With a square CCD, a=b so that reduces to (2 * a^2) = c^2
Now the 5MP number is a representation of the area of the CCD
in units of pixels (calculating the dimensions of a single pixel is
left
as an exercise for the reader). We want to compare the relative area
of a 0.5" CCD with a 0.67" one. Area is length times width, or a*b,
but in this case since it's a square, a*a, or a^2.
Solving for a^2 we get a^2 = c^2 / 2 by simply dividing both sides
by 2.
Given also:
(area of 0.5" CCD) / (area of 0.67" CCD) * (MP of 0.67" CCD) = MP
of 0.5" CCD
we get (0.5^2 / 2) / (0.67^2
2) * 5MP = 2.78458454
(Yote's formula was slightly wrong, but the "1/2" factors cancel each
other out so the answer was still correct).
However, after all that, dare I mention that the UZI doesn't have a
square CCD? Or if it does, then why don't we get 1600x1600 images
instead of 1600x1200? Does it even matter? That is, do all
rectangles with the same length diagonal have the same area
regardless of their width and height?
I started to work this out but it just got messy.
Maybe I should just make an 8x10 print of the formulas
and see if it simply
LOOKS GOOD . ;-)
Maybe it's because I'm tired or there's something wrong with the
math. Assuming the CCD is sqare my end result comes to 3.73MP. I
think you only got the relative size of one of the .67 CCD sides
compared to that of the .5 and forgot to square them to get the CCD
size
sqrt(.5/2) squared / sqrt(.67/2) squared x5M = .5/2
.67/2 x 5M =
3.73M
Maxven
now I get 2.81 Meg, still not enough.
sqrt(.5/2) / sqrt(.67/2) x 5M = 2.81M
(.5x.5)
(.67x.67) x 5 Meg = 2.78Meg
The way I read the specs, Oly gets 2 megs from a .5" chip. That
(as I understand it) is the diagonal measurement of the chip.
Nikon (in the new 5000) gets 5 megs from a .67" chip.
So, if you were to shave the Nikon chip down to .5" in order to use
the Canon 10xIS lens how many megs would you have left?
This might be the next step up that we could expect to see.
--
Yote
--
Yote
