HVX ccd size and pixel count, what is it?

If Andromeda and HVX work as advertised, which one would have the better HD image? Assuming that 1/3" chips 16:9 or 4:3 have the same area, it would seek HVX would have the advantage.

jmf
 
The HVX - obviously. Andromeda is producing some nice images, but they're going to be even more lens limited than the HVX, which will, at least, have a lens designed for the purpose of HD video.

Graeme
 
What we might see from Andromeda is the increased lattitude which I'd imagine the HVX just won't be able to do no? It's still 8 bit in the end.

Aaron
 
Yes, it's all in the DSP. It's how you take those 12 or whatever bits down to 8bit, which they'll eventually have to end up at for distribution anyway.

Graeme
 
Zig_Zigman said:
Slighty off topic, is the Varicam 2/3'' chip size?
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Yes -- as is their new HDX400, and basically all top-end professional cameras from the big manufacturers (Sony's CineAlta F900 and new F950 are both 2/3", for example). SDX, SPX800, DSR450, they're all 2/3".
 
Sure, I understand that, but wouldn't 10 bit be better because you MUST be introducing aliasing, and hence losing information, when downconverting to 8 from 10? If you can stay in 10 it must be better. Maybe it's a perception thing, and we just don't notice.

Thanks for the info though.

Aaron
 
You mean "quantisation" rather than "aliassing". The CCDs are analogue, so you must decide how many bits to digitise them at. You could decide to do so at 16bits, but I'm betting many of those bits would be noise. On a smaller chipped camera, I'm betting there's less benefit to 10bit quantisation than on a larger chip camera, where you can get a better dynamic range and lower noise.

However, under normal viewing conditions it's very hard to tell apart the picture from a top end DVCAM (8bit) from a Digital Betacam (10bit). As soon as you start to do colour correction, bits matter, but a lot of extra colour correction lattitude can be gained by working in 32bit floating point.

So yes, 10bit is better, but I'd prefer lower compression to extra bits above 8. Digital Betacam has low compression and 10bits, which is why it looks so good.

Graeme
 
Thanks Graeme. Ahh "Quantisation" that was the word I was looking for when I wrote aliasing. So perception does play a part here, and that's understandable as 24bit colour is really nice :)

Hadn't thought of the noise factor either - was assuming a clean signal but obviously it's never like that.

Cheers
Aaron
 
Maybe I'm stating the obvious, but the difference between 8bit with 256 levels
and 10bit with 1024 levels is HUGE!! Expect to see a 10bit P2 camera in 3-5 years.
 
24bit colour is 8bit colour on 3 channels - ie Y'cbCr or RGB. Asfor 1024 levels v 256 levels, it sounds huge, but visually, it's not so obvious.

Graeme
 
For the discussion of lp/mm required per electronic sensors, how would you figure film emulsions into the equation?
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For one, there is a distribution of grain sizes in film, which are inherently binary (i.e., once exposed, that whole grain goes "black"). If the size of the grains is smaller than the lens resolution then the film stock is "lens limited". It is not too hard to do this - and indeed, there are a wide variety of film stocks that cover a broad range. My background in actual "film" comes from an imaging physics perspective. Holographic plates are among the highest resolution stocks I'm aware of.
 
So we need a distribution of grain sizes to get out greyscale? So, and I'm not quite sure how to put this, do you not need a small area of size distributed grains to be the sort of equivalent to a sensor pixel so that the grain area can record a range of brightnesses. Is it the size of this grain area that would be the resolution in lp/mm, or is it some function of the size of the largest grain in the grain size distribution?

Graeme
 
So we need a distribution of grain sizes to get out greyscale?
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Not really - you can think of the grains as a whole bunch of little spheres. Practically, you can consider them all the same size as well, though technically they vary a bit. Getting a greyscale out of a film is essentially how you perceive greyscale on any printer (think laser printer toner or ink-jet dots)... all the dots are the same size, and you get black when all of the dots overlap, grey when there are some white spots.

I think the direct analogy to a CCD is that a film grain is the equivalent of an electron in a CCD well. When you expose all grains, the electron well is equivalently full, and the image is saturated. When you expose some finite number of grains in a given area, then you get a value between 0 and saturation.

The actual resolution of the film will then be a combination of the grain sizes and their spatial distribution, although practically you can probably consider the distribution to be uniform with very few gaps. Large grains = fast film and is subsequently low resolution and lattitude, small grains = slow film and higher resolution and lattitude. This doesn't scale particularly well to a CCD, since with a large pixel on a CCD you have essentially a large grain spatial distribution, but a lot of electrons and thus higher lattitude... this realization suggests that getting a CCD to behave "like film" is to defy its physical mechanism. A little further thinking suggests that with careful engineering you could make a CCD outperform film, and then fiddle with the output to bring it down to the film quality we're all familiar with.

In terms of "lp/mm" you have to be very specific as to how you define a "resolved line pair" in terms of contrast/SNR, the width of the lines, etc.
 
So you need a group of grains to get a greyscale? So it would be the size of a group that is necessary to get the contrast you want that would relate to the resolution of the film?? Or ami I still barking up the wrong tree?? Thanks for you help explaining this - I've got a good idea about CCDs, but not so about film.

Graeme
 
So you need a group of grains to get a greyscale?
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Yes.

So it would be the size of a group that is necessary to get the contrast you want that would relate to the resolution of the film?
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You're combining lattitude and the fundamental (binary) resolution of the medium to get a "resolution at a specific lattitude" here. Yet another reason why "line pair" isn't an SI unit ;)
 
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