Thrust Model

The thrust system in Moonlander is modeled as a first-order dynamic system. The actual thrust approaches the target thrust exponentially, with a time constant representing the engine's response delay.

Assumptions:

The engine cannot change its thrust instantaneously. Instead, it behaves like a first-order system (PT1 / first-order lag) with time constant $\tau$ .

This model class is the same as:
- RC circuits
- Thermal inertia
- Motor spin-up dynamics
- Low-pass filters

Physical / System Formulation:

The rate of change of thrust is proportional to the difference between commanded and actual thrust:

\dot{F} \propto (F_{cmd} - F)

Introducing the proportionality constant

\frac{1}{\tau}

\dot{F} = \frac{1}{\tau} (F_{cmd} - F)

Interpretation:
- $F_{cmd}$ : commanded thrust [N]
- $F$ : actual thrust [N]
- $\tau$ : engine response time / inertia [s]
Limited cases:
- $\tau \rightarrow 0$ : instantaneous response
- Large $\tau$ : slow engine response

Discrete Form (from ODE)
Start from:

\dot{F} = \frac{(F_{cmd} - F)}{\tau}

Assuming

F_{cmd}

is constant over a timestep

\Delta t

, the solution is:

F(t + \Delta t) = F(t) e^{-\Delta t / \tau} + F_{cmd} (1 - e^{-\Delta t / \tau})

Rearranged:

F_{k + 1} = F_k + (1 - e^{-\Delta t / \tau})(F_{cmd, k} - F_k)

Step-by-Step Solution of the Differential Equation

We solve the first-order ODE:

\dot{F} = \frac{F_{cmd} - F}{\tau}

Step 1: Rearrange
Bring into standard linear form:

\dot{F} + \frac{1}{\tau} F = \frac{F_{cmd}}{\tau}

Denote:

a = \frac{1}{\tau}, \quad b = \frac{F_{cmd}}{\tau}

Step 2: Homogeneous Solution
Solve

\dot{F}_h + a F_h = 0

Using exponential ansatz:

F_h(t) = C e^{-a t} = C e^{-t/\tau}

Step 3: Particular Solution
Since the forcing term is constant, try $F_p = k$ .
Substitute into ODE:

0 + a k = b \implies k = F_{cmd}

Step 4: General Solution

F(t) = F_h(t) + F_p = C e^{-t/\tau} + F_{cmd}

Step 5: Apply Initial Condition
At $t = 0$ , $F(0) = F_0$ :

F_0 = C + F_{cmd} \implies C = F_0 - F_{cmd}

Step 6: Evaluate at t + Δt

F(t + \Delta t) = (F_0 - F_{cmd}) e^{-\Delta t/\tau} + F_{cmd}

Expanding using

F_0 = F(t)

F(t + \Delta t) = F(t) e^{-\Delta t / \tau} + F_{cmd} (1 - e^{-\Delta t / \tau})

Step 7: Intuition
- The system always moves toward $F_{cmd}$ .
- The speed of convergence is governed by $\tau$ .
- Smaller $\tau$ → faster response.
- Larger $\tau$ → slower response.

Illustration

The figure below shows the thrust approaching the target over time.

$Thrust response curve$

Fuel Consumption

Specific impulse is defined as:

I_{sp} = \frac{F}{\dot{m} g_0}

Where:
- $I_{sp}$ : specific impulse [s]
- $F$ : thrust [N]
- $g_0 \approx 9.81$ : standard gravity [m/s²]

Solving for mass flow rate:

\dot{m} = \frac{F}{I_{sp} g_0}

Fuel mass decreases:

\dot{m}_f = -\frac{F}{I_{sp} g_0}

Typical values for common engines:

Engine Type	Fuel / Propellant	$I_{sp} \, [s]$
Liquid Rocket	LOX/LH2	450–465
Liquid Rocket	LOX/Kerosene	300–350
Solid Rocket	HTPB / Black Powder	200–300
Ion / Electric	Xenon, Hall / Electrostatic	1500–4000
Hybrid Rocket	HTPB + N₂O	250–300

Key Points

First-order exponential response to target thrust
Real-time fuel consumption derived from thrust and specific impulse
Time step update allows smooth and stable simulation