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D70 resolution: 3008x2000 - what's the 8 all about

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January 1, 2005 5:37:27 PM

Archived from groups: rec.photo.digital.slr-systems (More info?)

The D70 resolution is 3008x2000 pixels.

What is the extra 8 pixels all about?

Not that it matters I suppose, just curious.
January 2, 2005 12:47:14 AM

Archived from groups: rec.photo.digital.slr-systems (More info?)

paul wrote:
> The D70 resolution is 3008x2000 pixels.
>
> What is the extra 8 pixels all about?
>
> Not that it matters I suppose, just curious.


Why would you care? The Sony sensor used in the D70 doesn't have the 3:2
ratio anyway. A 35mm film is 36mmx24mm. Sony's chip is 23.7mmx15.6mm and
that makes the crop factor comes out to somewhere between 1.519x and
1.538x. Just for comparison, Canon's sensor is 22.5mmx15mm which is
exactly 1.6x.
Anonymous
January 2, 2005 2:06:00 AM

Archived from groups: rec.photo.digital.slr-systems (More info?)

On Sat, 01 Jan 2005 14:37:27 -0800, paul wrote:

> The D70 resolution is 3008x2000 pixels.
>
> What is the extra 8 pixels all about?
>
> Not that it matters I suppose, just curious.

3000 isn't divisible by 16, 3008 is.

The answer to your next question is to allow lossless rotation of JPEGs :-)

--
John Bean

The most likely way for the world to be destroyed, most experts agree, is by
accident. That's where we come in; we're computer professionals. We cause
accidents (Nathaniel Borenstein)
Related resources
Anonymous
January 2, 2005 6:37:09 AM

Archived from groups: rec.photo.digital.slr-systems (More info?)

In article <1ns37jweuk44d.dlg@waterfoot.net>,
John Bean <waterfoot@gmail.com> wrote:

> On Sat, 01 Jan 2005 14:37:27 -0800, paul wrote:
>
> > The D70 resolution is 3008x2000 pixels.
> >
> > What is the extra 8 pixels all about?
> >
> > Not that it matters I suppose, just curious.
>
> 3000 isn't divisible by 16, 3008 is.
>
> The answer to your next question is to allow lossless rotation of JPEGs :-)

And MY next question is "How does 3008x2000 equate to the advertised 6.1
megapixels?" Seems to me it should be 6.016 megapixels.

Merritt
Anonymous
January 2, 2005 12:54:27 PM

Archived from groups: rec.photo.digital.slr-systems (More info?)

Merritt Mullen wrote:
> In article <1ns37jweuk44d.dlg@waterfoot.net>,
> John Bean <waterfoot@gmail.com> wrote:
>
>> On Sat, 01 Jan 2005 14:37:27 -0800, paul wrote:
>>
>>> The D70 resolution is 3008x2000 pixels.
>>>
>>> What is the extra 8 pixels all about?
>>>
>>> Not that it matters I suppose, just curious.
>>
>> 3000 isn't divisible by 16, 3008 is.
>>
>> The answer to your next question is to allow lossless rotation of
>> JPEGs :-)
>
> And MY next question is "How does 3008x2000 equate to the advertised
> 6.1 megapixels?" Seems to me it should be 6.016 megapixels.
>
> Merritt

One presumes that marketing are allowed to round anything up!

Actually 3008 x 2000 is:-

6.016 million pixels

5875 KP (if K = 1024)

5.74MP (if M = 1024 x 1024)

So if the computing definition of mega were used, the camera is only
5.74MP!

(Some people complained a while back that not all "128MB" CF cards had the
same capacity, and indeed it seems that some manufacturers use the MB =
1000000 bytes and other use MB = 1024 * 1024 bytes).

Cheers,
David
Anonymous
January 2, 2005 2:41:09 PM

Archived from groups: rec.photo.digital.slr-systems (More info?)

John Bean wrote:

> On Sat, 01 Jan 2005 14:37:27 -0800, paul wrote:
>
>
>>The D70 resolution is 3008x2000 pixels.
>>
>>What is the extra 8 pixels all about?
>>
>>Not that it matters I suppose, just curious.
>
>
> 3000 isn't divisible by 16, 3008 is.
>
> The answer to your next question is to allow lossless rotation of JPEGs :-)

Lossless rotation can be maintained by merely keeping the number of pixels the
same, eg: 3000 x 1500 -> 1500 x 3000 (or whatever dimesnions you please).

The 3008/16 is, as you imply, simply adopting the digital 'rounding' boundary
that takes place with most digital devices.

Cheers,
Alan



--
-- r.p.e.35mm user resource: http://www.aliasimages.com/rpe35mmur.htm
-- r.p.d.slr-systems: http://www.aliasimages.com/rpdslrsysur.htm
-- [SI] gallery & rulz: http://www.pbase.com/shootin
-- e-meil: there's no such thing as a FreeLunch.
Anonymous
January 2, 2005 2:50:10 PM

Archived from groups: rec.photo.digital.slr-systems (More info?)

