Velocity Reviews > Field of View Question

# Field of View Question

jordan.marton@gmail.com
Guest
Posts: n/a

 06-28-2005
I have a digital camera with a CCD of 2/3 and an 8mm lens attached. Is
the following formula correct to determine FOV?

FOV=2arctan(.5*width of frame*distance to frame)

SECOND QUESTION, how would I determine the field of view if the camera
is not rotating around the origin, but rather on a boom that extends
out from the origin?

Thanks
Jordan

Bart van der Wolf
Guest
Posts: n/a

 06-28-2005

<(E-Mail Removed)> wrote in message
news:(E-Mail Removed) oups.com...
>I have a digital camera with a CCD of 2/3 and an 8mm lens
>attached. Is the following formula correct to determine FOV?
>
> FOV=2arctan(.5*width of frame*distance to frame)

If you *divide* by distance to frame, yes close enough (you actually
need the distance to the principal plane, which is hard to know except
for infinity focus where it equals the focal length).
Also, a 2/3 CCD might indicate either an aspect ratio or an odd
convention for physical size.

> SECOND QUESTION, how would I determine the field of view
> if the camera is not rotating around the origin, but rather on a
> boom that extends out from the origin?

It's the same, but stitching will cause problems (especially in the
foreground). The degree of overlap between images will determine the
number of images needed for 360 degree coverage.

Bart

Don Stauffer
Guest
Posts: n/a

 06-29-2005
http://www.velocityreviews.com/forums/(E-Mail Removed) wrote:
> I have a digital camera with a CCD of 2/3 and an 8mm lens attached. Is
> the following formula correct to determine FOV?
>
> FOV=2arctan(.5*width of frame*distance to frame)
>
> SECOND QUESTION, how would I determine the field of view if the camera
> is not rotating around the origin, but rather on a boom that extends
> out from the origin?
>
> Thanks
> Jordan
>

One mistake- the second times (*) sign is actually a division (2
atan(0.5*width/EFL), where efl is distance from lens node to image.

With a rectangular format, there are actually THREE fields of view.
The horizontal, vertical, and corner to corner (conical) fields of view
can each be computed. What you have shown is the horizontal FOV if you
mean the 36mm dimension of the format in 35mm equivalent. The width of
35 mm format is of course in the vertical dimension of the format.

So say you normally shoot landscape format. Then the width of the
format is the greater of the two dimensions and the formula you show is
for horizontal FOV (with the operator as division instead of multiplication.

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