Answers

U10.3 Cross-track error

Following a trajectory · University · about 35 min

What this lesson is about

Splitting the error into along and across the path, and a controller for the half that matters.

• The path frame
• Controlling it
• Choosing the gain
• The dead band, again

Questions 7 marks in all

1. [1 mark]A leg runs from (0, 0) to (30, 40) and the robot is at (10, 20). What does this print?

   import math
   ax, ay, bx, by = 0.0, 0.0, 30.0, 40.0
   x, y = 10.0, 20.0
   length = math.hypot(bx - ax, by - ay)
   ux, uy = (bx - ax) / length, (by - ay) / length
   nx, ny = -uy, ux
   along = (x - ax) * ux + (y - ay) * uy
   cross = (x - ax) * nx + (y - ay) * ny
   print(round(along, 2), round(cross, 2))

   Answer:

   22.0 4.0

   u = (0.6, 0.8) and the left normal is (-0.8, 0.6). along = 6 + 16 = 22 cm, cross = -8 + 12 = +4 cm, so the robot is 4 cm to the left of the path.

2. [1 mark]The line runs from (60, 30) to (60, 170) and the robot, facing +y, is at (70, 50). With correction = clamp(-0.9 x cross, 12) along the left normal, and drive()'s sideways term positive to the robot's right at 15 cm/s for 100, what sideways term should be sent?

   1. A-60, because the correction of +9 cm/s is along the left normal (-1, 0), which is the robot's left
   2. B+60, the correction of +9 cm/s converted to percent and sent straight in
   3. C+9, the correction in cm/s
   4. D-80, the clamped maximum
   Answer: A. cross = (70 - 60) x (-1) = -10, so correction = +9 cm/s along (-1, 0). The sideways term is 100 x 9 x nx / 15 = -60. Sending +60 straight in pushes the robot further from the line.

3. [1 mark]The holonomic cross-track loop has gain k = 0.9 per second and the robot moves along the path at 11 cm/s. Over how many centimetres of travel is one time constant of the correction spread? Give 1 decimal place.

   Answer: 12.2 (accept within 0.2). The time constant is 1 / k = 1.11 s, and 1.11 s at 11 cm/s is 12.2 cm of travel.

4. [1 mark]The robot sits 1 cm off the path with k = 0.9. What percentage command does the correction ask for, given 100 percent is 15 cm/s sideways?

   Answer: 6 (accept within 0.1). 0.9 x 1 = 0.9 cm/s, and 100 x 0.9 / 15 = 6 percent, well inside the 15 percent dead band, so nothing moves.

5. [1 mark]The holonomic cross-track loop is first order and cannot oscillate on its own. Why can it still wobble in practice?

   1. ADelays: the 0.25 s drive lag, the loop period and the dead band
   2. BAdding a sideways velocity makes the loop second order
   3. CThe cross-track error is signed
   4. DThe along-track error feeds into the cross-track error
   Answer: A. Velocity to position is one integrator. Everything that makes it oscillate is a delay, which adds phase lag.

6. [1 mark]What happens as the cross-track gain k is raised well above 0.9?

   Tick every answer that is true.

   1. AThe correction saturates at the available sideways speed, so the response stops getting faster
   2. BThe phase lost to the 0.25 s lag starts to matter and the approach becomes a wobble
   3. CThe loop becomes second order, like a car's steering
   4. DThe time constant 1 / k gets longer
   Answer: A, B. Only 15 cm/s of sideways speed exists, and the lag costs phase. The loop order does not change, and 1 / k gets shorter, not longer.

7. [1 mark]Why does the Stanley controller for a car divide the cross-track term by the speed?

   1. AAt higher speed the geometry closes a given error sooner, so the same error needs a gentler steering correction
   2. BFaster cars have larger cross-track errors, so the term needs shrinking
   3. CIt converts the error from centimetres to degrees
   4. DIt stops the steering saturating at low speed
   Answer: A. Without the speed scaling, a gain that is calm at walking pace becomes unstable on the motorway.

The task: drive the cross-track error to zero

from bugbot import *
import math
connect()

DT = 0.1
START = (85.0, 30.0)
AX, AY, BX, BY = 60.0, 30.0, 60.0, 170.0

from bugbot import *
import math
connect()

DT = 0.1
V_MAX, V_LAT = 20.0, 15.0
START = (85.0, 30.0)
AX, AY, BX, BY = 60.0, 30.0, 60.0, 170.0
CRUISE, K_CROSS = 11.0, 0.9

ux, uy = BX - AX, BY - AY
LENGTH = math.hypot(ux, uy)
ux, uy = ux / LENGTH, uy / LENGTH            # along the path
nx, ny = -uy, ux                             # across it

def band(u):
    """below about 15 percent the drive does nothing at all, so ask for a little more or for nothing"""
    if abs(u) < 5.0:
        return 0.0
    return max(17.0, min(100.0, u)) if u > 0 else min(-17.0, max(-100.0, u))

def steer(wx, wy):
    """a velocity in the world, through the inverse kinematics, as a command"""
    h = math.radians(heading())
    vx = wx * math.cos(h) - wy * math.sin(h)
    vy = wx * math.sin(h) + wy * math.cos(h)
    e = (0.0 - heading() + 180) % 360 - 180
    rot = 0.0 if abs(e) < 3 else max(-30.0, min(30.0, 2.0 * e))
    drive(band(100 * vy / V_MAX), band(100 * vx / V_LAT), band(rot))

cross = 0.0
for tick in range(500):
    px, py = position()
    x, y = START[0] + px, START[1] + py
    along = (x - AX) * ux + (y - AY) * uy
    cross = (x - AX) * nx + (y - AY) * ny
    plot("cross track", cross)
    if along > LENGTH - 5:
        break
    speed = min(CRUISE, 0.8 * (LENGTH - along))
    correct = max(-12.0, min(12.0, -K_CROSS * cross))
    steer(ux * speed + nx * correct, uy * speed + ny * correct)
    wait(DT)
stop()
wait(0.4)
print("cross track:", round(cross, 2))