how to calculate frame rate required when I need to scan surface of 1meter moving at 0.5 meter/second

[Nimish79]

I need to scan a moving object surface with a camera.

My question is: Surface area of object is 1 meter in lengh and my camera frame size is 2048*2048 , so what is the frame rate I need ?

My mapping is 1 mm = 10 pixels

so 1000 mm = 10000 pixels

Is is like 10000 pixels / 2048 pixels = 5 frames approx?

Is it right way to determine the frames/second ?

Thanks in advance

[muks]

See here

[muks]

Can you use a linescan camera for your application?

[Nimish79]

Thanks for the reply.

A good Line scan camera cost more and do not fit in my requirement as far as cost is there.

[Nimish79]

Thanks,

I have seen the link you suggested and it gives idea on band width calculation.

My question is 5 fps is sufficient to cover the surface I want to scan? or I may miss some surface area or some over lap may happen ?

[muks]

What is the speed in which the object is moving?

[Nimish79]

object is moving at 500mm per second.

and the surface area I need to scan is 1  meter. In my knowledge, If I will scan with 50 fps I will get 50 frames (mapping is 1 mm = 10 pixels)

but I will get frames which overlaps each other, to get unique frame I need to set 5 fps.

This is just my understanding and I am new to machine vision. so correct me if I am wrong here.

[BruceAmmons · Accepted solution]

You are confusing your frame rate and your total number of frames.

Frame rate:  Speed is 500 mm/s.  Each frame has 200 mm in it.  500 mm/s / 200 mm = 2.5 frames per second.

Number of frames: Length is 1000 mm.  Frame is 200 mm.  1000 mm / 200 mm = 5 frames.

So you need to take pictures at 2.5 fps for 2 seconds to get 5 frames.

Of course, this is assuming no overlap.  In practice, it is a good idea to have a little overlap so you don't have any defects right at the edge of your image.  This is also assuming you can move your object at exactly 500 mm/s.

Bruce

[Nimish79]

Thanks Bruce,

This clears my doubts and I have got my answer.