definition for overscan? (1 Viewer)

[Terry Flink]

Was wondering what is meant by overscan. Someone referenced it while explaining why some widescreen aspect ratios greater than 16x9 fill the screen. Apparently this is the reason 1.85:1 movies appear as 1.78:1...just wondering.

 

[Patrick Sun]

If your 16x9 TV had 0% overscan in either dimension (vertical or horizontal), then for a film with a 1.85 aspect ratio, you'd see very thin black bars at the top and bottom of the display area.

But if you have a small percentage of overscan (where you don't see the entire video display, but perhaps 95% of the video image), then a film with the 1.85 aspect ratio will seem to fill the entire 1.78 aspect ratio display area for a 16x9 TV. You'd be hard pressed to see the minimal black bars at the top and bottom of the display.

By having some overscan (i.e. 5%), imagine that the displayed video image is slightly stretched in the vertical dimension so that only 95% of the entire video image is displayed, which is probably just enough to shove the black bars out of the way from being displayed. And that gives the illusion that the 1.85 aspect ratio film fits just about perfectly within a 16x9 TV.

By having this amount of overscan in the horizontal dimension means that for a 1.85 aspect ratio film, you'd be missing just a bit of information on the sides of the display, but it's not all that earthshattering unless you have over 10% overscan or more.

 

[Michael TLV]

Greetings

The amount of image being cropped/chopped off on the sides of the TV.

for most sets, the number is in the order of 5% on all sides.

Notice that on computer monitors, you often have complete control over image size. If you make it too small, you see black bars on the sides. If you make it too big ... portions of the image are lost/hidden.

Overscan is needed because of the TV's inability to display acceptable edge to edge geometry and because not all signal sources are properly centered. some images are actually off set ... resulting in visible black bars on certain edges.

Regards

 

[Terry Flink]

Thanks for the quick responses. One of the reasons I love this forum!