2.6K Posts

March 19th, 2004 15:00

Its all over dell's site....but quickly:

WXGA: 1280x800 native resolution
WSXGA: 1680x1050
WUXGA: 1920x1400

37 Posts

March 19th, 2004 15:00

Tx....I wanted to ask the ? here, so that others if they were wondering would be helped as well. Again tx for the reply....

155 Posts

March 19th, 2004 20:00

According to http://translate.google.com/translate?hl=en&sl=de&u=http://www.billigdrucker.de/modules.php%3Fname%3DFachbegriffe%26file%3Daufloesungen&prev=/search%3Fq%3DWSXGA%26hl%3Den%26lr%3D%26ie%3DUTF-8%26oe%3DUTF-8%26sa%3DG

WUSXGA is 1920 x 1200. It is a wide screen version of the standard USXGA 1600 x 1200.

WXGA is 1366 x 768. It is a wide screen versions of the standard XGA 1024 x 768.

WSXGA is 1600 x 1024 (a wide screen version of standard SXGA 1280 x 1024).

WSXGA+ is 1680 x 1050 (a wide screen version of standard SXGA+ 1400 x 1050).

53 Posts

March 27th, 2004 10:00

whoops, accidently hit the old enter key too soon - let me try that again ... :)

I've never seen any of the wide screens with my own eyes, but i did some calcs and I'm not sure that it's entirely correct to say "WSXGA+ is a wide-screen version of the standard SXGA+" or that "WUXGA is a wide-screen version of the standard UXGA".

The UXGA is a 15.0" screen (diagonal) with a resolution of 1600 x 1200, and the width and height is 12.0" and 9.0" respectively. So in terms of physical pixels per inch of display the UXGA has 1600 / 12 = 133 pixels per inch.

The WUXGA is a 15.4" screen (diagonal) with a resolution of 1920 x 1200. I've never seen this screen, but using good old high-school pythagoras' theorem (and assuming the vertical and horizontal pixel/inch ratios are about the same), I assume the width and height are approx 13.06" and 8.06". (Ok getting too exact here, but it doesn't really matter for the sake of the argument. If one dimension is smaller, the other will be larger, which makes the point even more relevant)

So the physcial pixels per inch on the WUXGA screen is  1920 / 13.06 = 147 pixels per inch.

This pixels per inch value is an indication of how large or small text and icons will appear to the naked eye. The higher the value for a screen, the smaller a given object will appear on that screen.

So I come up with

WUXGA   15.4" 1920 x 1200    ...........  147 pixels per inch

WSXGA+ 15.4" 1680 x 1050   .............  128 pixels per inch

UXGA     15.0"   1600 x 1200  ..............  133 pixels per inch

 

So I think the UXGA is somewhere between the WSXGA+ and WUXGA in terms of how small things are. I have the UXGA which is about as small as I could handle, even with dpi changes and large icons etc.

If my assumptions are correct and the WUXGA really is a bit smaller, it would probably be a bit too small for me, but that's just a personal opinion.

Now if Dell would just bring out a 17" screen, I'd be very happy .... :)

 

53 Posts

March 27th, 2004 10:00

I've never seen any of the wide screens with my own eyes, but i did some calcs and I'm not sure that it's entirely correct to say "WSXGA+ is a wide-screen version of the standard SXGA+" or that "WUXGA is a wide-screen version of the standard UXGA".

The UXGA is a 15.0" screen (diagonal) with a resolution of 1600 x 1200, and the width and height is 12.0" and 9.0" respectively. So in terms of physical pixels per inch of display the UXGA has 1600 / 12 = 133 pixels per inch.

The WUXGA is a 15.4" screen (diagonal) with a resolution of 1920 x 1200. I've never seen this screen, but using good old high-school pythagoras' theorem (and assuming the vertical and horizontal pixel/inch ratios are about the same), I assume the width and height are approx 13.06" and 8.06". (Ok getting too exact here, but it doesn't really matter for the sake of the argument. If one dimension is smaller, the other will be larger, which makes the point even more relevant)

So the physcial pixels per inch on the WUXGA screen is  1920 / 13.06 = 147 pixels per inch.

This pixels per inch value is an indication of how large or small text and icons will appear to the naked eye. The higher the value for a screen, the smaller a given object will appear on that screen.

So I come up with

    

 

 

 

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