MontBlanc2012
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This note follows on from Changing the argument of periapsis - single-burn solution. In that note, a single-burn method for changing the argument of periapsis was examined. Here, we consider a two-burn method that is considerably more fuel efficient at moving the argument of periapsis (while keeping other orbital elements constant).
The two-burn approach set out below relies upon making two tangential burns - the first a prograde burn that rotates the argument of periapsis by exactly half the required angular change in the argument of periapsis while changing both the semi-major axis and orbital eccentricity; the second a retrograde burn that rotates the argument of periapsis by the same angular change while restoring the semi-major axis and eccentricity to their starting values. This scheme is illustrated below:
In this diagram, the blue ellipse is our starting ellipse. The periapsis radius is 6671 km; and the orbital eccentricity is 0.5. The black circle is the Earth with radius 6371 km. The red ellipse is the target ellipse with the same periapsis radius and orbital eccentricity as the blue ellipse except that the line of apsides has been rotated counter-clockwise by 90 degrees. The dotted orange ellipse is a transfer ellipse that is tangent to both the initial and target ellipses (at the points A and B respectively).
The sequence, then for transferring from the initial blue ellipse to the target red ellipse is as follows:
Now, this transfer sequence is not unique: There are multiple transfer ellipses that we could use that achieve the same rotation of the argument of periapsis. However, there is one transfer orbit that does this that requires the least amount of fuel. The question then becomes: what is the optimal transfer ellipse to use with which to effect the change in the argument of periapsis (while leaving the semi-major axis and orbital eccentricity unchanged?
An optimisation problem
The answer to this question is found by setting up and solving a simple non-linear optimisation problem. I'll set out the formal definition of the optimisation below - but to make life easier I've set up this optimisation problem in Microsoft Excel (using the in-built "Solver" utility) to allow the reader ready access to solving the optimisation problem.
In setting up the optimisation problem, we can take advantage of the fact that the problem has a natural symmetry in that each of the tangent burns rotates the argument of periapsis by exactly half of the target change in the argument of periapsis. Consequently, we only need to solve for the optimal prograde burn for transferring from the initial to the transfer orbit because the second tangent burn is just a mirror of the first.
Let's suppose that we want to change the argument of periapsis by [imath]\Delta\theta[/imath]. In the language of "Tangent ellipses", [imath]\Delta\theta[/imath] is just [imath]2\,\Delta\omega[/imath] where [imath]\Delta\omega[/imath] is the change in the argument of periapsis achieved by each of the prograde and retrograde burns. And from "Tangent ellipses", we know that:
[math] \sin (\frac{\Delta\theta}{4} )^2 = \frac{\left(r_{a,1}-r_{a,2}\right) \left(r_{p,1}-r_{p,2}\right)}{\left(r_{a,1}-r_{p,1}\right) \left(r_{a,2}-r_{p,2}\right)} [/math]
where [imath]r_{a,1}[/imath] and [imath]r_{p,1}[/imath] are the apoapsis and periapsis radii of the initial elliptical orbit; and [imath]r_{a,2}[/imath] and [imath]r_{p,2}[/imath] are the apoapsis and periapsis radii of the transfer orbit. This equation established a connection between [imath]r_{a,2}[/imath] and [imath]r_{p,2}[/imath] but we need some sort of additional information to solve the problem for unique values of [imath]r_{a,2}[/imath] and [imath]r_{p,2}[/imath].
This additional piece of information is found by minimising the fuel cost of the tangent transfer. This fuel cost, [imath]\Delta V[/imath], is given by:
[math]\Delta V = v_2 - v_1 [/math]
where:
[math] \begin{aligned} v_2 &= \sqrt{2\,\mu\,\left(\frac{1}{r^*} - \frac{1}{r_{a,2}+r_{p,2}}\right)} \\ v_1 &= \sqrt{2\,\mu\,\left(\frac{1}{r^*} - \frac{1}{r_{a,1}+r_{p,1}}\right)} \\ \\ r^* &= \frac{r_{a,1} \,r_{p,1} \,\left(r_{a,2}+r_{p,2}\right)-r_{a,2} \,r_{p,2} \,\left(r_{a,1}+r_{p,1}\right)}{r_{a,1}\, r_{p,1}-r_{a,2} \,r_{p,2}} \end{aligned}[/math]
So, our optimisation problem becomes:
Minimise [imath]\left|\Delta V\right| [/imath]
with respect to [imath]r_{a,2}[/imath] and [imath]r_{p,2}[/imath] subject to the requirement that:
[math] \sin (\frac{\Delta\theta}{4} )^2 = \frac{\left(r_{a,1}-r_{a,2}\right) \left(r_{p,1}-r_{p,2}\right)}{\left(r_{a,1}-r_{p,1}\right) \left(r_{a,2}-r_{p,2}\right)} [/math]
and
[imath]r_{a,2}>r_{a,1}>r_{p,2}>r_{p,1}>0[/imath]
Solving this problem then gives us the most [imath]dV[/imath] efficient transfer with which to effect the transfer from the initial ellipse to the final ellipse using the bi-tangent transfer described above.
