that can be verified in game. The bi-elliptic transfer

Let's say we have launched a satellite, the launch vehicle has delivered it to a circular orbit at an altitude of 100km. The intended orbit of the satellite is a circular orbit at an altitude of 10000km

For a simple Hohmann transfer an 830m/s would raise the apoapsis to the desired altitude, then at apoapsis a 374m/s burn would circularize. a total of 1204m/s required to make the transfer.

But lets say we make a longer initial burn, 917m/s instead of 830. this will raise our apoapsis almost to the edge of Kerbins SOI

Then when we reach that apoapsis we make a 71m/s prograde burn. this will raise the periapsis to the desired altitude.

Finally when we reach that periapsis we need a 192m/s burn to circularize.

The same transfer was made for 1180m/s

We burned for longer in the fast 100km orbit, and also we were going faster when we circularized. Whe the target orbits SMA is large enough compared to the initial orbit (*12), the gains in the circularization burn, more than make up for the extra cost in the initial burn.

This is the Oberth effect in action.

Edited by Rhomphaia
Link to comment
Share on other sites

I kind of doubt that the KSP engine models the kinetic energy change of the fuel, personally. Or even the kinetic energy of the fuel in the first place. Or even the energy of the fuel, really.

If you want to have doppler effect in a game, you have to code it. It's not there by default. Photons are not being simulated, nor are sound waves, nor are any other waves. No waves = no doppler.

Nor is the hall effect, as there are no electrons in KSPs physics model.

The Leidenfrost effect certainly isn't in KSP, as temperature is barely modeled, let alone state transitions.

The underlying physical framework of the universe is not what KSP is running! KSP is running a simulation, there's a very large difference. Not only that, but it's running an extremely simplified simulation. The planets don't have gravity for the same reason the real universe does (unless the "real universe" is a simulation too of course), nor do the atmospheres exist for the same reason or in the same way. Same goes for water. Same goes for electricity. Same goes for fuel. That's a doozy really!

Fuel is not burnt in KSP, and there is no exhaust. The engines exert a push of a given amount in a given direction when they are running, to do so there must be a positive value in variable X. For every Y milliseconds the engine runs at Z throttle, it lowers the value in X by Z * whateverTheEngineLowersTheFuelByAtFullThrottleAtThis"Atmospheric"Pressure.

There is no chemical reaction, there is no expansion chamber, there is no exhaust. You can tell easily that this is the case by the fact that KSP will run on a single CPU core.

Ever tried to do a liquid flow analysis on a consumer PC? Actually simulating those particles takes a hell of a lot of resources. Far, far, far more than a single consumer CPU core can manage in real time.

KSP is a wonderful game and I love it deeply, but don't confuse a game-type "simulation" with an actual model of the underlying physical framework of the universe.

Link to comment
Share on other sites

First off, I think we're talking past each other a bit here, which I'll get into later. But for the moment, I'm afraid I have to disagree with your math above. Those calculations have nothing to do with the Oberth Effect. What you've done is simple vector addition, demonstrating the physical "leverage" that causes you to make Ap changes at Pe anyway, Oberth or not. Also, you only added vectors at the point of the burn (Ap or Pe), without taking into account the resulting velocity change at the other end of the of the burn (Pe or Ap) caused by the burn.

I'm happy to discuss this further, my understanding of the Oberth effect certainly could be flawed. I'm not a rocket scientist by training or profession, I only play one in KSP.

The math I posted is certainly simplified, as it assumes an instantaneous burn among other things. But it is a valid approximation of the Oberth effect. It really is simple vector addition. Certainly, when at the other apsis of the orbit, the speed will be very different than when the burn occured, but, and this is the important part, the total orbital energy of the craft will be the same. Orbital energy is the sum of the potential energy (a function of mass, gravitational force, and altitude) and kinetic energy (a function of mass and the square of speed). So whatever energy is added at one apsis will still be present at the other apsis, it will just have a different distribution between potential and kinetic.

Anyway, back to Oberth being merely a fuel-efficiency thing we need not worry about.....

Energy isn't something you can measure directly, but only indirectly by its effects. That's why we have so many different types of energy: kinetic energy, thermal energy, gravitational potential energy, chemical potential energy, etc., each of which indirectly measures energy by looking changes in speed, temperature, altitude, etc.

Physically, spaceflight boils down simply to just moving from 1 orbit to another. You move from high orbits to low orbits, or parking to transfer orbits, etc., but you're always in an orbit (even if you're on the ground). All orbits have velocities directly associated with them, meaning spaceflight is all about changing the ship's velocity when and as needed to move from 1 orbit into the next. Because of this, kintetic energy is what's important to rockets; they convert the chemical potential energy in their fuel into kinetic energy to give them the velocity they need for the desired orbit. Thus, delta-V is the currency of spaceflight.

