Oberth Math

[KerikBalm]

So, anticipating stock refueling, I've been planning on sending craft to minmus, where they would refuel, and then go on an interplanetary voyage...

I figured it had to take less dV to depart from the edge of Kerbin's SOI, than from LKO.

Sure, without refeuling, I figured its more efficient to do 1 burn from LKO - than to first burn to a high orbit, then do a 2nd transfer burn

But I figured the 2nd transfer burn would surely be smaller than the transfer burn needed from LKO.

But using this:

http://alexmoon.github.io/ksp/

I see its quite different...

Minus orbits at 47,000 km

From a 100km LKO orbit to jool needs a dV of 1986 m/s

From a 47,000km orbit (ie, where minmus is), one needs 2584 m/s!!!

This doesn't make sense to me...

This should also mean that a burn from heliocentric orbit to Jool will take more dV than a burn from LKO.

If:

A = burn to heliocentric orbit/a orbit to the edge of kerbin's SOI

B = the transfer burn from High SOI/heliocentric orbit

C = The transfer burn from LKO

I would expect that A+B > C, but that B < C, and A<C

If I refuel after A, then my craft should be able get by with lower dV capabilties... but this seems to be wrong.

But then I also have to consider some other things... like the oberth effect from orbiting Minmus, doing a bi-elliptic transfer (from minmus, lower PE to just above kerbin, then burn from there, but then the transfer window timing is even more complicated), departing from Mun instead... and so on.

But... something just seems intuitively wrong here... I really expect B < C...

Can anyone help me with the math to show why this is? Supply actual numbers, rather than just "Oberth effect, it just works out that way" (which I've already concluded)

So then I revised my plan: Going outward? fuel at Duna.... the math works out there to save (1986-1344) = 642 m/s of dV

Going Inward? Stop by Eve's system, fuel at Gilly... bit Gilly has miniscule gravity for an oberth effect, and is in a high orbit... does it work?

Kerbin @100km -> moho requires 5031 m/s, but from gilly's orbit, 3785 m/s is needed... a large difference... goog

Then I try different orbits from Eve -> Moho... the higher orbits always require less dV, as I would expect from my initial intuition...

Next I try the same thing, Eve -> Jool

From a 150km orbit of eve, it requires 2,508 m/s

From a 1,000km orbit of eve, it requires 2,480

From a 85,000 orbit of eve (ie, right on the SOI), it requires 3,211 m/s!?

There seems to be an "optimal" orbital height to depart from (assuming you can refuel prior to departure), on Eve, this seems to be about 600 km...

How does one calculate this (though the savings over just above the atmosphere are so small, I wouldn't bother moving to such an orbit, and then topping off)

This situation will reverse if one uses a bi-elliptic transfer correct? - ie starting from a higher initial orbit will require less dV to reach the target orbit (in another SOI).

Edited April 22, 2015 by KerikBalm

[el_coyoto]

Not sure about the way the optimum departure altitude is computed, but this thread (among others) deals with the subject.

I especially like the graphic made by Starstrider42 :

Cheers!

[OhioBob]

The solar orbit that one requires to reach a particular planet at some particular launch window is largely defined by the hyperbolic excess velocity of the escape trajectory, i.e. the velocity leftover after the spacecraft reaches an infinite (hypothetically) distance from Kerbin. Excess hyperbolic velocity is given by the equation,

V_{∞}² = V_bo² - V_esc²

where V_{∞} is hyperbolic excess velocity, V_bo is the burnout velocity, and V_esc in the escape velocity.

In your example we have a ÃŽâ€v of 1986 m/s from a 100 km orbit. At 100 km orbital velocity is,

V_orb = SQRT(GM/r) = SQRT(3.5316E+12 / 700000) = 2246.1 m/s

And escape velocity is,

V_esc = SQRT(2GM/r) = SQRT(2 * 3.5316E+12 / 700000) = 3176.5 m/s

The burnout velocity is the orbital velocity plus the ÃŽâ€v,

V_bo = V_orb + ÃŽâ€v = 2246.1 + 1986 = 4232.1 m/s

Therefore V_{∞} equals,

V_{∞} = SQRT(V_bo² - V_esc²) = SQRT(4231.1² - 3176.5²) = 2795.0 m/s

If we now move out to an orbit of 47,000 km, V_{∞} remains (approximately) the same; only V_orb and V_esc change. We now have,

V_orb = SQRT(3.5316E+12 / 47000000) = 274.1 m/s

V_esc = SQRT(2 * 3.5316E+12 / 47000000) = 387.7 m/s

The burnout velocity is,

V_bo = SQRT(V_{∞}² + V_esc²) = SQRT(2795.0² + 387.7²) = 2821.8 m/s

And the required ÃŽâ€v is,

ÃŽâ€v = V_bo - V_orb = 2821.8 - 274.1 = 2547.7 m/s

As you can see, the above isn't exactly the same as the number that you obtained, but it clearly confirms that it does indeed cost more ÃŽâ€v to launch from the vicinity of Minmus than it does from low Kerbin orbit. The burnout velocity out near Minmus is considerably less but, since you're starting orbital velocity is so much lower, the ÃŽâ€v is more.