Solving the optimisation problem
Although the expression for the optimisation problem is reasonably succinct, this is not an easy problem to solve analytically so we have to resort to numerical procedures. And rather than set up optimisation code in a programming language sucks as C++, we can set up and solve this problem in Microsoft Excel. (A link to a spreadsheet tool with which to solve this bi-tangent transfer problems will be provided at the end of this note.)
To illustrate the use of this tool, let's consider application of the spreadsheet optimisation tool to the example problem.
First, we need to put in some basic parameters and initial guesses:
The basic inputs that we need to type in are:
Having provided this we provide some initial guesses for the transfer orbit apoapsis and periapsis radii such that [imath]r_{a,2}>r_{a,1}>r_{p,2}>r_{p,1}>0[/imath]. In our case, we use as initial guesses 30,000,000 m and 10,000,000 m respectively.
This specifies the problem that we need to solve. And now we let Excel find the minimum $dV$ solution. This requires that we go to the "Tools" menu and select "Solver" and hit the "Solve" button. And the Excel should quickly find the optimal solution and ask if you want to save the solution which, of course, you do.
If you go through this sequence Excel should provide the following information about your initial and transfer orbits:
This section of the spreadsheet says that your initial orbit has a semi-major axis of 13,342 km and orbital eccentricity of 0.5. It also now says that the optimal (least fuel cost) transfer is achieved by transferring to a transfer orbit with semi-major axis of 18,282.5 km and orbital eccentricity 0.3384.
OK, so what about the manoeuvres themselves? The next part of the spreadsheet provides information about the magnitude and timing of the prograde and retrograde burns.
The first block of the provides information about the manoeuvres if you wish to rotate your argument of periapsis by the target angle (in this case 90 degrees) in a counter-clockwise direction whereas the second block does the same for a clockwise for the argument of periapsis.
Let's take the counter-clockwise rotation first. To execute this manoeuvre the spreadsheet says that starting from the initial orbit, you wait until 2,398.3 seconds before apoapsis. Then, you execute a prograde burn of 983.06 m/s. This burn transfers you to the transfer orbit. When on your transfer orbit, you then wait until 5,005.8 seconds before periapsis. Then you execute a retrograde burn of exactly the same magnitude of 983.06 m/s to complete the rotation of your argument of periapsis and to change the semi-major axis and eccentricity of your orbit back to their initial values.
This process is modified to rotate the argument of periapsis in a clockwise direction. Here, starting on your initial orbit, you wait unit 5,270.2 seconds before periapsis and then execute a prograde burn of 983.06 m/s. This transfers you to the relevant transfer orbit. You then wait unit 7295.1 seconds before apoapsis on your transfer orbit whereupon you execute a retrograde burn of 983.06 m/s to again complete the rotation of your argument of periapsis and to change the semi-major axis and eccentricity of your orbit back to their initial values.
In either case, the total amount of fuel uses is 2 x 983.06 m/s.
A quick note on balancing burns
As with most burns in Orbiter, you should 'balance your burns' by timing it so that half of your burn occurs before the target time; and half after the target time. Orbit MFD gives the time to periapsis and the time to apoapsis; and BTC MFD gives you the duration of the burn. With these two pieces of information it is possible to execute burns with some precision.
The spreadsheet download
A link to the spreadsheet will follow shortly - once I work out where to put it so that it can be accessed by third parties.
