oberth effect

[Bobnova]

That assumes that Oberth is the vector math done earlier, rather than a change in the fuel's energy due to velocity. Two different effects.

[Red Iron Crown]

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)

I must report that this test doesn't work as I expected. I put three identical probes in circular orbits at 100, 1,000 and 10,000km. Set up the transfers and here are the results:

[table=width: 250]

[tr]

[td]Altitude[/td]

[td]Speed[/td]

[td]Transfer dV[/td]

[/tr]

[tr]

[td]100km[/td]

[td]2,248m/s[/td]

[td]1,035m/s[/td]

[/tr]

[tr]

[td]1,000km[/td]

[td]1,485m/s[/td]

[td]767m/s[/td]

[/tr]

[tr]

[td]10,000km[/td]

[td]571m/s[/td]

[td]578m/s[/td]

[/tr]

[/table]

Obviously my understanding is flawed. The higher total energy of the higher orbits is pretty clearly dominating the Oberth effect advantage of the lower orbits. I guess to see the effect properly, they shouldn't be in circular orbits but eccentric ones with the same orbital energy. I think this may be beyond my piloting skills as the apoapsis would have to be at the correct angle to prograde to make the transfer burn. Back to calculating...

[SSSPutnik]

I just made this, (The Dummies Guide is for humour, I had a bloody hard time understanding it too!)

I hope it's correct and simple, otherwise I might cry.

[maccollo]

A higher circular parking orbit will always require less delta V to transfer from than a lower orbit.

However, like you said, you have to get there first so you need the add the delta V it took to achieve those orbits.

To go from 100 km at Kerbin to 1000 km requires 730 m/s. So the total Delta V for doing that for a transfer is 1500 m/s.

To go to the 1000 km orbit requires 1200 m/s, so the total for that transfer adds up to 1878 m/s.

So that tell you that it's more efficient to burn straight for the transfer from a lower orbit that than it is to burn into a higher orbit and then transfer from there.

But that is not a fair comparison, because the starting energies are different. To really see what the oberth effect can do for us in a transfer we need to compare two different orbits with equal energy.

The specific orbital energy is simply µ/2a. Since the two orbits we want to compare both orbit Kerbin the semi major axis is all we need to compare the relative energies.

the semi major axis in the 10 000 km orbit around kerbin is 10 600.

For an eliptical orbit with a periapsis of 100 km to have the same energy the apoapsis needs to be 19 900 km.

The 10 000 km circular orbit required 578 m/s for the transfer.

The 100 km by 19 900 km orbit only requires 158 m/s.

Edited February 26, 2014 by maccollo

[Red Iron Crown]

OK, I did a much simpler test that demonstrates the effect a bit better. I put a probe into a 100x1000km orbit. I then set maneuver nodes to raise the apoapsis to various altitudes (which is equivalent to a transfer burn to a target at that altitude). Here are the results:

[table=width: 300]

[tr]

[td][/td]

[td][/td]

[td]Target Ap[/td]

[td][/td]

[/tr]

[tr]

[td]Burn at[/td]

[td]10,000km[/td]

[td]100,000km[/td]

[td]1,000,000km[/td]

[/tr]

[tr]

[td]Periapsis[/td]

[td]427[/td]

[td]513[/td]

[td]526[/td]

[/tr]

[tr]

[td]Apoapsis[/td]

[td]799[/td]

[td]925[/td]

[td]939[/td]

[/tr]

[tr]

[td]Ratio[/td]

[td]0.53[/td]

[td]0.55[/td]

[td]0.56[/td]

[/tr]

[/table]

I think I'm seeing Oberth at work here. The savings are decreasing as the target grows more distant, too.

It's amusing that this probe has no science parts but is actually doing real science.

[SSSPutnik]

It's amusing that this probe has no science parts but is actually doing real science.

Stick a Goo Container on so it can experience the glory of Oberth.

[Red Iron Crown]

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."

Delta V is not a measure of specific energy. Simple dimensional analysis shows it: Delta V is in m/s, specific energy is in J/kg, which is the same as m²/s².

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.

