Answers

U11.1 The pinhole camera

Vision · University · about 30 min

What this lesson is about

One pixel is one ray. What that buys you, and what it does not.

• The model
• What a pixel is worth
• The tangent matters
• A bearing is not a position
• Getting the depth back

Questions 7 marks in all

1. [1 mark]The BugBot's camera is 320 pixels wide with a 120 degree field of view. What does this print?

   import math
   f = (320 / 2) / math.tan(math.radians(120 / 2))
   bearing = math.degrees(math.atan((240 - 160) / f))
   print(round(f, 1), round(bearing, 1))

   Answer:

   92.4 40.9

   f = 160 / tan(60 degrees) = 92.4 px, and column 240 is atan(80 / 92.4) = 40.9 degrees right of the nose.

2. [1 mark]Which line works out the focal length in pixels correctly?

   1. Af = 160 / math.tan(math.radians(60))
   2. Bf = 160 / math.tan(60)
   3. Cf = 160 * math.tan(math.radians(60))
   4. Df = 320 / math.tan(math.radians(120))
   Answer: A. math.tan takes radians, so 60 must be converted first. Half the width goes with half the field of view, and the focal length divides by the tangent.

3. [1 mark]A blob's centre is at column 200. What is its bearing from straight ahead, in degrees to 1 decimal place, using f = 92.4?

   Answer: 23.4 (accept within 0.1). bearing = atan((200 - 160) / 92.4) = atan(0.433) = 23.4 degrees.

4. [1 mark]The linear guess bearing = (u - 160) x 0.375 degrees is used instead of the pinhole formula. How many degrees wrong is it at column 200? Give 1 decimal place.

   Answer: 8.4 (accept within 0.1). The linear guess gives 40 x 0.375 = 15.0 degrees, the pinhole gives 23.4, so it is 8.4 degrees short. They only agree at the centre and the very edge.

5. [1 mark]Why can a single pixel not tell you where an object is?

   1. ADepth appears only as a divisor, so every point along a ray lands on the same pixel
   2. BThe image is too low resolution to locate the object
   3. CThe principal point is not exactly at the centre
   4. DLens distortion moves the pixel
   Answer: A. Doubling X and Z gives the same u. A pixel names a direction, and the range has to come from a second constraint.

6. [1 mark]Which of these add the second constraint needed to get depth from a camera?

   Tick every answer that is true.

   1. AKnowing the object's real size
   2. BKnowing the object is on a flat floor, with the camera's height and tilt
   3. CA second camera a known distance away
   4. DThe same camera at two times, with the motion between them known
   5. EAveraging the bearing over many frames
   6. FA higher resolution sensor
   Answer: A, B, C, D. All four add a second ray or a known size. Averaging and more pixels make the bearing better but still give only a direction.

7. [1 mark]A tag is seen at column 220 and the detector reports it 50 cm away. What does this print?

   import math
   F = 92.4
   bearing = math.degrees(math.atan((220 - 160) / F))
   print(round(bearing, 1), round(50 * math.sin(math.radians(bearing)), 1), round(50 * math.cos(math.radians(bearing)), 1))

   Answer:

   33.0 27.2 41.9

   The bearing is atan(60 / 92.4) = 33.0 degrees, so the tag is 50 sin 33.0 = 27.2 cm right and 50 cos 33.0 = 41.9 cm ahead.

The task: turn a pixel into a bearing

from bugbot import *
import math
connect()

F = 92.4
set_cv("apriltag")
wait(0.3)

from bugbot import *
import math
connect()

F = 92.4                      # focal length in pixels: 160 / tan(60 degrees)

set_cv("apriltag")
wait(0.3)
tags = [t for t in apriltags() if t[0] == 7]
tag_id, cx, cy, dist = tags[0]

bearing = math.degrees(math.atan((cx - 160) / F))
across = dist * math.sin(math.radians(bearing))
print("bearing:", round(bearing, 2))
print("across:", round(across, 1))