Answers

U11.3 Range from apparent size

Vision · University · about 35 min

What this lesson is about

A known object gives a range, and the range gets worse as the square of itself.

• The relation
• How bad it gets, and how fast
• Detection versus measurement
• Where this is the right tool anyway

Questions 6 marks in all

1. [1 mark]An 8 cm ball appears 20 pixels wide on a camera with f = 92.4 px. How far away is it, in cm to 1 decimal place?

   Answer: 37.0 (accept within 0.1). d = f S / w = 92.4 x 8 / 20 = 36.96 cm.

2. [1 mark]How many centimetres of range is one pixel of width worth for the 8 cm ball at 50 cm and at 100 cm?

   F, S = 92.4, 8.0
   for d in (50, 100):
       print(d, round(d * d / (F * S), 1))

   Answer:

   50 3.4
   100 13.5

   One pixel is worth d^2 / (f S): 2500 / 739 = 3.4 cm at 50 cm and 10000 / 739 = 13.5 cm at 100 cm. Doubling the range quadruples the error.

3. [1 mark]Why does range from apparent size get so much worse with distance?

   1. AThe apparent width falls as 1 / d, so one pixel is a larger fraction of it, and the range error grows as d squared
   2. BThe detector adds more noise to far objects
   3. CThe lens distorts distant objects
   4. DFar objects are nearer the edge of the frame
   Answer: A. dd/dw = -d^2 / (f S). Nothing about the algorithm changes with range; the geometry does.

4. [1 mark]Beyond what range is one pixel of width worth more than 10 percent of the range itself, for the 8 cm ball and f = 92.4? Give the range in cm to 1 decimal place.

   Answer: 73.9 (accept within 0.2). d^2 / (f S) > 0.1 d when d > 0.1 f S = 0.1 x 92.4 x 8 = 73.9 cm.

5. [1 mark]A flat tag 40 cm away is seen at 60 degrees rather than square on, and its range is worked out from the width of its bounding box. What range is reported, in cm?

   Answer: 80 (accept within 0.5). At 60 degrees the tag's width is cos 60 = half its true width, so d = f S / w reads double: 80 cm.

6. [1 mark]The box width and the area of the far ball give ranges about 3 cm apart (123.2 and 126.1 cm), and repeated frames give the same numbers. Which statements are right?

   Tick every answer that is true.

   1. AThis is quantisation of the box edges, not noise
   2. BAveraging repeated frames will not remove it
   3. CThe area is a sum over many pixels, so it is quantised far more finely than an edge
   4. DAveraging a few dozen frames will remove the disagreement
   5. EThe pinhole model does not apply to a far ball
   Answer: A, B, C. A box edge is an integer, and the same error comes back every frame. Area is a finer measurement of the same size.

The task: range from apparent size

from bugbot import *
connect()

F = 92.4
R = 4.0
set_cv("blob", "red")
wait(0.3)

from bugbot import *
connect()

F = 92.4
R = 4.0                       # the balls are 8 cm across

set_cv("blob", "red")
wait(0.3)
seen = blobs()

ranges = []
for cx, cy, area, x0, y0, x1, y1, aspect in seen:
    w = x1 - x0
    ranges.append(2 * R * F / w)

print("near:", round(ranges[0], 1))
print("far:", round(ranges[1], 1))

# d = 2*R*F/w, so dd/dw = -d*d / (2*R*F): one pixel is worth this many cm out there
print("per pixel:", round(ranges[1] ** 2 / (2 * R * F), 1))