I have previously owned a D610 and I find the image quality of the D800 to be somewhat inferior. The following at 1:1 crops that are much more noisy than the d610.

The following were shot at ISO 800 and ISO 1600 respectively. At first I thought, it might be the digital amplification at non-native ISOs. But at present I am lost. What might be causing this loss of Image Quality?

Note: No in-camera NR, No In-camera sharpening, Base LR sharpening (25)

1/500 f4.5 ISO 1600 enter image description here

1/1000 f4.5 ISO 800 enter image description here

  • Could you explain the calculation to me? The images were therefore 0.67EV underexposed? How would you estimate the image to be 15EV?
    – ABCD312
    Aug 11 '18 at 19:09
  • Try remove the sharpening. Also I remember in camera raw it by default does some chroma noise reduction, which I believe is present here. Try also remove that and see what happens.
    – MetroWind
    Aug 11 '18 at 19:31
  • Shooting at EV 14.33 if the scene called for EV 15 would overexpose the image by 2/3 stop. Think about it. EV 0 is for 1 second at f/1 for ISO 100. Thats a much higher exposure of the same amount of light than EV 15 at 1/1000 at f/5.6! The scene, however, does not appear to be in direct sunlight, but rather in the shade or under cloudy skies, which would probably make EV 14.33 a bit too dark for the available light.
    – Michael C
    Aug 11 '18 at 20:44
  • @MichaelClark Oops, you're right. One (I?) should not post without any coffee-input and after 12 hours of work. Thanks for bringing up my nonsense - deleted it just now!
    – flolilo
    Aug 11 '18 at 22:03
  • EV 0 is for 1 second at f/1 .. is EV 0, but IT IS NOT ABOUT ISO 100. EV is about whatever ISO is being used to make those settings be of interest. The ducks exposure is 1/1000 f/4.5 at ISO 1600, which is EV 14.33 at any ISO (or equivalent exposure of Light Value EV 10.33 at ISO 100). Sunny 16 says the ducks in bright sun would be 1/1600 f/16 at ISO 1600. But the ducks are barely casting only the very vaguest shadows, which Sunny 16 says is about 2 EV down. However the 1/1000 f/4.5 is 4.33 EV more exposure than bright sun, so probably not underexposed. The rocks at top are pretty bright.
    – WayneF
    Aug 12 '18 at 0:09

The D610 is a 24MP camera.
The D800 is a 36MP camera.

When you are viewing images from both at 100%, you are magnifying the D800 images by a larger enlargement ratio than you are magnifying the images from the D610.

If you are viewing the images on a 23" HD (1920x1080) monitor, you are enlarging the D800 images to the same scale as a print about 77 x 51 inches. With the D610, your enlargement ratio is equivalent to a 63 x 42 inches. In terms of area, the D800 image is being enlarged by 48% more than the image from the D610! This also means the area of each sensel (pixel site) of the D610 is 1.48X the area of each sensel on the D800 sensor.

The higher enlargement ratio allows you to see the detail of the smaller pixels on the D800 sensor at the same size per pixel as the larger pixels of the D610. With smaller sensels (pixel sites), Poisson distribution noise, caused by the random distribution of photons, will be more noticeable.

If you normalize both images to the same display size you should see much less difference in the noise between the two respective sensors.

The D610, a camera introduced in October 2013, might also be expected to have a little more advanced sensor design and better noise reduction than the D800, introduced twenty months earlier in February of 2012.

Beyond that, the sensors in the D800 and the D610 might not be of similar design technology. Some cameras have sensors that do more on-die noise reduction while other cameras have sensors that do less on-die noise reduction and apply more noise reduction in the off-sensor processing pipeline.

Both the Nikon D800 and the D610 have Nikon Expeed 3 image processors. Both cameras have 'Sony' sensors (although it must be remembered that just because a sensor is manufactured by Sony Semiconductor for Nikon does not necessarily mean a sensor was designed by Sony Imaging for Nikon).

The example images also appear to be moderately underexposed, which will increase the noticeable amount of noise by decreasing the signal to noise (S/N) ratio. When read noise is constant and less light is captured in the exposure, the noise is a higher percentage of the total signal coming off the sensor.

Compare the D800 and the D610 at DxO Mark (the D600 is also included in the linked comparison):

With the size normalized (the 'Print' specification), there is virtually no difference between the two cameras with regards to S/N ratio or dynamic range.

enter image description here

enter image description here

Yet when we change 'Print' to 'Screen', in which the size is not normalized, we see a more noticeable difference:

enter image description here

enter image description here

The exact values for the D610 at ISO 1600 when we hover over the chart:

  • DR: 10.45 EV
  • S/N ratio: 29.3 dB

The exact values for the D800 at ISO 1600 when we hover over the chart:

  • DR: 10.03 EV
  • S/N ratio: 27.7 dB

A change of 1 EV is one stop of DR.
A change of 3 dB is one stop of S/N ratio.

At ISO 1600, the D610 is roughly 1/2 stop better in terms of DR and and 1/3 stop better in terms of S/N ratio compared to the D800 when images from both are viewed at 100%. This is pretty close to what would be predicted by a camera having sensels 1.5X the areal size of the other.

  • Fair point Michael, a couple of additions, corrections. The d610, whilst newer than the d800, is just a technicality since not much other than the shutter mechanism was improved on to the best of my knowledge. So the D800 and D600 are roughly from the same generation. The reason for my apprehension was also the infamous shutter-slap of the D800.
    – ABCD312
    Aug 11 '18 at 22:30
  • Here's a link to dpreview.com where they have an app that compares ISO performance: dpreview.com/reviews/… As you can see, the D610 has better ISO performance than the D800.
    – frank
    Aug 12 '18 at 0:55
  • 1
    @Chai The sensor may or may not be the same between the D600 and D610, but both have larger sensels than the D800 sensor, and that is where most of the difference between the D610 and D800 is coming from. DxO Mark shows enough difference between the D610 and D600 to assume that even if the sensor hardware was the same, they tweaked the amplification multipliers at each ISO setting a bit between the D600 and D610.
    – Michael C
    Aug 12 '18 at 4:31
  • I am aware and I agree michael, was just letting you know about the sensors.
    – ABCD312
    Aug 12 '18 at 14:07

I don't know, but wondered if maybe you applied too much sharpening?

This is my D800, at ISO 1600, and a 100% crop (with no noise processing):

enter image description here

  • I should have mentioned they are at base sharpening in Lightroom and Zero in camera sharpening.
    – ABCD312
    Aug 11 '18 at 18:45

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