Edited April 20, 2015 by OhioBob

[Eric S]

OhioBob's math should answer your why, but there's other things to answer.

Yes, boosting directly from a Minmus-altitude orbit to a transfer orbit takes more delta-v than boosting directly from LKO. On the other hand, if you're already in Minmus orbit, departing Minmus with a periapsis just above Kerbin's atmosphere and doing the transfer burn at periapsis rather than circularizing winds up taking less delta-v than either of the other two options (from that point, the fuel to burn up to Minmus then circularize would make the LKO plan cheaper).

However, this is not for the faint of heart, the timing on that maneuver can be tricky, as where Minmus is in its orbit controls what ejection angle your periapsis is at. Transfer windows are wide enough that you can easily hit some part of the window.

[OhioBob]

Yes, boosting directly from a Minmus-altitude orbit to a transfer orbit takes more delta-v than boosting directly from LKO. On the other hand, if you're already in Minmus orbit, departing Minmus with a periapsis just above Kerbin's atmosphere and doing the transfer burn at periapsis rather than circularizing winds up taking less delta-v than either of the other two options (from that point, the fuel to burn up to Minmus then circularize would make the LKO plan cheaper).

That's an excellent point. It should take only about 160-170 m/s to escape Minmus and establish an orbit with a periapsis close to Kerbin. Let's say we end up in a 46,400 x 100 km orbit; when we reach periapsis we'll be traveling 3153 m/s. However, the burnout velocity required to produce our transfer orbit to Jool is no different than we calculated before. At an altitude of 100 km we still require V_bo equal to 4232 m/s. Therefore, the ÃŽâ€v of our ejection burn at periapsis is only 1079 m/s.

[LethalDose]

Scott Manley did a great job demonstrating

into an interplanetary transfer exploiting the Oberth effect. This is the video that really clarified the Oberth effect to me.

OhioBob demonstrated the math very well above for this specific case. Another way to summarize and generalize the "Oberth maths" (and how I think about it) is that that escaping a planet's gravity is about specific orbital energy ("E", and otherwise labeled simply "orbital energy" from here), which is calculated using the equation:

E = 1/2 V² - mu/r

V is orbital velocity

mu is the gravimetric constant

r is the orbital radiuys

It's basically the difference between the kinetic and potential energy with the mass term dropped (don't worry, it's cool, and I can show you why it doesn't matter if you want. It doesn't make a difference for this application). For any given orbit, E is constant at all points (all r) on the orbit. So long as E < 0, you will have a "closed" orbit (you're captured) and when E =>0, you will escape the body you're orbiting. Put another way, you can escape a planet when your kinetic energy exceeds the gravitational potential energy the planet is exerting on it.

So you want to increase your kinetic energy to escape, and your KE is determined by velocity squared. You increase your KE and velocity by spending dV, and the change in E created by spending a fixed amount of dV at a given altitude is calculated:

dE = 1/2 V² - 1/2 (V + dV)² = 1/2 V² - 1/2 (V² + 2 V*dV + dV²) = V*dV +1/2 dV²

Since V appears in the expression of dE (the red term in the simplification), the amount of E changes when dV is spent is dependent on the vessel's velocity. Spefically, higher V leads to greater changes in E, so you want to be travelling as fast as possible when you spend the dV to escape.

So, since you tend have lower orbital velocities very high up (as high as Minmus) and you tend to have higher orbital velocities when orbiting at low altitudes, it should be clear that you want to spend dV at low altitudes/high speeds to maximize the effect of that dV.

Hope that helps.

Edited April 21, 2015 by LethalDose

[OhioBob]

Nice explanation, LethalDose.

E = 1/2 V² - mu/r

It is from this same equation that we derive hyperbolic excess velocity. (Note that "mu" is the Greek letter, μ, where μ = GM.)