On Sun, 02 Jan 2005 03:37:09 GMT, Merritt Mullen wrote:

> In article <1ns37jweuk44d.dlg@waterfoot.net>,
> John Bean <waterfoot@gmail.com> wrote:
>
>> On Sat, 01 Jan 2005 14:37:27 -0800, paul wrote:
>>
>>> The D70 resolution is 3008x2000 pixels.
>>>
>>> What is the extra 8 pixels all about?
>>>
>>> Not that it matters I suppose, just curious.
>>
>> 3000 isn't divisible by 16, 3008 is.
>>
>> The answer to your next question is to allow lossless rotation of JPEGs :-)
>
> And MY next question is "How does 3008x2000 equate to the advertised 6.1
> megapixels?" Seems to me it should be 6.016 megapixels.

You'll have to ask the whoever writes the advertising copy ;-)

FWIW my Pentax DS (with the same sensor) delivers 3008x2008 raw files, but
crops JPEGs to 3008x2000.

--
John Bean

In all large corporations, there is a pervasive fear that someone, somewhere
is having fun with a computer on company time. Networks help alleviate that
fear (John C. Dvorak)
January 2, 2005 3:37:22 PM

Archived from groups: rec.photo.digital.slr-systems (More info?)

"David J Taylor" <david-taylor@invalid.com> wrote in message
news:33pumkF43rv61U1@individual.net...
> Merritt Mullen wrote:
> > In article <1ns37jweuk44d.dlg@waterfoot.net>,
> > John Bean <waterfoot@gmail.com> wrote:
> >
> >> On Sat, 01 Jan 2005 14:37:27 -0800, paul wrote:
> >>
> >>> The D70 resolution is 3008x2000 pixels.
> >>>
> >>> What is the extra 8 pixels all about?
> >>>
> >>> Not that it matters I suppose, just curious.
> >>
> >> 3000 isn't divisible by 16, 3008 is.
> >>
> >> The answer to your next question is to allow lossless rotation of
> >> JPEGs :-)
> >
> > And MY next question is "How does 3008x2000 equate to the advertised
> > 6.1 megapixels?" Seems to me it should be 6.016 megapixels.
> >
> > Merritt
>
> One presumes that marketing are allowed to round anything up!
>
> Actually 3008 x 2000 is:-
>
> 6.016 million pixels
>
> 5875 KP (if K = 1024)
>
> 5.74MP (if M = 1024 x 1024)
>
> So if the computing definition of mega were used, the camera is only
> 5.74MP!
>
<rant>
It seems mega has two different meanings, in storage it's 1,024x1,024. But
in pixels it's the actual number. Same math my ISP uses for throughput speed
;)  I also would prefer CF cards speeds be in mbps not 40X, 80X etc..
</rant>
Anonymous
January 2, 2005 5:21:29 PM

Archived from groups: rec.photo.digital.slr-systems (More info?)

"Merritt Mullen" <mmullen8014@mchsi.com> wrote in message
news:mmullen8014-808F42.19370901012005@netnews.comcast.net...
> In article <1ns37jweuk44d.dlg@waterfoot.net>,
> John Bean <waterfoot@gmail.com> wrote:
>
> > On Sat, 01 Jan 2005 14:37:27 -0800, paul wrote:
> >
> > > The D70 resolution is 3008x2000 pixels.
> > >
> > > What is the extra 8 pixels all about?
> > >
> > > Not that it matters I suppose, just curious.
> >
> > 3000 isn't divisible by 16, 3008 is.
> >
> > The answer to your next question is to allow lossless rotation of JPEGs
:-)
>
> And MY next question is "How does 3008x2000 equate to the advertised 6.1
> megapixels?" Seems to me it should be 6.016 megapixels.

There are actually 6.24 million pixels on the sensor, of which 3008x2000 are
used in the max size JPG or in raw NEF files, which Nikon rounds to 6.1
million pixels.
Anonymous
January 2, 2005 5:31:36 PM

Archived from groups: rec.photo.digital.slr-systems (More info?)

In article <mmullen8014-808F42.19370901012005@netnews.comcast.net>,
Merritt Mullen <mmullen8014@mchsi.com> wrote:
>In article <1ns37jweuk44d.dlg@waterfoot.net>,
> John Bean <waterfoot@gmail.com> wrote:
>
>> On Sat, 01 Jan 2005 14:37:27 -0800, paul wrote:
>>
>> > The D70 resolution is 3008x2000 pixels.
>> >
>> > What is the extra 8 pixels all about?
>> >
>> > Not that it matters I suppose, just curious.
>>
>> 3000 isn't divisible by 16, 3008 is.
>>
>> The answer to your next question is to allow lossless rotation of JPEGs :-)
>
>And MY next question is "How does 3008x2000 equate to the advertised 6.1
>megapixels?" Seems to me it should be 6.016 megapixels.

