# How can Bielliptic transfers be more efficient than Hohmann transfers?

I don't have an exact citation or reference but I remember hearing that in some specific instances, that a bielliptic transfer can be more efficient than Hohmann transfers. The only possible instance I can think of where it would be more efficient would be changing inclinations to a target. Are there any other instances which make a bielliptic transfer maneuver more efficient than a Hohmann?

• Wkipedia says when the ratio of final to initial semi-major axis is 11.94 or greater, its possible that bi-elliptic transfera are more efficient, and cite Vallado, David Anthony (2001). Fundamentals of Astrodynamics and Applications. Springer. p. 318. ISBN 0-7923-6903-3 for it. Sadly, I do not currently have that book available. – Polygnome Jul 7 '17 at 23:06

Citing Wikipedia,

some bi-elliptic transfers require a lower amount of total delta-v than a Hohmann transfer when the ratio of final to initial semi-major axis is 11.94 or greater, depending on the intermediate semi-major axis chosen.

An intuitive approach to understanding it, is when you look at delta-V of insertion after Hohmann transfer. For near orbits, it's quite low, but as they grow apart, eccentricity of the transfer orbit increases.

Let's look at a bi-elliptic transfer "outwards" (from a lower to higher orbit) where it's more intuitive. Say, you want to go from LEO (7000km) to the Earth-Moon L5 point, almost 400,000 km. 400:7 is way more than 12:1 so it's a definite candidate for a bi-elliptic transfer.

Your transfer orbit will be a narrow spike, an very elongated ellipse. Speed near apogeum will be a crawl, a scarce couple meters per second. Meanwhile, the Moon in its circular orbit, goes at over 1000m/s, and so does it's L5 Lagrangian point. You'd need to accelerate your probe by nearly 1000m/s on arrival.

And while making that "spike" reach Moon orbit is costly, we can't skip that cost - but extending it much farther, to near Hill sphere. is nearly free - that's just a couple meters per second more. Circularizing an orbit with perigee near Moon orbit, from apogee near Earth's Hill sphere will still cost a bit - but considerably less than the 1000m/s needed to enter Moon orbit; not only motion at apogee this high is way slower, you're not performing a full circularization, leaving a considerable eccentricity. Then you'll still need to perform third burn, to match Moon orbit at L5, dropping that apogee, but even adding it, you're still spending less than you spent on bringing the perigee from LEO to Moon altitude.

Another way to think of it is: Hohmann orbit leaves you at apogee with speed lower than the target by x. Arrival "from infinity" / "from outside the system" into a flyby trajectory at altitude of target orbit leaves you with speed higher than the target by y. If the radii are much different, the difference between speed for Hohmann transfer and between escape is peanuts$_1$. If you arrive from outside the system, and want to choose altitude of your fly-by, when you're still very far, that costs peanuts$_2$.

Now if y + peanuts$_1$ + peanuts$_2$ < x, Bi-elliptic transfer takes less delta-V.

And while Hohmann transfer takes less time to transfer, finding the optimal time to perform it, especially with elliptic, non-coplanar orbits can get quite tricky. With bi-elliptic, all maneuvers at the apogee cost peanuts, and you can take significant liberties as to where that apogee will be. As result, the bi-elliptic transfer may take less than Hohmann transfer - if you count time until optimal alignment of departure and arrival objects towards the transfer time.

• More specifically, if the speed of the outer orbit (assuming it's circular) is $v$, then $y$ is approximately $(\sqrt{2}-1)v\approx 0.4v$. And if the orbits' radii are very different, then $x$ is close to $v$. – Litho Jul 8 '17 at 7:33