Re: Video Rules?
On Sun, 22 Aug 1999, Karl F. Larsen wrote:
> Are there any rule of thumb on how much ram the video card must
> have to achieve a given resolution? I have a 1 meg card now and get 8 bit
> 16 color when using a 1024x480 screen.
It's actually pretty simple. The formula is:
(horizontal resolution) x (vertical resolution) x (bytes per pixel)
Note: that's *bytes* per pixel. It's 8 bits per byte, so...
8 bpp = 1 byte/pixel
16 bpp = 2 bytes/pixel
24 bpp = 3 bytes/pixel
32 bpp = 4 bytes/pixel
So, for example, 1024x768@16bpp needs 1024 x 768 x 2 = 1,572,864 bytes.
Now, it's 1024 bytes in a kilobyte, and 1024x1024 = 1,048,576 bytes in a
megabyte. So, for 1024x768@16bpp, you need 1,572,864/1,048,576 = 1.5MB.
Now, for 1600x1200@32bpp, you need (1600x1200x4)/(1024x1024) = 7.32MB.
So, basically, a video card with 8MB has enough memory for almost any
currently realistic resolution. (3D video cards need more, to store
textures and such.)
With a 1MB card, you can get:
1024x768 @ 8bpp (786,432 bytes)
800x600 @ 16bpp (960,000 bytes)
640x480 @ 24bpp (921,600 bytes)
> But windows 98 seems to get far better resolution with this card.
> I wonder how?
What type of video card is it? (E.g., Trident, Matrox, ATI...)
> I will buy a new video card if I can figure out how much more ram
> to get. The monitor is not special but it's a 13 inch called a 15 inch
> screen. Odd.
*All* the manufacturers do that these days. In their terms, you would
have a "15 inch (13 inch viewable)" monitor.
Sincerely,
Ray Ingles (248) 377-7735 ray.ingles@fanucrobotics.com
"Transported to a surreal landscape, a young girl kills the first woman
she meets and then teams up with three complete strangers to kill again."
- TV listing for the Wizard of Oz in the Marin Independent Journal
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