The full sensor imaging area is 3040x2024 (6,152,960 pixels).
Anonymous
January 2, 2005 6:59:47 PM

Archived from groups: rec.photo.digital.slr-systems (More info?)

In article <cr7nbn07ms@news1.newsguy.com>, leo <someone@somewhere.net>
wrote:

> paul wrote:
> > The D70 resolution is 3008x2000 pixels.
> >
> > What is the extra 8 pixels all about?
> >
> > Not that it matters I suppose, just curious.
>
>
> Why would you care? The Sony sensor used in the D70 doesn't have the 3:2
> ratio anyway. A 35mm film is 36mmx24mm. Sony's chip is 23.7mmx15.6mm and
> that makes the crop factor comes out to somewhere between 1.519x and
> 1.538x. Just for comparison, Canon's sensor is 22.5mmx15mm which is
> exactly 1.6x.

why post an answer if you are totally clueless?

Lourens
January 2, 2005 6:59:48 PM

Archived from groups: rec.photo.digital.slr-systems (More info?)

Lourens Smak wrote:
> In article <cr7nbn07ms@news1.newsguy.com>, leo <someone@somewhere.net>
> wrote:
>
>
>>paul wrote:
>>
>>>The D70 resolution is 3008x2000 pixels.
>>>
>>>What is the extra 8 pixels all about?
>>>
>>>Not that it matters I suppose, just curious.
>>
>>
>>Why would you care? The Sony sensor used in the D70 doesn't have the 3:2
>>ratio anyway. A 35mm film is 36mmx24mm. Sony's chip is 23.7mmx15.6mm and
>>that makes the crop factor comes out to somewhere between 1.519x and
>>1.538x. Just for comparison, Canon's sensor is 22.5mmx15mm which is
>>exactly 1.6x.
>
>
> why post an answer if you are totally clueless?
>
> Lourens


Why not. As the ratio indicates, there is more width info than height in
the sensor (0.3mm wider). It might even translate into 8 pixels...
Anonymous
January 2, 2005 7:45:58 PM

Archived from groups: rec.photo.digital.slr-systems (More info?)

In message <mmullen8014-808F42.19370901012005@netnews.comcast.net>,
Merritt Mullen <mmullen8014@mchsi.com> wrote:

>And MY next question is "How does 3008x2000 equate to the advertised 6.1
>megapixels?" Seems to me it should be 6.016 megapixels.

There is a border of RAW image data around the edges that most cameras
making JPEGs or RAW converters will toss away in the process. There are
also areas of pixels that are not exposed to light at all.
--

<>>< ><<> ><<> <>>< ><<> <>>< <>>< ><<>
John P Sheehy <JPS@no.komm>
><<> <>>< <>>< ><<> <>>< ><<> ><<> <>><
Anonymous
January 2, 2005 7:52:38 PM

Archived from groups: rec.photo.digital.slr-systems (More info?)

Alan Browne wrote:
[]
> Lossless rotation can be maintained by merely keeping the number of
> pixels the same, eg: 3000 x 1500 -> 1500 x 3000 (or whatever
> dimesnions you please).
> The 3008/16 is, as you imply, simply adopting the digital 'rounding'
> boundary that takes place with most digital devices.

Alan, I think you will find that lossless rotation requires that the image
dimensions be a multiple of 16 (and not 8 as I incorrectly stated
earlier).

Cheers,
David
Anonymous
January 2, 2005 7:52:39 PM

Archived from groups: rec.photo.digital.slr-systems (More info?)

David J Taylor wrote:

> Alan Browne wrote:
> []
>
>>Lossless rotation can be maintained by merely keeping the number of
>>pixels the same, eg: 3000 x 1500 -> 1500 x 3000 (or whatever
>>dimesnions you please).
>>The 3008/16 is, as you imply, simply adopting the digital 'rounding'
>>boundary that takes place with most digital devices.
>
>
> Alan, I think you will find that lossless rotation requires that the image
> dimensions be a multiple of 16 (and not 8 as I incorrectly stated
> earlier).

I fail to understand why. A matrix of 7 x 11 has the same quantity of
information as a matrix of 11 x 7. Assuming the rotation is correctly done (and
after conversion of individual R,G,B sensor to RGB pixels) then it makes no
matter at all.

Cheers,
Alan


--
-- r.p.e.35mm user resource: http://www.aliasimages.com/rpe35mmur.htm
-- r.p.d.slr-systems: http://www.aliasimages.com/rpdslrsysur.htm
-- [SI] gallery & rulz: http://www.pbase.com/shootin
-- e-meil: there's no such thing as a FreeLunch.
January 2, 2005 7:52:40 PM

Archived from groups: rec.photo.digital.slr-systems (More info?)

