Related Links:
Angle of View (formula)
Depth of Field Calculator (Local)
Dudak's FOV Calculator [8/2002]
Focal Length Comparison Chart [4/2001]
Focal length converter
Hyperfocal Distance Formula
Java based Focal Length Comparisons (35mm vs MF..) [7/2001]
Michael K. Davis' Equiv. Focal Length Spreadsheet (Local archive)
Zeiss Lens Interactive Lens FOV Demo
(click on Changing Lenses - Interactive) [3/2001]
From Rui
Salgueiro's Field of View Calculator
Query Results for 6x6 Format:
You submitted the following name/value pairs:
focal = 40 50 75 105 135 150 200 300 400 600 800 1200 2400
format = 56 56
width = 56
length = 56
Width = 56 mm, Length = 56 mm, Diagonal = 79.196 mm
f Hor Vert Diag H/V
40.0 69.9840 69.9840 89.4212 1.0000
50.0 58.4977 58.4977 76.7556 1.0000
75.0 40.9446 40.9446 55.6657 1.0000
105.0 29.8628 29.8628 41.3253 1.0000
135.0 23.4349 23.4349 32.6948 1.0000
150.0 21.1470 21.1470 29.5759 1.0000
200.0 15.9392 15.9392 22.3983 1.0000
300.0 10.6643 10.6643 15.0384 1.0000
400.0 8.0083 8.0083 11.3071 1.0000
600.0 5.3437 5.3437 7.5517 1.0000
800.0 4.0091 4.0091 5.6674 1.0000
1200.0 2.6733 2.6733 3.7800 1.0000
2400.0 1.3368 1.3368 1.8905 1.0000
From: cghol@aol.com (CGhol)
Newsgroups: rec.photo.equipment.medium-format
Subject: Re: MF focal lengths
Date: 12 Mar 1998
How to Convert Lens Focal Length
Keppler & H Martin
POP Photography Oct 1997
Advanced
Focal Photo
Length System 35mm 6x4.5cm 6x6cm 4x5
16mm 96 115
18mm 88 100
21mm 80 90
24mm 72 84
28mm 64 75 115
35mm 53 64 95 105
50mm 39 46 74 81
65mm 30 37 60 66 110
75mm 26 32 53 59 92
85mm 23 28 48 53 87
90mm 22 27 46 51 83
105mm 18 23 40 44 75
135mm 14 18 31 35 62
150mm 12 16 28 32 56
180mm 11 13 24 27 48
200mm 10 12 21 24 44
210mm 9 11 20 23 42
250mm 8 10 18 19 36
300mm 7 8 14 17 30
400mm 5 6 11 14 23
500mm 4 5 9 10 18
600mm 3.2 4 7 8 15
700mm 2.8 3.5 6 7 12
800mm 2.4 3 5 6 11
1000mm 2 2.5 3.5 4 9
All degrees represent measurements across the diagonal of the camera
format, which is the "official" unofficial means of international
measurement.
Cliff Gholson
CGhol@aol.com
Look for the good in someone and use it to their advantages.
Hi
I will give you a hint. If the focal length equals the film size then the
distance from the object will equal the film coverage. Example: If I
photograph a 15 ft car with a 35mm lens on a 35mm camera, I will need to
be 15 ft from the car to include the whole car in the field.
Larry
The things you do when you're sick at home and/or bored at work...
Anyway, I created an Excel spreadsheet and have put it up at
http://www.stanford.edu/~dru/Lens_Comparison.xls. It's supposed
to be Excel 5.0/95/97/98 compatible. I created it with Excel 98 on
my Mac.
Anyway, to use it, just change the boldface number in the format
you're interested in to see the "equivalent" focal length in all
the other formats. I used diagonal of the image area on the film
as the basis of comparison. Not everyone agrees with that method,
but if you do, here's a useful tool. I had to guess at 645 being
56x43mm and 6x9 as 56x88mm. If that isn't right, I can fix it. I
also know that for medium format, non-6x6, not every manufacturer
agrees on what a centimeter is. Since this is an Excel
spreadsheet, just figure out the diagonal of your camera and put
it in the diagonal column, and all the other rows and columns will
correct themselves accordingly. It compares APS, 35mm, 645, 66,
67, 69, 4x5, 5x7 and 8x10 formats.
