1/3 inch chip 16:9 versus 4:3
 

Is a 16:9 chip wider than a 4:3 chip, for the same nominal size? In other words, will a 1/3 inch chip for a 4:3 camera have the same depth of field as a 1/3 inch chip for 16:9?

Patrick

I believe you will not see a difference in DOF between the two chip aspects.

From what I have gleened is a 16x9 chip is just a 4x3 chip with the top and bottom parts "turned off".

At least this is how the SD cameras worked when 16x9 chips were first coming out.

Maybe a wider chip would be outside the lens projection area.

There are two ways to achieve 16x9, but first a disclaimer : I'm no technician. I really don't know what I'm talking about, but this is what my understanding is. Anyone please correct me if I'm wrong. :)

The first SD 16x9 cameras were just cropping off the top and bottom of the picture, resulting in an effective 720x300 or so resolution. The clip would still be 720x480 anamorphic widescreen, but the actual resolution would be lost.

Nowadays, HD or newer SD cameras seem to work natively in 16x9. (I'm talking about HD cameras in SD mode) They capture full 720x480 at a widescreen aspect ratio. To achieve 4x3, they would effectively do the opposite of the older cameras - chop the sides. I have yet to find a way to record 4x3 (never even wanted to look ;D ) on my z1, but with the resolution of HDV nothing is lost by a downres and crop.

Carl

Yes. Draw a nominal circle, then draw boxes with 16:9 and 4:3 ratio dimensions, such that in both cases all four corners of the box lie on the circle. You'll find that the 16:9 box is wider but not as tall as the 4:3 one.

The circle represents effective lens coverage, and ' 1/3" ' etc refers to that diameter. It's a hang over from tube days, and the diameter of the image tubes.

I think I saw a table somewhere that explained the 1/3", 1/2" size thing - as you said, 1/3" would be the diameter of the tube - but as I recall, a 1/3" sensor is significantly smaller than 1/3"

ie the tube size is quite a bit bigger than the sensor itself. If I find the table I'll post it.

That's one way. Further explained at http://dvinfo.net/canonxl2/articles/article06.php

Indeed. Further explained at http://dvinfo.net/canonoptura/articl...eage.php#opccd

Thanks for those responses, so my widescreen chip is actually no wider than a 4:3 chip, and depth of field is exactly the same. Except that the picture is only 3/4 of the height. So, if I fill a 4:3 frame with a head at 45mm focal length, I would need to zoom out to 33.75mm to avoid cropping it, and then the depth of field would be increased! I am clearly going to have to start cropping more aggressively, but I guess that is just part of learning to make nice pictures in a 16:9 frame.
Patrick

That's not the right conclusion to draw - it depends on individual cases. What you say may may have been true for early 16:9 cases, but less so nowadays, especially with the move to HD and it being 16:9, period, increasingly chips are thought of as 16:9 first. I think you'll find nowadays the case I described (bit wider but not as tall) is more likely to be the case.

It would be so much easier if the manufacturers actually stated the diagonal of the chip in millimetres.
If David Heath is right, and, if I have not made a mistake with Pythagoras's theorem, a 1/3 ins chip for 16:9 would be almost exactly the same width as a 1/2.7 ins chip for 4:3.
For this slightly wider sensor, a 49mm lens would give the same final image size as 45mm lens on a 1/3 ins chip. But to avoid vertical cropping you would need to zoom out to 3/4 of 49mm, i.e. 37mm.
So the conclusion is the same, that to maintain the same depth of field as 4:3, when shooting 16:9, you have to vertically crop the picture.
(Which is pretty obvious when one thinks about it).
Patrick

Well actually...There are many cameras that capture a 1920x1440 area (4:3) and then just crop the bottom and top parts off. Like the hv20 or hc1. You can use photomode to get the full 1920x1440 4:3 area which is not recorded in videomode.