"Alan Browne" <alan.browne@freelunchVideotron.ca> wrote in message
news:cr99b9$lu1$1@inews.gazeta.pl...
> David J Taylor wrote:
>
> > Alan Browne wrote:
> > []
> >
> >>Lossless rotation can be maintained by merely keeping the number of
> >>pixels the same, eg: 3000 x 1500 -> 1500 x 3000 (or whatever
> >>dimesnions you please).
> >>The 3008/16 is, as you imply, simply adopting the digital 'rounding'
> >>boundary that takes place with most digital devices.
> >
> >
> > Alan, I think you will find that lossless rotation requires that the
image
> > dimensions be a multiple of 16 (and not 8 as I incorrectly stated
> > earlier).
>
> I fail to understand why. A matrix of 7 x 11 has the same quantity of
> information as a matrix of 11 x 7. Assuming the rotation is correctly
done (and
> after conversion of individual R,G,B sensor to RGB pixels) then it makes
no
> matter at all.
>
Are the photosites in the sensor grid perfect squares? a 7x11 grid of
rectangles is not the same 7x11 as it would be at 11x7
Anonymous
January 2, 2005 7:52:41 PM

Archived from groups: rec.photo.digital.slr-systems (More info?)

Darrell wrote:

> "Alan Browne" <alan.browne@freelunchVideotron.ca> wrote in message
> news:cr99b9$lu1$1@inews.gazeta.pl...
>
>>David J Taylor wrote:
>>
>>
>>>Alan Browne wrote:
>>>[]
>>>
>>>
>>>>Lossless rotation can be maintained by merely keeping the number of
>>>>pixels the same, eg: 3000 x 1500 -> 1500 x 3000 (or whatever
>>>>dimesnions you please).
>>>>The 3008/16 is, as you imply, simply adopting the digital 'rounding'
>>>>boundary that takes place with most digital devices.
>>>
>>>
>>>Alan, I think you will find that lossless rotation requires that the
>
> image
>
>>>dimensions be a multiple of 16 (and not 8 as I incorrectly stated
>>>earlier).
>>
>>I fail to understand why. A matrix of 7 x 11 has the same quantity of
>>information as a matrix of 11 x 7. Assuming the rotation is correctly
>
> done (and
>
>>after conversion of individual R,G,B sensor to RGB pixels) then it makes
>
> no
>
>>matter at all.
>>
>
> Are the photosites in the sensor grid perfect squares? a 7x11 grid of
> rectangles is not the same 7x11 as it would be at 11x7

On the sensor in the D70 and Maxxum 7d I'm not sure they are really perfect
squares. The data on dpreview suggests perhaps not. (pixel dimensions v.
sensor dimensions. It's not clear what is 'cropped' by the camera when reading
the device).

In the Canon 20D they do appear to be square..


--
-- r.p.e.35mm user resource: http://www.aliasimages.com/rpe35mmur.htm
-- r.p.d.slr-systems: http://www.aliasimages.com/rpdslrsysur.htm
-- [SI] gallery & rulz: http://www.pbase.com/shootin
-- e-meil: there's no such thing as a FreeLunch.
Anonymous
January 2, 2005 8:06:51 PM

Archived from groups: rec.photo.digital.slr-systems (More info?)

Alan Browne wrote:
> David J Taylor wrote:
>
>> Alan Browne wrote:
>> []
>>
>>> Lossless rotation can be maintained by merely keeping the number of
>>> pixels the same, eg: 3000 x 1500 -> 1500 x 3000 (or whatever
>>> dimesnions you please).
>>> The 3008/16 is, as you imply, simply adopting the digital 'rounding'
>>> boundary that takes place with most digital devices.
>>
>>
>> Alan, I think you will find that lossless rotation requires that the
>> image dimensions be a multiple of 16 (and not 8 as I incorrectly
>> stated earlier).
>
> I fail to understand why. A matrix of 7 x 11 has the same quantity of
> information as a matrix of 11 x 7. Assuming the rotation is
> correctly done (and after conversion of individual R,G,B sensor to
> RGB pixels) then it makes no matter at all.

If the image is JPEG, then the indivdual blocks are 16 x 16. The size
quantisation restriction only applies to JPEG images, but of course almost
all cameras do allow JPEG images as output.

If the image is not JPEG then it can, of course, be losslessly rotated at
any size - as you say.

Cheers,
David
Anonymous
January 2, 2005 8:06:52 PM

Archived from groups: rec.photo.digital.slr-systems (More info?)

