Hohmann vs Ballistic

[zeppelinmage]

This is more of a curiosity...

What's a better option for interplanetary orbital transfers? Ballistic, or Hohmann? Is it better to wait for a window and send your ship on a ballistic course, or to say "meh" and launch anytime, performing a solar Hohmann transfer when it is ideal.

Here's my thinking:

Hohmann transfer:

- don't have to wait for a window

- "easier" to match inclination

- total flight time longer

- potentially less course corrections?

Ballistic:

- Need a transfer window

- all inclination changes can be performed at the initial burn

- "fire and forget" method.

Mainly what I'm wondering is if there is any Delta-V savings at all with either method (Oberth, maybe?), and which would take more time. I suspect the Hohmann transfer would be more temporally expensive as you'd have to orbit the sun at least once to get an encounter... but then, you'd have to wait for a window anyway if you waited in LKO for a ballistic trajectory.

[Kasuha]

Hohmann transfer is a method to change your orbital paramteres. It does not guarantee you rendezvous with anything but it can be phased so you rendezvous in several orbits.

Ballistic transfer is a method to rendezvous with another object on a different orbit. It usually costs more dv but guarantees you rendezvous within single orbit time.

Rendezvous using Hohmann transfer - if done right - costs less dv but takes more time than Ballistic.

Edited February 3, 2014 by Kasuha

[zeppelinmage]

The idea is the same.

Ballistic is the "straight" path to your target planet.

The other option would be to escape to solar orbit first, then perform a Hohmann transfer to your target's orbit.

[SRV Ron]

Hoffman transfer from low Kerbal orbit directly to an intercept is the most efficient of all. Finding that transfer window is the most difficult part for planning that maneuver.

Lacking the mods to calculate that move, one can place a probe into solar orbit and using the Add Maneuver Mode, find the intercept window. Then, when that window arrives, take advantage of that window to launch your manned flights to those locations.

[Kasuha]

The idea is the same.

Ballistic is the "straight" path to your target planet.

The other option would be to escape to solar orbit first, then perform a Hohmann transfer to your target's orbit.

In general, ballistic transfer is every transfer where you apply impulses only at certain points and spend most time of the transfer coasting.

Transfer which is not considered ballistic is transfer where impulse is applied over considerable time. These transfers are used e.g. for ion probes.

Meaning, every Hohmann transfer is ballistic. Not every ballistic transfer is Hohmann transfer, though.

Example of ballistic transfer which is not Hohmann transfer:

http://en.wikipedia.org/wiki/File:Orbital_General_Transfer.svg

I think you're getting confused by terminology used in Alexmoon's calculator. In fact, all it calculates are ballistic transfers, the only difference is number of impulses applied - two or three.

Hohmann transfer is special case of ballistic transfer where two impulses are applied and both are in direction parallel to current orbital velocity. From this point of view (with some generalisation), all "ballistic" transfers provided by Alexmoon's calculator are Hohmann transfers.

_______________________________________

Now with terminology out of the way, what I would consider "Hohmann" interplanetary transfer would be achieving escape orbit which touches the target orbit in single point, then adjusting orbital period at that point to achieve rendezvous with the target at that point some next orbit. Meaning nothing prevents you doing the initial burn in your intial body's vicinity, and your final braking burn in your destination body's vicinity, using Oberth effect to your advantage. These transfers take usually considerably more time than transfers calculated by Alexmoon's calculator, and when done correctly, may cost considerably less dv.

Alexmoon's calculator calculates values for transfer for any time. The "window" is the time when the transfer can be performed for minimal (within certain period) value of dv. The "hohmann" transfer described above has similar windows, there are places where the transfer costs less dv and places where the same transfer costs more dv. For instance when doing Hohmann transfer to Moho, starting opposite to Moho's periapsis will cost you way less dv than starting at the opposite point.

Edited February 3, 2014 by Kasuha

[zeppelinmage]

Okay, forget the terminology. I'm just trying to compare methods of reaching a target destination. I just wanted a label that made sense for each. Since you could technically apply "Hohmann" and "Ballistic" to both.

At any rate, this is what I'm trying to find out. I just loaded up a "test" flight to Jool, with maneuver nodes.

