4
$\begingroup$

When I was calculating what delta-v was required to transfer between two orbits of radius 1 and 2, with a relative inclination of 20 degrees, I found, as expected, that it was best to combine the plane change with arriving at the outer orbit. (0.411756 compared to 0.533744 doing it at the inner one). However, doing a little of the plane change at the first burn got me a slightly better result. (0.391922 for a 4.42 degree angle). What is the reason for this? And how do I calculate the optimum split? enter image description here

$\endgroup$
5
  • $\begingroup$ A graph with unlabeled axes is worse than no graph at all. xkcd.com/833 $\endgroup$ Dec 15, 2015 at 16:35
  • $\begingroup$ @Russell Borogove I have no equation for calculating the points on the graph, so I used a Riemann sum. That mess up your x-axis a lot. But I see your point. $\endgroup$ Dec 15, 2015 at 16:38
  • 1
    $\begingroup$ It cuts plane change expense if the Hohmann launch window lies near an ascending or descending node. See hopsblog-hop.blogspot.com/2013/01/deboning-porkchop-plot.html $\endgroup$
    – HopDavid
    Dec 15, 2015 at 23:43
  • $\begingroup$ @HopDavid I am indeed doing the calculation for a transfer starting from the line of nodes. This is intended for an orbit to orbit transfer, where it is no need for intercepting a target. Therefore a perfect Hohmann transfer is possible. $\endgroup$ Dec 15, 2015 at 23:48
  • $\begingroup$ @HopDavid Your blog post is valuable, bordering to essential, for understanding realistic transfers between two objects though. $\endgroup$ Dec 15, 2015 at 23:53

2 Answers 2

4
$\begingroup$

Using @Hohmanfan's expression, the total $\Delta V$ for the Hohmann transfer with inclination change $\Delta i$ is assumed as \begin{equation} \Delta V = \sqrt{\left(\zeta_{1}s_{x}\right)^2 + \left(\zeta_{1}c_{x}-\lambda_{1}\right)^2} + \sqrt{\left(\lambda_{2}s_{ix}\right)^2 + \left(\lambda_{2}c_{ix}-\zeta_{2}\right)^2} \end{equation} where the coefficients are used for analytical simplicity and are given as \begin{equation} \begin{split} \zeta_1 &= \sqrt{\frac{2}{r_1}-\frac{2}{r_1+r_2}}\\ \zeta_2 &= \sqrt{\frac{2}{r_2}-\frac{2}{r_1+r_2}}\\ c_{i} &= \cos{(i)}\\ s_{i} &= \sin{(i)}\\ \end{split} \quad\quad \begin{split} \lambda_1 &= \frac{1}{\sqrt{r_1}}\\ \lambda_2 &= \frac{1}{\sqrt{r_2}}\\ c_{ix} &= \cos{(i-x)}\\ s_{ix} &= \sin{(i-x)} \end{split} \end{equation} Using Matlab, we can take the first and second partials of $\Delta V$ with respect to $x$, the initial inclination change to find $\frac{\partial\Delta V}{\partial x}$ and $\frac{\partial^2\Delta V}{\partial x^2}$, which are not presented here. @Hohmanfan's expression for the first partial gives us a nonlinear transcendental function that likely does not have an analytical solution in general. However, some simple root finding techniques (Newton-Raphson, etc.) are quite adequate here.

For the given $r_1 = 1, r_2 = 2$ scenario, I found the minimizing $x$ for a range of total inclination changes $i$. This is accomplished by iterating through the total inclination change vector, finding the min $\Delta V$ and the corresponding minimizing $x$, and then verifying local optimality by checking the value of the first derivative is zero, and the sign on the second derivative is positive. Figure 1 shows the minimizing $x$ and minimum $\Delta V$ for a given total inclination change $\Delta i$. Matching the OP's intuition, it appears that the largest portion of the plane change is saved at the end of the transfer, while there is some benefit to performing an initial inclination change with the first burn. The graph also makes intuitive sense when looking at the coplaner case hen $\Delta i = 0$, which is the case of a coplaner Hohmann transfer. Figure 1: Minimizing inclination change and minimum burn value.

