Howdy from Texas,

I've checked through a few brochures and have been unable to find the answer to this. Does anyone know for certain exactly how the CCD target area for 16:9 widescreen video is defined in the PDX10 and TRV950? I know they're different, but what are they exactly in terms of pixel dimensions... 960x480, or what?

Funny you should ask, I just performed this experiment by framing the image in 4:3 video mode, switching to 16:9 mode and finally switching to photo mode. By overlaying these three images (putting them in photoshop on separate 50% transparent layers) I arrived at the following: http://www.greenmist.com/dv/res/f.JPG

Now these results only apply to the PDX-10. The TRV-950 has been crippled in firmware for reasons known only to Sony. The following site shows a similar experiment, although he admits the TRV-950 frame sizes involve some speculation, and one camera as PAL while the other was NTSC. http://www.techshop.net/PDX-10/

Your mileage may vary...

Check the diagram located at:

www.sonybiz.net -> Products -> Broadcast and professional video -> Camcorders -> DVCAM -> PDX10

Menu: About this product - Benefits

Chapter: "Superior 16:9 picture quality"

Location: "click here for a diagram"

Sorry, can't post a direct link (dynamical web pages).

------------------------------

*If* the diagram is accurate and the 16:9 area uses all available CCD width then we know the number of horizontal pixels:

Megapizel count is known and is = width times height
width:height = 4:3

From these equations one gets the width of the CCD and the 16:9 area. The height is then 9/16 of the width.

At least this gives an upper limit for the 16:9 area in terms of pixels.

4:3 remains a mystery.

That brings up a window "about:blank" which is empty on my computer! Sony's websites are so dysfunctional...

In my browser the "about:blank" only remains there for one second. Then a diagram is presented.

The diagram also tells that the 16:9 mode uses "+44% more of megapixel areas" than the 4:3 mode

Now that we know the dimensions (and thus the pixel count) of 16:9 and also that it is 1,44 times the 4:3 pixel count, we can calculate the pixel count of the 4:3 area.

Given that its aspec ratio is known, we get the dimensions.

The calculations are left as an exercise...

Caveats:

1) Does the 16:9 mode *truly* use 100% of the CCD width?
2) Is the CCD megapixel count accurate?
3) How accurate is that +44% ?

See the link to my image in the posting above. I was careful to be as precise as possible, the camera was locked down on a tripod and all three images aligned perfectly in Photoshop.

Yes, 16:9 mode on the PDX-10 definitely uses the full available chip width. However the 4:3 mode uses a slightly taller area than 16:9, presumably because the designers did not want to limit its vertical resolution to only 706 pixels. This raises the larger question however, of why didn't they use the WHOLE 4:3 area of the chip? Perhaps the circuitry wouldn't be able to read that many pixels in real time, or maybe they just found that it didn't increase quality? Maybe they wanted the telephoto end of the zoom to produce a larger image? Or they wanted bragging rights in order to say that switching to 16:9 mode widens the field of view? Only Sony knows....

I don't understand your "44% more" calculation. Doing your "exercise"... 1.78-1.33=0.45 and 0.45/1.33=0.338, therefore there should be only ~34% more pixels needed to capture optimum 16:9, based on the comparative areas of the images. Using the actual pixel dimensions from my tests, the PDX-10 is using 12.9% more pixels in 16:9 mode vs 4:3 mode. Now of course this all gets downsampled to 720x480 for both modes in the end, due to the anamorphic squeeze. But it's a huge improvement over shooting 16:9 on a standard 4:3 camcorder, where you are actually using 25% LESS pixels in widescreen mode.

And another discrepancy: the specs say approx 1,070,000 gross pixels or approx 1,000,000 for stills. 1152x864=995,328. I guess some pixels are not used, and they also rounded up the effective number of pixels for stills. It is also spec'ed as approx 690,000 pixels for video but my numbers indicate 664,346 for 4:3 mode and 749,952 for 16:9. Maybe they averaged the two and rounded down? ;-)

OK, I cannot resist... Let's do some math.

