PYTHON · NUMPY

Two-impulse rendezvous

You solve for a two-impulse transfer and shorten the flight time. What happens to the total delta-v?

Solve for the burn that flies a chaser to the target in a chosen time, price both burns, and find the faster-costs-more tradeoff.

01Challenge

Given a relative state and a transfer time, solve for the departure velocity that arrives at the target, then null the arrival velocity for a soft berth. Sweep the transfer time and read the delta-v tradeoff.

Predict firstYou solve for a two-impulse transfer and shorten the flight time. What happens to the total delta-v?

02Model

Rendezvous is a targeting problem. You have a relative position now and you want to be at the target after a chosen transfer time. The CW state-transition matrix gives that as a linear equation: the position blocks say final position equals rr times start plus rv times departure velocity. Set the final position to zero and invert the rv block, and you have the exact departure velocity. Fire that, coast, and you arrive; then a second burn cancels the arrival velocity so you berth softly instead of colliding. The interesting result is the tradeoff: a slower transfer usually costs less delta-v, so the certified approach corridor the autonomy flies is a compromise between fuel and time, not a race.

03Guided practice

1. The solve. From 0 = rr@r0 + rv@v0, the departure velocity is v0 = -inv(rv) @ (rr@r0).

2. Two burns. dv1 injects onto the transfer (from your current relative velocity); the arrival velocity is vr@r0 + vv@v0, and dv2 nulls it.

3. The tradeoff. Sweep the transfer time and print the total delta-v; slower is usually cheaper.

04Feedback

PASS WHEN PASS once the table prints and the total delta-v falls as the transfer time grows from 5 to ~40 minutes.

If your run fails, check:

If inv(rv) fails, use np.linalg.solve(rv, rr@r0) instead of forming the inverse.
Delta-v here is in m/s and should be small (well under 1 m/s for a 100 m rendezvous); if it's huge, check units.
The arrival velocity uses the vr and vv blocks, not rr and rv.

05Retrieve & space

How do you solve for the departure velocity? (set final position to zero and invert the rv block of the CW state-transition matrix).
Why is a second burn needed? (to null the arrival velocity so the chaser berths softly instead of colliding).
What is the fuel-versus-time tradeoff? (slower transfers usually cost less delta-v).

06Mastery & project

You can solve a two-impulse rendezvous with CW and explain why the certified corridor trades fuel against time.

When you plan a maneuver, price it in delta-v as a function of time, not just feasibility: the cheapest safe transfer is usually not the fastest.

GO FURTHER · THE LIVE PLAYGROUND

On-orbit capture (live) ↗ · See the corridor the solved maneuver flies.

← Relative motion: Clohessy-Wiltshire Estimating a non-cooperative target →

PYTHON · NUMPY

Two-impulse rendezvous

Implement two-impulse targeting and sweep transfer time against total delta-v.

You read

the arrays and values already in scope

You change

the code you write in each cell

Fixed

the dataset and the checks that grade you