RE: Digital photo frames comparison

["Paul Gordon" <paul@xxxxxxxxxxxxxxxxxxxxx>]

Because I was thinking that I need to work out the number of pixels in
a
known area to compare one display against another.
My thinking was thus:

If the resolution is known, lets say 640x480, then I can simply multiply
those two number to give me the total number of pixels on the whole
display - in this example 307,200. OK so far?

I'm then thinking that two different displays with the same resolution,
but a different size panel will be vastly different, - 640x480 on a
5-inch screen will probably look quite good, on an 8-inch screen will
likely look quite pants, so I need to then calculate the number of
pixels in a common amount of screen area to see which panel has the
greatest pixel density...

Therefore I reckon if I know the surface area of the screens, lets say
one is 5-inches by 3-inches = 15 square inches, the other is 8-inches by
5-inches = 40 square inches, then I can calculate:

307,200 divided by 40 = 7680 pixels per square inch
307,200 divided by 15 = 20,480 pixels per square inch

Thus I can see that the smaller screen has nearly 3 times the number of
pixels in a given area (1 square inch), and thus is likely to be
"better" (by which I really just mean sharper I suppose).

Yes, I had realised that there are both square and non-square pixels out
there, but I was deliberately avoiding that consideration to keep things
relatively simple!

The only bit I needed to fill in was how to deduce the X & Y dimensions
from the diagonal using Pythagoras' theorem.

Does anyone think this logic is flawed?

Cheers.

Paul G.



> -----Original Message-----
> From: ukha_d@xxxxxxx [mailto:ukha_d@xxxxxxx] On Behalf
Of
> Andy Laurence
> Sent: 12 November 2006 17:42
> To: UKHA_D Group
> Subject: RE: [ukha_d] Digital photo frames comparison
>
> Steve Cuthbertson <mailto:steve@xxxxxxx> wrote:
> > On Sun, 12 Nov 2006 14:52:58 +0000, I wrote:
> >
> > Lots of deleted arithmetic...
> <snip>
>
> More importantly, as the diagonal of the screen is proportional to the
> area, why bother working out the area for comparisons?
>
> Cheers,
> Andy
>
>
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