The two-burn approach set out below relies upon making two tangential burns - the first a prograde burn that rotates the argument of periapsis by exactly half the required angular change in the argument of periapsis while changing both the semi-major axis and orbital eccentricity; the second a retrograde burn that rotates the argument of periapsis by the same angular change while restoring the semi-major axis and eccentricity to their starting values. This scheme is illustrated below:
In this diagram, the blue ellipse is our starting ellipse. The periapsis radius is 6671 km; and the orbital eccentricity is 0.5. The black circle is the Earth with radius 6371 km. The red ellipse is the target ellipse with the same periapsis radius and orbital eccentricity as the blue ellipse except that the line of apsides has been rotated counter-clockwise by 90 degrees. The dotted orange ellipse is a transfer ellipse that is tangent to both the initial and target ellipses (at the points A and B respectively).
The sequence, then for transferring from the initial blue ellipse to the target red ellipse is as follows:
- Assuming that one orbits in a conventional counter-clockwise direction, the first step is to wait until one has passed periapsis on the initial blue ellipse until the orbital radius is [imath]r^*[/imath]. Then, execute a prograde burn to transfer to the higher dotted orange transfer orbit.
- After passing through apoapsis on the transfer orbit, wait until the orbital radius is again [imath]r^*[/imath]. Then execute a retrograde burn to transfer to the red target ellipse.
Now, this transfer sequence is not unique: There are multiple transfer ellipses that we could use that achieve the same rotation of the argument of periapsis. However, there is one transfer orbit that does this that requires the least amount of fuel. The question then becomes: what is the optimal transfer ellipse to use with which to effect the change in the argument of periapsis (while leaving the semi-major axis and orbital eccentricity unchanged?
An optimisation problem
The answer to this question is found by setting up and solving a simple non-linear optimisation problem. I'll set out the formal definition of the optimisation below - but to make life easier I've set up this optimisation problem in Microsoft Excel (using the in-built "Solver" utility) to allow the reader ready access to solving the optimisation problem.
In setting up the optimisation problem, we can take advantage of the fact that the problem has a natural symmetry in that each of the tangent burns rotates the argument of periapsis by exactly half of the target change in the argument of periapsis. Consequently, we only need to solve for the optimal prograde burn for transferring from the initial to the transfer orbit because the second tangent burn is just a mirror of the first.
Let's suppose that we want to change the argument of periapsis by [imath]\Delta\theta[/imath]. In the language of "Tangent ellipses", [imath]\Delta\theta[/imath] is just [imath]2\,\Delta\omega[/imath] where [imath]\Delta\omega[/imath] is the change in the argument of periapsis achieved by each of the prograde and retrograde burns. And from "Tangent ellipses", we know that:
[math] \sin (\frac{\Delta\theta}{4} )^2 = \frac{\left(r_{a,1}-r_{a,2}\right) \left(r_{p,1}-r_{p,2}\right)}{\left(r_{a,1}-r_{p,1}\right) \left(r_{a,2}-r_{p,2}\right)} [/math]
where [imath]r_{a,1}[/imath] and [imath]r_{p,1}[/imath] are the apoapsis and periapsis radii of the initial elliptical orbit; and [imath]r_{a,2}[/imath] and [imath]r_{p,2}[/imath] are the apoapsis and periapsis radii of the transfer orbit. This equation established a connection between [imath]r_{a,2}[/imath] and [imath]r_{p,2}[/imath] but we need some sort of additional information to solve the problem for unique values of [imath]r_{a,2}[/imath] and [imath]r_{p,2}[/imath].
This additional piece of information is found by minimising the fuel cost of the tangent transfer. This fuel cost, [imath]\Delta V[/imath], is given by:
[math]\Delta V = v_2 - v_1 [/math]
where:
[math] \begin{aligned} v_2 &= \sqrt{2\,\mu\,\left(\frac{1}{r^*} - \frac{1}{r_{a,2}+r_{p,2}}\right)} \\ v_1 &= \sqrt{2\,\mu\,\left(\frac{1}{r^*} - \frac{1}{r_{a,1}+r_{p,1}}\right)} \\ \\ r^* &= \frac{r_{a,1} \,r_{p,1} \,\left(r_{a,2}+r_{p,2}\right)-r_{a,2} \,r_{p,2} \,\left(r_{a,1}+r_{p,1}\right)}{r_{a,1}\, r_{p,1}-r_{a,2} \,r_{p,2}} \end{aligned}[/math]
So, our optimisation problem becomes:
Minimise [imath]\left|\Delta V\right| [/imath]
with respect to [imath]r_{a,2}[/imath] and [imath]r_{p,2}[/imath] subject to the requirement that:
[math] \sin (\frac{\Delta\theta}{4} )^2 = \frac{\left(r_{a,1}-r_{a,2}\right) \left(r_{p,1}-r_{p,2}\right)}{\left(r_{a,1}-r_{p,1}\right) \left(r_{a,2}-r_{p,2}\right)} [/math]
and
[imath]r_{a,2}>r_{a,1}>r_{p,2}>r_{p,1}>0[/imath]
Solving this problem then gives us the most [imath]dV[/imath] efficient transfer with which to effect the transfer from the initial ellipse to the final ellipse using the bi-tangent transfer described above.