Thanks to http://www.ptw.com/~stonehvn/emulsion.htm for the
correct sizes of APS and sheet film (or, rather, the 6.3mm
reduction of image size from the borders of 4x5 film, which I just
estimated as being the same for other large format films, but I
assume someone will correct me if I'm wrong.)
Drew W. Saunders
Drew.Saunders@leland.stanford.edu
http://www-leland.stanford.edu/~dru
"In cyberspace, everyone can hear you whine."
From: Drew W Saunders dru@leland.stanford.edu
[1] Re: Lens Comparison Spreadsheet
Date: Mon May 11 1998
I just updated it to allow the user to change the height and width
of the film, and the diagonal, and therefore all lens comparisons,
will change accordingly. This will be useful if, for example, you
have a 6x9 camera that doesn't take a 56x86mm image (which I
arrived at by pure guessing anyway), you can change the width to
whatever your camera uses.
Same location, as specified above. So far about 18 folks have
downloaded it, which is 18 more than I had anticipated.
Drew W. Saunders
rec.photo.equipment.large-format
From: guran.c.ellqvist@telia.se
[1] Re: Exposure compensation chart for bellows extension
Date: Tue May 12 1998
There is something that is wrong with the above formula. When the
extension is
twice the focal length the exposure compensation should be 2 fstops.
The formula I have in the back of my head is related to magnification,
Bellow factor = sqr(magnification + 1)
When magnification is 1 (lifesize) the bellowfactor is 4, that is there
should be four times as much light passing the lens. This equals to
opening up 2 steps. Compare an earlier post by Peter de Smidt where he has
a formula with bellow draw and the correct result.
/Guran
[Ed. note: see Links Page for archive of lens
comparison spreadsheet...]
From: Drew W Saunders dru@leland.stanford.edu
Newsgroups: rec.photo.equipment.large-format
Subject: Re: Lens length equivalents between 35mm and 4x5 formats?
Date: 9 Jun 1998
drude@nospam.microsoft.com writes: >Does anyone know the little bit of math needed in order to find equivalent >lens length between 35mm and 4x5? > >So far all I've been able to find is estimations and I was hoping for >something a little more solid.
As many others have said, comparing a 2:3 ratio format (24x36mm)
to a 4:5 ratio format is problematic. One convenient "equalizer"
is to use the diagonal. If you want to compare (almost) every
format from APS to 8x10 based on diagonals, and can handle Excel
files (a safe assumption for someone with a "microsoft.com"
address), I have a spreadsheet that does these comparisons (yes, I
was REALLY bored at work one day) at
http://www.stanford.edu/~dru/Lens_Comparison.xls.
If you prefer to measure by width, it would be easy to change the
"diagonal" column to be the same as width. You could also change
the width of 35mm to 30mm instead of 36 if that's how you always
crop it, etc. All the calculations would fall in place.
P.S. If anyone has measured the exact usable image size of 5x7 and
8x10 film I'd like to know. I found a web page that listed 4x5 as
95.3mm x 120.7mm and used that measurement to guess at the borders
of 5x7 and 8x10, but if someone has specs of film holders and/or a
really accurate ruler, I'd gladly update the spreadsheet. Use my
address in my .sig, of course, sorry for the spam block.
Drew W. Saunders
Drew.Saunders@stanford.edu
Date: Tue, 11 Aug 1998
From: Peter Klosky PKlosky@grpwise-east.trw.com
To: hasselblad@kelvin.net
Subject: Re: Hasselblad Xpan -Reply
Joe,
Here's a little C program I use for angle calculations. I fed your data
into it, and came out with the 45 being halfway between your 38 and a 40.
In other words, the 45 is just ever so slightly less wide than the 38.
While I was at it, I went ahead and added popular lengths for the Hassy
and typical 35mm lenses. The output follows the program. Note that the
output includes diagonal as well as the more realistic width.