"Ballistic":

Delta-V: 2086.1 m/s

Flight time: 272 days

Wait time: 100 days

"Hohmann":

Total Delta-V: 4050.7

Flight time: 369 days

Wait time: 0 days

(Three maneuvers: 1: exit Kerbin SOI, 2: inclination change, 3: Jool encounter.)

* Both flights leave Kerbin in the same direction - prograde into Kerbin's orbital path.

* I computed the "ballistic" path first, so technically the "100 days" occurred before launch of the "Hohmann" option. But that shouldn't affect the final comparison, IMO.

Now I acknowledge my KSP skills are less than ideal, and these are likely pretty gross estimations. But for this particular flight, it costs nearly twice the delta-V to save three days. My question is... is this true in all cases?

Also, the "Hohmann" option could concievably include gravitational slingshots. The cost to exit Kerbin SOI was 1244 m/s. Would it save anything to launch "right away", perform slingshots over those "extra" 100 days, and arrive in Jool spending only 1200 m/s?

[zeppelinmage]

Just noticed your edit.

Now with terminology out of the way, what I would consider "Hohmann" interplanetary transfer would be achieving escape orbit which touches the target orbit in single point, then adjusting orbital period at that point to achieve rendezvous with the target at that point some next orbit. Meaning nothing prevents you doing the initial burn in your intial body's vicinity, and your final braking burn in your destination body's vicinity, using Oberth effect to your advantage. These transfers take usually considerably more time than transfers calculated by Alexmoon's calculator, and when done correctly, may cost considerably less dv.

Alexmoon's calculator calculates values for transfer for any time. The "window" is the time when the transfer can be performed for minimal (within certain period) value of dv. The "hohmann" transfer described above has similar windows, there are places where the transfer costs less dv and places where the same transfer costs more dv. For instance when doing Hohmann transfer to Moho, starting opposite to Moho's periapsis will cost you way less dv than starting at the opposite point.

I'm actually not using that calculator. Rather Kerbal Alarm Clock for transfer windows. I'm just comparing options, which would be better in terms of "flight time" vs "delta-v". Also, your first option would be more akin to a bi-elliptic transfer, would it not?

[Kasuha]

Also, your first option would be more akin to a bi-elliptic transfer, would it not?

Definitely not, bi-elliptic transfer consists of three impulses, first of which is rising the apoapsis to very high value.

Example of bi-elliptic transfer:

http://en.wikipedia.org/wiki/File:Bi-elliptic_transfer.svg

(not a very good example, bi-elliptic transfer can only save dv when there is great difference between initial and destination orbit radius)

In my approach the trajectory is always between the two orbits.

[Jasmir]

Hi Zeppelinmage,

a Hohmann-Transfer is a well-definined thing to alter orbits (not planes!) with low energy use. See here: http://en.wikipedia.org/wiki/File:Hohmann_transfer_orbit.svg

If you want to rendevouz with a object a other orbit, you can archive it with a timed Hohmann-Transfer. It still has a low energy use, but you has to wait for a transfer-window.

The picture http://en.wikipedia.org/wiki/File:Orbital_General_Transfer.svg instead shows a random -high energy- target interception. For this, you don't have do wait for a transfer-window, but you will waste a lot of fuel. Don't do that.

If you needes to change your plane, things get complicated. In general, you want to do this on a point where your speed is lowest. That means -if you go to jool- do the planechange at the apoapsis of you transferorbit, during your injection-burn in the targetorbit. Or if you go to moho, do it on your ejectionburn on your kerbin-departure. The problem is, to find a transferwindow that allows that. To make it easyer, most missions do a midway-coursecorrection.

[Tank Buddy]

I just wait for launch windows. It makes ship requirements much less demanding.

[zeppelinmage]

I know the basics; my interplanetary flights have all been "direct" flights at transfer windows with mid-course corrections. Akin to my "Ballistic" picture above.

I'm wondering if there's any reason to use a different method; the "Hohmann" picture above. In my example, I spent twice the DV to save three days' flight time via solar orbit. Again, I admit I am less than skilled in optimizing flights, and my numbers may not be 100% representative.

I understand all maneuvers should be performed at ideal speed locations (periapsis/apoapsis); but if your orbit is largely circular, wouldn't that difference would be negligible? (all my orbits above are close to circular.) Interestingly, the "Ballistic" option makes all necessary maneuvers at once (acceleration and inclination change) for a direct injection into the target SOI, and costs far less. Doing straight prograde accelerations with a separate normal/anti-normal inclination change via solar orbit costs twice as much.