Careful here, you need to remember that delta-V is not a measure of energy, it is a measure of potential velocity change. Delta-V is in units of m/s, energy is in units of Joules, which is shorthand for kg*m2/s2. Obviously, these are not interchangeable units. It is possible to consume delta-V without changing orbital energy, by burning in a direction perpendicular to prograde (normal, antinormal, radial, antiradial and all points between). The eccentricity or inclination will change, but the sum of potential and kinetic energy will not.

I'm sure you are familiar with the Tsiolkovsky rocket equation for calculating delta-V, and that the only factors used in calculating delta-V are initial mass, final mass and specific impulse. You'll note that speed does not appear in that equation, so delta-V obviously does not vary with speed, ever. So it cannot be that the Oberth effect, which is speed related, changes the amount of delta-V in a given craft.

It can be easily demonstrated that the amount of energy extractable from a given amount of delta-V varies with speed, and this is the heart of the Oberth effect. Consider the following:

Two identical 1-ton craft perform a 100m/s prograde burn consuming 0.1t of propellant, with one (F) having an initial speed of 200m/s and the other (S) having an initial speed of 100m/s. Let's calculate their change in kinetic energy:

Delta-KEF = EF-Final - EF-Initial

= 1/2mFinalvFinal2 - 1/2mInitialvInitial2

= 1/2*0.9*3002 - 1/2*1.0*2002

= 40,500 - 20,000

= 20,500 KJ

Delta-KES = ES-Final - ES-Initial

= 1/2mFinalvFinal2 - 1/2mInitialvInitial2

= 1/2*0.9*2002 - 1/2*1.0*1002

= 18,000 - 5,000

= 13,000 KJ

From this, it should be clear that the faster a craft is going, the more kinetic energy is extracted from a given amount of delta-V. This is because of the non-linear, geometric relationship between speed and kinetic energy.

Let's refine that example. Instead of two craft, make it one craft in an eccentric orbit with a speed 100m/s at apoapsis and 200m/s at periapsis. The same calculations above apply. But, you may ask, how can the energy delivered by the burn vary for the same craft in one orbit? Where does the energy come from? Doesn't that violate conservation of energy?

The answer is that our fuel contains more than just chemical energy. It contains potential energy, kinetic energy, and chemical energy. Let's look at these more closely:

Chemical energy is the most familiar, and is a fixed quantity. This is extracted to complete the burn.

Potential energy is based on altitude, and cannot be changed in our instantaneous burn, so we cannot use this energy.

Kinetic energy is based on the square of speed, which can be utilized via the Oberth effect.

At apoapsis, more of the fuel's total energy is "locked away" in potential energy, where it cannot be used to change our orbital energy. Less of it is kinetic energy, which can be used.

At periapsis, less of the energy is in potential energy where it cannot be used and more of it is kinetic energy, which can be.

So, while total energy remains the same at both apses, the energy that can be used to change orbital energy is greatest at periapsis, when the largest fraction of the fuel's total energy is kinetic.

However, it's essential to understand that the delta-V required to move from an arbitrary point along Orbit A into desired Orbit B is exclusively a function of vector math. The specifics of the ship involved have no effect at all on the delta-V required; all ships of whatever size and TWR require the same delta-V to make the same trip between the same 2 end points. You can easily see this is true in the game by switching to a new ship and plotting a maneuver for it before doing anything else to it. You will see the amount of delta-V required but the burn time will say "N.A." because the game doesn't know the TWR of the ship until you burn its engines. Thus, you don't need to know anything about the ship to determine the amount of delta-V needed for a maneuver.

But anyway, the delta-V required to do any given burn is determined solely by vector math. When creating, moving, and tweaking thrust direction gizmos on maneuver nodes, the game is doing vector math under the hood and you're just seeing the sums of the vectors.

I agree that my math is not complete, as I was trying to make an example in which all the numbers were simple and no calculus was involved.

However, it's generally good form when criticizing someone's math to provide better math to replace it. All you've provided is some vague handwaving about vector math (and claiming that vector addition is not vector math!). So I will ask you respectfully to either provide better math for my example or accept it as a reasonable approximation.

Now, given this, there's Oberth. Oberth has nothing to do with the amount of delta-V required by a burn; that's solely a vector math thing. Obeth therefore can only come into play during the burn. Oberth can't change the fixed amount of delta-V required by the burn, so the only benefit he can provide is making the burn more efficient. The ONLY way to make a burn more efficient is by allowing the ship to consume less fuel to reach its required delta-V total than the stats of its engine/fuel system would dictate on their own.