Peter
#include#define PI 3.14159 double ang ( fl, fw ) double fl, fw; { double hw = fw / 2.0; /* Base angle half width. */ double r = hw / fl; /* Ratio of opposite over adjacent. */ /* print twice half angle, converted to degrees from radians. */ return 2.0 * 180.0 * atan(r) / PI; } main(){ printf ( "frame, focal length in mm, diagonal, width and height in degrees.\n"); rpt ( 24.0, 65.0, 45 ); rpt ( 57.0, 57.0, 38 ); rpt ( 57.0, 57.0, 40 ); rpt ( 57.0, 57.0, 50 ); rpt ( 57.0, 57.0, 60 ); rpt ( 57.0, 57.0, 80 ); rpt ( 57.0, 57.0, 100 ); rpt ( 57.0, 57.0, 120 ); rpt ( 57.0, 57.0, 150 ); rpt ( 57.0, 57.0, 180 ); rpt ( 57.0, 57.0, 250 ); rpt ( 57.0, 57.0, 350 ); rpt ( 57.0, 57.0, 500 ); rpt ( 45.0, 57.0, 60 ); rpt ( 24.0, 36.0, 17 ); rpt ( 24.0, 36.0, 20 ); rpt ( 24.0, 36.0, 24 ); rpt ( 24.0, 36.0, 35 ); rpt ( 24.0, 36.0, 50 ); rpt ( 24.0, 36.0, 85 ); rpt ( 24.0, 36.0, 100 ); rpt ( 24.0, 36.0, 135 ); rpt ( 24.0, 36.0, 200 ); } rpt ( fh, fw, f ) double fh, fw; int f; { /* Field width, height. Focal length. */ double fd; /* Field diagonal. */ double fl = f; fd = sqrt( ( fh * fh ) + ( fw * fw ) ); printf ( "%10.2f by %10.2f frame. ", fh, fw ); printf("%d %10.2f %10.2f %10.2f\n", f, ang( fl, fd ), ang ( fl, fw ), ang ( fl, fh ) ); } frame, focal length in mm, diagonal, width and height in degrees. 24.00 by 65.00 frame. 45 75.18 71.68 29.86 57.00 by 57.00 frame. 38 93.37 73.74 73.74 57.00 by 57.00 frame. 40 90.44 70.94 70.94 57.00 by 57.00 frame. 50 77.74 59.37 59.37 57.00 by 57.00 frame. 60 67.78 50.82 50.82 57.00 by 57.00 frame. 80 53.48 39.22 39.22 57.00 by 57.00 frame. 100 43.90 31.82 31.82 57.00 by 57.00 frame. 120 37.13 26.72 26.72 57.00 by 57.00 frame. 150 30.08 21.52 21.52 57.00 by 57.00 frame. 180 25.24 17.99 17.99 57.00 by 57.00 frame. 250 18.32 13.01 13.01 57.00 by 57.00 frame. 350 13.14 9.31 9.31 57.00 by 57.00 frame. 500 9.22 6.52 6.52 45.00 by 57.00 frame. 60 62.36 50.82 41.11 24.00 by 36.00 frame. 17 103.68 93.27 70.44 24.00 by 36.00 frame. 20 94.49 83.97 61.93 24.00 by 36.00 frame. 24 84.06 73.74 53.13 24.00 by 36.00 frame. 35 63.44 54.43 37.85 24.00 by 36.00 frame. 50 46.79 39.60 26.99 24.00 by 36.00 frame. 85 28.56 23.91 16.07 24.00 by 36.00 frame. 100 24.41 20.41 13.69 24.00 by 36.00 frame. 135 18.21 15.19 10.16 24.00 by 36.00 frame. 200 12.35 10.29 6.87
"Joe McCary - Photo Response" mccary@erols.com 08/10/98 08:14pm
Does anyone know the angle of the image on a 66mm neg with a 45mm lens? Is
it more or less than the 38mm lens on 6x6?
Joe
However, let me point out a small detail. It is not possible to use a
66mm
section out of a Normal hasselblad neg, because they measure 56 or 57
mm, a
difference of about 12 %.
