I want to upgrade my 28" 4:3 CRT to a flat screen. I believe that the Toshiba WL56 LCD's are recomended by many and am looking at the 27" and 32" models. I need however to compare the visual image (not screen) sizes with my existing 28" 4:3 CRT TV. The diagonal measurement does not help, because the formats are different. Is it possible to find out what the visual image sizes (height and width) are of 26" and 32" 16:9 screens (so I can then compare them to what my 4:3 image size measures as)? Thanks

This is what Pythagoras' theorem was invented for For a 4:3 set, the height is 0.6 times the diagonal, and the width is 0.8 times the diagonal. For a 16:9 set, the height is 0.49 times the diagonal, and the width is 0.87 times the diagonal. So to achieve the same screen height (so objects appear the same size), a 16:9 set needs a diagonal measurement approximately 20% greater than a 4:3 set.

That's right, one of those things in school that seemed pointless yet now in the world of AV finally finds a use!! (a*a) + (b*b) = (c*c) where c is the hypotenuse while a and b are the sides of the right angled triangle. (Hpotenuse=the long side of the triangle) We know that a widescreen is 16 long by 9 high, so if this triangle is taken, we get. (16*16)+(9*9)=256+81=337 therefore the diagonal would be the square root of 337, or 18.3575. To work out the height and width, thake the screen size and divide by 18.3575, then multiply the result by 16 for the width and 9 for the height. e.g. 42/18.3575= 2.28279, width is 16*2.28279 =36.52464 height is 9*2.28279 = 20.5451 e.g. 32/18.3575= 1.743157 width is 16*1.743157 =27.8905 height is 9*1.743157 = 15.6884 *EDIT* Sure there's a simpler method somewhere but this will work. Use 18.35755975 instead of 18.3575 for greater accuracy, 18.3575 should probably have been 18.3576 for the example anyway.

However, don't forget to apply the salesman's theorem: whenever the customer can be mislead by specifications, do that. Certainly when the competitor is doing it as well. As a result of this, picture diagonals are unreliable figures. On a CRT they are nearly always specified for the outer measurement of the glass tube, and the visible area is several cm (between 1 and 2 inch) less than that! So your 28" CRT TV is more like 26.5" diagonal. Check it with a ruler. For LCD, there is usually no such difference (yet...). So a 28" LCD is significantly larger than a 28" CRT! Or, when you want to change to 16:8, you would think you need 1.2*28 = 33.6" but it is more like 1.2*26.5 or about 32".

Well, I can't say how it compares to a 26" LCD, but I measured my 32" set's 4:3 size a while back, and it was just over 26" if I remember correctly. My 28" CRT was around 21" diagonal.

It's started already! Go check out Currys and Comet. I was in there the other week and a 26" LCD TV had a viewable area of 24.6". The figures may not be accurate due to my poor memory, but it was that sort of difference! Dave

You must have a very small 28" inch TV then because the diagnonal on mine's around 26.5". You're sure you didn't measure the width?

Yes sorry, I meant I got about 21" diagonal with 4:3 content. Depending on how much is actually visible on your set, I'd say a 32" LCD will either be about the same size, or perhaps very slightly smaller with 4:3 content.

Many thanks to all for your assistance and reminders of the hellish hours spent in Maths classes. Thanks to Excel I have used your methodologies to calculate the figures and guess what - it failed. UNTIL I went home and discovered that my "29"" Sony Trinitron actiually measured about 26.5". Then the maths produced exactly the right result!