Oberth most certainly does change the amount of delta-V required, that's the whole point!

Here's the thing: delta-V doesn't change your orbit directly, it changes your speed, which affects your total orbital energy. This is an important distinction, because differing amounts of energy are contained in a m/s depending on how fast you're going (see my math example above again).

This is trivial to test in KSP. Put identical ships into circular orbits at differing altitudes (and thus differing speeds). You will agree, I'm sure that the higher orbit is a higher energy orbit (it cost more dV to get there) even though its speed is lower. Now plot an interplanetary transfer to the same target from both orbits and you'll see that the higher orbit requires more delta-V, even though it's a higher energy orbit and is attempting to transfer to the same target. The reason for this is that more of its fuel's energy is trapped as potential energy which doesn't help us, while the lower orbiting ship's fuel has more kinetic energy which does help us. (Incidentally, this also demonstrates that Oberth is alive and well in KSP)

You might be wondering how delta-V maps work if transfers cost different amounts from different orbits. The answer is to look closely at the map. It always specifies the altitude of the low orbit that is the first node away from a body. Which means the velocity is known, and thus the Oberth effect can be taken into account. You'll also notice that there are no circularization nodes between two bodies, only transfers and escape/captures. The map is assuming you will make your entire transfer burn from the low orbit to fully harness Oberth's advantage.

Assuming the engine runs full throttle the whole burn, the rate of fuel consumption is constant throughout due to the engine's fixed vacuum Isp. Therefore, the only way to reduce the amount of fuel consumed in the burn is to make the burn time shorter. But the amount of delta-V required is still the same regardless of burn time. Therefore, the only way to decrease burn time for the fixed amount of delta-V is to accelerate faster, so you "delta your V" faster. Which essentially means Oberth, if he exists in the game, is providing you with extra thrust.

The Oberth Effect wiki page says this right off the top:

The Oberth effect occurs because the propellant has more usable energy due to its kinetic energy on top of its chemical potential energy. The vehicle is able to employ this kinetic energy to generate more mechanical power

IOW, a given ship with a given engine burns X amount of fuel and gets Y amount of change in Ek purely from the combustion. Oberth then adds an extra amount Z of Ek change on top of this. Thus, the total Ek change is Y (from the fuel alone) + Z (from Oberth). The mass change in burning the fuel is the same for both so both Y and Z can only be measured as changes in velocity AKA delta-V. So there it is: Oberth gives you more delta-V per unit of fuel burned than you get from the fuel alone. Therefore, Oberth is a fuel-efficiency thing.

You are assuming that a larger change in kinetic energy equals a larger change in velocity, which is not so. Two equal changes in velocity can have differing changes in kinetic energy, depending on the initial velocity, because kinetic energy is proportional to the square of velocity. The fuel consumed in both changes is the same, though.

Sorry for this very long and wordy post, but you obviously put a lot of time and thought into yours and I felt I should do the same. If you see where I've made an error or obvious misstep, please point it out. It is entirely possible that I don't understand this as well as I think I do.

Link to comment
Share on other sites

I think you're a little confused about what exactly the Oberth effect is. You're talking about how it increases fuel efficiency, and you're technically right about that, but not in the way you explained. The Oberth effect can't magically increase the chemical energy available in the fuel, or the exhaust velocity of your engine. It doesn't do that in real life, and it doesn't do that in KSP. The reason burns higher velocity are more efficient is that your rocket is extracting the maximum total energy, chemical and kinetic, from your fuel.

I thought that's what I said :).

Additionally, saying that the Oberth effect increases the dV from a burn is incorrect, because you're wrongly equating orbital energy and dV. If my spaceship is capable of an average acceleration of 10m/s^2, and I burn my engine for 10 seconds, I will be going 100m/s faster than I was before the burn. It doesn't matter what my initial speed was, accelerating at a given rate for a given time will always result in the same change in velocity. However, the same change in velocity does not always give you the same change in orbital energy.

I never said Oberth increased, or in fact had anything to do with, the delta-V of the burn. The delta-V of the burn is the amount that vector math dictates is needed to move from your current orbit to the desired orbit. For any given burn, this is a fixed amount due to the geometry of the situation. The question is, how much fuel does the ship need to burn to change its velocity by the amount needed?

You cannot measure energy directly. You can only infer its presence from the changes it causes. In the case of kinetic energy, you measure it mostly by noting the change in velocity before and after, because in most cases, the mass change is way less significant than the square of the velocity. So if Oberth does anything at all for you, it will give you a greater change in velocity than the engine itself would have from burning a given amount of fuel. In real life, Oberth does this by sucking Ek out of the exhaust and applying it to the ship, which makes the ship go faster, which is the measurable result.