----------------------------------------------------------------------
[ Part 2: "WordPerfect 4.2" ]
frame, focal length in mm, diagonal, width and height in degrees.
24.00 by 65.00 frame. 45 75.18 71.68 29.86
57.00 by 57.00 frame. 38 93.37 73.74 73.74
57.00 by 57.00 frame. 40 90.44 70.94 70.94
57.00 by 57.00 frame. 50 77.74 59.37 59.37
57.00 by 57.00 frame. 60 67.78 50.82 50.82
57.00 by 57.00 frame. 80 53.48 39.22 39.22
57.00 by 57.00 frame. 100 43.90 31.82 31.82
57.00 by 57.00 frame. 120 37.13 26.72 26.72
57.00 by 57.00 frame. 150 30.08 21.52 21.52
57.00 by 57.00 frame. 180 25.24 17.99 17.99
57.00 by 57.00 frame. 250 18.32 13.01 13.01
57.00 by 57.00 frame. 350 13.14 9.31 9.31
57.00 by 57.00 frame. 500 9.22 6.52 6.52
45.00 by 57.00 frame. 60 62.36 50.82 41.11
24.00 by 36.00 frame. 17 103.68 93.27 70.44
24.00 by 36.00 frame. 20 94.49 83.97 61.93
24.00 by 36.00 frame. 24 84.06 73.74 53.13
24.00 by 36.00 frame. 35 63.44 54.43 37.85
24.00 by 36.00 frame. 50 46.79 39.60 26.99
24.00 by 36.00 frame. 85 28.56 23.91 16.07
24.00 by 36.00 frame. 100 24.41 20.41 13.69
24.00 by 36.00 frame. 135 18.21 15.19 10.16
24.00 by 36.00 frame. 200 12.35 10.29 6.87
[ Part 3: "WordPerfect 4.2" ]
#include
#define PI 3.14159
double ang ( fl, fw ) double fl, fw; {
double hw = fw / 2.0; /* Base angle half width. */
double r = hw / fl; /* Ratio of opposite over adjacent. */
/* print twice half angle, converted to degrees from radians. */
return 2.0 * 180.0 * atan(r) / PI;
}
main(){
printf (
"frame, focal length in mm, diagonal, width and height in degrees.\n");
rpt ( 24.0, 65.0, 45 );
rpt ( 57.0, 57.0, 38 );
rpt ( 57.0, 57.0, 40 );
rpt ( 57.0, 57.0, 50 );
rpt ( 57.0, 57.0, 60 );
rpt ( 57.0, 57.0, 80 );
rpt ( 57.0, 57.0, 100 );
rpt ( 57.0, 57.0, 120 );
rpt ( 57.0, 57.0, 150 );
rpt ( 57.0, 57.0, 180 );
rpt ( 57.0, 57.0, 250 );
rpt ( 57.0, 57.0, 350 );
rpt ( 57.0, 57.0, 500 );
rpt ( 45.0, 57.0, 60 );
rpt ( 24.0, 36.0, 17 );
rpt ( 24.0, 36.0, 20 );
rpt ( 24.0, 36.0, 24 );
rpt ( 24.0, 36.0, 35 );
rpt ( 24.0, 36.0, 50 );
rpt ( 24.0, 36.0, 85 );
rpt ( 24.0, 36.0, 100 );
rpt ( 24.0, 36.0, 135 );
rpt ( 24.0, 36.0, 200 );
}
rpt ( fh, fw, f )
double fh, fw; int f; { /* Field width, height. Focal length. */
double fd; /* Field diagonal. */
double fl = f;
fd = sqrt( ( fh * fh ) + ( fw * fw ) );
printf ( "%10.2f by %10.2f frame. ", fh, fw );
printf("%d %10.2f %10.2f %10.2f\n", f,
ang( fl, fd ), ang ( fl, fw ), ang ( fl, fh ) );
}
From: bg174@FreeNet.Carleton.CA (Michael Gudzinowicz)
Newsgroups: rec.photo.technique.nature
Subject: Re: Viewing angle formula??
