4
\$\begingroup\$

I have Nikon lenses with distance scales on them so I can set the lenses to the hyperfocal distance when I need.

Say now the focal length is 28mm and the aperture is f/16. The corresponding hyperfocal distance is 1.66 m. Now on the distance scale I have only the marks 1m and 2m and nothing in between them.

So how should I set the lens distance, closer to the 2m mark? Is halfway between the 1 and 2m marks the distance 1.5m? I am asking because it does not seem to be a linear scale. On the other hand I have found not tips on how to interpret it.

\$\endgroup\$
4
  • 3
    \$\begingroup\$ What lens are we talking about? Does the lens have a DoF scale? If so, the easy way to set hyperfocal is to place the infinity mark on the distance ring above the 16 on the DoF scale (since you're using f/16). \$\endgroup\$
    – inkista
    Commented Aug 12, 2014 at 20:09
  • \$\begingroup\$ One I use oft is the 35-70mm f2.8 D. Using the FL 35mm and the aperture f8 I should set focus distance to hyperfocal distance = 5.14. In this case it is easy, as I have the marks 5m and then infinity on the scale and I can use 5m as a good approximation. But in other cases the calculated hyperfocal distance has a value that falls between two marks, and I do not know how to approximate. If for example on the same lense I have to set the focus dist. to 4, should it be exactly halfway between the available 3m and 5m marks? Or closer to 5m mark as the scale is (apparently) not linear? Thank you. \$\endgroup\$
    – az1mov
    Commented Aug 12, 2014 at 20:32
  • 2
    \$\begingroup\$ The scale is hyperbolic - it maps the bounded range of distance scale positions onto the infinite range of focus distances [mfd, ∞] where mfd is the minimum focus distance. Unfortunately I'm too tired right now to figure out the math to calculate the distance scale position corresponding to an arbitrary focus distance :/ \$\endgroup\$
    – JohannesD
    Commented Aug 12, 2014 at 22:46
  • \$\begingroup\$ Did you have some rest? :) Can you explain it now or point me out to some resource in Internet? \$\endgroup\$
    – az1mov
    Commented Aug 13, 2014 at 12:40

1 Answer 1

2
\$\begingroup\$

Okay, I'm pretty sure this checks out, but haven't yet sanity-checked it with any physical lenses.

Let's say 0 and 1 represent the distance scale / focus ring position at the MFD mark and infinity, respectively. Then 0.5, for instance, would mean turning the ring halfway between the extremities.

Then, we have a hyperbolic curve like this:

a = 1 - dmfd / df

where dmfd is the minimum focus distance and df is the desired focus distance. a is the number between 0 and 1 indicating how far to turn the focus ring. As you see, if we have dmfd=df, then a=0 as it should, and if df=∞, then a=1, also as expected.

Let's say your MFD is 0.3m and you want to focus at 1.6m. Then

a = 1 - 0.3m / 1.6m = 0.8125,

indicating you should turn the focus ring about four fifths of the way from MFD to infinity; if there are 1m and 2m marks in the distance scale they should be at a=0.7 and a=0.85 respectively, so the correct position would be about three fourths of the way from the 1m mark towards the 2m mark.

\$\endgroup\$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.