Calculating field of view for a 1/7" CMOS sensor webcam

Hi,

I have a Swann Netmate USB webcam (
http://www.swann.com.au/new brochures/IT Range/USB/NetMate_SW-U-NM.pdf
) , and I need to work out it's field of view. I know the sensor is a
1/7" CMOS, and the focus is 10cm-infinity. Doe anyone know if there
is a way to calculate the FOV from this information? Or how else I can
do it?

Thanks,
Jason

 

You need to know the focal length.

You could try taking a picture with it and measuring the coverage.

 

Thanks - I was hoping there might be an easier method (or a CMOS
calculator like http://www.csgnetwork.com/foclencalcl.html ). Or that
someone out there might know...

 

The critical infomation that seems to be missing is the focal length of
the lens, without this there is not way to know what the field of view
will be.

Scott

 

There are still several thing's you'd need to know.
About the sensor: its size and pixel density.
The lens: focal length. Obviously having a fixed focal length will
help here.
The subject: size and distance from the lens.
These (and probably a few I haven't thought of) will all affect how
many pixels the subject covers in any direction on the sensor.

 

A 1/7" sensor has a diagonal of 16/7 = 2.3 mm. Therefore the normal lens
is 2.3 mm (or 50 mm equivalent of the 35 mm format).

 

: On 23 Dec 2005 22:23:17 -0800, wrote:

: >Thanks - I was hoping there might be an easier method (or a CMOS
: >calculator like http://www.csgnetwork.com/foclencalcl.html ). Or that
: >someone out there might know...

: There are still several thing's you'd need to know.
: About the sensor: its size and pixel density.
: The lens: focal length. Obviously having a fixed focal length will
: help here.
: The subject: size and distance from the lens.
: These (and probably a few I haven't thought of) will all affect how
: many pixels the subject covers in any direction on the sensor.

You are correct if you are wanting to calculate the number of pixels a
particular subject will cover. But the OP was asking about the field of
view (angle) and this requires a few less variables.

The information you need is the focal length of the lens and the
dimensions (in mm) of the sensor. If you are using a camera with
replacable lenses (SLR) the focal length of the lens is on the lens (28mm,
50mm, 300mm, etc). On many cameras with an integral lens (such as P&S
cameras) the focal length is many times imprinted on the ring that secures
the front lens element into the camera case. In the case of zoom lenses
the focal length will be expressed as a range of numbers (28-84mm). If the
numbers are not on the lens you will have to try to find the number in the
"specifications" section of the manual.

The sensor dimensions (length, height, and sometimes diagonal) may be
found in the specifications section of the manual. If it is not there you
may have to search the web or inquire the numbers from the manufacturer.
These numbers should be in mm.

Now comes the hard part. plug the numbers above into the following
equation.

A=2*arctan(D/(L*2))

D= the appropriate (horizontal or vertical) dimension of the sensor
L= Lens Focal Length (mm)
A= Angle of view (Field of View)

This is fairly easy with an inexpensive "scientific" calculator. On many
of these the "arctan" function is marked as tan-1.

So for an example:

A dimension of 23.5mm (horiz)
A Lens length of 100mm

work the equation from the inside out in the following steps

(L*2)= 200
(D/(L*2))= .1175
Tan-1(D/(L*2))= 6.701525845
2*Tan-1(D/(L*2))= 13.40305169

So the Angle of view in the horizontal direction with this combination of
lens and sensor would be 13.4 degrees.

Randy

==========
Randy Berbaum
Champaign, IL

 

Thanks to everyone for your replies.

There is no manual or information on the web for this camera, and the
manufacturers (Swann) have not responded (but they are probably closed
over the Christmas-New Year period).

I have tried to measure the (horizontal) field of view by measuring an
object and manually calculating the angle - but I'm not sure how
accurate this is? I placed the webcam exacly 40 cm away, from another
ruler. From 40 cm away, I was able to view 24cm of the distant ruler.
So half of the FOV will be tan x = half of the viewing width divided
by the distance = (24/2) /40 and therefore x=17 degrees. Therefore
the (horizontal) FOV should be 2*x, or 34 degrees, which is narrow, but
not impossible for a cheap mini-webcam? Do these calculations look
reasonable or have i missed something?

Thanks,
Jason

 

Those are the right calculations. A 34 degree horizontal FOV isn't that
odd; it corresponds to a 60 mm lens on a 35mm camera - just slightly
narrower than "normal".

You can also estimate the lens focal length of your camera with this
data. The ratio of the width of the field (24) to the distance of the
ruler (40) is the same as the ratio of the sensor width to the lens
focal length. A "1/7 inch" sensor actually has a diagonal of about 16/7
= 2.3 mm, with a horizontal width of 4/5 of that (assuming 4:3 aspect
ratio), or 1.83 mm wide. So the lens FL is about 40/24*1.83 = 3.05 mm.
Probably really 3 mm, since the difference is under 2%.

Dave