英文:
How to convert the distance between object and camera from pixel for meter?
问题
我需要将物体与摄像机之间的距离从像素转换为米/厘米，逐帧计算移动物体的速度。在第一帧中，物体离摄像机的距离是4米，焦距FL为8毫米，物体的宽度为0.0373米。我首先做的是计算物体的投影：
focal_length_px = 8000/4.8 # 焦距(px): 8mm / (4.8 µm / px) = 1667px =: f
Object :
object_reall_width = 0.0373 # 宽度: 0.0373m
distance = 4 # 距离: 4m
obj_width_px = (object_reall_width/distance) * focal_length_px # 物体投影(px): (0.0373 m / 4 m) * focal_length_px = 15.5 px
物体检测后，我想计算距离。我使用了此链接中的distance_finder()代码。
def distance_finder(focal_length, real_obj_width, obj_width_in_frame):
distance = (real_obj_width * focal_length) / obj_width_in_frame
return distance
输出是：
obj_width_in_frame = 17
obj_dst = distance_finder(focal_length_px, obj_width_px, obj_width_in_frame) # obj_dst = 1523.6928104575165
我的问题是：如何将来自摄像机的帧中的物体距离（obj_dst = 1523.69）从像素转换为米/厘米？
然后计算帧中物体的速度。
英文:
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.
答案1
得分: 2
obj_dst = 1523.69
# [厘米] = [像素]* 2.54 / DPI
distance = obj_dst*2.54/96
# 96 是 DPI，2.54 是英寸到厘米的转换
print(distance, "厘米")
英文:
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")
答案2
得分: 2
我们之前讨论过数学。现在你只需要重新排列一些方程。
针孔相机模型。
- 焦距（毫米）：8毫米
- 传感器像素间距：4.8微米/像素
- 焦距（像素）：8毫米 /（4.8微米/像素）= 1667像素 =：f
物体：
- 实际宽度：0.0373米
- 表观宽度：17像素
解算物理距离，我们称之为 d。
(0.0373米 / d) * f = 17像素，我们已经有了（用d替换先前的4米，插入已知的17像素）d =（0.0373米）/（17像素 / f）= 0.0373米 * f / 17像素
=> d约为3.66米
这个方程已经在你的函数 distance_finder() 中实现。它是相同的表达式。
英文:
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
This equation is already implemented in your function distance_finder(). It is the same expression.
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