[opendtv] Re: The Math on Screen sizes

  • From: "John Willkie" <johnwillkie@xxxxxxxxxx>
  • To: <opendtv@xxxxxxxxxxxxx>
  • Date: Thu, 10 Jun 2004 21:48:25 -0700

Using those figures, I get similar results: 38.55 inch line width,  or a 44
inch diagonal measure 16:9 screen, and a 48 inch or so 4:3 screen.  Of
course, I haven't accounted for having more than one dot of each color per
pixel, nor the actual pixel configuration.

The issue, then, is how many sets have such closely spaced phosphors, and
how many sets have monospaced pixels.  Neither appears to be universal.

John Willkie

-----Original Message-----
From: opendtv-bounce@xxxxxxxxxxxxx
[mailto:opendtv-bounce@xxxxxxxxxxxxx]On Behalf Of John Golitsis
Sent: Tuesday, June 08, 2004 12:24 PM
To: opendtv@xxxxxxxxxxxxx
Subject: [opendtv] Re: The Math on Screen sizes


Fine.  So then what math are you using to calculate your massive screen
widths?

My math would be:

.51mm x 1920 = 979.2mm or about 98cm, or about 38.5 inches.  Since we're
being
precise here, those measurements aren't exactly correct because it would
clip
exactly one 'dot' (half a dot each side).

To be honest, I've lost where this thread is/was going.  None of this stuff
has
anything to do with 4:3 versus 16:9.

----- Original Message -----
From: "John Willkie" <johnwillkie@xxxxxxxxxx>


> Horizontal or vertical spacing of imaging elements.  Like homes being
framed
> with 2 x 4's being spaced at 16 inches on center.  The pitch -- the
spacing
> between the same point on the members -- is 16 inches.



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