06. A simple two-dimensional rocket

A vehicle can change its acceleration direction but not its magnitude. Starting at rest, how can it reach horizontal flight at height 1 m as quickly as possible? This is a simplified rocket guidance problem without gravity or air.

Item	Definition
States	Horizontal and vertical velocities $u,v$; position $x,y$
Control	Acceleration angle $\beta$ from horizontal
Objective	Minimize $t_f$
Phases	One phase, 20 nodes, free final time
New concepts	Two-dimensional motion and direction as a control

Situation

Orbital ascent requires building height, then horizontal speed, and reaching the target with no vertical velocity. Here we isolate the steering question:

• Motion is planar, without gravity or drag.
• Acceleration magnitude is fixed at $a=1$ m/s².
• The vehicle starts at rest and must finish at $y=1$ m with $u=1$ m/s and $v=0$.
• Final horizontal position is free.

Accelerating only upward gains height but leaves vertical speed and no horizontal speed. Accelerating only sideways gains no height. Steering must meet both requirements together.

Formulation

Dynamics

$$\dot u=a\cos\beta,\qquad\dot v=a\sin\beta,\qquad\dot x=u,\qquad\dot y=v.$$

Angle $\beta=0$ points horizontally forward; $90^\circ$ points up; negative angles point downward.

Boundary conditions

$$u(0)=v(0)=x(0)=y(0)=0,\qquad u(t_f)=1,\quad v(t_f)=0,\quad y(t_f)=1.$$

There is no condition on $x(t_f)$.

Constraints, objective, and parameter

$$-\frac\pi2\le\beta\le\frac\pi2,\qquad J=t_f,\qquad a=1\ \mathrm{m/s^2}.$$

The acceleration direction stays in the forward half-plane.

Initial guess

For time 0 to 2 s, let $s=t/2$:

• $u=s$ and $x=s^2$.
• $y=3s^2-2s^3$ and $v=3s(1-s)$.
• $\beta=\arctan\bigl(3(1-2s)\bigr)$, turning from upward to downward.

Results

Trajectory in blue, node positions in white, and acceleration directions as orange arrows. A black circle and square mark the start and finish.

Position x and height y (top), horizontal and vertical velocities (middle), and acceleration angle β (bottom). Red dotted lines mark ±90°.

The vehicle's trajectory and direction beside velocity and angle histories. The black vertical cursor marks the current time.

• Minimum duration: 2.085 s.
• Final horizontal position: 1.043 m.
• Angle varies smoothly from 1.35 rad (about 77°) to -1.34 rad (about −77°).

Explanation

Gain height, then cancel vertical speed

Early acceleration points almost upward to build vertical speed. Later it points downward to cancel that speed exactly at $y=1$. Horizontal speed grows most rapidly around the middle, when $\beta$ passes through zero. The fixed acceleration is allocated between reaching the height target and reaching 1 m/s horizontally. Without gravity, the upward and downward portions are nearly symmetric.

The linear tangent law

For this problem, optimal steering has the form

$$\tan\beta(t)=c_1-c_2t.$$

This linear tangent law underlies practical rocket guidance. The free final horizontal position makes the denominator constant in this case. The computed $\tan\beta$ is almost linear, while $\beta$ itself forms an S-shaped curve, changing slowly near its endpoints and quickly near the middle. The initial guess has the same general form.

Indirect methods derive this property from optimality conditions. OpenGoddard obtains it directly from the dynamics and constraints without requiring the steering law in advance.

Try it

• Set PARAMETERS["a"] to 2 and compare duration, trajectory, and steering.
• Add a final horizontal-position equality. The tangent of the steering angle is no longer a linear function of time.
• Change the node count from 20 to 10 or 40 and compare the stable digits of the minimum duration.

Run

uv run python examples/06_simple_rocket_2d.py

View the Python source on GitHub.