# Homework Help: Minimal altitude to start a “Suicide Burn” - variable thrust

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1. Nov 23, 2016

### Daquicker

1. The problem statement, all variables and given/known data
I'm trying to calculate the absolute lowest safe altitude above a planet's surface to start firing the thrusters on a simulated lander in order tot just reach vCraft = 0 at the planet's surface.

The force the craft's thrusters responsible for desceleration generate is however dependent on the density of the atmosphere they're in, which in turn depends on the altitude above seaLevel they're currently at.

I'm looking for a general solution that returns an altitude H.

general parameters of the problem:
• Drag is not to be taken into account.
• Craft mass does not vary during descent. (Electricity fuels the thrusters so no solid/liquid fuel being burnt).
• Craft falls straight down along the gravity vector of the planet.
• Thrusters providing desceleration are pointed straight down along the gravity vector.
• The planet does not rotate.
• Planet g does not change. (At applicable altitudes change will be negligible).

Data accesible from telemetry:

general:

• CraftAltitude above surface: xSurf is known at any time (in m)
• CraftAltitude above seaLevel: xSea is known at any time (in m)
• CraftVelocity (downwards): vCraft is known at any time (in m/s)
• CraftMass: mCraft is known at any time (in Kg)
• CraftMaxThrust at SeaLevel: xMaxThrustSea is known at any time (in Newton)
• Planet g: is known at any time (in m/s^2)
for this example:
• Atmosphere density: Altitude = 0m -> AtmDens = 1 | Altitude = 14400m -> AtmDens = 0
• Thruster efficiency: AtmDens = 1 -> ThrustEff = 1 | AtmDens = 0.3 -> ThrustEff = 0

2. Relevant equations
• Work = m * g * d | m: mCraft, g: aCraft, d: xSurf
• Kinetic Energy = (m * v^2) / 2 | m: mCraft, v: vCraft

3. The attempt at a solution

I started with the following simple and common idea:
Get an altitude from the Work required to dissipate the craft's Kinetic Energy

=> W = KE => d = (v^2) / (2 * g)

This however, fails to account for the change in g due to the change in the craft's altitude above sea level (xSea), instead always using the theoretical maximum thrust. The consequence being it gravely underestimates the required braking distance.

I then came up with the following function which describes the craft's instantanious thrust at an altitude xSea:

CraftMaxThrust(xSea) = cMaxThrustSea - (cMaxThrustSea / 100) * ((100 / 0.7) * (1 - (1 - xSea / 14400)))

The acceleration at an altitude xSea provided by the thrusters would then become:

a(xSea) = CraftMaxThrust(xSea) / mCraft

So I figured I needed some type of differential equation to adjust for this shift in force.
This is where I get stuck, all I can come up with is that it should probably be something amongst the lines of:

0 = vCraft + ∫(H->0) |some derivative of a(xSea)?| a(xSea)

With vCraft being initial velocity, a(xSea) adjusted with the planet's g, and H the altitude I've been looking for all along.

Eventhough it's not an actual homework assignment, any help would be greatly appreciated, it's been evading me for a couple of days now.

Last edited: Nov 23, 2016
2. Nov 23, 2016

### haruspex

A fundamental principle of rocketry is that there has to be something to thrust. Electricity alone is not going to generate an upward force.

3. Nov 23, 2016

### Daquicker

The electricity powers an "atmospheric thruster", it's more or less really big "fans". Anyways, it's not supposed to be real world, so this can safely be ignored.

Edit: on a rather unrelated sidenote, you did remind me of this article: http://arc.aiaa.org/doi/10.2514/1.B36120

4. Nov 23, 2016

### haruspex

Ok, so deceleration will be constant, right? What equations do you know for constant acceleration?

5. Nov 23, 2016

### Daquicker

No, deceleration will not be constant, it will increase as the craft's altitude above planetSeaLevel decreases. If it was constant I could have used EKin = Work and gotten a d from that like i wrote in my post?

6. Nov 23, 2016

### haruspex

That looks unnecessarily complicated. You're just saying the thrust is a linear function of height, right?
Might be easier to think of it backwards, taking off at max thrust, turning thrusters off at some height h, having reached escape velocity, or whatever.
At height x, acceleration is A-Bx. You can write that as a differential equation, yes?

7. Nov 23, 2016

### Cutter Ketch

Your equation for thrust vs height is wrong if for no other reason than it doesn't depend on h. h doesn't appear in the equation!

8. Nov 23, 2016

### Daquicker

I edited it so the variable names match.

