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Navigation · Theory · M5

M5-T5 — Stanley and Cross-Track Control

Steer the front wheels to remove the heading error, plus a term that brings the front axle back onto the line.

  • Lesson 5 of 7
  • Module 5: Trajectories and Path Following
  • Theory
  • About 8 minutes to read

Pure pursuit needs a lookahead and a point ahead. Stanley, the controller of the car that won the 2005 DARPA Grand Challenge, needs neither. It measures two errors at the front axle and steers against both. Its behaviour is easy to analyse, which is why it is the textbook companion to pure pursuit.

Learning objectives

After this lesson, you should be able to:

  • Write the Stanley steering law from the heading error and the cross-track error.
  • Explain the role of speed and of the softening constant, and show that the cross-track error decays exponentially.
  • State what reference point Stanley uses and what that does on a bend.
  • Compare Stanley with pure pursuit on what each needs from the path and the speed.

The intuition

Suppose the front wheels point exactly along the path but sit a little to its left. To get back, you steer slightly right. The further away you are, the harder you steer, but never so hard that you turn across the path: the steering from the offset alone tops out at 90°. Now suppose you are on the line but pointing away from it. Steer back by exactly the angle you are off. Stanley adds the two.

The offset term is divided by the speed. At high speed a small steering angle moves the robot sideways quickly, so less steering is needed for the same correction; at low speed it takes more.

Stanley geometryA horizontal path. The front axle of the robot is below the path, with its heading turned slightly down from the path direction. The perpendicular distance to the path is the cross-track error e and the angle between the heading and the path is the heading error psi.path e path direction (parallel)heading steering δ = ψ + atan(k e / (k_s + v)) front axle ψ e > 0: the path is on the robot’s left. ψ > 0: the path points to the left of the heading.
Stanley measures a heading error ψ and a cross-track error e at the front axle and steers by ψ plus a term that grows with e and shrinks with speed.

The mathematics

The law

Stanley steering law
δ=ψ+arctan⁡(k⋅eks+v)\delta =\psi +\arctan (\frac{k\cdot e}{k_{s}+v})
then clamp ∣δ∣≤δmax⁡\text{then clamp}\,\lvert \delta \rvert \le \delta _{\max}
ψ
heading error: path direction minus robot heading, wrapped into −π..π.
e
cross-track error: the distance of the front axle from the nearest point of the path, positive when the path is on the robot’s left.
v
speed in m/s.
k
gain of the cross-track term, in 1/s. Larger values converge faster.
ks
softening constant, in m/s, so that the formula stays finite at v = 0.

The sign convention is what makes the law work. With e positive when the path is on the left, both terms steer left when the robot needs to go left. Using the opposite convention for e needs a minus sign in front of the arctangent.

Why the error decays

When the heading error is zero, the front axle moves sideways at a rate ė = −v·sin(δ). With δ = atan(k e / (ks + v)):

Cross-track error dynamics
e˙=−v⋅sin⁡(arctan⁡(k eks+v))≈−(k⋅vks+v)⋅e  for small e\dot{e}=-v\cdot \sin (\arctan (\frac{k\,e}{k_{s}+v}))\approx -(\frac{k\cdot v}{k_{s}+v})\cdot e\;\mathrm{for\ small\ e}

So the error decays exponentially with rate k·v/(ks + v), which is close to k at high speed. For a large error the sine saturates at 1 and the robot closes in at a speed of at most v.

Measuring the errors

Find the nearest point of the path as a polyline (not the nearest waypoint) and take the direction of that segment. With sparse waypoints, a distance to the nearest waypoint can be metres off (lab M5-L4, stage 2). The heading difference must be wrapped; subtracting angles without wrapping makes the robot steer hard the wrong way near ±π (stage 1).

The reference point

Stanley is defined for the front axle of a car-like robot. If the pose you track is the rear axle (as in most odometry), compute the front axle position as the rear axle plus the wheelbase along the heading. If the front axle follows a circle of radius Rf, the rear axle follows a smaller one:

Front and rear axle on a bend
Rrear=Rf2−wheelbase2R_{\text{rear}}=\sqrt{R_{f}^{2}-\mathrm{wheelbase}^{2}}
inward shift≈wheelbase22⋅Rf\text{inward shift}\approx \frac{\mathrm{wheelbase}^{2}}{2\cdot R_{f}}

A worked example

One command. Heading error ψ = 0.10 rad, cross-track error e = 0.30 m, speed v = 1.0 m/s, k = 1.5 1/s and ks = 0.5 m/s:

Worked Stanley step
arctan⁡(1.5⋅0.300.5+1.0)=arctan⁡(0.30)=0.2915 rad\arctan (\frac{1.5\cdot 0.30}{0.5+1.0})=\arctan (0.30)=0.2915\,\mathrm{rad}
δ=0.1000+0.2915=0.3915 rad=22.4∘\delta =0.1000+0.2915=0.3915\,\mathrm{rad}=22.4^{\circ}

At a standstill (v = 0) with the same errors the cross-track term is atan(0.45 / 0.5) = 0.733 rad = 42.0°. That is beyond a 35° steering limit, so the command is clamped. Without ks the term would be atan(∞) = 90° for any error at all.

