New Pentax 70-200 on the latest roadmap?

Started 10 months ago | Discussions thread
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Re: I'll get some corrections on this for sure but...
In reply to Ian Stuart Forsyth, 10 months ago

Ian Stuart Forsyth wrote:

GossCTP wrote:

Jim in Hudson wrote:

I thought converting the aperture, going between formats, is only relevant to determining equivalent DOF and essentially nothing else. Do I have that wrong?

I'm sure I'll get corrected on this, but hey, that's how we learn...

Imagine you have two formats. One is the size of a sheet of plywood, and the other the size of a postage stamp. Both have the same F/stop and are therefore receiving the same intensity of light per unit of area. They have equal pixel pitch, so per-pixel noise is the same (may not hold in real world, but keeping it simple). But the 4'x8' sheet of plywood has about 30 billion more pixels than the postage stamp size. At the same print size, the large format will look much cleaner, even if shot with a much higher ISO, because the noise will average out.

perfect !

Except the model is incorrect because at current time APS-c out pixels FF

So in your model the postage stamp has 30 billion more pixels than the 4'x8' sheet for the same reason it happens with sensors bigger pixels more heat = less density.

Which turns everything below on it head.....

Aperture and F/stop are two different things that get used interchangeably. The size of the aperture plays into DOF, diffraction, and the actual amount of light that enters the camera. The F stop is a mathematical formula for determining the theoretical intensity of light on the film plane.

again perfect !

Lets take two "equivalent" lenses, a 200mm FF and a 135 APS-C

F/stop = focal length / aperture

f/4 = 200mm / 50mm

f/2.7 = 135mm / 50mm

perfect !

But I don't recognize this as equivalent at all.

I was shooting alongside a FF shooter this weekend as were in the UK the light was appalling, he was armed with 70-200 f2.8 and the 100-400 Pump , Me 70-200 f2.8 , da*300 F4 and Bigma.

Due to light level the use of both the Bigma and 100-400 was not possible (with the exception I could get working images due to IS (but that's another story) )

So I was shooting in high day the da*300 and him his 70-200 hes was taking very few shots as 200mm would only cover at a reasonable target size 3 of the lanes whereas with the 300 I could cover about 1/2 the 50 M pool.

As daylight fell I resorted to my 70-200 f2.8 , Both our exposures were approximate the same (obvious it the same lens) only I still had coverage of nearly a 1/3 of the pool to his 3 lanes directly on front.

At the end of the day we compared

Shots taken me 360 him 70

Acceptably sharp (LCD viewing) me 340 (I chimp as I shoot) him 70 (so did he)

decent composition me 300 him 10

Best images him (2)

good images me (200)

usable image me (315) him 50

The only way he would have been able to improve on my hit rate is if he'd had a £10,000 600mm f4

For anything else I could match and do better at less cost

ie him 400 f2.8 (£6500) me 300 F2.8 (£2000)

The reason why the two formats shoot the same DOF here is because the physical apertures are the same size (50mm). The same amount of light is falling on both the sensors, but in the FF camera it is less intense and spread over a larger area. Like our first plywood example, the larger sensor can compensate with a higher ISO and still maintain similar noise despite lower light intensity.


Now, instead of giving the equal apertures, let's give them equal F/stops

F/stop = focal length / aperture

f/4 = 200mm / 50mm

f/4 = 135 / 33.75mm

Now the APS-C has the same intensity of light per unit area as the FF sensor. The APS sensor has less overall light received because it is smaller. The physical aperture is smaller, and therefore the DOF is greater and the theoretical maximum resolution is lower due to diffraction.

perfect ! and also good to include diffraction

I will also add that the 200 F/4 will have less noise as it captured 2.25 times more light

Wrong 100%

F-stop is ratio of focal length and entrance pupil of the lens, the physical aperture has nothing to-do with it at all

The entrance pupil is the diameter of the aperture as viewed through he front of the lens.

DoF is related to the CoC this varies depending on the magnification factor required to project the same size image.

The two are not related at all in any way shape or form and to try and link them necessitate breaking the rules of physics or making assumptions that just aren't true.

As we tend to use 35mm as the 'standard' image to project so generally you use

CoC for 35mm / camera lens focal multiplier to establish a working CoC

Which is the only point 'equivalent' has any meaning because

the focal multiplier

35mm equivalent length \ actual focal length

If we based it on the CoC of MF then you would use MF equivalence instead, It is only a way of standardizing to a set image size so you can equate the DoF gained or lost at a 'Standard' viewing distance.

Hopefully I did it justice...

you did

IMO you have both fallen in to the same misconception where people believe focal length effect DoF and use the argument but when I zoom DoF narrows, but in reverse (because you behind the exit pupil.

Given my point on magnification and smaller sensors you can see both you and they are mistaken.

as Cambridge tutorials point out.

contrary to popular belief. Even though telephoto lenses appear to create a much shallower depth of field, this is mainly because they are often used to magnify the subject when one is unable to get closer. If the subject occupies the same fraction of the image (constant magnification) for both a telephoto and a wide angle lens, the total depth of field is virtually* constant with focal length!

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