RE: Digital photo frames comparison

["Steve Morgan" <smorgo@xxxxxxxxxxxxx>]

I've uploaded a small spreadsheet to the files area:

http://tinyurl.com/yh2dgv

Enter the aspect ratio (normal 4 wide to 3 high) for a non-widescreen
display
Enter the diagonal and it will calculate height and width.

Hope that helps.

Steve

> -----Original Message-----
> From: ukha_d@xxxxxxx [mailto:ukha_d@xxxxxxx]=20
> On Behalf Of Paul Gordon
> Sent: 12 November 2006 10:04
> To: UKHA_D Group
> Subject: [ukha_d] Digital photo frames comparison
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> Now I think we all know that the Philips digital frame that=20
> has been discussed here previously is pretty much (one of)=20
> the best around at the moment. However the price of that one=20
> has been slowly creeping up, to the point that the best price=20
> I know of at the moment is =A3130. There are new ones appearing=20
> every day, at different price points, and with different=20
> specifications... the quandary is, how to compare them...
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> SWMBO started off looking *just* at the display resolution=20
> (640x480, 720x480 and so on...), but then I thought, hang on,=20
> - that doesn't take into account the screen size as well... -=20
> to truly do a qualitative comparison, one would need to work=20
> out the number of pixels in a given area - pixels per square=20
> inch for example. - Anyone disagree?
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> First problem, - how to work out the area of any given=20
> display? - most products seem to give just one measurement,=20=20
> - this I presume is the diagonal which is traditionally the=20
> measurement given for all kinds of screens. OK, from that it=20
> is possible to work out the area using Pythagoras' theorem. -=20
> Rats, I knew I should have paid attention during O'level=20
> maths! - Although, I can honestly say that in the 25 years=20
> since I left school this is the FIRST time I've ever needed=20
> to use anything they were teaching in maths class! - So I'll=20
> admit it, whilst I can rattle off the rule "In a right angled=20
> triangle the square of the hypotenuse is equal to the sum of=20
> the squares of the other two sides." Yadda-yadda... - how in=20
> 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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> ** Sponsored by http://www.BERBLE.com ** all the Cool
Stuff,=20
> in one Place=20
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