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-   -   XL2 CMOS Dimensions? (https://www.dvinfo.net/forum/canon-xl-gl-series-dv-camcorders/130861-xl2-cmos-dimensions.html)

John Sandel September 15th, 2008 01:27 PM

XL2 CCD Dimensions?
 
Anybody know the height & width of the XL2's chip(s)—just one—in millimeters?

I can find the height & width in pixels. I can find the diagonal in millimeters & inches.

I've searched the XL2 manual, this thread, this forum, other fora, Canon's website and the web generally, to no avail.

… anybody?

Jarrod Whaley September 15th, 2008 02:48 PM

Well, for one thing, the chips are CCD's, not CMOS.

As for the height and width: can't you just do a little basic high-school algebra to figure out the dimensions based on the diagonal measurement and the 4:3 AR of the chips?

Just out of curiosity, why do you need to know this anyway?

John Sandel September 15th, 2008 03:00 PM

Quote:

Originally Posted by Jarrod Whaley (Post 935144)
Well, for one thing, the chips are CCD's, not CMOS.
As for the height and width: can't you just do a little basic high-school algebra to figure out the dimensions based on the diagonal measurement and the 4:3 AR of the chips?
Just out of curiosity, why do you need to know this anyway?

1. CCDs, yes, thanks. Call them "chips" and we understand each other.

2. No. I was an English major.

3. That's a secret.

Greg Boston September 15th, 2008 03:15 PM

Quote:

Originally Posted by Jarrod Whaley (Post 935144)
As for the height and width: can't you just do a little basic high-school algebra to figure out the dimensions based on the diagonal measurement and the 4:3 AR of the chips?

This might help...

http://www.dvinfo.net/canonxl2/articles/article06.php


-gb-

John Sandel September 15th, 2008 05:18 PM

Thanks, Greg. I remember when Chris posted that. It doesn't give the dimensions I want, which are proving elusive.

Chris Hurd September 15th, 2008 05:26 PM

The width and height of the imaging area only, or the entire face of the chip?

John Sandel September 15th, 2008 05:37 PM

Imaging area, please. I thought you'd pop up.

Chris Hurd September 15th, 2008 05:42 PM

It's *roughly* 4.8mm by 3.6mm. From that you'll have to do some basic math to account for the unused pixel area above and below the 16:9 active portion. Best I can offer here, sorry.

John Sandel September 15th, 2008 06:48 PM

Addition and subtraction I can do. Did you measure it by hand or do you have a reference for that?

Chris Hurd September 15th, 2008 09:26 PM

That's a secret.

John Sandel September 15th, 2008 09:31 PM

(Ow.)

Thanks. You guys always come through.


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