Re: Digital photo frames comparison

["David Gumbrell" <david.gumbrell@xxxxxxxxx>]

For photo display, as or even more important than the resolution/pixel
size=
,
will be the absolute brightness and also the contrast ratio. The other
problem that happens with LCDs is uneven illumination across the screen and
also off-axis viewing. I think this is going to be an example of one of
those times where you get what you pay for !

Cheers,

David

On 11/12/06, Paul Gordon <paul@xxxxxxx> wrote:
>
>   Now I think we all know that the Philips digital frame that has been
> discussed here previously is pretty much (one of) the best around at
the
> moment. However the price of that one has been slowly creeping up, to
the
> point that the best price I know of at the moment is =A3130. There are
ne=
w
> ones appearing every day, at different price points, and with
different
> specifications... the quandary is, how to compare them...
>
> SWMBO started off looking *just* at the display resolution (640x480,
> 720x480 and so on...), but then I thought, hang on, - that doesn't
take i=
nto
> account the screen size as well... - to truly do a qualitative
comparison=
,
> one would need to work out the number of pixels in a given area -
pixels =
per
> square inch for example. - Anyone disagree?
>
> First problem, - how to work out the area of any given display? - most
> products seem to give just one measurement, - this I presume is the
diago=
nal
> which is traditionally the measurement given for all kinds of screens.
OK=
,
> from that it is possible to work out the area using Pythagoras'
theorem. =
-
> Rats, I knew I should have paid attention during O'level maths! -
Althoug=
h,
> I can honestly say that in the 25 years since I left school this is
the
> FIRST time I've ever needed to use anything they were teaching in
maths
> class! - So I'll admit it, whilst I can rattle off the rule "In a
right
> angled triangle the square of the hypotenuse is equal to the sum of
the
> squares of the other two sides." Yadda-yadda... - how in the heck
do I us=
e
> that to deduce the area of a 5.7 inch display?
>
> Any mathematicians care to help me out?
>
> Cheers
>
> Paul G.
>
> [Non-text portions of this message have been removed]
>
>=20=20
>


[Non-text portions of this message have been removed]



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