If r₁ and v₁ are the distance and velocity of the spacecraft when it is near the planet, and r₂ and v₂ are the distance and velocity of the spacecraft when it is far from the planet, then, since energy is constant, we have

v₁²/2 - μ/r₁ = v₂²/2 - μ/r₂

To obtain the hyperbolic excess velocity we want r₂ to be some distance far from the planet. As r₂ approaches infinity, -μ/r₂ goes to zero, therefore we have

v₁²/2 - μ/r₁ = v₂²/2

Multiplying by 2 we get

v₁² - 2μ/r₁ = v₂²

Escape velocity occurs when we give the spacecraft just enough kinetic energy to overcome the negative gravitational potential energy, that is

V_esc² = 2μ/r

Therefore we can substitute as follows

v₁² - V_esc² = v₂²

Since v₁ is the velocity near the planet we can call this the burnout velocity. And since v₂ is the velocity at some infinite distance, it represents the hyperbolic excess velocity. Thus the final equation is

V_bo² - V_esc² = V_{∞}²

Edited April 21, 2015 by OhioBob

[LethalDose]

Can I ask for a clarification, OhioBob?

When you refer to "hyperbolic excess velocity", is that effectively the velocity of the vessel when you leave the SoI? I understand it should be the velocity when the body is infinitely far from the parent and is no longer being affected by the parent's gravity, but with the patched conics, the parent no longer exerts gravity after leaving the SoI.

If this is true, the "hyperbolic excess velocity" is essentially the target velocity to transfer from one body to another, right?

[impyre]

One thing I haven't seen considered here is TWR differences. While burning from that high up would take more fuel than if done from lower down, the much weaker gravity means a much smaller engine can push much more fuel. Since the fuel-mass to ship-mass ratio is improved you have higher starting dV. I'm not a math wizard, so I can't be sure if it's even possible that this can offset the costs of an inefficient burn (let alone save anything), but I thought it was an interesting idea (and sortof on topic). Any thoughts?

[OhioBob]

Can I ask for a clarification, OhioBob?

When you refer to "hyperbolic excess velocity", is that effectively the velocity of the vessel when you leave the SoI? I understand it should be the velocity when the body is infinitely far from the parent and is no longer being affected by the parent's gravity, but with the patched conics, the parent no longer exerts gravity after leaving the SoI.

If this is true, the "hyperbolic excess velocity" is essentially the target velocity to transfer from one body to another, right?

Hyperbolic excess velocity, as I explained it, it the theoretical velocity that a spacecraft would retain at an infinite distance (i.e. the real life definition). As you correctly describe, this isn't completely relevant to KSP because of the use of patched conics. Likewise, the mathematically derived definition of escape velocity isn't applicable to KSP either. To "escape" in KSP we just have to raise the apoapsis of the orbit until it touches the sphere of influence, thus it's possible to escape with what is technically an elliptical orbit. In KSP, the residual velocity that a spacecraft retains after crossing the SOI is greater than the calculated value of V_{∞}. The "effective V_{∞}" is simply the spacecraft's velocity relative to Kerbin as it crosses the SOI. Similarly, "effective V_esc" is the velocity needed to raise the apoapsis to exactly the radius of the SOI. The relative velocity that we require at the SOI is the difference between the orbital velocities of Kerbin and our planned interplanetary transfer orbit.

- - - Updated - - -

One thing I haven't seen considered here is TWR differences. While burning from that high up would take more fuel than if done from lower down, the much weaker gravity means a much smaller engine can push much more fuel. Since the fuel-mass to ship-mass ratio is improved you have higher starting dV. I'm not a math wizard, so I can't be sure if it's even possible that this can offset the costs of an inefficient burn (let alone save anything), but I thought it was an interesting idea (and sortof on topic). Any thoughts?

One advantage of performing the burn from a high orbit is that the spacecraft will traverse a much smaller sweep angle in the same unit of time. This means longer burns are possible from a high orbit with fewer detrimental effects, such as loss of ejection accuracy. It's therefore possible to use a smaller engine and lower TWR from high orbit.

[GoSlash27]

The short version:

It takes kinetic energy to get to Jool, not potential energy. Being at Minmus' altitude gives you a lot of potential energy, but very little kinetic energy.

Kinetic energy in this case is all about velocity, and you're going faster in LKO than you are at Minmus' altitude, so that's velocity you don't have to make up for by burning fuel.

As others have pointed out, you can save even more DV by leaving from a highly elliptical orbit with a low periapsis over Kerbin, which will give you even more velocity when you do your transfer burn.

HTHs,

-Slashy

[LethalDose]

Hyperbolic excess velocity, as I explained it, it the theoretical velocity that a spacecraft would retain at an infinite distance (i.e. the real life definition). As you correctly describe, this isn't completely relevant to KSP because of the use of patched conics. Likewise, the mathematically derived definition of escape velocity isn't applicable to KSP either. To "escape" in KSP we just have to raise the apoapsis of the orbit until it touches the sphere of influence, thus it's possible to escape with what is technically an elliptical orbit. In KSP, the residual velocity that a spacecraft retains after crossing the SOI is greater than the calculated value of V_{∞}. The "effective V_{∞}" is simply the spacecraft's velocity relative to Kerbin as it crosses the SOI. Similarly, "effective V_esc" is the velocity needed to raise the apoapsis to exactly the radius of the SOI. The relative velocity that we require at the SOI is the difference between the orbital velocities of Kerbin and our planned interplanetary transfer orbit.

