Bill_1856

14-Nov-2003, 15:00

I don't seem to be able to find a chart for it. Thank you.

View Full Version : What is hyperfocal distance for 210mm lens @f:64 (4x5 format)

Bill_1856

14-Nov-2003, 15:00

I don't seem to be able to find a chart for it. Thank you.

Jay DeFehr

14-Nov-2003, 15:10

24.32' Here's a calculator:

http://www.mountainstorm.com/HyperFocal/HyperFocal.html

http://www.mountainstorm.com/HyperFocal/HyperFocal.html

Witold Grabiec

14-Nov-2003, 15:40

It should be around 24 ft.

Download this free f/calc applet from:

http://www.tangentsoft.net/fcalc/

This way you've got one on your PC for a quick reference.

Download this free f/calc applet from:

http://www.tangentsoft.net/fcalc/

This way you've got one on your PC for a quick reference.

jerry brodkey

14-Nov-2003, 15:48

I had downloaded fcalc and so my fcalc gives a value of 15 feet. The circle of confusion suggested by fcalc is 0.15mm. One of these programs seems off.....

Bill_1856

14-Nov-2003, 16:21

Thanks, Guys.

Bob Salomon

14-Nov-2003, 16:23

And you will be well into diffraction too.

Witold Grabiec

14-Nov-2003, 17:38

Bill,

I've just checked f/calc which I downloaded a couple of days ago. It gives me 24 ft. Make sure you have all veriables in order. Mine gives me a COC of ~ 0.09. COC of 0.15 does indeed give you some 15 ft on HD, but I had to enter this value manually.

By the way f/calc calculates COC as f/1730 and author calls it a "zeiss formula". COC is the key factor that affects HD or DOF calculations. As you can see from above it will make a substantial difference. I'm assuming however, that you need the HD to get maximum DOF. The higher the COC the more DOF you get, thus the closer the HD will be. Your COC of 0.15 is a bit high as it is usually listed in the 0.10 range, value also stated in the Applied Photographic Optics.

I've just checked f/calc which I downloaded a couple of days ago. It gives me 24 ft. Make sure you have all veriables in order. Mine gives me a COC of ~ 0.09. COC of 0.15 does indeed give you some 15 ft on HD, but I had to enter this value manually.

By the way f/calc calculates COC as f/1730 and author calls it a "zeiss formula". COC is the key factor that affects HD or DOF calculations. As you can see from above it will make a substantial difference. I'm assuming however, that you need the HD to get maximum DOF. The higher the COC the more DOF you get, thus the closer the HD will be. Your COC of 0.15 is a bit high as it is usually listed in the 0.10 range, value also stated in the Applied Photographic Optics.

jerry brodkey

14-Nov-2003, 19:28

We probably have different versions of fcalc. My version is running on linux and when I click on 4X5 it puts in the value of 0.15 for the COC. When I manually enter 0.10 I get the same value as you get.....

Bill_1856

14-Nov-2003, 19:35

Thanks again, Guys. Please don't spend a lot of time on this. Basically I wanted the figure to check myself visually (since the GG image is so difficult to see distinctly when stopping the Dagor down to f:64). Visually, I got 25' which is "close enough for government work" as they say.

Witold Grabiec

14-Nov-2003, 20:30

Jerald,

It might help if you emailed the author of f/calc (something he's asking for) about the error. Yu can find his contact in the Help section of f/calc. I say error because the COC value of ~ 0.10 is the most frequently quoted for 4x5 in major publications on this subject (and my Windows version actually gets it right). It is likely a simple typo in the code. I would guess that all of your COC are off. For 6x6 you should be getting ~ 0.05.

It might help if you emailed the author of f/calc (something he's asking for) about the error. Yu can find his contact in the Help section of f/calc. I say error because the COC value of ~ 0.10 is the most frequently quoted for 4x5 in major publications on this subject (and my Windows version actually gets it right). It is likely a simple typo in the code. I would guess that all of your COC are off. For 6x6 you should be getting ~ 0.05.

jerry brodkey

14-Nov-2003, 21:11

Witold, Several years ago I tried fcalc and found that he was using a wrong formula for magnification. I wrote him an email and asked about this problem. He wrote back and said he didn't care. I got so mad I took it off my computer. When people started talking about it recently I put it back on to see if it had changed. I don't know if he has updated the magnification formula or not and frankly I'll get rid of the program again. If I need to find the hyperfocal distance I'll just plug the values into HD = f*f/Nc.

