D500 + 300/4E Are Equal To :

<p>Help me figure out the equivalence on FX if I use these combo:<br>

D500 (on 1.3x format) + nikon 300/4E + TC 1.4x<br>

are equivalent to = 300 x 1.5 x 1.3 x 1.4 = 819 mm F5.6 on FX Format.<br>

Correct?<br>

Thanks in advance.</p>

 

<p>Getting complicated there Edward! Nikon DX is 1.5 so 300mm x 1.5 gives you 450mm FX equivalent, multiply that by 1.4 for the teleconverter and you get 630mm FX equivalent multiply that by 1.3 for in camera crop of your image and you get 819mm. But with a camera like the D500 you might be as well to leave it in full DX mode unless you are really going to need that kind of focal length as I doubt it will be very much slower and it might give you better options for cropping afterwards.</p>

In terms of angle of view, yes; in terms of signal to noise ratio, resolution, depth of field: not equivalent.

<p>Thanks Allan and Ilkka. I just recently purchased a D500 and how it basically put my D800e on dry box most of the time. Thinking of getting a 300/4E and how it fits perfectly with D500 for 450mm range.</p>

<blockquote>D500 (on 1.3x format) + nikon 300/4E + TC 1.4x<br />

are equivalent to = 300 x 1.5 x 1.3 x 1.4 = 819 mm F5.6 on FX Format.</blockquote>

 

<p>Where did the f/5.6 come from? 300x1.5x1.3x1.4 = 819mm. f/(4 x 1.5 x 1.3 x 1.4) = f/10.92 from a depth of field/noise perspective. f/(4 x 1.4 and ignore all cropping) from an ISO setting perspective, yes.<br />

<br />

Put another way, the 300mm f/4 has a 75mm effective front aperture. That controls the cone of confusion and the light gathering ability. If it's effectively on the end of an 819mm lens, you've got 75=f/aperture, or aperture = 819/75 = 10.92. And you'll probably be a bit diffraction-limited.<br />

<br />

The D500 is exactly like using the DX crop of the D810 (or cutting the same pixels out of your final image, thereby enlarging the noise relative to the total image size) - although the D500 sensor is a bit better than the D800/D810 per unit area in terms of noise response (by technology moving on), and you have a few more pixels. Likewise the 1.3x crop on the D500 is the same as cutting a 1.3x smaller image out of the D500 image. No magic.<br />

<br />

The reason a genuine 800mm f/5.6 lens is expensive is, in large part, because its front elements are roughly 800/5.6=142mm across. Optically perfect glass with an area of 160 cm² costs a lot more than optical glass with 44 cm² area (in the 300 f/4), and a compact camera with an "800mm-equivalent" f/5.6 lens on it can't do the same thing as an actual 800mm f/5.6. Of course, an actual 800mm f/5.6 has a few downsides too, especially if you want to move it.<br />

<br />

I notice the Sigma 800mm f/5.6 is still available. :-)</p>

<p>Andrew, the only thing in the equation that alters the aperture is the 1.4x converter. So the f/4 lens + converter combo will have an effective aperture of f/5.6. That's regardless of any format cropping (digital zooming - call it what you will).</p>

<p>OK, the depth-of-field will be equivalent to an 800mm f/11 (ish) lens on full-frame, but with its improvements in sensor S/N the D500's noise won't equal that of the D800E pushed two stops further in ISO.</p>

<p>I'm surprised at you Andrew! Not had your Weetabix this morning?</p>

<p>S/N ratio on D500 and D7200 are exactly the same down to ISO 12800. D500 is only improved below ISO 12800, actually about 1/3rd of a stop at ISO 25600. (source dxomark).</p>

<p>D800 has exactly the same S/N ratio per square area of sensor as those two cameras. But the FX sensor is bigger of course, about 2.2 times. That equals about 1.1 stop of noise advantage to the D800.</p>

<p>When we crop DX 1.3x the total crop becomes 1.5x1.3=1.95. Essentially we are now shooting m43 sensor size, 110 film for you old timers. 1.95x crop means FX area is 3.8 times larger which equates 1.9 stop noise advantage to D800 - if we are shooting down to ISO 12800 on the D500. Say roughly 1.5 stops if we are lower. It's hard to beat the advantage of a larger sensor.</p>

