How to convert the distance between object and camera from pixel for meter?

Question:

I need to convert the distance between object and camera from pixels to meter/cm frame by frame and then calculate the speed of moving object. In the first frame, the distance of object from camera is 4 meter, and Focal length FL 8mm, width of the object is 0.0373 meters. What I did first to calculate the projection of object:

focal_length_px = 8000/4.8  #Focal length(px): 8mm / (4.8 µm / px) = 1667px =: f

Object :

    object_reall_width = 0.0373 # Width: 0.0373m
    distance = 4       # Distance: 4m
    obj_width_px = (object_reall_width/distance) * focal_length_px  # Projection of object(px): (0.0373 m / 4 m) * focal_length_px = 15.5 px

After object detection, I want to calculate the distance. I used the distance_finder() code in this link.

def distance_finder(focal_length, real_obj_width, obj_width_in_frame):  
    distance = (real_obj_width * focal_length) / obj_width_in_frame  
    return distance

and the output is:

obj_width_in_frame = 17
obj_dst = distance_finder(focal_length_px, obj_width_px, obj_width_in_frame)  # obj_dst = 1523.6928104575165

My question is: how to convert the object distance in the frame from the camera (obj_dst = 1523.69) from pixels to meters/cm?

and then calculate the speed of object in the frame.

Asked By: S.EB

||

Answers:

obj_dst = 1523.69 
# [cm] = [px]* 2.54 / DPI 
distance = obj_dst*2.54/96
# 96 is DPI, 2.54 is inch to cm
print(distance,"cm")
Answered By: kimhyunju

We discussed the math before. Now you just need to rearrange some equations.

Pinhole camera model.

  • Focal length (mm): 8 mm
  • Sensor pixel pitch: 4.8 µm/px
  • Focal length (px): 8 mm / (4.8 µm/px) = 1667 px =: f

Object:

  • Physical width: 0.0373 m
  • Apparent width: 17 pixels

Solve for physical distance, let’s call it d.

  • (0.0373 m / d) * f = 17 px, which we already had (replace previous 4 meters by d, insert 17 px as known)
  • d = (0.0373 m) / (17 px / f) = 0.0373 m * f / 17 px

=> d ~= 3.66 meters

Answered By: Christoph Rackwitz