How you can write that and not see that Oberth reduces dV requirements is beyond me. Changing the amount of fuel spent is changing the amount of dV spent.

[Red Iron Crown]

Stick a Goo Container on so it can experience the glory of Oberth.

Ha! Changing the probe name from "Oberth 1" to "Oberth's Glory".

[Red Iron Crown]

Thinking about the Wiki quote, this occurs to me:

When doing a burn, the thrust, mass and therefore acceleration are the same whether the burn is done at high speed. The mechanical power (in Watts, or J/s) is higher at higher velocity due to Oberth. So for any duration burn, the change in delta-V is the same but the amount of energy gained is different the depending on speed at which the burn is made.

[SSSPutnik]

Thinking about the Wiki quote, this occurs to me:

When doing a burn, the thrust, mass and therefore acceleration are the same whether the burn is done at high speed. The mechanical power (in Watts, or J/s) is higher at higher velocity due to Oberth. So for any duration burn, the change in delta-V is the same but the amount of energy gained is different the depending on speed at which the burn is made.

Yes correct, it's completely a kinetic energy thing. (And most of kE = velocity) I dunno about acceleration though, since you are gaining extra speed I would have thought your acceleration was higher...

You do get less of a boost using a long low TWR burn, because it means a lot of the burn isn't happening at peak velocity. (by the time your itty bitty ion engine has finished burning your periapsis probably happened an hour ago).

You get the most Oberth bang for buck if you instantly did the burn at your peak speed. (It might turn your crew into jelly though).

When your ship starts moving from the ground, a lot of that thrust is investing kE in the ship (and it's fuel).

Edited February 26, 2014 by SSSPutnik

[MockKnizzle]

Yes correct, it's completely a kinetic energy thing. (And most of kE = velocity) I dunno about acceleration though, since you are gaining extra speed I would have thought your acceleration was higher...

You're gaining extra orbital energy, not extra speed. The change in speed (dV) is the same, but that same dV changes your orbit more when you're going faster.

[SSSPutnik]

No... The speed does change. (If you change your orbit, the speeds are going to have to change). You are spending a set amount of deltaV but you get a bonus velocity change from that expenditure with Oberth.

[Jouni]

I'm starting to believe that the only reason anyone ever uses the term "Oberth effect" is to confuse people.

[SSSPutnik]

It's a ripper eh!

[MockKnizzle]

No... The speed does change. (If you change your orbit, the speeds are going to have to change). You are spending a set amount of deltaV but you get a bonus velocity change from that expenditure with Oberth.

Obviously your speed changes by the amount of dV you spend, but you do not get any "bonus speed" out of your burn. If you burn 100m/s worth of dV when you're going fast near a planet, you don't magically get an extra 5 or 10m/s out of your fuel.

You do, however, get bonus energy out of your fuel that goes towards increasing your ship's orbital energy, more so than if you burned that fuel when you ere going slower.

[SSSPutnik]

Let's do a silly car example. Yes you can get Oberth in a car if you throw something out!

The car is travelling at 25ms.

You have a gun, it fires 0.005 kilogram bullets at 400ms.

In the car, sitting in the magazine, the bullet has a kE = 1.56J = 0.5 x 0.005kg x 25ms^2

Potential energy

pE1=400=(0.5 x 0.005 x 400^2), you would get this if you fired whilst not moving in any direction.

pEtotal= 401.56J=(400+1.56)

If I lean out the car window and fire forward, the bullet travels at 425ms. (400ms bullet + 25ms car).

The kE of the bullet is 0.5 x 0.005 x 425^2 = 451.56J

If I turn around and fire backwards, the bullet travels at 375ms. (400ms bullet - 25ms car).

The kE of the bullet is 0.5 x 0.005 x 375^2 = 351.56J

So...

Firing moving forward we get 50J more than the potential energy we had.

Firing backwards we got 50J less than the potential energy we had.

(Firing straight up we'd get 0 difference).

Energy cannot be destroyed. So where does the difference go? Back into the car.

Roughly, firing backwards, the car gains an extra 50J of energy, on top of the thrust/recoil of the gun.

Firing forwards, the car loses that 50J, slowing even more than the thrust of the gun.