<p> </p>

<p>Oh yes, the aperture remains f/5.6, which is why the ISO figures and metering are all correct. Just not necessarily helpful. Similarly, the lens with the teleconverter becomes a 1.4x300mm lens, and the crop doesn't affect this at all.<br />

<br />

Saying that the crop is <i>equivalent</i> to using a 1.5x1.3 teleconverter, and so we can talk about an 819mm lens, means that we should also talk about the <i>equivalent</i> aperture of this set-up. This doesn't affect ISO, but since enlarging the crop enlarges the noise, it would be pretty "equivalent" to say that the 1.5x1.3mm crop on the 1.4x300mm f/(4 x 1.4) lens is like shooting at f/10.92 with the full-frame camera at an ISO multiplied by (1.5x1.3)². ISO as a sensitivity measure makes sense because you shouldn't have to change exposure just because you're cropping an image, but effectively it works because noise, focal length and aperture cancel out.<br />

<br />

In actuality, the D500's sensor will do a bit better than a D8x0 sensor if those are what you're comparing, as RJ says, unless I'm misinterpreting DxO's numbers: the D810 has worse per-pixel ("screen") performance than the other cameras, and it's achieving this with larger pixels (the DX crop is about 16MP); dynamic range at base ISO is only slightly better on a D810 than a D7200 (and only because the D810 has a lower base ISO), and the D500 has the best tonal range score of the three. I don't think DxO scores are normalised per sensor area, though I'll feel silly if I'm wrong. But within one sensor, a 1.3x vs 1.0x crop of the D500 sensor will behave directly like this. I'm looking forward to a D810 replacement with some mild sensor improvements (to go with the D5's AF!)<br />

<br />

Edit: Pete, where do you get the per-area figures? I'm extrapolating, and may be wrong.</p>

<p><em>I don't think DxO scores are normalised per sensor area, though I'll feel silly if I'm wrong.</em></p>

<p>The main scores are derived from the "print" data which is normalized to 8MP if I recall correctly (not by area!). There is also "screen" data which is by pixel AFAIK, but it's not used to calculate the scores.</p>

<p>The scores are intended to reflect the quality you get when you use the whole sensor area for the picture, purely considering signal to noise aspects and not resolution. Whether the scores reflect the data sensibly that is a totally different matter and I wouldn't say they do. But the data itself is interesting.</p>

<p>If you need per area figures, that requires a different normalization, but it shouldn't be too difficult to calculate from the data given.</p>

<p>Agreed, Ilkka - it's not hard to normalise by area, I'm just sanity checking that it's not already done.<br />

<br />

Comparing a D810, D7200 and D500 on DxO, the D810 doesn't "win" on dynamic range except by using ISO 64 (the D500 is ahead from ISO 800 up; the D7200 is ahead below that); it doesn't "win" on tonal range (the D500 has the edge at every ISO); it only "wins" on colour sensitivity - and even then, it's a wash if you look pixel-by-pixel (which suggests the advantage is less than the sensor size). Given the sensor size advantage, I'd have expected the D810 to be ahead everywhere.<br />

<br />

Slightly to my surprise, given how good the sensors are, this seems to mean there's more to come from sensor designs than we're seeing in the D810 - and not just the BSI benefits at higher ISO seen by the A7R2. So although I'd thought the advances in sensor design seemed to have levelled off, based off my "a D800 sensor is a D7000 sensor stretched to full frame" premise, it looks like there's hope for a little more. And yet another reason to want to upgrade (to a currently-hypothetical body)...</p>

I'm not sure how the D500's tonal range graph could be correct given the other data.

<p>There is really no advantage to the 1.3x in camera crop, because the image on the sensor itself does not change. Cropping in post processing achieves the same effect. Personally I think it is more of a marketing thing than anything else.</p>

Well in some cameras the crop mode allows a slightly faster frame rate (AFAIK not the D500) and increases the number

of shots you can get in a burst. It does make for smaller files to store if you know you are going to crop anyway and won't

be using the extra image area.

