About long burn time efficiency with low TWR

[Spricigo]

 

Oberth Effect...

I'm not moving, my velocity is 0 and my Kinetic energy is 0. If now my velocity change by dV my Kinetic energy will be 1/2*m*dV², That is a change in energy of 1/2*m*dV². 

A friend of mine is now moving, his velocity is V and his Kinetic energy is of 1/2*m*V²   (we have equal mass). If his velocity change by dV his Kinetic energy will be 1/2*m*(V+dV)² or 1/2*m*(V²+2.V.dV+dV²) , that is a change of 1/2*m*(2.V.dV+dV²) or  m*.V*dV    +   1/2*m*dV²

We both changed or Knetic Energy by 1/2*m*dV² . But he got extra m*.V*dV   from the Oberth Effect.

It become easier to see with a diagram:

Notice that is nothing new,  just a mathematical/graphical way to tell what @Snark, @OhioBob, @Starman4308, @Kosmo-not, @foamyesque and @OHara already said in plain words. Also seems to be what the OP was talking about, given what he marked as best/correct answer.

Well, I think those are smart guys, however a claim is as good ass the evidence behind it. Fortunately there is that enough evidence for their claim, and I make mine those words:

On 13/01/2018 at 3:54 AM, Geschosskopf said:

Go see the oft-cited wiki page if you don't believe me  

yes, lets see it:

Quote

In astronautics, a powered flyby, or Oberth maneuver, is a maneuver in which a spacecraft falls into a gravitational well, and then accelerates when its fall reaches maximum speed.^[1][2] The resulting maneuver is a more efficient way to gain kinetic energy than applying the same impulse outside of a gravitational well. 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. In practical terms, this means that the most energy-efficient method for a spacecraft to burn its engine is at the lowest possible orbital periapse, when its orbital velocity (and so, its kinetic energy) is greatest.^[1] In some cases, it is even worth spending fuel on slowing the spacecraft into a gravity well to take advantage of the efficiencies of the Oberth effect.^[1] The maneuver and effect are named after Hermann Oberth, the Austro-Hungarian-born German physicist and a founder of modern rocketry, who first described them in 1927.^[3]

The Oberth effect is strongest at a point in orbit known as the periapsis, where the gravitational potential is lowest, and the speed is highest. This is because firing a rocket engine at high speed causes a greater change in kinetic energy than when fired at lower speed. Because the vehicle remains near periapse only for a short time, for the Oberth maneuver to be most effective, the vehicle must be able to generate as much impulse as possible in the shortest possible time. Thus, the Oberth maneuver is much more useful for high-thrust rocket engines like liquid-propellant rockets, and less useful for low-thrust reaction engines such as ion drives, which take a long time to gain speed. The Oberth effect also can be used to understand the behavior of multi-stage rockets: the upper stage can generate much more usable kinetic energy than the total chemical energy of the propellants it carries.^[3]

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.

and again

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 is strongest at a point in orbit known as the periapsis, where the gravitational potential is lowest, and the speed is highest. This is because firing a rocket engine at high speed causes a greater change in kinetic energy than when fired at lower speed.

...

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.

 

 

[Geschosskopf]

4 hours ago, OhioBob said:

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

True.  My bad.  Still, you can't change velocity without acceleration, so ΔV is an indicator of your ability to accelerate, especially when the parameters of thrust, mass, and rate of mass change are known.

Anyway, you all win.  There's no point in belaboring this point any more.

[Wcmille]

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. 

[Laie]

15 hours ago, Wcmille said:

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.

 

this. very much this.

[CanOmer]

Mun was on the way of my planned route. So I made the burn an hour earlier. Nuclear engines made 8 min (2400 Delta-V) burn from Kerbin orbit to Moho. Insertion burn to enter Moho orbit was supposed to be around 2800 Delta-V according to Transfer Window Planner. But after the Kerbin departure burn, now maneuver node says it requires 4550 Delta-V. Ho can long burn time and an hour early burn create such difference?

Edited January 17, 2018 by CanOmer

[OHara]

8 hours ago, CanOmer said:

I made the burn an hour earlier. Nuclear engines made 8 min (2400 Delta-V) burn from Kerbin orbit to Moho. ... Ho can long burn time and an hour early burn create such difference?

That is a very different question to the first one you asked.  If you want more than a brief answer, I suggest you start a new question thread, and link to it here, or else this thread will be confusing for people reading it later --  especially if the answer-voting system re-orders the questions and answers.

An hour early would not make much difference, if it is exactly two (or exactly three...) orbits around Kerbin.  It looks like you were a little later than exactly two orbits early, because your exit from Kerbin looks to be angled in toward the sun, so you reach Moho 10 days early compared to the planned 90 days, so you intersect Moho's orbit at and angle, compared to the roughly Hohmann transfer that is usually similar to the best transfer.  If you reached Moho later, 60° counterclockwise in its orbit where the orbits are tangent to each other, you would be travelling parallel to Moho and the relative velocity much slower, so the burn much less.

The 8-minute burn might have contributed to the bad angle out of Kerbin, especially if you followed the prograde marker. Often you need to make corrections at the end of a long ejection burn, to meet your target and meet it at the desired place in its orbit.

[Tyko]

All the fantastic math advice aside...I've found empirically that a single ejection burn of of 4-5 minutes can be accomplished at 100-200K altitude without an appreciable loss of DV. Anything longer than that and I break it up into a series of 4-5 minute burns.

I'm measuring my loss by comparing my actual DV expended versus the planned DV expenditure from TWP. In my comparisons it's within a few percentage points.

Not as exact as the math above, but a useful rule of thumb.

[Norcalplanner]

And the more delta V the ship has on tap and the longer the burns, the more importance is attached (for me anyway) to the accuracy of the burn, which translates to a higher departure orbit. A Karborundum-powered ship making a 20 km/s burn does much better when departing from a 2,000 km orbit.