David J Taylor wrote:

> Alan Browne wrote:
>
>>David J Taylor wrote:

>>>Alan, I think you will find that lossless rotation requires that the
>>>image dimensions be a multiple of 16 (and not 8 as I incorrectly
>>>stated earlier).
>>
>>I fail to understand why. A matrix of 7 x 11 has the same quantity of
>>information as a matrix of 11 x 7. Assuming the rotation is
>>correctly done (and after conversion of individual R,G,B sensor to
>>RGB pixels) then it makes no matter at all.
>
>
> If the image is JPEG, then the indivdual blocks are 16 x 16. The size
> quantisation restriction only applies to JPEG images, but of course almost
> all cameras do allow JPEG images as output.
>
> If the image is not JPEG then it can, of course, be losslessly rotated at
> any size - as you say.

Now I'm really confused. Nothing prevents me from editing a JPG image to 1
pixel in either dimension. I can easilly come out with a 111 x 213 pixel image
in JPG. From there, I can rotate it to 213 x 111 and it contains the same info.
I can then save it at full quality. I've lost nothing.

Getting back on track, the reason we end up with 3008x2000, IMO, (D70, Maxxum
7d, et al) is that the hardware and software can be designed in even boundaries
of 16 (or 8) and that is convenient to the engineers in terms of performace/$.

--
-- r.p.e.35mm user resource: http://www.aliasimages.com/rpe35mmur.htm
-- r.p.d.slr-systems: http://www.aliasimages.com/rpdslrsysur.htm
-- [SI] gallery & rulz: http://www.pbase.com/shootin
-- e-meil: there's no such thing as a FreeLunch.
Anonymous
January 2, 2005 8:06:53 PM

Archived from groups: rec.photo.digital.slr-systems (More info?)

In article <41D84A38.3020302@freelunchVideotron.ca>,
Alan Browne <alan.browne@freelunchVideotron.ca> wrote:
>
>Now I'm really confused. Nothing prevents me from editing a JPG image to 1
>pixel in either dimension. I can easilly come out with a 111 x 213 pixel image
>in JPG. From there, I can rotate it to 213 x 111 and it contains the same info.
> I can then save it at full quality. I've lost nothing.

Yes you have. If you don't save it as a JPG, you've lost the space savings.
If you do re-save it as a JPG, you're going to re-run the compression algorithm.
That's not being run on the same (original, uncompressed) image data that was
the input to the first JPG compression, so it's extremely unlikely to come up
with the same final result. That means you've lost some more image quality.
The same holds true even if you don't rotate the image; you can still end up
with a loss of quality just by re-saving a JPG image with no editing changes.
That's why some image editing software checks for this case, and will simply
copy the input file to the output in that situation.
Anonymous
January 2, 2005 9:11:11 PM

Archived from groups: rec.photo.digital.slr-systems (More info?)

Darrell wrote:
[]
> <rant>
> It seems mega has two different meanings, in storage it's
> 1,024x1,024. But in pixels it's the actual number. Same math my ISP
> uses for throughput speed ;)  I also would prefer CF cards speeds be
> in mbps not 40X, 80X etc.. </rant>

You will be delighted to hear that both my SanDisk Ultra II CF and my
wife's Kingmax Platinum SD cards do indeed carry a read and write MB/s
rating....

David
Anonymous
January 2, 2005 9:36:16 PM

Archived from groups: rec.photo.digital.slr-systems (More info?)

On Sun, 2 Jan 2005 17:06:51 -0000, David J Taylor wrote:

> Alan Browne wrote:
>> David J Taylor wrote:
>>
>>> Alan Browne wrote:
>>> []
>>>
>>>> Lossless rotation can be maintained by merely keeping the number of
>>>> pixels the same, eg: 3000 x 1500 -> 1500 x 3000 (or whatever
>>>> dimesnions you please).
>>>> The 3008/16 is, as you imply, simply adopting the digital 'rounding'
>>>> boundary that takes place with most digital devices.
>>>
>>>
>>> Alan, I think you will find that lossless rotation requires that the
>>> image dimensions be a multiple of 16 (and not 8 as I incorrectly
>>> stated earlier).
>>
>> I fail to understand why. A matrix of 7 x 11 has the same quantity of
>> information as a matrix of 11 x 7. Assuming the rotation is
>> correctly done (and after conversion of individual R,G,B sensor to
>> RGB pixels) then it makes no matter at all.
>
> If the image is JPEG, then the indivdual blocks are 16 x 16. The size
> quantisation restriction only applies to JPEG images, but of course almost
> all cameras do allow JPEG images as output.
>
> If the image is not JPEG then it can, of course, be losslessly rotated at
> any size - as you say.

Thank you David, exactly correct :-)

--
John Bean

Why is it drug addicts and computer afficionados are both called users?
(Clifford Stoll)
Anonymous
January 2, 2005 9:44:56 PM

Archived from groups: rec.photo.digital.slr-systems (More info?)

leo <someone@somewhere.net> wrote in news:cr9d56021pt@news3.newsguy.com:

> Why not. As the ratio indicates, there is more width info than height in
> the sensor (0.3mm wider). It might even translate into 8 pixels...