------------------------------------------------------------------------
FIRST WE DETERMINE THE 16:9 DIMENSIONS

X(ccd) / Y(ccd) = 4/3
X(wide) / Y(wide) = 16/9
X(ccd)=X(wide)
X(ccd) Y(ccd) = 1070 000

First and last gives X(ccd) X(ccd) = 1070 000 * 4/3
=> X(ccd)= 1194,431...

This can't be true. Either the full width is not beeing used or the megapixel count is not accurate. This rough value (1194) gives a hint though.

We know the PAL standard: 768 * 576 pixels

It makes sense to use 1,5 or 2,0 times the pixels.
I'm assuming that they chose 1,5 times.

Thus:

X(ccd)=1152

and
<=> Y(ccd)= 1152*3/4=864
which is, in fact, 1,5 * the PAL vertical resolution 576

It seems clear now that the usable CCD size is 1152 * 864, which means that

X(wide)=1152
Y(wide)=1152*9/16=648
Pixels(wide)=1152*648= 746496

This result is almost identical to the measured values.

-----------------------------------------------------------------------
NEXT THE 4:3 DIMENSIONS

The +44% sentence leads to:
1,44 * Pixels(4:3) = Pixels(wide)
<=> Pixels(4:3) = 746496/1,44=518400 pixels

X(4:3) Y(4:3) = 518400
X(4:3) / Y(4:3) = 4/3

<=> X(4:3) X(4:3) = 518400 * 4/3 = 691200
<=> X(4:3) = 831,384...
<=> Y(4:3) = 3/4 * 831,384... = 623,538...

OK, so the +44% wasn't accurate. Let's assume it's reasonably accurate. A bit experimenting shows that

X(4:3)=831
Y(4:3)=623

gives the best match for 4:3

These values are different from the measured ones. BOTH the measured X(4:3) AND Y(4:3) are about 1,13... times larger. Probably because I used an inaccurate +44%...

--------------------------------------------------------------------------
RESULT

16:9 size confirmed:
X=1152
Y=648 (if exactly 16:9), Y=651 (measured)

4:3 size seems logical too:
X=941 (measured)
Y=706 (measured)
Calculations suggest that there is no considerable error here.

These results are exact enough considering that this is a videography forum. I'm not the manufacturer...

They want either

1x or 1.5x or 2x the PAL dimension...

I'm in a bit of a hurry. Must run. I'll be back in 48 hours to read the rest... Bye.

Thanks, all! This has been most interesting!

Chris: ...probably a lot more than you bargained for as well!

Phew Ralf, I feel like I'm back in my high school algebra class :-) Just a few things:

1. I certainly might have been off by 3 pixels with the 951 measurement, though I tried to be careful.

2. Remember, I'm using an NTSC camera. The PAL mapping might be a little different. If you look at the techshop link above, he does the same sort of test with a PAL PDX-10, and the 16:9 dimensions are pretty close, but in 4:3 mode he measures 960x720. I don't think there's any way my measurements are off by that much, so there probably is a difference in the PAL model.

3. I thought PAL DV was 720 x 576, not 768 x 576... but never having the good fortune to use such equipment I could very well be wrong about that.

> This raises the larger question however, of why didn't they
> use the WHOLE 4:3 area of the chip? Perhaps the circuitry
> wouldn't be able to read that many pixels in real time

This is probably the case. Part of the image processing circuitry is probably inside the CCD chip assembly (they do that to make it simpler and less expensive to build) itself and it must have a limit in the bandwidth it can process and spit out.

The CCD has more pixels than the output image. A many-to-one mapping must thus be made.

ONE PIXEL
A 1-to-1 mapping is trivial.

TWO PIXELS
A 2-to-1 mapping is simple: Take two values. Generate a new value from them (average or something).

THREE PIXELS
A 3-to-1 is simple too.