There is a lesson here. People write programs and you really have no idea if their formulas and methods are correct. It is better to calculate what you need yourself.Then you know it's correct.

There is a lesson here. People write programs and you really have no idea if their formulas and methods are correct. It is better to calculate what you need yourself.Then you know it's correct.

Witold Grabiec

14-Nov-2003, 21:36

Yep, I was not aware of this part. If that's how he responded then, oh well. The help section kind of goes over what he uses for calculations. I'll be checking on that in the near future. He has some other strange things there too.

I agree the HD formula, as all other for what we might need in LF calculations, is simple and easy. This is just a little applet that cost nothing to download.

I agree the HD formula, as all other for what we might need in LF calculations, is simple and easy. This is just a little applet that cost nothing to download.

Darin Cozine

15-Nov-2003, 00:17

OK, I'm a bit confused..

I used the mountainstorm website and plugged in my 65mm lens and f-16 for 4x5 format. The calculator gave me 9.3 feet. <br/>

Question 1: Do I focus at full aperture on 9.3 feet, then stop down to f16 to get infinity in focus?<br/>

Question 1a: If so, how close can objects get and still be in focus?

<br/><br/>

Question 2: Why does the format make a difference in the hyperfocal distance?

I used the mountainstorm website and plugged in my 65mm lens and f-16 for 4x5 format. The calculator gave me 9.3 feet. <br/>

Question 1: Do I focus at full aperture on 9.3 feet, then stop down to f16 to get infinity in focus?<br/>

Question 1a: If so, how close can objects get and still be in focus?

<br/><br/>

Question 2: Why does the format make a difference in the hyperfocal distance?

Witold Grabiec

15-Nov-2003, 06:07

Darin,

HD is a special case of Depth of Field, where the far focusing end of DOF falls on infinity and the near focus end is the Hyperfocal Distance, or the distance you focus your lens on, or the distance of closest focus. So for (and based on your figures):

1. Yes

1a. 9.3 ft

2. DOF and thus HD calculations depend on the ASSUMED size of the Circle of Confusion (COC), that will give adequate sharpness, which determines how well the detail is resolved in a given plane of "focus". This depends on the viewing distance of the print, which is considered as the diagonal of the negative. As the negative size goes up, so does the COC.

As you well know, a standard lens for ANY format is the one with a focal length approximately equal to the diagonal of the negative. A term "cone of vision" is derived from the structure of the human eye and how an image is "printed" on its retina. It is stated, that a human eye resolves about 5 lp mm (lines per mm) which translates to a dot size of 0.2 mm which, in turn, also translates to an 8x10 inch print with the smallest dot of 0.2 mm viewed at its diagonal. (And by the way, photographic paper is said to be capable of around 6 lp mm, so paper is not an issue in these considerations as it resolves better)

Imagine a cone projecting out from your eye, with its apex at the eye ball and lets see if I can explain this important concept:

a - comfortable viewing angle of around 45-55 degrees (as per standard lens) gives an approximate comfortable viewing distance equal to negative's diagonal, this would be the angle of the "cone of vision" or COV, comfort being stated as seeing it in a natural perspective

b - when the negative is enlarged (or you're viewing a larger print), it will then have to be viewed from further away in order for it to match this COV, so as to make a "point of detail" on the print come to a point at the eye ball.

c - after the light passes the eye ball entrance, it will then project on the retina and the ~50 degree angle of projection will end up being that 0.2 mm dot on the retina

d - now, if we take the value of 0.2 mm and print multiplication factor (based on 8x10 standard as this applies to), we will arrive at the COC size that matches our eye's resolving ability (based of course on the viewing distance of prints diameter, if you view this print at a closer range you will see it more out of focus)

e - to clarify point "d", the 0.2 mm COC is as for an 8x10 viewed at its diagonal, so if you make a contact print from an 8x10 negative, then the COC would be 0.2 mm, if you use a smaller negative you need to work backwards ( a 35 mm frame is about 8 times smaller, so the COC for a 35 mm negative is 0.2 / 8 = ~ 0.025 mm) and if you use a larger one, you would multiply (an 11x14 negative is some 2 times larger, so the COC for it would come to around 2 x 0.2 = 0.4 mm)

This is a subjective matter because no human eye is created equal. So some may accuse of you of producing out of focus prints, while others may disagree. The maximum researched threshold for a human eye resolving ability has been found at 9 l pm or 0.1 mm dot, almost twice the average.

HD is a special case of Depth of Field, where the far focusing end of DOF falls on infinity and the near focus end is the Hyperfocal Distance, or the distance you focus your lens on, or the distance of closest focus. So for (and based on your figures):

1. Yes

1a. 9.3 ft

2. DOF and thus HD calculations depend on the ASSUMED size of the Circle of Confusion (COC), that will give adequate sharpness, which determines how well the detail is resolved in a given plane of "focus". This depends on the viewing distance of the print, which is considered as the diagonal of the negative. As the negative size goes up, so does the COC.

As you well know, a standard lens for ANY format is the one with a focal length approximately equal to the diagonal of the negative. A term "cone of vision" is derived from the structure of the human eye and how an image is "printed" on its retina. It is stated, that a human eye resolves about 5 lp mm (lines per mm) which translates to a dot size of 0.2 mm which, in turn, also translates to an 8x10 inch print with the smallest dot of 0.2 mm viewed at its diagonal. (And by the way, photographic paper is said to be capable of around 6 lp mm, so paper is not an issue in these considerations as it resolves better)

Imagine a cone projecting out from your eye, with its apex at the eye ball and lets see if I can explain this important concept:

a - comfortable viewing angle of around 45-55 degrees (as per standard lens) gives an approximate comfortable viewing distance equal to negative's diagonal, this would be the angle of the "cone of vision" or COV, comfort being stated as seeing it in a natural perspective

b - when the negative is enlarged (or you're viewing a larger print), it will then have to be viewed from further away in order for it to match this COV, so as to make a "point of detail" on the print come to a point at the eye ball.

c - after the light passes the eye ball entrance, it will then project on the retina and the ~50 degree angle of projection will end up being that 0.2 mm dot on the retina

d - now, if we take the value of 0.2 mm and print multiplication factor (based on 8x10 standard as this applies to), we will arrive at the COC size that matches our eye's resolving ability (based of course on the viewing distance of prints diameter, if you view this print at a closer range you will see it more out of focus)

e - to clarify point "d", the 0.2 mm COC is as for an 8x10 viewed at its diagonal, so if you make a contact print from an 8x10 negative, then the COC would be 0.2 mm, if you use a smaller negative you need to work backwards ( a 35 mm frame is about 8 times smaller, so the COC for a 35 mm negative is 0.2 / 8 = ~ 0.025 mm) and if you use a larger one, you would multiply (an 11x14 negative is some 2 times larger, so the COC for it would come to around 2 x 0.2 = 0.4 mm)

This is a subjective matter because no human eye is created equal. So some may accuse of you of producing out of focus prints, while others may disagree. The maximum researched threshold for a human eye resolving ability has been found at 9 l pm or 0.1 mm dot, almost twice the average.

phil sweeney

15-Nov-2003, 06:25

Darin,

as format size goes up depth of field goes down. The CoC is 1/1500 (or 1/1720 or whichever the user decides) multiplied by the diagonal of the film. 4 x 5 film diagonal is 6.403 inches. CoC of 0.1 mm is a good all around number for enlarging.

Focus on the hyperfocal distance of 9.3 feet (your example) and about 1/2 times 9.3 (4.65 feet) to infinity should be in focus at f16.

as format size goes up depth of field goes down. The CoC is 1/1500 (or 1/1720 or whichever the user decides) multiplied by the diagonal of the film. 4 x 5 film diagonal is 6.403 inches. CoC of 0.1 mm is a good all around number for enlarging.