The correct answer is already given. 3008 is possible to divide by 16.
Thats it - nothing more to see here - please move along.


/Roland
Anonymous
January 2, 2005 10:42:15 PM

Archived from groups: rec.photo.digital.slr-systems (More info?)

Alan Browne wrote:
[]
> Now I'm really confused. Nothing prevents me from editing a JPG
> image to 1 pixel in either dimension. I can easilly come out with a
> 111 x 213 pixel image in JPG. From there, I can rotate it to 213 x
> 111 and it contains the same info. I can then save it at full
> quality. I've lost nothing.

If you save it as JPEG that will normally produce an extra loss compared
to the original image. However, if you have a JPEG which is exactly a
multiple of 16 in its X and Y dimension it can be losslessly rotated by 90
degrees.

Cheers,
David
Anonymous
January 2, 2005 10:42:16 PM

Archived from groups: rec.photo.digital.slr-systems (More info?)

David J Taylor wrote:

> Alan Browne wrote:
> []
>
>>Now I'm really confused. Nothing prevents me from editing a JPG
>>image to 1 pixel in either dimension. I can easilly come out with a
>>111 x 213 pixel image in JPG. From there, I can rotate it to 213 x
>> 111 and it contains the same info. I can then save it at full
>>quality. I've lost nothing.
>
>
> If you save it as JPEG that will normally produce an extra loss compared
> to the original image. However, if you have a JPEG which is exactly a
> multiple of 16 in its X and Y dimension it can be losslessly rotated by 90
> degrees.

I think I see what you're saying, though I don't know why truncation should
affect the lossiness as the image is edited as an image not as a JPG. (JPG
compression is a function of storage, not editing).

I also assume that at its least lossy setting JPG is lossless (which is not
quite true, but close enough).

In terms of a RAW image it's a non-issue.

So, I believe the reason the engineers chose 3008:2000 is simply for the usual
computer design reasons, not image related reasons. This also follows with the
other saveable image sizes Maxxum 7d/Nikon d70: (2256x1496, 2240x1488,
1504x1000) ... 1496 and 1000 do not divide by 16, but all the dimensions
mentioned divide by 8.

Cheers,
Alan


--
-- r.p.e.35mm user resource: http://www.aliasimages.com/rpe35mmur.htm
-- r.p.d.slr-systems: http://www.aliasimages.com/rpdslrsysur.htm
-- [SI] gallery & rulz: http://www.pbase.com/shootin
-- e-meil: there's no such thing as a FreeLunch.
Anonymous
January 2, 2005 11:29:16 PM

Archived from groups: rec.photo.digital.slr-systems (More info?)

Alan Browne <alan.browne@freelunchVideotron.ca> wrote:

>> Alan, I think you will find that lossless rotation requires that the image
>> dimensions be a multiple of 16 (and not 8 as I incorrectly stated
>> earlier).
>
> I fail to understand why. A matrix of 7 x 11 has the same quantity of
> information as a matrix of 11 x 7. Assuming the rotation is correctly
> done (and after conversion of individual R,G,B sensor to RGB pixels)
> then it makes no matter at all.

It matters for JPEGs, not for images in general. JPEGs can only be
rotated losslessly (without decompressing and then recompressing the
data) if the dimensions are divisible by 16.

--
Jeremy | jeremy@exit109.com
Anonymous
January 2, 2005 11:29:17 PM

Archived from groups: rec.photo.digital.slr-systems (More info?)

Jeremy Nixon wrote:

> Alan Browne <alan.browne@freelunchVideotron.ca> wrote:
>
>
>>>Alan, I think you will find that lossless rotation requires that the image
>>>dimensions be a multiple of 16 (and not 8 as I incorrectly stated
>>>earlier).
>>
>>I fail to understand why. A matrix of 7 x 11 has the same quantity of
>>information as a matrix of 11 x 7. Assuming the rotation is correctly
>>done (and after conversion of individual R,G,B sensor to RGB pixels)
>>then it makes no matter at all.
>
>
> It matters for JPEGs, not for images in general. JPEGs can only be
> rotated losslessly (without decompressing and then recompressing the
> data) if the dimensions are divisible by 16.

I'm beginning to understand that through the force of people saying so and not
because I've looked at the mechanics.

You bring up 'uncompressing and recompressing' and that really is it... a JPG
image is only such when stored. When it is 'in the camera' prior to storage or
in photoshop it is not JPG... it is 'image'. So it can be rotated without loss
regardless of dimensions. It's all semantics, I suppose.

Cheers,
Alan.

