These manoeuvres typically come in two flavours:
- Manoeuvres adjusting the orbital period, meeting up with the target at the origin after some number of revolutions.
- Manoeuvres entering a transfer orbit, encountering the target at some different part of the orbit ("point and thrust" in its extreme case)
This is a three-variable solution space, with phase angle, delta-v and transfer time as its axis.
For any given phase angle, there are always two extreme solutions, one requiring zero delta-v but infinite transfer time, and one requiring zero transfer time but infinite delta-v. This motivates finding a compromise, since both time and delta-v is often only available in a quantity less than infinite.
Definitions:
To keep things simple, I'm assuming a two circular orbits with equal radius. These have unit radius, unit velocity. Scale to your actual use case.
Ahead in the orbit is the positive direction of the phasing angle.
Orbital period adjustments
These rely on the neat fact that if you do some impulsive manoeuvre, you will eventually reach the exact same location again, and will be able to "undo" the manoeuvre to get back the original orbit.
Phasing using this method relies on the intermediate orbit having a different orbital period.
As the rendezvous point is fixed, the transfer time is quantised, and is on the form:
$$t = 2\pi n - \theta$$
Where $n$ is a whole number and $\theta$ is the phasing angle.
increased orbital period
This helps increase the phasing angle. Useful when it's already close to $2\pi$, and you want it to roll around to 0. That is, a "trailing" target.
This can be done by raising the apoapsis, in fact, the optimal way of increasing orbital period this is by raising the apoapsis.
Adds the constraint $n \geq 2$.
We will be in the transfer orbit for $\lfloor{t\rfloor}$ orbits, meaning the orbital period of it is $\frac{t}{\lfloor{t\rfloor}}$. Clearly, this approaches 1 when $t$ is very large.
Orbital period is calculated from the semi-major axis by $2\pi\sqrt{a^3}$, so from that we can find the needed apoapsis height:
$$r_A = 2\sqrt[3]{\left(\frac{t}{\lfloor{t\rfloor}}\right)^2} - 1$$
From the vis-viva equation, the total delta-v is:
$$\Delta v = 2\sqrt{2 - \frac{1}{\sqrt[3]{\left(\frac{t}{\lfloor{t\rfloor}}\right)^2}}} - 2$$
The worst case scenario for this is when the target is slightly leading, and one tries to reach it as fast as possible by changing into a transfer orbit with an orbital period close to 2.
$$\Delta v_{worst} = \frac{4}{\sqrt(3)} - 2 \approx 0.31$$
That would however be a very misguided use of resources, since the following method should have been used instead in this case:
reduced orbital period
This helps reducing the phasing angle. Useful when it's already low, that is, a "leading" target.
This can be done by lowering the periapsis. Unfortunately, lowering the periapsis isn't always the optimal way of reducing orbital period. This follows as a corollary to the optimality of increasing apoapsis to increase orbital period.
Additionally, there's often the constraint that some lower limit exists for the periapsis, since there's often a planet in the way. These two complications makes analysis slightly more difficult.
Nevertheless, one may still use a similar strategy as in the case with the increased orbital period, this time lowering the periapsis into an elliptical transfer orbit. The orbital period is now $\frac{t}{\lceil{t\rceil}}$. Unlike the previous strategy, $n$ is not required to be 2 or greater, but as an additional constraint, the orbital period is not allowed to be lower than $\frac{1}{2\sqrt{2}}$.
$$\Delta v = 2 - 2\sqrt{2 - \frac{1}{\sqrt[3]{\left(\frac{t}{\lceil{t\rceil}}\right)^2}}}$$
You should generally check the delta-v for both the increased and reduced orbital period strategies, as they have different costs but similar transfer times (except in the $n = 1$ case where only the reduced period solution may exist). The best choice depends on the phasing angle.
reduced orbital period, improved
The most efficient way of reducing orbital period is to first lower the periapsis, and then circularise. In other words, a Hohmann transfer (bi-elliptic transfers are not useful for this particular problem).
This increases the manoeuvre from two impulses to four:
- Start Hohmann transfer to towards lower circular orbit
- Break into lower orbit.
- Start Hohmann transfer back.
- Enter original orbit at target phasing angle.
Unlike the two previous strategies, transfer time is not quantised, since the transfer back into the original orbit can be done at an arbitrary phase angle, because the intermediate orbit is circular.
The free variable here is the periapsis radius ($r_P$). Once it's selected, the transfer time and delta-v cost follows. It's a trade-off between slow but cheap transfers at a higher periapsis radius, and faster but more expensive transfers at a lower periapsis radius.
