8Mp Vs 5Mp

A rough back-of-the-napkin calculation shows that a 400 sq mm sensor with
6 megapixels gives about 67 millionths of a square millimeter per pixel,
while a 48 sq mm sensor with 8 megapixels gives six millionths of a square
millimeter per pixel.

That rough back-of-the-napkin calculation shows some remarkable acumen; did you
actually do it roughly on the back of a napkin or was that just a figure of
speech? :-p

A rough back-of-the-napkin calculation shows that a 400 sq mm sensor with
6 megapixels gives about 67 millionths of a square millimeter per pixel,
while a 48 sq mm sensor with 8 megapixels gives six millionths of a square
millimeter per pixel.

That rough back-of-the-napkin calculation shows some remarkable acumen; did you
actually do it roughly on the back of a napkin or was that just a figure of
speech? :-p

"Paul Wylie" wrote in message
...
Jim Townsend wrote:
[...]
The sensor is almost 3 times bigger in
size. This allows for MUCH larger sensor sites and as a result,
less amplification is required and you get less noise.

Small quibble here. The 2/3" sensor has an area of about 48 sq mm, using
your dimensions, and the APS-C sensor has an area of about 400 sq mm,
again using your dimensions.

This results in a sensor that's about 8.3 times bigger, not three times
bigger.

A rough back-of-the-napkin calculation shows that a 400 sq mm sensor with
6 megapixels gives about 67 millionths of a square millimeter per pixel,
while a 48 sq mm sensor with 8 megapixels gives six millionths of a square
millimeter per pixel.

Thus, the DSLR's combination of larger sensor and fewer photosites adds up
to each photosite having about 11 times the area available to each
photosite in the sensor of an 8MP prosumer digicam.

Damn. I knew there was a big difference, but until I did those
calculations, I had no clue how dramatic it was.

No, I don't think that's right. A 2/3" diagonal sensor in a 4:3 ratio has a
width of

4/5 * .66666 * 2.54 = 13.5mm and a height of
3/5 * .66666 * 2.54 = 10.15mm

So the area is 137 sq mm not 48. That's about 1/3 the size of a 400 sq mm
sensor.

Given the 8MP vs 6MP difference, each 6MP APS pixel is about 4 times the
size of each 8MP 2/3" pixel. All else being equal, that would mean about 2
extra stops of 'film speed' -- a DSLR at ISO 1600 is about as noisy as an
8MP compact at ISO 400.

Mark

"Paul Wylie" wrote in message
...
Jim Townsend wrote:
[...]
The sensor is almost 3 times bigger in
size. This allows for MUCH larger sensor sites and as a result,
less amplification is required and you get less noise.

Small quibble here. The 2/3" sensor has an area of about 48 sq mm, using
your dimensions, and the APS-C sensor has an area of about 400 sq mm,
again using your dimensions.

This results in a sensor that's about 8.3 times bigger, not three times
bigger.

A rough back-of-the-napkin calculation shows that a 400 sq mm sensor with
6 megapixels gives about 67 millionths of a square millimeter per pixel,
while a 48 sq mm sensor with 8 megapixels gives six millionths of a square
millimeter per pixel.

Thus, the DSLR's combination of larger sensor and fewer photosites adds up
to each photosite having about 11 times the area available to each
photosite in the sensor of an 8MP prosumer digicam.

Damn. I knew there was a big difference, but until I did those
calculations, I had no clue how dramatic it was.

No, I don't think that's right. A 2/3" diagonal sensor in a 4:3 ratio has a
width of

4/5 * .66666 * 2.54 = 13.5mm and a height of
3/5 * .66666 * 2.54 = 10.15mm

So the area is 137 sq mm not 48. That's about 1/3 the size of a 400 sq mm
sensor.

Given the 8MP vs 6MP difference, each 6MP APS pixel is about 4 times the
size of each 8MP 2/3" pixel. All else being equal, that would mean about 2
extra stops of 'film speed' -- a DSLR at ISO 1600 is about as noisy as an
8MP compact at ISO 400.

Mark

Paul Wylie wrote:

Small quibble here. The 2/3" sensor has an area of about 48 sq mm, using
your dimensions, and the APS-C sensor has an area of about 400 sq mm,
again using your dimensions.

Well, to be accurate a 2/3" sensor is 6.6 x 8.8 mm, which is 58 sq mm,
not 48. The Nikon D70 uses a 23.7 x 15.6 mm sensor, with an area of 370
sq mm.

This results in a sensor that's about 8.3 times bigger, not three times
bigger.

370 / 58 = 6.4

A rough back-of-the-napkin calculation shows that a 400 sq mm sensor with
6 megapixels gives about 67 millionths of a square millimeter per pixel,
while a 48 sq mm sensor with 8 megapixels gives six millionths of a square
millimeter per pixel.

Thus, the DSLR's combination of larger sensor and fewer photosites adds up
to each photosite having about 11 times the area available to each
photosite in the sensor of an 8MP prosumer digicam.

58 / 8 = 7.3 sq micrometer
370 / 6 = 62 sq micrometer
62 / 7.3 = 8.5

It's a factor 8.5, not 11. With the 300D, which has 6.3 MP, it's 8.

Damn. I knew there was a big difference, but until I did those
calculations, I had no clue how dramatic it was.

Thermal noise is proportional to the square root of the areas, meaning
that you end up with noise levels approx. 2.8 times higher in the 8MP
2/3" camera. 2.8 is about 9 dB - we are talking about 9 additional dB
signal to noise ratio, at the cost of a sensor eight times the size.

Add to that that DSLRs are frequently used at higher ISOs - 200, 400 or
higher, because apertures are not so big (affordable zooms starting
around F4 at wide angle) and DSLR lenses reach their optimum at F8
anyway, and the advantage is not so huge anymore.

In fact, if you use these 8MP prosumer cameras at the lowest ISO, noise
is quite acceptable.

I'm about to get an Olympus 8080 (instead of a DSLR) and will keep it
until compact DLSRs with live preview and foldable LCD screens hit the
market.
--

Alfred Molon
------------------------------
http://groups.yahoo.com/group/Olympus_405080/
Olympus 5050 resource - http://www.molon.de/5050.html
Olympus 5060 resource - http://www.molon.de/5060.html
Olympus 8080 resource - http://www.molon.de/8080.html

Paul Wylie wrote:

Small quibble here. The 2/3" sensor has an area of about 48 sq mm, using
your dimensions, and the APS-C sensor has an area of about 400 sq mm,
again using your dimensions.

Well, to be accurate a 2/3" sensor is 6.6 x 8.8 mm, which is 58 sq mm,
not 48. The Nikon D70 uses a 23.7 x 15.6 mm sensor, with an area of 370
sq mm.

This results in a sensor that's about 8.3 times bigger, not three times
bigger.

370 / 58 = 6.4

A rough back-of-the-napkin calculation shows that a 400 sq mm sensor with
6 megapixels gives about 67 millionths of a square millimeter per pixel,
while a 48 sq mm sensor with 8 megapixels gives six millionths of a square
millimeter per pixel.

Thus, the DSLR's combination of larger sensor and fewer photosites adds up
to each photosite having about 11 times the area available to each
photosite in the sensor of an 8MP prosumer digicam.

58 / 8 = 7.3 sq micrometer
370 / 6 = 62 sq micrometer
62 / 7.3 = 8.5

It's a factor 8.5, not 11. With the 300D, which has 6.3 MP, it's 8.

Damn. I knew there was a big difference, but until I did those
calculations, I had no clue how dramatic it was.

Thermal noise is proportional to the square root of the areas, meaning
that you end up with noise levels approx. 2.8 times higher in the 8MP
2/3" camera. 2.8 is about 9 dB - we are talking about 9 additional dB
signal to noise ratio, at the cost of a sensor eight times the size.

Add to that that DSLRs are frequently used at higher ISOs - 200, 400 or
higher, because apertures are not so big (affordable zooms starting
around F4 at wide angle) and DSLR lenses reach their optimum at F8
anyway, and the advantage is not so huge anymore.

In fact, if you use these 8MP prosumer cameras at the lowest ISO, noise
is quite acceptable.

I'm about to get an Olympus 8080 (instead of a DSLR) and will keep it
until compact DLSRs with live preview and foldable LCD screens hit the
market.
--

Alfred Molon
------------------------------
http://groups.yahoo.com/group/Olympus_405080/
Olympus 5050 resource - http://www.molon.de/5050.html
Olympus 5060 resource - http://www.molon.de/5060.html
Olympus 8080 resource - http://www.molon.de/8080.html

Mark Weaver wrote:

"Paul Wylie" wrote in message
...
Jim Townsend wrote:
[...]
The sensor is almost 3 times bigger in
size. This allows for MUCH larger sensor sites and as a result,
less amplification is required and you get less noise.

Small quibble here. The 2/3" sensor has an area of about 48 sq mm, using
your dimensions, and the APS-C sensor has an area of about 400 sq mm,
again using your dimensions.

This results in a sensor that's about 8.3 times bigger, not three times
bigger.

A rough back-of-the-napkin calculation shows that a 400 sq mm sensor with
6 megapixels gives about 67 millionths of a square millimeter per pixel,
while a 48 sq mm sensor with 8 megapixels gives six millionths of a square
millimeter per pixel.

Thus, the DSLR's combination of larger sensor and fewer photosites adds up
to each photosite having about 11 times the area available to each
photosite in the sensor of an 8MP prosumer digicam.

Damn. I knew there was a big difference, but until I did those
calculations, I had no clue how dramatic it was.

No, I don't think that's right. A 2/3" diagonal sensor in a 4:3 ratio has a
width of

4/5 * .66666 * 2.54 = 13.5mm and a height of
3/5 * .66666 * 2.54 = 10.15mm

So the area is 137 sq mm not 48. That's about 1/3 the size of a 400 sq mm
sensor.

Mark,

see here for the sensor sizes:

http://www.dpreview.com/learn/?/Glos.../Sensor_Sizes_
01.htm
--

Alfred Molon
------------------------------
http://groups.yahoo.com/group/Olympus_405080/
Olympus 5050 resource - http://www.molon.de/5050.html
Olympus 5060 resource - http://www.molon.de/5060.html
Olympus 8080 resource - http://www.molon.de/8080.html

Mark Weaver wrote:

"Paul Wylie" wrote in message
...
Jim Townsend wrote:
[...]
The sensor is almost 3 times bigger in
size. This allows for MUCH larger sensor sites and as a result,
less amplification is required and you get less noise.

Small quibble here. The 2/3" sensor has an area of about 48 sq mm, using
your dimensions, and the APS-C sensor has an area of about 400 sq mm,
again using your dimensions.

This results in a sensor that's about 8.3 times bigger, not three times
bigger.

A rough back-of-the-napkin calculation shows that a 400 sq mm sensor with
6 megapixels gives about 67 millionths of a square millimeter per pixel,
while a 48 sq mm sensor with 8 megapixels gives six millionths of a square
millimeter per pixel.

Thus, the DSLR's combination of larger sensor and fewer photosites adds up
to each photosite having about 11 times the area available to each
photosite in the sensor of an 8MP prosumer digicam.

Damn. I knew there was a big difference, but until I did those
calculations, I had no clue how dramatic it was.

No, I don't think that's right. A 2/3" diagonal sensor in a 4:3 ratio has a
width of

4/5 * .66666 * 2.54 = 13.5mm and a height of
3/5 * .66666 * 2.54 = 10.15mm

So the area is 137 sq mm not 48. That's about 1/3 the size of a 400 sq mm
sensor.

Mark,

see here for the sensor sizes:

http://www.dpreview.com/learn/?/Glos.../Sensor_Sizes_
01.htm
--

Alfred Molon
------------------------------
http://groups.yahoo.com/group/Olympus_405080/
Olympus 5050 resource - http://www.molon.de/5050.html
Olympus 5060 resource - http://www.molon.de/5060.html
Olympus 8080 resource - http://www.molon.de/8080.html

You have me confused...
I'm considering the D-70 as a "real" companion to my Sony 717 (which I'll
use as a EDC), and the reason is the ability to use higher speeds with less
noise as the light drops (since most of my landscape work is just before
dusk).

The max f-stop for me will be 1.8 to 2.8 because I'll be using primes, not
zooms.

Like you pointed out, I'll still be stopping down to f8, but I'd be doing
that regardless of what cameras I'm using, so I think you DO benefit more
from the low light potential more than the number of megapixels.

"Alfred Molon" wrote in message
...
In fact, if you use these 8MP prosumer cameras at the lowest ISO, noise
is quite acceptable.

I'm about to get an Olympus 8080 (instead of a DSLR) and will keep it
until compact DLSRs with live preview and foldable LCD screens hit the
market.
--

Alfred Molon
------------------------------
http://groups.yahoo.com/group/Olympus_405080/
Olympus 5050 resource - http://www.molon.de/5050.html
Olympus 5060 resource - http://www.molon.de/5060.html
Olympus 8080 resource - http://www.molon.de/8080.html

You have me confused...
I'm considering the D-70 as a "real" companion to my Sony 717 (which I'll
use as a EDC), and the reason is the ability to use higher speeds with less
noise as the light drops (since most of my landscape work is just before
dusk).

The max f-stop for me will be 1.8 to 2.8 because I'll be using primes, not
zooms.

Like you pointed out, I'll still be stopping down to f8, but I'd be doing
that regardless of what cameras I'm using, so I think you DO benefit more
from the low light potential more than the number of megapixels.


And in turn, you have me confused there...

This must be the first time i hear someone speak of using an SLR as a
"companion" to a P&S, it's usually the other way round. BTW, what's EDC short
for? this may be the crux of my confusion.

Also, why are you considering getting a digital SLR for a specific usage of
just-before-dusk landscape? For a fraction of what it'd cost you to get the
D-70 and suitable lenses, I suggest you get yourself a medium format camera kit
and If you need to digitze it get the Epson 4870 scanner and that'll give you
circa 22-megapixel on 645 and circa 33-megapixel on 6x6, and more if you use
6x7 or 6x9.