Review of the Problem
In last weeks post, which you can find here, we went over what is a suicide burn, why it is beneficial, and what are the governing equations.
We plugged those nonlinear governing equations into Maple to get a solution and it gave us one. The problem was that it was long, and didn’t give us insight into the problem. This post will focus on a different way of approaching the problem analytically to mitigate this problem.
Zero Horizontal Velocity
The first way get an analytical solution is to simplify the problem. Lets assume that we have zero horizontal velocity. This simplifies our equations of motion to a single one dimensional linear time invariant ordinary differential equation.
We can solve this equation analytically by integrating twice and using our initial conditions to solve for its position and velocity at any time.
We know that the final conditions of velocity and position must be equal to 0, so lets set both of the above equations to zero and solve for our initial conditions in terms of the force of thrust, the force of gravity, and time. We get the following two relationships
Now, thrust and gravity, in our case, are both, constants. This means the only variables we can adjust are time, initial position, and initial velocity. 2 Equations, 3 unknowns. This means that if we have one of these values, we can compute the other two. Lets pick initial velocity. Below is a graph of different initial positions needed for different initial velocities. Any successful suicide burn must start on this blue line.
But what if our current position is not on this line? Well, if our engine is off than the force of thrust is zero and we will be falling, which is the non-technical way of saying we are trading vertical position for velocity. If we have an unlimited burn time, than any point that begins above our suicide burn line will eventually cross our burn line, at which point a suicide burn can be started. Additionally, in the above equation we only considered negative initial velocities, but because we are allowing our craft to drift before beginning our burn we can allow positive initial velocities. Any set of initial conditions above the two curves shown below allow you to complete a suicide burn.
In reality, you have a limited amount of fuel meaning that there is some finite set of initial condition that you can begin in but we are in math land so let’s ignore that.
Note: Here’s a question for you. Assuming infinite fuel, which is larger , the set of all possible initial conditions through which a suicide burn can occur, or the set of all possible initial conditions after which a suicide burn cannot occur? Email your answer (and a reason why) to ari@gereshes.com and I will add your name to a list at the bottom of this post! (Hint: Both are infinitely large. One is a larger infinity though.) (If too many people get this right I’ll only put up the first 25 or so to send me their answers)
Two Burns Define a Surface
We’ve now got a pretty powerful result in the last section for 1D suicide burns. It’s both analytic and simple. If we remember from part 1 of this post, the pure analytical solution for 2D suicide burns is convoluted. Let’s see if we can reduce its complexity by separating it into 2.5 easier problems. Instead of one single burn where the angle of the lander changes continuously, lets allow the lander to switch from being completely horizontal to completely vertical instantaneously. This allows us to solve the 2-D case as 2 1-D cases. Earlier I said it would be 2.5 easier problems. That extra half a problem is because we also have to account for an extra distance and velocity that the lander acquired in the vertical direction while thrusting in the horizontal direction.
Before we begin lets just define the sign of two units. The force of gravity is always negative, while the force of thrust is always positive. Additionally all equations shown below can be derived by simply integrating the 1D equations of motion and trivial algebraic manipulations.
Problem 1
First let’s tackle the horizontal burn. We need to drop the final x velocity to 0 so we can use the equation shown below
This gives us the time that the horizontal portion of the burn will take.
Problem 2
Now let’s find the distance that the lander fell in the time that it was thrusting horizontally. There’s only a single force from gravity to worry about and it can be represented here as the following
Problem 2.5
Finally, we need to now combine this new falling distance with the one dimensional fall from the beginning of this post. Again we plug in our final conditions to the constant motion equations and get the following equations to solve for initial height and burn time.
In the Gif below you can see a surface from which a suicide burn can occur. The x and y axis are the initial vertical and horizontal velocity, and the z axis is the initial height needed. Again, like with the above 1D case, any set of ordinates above this surface with the engine off will pass through this surface.
Now, how do these analytical results hold up against numerical integration from part 1? Quite well. Using the numerical integration as the truth, we get a percent error of <5% with most error being around 1%. This difference is probably due to compounding error through the numerical integration.
In effect, we solved a continuous nonlinear problem analytically by making it a non continuous function.
A More Optimal Burn Strategy
Now, if you remember from the first post, the burn strategy was to thrust directly opposite from the velocity of the lander. In this part, I quietly abandoned this strategy. Well, I didn’t quite abandon it, I just added two constraints. The initial angle of the lander must be either 90 or -90 degrees and the burn could not start unless the velocity vector was between this range and pointed towards the moon. What this equates to, is the lander would not spend time trying to fight an initial upward velocity, and would instead let the moons gravity fight this velocity. This is a slightly more optimal strategy but it has another feature that I did not use in this post. If you integrate the theta from time 0 to the final time, it will always equal ±90 degrees. This can make some analytical solutions easier but I chose to go through the path of solving 2.5 easier 1D problems instead of the full 2D case.
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