Hope that helps....

Edited February 26, 2014 by SSSPutnik

[SSSPutnik]

Obviously your speed changes by the amount of dV you spend, but you do not get any "bonus speed" out of your burn. If you burn 100m/s worth of dV when you're going fast near a planet, you don't magically get an extra 5 or 10m/s out of your fuel.

You do, however, get bonus energy out of your fuel that goes towards increasing your ship's orbital energy, more so than if you burned that fuel when you ere going slower.

What's "orbital energy" ? Speed....

^Oops apologies I am wrong here. It's a combination of potential and kinetic energy.

Thanks MockKnizzle and maccollo.

From the Rho Project page....

"Given that you are going to travel a parabolic orbit around the Sun that has an escape velocity of 200 km/s at periapsis, you have an initial velocity of 3.2 km/s, and you wish to exit the Oberth Maneuver with a final velocity of 50 km/s, calculate the required ÃŽâ€v burn at periapsis.

Vesc = 200 km/s = 200,000 m/s. Vh = 3.2 km/s = 3200 m/s.Vf = 50 km/s = 50,000 m/s.

ÃŽâ€v = sqrt(Vf2 + Vesc2) - sqrt(Vh2 + Vesc2)

ÃŽâ€v = sqrt(50,0002 + 200,0002) - sqrt(32002 + 200,0002)

ÃŽâ€v = sqrt(2,500,000,000 + 40,000,000,000) - sqrt(10,240,000 + 40,000,000,000)

ÃŽâ€v = sqrt(42,500,000,000) - sqrt(40,010,240,000)

ÃŽâ€v = 206,000 - 200,000

ÃŽâ€v = 6,000 m/s = 6 km/s

So by burning 6 km/s of ÃŽâ€v, you get an actual ÃŽâ€v increase of 46.8 km/s. That's 40.8 km/s for free. Sweet!"

See... You do get bonus speed... The increased energy is increased kinetic energy...

Going back to my car... Say I had some crazy magic car that could do 1000ms.

The potential energy of the bullet becomes 2500+400 = 2900J

Firing forward = 4900J

Firing backwards = 900J

I get an extra 2000J of Oberth kinetic energy vs measly 50J in my original example.

(Enough to push a 1 ton car an extra 2ms!)

This is why Oberth effect is greatest at the highest speed, (periapsis).

Edited February 26, 2014 by SSSPutnik

[maccollo]

What's "orbital energy" ? Speed....

kinetic energy+potential energy

[SSSPutnik]

kinetic energy+potential energy

It manifests as increased speed. See above...

[maccollo]

It manifests as increased speed. See above...

Yes, potential energy gets converted into kinetic energy as you fall towards periapsis, but saying orbital energy is just speed is an incorrect description.

Edited February 26, 2014 by maccollo

[SSSPutnik]

Is it? How? They should be interchangeable... Not having a go, I need more enlightenment.

Ahh OK, sorry I get you, yes, your potential energy vs your actual speed varies depending on orbital position.

(Apoapsis=slow and high potential energy, Periapsis fast and low potential).

Edited February 26, 2014 by SSSPutnik

[Jouni]

From the Rho Project page....

"Given that you are going to travel a parabolic orbit around the Sun that has an escape velocity of 200 km/s at periapsis, you have an initial velocity of 3.2 km/s, and you wish to exit the Oberth Maneuver with a final velocity of 50 km/s, calculate the required ÃŽâ€v burn at periapsis.

[..]

So by burning 6 km/s of ÃŽâ€v, you get an actual ÃŽâ€v increase of 46.8 km/s. That's 40.8 km/s for free. Sweet!"

See... You do get bonus speed... The increased energy is increased kinetic energy...

You get bonus speed, but not immediately after the burn. Before the burn, you are at periapsis, moving approximately at the escape velocity of 200 km/s, because the extra energy from the initial velocity of 3.2 km/s is negligible. After the 6 km/s burn, you are moving at 206 km/s, so you didn't get any bonus speed at this point. On the other hand, your kinetic energy increased so much that you are still moving at 50 km/s (instead of the 3.2 km/s you would be moving without the burn), when you are so far away from the Sun that your potential energy is essentially as high as it's ever going to be.