Solving the optimisation problem
Although the expression for the optimisation problem is reasonably succinct, this is not an easy problem to solve analytically so we have to resort to numerical procedures. And rather than set up optimisation code in a programming language sucks as C++, we can set up and solve this problem in Microsoft Excel. (A link to a spreadsheet tool with which to solve this bi-tangent transfer problems will be provided at the end of this note.)
To illustrate the use of this tool, let's consider application of the spreadsheet optimisation tool to the example problem.
First, we need to put in some basic parameters and initial guesses:
The basic inputs that we need to type in are:
- The gravitation parameter for the central body - we need to specify the strength of the gravitational field that we are moving in. This is given by the product 'GM'. For the planets these values can be obtained from Wikipedia (see "Standard gravitational parameter");
- We need to provide the periapsis radius and apoapsis radius of our initial orbit. for our example, this is 6,671,000 m and 20,013,000 m respectively; and
- We also want to supply the change in the argument of periapsis that we want to achieve by our bi-tangent transfer.
Having provided this we provide some initial guesses for the transfer orbit apoapsis and periapsis radii such that [imath]r_{a,2}>r_{a,1}>r_{p,2}>r_{p,1}>0[/imath]. In our case, we use as initial guesses 30,000,000 m and 10,000,000 m respectively.
This specifies the problem that we need to solve. And now we let Excel find the minimum $dV$ solution. This requires that we go to the "Tools" menu and select "Solver" and hit the "Solve" button. And the Excel should quickly find the optimal solution and ask if you want to save the solution which, of course, you do.
If you go through this sequence Excel should provide the following information about your initial and transfer orbits:
This section of the spreadsheet says that your initial orbit has a semi-major axis of 13,342 km and orbital eccentricity of 0.5. It also now says that the optimal (least fuel cost) transfer is achieved by transferring to a transfer orbit with semi-major axis of 18,282.5 km and orbital eccentricity 0.3384.
OK, so what about the manoeuvres themselves? The next part of the spreadsheet provides information about the magnitude and timing of the prograde and retrograde burns.
The first block of the provides information about the manoeuvres if you wish to rotate your argument of periapsis by the target angle (in this case 90 degrees) in a counter-clockwise direction whereas the second block does the same for a clockwise for the argument of periapsis.
Let's take the counter-clockwise rotation first. To execute this manoeuvre the spreadsheet says that starting from the initial orbit, you wait until 2,398.3 seconds before apoapsis. Then, you execute a prograde burn of 983.06 m/s. This burn transfers you to the transfer orbit. When on your transfer orbit, you then wait until 5,005.8 seconds before periapsis. Then you execute a retrograde burn of exactly the same magnitude of 983.06 m/s to complete the rotation of your argument of periapsis and to change the semi-major axis and eccentricity of your orbit back to their initial values.
This process is modified to rotate the argument of periapsis in a clockwise direction. Here, starting on your initial orbit, you wait unit 5,270.2 seconds before periapsis and then execute a prograde burn of 983.06 m/s. This transfers you to the relevant transfer orbit. You then wait unit 7295.1 seconds before apoapsis on your transfer orbit whereupon you execute a retrograde burn of 983.06 m/s to again complete the rotation of your argument of periapsis and to change the semi-major axis and eccentricity of your orbit back to their initial values.
In either case, the total amount of fuel uses is 2 x 983.06 m/s.
A quick note on balancing burns
As with most burns in Orbiter, you should 'balance your burns' by timing it so that half of your burn occurs before the target time; and half after the target time. Orbit MFD gives the time to periapsis and the time to apoapsis; and BTC MFD gives you the duration of the burn. With these two pieces of information it is possible to execute burns with some precision.
The spreadsheet download
A link to the spreadsheet will follow shortly - once I work out where to put it so that it can be accessed by third parties.
Last edited:
