FREE Hyperfocal Distance Charts !!

Started Nov 1, 2012 | Discussions thread
Regular MemberPosts: 140
Re: FREE Hyperfocal Distance Charts !!
In reply to EOS5D, Nov 1, 2012

EOS5D wrote:

Hi all, just letting you know I created a spreadsheet to produce accurate (not to mention, FREE!!) Hyperfocal Distance Charts for all Canon, Nikon, Leica and Sony digital SLR's, and have uploaded them to my website.


I just found your post here and in some other forums on dpreview. And I realised you have these charts on your website since several years. I have to admit, when I saw you saying "for all ..... cameras" I expected to find the CoC somehow related to pixel pitch. It is not, and I'm glad you're not saying so. It is true, CoC is related to image enlargement and viewing distance. Pixel pitch only sets a lower limit to useful CoC.

But I do have a few other comments. You call your charts "accurate", but they are not. They can't. There is very little accuratesse in the optical and physiological models, that the HfD is based on. They are nothing but an estimate and anything beyond 1-2 significant digits is just pseudo-exact.

It is good that you mention smaller CoC than the standard 1/1500 of the sensor diagonal for larger prints viewed from a shorter distance. But with cameras like D800 and beyond, people will tend to crop by more than the factor 2, that you have in your charts. While this has not too much effect in the far background, it might spoil the perceived image sharpness at the near end of the DoF, while any object, that happened to be just the HfD away, will remain sharp(er).

And finally: The HfD is rather easy to calculate/estimate. If you don't consider close-ups, it is simply HfD=f²/N/CoC. To me it would be a nightmare to even try that with "inches and furlongs", but in the SI system it is rather easy to do mentally. The focal length f is usually given in [mm] and the CoC in [μm], e.g. 30μm for standard FF. That deletes all the "10^x" from the formula. (It is (10^-3)²=10^-6 in the numerator and 10^-6 in the denominator, and they cancel out). Simply calculate/estimate focal length in [mm] (e.g. "60") over f-number (e.g. "10"). That gives you a "6". Then calculate/estimate focal length in [mm] (again "60") over CoC in [μm] (e.g. "30"). That gives you a "2". Now multiply the two, that gives you 2x6=12. Voila, there you are, the HfD is 12m! And there is no need to calculate it more "accurately".

Gruß, and please excuse my english, Matthias

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