In KSP physics, you would be entering Sun's SoI at 3.2 km/s, moving at 200 km/s at periapsis, burning to increase that to 206 km/s, and leaving Sun's SoI at 50 km/s.

[SSSPutnik]

In my car example the car would gain speed as you can't increase the potential energy.

[MockKnizzle]

Your car example isn't actually an illustration of the Oberth effect, it's just a conservation of momentum problem, and you don't get any extra speed when traveling faster. I'm gonna tweak your numbers a bit so I don't have to type as many things into my calculator.

Say we have a car of mass 1000kg and a bullet (cannonball, really) of mass 1kg. The car and bullet are traveling to the right (positive direction) at a speed of 25m/s. The bullet is fired backwards from the car with a relative muzzle velocity of 400m/s. How fast is the car traveling after the shot?

Since the momentum of the car-bullet system is conserved, the math is as follows:

p_total = p_car + p_bullet

m_total * V_total = m_car * V_car + m_bullet * V_bullet

(1000kg + 1kg) * 25m/s = 1000kg * V_car + 1kg * (25m/s - 400m/s)

Solving for V_car we get 25.4m/s after the shot. Since we started with our car traveling at 25m/s, shooting the bullet has given us a dV of 0.4m/s.

Now, say our car is suddenly traveling at 1000m/s (nearly mach 3 in a land vehicle, quite impressive) and we do the same thing:

p_total = p_car + p_bullet

m_total * V_total = m_car * V_car + m_bullet * V_bullet

(1000kg + 1kg) * 1000m/s = 1000kg * V_car + 1kg * (1000m/s - 400m/s)

This time, we get V_car = 1000.4m/s.

See? Throwing the same reaction mass out at the same exhaust velocity has given us the same instantaneous dV, regardless of our initial speed.

The Oberth effect doesn't have to do with the instantaneous dV you get from a burn. It has to do with how much your orbital energy (and thus the SMA of your orbit, which is directly related to your orbital energy) changes when you apply the same dV at higher initial speeds.

Edited February 26, 2014 by MockKnizzle

[Jouni]

Let's look at the car example. I'm using variables instead of concrete numbers, because the changes in car's velocity would be negligible.

Case 1: We have a car of mass m₁ carrying a bullet of mass m₂. Initially the car is stationary, so there is no kinetic energy at this point. We fire the bullet at velocity v₂, causing the car to start moving at velocity v₁. Because of the conservation of momentum, m₁v₁ = -m₂v₂. The total kinetic energy of the system after firing is 0.5m₁v₁² + 0.5m₂v₂², which is entirely attributable to whatever mechanism we used for firing the bullet.

Case 2: The situation is otherwise the same, but the car and the bullet are initially moving at velocity v₀. Hence the initial kinetic energy of the system is 0.5(m₁+m₂)v₀². After firing the bullet, the velocity of the car is v₀+v₁, while the bullet is moving at velocity v₀+v₂. The total kinetic energy of the system is now

0.5m₁(v₀+v₁)² + 0.5m₂(v₀+v₂)² = 0.5(m₁+m₂)v₀² + m₁v₀v₁ + m₂v₀v₂ + 0.5m₁v₁² + 0.5m₂v₂².

The first term is equal to our initial kinetic energy, while the last two terms are equal to the energy gained from firing the bullet. This leaves us with the second and the third terms. Because of the conservation of momentum, m₁v₁ = -m₂v₂, so we can represent these terms as

m₁v₀v₁ + m₂v₀v₂ = v₀(m₁v₁-m₁v₁) = 0.

Regardless of the initial velocity v₀, the velocity of the car changes by v₁, the velocity of the bullet by v₂, and the kinetic energy of the system increases by 0.5m₁v₁² + 0.5m₂v₂². If v₀ is nonzero, the one of the car and the bullet that gains velocity in the direction of v₀ gains more kinetic energy than in the stationary case. The other object gains less kinetic energy than in the stationary case, or possibly even loses it. This is the Oberth effect.

This obviously happened in an one-dimensional world, because further dimensions exist just to confuse people.