Additionally, Fig. 2 displays the minimum $\Delta V$ as a function of the minimizing inclination angle change $x$. We can see there is some benefit to the initial plane change, but for very large inclination changes, the burn is incredibly expensive and the effect of the initial change is much less significant.

enter image description here

$\endgroup$
2
  • $\begingroup$ I think one of the reasons for the lack of an analytical solution is that there is an infinite amount of solutions for the angle x. (Like 4, 364, 724, 1084......) $\endgroup$ Dec 16, 2015 at 15:22
  • 1
    $\begingroup$ Note that it is even cheaper to do a bi-elliptical transfer if the inclination, in this case, is larger than 52 degrees. $\endgroup$ Dec 20, 2015 at 12:24
3
$\begingroup$

I think I see what is going on. Doing a plane change is in general better to do farther away from the planet, where the velocity is lower.
However, combining parts of the inclination change with the initial burn to enter the transfer orbit saves you delta-v because adding a small thrust vector normal to the main burn is for a large ratio between them virtually free. To illustrate this, the highlighted part of the hypotenuse is enough for adding the thrust vector of the smaller side.
enter image description here
This advantage become smaller if we add more side way thrust, eventually reaching an equilibrium.

Finding this equilibrium is harder. As a point of departure, the delta-v of a Hohmann transfer with inclination change is.

$$\mathbf{\sqrt{ \left(\operatorname{sin} \left( x \right) \; \sqrt{\frac{2}{r1} - \frac{2}{r1 + r2}} \right)^{2} + \left(\operatorname{cos} \left( x \right) \; \sqrt{\frac{2}{r1} - \frac{2}{r1 + r2}} - \frac{1}{\sqrt{r1}} \right)^{2}} + \sqrt{ \left(\frac{\operatorname{sin} \left( inc - x \right)}{\sqrt{r2}} \right)^{2} + \left(\frac{\operatorname{cos} \left( inc - x \right)}{\sqrt{r2}} - \sqrt{\frac{2}{r2} - \frac{2}{r1 + r2}} \right)^{2}}}$$

Where r1 and r2 are the radii of the departure and target orbit respectively, inc the total angle of the plane change, and x is the plane change done on departure. The issue is then solving for x when the derivative equals zero. That is the nasty equation:

$$\mathbf{\frac{\left(-\sqrt{2} \; \sqrt{\frac{1}{r2} - \frac{1}{r1 + r2}} + \frac{\operatorname{cos} \left( inc - x \right)}{\sqrt{r2}} \right) \; \frac{\operatorname{sin} \left( inc - x \right)}{\sqrt{r2}} - 2 \; \operatorname{sin} \left( inc - x \right) \; \frac{\operatorname{cos} \left( inc - x \right)}{r2}}{\sqrt{ \left(\frac{\operatorname{sin} \left( inc - x \right)}{\sqrt{r2}} \right)^{2} + \left(-\sqrt{2} \; \sqrt{\frac{1}{r2} - \frac{1}{r1 + r2}} + \frac{\operatorname{cos} \left( inc - x \right)}{\sqrt{r2}} \right)^{2}}} + \frac{2 \; \left(\frac{1}{r1} - \frac{1}{r1 + r2} \right) \; \operatorname{sin} \left( x \right) \; \operatorname{cos} \left( x \right) - 2 \; \sqrt{2} \; \left(\sqrt{2} \; \sqrt{\frac{1}{r1} - \frac{1}{r1 + r2}} \; \operatorname{cos} \left( x \right) - \frac{1}{\sqrt{r1}} \right) \; \sqrt{\frac{1}{r1} - \frac{1}{r1 + r2}} \; \operatorname{sin} \left( x \right)}{\sqrt{ \left(\sqrt{2} \; \sqrt{\frac{1}{r1} - \frac{1}{r1 + r2}} \; \operatorname{sin} \left( x \right) \right)^{2} + \left(\sqrt{2} \; \sqrt{\frac{1}{r1} - \frac{1}{r1 + r2}} \; \operatorname{cos} \left( x \right) - \frac{1}{\sqrt{r1}} \right)^{2}}}} = 0$$

@jah138 posted an answer on how to go further with this.

$\endgroup$
4
  • 1
    $\begingroup$ Allow me to introduce you to MathJax formatting. $\endgroup$
    – kim holder
    Dec 15, 2015 at 17:10
  • $\begingroup$ @kimholder That is going to be a lot better! I am converting it as soon as possible. Although this is perhaps not the easiest equation to start with... $\endgroup$ Dec 15, 2015 at 17:11
  • 2
    $\begingroup$ No... i thought about doing it and then said...no...:P $\endgroup$
    – kim holder
    Dec 15, 2015 at 17:15
  • $\begingroup$ Turned out it was just like LaTeX anyway :) $\endgroup$ Dec 15, 2015 at 17:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.