While the escape velocities are calculated slightly differently in the game due to patched conics, the differences are pretty small. For example, the "proper" escape velocity from a 70 km orbit around Kerbin is 3246.9 m/s (dV = 951.0 m/s) where the "effective" escape velocity is 3234.1 m/s (dV = 938.1 m/s). That's a 13 m/s difference in nearly 1 km/s burn, just above a 1% difference.

Now since these escape velocities are practically interchangeable, V_inf is a reasonable approximation of the relative velocity to the parent at the SoI change, and we can use your presented equations to answer some of the OP's questions that have kind of gotten ignored, specifcally this passage:

A = burn to heliocentric orbit/a orbit to the edge of kerbin's SOI

B = the transfer burn from High SOI/heliocentric orbit

C = The transfer burn from LKO

I would expect that A+B > C, but that B < C, and A<C

If I refuel after A, then my craft should be able get by with lower dV capabilties... but this seems to be wrong.

But then I also have to consider some other things... like the oberth effect from orbiting Minmus, doing a bi-elliptic transfer (from minmus, lower PE to just above kerbin, then burn from there, but then the transfer window timing is even more complicated), departing from Mun instead... and so on.

But... something just seems intuitively wrong here... I really expect B < C...

Can anyone help me with the math to show why this is? Supply actual numbers, rather than just "Oberth effect, it just works out that way" (which I've already concluded)

We can translate his A, B & C terms into equivalent terms in your equation:

V_bo² - V_esc² = V_{∞}²

A is V_esc

B* is V_{∞}

C is V_bo

*Note this is using the OP's "Heliocentric orbit" definition

Your equation can be rewritten: A² - C² = B²

Which can be directly rearranged to the Pythagorean theorem: A²+ B² = C²

In this light, the triangle inequality theorem should support the OP's intuition on the following points:

• A + B > C
  
• C > B
  
• C > A
  

Specifically the second point, because the OP states:

I really expect B < C...

However, this would seem to be contradicted by Alexmoon's transfer calculator, which I confirmed gives values similar to those in the original post:

From a 100km LKO orbit to jool needs a dV of 1986 m/s

From a 47,000km orbit (ie, where minmus is), one needs 2584 m/s!!!

The burn is almost identical to the Minmus orbit at an even higher starting altitude of 75 Mm. And, by my calculations, the dV cost of a burn from a Heliocentric orbit at Kerbin's altitude (which should be equivalent to the hyperbolic excess for the transfer) is greater still at 2760 m/s.

I think what is being missed here is the speed of the vessel relative to the sun due to it's prograde orbit around Kerbin. When orbiting around Kerbin at 100 km, a vessel has an prograde orbital velocity relative to the sun of 9284.5 m/s (Kerbin's orbital speed) + 2246.1 (Orbital speed around Kerbin at 100 km) = 11530.6 m/s. At 47,000 km, orbital velocity around kerbin is ~ 272 m/s, almost 2 km/s slower than LKO, and when starting from a slower speed relative to the sun you have to spend more dV to accelerate to initiate the transfer.

I found it illuminating to to fully comparing the two sets of numbers.

For a 100 km LKO burn:

V_bo²= (2760 m/s)² + (3176 m/s)²

V_bo = 4208 m/s

dV = V_bo - V_orb = 4208 m/s - 2246 m/s = 1962 m/s

For a "near Minmus" burn (47,000 km):

V_bo²= (2760 m/s)² + (385 m/s)²

V_bo = 2787 m/s

dV = V_bo - V_orb = 2787 m/s - 272 m/s = 2515 m/s

So while the target burnout velocity is smaller in Minmus escape scenario (i.e. C is reduced because A is reduced, and C > B remains true) , it actually takes more dV to achieve that velocity because the starting orbital speed is lower. Essentially, I think A, B, C's presented by the OP work provided that you remember they're true for velocities, but are not true for the delta Vs in this situation, and I think this may have been where the OP was getting hung up. Certainly was for me.

And this is math doing a direct burn from Minmus' altitude, not dropping down to LKO from Minmus before performing an ejection burn. The latter is still an awesome idea that works well.

Okay, I feel way better about this now.

P.S. the numbers are a little rough because I was dropping decimal places for expediency. Also they won't exactly match Alexmoon's calculator because he tweaks ejection phase angles, which I'm not doing. Alexmoon's numbers should be more accurate.

- - - Updated - - -

One thing I haven't seen considered here is TWR differences. While burning from that high up would take more fuel than if done from lower down, the much weaker gravity means a much smaller engine can push much more fuel. Since the fuel-mass to ship-mass ratio is improved you have higher starting dV. I'm not a math wizard, so I can't be sure if it's even possible that this can offset the costs of an inefficient burn (let alone save anything), but I thought it was an interesting idea (and sortof on topic). Any thoughts?

It's true that increasing TWR can reduce dV costs. However, It's basically assumed, for everything presented here, that the changes in speed are instantaneous (essentially TWR is infinite). In reality, all the transfers and burns presented with have slightly higher dV costs than what we've calculated and presented, and as OhioBob pointed out, lower TWR is going to have less effect at high altitudes, but it's never really going to save you anything worth noting.

Edited April 21, 2015 by LethalDose

[OhioBob]

Another potentially big problem with performing the transfer from a high orbit is the timing. The burn has to be performed when the 'angle to prograde' is correct. Minmus has an orbital period of over 56 days. When the ideal time for the ejection burn comes, we could be on the opposite side of Kerbin and 28 days away from where we need to be. Being off by that much means we're likely to miss the launch window altogether. And even if we're off by only a day or two, it still means we're performing a less than optimal transfer.

- - - Updated - - -

Essentially, I think A, B, C's presented by the OP work provided that you remember they're true for velocities, but are not true for the delta Vs in this situation, and I think this may have been where the OP was getting hung up.

I agree that's likely the root of the misunderstanding.

[LethalDose]

Another potentially big problem with performing the transfer from a high orbit is the timing. The burn has to be performed when the 'angle to prograde' is correct. Minmus has an orbital period of over 56 days. When the ideal time for the ejection burn comes, we could be on the opposite side of Kerbin and 28 days away from where we need to be. Being off by that much means we're likely to miss the launch window altogether. And even if we're off by only a day or two, it still means we're performing a less than optimal transfer.

I don't think the timing is as big of an issue as you've presented it. Yes, the orbital period is 56 days, but these are 6 hr kerbal days. It orbits Kerbin every fourteen 24 hr days, so it's not that bad. And with some appropriate planning, you can completely mitigate the issue. The transfer windows are also typically wide enough to accomodate some fudging by a few days on each side.

I've never done direct interplanetary transfers from Minmus, but I have done a handful of Minmus-to-LKO IP transfers, which are even trickier to time. I have never had a problem with Minmus being in such a bad position that I completely missed a transfer, or even had a transfer that would have cost more dV than a direct LKO escape.

Edited April 21, 2015 by LethalDose

[Eric S]

Another potentially big problem with performing the transfer from a high orbit is the timing. The burn has to be performed when the 'angle to prograde' is correct. Minmus has an orbital period of over 56 days. When the ideal time for the ejection burn comes, we could be on the opposite side of Kerbin and 28 days away from where we need to be. Being off by that much means we're likely to miss the launch window altogether. And even if we're off by only a day or two, it still means we're performing a less than optimal transfer.

All my motherships that I've done in over two years of play go to the Mun and Minmus as a dry run to minimize the chance that the craft will have phantom forces someplace where I can't easily rescue the crew. Not one of them has ever missed a transfer window, even tight transfer windows like Moho. The 5-50 m/s delta-v cost from not being in the optimal spot is lost in the noise of the overall mission and is a small fraction of the reduction of the delta-v requirement of planetary transfers done from Minmus, provided you're refueling. DISCLAIMER: I've been mostly playing career mode since it went live, restarting the save after I finish the tech tree, so we're not talking dozens of motherships. Not a huge data set, it's possible I've just gotten lucky every time, but keep in mind that except for Eeloo and Moho, transfer windows do tend to be in the 56 day range (using 6 hour days).

[LethalDose]

The more I look at this:

V_{∞}² = V_bo² - V_esc²

the more I love it. At least when I rearrange it to:

V_bo² = V_esc² + V_{∞}²

This method is so much more direct and easier than the other methods I've used to figure out IP transfer costs. Srsly, thank you for bringing this to my attention.

Edited April 21, 2015 by LethalDose

[OhioBob]

I don't think the timing is as big of an issue as you've presented it. Yes, the orbital period is 56 days, but these are 6 hr kerbal days. It orbits Kerbin every fourteen 24 hr days, so it's not that bad. And with some appropriate planning, you can completely mitigate the issue. The transfer windows are also typically wide enough to accomodate some fudging by a few days on each side.

I've never done direct interplanetary transfers from Minmus, but I have done a handful of Minmus-to-LKO IP transfers, which are even trickier to time. I have never had a problem with Minmus being in such a bad position that I completely missed a transfer, or even had a transfer that would have cost more dV than a direct LKO escape.

All my motherships that I've done in over two years of play go to the Mun and Minmus as a dry run to minimize the chance that the craft will have phantom forces someplace where I can't easily rescue the crew. Not one of them has ever missed a transfer window, even tight transfer windows like Moho. The 5-50 m/s delta-v cost from not being in the optimal spot is lost in the noise of the overall mission and is a small fraction of the reduction of the delta-v requirement of planetary transfers done from Minmus, provided you're refueling. DISCLAIMER: I've been mostly playing career mode since it went live, restarting the save after I finish the tech tree, so we're not talking dozens of motherships. Not a huge data set, it's possible I've just gotten lucky every time, but keep in mind that except for Eeloo and Moho, transfer windows do tend to be in the 56 day range (using 6 hour days).

In most cases, a few days one side or the other of the optimum departure isn't going to make a big difference. The big exception I think is Moho, which has an orbital period of just 102 days. Let's say we miss the ideal ejection time by 10 days, then Moho has moved 35 degrees from the optimal phase angle. This is quite significant and will result in a suboptimal transfer, likely costing several hundred m/s in ÃŽâ€v. Don't forget that it's just not the ejection burn that matters, it's the orbit insertion on the other end as well (unless you're doing an aerocapture). The penalty that you pay for a suboptimal transfer may come in the orbit insertion rather than the ejection burn.

One technique that can be used to control the timing of the burn is to adjust the orbital period so that you arrive at the right place at the right time. What I've done is to identify the location on my orbit where the ejection burn will take place several orbits from now*. When I reach that spot I compare my orbital period to the "time to ejection burn". I then perform a short burn to change my orbital period so that the number of orbits remaining until the ejection burn is some whole integer. In this way I'll be exactly where I need to be when it comes time to perform the ejection burn.

* One tricky part of this is that you must recognize that the 'angle to prograde/retrograde' that you want for your burn several days from now is not the current angle. Let's say we're orbiting Kerbin and we want to perform an ejection burn 20 days from now that has an angle to prograde of 100 degrees. In the 20 days between now and then, Kerbin will revolve 17 degrees around the sun. Therefore, the location of the maneuver node at the present time is 100 - 17 = 83 degrees to prograde.

- - - Updated - - -

The more I look at this:

V_bo² - V_esc² = V_{∞}²

the more I love it. At least when I rearrange it to:

V_bo² = V_esc² + V_{∞}²

This method is so much more direct and easier than the other methods I've used to figure out IP transfer costs. Srsly, thank you for bringing this to my attention.

Yes, I too find it very useful. It, along with the vis-viva equation and Tsiolkovsky rocket equation, is one of my most frequently used equations.

[LethalDose]

In most cases, a few days one side or the other of the optimum departure isn't going to make a big difference. The big exception I think is Moho, which has an orbital period of just 102 days. Let's say we miss the ideal ejection time by 10 days, then Moho has moved 35 degrees from the optimal phase angle. This is quite significant and will result in a suboptimal transfer, likely costing several hundred m/s in ÃŽâ€v. Don't forget that it's just not the ejection burn that matters, it's the orbit insertion on the other end as well (unless you're doing an aerocapture). The penalty that you pay for a suboptimal transfer may come in the orbit insertion rather than the ejection burn.

I think its fair to point out that Moho is the single glaring exception where your orbital position would matter in an ejection from a higher-than-LKO parking orbit, but I feel like we're all in agreement that the method is a bad idea anyway. The only reason we brought up the counterpoint was to point out that the problems with the method are not so much the positional difficulties, but increased dV costs.

Also Moho is almost always the exception because Moho is a freaking nightmare of a planet to reach, regardless of method.

But since you brought it up, this paragraph makes gives me the impression that you're doing direct Hohmann transfers to Moho.

And that makes me very sad.

I've found the best approaches to Moho do not determine transfer windows based on relative locations between Kerbin and Moho, but between Kebin and Moho's AN/DN. Moho orbits so fast that you don't need to line up with the planet on your first approach, you can just slow yourself down at the PE (which should be near to Moho's orbit) so you can catch it on the next orbit. This also reduces the TWR requirements for the capture burn, since you've already killed some orbital velocity relative to Moho. Direct Moho transfers are just bad ideas.

And if you're trying to use such a complex method to approach Moho in the first place, you're either (a) smart enough to figure out when to launch from Kerbins SoI OR ( screwed from the get-go because you're trying to imitate what you saw someone else do without really understanding what's going on.

[OhioBob]

I've flown only three missions to Moho. The first was a simple one-way unmanned lander. I tried a trajectory like this:

The idea is to start out in a zero-inclination orbit and then perform a plane change when 90^o from the planned intercept point. I'm still not sure where I miscalculated but, when I got to Moho, it took so much more ÃŽâ€v than planned for orbit insertion that I had to scrub the landing. It was one of my worst failures. I ended up just having to collect what science I could from orbit.

The second mission was essentially a repeat of the first in terms of objectives, and the third mission was a manned landing and return. Both of those used a direct transfer, however I strategically waited for launch windows in which Moho was ideally located near one of its nodes at the time of encounter. This allowed for a low-inclination, low-ÃŽâ€v transfer (low for Moho). I have some recollection that it was more difficult than usual to get the maneuver node set up just right to produce the encounter I wanted, but it was far from an impossible task. (This was also before I started using Precise Node.) Once the injection was completed, I don't recall there being anything especially difficult about flying the mission. I may have had to perform a course correction at some point, but that's nothing unusual.

I would never attempt a direct transfer to Moho when the encounter occurs far above or below the ecliptic. Such a transfer would be needlessly costly in terms of ÃŽâ€v.

Edited April 22, 2015 by OhioBob

[impyre]

I think everyone who responded to me was missing the point entirely. Simply put... if you stick a an LV-T45 engine behind some FL-T800 fuel tanks, you can comfortably push around three of them at 70km orbit. More if a lower TWR is acceptable. Using [a=g*(R/R+h)^2] (from ksp wiki basic orbiting math tutorial), acceleration due to gravity at minmus altitude is about .001556 m/s. The ratio of acceleration due to gravity at 70km to acceleration due to gravity at 47,000km is about 5012:1. Since F=ma shows that the relationship between acceleration and force is linear, this shows that TWR follows the same ratio (inversely proportional). This means that our little theoretical craft (just an engine and three tanks) has gone from a cozy TWR of 1.7 (at 70km) to a whopping TWR of 8547. This little LV-T45 (up at minmus' orbital altitude) could push well over ten times the number of tanks and still maintain a healthy TWR. (Actually, it could push a great deal more... and still maintain a comfortable TWR) Assuming 30 tanks are used on our theoretical freighter if it leaves from minmus altitude, and 3 if from 70km... it has 5835.8 m/s dV with 3 tanks (@70km) and 7661.6 m/s dV with 30 tanks from minmus.

These numbers are really just for illustrative purposes. If your computer could handle it, you could launch your minmus freighter with 100 tanks (or 1000). And I obviously didn't account for things like solar panels (who needs power anyway?)... but the idea is sound. Higher TWR = more fuel per engine... and more fuel per engine = more dV. I believe this demonstrates that launching from higher up can more than offset the extra requirements.

Of course, the real problem with leaving from that high up (as ever before) is the potential for missing launch windows due to being in a poor spot in your parking orbit... and launch windows trump pretty much everything else. I usually set up my parking orbits (and gas station) somewhere around geosynchronous orbit. Keeps line of sight with KSC (handy for rt2), gets some advantage of lower gravity, close enough to the mun (from which I launch mined/refined fuel), and Kerbin (from whence cometh my kerbals), all the while allowing me to still have potentially multiple launch opportunities within a single window.

[OhioBob]

impyre,

TWR as a function of local gravity I find useful only when I'm launching from the surface of a planet. Once in space, TWR is not important. What is important is acceleration. I may sometimes incorrectly use the term TWR when referring to a spacecraft in orbit, but what I'm really talking about is the acceleration expressed in units of standard gravity. I can't speak for others, but I assume they are doing the same thing. Stating that the TWR is 8547 I find to be completely useless information. What I want to know is that the acceleration is 13 m/s² or 1.3 g. Acceleration is a function of only mass and thrust, so a LV-T45 isn't going to push around a stack of tanks any easier at 47,000 km than it will at 70 km.

When we start adding tanks the ÃŽâ€v goes up because we're increasing the mass ratio, but that has a diminishing margin of return. We can only add so many tanks before the gain in ÃŽâ€v is no longer worth it. The downside of adding tanks is that we decrease the acceleration and increase the duration of the burn. The only really valid part of your argument that I see is that we can more easily endure a long burn from a high orbit. For example, burning 30 tanks of propellant (120 t) with a single LV-T45 will take about 36 minutes. It is impractical to perform a 36 minute burn from low Kerbin orbit, but it is certainly feasible from a high orbit.

Edited April 22, 2015 by OhioBob

[impyre]

Aww.... now I wanna delete my post. Sometimes I forget that weight doesn't equal mass. I'll leave it anyhow so that your post makes sense. Thanks for straightening me out.

[KerikBalm]

Thanks OhioBob...

I was thinking that basically you trade KE for PE (a higher heliocentric Apopasis), so starting with a higher PE is better. Of course Oberth effect means your KE increases more per unit fuel if you are going faster. Its nice to see the math that an be used to determine where the break even point is.

