RE: Digital photo frames comparison

["Paul Gordon" <paul@xxxxxxxxxxxxxxxxxxxxx>]

Thanks Simon, SWMBO hadn't found that one... - It looks pretty good...

Paul G.


> -----Original Message-----
> From: ukha_d@xxxxxxx [mailto:ukha_d@xxxxxxx] On Behalf Of
> Simon Coates
> Sent: 12 November 2006 10:17
> To: UKHA_D Group
> Subject: Re: [ukha_d] Digital photo frames comparison
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> I've been considering buying this one
> Digi-Vue 8 Photo Frame Misco No.: 129656
> http://www.misco.co.uk/productinformation/~129656~WW~/Digi-
> Vue%208%20%20Photo%20Frame.htm
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> Looks to be best compromise between price, spec and quality.
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> Regards
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> Simon
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>   ----- Original Message -----
>   From: Paul Gordon
>   To: UKHA_D Group
>   Sent: 12 November 2006 10:04
>   Subject: [ukha_d] Digital photo frames comparison
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>   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...
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>   SWMBO started off looking *just* at the display resolution (640x480,
> 720x480 and so on...), but then I thought, hang on, - that doesn't
take
> into 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?
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>   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
> diagonal 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! - Although, 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 equa=
l
> to the sum of the squares of the other two sides." Yadda-yadda...
- how i=
n
> the heck do I use that to deduce the area of a 5.7 inch display?
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>   Any mathematicians care to help me out?
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>   Cheers
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>   Paul G.
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>   [Non-text portions of this message have been removed]
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> [Non-text portions of this message have been removed]
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