--
-- r.p.e.35mm user resource: http://www.aliasimages.com/rpe35mmur.htm
-- r.p.d.slr-systems: http://www.aliasimages.com/rpdslrsysur.htm
-- [SI] gallery & rulz: http://www.pbase.com/shootin
-- e-meil: there's no such thing as a FreeLunch.
Anonymous
January 2, 2005 11:31:56 PM

Archived from groups: rec.photo.digital.slr-systems (More info?)

Alan Browne <alan.browne@freelunchVideotron.ca> wrote:

> Now I'm really confused. Nothing prevents me from editing a JPG image to 1
> pixel in either dimension. I can easilly come out with a 111 x 213 pixel
> image in JPG. From there, I can rotate it to 213 x 111 and it contains the
> same info.
> I can then save it at full quality. I've lost nothing.

Yes, you have, if you saved it as a JPEG. You have re-compressed the image.
If you load a JPEG image, do something to it, and then save it again as a
JPEG, that is a lossy operation due to the compression.

However, a JPEG image with dimensions divisible by 16 can be rotated without
decompressing and recompressing the image at all -- that is, without ever
loading it into an image editor in the first place, requiring no further
step of re-saving as a JPEG and thus recompressing the data.

--
Jeremy | jeremy@exit109.com
Anonymous
January 2, 2005 11:40:36 PM

Archived from groups: rec.photo.digital.slr-systems (More info?)

On Sun, 02 Jan 2005 15:27:39 -0500, Alan Browne wrote:

> David J Taylor wrote:
>
>> Alan Browne wrote:
>> []
>>
>>>Now I'm really confused. Nothing prevents me from editing a JPG
>>>image to 1 pixel in either dimension. I can easilly come out with a
>>>111 x 213 pixel image in JPG. From there, I can rotate it to 213 x
>>> 111 and it contains the same info. I can then save it at full
>>>quality. I've lost nothing.
>>
>>
>> If you save it as JPEG that will normally produce an extra loss compared
>> to the original image. However, if you have a JPEG which is exactly a
>> multiple of 16 in its X and Y dimension it can be losslessly rotated by 90
>> degrees.
>
> I think I see what you're saying, though I don't know why truncation should
> affect the lossiness as the image is edited as an image not as a JPG. (JPG
> compression is a function of storage, not editing).

Lossless rotation is done on the JPEG data itself, not on a decoded image.
The JPEG data is not decoded/rotated/re-encoded as it must be for any JPEG
whose dimensions are not multiples of 16. That's why it's lossless.

> So, I believe the reason the engineers chose 3008:2000 is simply for the usual
> computer design reasons, not image related reasons.

Not so. As I said earlier the Pentax *istDS uses 3008x2008 for raw images
but 3008x2000 for JPEG, for the reasons I've stated. This was a change from
the older *istD, which used 3008x2008 for both - much to the annoyance of
some users.

--
John Bean

Work is the curse of the drinking classes (Oscar Wilde)
Anonymous
January 3, 2005 12:18:05 AM

Archived from groups: rec.photo.digital.slr-systems (More info?)

Alan Browne <alan.browne@freelunchVideotron.ca> wrote:

> You bring up 'uncompressing and recompressing' and that really is it... a JPG
> image is only such when stored. When it is 'in the camera' prior to storage
> or in photoshop it is not JPG... it is 'image'. So it can be rotated without
> loss regardless of dimensions. It's all semantics, I suppose.

The normal case is that someone takes a bunch of pictures, loads them onto his
computer, looks at a screen full of thumbnails, and clicks a "rotate" button
to make the vertical compositions vertical. If the saved images are JPEGs,
this can be done with no recompression loss if the image dimensions are
divisible by 16.

If you don't use JPEG, this doesn't matter to you.

--
Jeremy | jeremy@exit109.com
Anonymous
January 3, 2005 3:26:44 AM

Archived from groups: rec.photo.digital.slr-systems (More info?)

On 2005-01-02, Roland Karlsson <roland_dot_karlsson@bonetmail.com> wrote:
> leo <someone@somewhere.net> wrote in news:cr9d56021pt@news3.newsguy.com:
>
>> Why not. As the ratio indicates, there is more width info than height in
>> the sensor (0.3mm wider). It might even translate into 8 pixels...
>
> The correct answer is already given. 3008 is possible to divide by 16.
> Thats it - nothing more to see here - please move along.
>
>
> /Roland

Anyone ask to see Roland's badge number?

;) 

Will D.
Anonymous
January 3, 2005 8:59:22 AM

Archived from groups: rec.photo.digital.slr-systems (More info?)

In article <RqOdnQgakYJw7kXcRVn-tw@wavecable.com>,
"C J Campbell" <christophercampbellNOSPAM@hotmail.com> wrote:

> "Merritt Mullen" <mmullen8014@mchsi.com> wrote in message

> > And MY next question is "How does 3008x2000 equate to the advertised 6.1
> > megapixels?" Seems to me it should be 6.016 megapixels.