Link to comment
Share on other sites

I thought that's what I said :).

So if Oberth does anything at all for you, it will give you a greater change in velocity than the engine itself would have from burning a given amount of fuel. In real life, Oberth does this by sucking Ek out of the exhaust and applying it to the ship, which makes the ship go faster, which is the measurable result.

No, The Oberth effec does not let you increase the DeltaV available to your ship. rathe an understanding of the effect lets you minimize the deltaV requirements for a given maneuver. Check my previous post for a working example

Link to comment
Share on other sites

Fuel is not burnt in KSP' date=' and there is no exhaust.[/i'] The engines exert a push of a given amount in a given direction when they are running, to do so there must be a positive value in variable X. For every Y milliseconds the engine runs at Z throttle, it lowers the value in X by Z * whateverTheEngineLowersTheFuelByAtFullThrottleAtThis"Atmospheric"Pressure.

Not that it matters, because Oberth manifests itself anyway simply because the game understands and implements basic Newtonian physics. It does not require additional simulation.

Force = Mass * Acceleration. Engines provide the force, the ship has mass, and therefore the ship is given an acceleration.

Work = Force * Distance. Engine provides force, the ship is moving (at an ever increasing rate.) Work is being done even if the game does not explicitly calculate it.

The engine "burns" a certain amount of fuel every second. During that second, the ship will cover some finite distance (S = V*t + 0.5*a* t^2). Therefore, that amount of fuel can be equated to a certain distance traveled, as a function of velocity and acceleration.

Since engine force is constant, and fuel use per distance traveled is a function of velocity and acceleration, then the faster you are moving the more distance you travel per second. And since fuel use rate is also constant, then the faster you travel, the farther you travel per unit of fuel consumed.

Therefore! Since work is force multiplied by distance, and the faster you are going the larger the distance per unit fuel consumed, then the faster you are going the more work is being done per unit fuel consumed. And you never had to actually calculate energy or work for this to happen - it just happens.

Let's glue all of that together:

Work = delta-Ek = delta-(0.5 * Mass * Velocity^2) = Force * Distance = Force * (Velocity * Time + 0.5 * Acceleration * Time^2)

It's pretty easy to see that the change in kinetic energy (and therefore the change in velocity) is dependent upon your current velocity. That's Oberth writ large and you don't need to do any special simulation to make it work.

Oberth most certainly does change the amount of delta-V required, that's the whole point!

No, no it does not. You seem to be confusing delta-V as some sort is discrete physical quantity, rather than a convenient way to express specific energy (energy per unit mass) of the craft. If you are in an orbit at R1 and want to be in an orbit at R2, you need to change the specific energy of your craft by some amount. You effectively do that by changing your velocity (since conservation of momentum prohibits you from magically decreasing your mass without consequence) which is expressed "delta-V" - literally "the change in velocity." Oberth can't affect the amount you need to change your specific energy, only how much fuel you need to spend to make it happen.

=Smidge=

Link to comment
Share on other sites

I never said Oberth increased, or in fact had anything to do with, the delta-V of the burn. The delta-V of the burn is the amount that vector math dictates is needed to move from your current orbit to the desired orbit. For any given burn, this is a fixed amount due to the geometry of the situation. The question is, how much fuel does the ship need to burn to change its velocity by the amount needed?

You cannot measure energy directly. You can only infer its presence from the changes it causes. In the case of kinetic energy, you measure it mostly by noting the change in velocity before and after, because in most cases, the mass change is way less significant than the square of the velocity. So if Oberth does anything at all for you, it will give you a greater change in velocity than the engine itself would have from burning a given amount of fuel. In real life, Oberth does this by sucking Ek out of the exhaust and applying it to the ship, which makes the ship go faster, which is the measurable result.

We need a /kerbalheaddesk ... the bolded parts are directly contradictory.

The Oberth effect does not increase delta-v in the slightest, but alters the NEEDED delta-v by altering the energy state of the orbit by a greater amount when at periapsis (and therefore at a higher speed relative to the body being orbited). You know what, if you want a rehash of the math, look at somebody else's post...

Link to comment
Share on other sites

This thread is quite old. Please consider starting a new thread rather than reviving this one.

Join the conversation

You can post now and register later. If you have an account, sign in now to post with your account.
Note: Your post will require moderator approval before it will be visible.

Guest
Reply to this topic...

×   Pasted as rich text.   Paste as plain text instead

  Only 75 emoji are allowed.

×   Your link has been automatically embedded.   Display as a link instead

×   Your previous content has been restored.   Clear editor

×   You cannot paste images directly. Upload or insert images from URL.

 Share

×
×
  • Create New...