How fast does it converge? At v = 1.0 m/s the rate is 1.5 × 1.0 / 1.5 = 1.0 per second, a time constant of one second, one metre of travel. Starting 0.5 m off the line, three time constants (3 s, 3 m) leave 0.5 × e−3 = 0.025 m. Integrating the full nonlinear equation gives 0.026 m. At 2 m/s the rate is 1.2 per second: a time constant of 0.83 s, but 1.7 m of travel, so a faster robot needs more room to settle.

On a bend. Scout’s tug has a wheelbase of 0.8 m. If its front axle runs along a bend of radius 1.5 m, the rear axle runs on a radius of sqrt(1.5² − 0.8²) = 1.269 m, which is 0.231 m inside (the approximation gives 0.213 m). In the simulation of lab M5-L4, with Stanley on a 1.5 m bend at 1.5 m/s, the rear axle peaks 0.18 m from the centre line, against 0.09 m for pure pursuit with a 1.5 m lookahead. Both are small on a 2 m corridor. If the pose reference is the rear axle, track the rear axle or plan with the offset in mind.

Engineering implications and limitations

Pure pursuit or Stanley?

Pure pursuitStanley
Main tuning knobLookahead distance (depends on speed and latency)Gain k and softening ks
Reference pointRear axle (or robot centre)Front axle
Path it needsAny polyline, even roughA path with continuous heading, or the heading error jumps at corners
Behaviour on bendsCuts them as the lookahead growsFront axle accurate, rear axle inside
Typical homeDifferential-drive indoor robotsCar-like vehicles at higher speed

Smooth the path first

At a polyline vertex the path heading jumps by the corner angle. Stanley’s heading error then jumps too, and the command goes to the steering limit at once. Smooth the path (lesson M5-T2) before using Stanley.

Reverse and low speed

The law is defined for forward driving. In reverse the signs change, and near zero speed ks alone decides how hard the robot steers. Test the controller at the slowest speed the stack will use.

Pick gains from a step response

Start the robot 0.5 m off the line on a straight, and read the error. A time constant of about a second is a reasonable first target; ringing means the gain is too high for the latency (lesson M5-T7).

Common misconceptions

“Stanley needs no tuning.”

It has two gains, and the right values depend on speed, latency and the steering rate limit.

“The cross-track term can steer by any amount.”

The arctangent bounds it by 90°, and the clamp bounds the total by the steering limit.

“Stanley follows the rear axle.”

It follows the front axle. Tracked at the rear axle the result is inside every bend.

“Heading error is just the difference of two headings.”

It is the shortest turn between them. Without wrapping, a pair of headings either side of ±π looks like a 2π error.

Engineering takeaways

  1. Stanley: δ = ψ + atan(k·e / (ks + v)), clamped to the steering limit.
  2. Positive e means the path is on the left; both terms then steer left.
  3. With zero heading error the cross-track error decays exponentially at about k·v/(ks + v).
  4. It controls the front axle, so a rear-axle pose cuts bends by about wheelbase²/(2R).
  5. It needs a path with continuous heading; smooth the path first.

References and further reading

The controller of the Stanley vehicle, and a comparison with other path-tracking methods.

  • G. M. Hoffmann, C. J. Tomlin, M. Montemerlo and S. Thrun, “Autonomous Automobile Trajectory Tracking for Off-Road Driving: Controller Design, Experimental Validation and Racing”, American Control Conference, 2007.
  • S. Thrun et al., “Stanley: The Robot that Won the DARPA Grand Challenge”, Journal of Field Robotics, 23(9), 2006.
  • J. M. Snider, “Automatic Steering Methods for Autonomous Automobile Path Tracking”, Carnegie Mellon University Robotics Institute, Technical Report CMU-RI-TR-09-08, 2009.

Knowledge check

Three conceptual questions. Write an answer, then reveal the explanation. Your answers stay in your browser.

Question 1

ψ = 0, e = 0.5 m, v = 2.0 m/s, k = 1.0, ks = 0.5. What is the steering angle?

Explanation

atan(1.0 × 0.5 / 2.5) = atan(0.2) = 0.197 rad = 11.3°.

Question 2

Heading is 3.0 rad and the path direction is −3.0 rad. What is ψ?

Explanation

The difference −6.0 rad wraps to −6.0 + 2π = 0.283 rad. The robot needs a small left turn, not a large right one.

Question 3

Why is the softening constant ks needed?

Explanation

Without it, the cross-track term is atan(k·e / v), which is 90° for any error at v = 0, and 0/0 when both are zero. ks keeps the law finite and gentle at a standstill.

What’s next

Next: controllers that use a model of the robot to choose the steering, and can respect limits.

Practice this lesson

Compare Pure Pursuit and Stanley

Run both controllers on the same route and decide which suits a long straight corridor with tight bends.

Next lesson

M5-T6 — LQR and MPC for Path Tracking

Describe how a linearised error model gives LQR gains and how MPC adds constraints and a horizon.

Nothing below is required to finish this lesson.