Focus on the hyperfocal distance of 9.3 feet (your example) and about 1/2 times 9.3 (4.65 feet) to infinity should be in focus at f16.

Witold Grabiec

15-Nov-2003, 07:51

Looking at my expalnation above, I was so proud of for a moment, reveals some contradictions. Without going into detailed calculations: - as stated, a COC of 0.2 mm is used as minimum perceiveable detail when viewed at a comfortable viewing distance - comfortable viewing distance refers to a comfortable "near distance of distinct vision" which was established to be around 250 mm (this is considered closest an eye can focus on comfortably), in other words if a dot of 0.2 mm is viewed at 250 mm it should be resolved by an average human eye, which ultimately approximates an 8x10 with a minimum dot of 0.2 mm viewed at its diagonal

- this lead to establishing what a standard lens should be, given a negative size, and all of it fills nicely in the concept of a Cone of Vision as stated earlier, and the ~ 50 degree angle of view for such,

- depending on the final print size, it is easiest to simply compare it's magnification ratio against an 8x10 and adjust the 0.2 mm accordingly

All other numbers are correct, as are examples of COCs. And the 4x5 does lead to a COC of 0.1 mm based on this.

Since there is never a gurantee a print will be viewed at its "proper" distance, there is never a guarantee it will be perceived as intended. This is especially the case with larger prints.

While I am now comfortable with above, I welcome any comments on this.

- this lead to establishing what a standard lens should be, given a negative size, and all of it fills nicely in the concept of a Cone of Vision as stated earlier, and the ~ 50 degree angle of view for such,

- depending on the final print size, it is easiest to simply compare it's magnification ratio against an 8x10 and adjust the 0.2 mm accordingly

All other numbers are correct, as are examples of COCs. And the 4x5 does lead to a COC of 0.1 mm based on this.

Since there is never a gurantee a print will be viewed at its "proper" distance, there is never a guarantee it will be perceived as intended. This is especially the case with larger prints.

While I am now comfortable with above, I welcome any comments on this.

Witold Grabiec

15-Nov-2003, 08:05

To clarify the Hyperfocal Distance concept:

Hyperfocal Distance is what you set your lens on AND, this is the closest subject in focus (not half of that) with far end of the resulting Dpeth of Field falling on infinity. So again, if your HD is 9.3 ft, then everything closer then 9.3 ft will be outside of your depth of field limits (as per chosen COC value, and other factors).

Hyperfocal Distance is what you set your lens on AND, this is the closest subject in focus (not half of that) with far end of the resulting Dpeth of Field falling on infinity. So again, if your HD is 9.3 ft, then everything closer then 9.3 ft will be outside of your depth of field limits (as per chosen COC value, and other factors).

Witold Grabiec

15-Nov-2003, 08:26

I stand corrected, it must be this damn coffee this morning:

Phil is correct - HD is put to work so you get your far end of DOF on infinity and the near end will be half of HD. Talk about Circle of Confusion huh?

Phil is correct - HD is put to work so you get your far end of DOF on infinity and the near end will be half of HD. Talk about Circle of Confusion huh?

Darin Cozine

15-Nov-2003, 14:15

Thank you all for your clarifacatinos (and confusions :)

Alan Davenport

16-Nov-2003, 16:24

Talk about Circle of Confusion huh? LOL!

Maybe I'm out of line here, but I'm mostly wondering how many people ever worry about hyperfocal distance with a large format camera. Once you get the camera on a tripod and start using movements, is HD focusing really meaningful? Are that many folks still using press cameras handheld?

Maybe I'm out of line here, but I'm mostly wondering how many people ever worry about hyperfocal distance with a large format camera. Once you get the camera on a tripod and start using movements, is HD focusing really meaningful? Are that many folks still using press cameras handheld?

phil sweeney

17-Nov-2003, 03:57

Alan: probably not often however, for me, I do when using a 90mm lens for 4 x 5. See this small article on prefocus.

<html>

Depth of field with a 90mm lense on a 4 x 5 (http://home.att.net/~shipale/DOF.html)

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</html>

<html>

Depth of field with a 90mm lense on a 4 x 5 (http://home.att.net/~shipale/DOF.html)

</body>

</html>

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