Where does that extra DV come from? Is it simply due to changing frame of reference (SOI)? Wouldn't my vessel's velocity relative to the sun be the same, regardless of what SOI I'm in? Therefore, the required delta-v should be the same?

[Kasuha]

Doing straight prograde accelerations with a separate normal/anti-normal inclination change via solar orbit costs twice as much.

Where does that extra DV come from?

In a triangle, one side is always shorter than the sum of the two others. That's also the reason why doing multiple orbital maneuvers in a single burn usually costs less dv than doing them separately.

Also, burning in a vicinity of a large body (most importantly Kerbin, Eve, Jool, but it applies to all bodies in the system, just at different rates) is adding more kinetic energy to your orbit than burning in interplanetary space (unless you're close to Sun). This is called Oberth effect.

However, many maneuvers cost different dv based on where in the orbit you perform them. If you clump maneuvers together, you're very likely making some of them in less-than-optimal place and that lowers the saving. That's also why in Alexmoon's calculator, the plane-change trajectory is sometimes more efficient than ballistic. Because otherwise the "ballistic" transfer has the plane-align maneuver distributed among the two remaining burns, ejection and injection, and thanks to principles above the sum is lower than doing them separately and in their optimal location.

[zeppelinmage]

a²+b²=c².

I didn't think Oberth would make a 2000 m/s difference... and the trig wouldn't double the cost either.

[LethalDose]

a²+b²=c².

remember the Pythagorean theorem is only for right triangles. What Kashua is talking about when he says the sum of two sides of a triangle exceed the length of third is a separate theorem is true regardless of the angle between the sides. You can see it, though, in the Pythagorean though, because if a²+b²=c², then a + b > c.

I didn't think Oberth would make a 2000 m/s difference... and the trig wouldn't double the cost either.

the Oberth effect can have a massive effect on the amount of fuel that is used to change orbits. The way this makes sense in my head is that, when a vessel is trying to take advantage of the Oberth effect, it's burning when it's close to a planetary body when the proportion of it's orbital energy is kinetic energy. So when you burn "x" amount of dV, you're getting x² more kinetic energy out of it (from KE = 1/2mV²). It's been my experience that Oberth leads to huge savings in dV when used properly, dropping dV requirements by more than half. I forget the exact math, but I want to say I've used Oberth to reduce some interplanetary dV costs to 1/3 or 1/4 the costs of just using direct transfers.

Edited February 4, 2014 by LethalDose

[zeppelinmage]

remember the Pythagorean theorem is only for right triangles. What Kashua is talking about when he says the sum of two sides of a triangle exceed the length of third is a separate theorem is true regardless of the angle between the sides. You can see it, though, in the Pythagorean though, because if a²+b²=c², then a + b > c.

I focused on Pythagoras since prograde vs normal is a right angle.

the Oberth effect can have a massive effect on the amount of fuel that is used to change orbits. The way this makes sense in my head is that, when a vessel is trying to take advantage of the Oberth effect, it's burning when it's close to a planetary body when the proportion of it's orbital energy is kinetic energy. So when you burn "x" amount of dV, you're getting x² more kinetic energy out of it (from KE = 1/2mV²). It's been my experience that Oberth leads to huge savings in dV when used properly, dropping dV requirements by more than half. I forget the exact math, but I want to say I've used Oberth to reduce some interplanetary dV costs to 1/3 or 1/4 the costs of just using direct transfers.

KE = 1/2mV²

That makes much more sense. And reading the wiki...

However, integrating this is often unnecessary if the burn duration is short. For example as a vehicle falls towards periapsis in any orbit (closed or escape orbits) the velocity relative to the central body increases. Briefly burning the engine (an "impulsive burn") prograde at periapsis increases the velocity by the same increment as at any other time (). However, since the vehicle's kinetic energy is related to the square of its velocity, this increase in velocity has a disproportionate effect on the vehicle's kinetic energy; leaving it with higher energy than if the burn were achieved at any other time.

^{So Oberth can basically halve the delta-v requirement relative to not having a body present at all. Which explains the above, and the fact it is always advantageous to perform as many burns as possible as close to your parent body as possible.}