10D hyperfocal distance question

[miller 0]

Please excuse my confusion but:

Due to the 1.6x scaling factor associated with the 10D, when determining your hyperfocal distance for a given lense and aperature (let's say a 20mm lense), do you use 20mm in the hyperfocal distance equation, or do you use 32mm (1.6 x 20mm).

Should seem obvious, but I'm a little slow on the uptake today.

Thanks,

Mike

[quade 3]

If you only wanted your image as sharp as 35mm film, then you'd have to scale it with the conversion factor, but since the resolution of the imager is greater than 35mm film, I think you'd actually need to do some refiguring there as well.

I'd be a great question for the techs at Canon. I just emailed them and am waiting for their reply.

quade -
The World's Most Boring Skydiver

[Deuce 1]

"Hello, Mr. Canon? Q-doggy here. Find out what the hyperfocal conversion is on all Canon lenses when used with the 10D, got it?! Have your people call my people ASAP!"

[quade 3]

Ok, here's the real question . . .

Quote

When calculating the hyperfocal distance for 35mm film the circle of confusion is normally taken to be 0.030mm. Since the Canon 10D imager is smaller than 35mm film and taking into account pixel size, what CoC should be used?

BTW, here is the formula.

Hyperfocal Distance
Setting focus at the Hyperfocal Distance gives maximum depth of field from H/2 to infinity.

H = (L x L) / (f x d)
Where:
H = Hyperfocal Distance (in millimeters)
L = lens focal length (ie, 35mm, 105mm)
f = lens aperture f-stop
d = diameter of circle of least confusion (in millimeters)
for 35mm format d = 0.03
for 6x6cm format d = 0.06
for 4x5in format d = 0.15

quade -
The World's Most Boring Skydiver

[quade 3]

Interesting stuff.

I just did some googling and came up with some different web pages on the subject. Some different calculators et al.

http://dfleming.ameranet.com/digital_coc.html is one that suggests that you can use their calculator without doing a conversion factor, but changing the CoC number. What I find interesting is that the CoC number is almost the exact reciprocal of the lens conversion factor . . . so it all ends up being the same thing as if you'd stayed with the 35mm CoC and simply used the conversion factor!

Again, I personally don't think that either of these methods is exactly right. Just doesn't have the right gut feeling to it.

Still waiting on word from Canon.


BTW, for anyone that doesn't understand what the heck we're talking about . . . HERE is a nice explanation.

quade -
The World's Most Boring Skydiver

[bmcd308 0]

Recognizing that it is dangerous to post this before Quade posts the real facts from Canon, I'm going to speculate that the 1.6x cropping factor won't matter.

My logic for this is that the 1.6x cropping factor happens in the camera, not the lens. The only thing that is really changing is the field of view, because the image sensor in the 10D is smaller than a frame of film. All the image sensor sees, since it is smaller, is the image from the center of the lens, while a piece of film will see more of the light coming through the lens.

Taking a picture with a given lens on a 10D and printing it at 4x6 is just like taking it with 35MM film and then cutting the image down a little with scissors (to reflect the same field of view as the sensor on the 10D) and blowing it back up to 4x6. So you are not really changing the way light behaves through the lens, you are really just only getting the center of the image that would be on the film in your file.

While 10D is probably capable of smaller circle of confusion numbers than film, I don't think it is going to make much difference in practice, since you'll want the slightly smaller circle of confusion because you'll have to "enlarge" at any given print size.

If this makes no sense, send me PMs that call me dirty names.

Brent

----------------------------------
www.jumpelvis.com

[quade 3]

Canon still hasn't replied to me yet but there are a couple of things that are "sort of" clear.

Yes, the numbers are different on at least two levels and you can approach the problem by either scaling the CoC number or the focal length to compensate for the imager size.

Why?

You're right, the key factor here isn't the lens, never was, but rather the concept of what is considered to be "acceptable focus" for enlargements.

At some point in the distant past, somebody figured out that "acceptable focus" meant that the CoC was 1/1440th the distance from corner to corner on any 35mm format image. Looked at another way (and imprecisely, but this is ONLY for illustarion purposes) an image had to have about 1440 "pixels" diagonally in order for it to be enlarged to "normal" size prints (or any prints really but viewed from "appropriate" or "typical" distances) and still have "acceptable focus".

Lot's of quotation marks up there indicating euphemisisms are being used, but again, simply for illustrative purposes.

So, let's take a look at the D60 for instance (or, um 10D - same imager).

The D60 (or 10D) has an imager with 3072x2048 pixels.

Let's see . . . Sqrt(3072^2 + 2048^2) = 3652 and change.

Nyquist suggest that you'd want twice the number of the pixels "required" or at least 2880. Since the D60 (or 10D) has about 1.25 times that, it's a bit sharper than what would be considered to be "acceptable" in most cases.

So, I'm going to say that that the CoC number could actually be (should actually be) quite a bit smaller than the simple scaling factor of the imager would suggest if you really wanted to maximize the sharpness of your enlargments.

Does any of this really matter?

Only if you were absolutely fanatical about the precision of your focusing for lenses with focal lengths longer than what we're typically using for freefall photography.

Myself, I'm now convinced that simply scaling either the focal length of the lens by 1.6 or the CoC by 0.625 is "close enough for jazz". Just don't do both.

quade -
The World's Most Boring Skydiver