On the other hand, if you're already in Minmus orbit, departing Minmus with a periapsis just above Kerbin's atmosphere and doing the transfer burn at periapsis rather than circularizing winds up taking less delta-v than either of the other two options (from that point, the fuel to burn up to Minmus then circularize would make the LKO plan cheaper).

Scott Manley did a great job demonstrating

into an interplanetary transfer exploiting the Oberth effect. This is the video that really clarified the Oberth effect to me.

It takes kinetic energy to get to Jool, not potential energy. Being at Minmus' altitude gives you a lot of potential energy, but very little kinetic energy. Kinetic energy in this case is all about velocity, and you're going faster in LKO than you are at Minmus' altitude, so that's velocity you don't have to make up for by burning fuel.

As others have pointed out, you can save even more DV by leaving from a highly elliptical orbit with a low periapsis over Kerbin, which will give you even more velocity when you do your transfer burn.

I find all this a little annoying, because in the first post I already said:

doing a bi-elliptic transfer (from minmus, lower PE to just above kerbin, then burn from there, but then the transfer window timing is even more complicated)

Another way of thinking about it is to consider how much velocity you lose between perapsis, and crossing the SoI

Lets suppose that you need a velocity relative to Kerbin of 2.5km/s to planet Y.

Lets suppose that orbital velocity of kerbin is 2km/s, and 3km/s is "escape velocity" (ie pushes your apoapsis beyond the soi... although in KSP, that can still be an elispe that is very close to going hyperbolic)

If you departed from orbit at the very edge of the SOI, you'll only have a few hundred m/s of orbital velocity at best. You'll still need to burn over 2km/s

If you simply burned to escape velocity (1 km burn, taking 2km to 3km/s), then at the edge of the SOI, you'd have a few m/s relative velocity... you've lost pretty much 3km/s of relative velocity due to gravity.

If you burned again, you'd need to burn a full 2.5km/s (for a total of 3.5 km/s) - the worst way of doing it (yet I've seen people suggest this for beginners to get yo easy targets like Duna)

If you instead burn for a dV of 2000 m/s, for a total velocity of 4km/s... you'll have lost far less than 3km/s by the time you leave kerbin's SOI.

If you consider it like gravity drag... gravity acts on you for a far shorter time... lets suppose you only lose 1,500 m/s to gravity drag, and you've got that 2.5 km/s that you need.

My mistake was I wasn't really considering the variability in how much relative velocity you lose between a PE just above kerbin, and crossing the SOI.

Which I guess is pretty much the same as the calculation of excess hyperbolic velocity.

*edit*

Well, some good news for Jool refueling, I calculated (ie ran the transfer window calculator) the Laythe orbit -> Duna transfer, and it takes significantly less dV to depart from Laythe's orbital height (27184 km), than from low Jool Orbit (100km)

So I can have a craft capable of a dV of less than 1,500 m/s, and it can shuttle back and forth from Duna to the Jool system.

Jool @ 100 km to aerocapture at duna -> 2,892 m/s

Jool @ 27184km to aerocapture at duna -> 1,374 m/s

That number should be even lower from low laythe orbit (it should be a comparable advantage to starting in LKO, vs at the edge of kerbin's SOI, as laythe is very similar-> 80% the surface gravity, 83.3% the radius)

And it should be even lower if a bi-eliptic transfer is used (dropping PE to just above Jool), and then there's the chance of gravity assists and such.

Once fueling stations are setup... a craft capable of a dV of 1,500 m/s should be able to go all the way from orbit above eve, to kerbin and its moons, to duna, to jool, to Eeloo.

Only Moho and Dres would not be reachable.

Edited April 22, 2015 by KerikBalm

[LethalDose]

I find all this a little annoying, because in the first post I already said:

Well, What you're describing really isn't a bi-elipitcal transfer, but sorry to have wasted your time. I won't make the mistake of trying to help you again in the future...

Edited April 22, 2015 by LethalDose

[OhioBob]

I don't know if anybody will find it useful or not, but I just derived the following equation

V_soi² = 2μ * (1/R_soi - 1/R_bo) + V_bo²

where V_soi is the velocity at the sphere of influence, V_bo is the burnout velocity, R_soi is the radius of the sphere of influence, R_bo is the radius at burnout, and μ is the gravitational parameter.

This equation is the KSP version of V_{∞}² = V_bo² - V_esc² except is takes into account that patched conics are used by giving the velocity at the sphere of influence rather than at infinity. Note that the equations are the same if we set R_soi = ∞. In the KSP universe, V_soi is functionally equivalent to V_{∞} because it represents the velocity remaining after escaping the planet's gravity.