> There are actually 6.24 million pixels on the sensor, of which 3008x2000 are
> used in the max size JPG or in raw NEF files, which Nikon rounds to 6.1
> million pixels.

Yes, I read that somewhere myself, but I still don't see how 3008 x 2000 =
6.016 million pixels "rounds" to 6.1 million pixels. Seems to me it
rounds to 6.0 million pixels.

Can you clarify further?

Thanks,

Merritt
Anonymous
January 3, 2005 8:59:23 AM

Archived from groups: rec.photo.digital.slr-systems (More info?)

"Merritt Mullen" <mmullen8014@mchsi.com> wrote in message
news:mmullen8014-8DF447.21592202012005@netnews.comcast.net...
> In article <RqOdnQgakYJw7kXcRVn-tw@wavecable.com>,
> "C J Campbell" <christophercampbellNOSPAM@hotmail.com> wrote:
>
> > "Merritt Mullen" <mmullen8014@mchsi.com> wrote in message
>
> > > And MY next question is "How does 3008x2000 equate to the advertised
6.1
> > > megapixels?" Seems to me it should be 6.016 megapixels.
>
> > There are actually 6.24 million pixels on the sensor, of which 3008x2000
are
> > used in the max size JPG or in raw NEF files, which Nikon rounds to 6.1
> > million pixels.
>
> Yes, I read that somewhere myself, but I still don't see how 3008 x 2000 =
> 6.016 million pixels "rounds" to 6.1 million pixels. Seems to me it
> rounds to 6.0 million pixels.
>
> Can you clarify further?

Nope. No doubt some lawyer will sue the camera companies for false
advertising, claiming to represent all us consumers as a class. As part of
the settlement, he will be paid millions of dollars, and we will get a
coupon for $10 off on our next purchase of a camera. It will be hailed as "a
great victory for consumers." At least, that is what happened to the
manufacturers of disk drives and computer screens.
Anonymous
January 3, 2005 9:07:21 AM

Archived from groups: rec.photo.digital.slr-systems (More info?)

In article <cr9i6o$stq$1@panix5.panix.com>,
johnf@panix.com (John Francis) wrote:

> In article <mmullen8014-808F42.19370901012005@netnews.comcast.net>,
> Merritt Mullen <mmullen8014@mchsi.com> wrote:

> >And MY next question is "How does 3008x2000 equate to the advertised 6.1
> >megapixels?" Seems to me it should be 6.016 megapixels.
>
> The full sensor imaging area is 3040x2024 (6,152,960 pixels).

Then, to summarize what has been posted so far:

Total number of pixels = 6.24 million

Full sensor imaging area = 6.15 million

Advertised full image size = 6.1 million

Actual stored image = 6.016 million

Do I have that correct? And does any of the above really matter except
for the last number (3008 x 2000 full image size)?

Merritt
Anonymous
January 3, 2005 9:28:11 PM

Archived from groups: rec.photo.digital.slr-systems (More info?)

"Will D." <willd@no.spam> wrote in news:10th4a4r8s0rr05@corp.supernews.com:

> Anyone ask to see Roland's badge number?

Hehe ...


/Roland
Anonymous
January 10, 2005 10:36:13 PM

Archived from groups: rec.photo.digital.slr-systems (More info?)

Alan Browne <alan.browne@freelunchVideotron.ca> wrote:

>I'm beginning to understand that through the force of people saying
>so and not because I've looked at the mechanics.

>You bring up 'uncompressing and recompressing' and that really is it... a JPG
>image is only such when stored. When it is 'in the camera' prior to storage or
>in photoshop it is not JPG... it is 'image'. So it can be rotated without loss
>regardless of dimensions. It's all semantics, I suppose.

You're right, it's because of compressing, but no you don't understand
it; it has nothing to do with semantics.

I'll have a stab at explaining it:

The original image comes out of the camera in JPG. That is lossy
compression with effects; it has artifacts. For illustration, let's
say that horizontal high-contrast areas cause a shadow in the lighter
region.

You open that image (uncompressing it) in your viewer, and edit the
pixels. What you're looking at has loss and artifacts. You do a
pixel-for-pixel rotation (also duly rotating the artifacts). The
result in your editing window, as you point out, is exactly the
same as the original rotated.

Now, you save that edited image as JPG. The saving process compresses,
and introduces new artifacts due to the new horizontal high-contrast
areas (as well as the previous, rotated artifacts of course). This
compression loses more info. When you open the newly formed JPG,
it no longer looks exactly like the original, rotated. It's a lossless
rotation only so long as you don't save it in JPG.

On the other hand, lossless rotation is a mathematical operation on
the original JPG, effectively transforming it so that it uncompresses
into a rotated version. There is no additional loss, no additional
artifacts.

Hope that helps?

--
Ken Tough
!