Pixel density question.

Phil Young

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Hoppy (and others)

Here is an open question to be explained if you please :) :)

When trying to work out pixel density against a Nikon D800 I asked a physics Bsc to work out how many pixels I would need on my 1.5x crop to get the same amount of pixels on that sensor to equal the same pixel density of the D800 36mp.

The 2 methods used came the same answer of 22mp but everyone says the D7000 has the same at 16.2

Can anyone explain?

Cheers,
Phil.
 
Last edited:
Phil Young said:
Hoppy (and others)

Here is an open question to be explained if you please :) :)

When trying to work out pixel density against a Nikon D800 I asked a physics Bsc to work out how many pixels I would need on my 1.5x crop to get the same amount of pixels on that sensor to equal the same pixel density of the D800 33mp.

The 2 methods used came the same answer of 22mp but everyone says the D7000 has the same at 16.2

Can anyone explain?

Cheers,
Phil.

Your maths, methods and figures are wrong is the short answer. :) multiple the number mp on your d7000 by the increase in sensor size of the d800 and you'll get the number of mp on the d800 (36 not 33)
 
Your maths, methods and figures are wrong is the short answer. :) multiple the number mp on your d7000 by the increase in sensor size of the d800 and you'll get the number of mp on the d800 (36 not 33)

Oops sorry, 36.

It comes up with 24.3 for both methods (and these are not my methods by someone who has a degree in physics).

What should the calculation be exactly??
 
Your friend forgot (or was unaware) that the a 1.5 crop sensor is 1.5 times smaller in BOTH width and height, so to convert you have to divide the 36 MP by 1.5 TWICE.

(36 / 1.5) / 1.5 = 16
 
Phil Young said:
Oops sorry, 36.

It comes up with 24.3 for both methods (and these are not my methods by someone who has a degree in physics).

What should the calculation be exactly??

A full frame sensor is 2.25 times as big as a 1.5 crop sensor. As above its area and not crop factor affects this
 
Yes - It's not the diagonal measurement it's how many times bigger the area of the sensor is.

36.3Mp D800 sensor is 2.25 times larger than a D700 so 36.3/2.25 = 16.1MP

edit - oops beaten to it.
 
Is there not an easy answer to the OP's question without delving into maths.
Switch a D800/E onto DX mode and it will provide a 15.3 MP image.
 
I think that's a fairly simple way of explaining and is exactly the answer I was looking for.

Good on that Phil.

The D800/E can also be set to produce a 1.2 crop factor giving a 25.1 MP image,
and a 5:4 Image area producing 30.2 MP.
 
As I understand, Nikon do not have a camera that can resize their RAW file?

...Or then dx mode is not just resizing an image but actually does only use 1.5x less the normal sensor size?

You're right in that theres no Nikon equivilent of canon's sraw.

But the various crop modes only use the areas of the sensor needed. So you can't get the FX area back on a DX crop mode photo. RAW files in DX mode (14 bit compressed lossless) are about 20mb
 
People, I know this isn't directly related to the OP's question and maybe I'm being a bit of a pedant but can we stop with the "1.5x less" or "1.5 times less"? Unless you can show me how to write that in a formula, of course.

Mathematically it should be expressed as "0.667 times", "0.667x" or "66.667% of".

Okay, rant over ... carry on as normal :)

PS: Yes I know I've truncated the numbers above, but how many 6s do you want after the decimal point?
 
People, I know this isn't directly related to the OP's question and maybe I'm being a bit of a pedant but can we stop with the "1.5x less" or "1.5 times less"? Unless you can show me how to write that in a formula, of course.

Mathematically it should be expressed as "0.667 times", "0.667x" or "66.667% of".

Okay, rant over ... carry on as normal :)

PS: Yes I know I've truncated the numbers above, but how many 6s do you want after the decimal point?

alright then 1/1.5x :D
 
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