<p>I agree with Ilkka: I've been known to use a D800 in crop mode just because I knew I was going to have to crop eventually (I didn't have a long enough lens on me), and I wanted to save card space. If small raw worked "properly" (binning but otherwise lossless) I'd use that in low light for the same reasons. I'm potentially going to be using a D810 in crop mode soon, and I've been known to use the 1.2x crop for the extra fps it gives me (although I <i>have</i> now arranged to hire a D500 for my cheetah run problem). But yes, image wise the crop doesn't do any thing you couldn't do yourself. Since you're dropping down to 12MP with a 1.3x crop on a D500, I'd start to get nervous about large images and subsequent cropping - I'm a little less scared of the 16-25MP that the crop modes on the D810 give you.<br />

<br />

And yes, DxO's results are odd - though maybe Nikon are doing something funny in the encoding on the D810. Not that I any way feel a lack in the abilities of the D810 (though I'd take a bit more high ISO given the option), but it's always nice to have a step forward.</p>

<p>I suspect it is an error on dxomark's part. It doesn't make sense that the supposed extra tonal range would not show in the 18% gray SNR or the dynamic range (relative to other cameras). Also the color sensitivity is not in any way shining out as you'd expect if there is extra information.</p>

<p>> Edit: Pete, where do you get the per-area figures? I'm extrapolating, and may be wrong.<br /> You have to calculate those by yourself. Dxomark's "print" numbers take the whole sensor into account by using a common print size (8x12" 300 dpi), while dxomark's "screen" numbers only take one pixel into account.</p>

<p>But what I did was to calculate that the difference between FX and DX is 2.25 times area which equals to about 1.1 stop - calculated by ln(2.25)/ln(2).</p>

<p>If you take both the D500 and D800 and look under Measurements > SNR 18% and "print" comparison.<br /> http://www.dxomark.com/Cameras/Compare/Side-by-side/Nikon-D500-versus-Nikon-D800___1061_792<br /> If you pick a noise level like 24 dB and you follow that from the D500 to the D800 (left to right) you can see how many stops better the D800 is. As you'll see, it's a little more than one stop until you get under 24dB. Since the theoretical difference between FX and DX is 1.1 stops (and that is the difference we are seeing) it means that D500 and D800 are equal in noise per area.</p>

<p>It would also be possible to take the "screen" numbers and compensate for the difference in pixel size between the two cameras. I haven't tried it that way before though but it should give the same result. Let's try.</p>

<p>The D500 has 5599 pixels on the long side (DX: 24mm) and the D800 7424 pixels (FX: 36mm). That means the D500 has 233 pixels per mm and D800 has 206 pixels per mm. So the D800 pixels are 13% bigger and 1.28 times larger in area. That equals 0.35 stop, calculate by ln(1.28)/ln(2).</p>

<p>So the D800 has 1/3 of a stop advantage per pixel because it has larger pixels. Looking at the "screen" comparison D500 versus D800 the difference is about 1/3rd of a stop until we get down to about ISO 12800.</p>

<p>So we are seeing the same thing as our earlier comparison. Per square area the sensors noise levels are about the same down to ISO 12800 or so. Then the D500 is about 1/3rd of stop better (per area).</p>

<p> </p>

<p>I am not aware photography technology is actually rocket science. :)<br>

I got lost after...5th posting.</p>

<p>Thanks for explaining, Pete. I think the short answer is that you are looking at the DxO SNR18% entry (where the D810 <i>does</i> have a clear advantage) and I'd been looking at dynamic range, tonal range and colour sensitivity. For DR the results are very similar (and where the D500 isn't ahead of the D810 the D7200 <i>is</i>); for the tonal range measurements, the D500 is substantially ahead of the D810; for colour sensitivity the D810 has a small advantage that, per pixel, balances out (so the 25MP 1.2x crop of the D810 about matches the 24MP 1.5x crop of the D7200). The SNR score, it's true, is much more in the D810's favour. I agree with Ilkka that it's a bit odd that the scores differ so widely, and without knowing more about what's being measured it's hard to know what's happening - but it does seem there's a bit of sensor advance available beyond the D810's, so I'm optimistic for a small improvement in the next generation. I did similar maths to yours, I'm just glad (for my sanity) that they were necessary and DxO hadn't already applied a sensor size scaling factor.<br />

