Answers

U3.6 An error budget

Odometry and drift · University · about 25 min

What this lesson is about

Predicting how far out you will be before you drive, which is what covariance is for.

• A budget for a 2 m straight run
• Adding the pieces up
• This is covariance, informally

Questions 6 marks in all

1. [1 mark]This adds up the lesson's error budget for a 2 m run with a calibrated gyro, then again with the noise term removed. What does it print?

   import math
   pieces = {"heading": 0.4, "scale": 6.0, "noise": 0.7, "slip": 2.0}
   total = math.sqrt(sum(v * v for v in pieces.values()))
   print("total:", round(total, 1), "cm")
   print("without noise:", round(math.sqrt(total ** 2 - 0.7 ** 2), 1), "cm")

   Answer:

   total: 6.4 cm
   without noise: 6.3 cm

   0.4² + 6² + 0.7² + 2² = 40.65, whose square root is 6.4. Removing the 0.7 cm noise term leaves 6.3: the total moves by less than half a millimetre, so a term much smaller than the largest is not worth working on.

2. [1 mark]Two independent errors of 3 cm each. What is the total, in cm to one decimal place?

   Answer: 4.2 (accept within 0.05). Independent errors add in quadrature: the square root of 3² + 3² = 18 is 4.2 cm, not 6.

3. [1 mark]An uncalibrated gyro bias of 0.19 deg/s acts for 10 s while the robot drives 200 cm at a steady speed, so the heading error grows steadily from 0 to 1.9 degrees. Using d × e with the average heading error, how far to the side does it end up, to the nearest cm?

   Answer: 3 (accept within 0.5). The heading error averages 0.95 degrees over the run, which is 0.0166 radians, and 200 × 0.0166 = 3.3 cm sideways. Using the final 1.9 degrees for the whole run would double it: the robot was only that far out at the very end.

4. [1 mark]Flow noise contributes 0.7 cm over a run of 100 steps. Following the random walk, what does it contribute over 400 steps, in cm?

   Answer: 1.4 (accept within 0.05). Four times as many steps multiplies a random walk by the square root of 4, which is 2: 0.7 × 2 = 1.4 cm.

5. [1 mark]With the gyro calibrated, the budget reads heading 0.4 cm, scale 6 cm, noise 0.7 cm and slip 2 cm. What should be worked on next?

   1. AThe flow scale error
   2. BThe noise, by averaging more readings
   3. CThe slip
   4. DThe gyro bias again
   Answer: A. Scale is by far the largest term in quadrature. Halving the noise would change the total by less than a millimetre.

6. [1 mark]For that calibrated budget after a long straight run, what shape is the region the robot is likely to end up in?

   1. AAn ellipse, longest along the direction of travel, because the scale error dominates
   2. BA circle, because the errors are added in quadrature
   3. CAn ellipse, longest across the direction of travel, because heading error dominates
   4. DA line along the direction of travel, because heading error is zero after calibration
   Answer: A. Scale error stretches the estimate along the path and heading error throws it sideways. Here the 6 cm scale term outweighs the 0.4 cm heading term, so the ellipse is long and thin along the run.

The task: an error budget

from bugbot import *
import math
connect()

DISTANCE = 100.0

from bugbot import *
import math
connect()

DISTANCE = 100.0
HEADING_ERR = 0.7          # degrees: 0.19 deg/s over the 7 s drive, averaged
SCALE_ERR = 0.03           # the flow sensor reads about 3 percent high

sideways = DISTANCE * math.radians(HEADING_ERR)
along = DISTANCE * SCALE_ERR
print("predicted error:", round(math.hypot(sideways, along), 1))

forward(75)
wait(7.0)
stop()
wait(0.6)
ox, oy, oh = odometry()
tx, ty = position()
print("actual error:", round(math.hypot(ox - tx, oy - ty), 1))