9. Nov 23, 2016

### haruspex

10. Nov 23, 2016

### Daquicker

The simulation this has to work under is rather symplistic, the atmosphere does scale linearly. The thing I'm stuck on is building up this differential equation. Most, if not all of the information I come across describes acceleration in terms of either time, or velocity. Doing it in terms of position doesn't seem to be very popular.

11. Nov 23, 2016

### haruspex

Did you read the last part of my post #6?

12. Nov 24, 2016

### Daquicker

Yes, and I agree looking at the problem "from the bottom up" seems like the sensible thing to do as since it gives you a set starting point. Describing a in terms of altitude (xSea) with form A-Bx because it is a linear function is also ok, can do that.

=> a(xSea) = gPlanet - (cMaxThrustSea - (cMaxThrustSea / 100) * ((100 / 0.7) * (1 - (1 - xSea / 14400)))

with: xSea variable for altitude, cMaxThrust a constant for the particular craft (filling it in would shorten the function but also make it less generic, which is not what I need), and gPlanet the acceleration due to the planet's gravity. This Will return some negative acceleration value given that the thrust is sufficient to overcome the gravity of the planet.

It's the next part I can't get done atm.

The only thing I can come up with is:

vCraft^2 = vCraftInit + 2∫(xSurf -> X) a(X) dx

But I think this isn't correct.

13. Nov 24, 2016

### haruspex

So write that as a differential equation.

14. Nov 24, 2016

### Daquicker

I'll give it another shot tomorrow after a couple hours of sleep, this isn't working after a 26h day.
Thankyou for your time so far though :)

15. Nov 24, 2016

### Daquicker

The only thing I can come up with is:

vCraft^2 = vCraftInit + 2∫(xSurf -> X) a(X) dx

But I think this isn't correct.

16. Nov 24, 2016

### haruspex

You are making it quite unnecessarily complicated to write and read the equations by using a computer language style. Much easier in normal algebra.
If x is the distance variable, and t is time, how does one normally write the acceleration in calculus notation?
If the acceleration is presumed to be A-Bx for some constants A and B, what differential equation does that give you?

17. Nov 24, 2016

### Daquicker

With x: distance, t: time, a: acceleration, v: velocity

as far as I know

with a constant -> integrate a to get v
=> v = ∫ a dt
=> v = v0 + at -> integrate v to get x
=> x = ∫ v dt
=> x = ∫ (v0 + at) dt
=> x = x0 + v0t + (at2)/2
<=> a = 2 * (x - x0 - v0t) / t2

with a variable:
a(t)
=> v(t) = v0 + ∫(0->t) a dt'
=> x(t) = x0 + ∫(0->t) v dt'

Sadly I still fail to see how this helps me, I'm probably just missing something really big...

Last edited: Nov 24, 2016
18. Nov 24, 2016

### haruspex

As you pointed out, acceleration will not be not constant.
When I ask a sequence of questions, I am trying to guide you through one question at a time.
Please make an attempt at this very general question:
Have you actually studied differential equations?

19. Nov 24, 2016

### Daquicker

Last time I had to use one was at least 4 years ago, which is clearly showing.

anyways, with position x(t)
=> v(t) = dx/dt
=> a(t) = dv/dt = d2x/dt2

20. Nov 24, 2016

### haruspex

Right.
Believing that the acceleration should be of the form A-Bx, what differential equation can you write?

21. Nov 24, 2016

### Daquicker

$\frac {d^2x}{dt} = A - Bx$
$<=> \frac {d^2x}{dt} + Bx = A$

Then get a characteristic equation for $\frac {d^2x}{dt} + Bx = 0$ I suppose?
Char. Eq.: $r^2 + Bx = 0 <=> r = \sqrt{Bx}$ or $r = - \sqrt{Bx}$
$=> x(t) = C_1 e^{\sqrt{Bx}t} + C_2 e^{- \sqrt{Bx}t}$

Then as since A is a constant, try x(t) = Y with Y a constant, as the "particular solution" of the entire equation?
$D^2(Y) = 0 => BY = A <=> Y = \frac {A}{B}$

$x(t) = C_1 e^{\sqrt{Bx}t} + C_2 e^{- \sqrt{Bx}t} + \frac {A}{B}$

This anywhere near correct?

22. Nov 24, 2016

### haruspex

Not bad, but you made a sign error. You should have got √(-B). More simply, try trig functions.
You might have recognised that the homogeneous part of the equation is SHM.