Re: Nikon D40 with 300mm lens AND teleconverter (Nikkor AF-S Teleconverter TC-20E II)

Chris Malcolm wrote:
Paul Furman <paul-@xxxxxxxxxxxxx> wrote:

[snipped interesting ramblings]

While I ramble on this, one more thought: the diagonal distance is not intuitive and perhaps that's where a lot of problems come from. It seems the accepted standard for a 'normal' focal length might be a bit on the long side, so if reinventing a sensible terminology from scratch it might make more sense to talk about the width of the sensor rather than the diagonal, which for 35mm would be a 36mm lens, for APS it would be a 24mm lens. That would make 4:3 compacts' normal lenses a little less wide than the diagonal paradigm defines as well as the squarish large & medium formats.

The thing about the diagonal is that its actually the diameter of the
image circle of the lens. Our circular lenses oddly enough don't
really produce rectangluar images. The rectangle is cropped out of the
circle. So the diagonal of a 4:3 image for a given lens will the same
as the diagonal of a 3:2 image in another camera (or the same one if
it produces both formats without cropping from one to the other).

Stating the actual focal length when presenting a photo really does not make sense, just like it doesn't make sense to talk about aperture as diameter or area but rather the ratio to the focal length.

But it does if you specify the sensor size as well.

Stating the magnification is a little too abstract given the ambiguity of what is 'normal' and that it relates to the print, not what you see with your eye.

What we really want to know is the focal length divided by the
diagonal of the sensor size. That gives us a magnification factor
related to sensor size. That was also the same as a binocular or
telescope magnification factor back in the old days when people often
used straight contact prints from negatives and called anything else
an "enlargement". In a contact print the print is the same size as the
sensor. Nowadays when we all "enlarge" onto screen or paper we simply
have to factor in our personal enlargement factor.

That explains the 1:1 macro terminology. In digital, macro would make more sense expressed in pixels per inch on the sensor, or actually divide that by 300dpi for print and you get ACTUAL magnification as if measuring the object & print with a ruler. 3 dimensional photos are harder to measure :-)

For a 10MP Nikon DSLR a 1:1 macro is 12x printed at 300ppi.
3872 pixels wide in a 24mm = 0.94 inch sensor /300 = 12.2


Stating the angle of view is sensible though it's not something most people are accustomed to judging. That's why we say 'see the boat at 2:00 on the horizon' rather than 30 degrees north-northwest as a surveyor would. But this might be the most appropriate terminology if it were being reinvented: the most useful standards to memorize. The only problem is the math becomes unmanageable and that's where magnification power based on a normal field of view is more practical. I don't even think I could figure out the angle of view given a focal length, sensor size & a calculator, let alone do it in my head. What would a 50mm view on 35mm angle be in degrees? If anyone bothers, I'm more interested in the width not the diagonal... and the height is not important because human eyes work in landscape orientation pbtbtbt! :-)

In the simple standard case of a lens producing a rectilinear image
the angle from focal point of lens to image width is the same as the
angle of view outwards. So we can calculate angle of view from the
image forming geometry.

Focal length gives you the perpendicular distance from the focal point
of the lens to the centre of the image plane. The width stretches out
on each side by an amount of half the width, i.e. 18mm in the 36x24mm
image on 35mm film. So we have a right angle triangle say 50mm high
and 18mm wide. The subtended angle at the lens of the half image to
one side of the vertical is therefore atan(18/50). The whole width of
the rectangle is therefore twice that, 2 x atan(18/50) = 39.6 degrees.

I take it that everyone reading this has a computer or a calculator?
In fact I can plug that formula into a spread*** in my mobile phone
and make it produce tables of horizontal view angles and focal lengths
for any sensor size and format I like. It also takes photographs :-)

Thanks! Here's that formula plugged into a chart. When you say "horizontal view angle" I assume that's width of a landscape frame which looks about right since a 10.5mm digital fisheye is 180 degrees diagonal as I recall... although you suggested that math may not work with a fisheye projection :-)

Lens characteristics for Nikon DX sensor (1.5x crop factor)
focal length
x-power*
35mm eq.
angle**
10 0.3 15 175
10.5 0.4 16 170
12 0.4 18 157
14 0.5 21 142
16 0.5 24 129
18 0.6 27 118
20 0.7 30 108
24 0.8 36 93
28 0.9 42 81
30 1.0 45 76
35 1.2 53 66
45 1.5 68 52
50 1.7 75 47
58 1.9 87 41
70 2.3 105 34
75 2.5 113 32
85 2.8 128 28
90 3.0 135 27
105 3.5 158 23
135 4.5 203 18
150 5.0 225 16
200 6.7 300 12
300 10 450 8
420 14 630 6
600 20 900 4
840 28 1260 3

*focal length/30
(diagonal of sensor or conventional normal FL)

**angle of view for the width of a landscape frame
(2 x atan(12/FL)*100 where 12 is 1/2 the sensor width)

.