Even an 1.5-to-1 (3-to-2) mapping is simple: Take three pixels. Generate one value from the two leftmost pixels and another value from the two rightmost pixels. Then take next three pixels and repeat. This results in 3-to-2 or 1.5-to-1 downsampling of resolution. Somehow I feel/guess that such a method is applied in the PDX10...

FOUR PIXELS
We have a few new sampling alternatives: 4-to-1 and 4-to-3

AND SO ON...

The CCD has fixed dimensions. The available mappings thus generate a finite set of alternatives.

The CCD and the chosen output resolution happen to be in a 3:2 ratio roughly (for PAL format). This same 3:2 mapping would give a different CCD area for NTSC... Widescreen and traditional width pictures also result in different used CCD areas.

If the mapping is implemented in hardware, then it might be impossible to choose different mappings for different image formats and widths. It would definitely be complicated even if it was done in software.

How would you implement an 1.743956737 pixels to one mapping? It might not be possible at all because there simply isn't any usable x-to-y mapping.

There is no standard for still image resolution. A one-to-one mapping can thus be used and the whole CCD area is available.

To make things more complicated some devices use a 13.5 MHz sampling rate for horizontal pixels (720 pixels, PAL) while others use 14.75 MHz (768 pixels, PAL).

I read that DV camcorders, DVD players, TVs etc. use 13.5 MHz and computers (PhotoShop) use 14.75 MHz. Thus they present the picture using different number of horizontal pixels. This has something to do with the pixel format. 13.5 MHz for non-square pixels on TV and 14.75 MHz for square pixels in PhotoShop...

http://www.uwasa.fi/~f76998/video/modes/

Only the number of vertical lines is constant: 576 in PAL and 480 in NTSC. PDX10 generates, guess what, 540 TV-lines. Maybe the "Safe zone" plays a role here?

Breaking the signal to horizontal pixels depends on frequency / display type.

This is becoming complicated and I'm not an broadcast engineer. I do not know what I'm writing about, so to say. I just want to tell what I've found out. This qustion about resolution seems to be much more complicated than it first seemed to be.

> How would you implement an 1.743956737 pixels to one
> mapping? It might not be possible at all because there simply
> isn't any usable x-to-y mapping.

Interesting that you mention it, a pretty sophisticated resampling must be going on if the CCD chips have square pixels, since DV has non-square pixels, both in 4:3 and in 16:9 modes.

Well, maybe the CCD pixels are non-square then.

Does the CCD even have pixels?

Doesn't it just have monochromatic light detectors than have some unknown form factor. The Fuji SuperCCD SR chip, for example, has two different types of light sensors - larger and smaller. With any CCD at least three images must be taken so that color values can be recorded. Some camcorders use three CCDs while others use some sort of filtering / pixel shifting etc.

Now when there are lots of light detectors then the data they give can be processed in such a way that the image looks good on the _output_device_of_choise_ (a TV set in the case of DV tape).

A TV screen and a computer display generate the picture in slightly different ways (various forms and arrangements of those light emitting phosphor dots than make pixels). So there must be some kind of adaptation taking place somewhere...

1) in this ccd I suppose that the distance between pixels is the same in vertical and orizzontal, so is like square sensor detection.

2) the usable ccd area is 1152 x 864 =995328

3) in NTSC 4/3 are used 960 x 720 = 691200 remapped in

960>720 (4>3) and 720 > 480 (3>2)

4) in PAL 4/3 are used 960 x 720 = 691200 remapped in

960>720 (4>3) and 720 > 576 (5>4)


Francesco

> How would you implement an 1.743956737 pixels to one
> mapping? It might not be possible at all because there simply
> isn't any usable x-to-y mapping.

may be 1,75 to one similar to 7 to 4

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Ciao

in 16/9 i think that the dvcam use

1152 x 648 =746496

1152 > 720 is like 8>5 (anamorfic)

PAL 648>576 10> 9
NTSC 648>486 4> 3


Francesco