Delta-v cost, per usual Hohmann calculations (notably, it's independent of phasing angle):
$$\Delta v = 2\left(1 - \sqrt{2 - \frac{2}{1 + r_P}} + \sqrt{\frac{2}{r_P} - \frac{2}{1 + r_P}} - \sqrt{\frac{1}{r_P}}\right)$$
Transfer time is a bit more involved. The phase angle is reduced both in the two Hohmann transfer legs, and during the time spent in the lower circular orbit:
- Transfer legs. The orbital period of the full elliptical orbit is $2\pi\sqrt{\frac{(r_P + 1)^3}{8}}$. This then reduces the phasing angle by $2\pi - 2\pi\sqrt{\frac{(r_P + 1)^3}{8}}$.
- Circular orbit. While the angular velocity of the original orbit is 1, it's higher in the lower orbit. This leads to a reduction of the phasing angle of $\frac{1}{\sqrt{r_P^3}} - 1$ per unit time.
Total transfer time is therefore:
$$t = 2\pi\sqrt{\frac{(r_P + 1)^3}{8}} + \frac{\theta - 2\pi + 2\pi\sqrt{\frac{(r_P + 1)^3}{8}}}{\frac{1}{\sqrt{r_P^3}} - 1}$$
At very cheap, long transfer time uses of this (that is, the time spent in the inner circular orbit is much higher than the Hohmann transfers), the equations can be simplified:
$$\Delta v \approx 2\sqrt{\frac{1}{r_P}} - 2$$
$$t \approx \frac{\theta}{\frac{1}{\sqrt{r_P^3}} - 1}$$
Meeting up at a different location
Sensible direct trajectories
TODO
(also quite an important TODO for fast phasing at reasonable cost. It's what's illustrated in the image in your question)
(outline: Pick some target $r_P < 1$ and $r_A > 1$, so the spacecraft goes through a slower of faster arc before reaching $r=1$ again.
The transfer time is not straight forward, and I don't quite have the equations for picking an optimal $r_P$ and $r_A$ at your delta-v budget.
The cost is however given by:
$$v_{horizontal} = r_p \sqrt{\frac{2}{r_P} - \frac{2}{r_P + r_A}}$$
$$v_{vertical} = \sqrt{2 - \frac{2}{r_P + r_A} - v_{horizontal}^2}$$
$$\Delta v = 2\sqrt{v_{vertical}^2 + (v_{horizontal} - 1)^2}$$
:outline end)
A class of such trajectories, nadir catch-up burns.
These are the direct phasing manoeuvres that meet up on the exact oppsite side of the orbit. These are easy to calculate, hence why they are included here before completing this section.
Given some delta-v spent burning directly towards the parent body, the orbit's semi-major axis is given by $a = \frac{1}{2 - \left(1 + \left(\frac{\Delta v}{2}\right)^2\right)}$
And the periapsis by $r_P = a - \sqrt{a^2 - a}$
The transfer time in the fast arc is then:
$$t = 2 \sqrt{a^3} \tan^{-1}\left(\frac{a}{(a - r_P) \sqrt{r_P (2a - r_P)}}\right)$$
From which we can find the phase angle:
$$\theta = \pi - t$$
Torchships
In the extreme case, where available acceleration is arbitrarily high and delta-v is much higher than the orbital velocity, the problem approaches a simple relationship to distance:
$$t = \frac{2\sqrt{\sin(\theta)^2 + (1 - \cos(\theta))^2}}{\Delta v}$$
Low-thrust phasing
Certain spacecraft, such as those propelled by ion engines, have very limited thrust. They can therefore not perform impulsive burns, and instead follow constant thrust spirals.
In such low-thrust spirals, the spacecraft is always in an approximately circular orbit. The delta-v cost between two such circular orbits is remarkably simple:
$$\Delta v = v_1 - v_2$$
That is, simply the difference between the orbital velocities.
A low-thrust spacecraft would spiral up (or down) until half the phasing angle has been reached, and then return in a similar spiral transfer to cover the other half.
The relationship between transfer time and phasing angle is here highly non-linear. The change in phasing angle at any instant is $\theta' = \omega - 1$, where $\omega$ is the current angular velocity, which itself is given by $\omega = \frac{v}{r}$, where $v$ and $r$ is the current velocity and orbital radius.
Integrating over time is then:
$$\Delta \theta = \int 1 - (1 - at)^3 dt$$
Where $a$ is acceleration. With the base orbit as the origin, it yields the result:
$$\Delta \theta = \frac{\frac{t^4}{4} - t^3 + \frac{3t^2}{2}}{a}$$
(positive $a$ is here spiralling outwards)
The practical result of this is that spiralling outwards is the optimal for phasing angles up to ~80 degrees in the trailing direction, while all other phasing angles are achieved faster by spiralling inwards:
