3+ meg Uzi... Can someone do the math?

Yes, I am. CCDs are not linear, they are 2 dimensional. You must take a ratio of area. This is why, as you go from 1.5 to 2 MP its not such a big deal. UZI has 37% more pixels than E-100 but only 17% more resolution in each direction. Double the resolution, quadruple the pixels.

When we start using holographic cameras in 3D we'll have to cube it all and the PentaHixels will be off the chart....

PentaHixels - remember that, 'cause I just invented it!

1 PentaHixel = 2^50 Holographic Pixels, also known as a "Yote"

Smoke24 said:

Bob,
I figured this way: .5/.67=.746X5=3.73MP
The .5 CCD is 74.6% of the size of the .67 and is a diagonal
measurement and not square. So if you actually use only 1/2" of the
2/3 CCD you should get the 3.73 megapixels. Don't know where Yote
got to using the square root, maybe he's a better mathmetician than
me?
The Smokester
--
http://www.pbase.com/smoke24/galleries

Click to expand...

--Yote

Hmm... I don't know.... I'd rather have a smoke than a Yote. ;-)

Maxven

P.S.: Yote's calculations are correct as I wrote before. I forgot to sqare a Yote in order to get a correct newby.

Yote said:

When we start using holographic cameras in 3D we'll have to cube it
all and the PentaHixels will be off the chart....

PentaHixels - remember that, 'cause I just invented it!

1 PentaHixel = 2^50 Holographic Pixels, also known as a "Yote"

Smoke24 said:

Bob,
I figured this way: .5/.67=.746X5=3.73MP
The .5 CCD is 74.6% of the size of the .67 and is a diagonal
measurement and not square. So if you actually use only 1/2" of the
2/3 CCD you should get the 3.73 megapixels. Don't know where Yote
got to using the square root, maybe he's a better mathmetician than
me?
The Smokester
--
http://www.pbase.com/smoke24/galleries

Click to expand...

--
Yote

Click to expand...

Ok here, gimme a second... I fully agree the measurements are diagonal - that's why I have them as "c" in the equations. I don't follow where
you say you can't square the 0.5 or .67 -- if you mean to simply say
it's 0.5 x 0.5 to get the area then I absolutely agree. I don't believe I am
doing that, even though given the special circumstance of it being a square
appears to yield the SAME result if you did. In fact, that's exactly what
it looks like you did in your original post:

Yote said:

Yote said:

(.5x.5) (.67x.67) x 5 Meg = 2.78Meg

Click to expand...

Click to expand...

I also agree that changing the proportions of the CCD from 1:1
alters the area somewhat.

What part of:

(0.5^2 / 2) / (0.67^2 2) * 5MP = 2.78458454 (same number as yours,
just more digits of precision)

isn't correct? I did not square "a", I substituted (c^2/2) for "a^2" to
get the area, and then multiplied the ratio of that area to the area
of the 0.67" CCD times 5MP.

Using your own number, a = sqrt(.25/2), a=0.353553391
that's one side. the area of the square then is a*a = 0.125
similarly you can compute the length of one side of the 0.67"
CCD to be 0.473761543, which gives an area of 0.22445,
yielding a ratio of 0.556916908 between the two, times 5MP
gets you back to 2.78458454.

I'm simply curious at this point... I want to get it straight in my head.

We can take this offline if it's putting everyone else to sleep.

Yote said:

Inigo, the measurements are diagonal, so you can't square the .5
and .67. First, calculate the sides which is 2a^2 = .5^2 so a =
sqrt(.25/2).

Thats why yours came out a little low.

I did the 4:3 ratio. Same thing but not a*a, its a*b. Ratio ends up
.12/.2135 and we get a slightly smaller number.... 2.809 Mp.

The .5 inch CCD is a 3:4:5 right triangle. The .667 is of course in
the same ratio w/different numbers - .4002x.5336 yada yada yada

Click to expand...

I am correct. You are correct. Same calc. Difference is using .67 and .667.
Square it and round off gets worse.

Inigo Montoya said:

Inigo Montoya said:

Inigo Montoya said:

(.5x.5) (.67x.67) x 5 Meg = 2.78Meg

Click to expand...

Click to expand...

I also agree that changing the proportions of the CCD from 1:1
alters the area somewhat.

What part of:

(0.5^2 / 2) / (0.67^2 2) * 5MP = 2.78458454 (same number as yours,
just more digits of precision)

isn't correct? I did not square "a", I substituted (c^2/2) for
"a^2" to
get the area, and then multiplied the ratio of that area to the area
of the 0.67" CCD times 5MP.

Using your own number, a = sqrt(.25/2), a=0.353553391
that's one side. the area of the square then is a*a = 0.125
similarly you can compute the length of one side of the 0.67"
CCD to be 0.473761543, which gives an area of 0.22445,
yielding a ratio of 0.556916908 between the two, times 5MP
gets you back to 2.78458454.

I'm simply curious at this point... I want to get it straight in my
head.

We can take this offline if it's putting everyone else to sleep.

Yote said:

Inigo, the measurements are diagonal, so you can't square the .5
and .67. First, calculate the sides which is 2a^2 = .5^2 so a =
sqrt(.25/2).

Thats why yours came out a little low.

I did the 4:3 ratio. Same thing but not a*a, its a*b. Ratio ends up
.12/.2135 and we get a slightly smaller number.... 2.809 Mp.

The .5 inch CCD is a 3:4:5 right triangle. The .667 is of course in
the same ratio w/different numbers - .4002x.5336 yada yada yada

Click to expand...

Click to expand...

--Yote

I think the Yote is correct too. I was trying to pack in more pixels into that smaller area, ya know, cheat a little. I don't remember a 15" monitor or even a 17" although I had both, I had a 21" Mitsubishi till lightening got it, now I'm back on my trusty old Mitsubishi 19". Got a 15" Viewsonic flat panel that measures 17" diagonal but it is too darn small and doesn't give as true color so I use it on my test bench.--www.pbase.com/smoke24/galleries

After a good night's sleep and a cup of coffee I felt like giving it a try.

I started with the assumption that the heigth of the chip must be .75 of the width (1200/1600). Used the diagonal measurement (.5 and .67) and solved for x (and .75x).

I came up with the Oly chip being .12 square inches, the Nikon being .22 square inches.

That being more or less correct I figure that the Oly chip packs 16.7 megs per square inch, the Nikon 22.7.

Given that the Oly C2100 chip has .55 of the area of the Nikon, you should be able to trim down a Nikon chip to .3" x.4" and it would have about 2.75 megs.

Well, not 3 megs (unless you're in marketing), but things area getting better.

Now, let's see how Nikon handles the noise issue with it's denser chip.

b

the 2100 CCD is 0.55" diag CCD.
Leo

BobTrips said:

The way I read the specs, Oly gets 2 megs from a .5" chip. That
(as I understand it) is the diagonal measurement of the chip.

Nikon (in the new 5000) gets 5 megs from a .67" chip.

So, if you were to shave the Nikon chip down to .5" in order to use
the Canon 10xIS lens how many megs would you have left?

This might be the next step up that we could expect to see.

Click to expand...