<br />

Edward: Rocket science isn't generally that complicated. :-) Still, apologies for the diversion. All I was trying to say was:</p>

<ul>

<li>Your "819mm lens" will give you the same field of view as an 819mm lens on a full-frame camera.</li>

<li>For the purposes of setting ISO, the combination is f/5.6 (which is what 1.4x teleconverters do to f/4 lenses).</li>

<li>The "equivalent full frame camera" only has 12MP (that's the result of your 1.3x crop on a 20MP D500).</li>

<li>The amount of light hitting the total sensor area that you're using will be lower, so it's like a full-frame camera shooting at (1.5x1.3)² times higher ISO (give or take the discussion about sensor technology advances, ISO 100 on the cropped D500 will behave like ISO 380 on the full-frame camera with the f/5.6 combination).</li>

<li>For depth of field/background isolation, the 300mm f/4 combined with the teleconverter and cropped sensor acts like an 819mm f/10.92 lens on a full-frame camera (f/(4 x 1.4 x 1.5 x 1.3)), wide open - nearly two stops down from an actual 800mm f/5.6.</li>

</ul>

<p>Which you may have known already, I just have a nervous tick when people start saying things like "819mm f/5.6 equivalent". It's only equivalent in some ways; scaling the aperture and then scaling the ISO as well so that the exposure balances out as the same gives you "equivalence" that also explains what's happening to depth of field and - roughly - to sensor noise. Which you may not care about if you just want to know how much reach you're getting, but you may care about if you want to work out how much you can isolate a subject, how to stop the image being grainy, and why people spend so much on actual 800mm f/5.6 lenses. :-)</p>

<p>I have enjoyed this graduate course in pixel peeping.</p>

<p>Thanks Andrew for elaborate summation. However, I am not getting with 1.3x cropped format (as well as 1.5x cropped format on d800/e) affecting DOF. I understand about the S/N ratio, and pixels size and all that but how is DOF different in those original vs cropped format? It is basically the same image, only cropped with photoshop.<br /> I think it is more like 819mm / (4x1.5 x 1.4) = 819 F/8.4, as far as DOF concerned.</p>

Well the thing is that the cropped 420/5.6 image is the same at the pixel level as it was before cropping but when it is magnified to the same display size as a FX+800mm image shot at f/5.6 the depth of field of the FX+800/5.6 image is shallower. You can try this out by using shorter lenses. Most of us will never own an actual 800mm f/5.6 lens and probably would not prefer its handling but the

output is reflective of the cost to some degree. ;-)

<p>E.J., you're welcome. Now I just need a course in how to take decent photos...<br />

<br />

Edward: Glad if I'm helping!<br />

<br />

I was part way through a long explanation, but - what Ilkka said. Just talking about the 1.5x crop, you're magnifying the image from a smaller sensor more than you are a larger sensor, which means you get less depth of field (more blur) with the smaller sensor by a factor of 1.5, compared with using the same lens at the same place. But you're using a shorter lens on the crop camera than the "equivalent" lens on the full-frame, which (at the same relative aperture) means you've got a narrower cone of confusion (your physical aperture is smaller) and you're magnifying it less at the sensor plane (because the actual lens you're using is shorter). You end up scaling up by 1.5, but scaling down by 1.5 squared - so you end up with the DoF of 1.5x less aperture.<br />

<br />

Thinking another way, for the same field of view and same camera position, the depth of field depends entirely on the real-world size of the aperture of the lens and where it is, because this is what defines the size of the "cone of confusion" between the lens and focal plane - what happens behind the lens doesn't affect this at all. You can think of the crop, teleconverter use and focal length of the lens as cancelling out if you're using them to get the same angle of view - by having a smaller sensor, cropping it, and using a teleconverter, we've got to the equivalent of an 819mm lens. The front element is still the same 75mm it was when you bought the 300mm f/4, so you've got - in full-frame terms - an 819mm lens with a 75mm aperture, and that's f/(819/75), or 10.92. There's no difference between the way to handle 1.3x and 1.5x crops - there's nothing special about cropping an area of the image compared with having a physically larger sensor, and the 1.5x crop on a D800 is <i>exactly</i> like a DX camera.</p>