Date: 21 Nov 1998
GPSbouvier gpsbouvier@aol.com writes:
>Is there a formula to calculate the viewing angle of a lens for 35 mm >photography? My best guess is that my normal lens sees an angle of about 30 >degrees edge to edge of the frame.
view_angle = 2 * inverse_tangent(1/2 * format_dimension / focal_length)
The inverse_tangent is also called the arctangent, and the common function
in spreadsheets and programming languages is atan(), which usually returns
a value in radians. Multiply by 180/PI to convert to degrees.
For your 50 mm lens, and the full 36 mm width of the 35 mm frame:
view_angle = 2 * atan(1/2 * 36 / 50) = 39.6 degrees
For the short 24 mm side, the angle is 27 degrees.
>In planning a shot, I need to find the viewing angle for a wide angle lens an >a telephoto.
For non-macro work, you can use the ratios of focal length and format, and
distance and field of view.
focal_length / format_dimension = subject_distance / field_of_view
For instance, your 50 mm lens focused at 50 feet, would have a field of
36 feet in the long dimension and 24 feet on the short side.
If your subject were 100 feet away, and you wanted to frame a 10' X 15'
area, the focal length required would be 24 mm * 100' / 10' = 240 mm.
>Obviously a wide angle lens has a wide viewing angle and a telephoto lens has >narrow angle, but I need to be more precise. Thanks.
Date: Tue, 26 Jan 1999
From: Randy Vaughn-Dotta randyd@csufresno.edu
Subject: Re: Lens Link
Mike,
This is a location I favor for such information.
http://www.mat.uc.pt/~rps/photos/angles.html
Randy
>Greetings all,
>
>I remember a link someone had that would give one the field of view for a
>given lens/format i.e. 135mm lens on 4x5 would yield a verticle angle of X and
>a horizontal angle of Y.
>
>My system crashed and I've lost all my bookmarks (as well as many other items
>{;-( ) - I don't remember where this link was - anyone have an idea? I don't
>care if its the same one I had or not as long as I can check some
>relationships out.
>
>Thanks all,
>-Mike
Randy Vaughn-Dotta, Photographer
Academic Innovation Center
Date: 10 May 1999
From: Charles Steinmetz csteinmetz@redneck.efga.org
Newsgroups: rec.photo.equipment.large-format
Subject: Re: How to calculate Angle of view?
Mac wrote:
>I'm trying to determine what
>different WA lenses would compare to in 35mm. I'd like to compare based
>on vertical and then horizontal angle of view.
>
>If you have a formula, I'd appreciate it.
The general formula for angular coverage is:
2 * tan^-1(L/2F)
where
L = any film-plane dimension (height, width, diagonal)
F = lens focal length, in same units as L
To compare coverage between formats: Find the ratio of L1/L2, where
L1 and L2 are the same dimension (height, width, diagonal) of the two
formats. Lens focal length for equivalent coverage scales according
to the ratio of film dimensions.
Easy example: Every dimension on an 8x10 is twice as large as on 4x5.
To match the coverage of a given lens on 4x5, use a lens
twice as long on 8x10.
Your example:
L1 = 24mm = height of 35mm frame
L2 = 56mm = height of 6cm frame
L1/L2 = 0.429 = ratio of heights
300mm * 0.429 = 128.6mm = lens that covers same height on 35mm
as 300mm covers on 6x17cm
L1 = 36mm = width of 35mm frame
L2 = 166mm = width of 17cm frame
L1/L2 = 0.217 = ratio of widths
300mm * 0.217 = 65mm = lens that covers same width on 35mm
as 300mm covers on 6x17cm
Best regards,
Charles
Date: Mon, 10 May 1999
From: Bob Wheeler bwheeler@echip.com
Newsgroups: rec.photo.equipment.large-format
Subject: Re: How to calculate Angle of view?
Two points.
(1) Equivalent lenses in different formats do not
depend on any particular angle of view, but in
fact hold for all angles of view. That is,
equivalent lenses for a 23o angle of view are the
same as for a 40o angle of view, as is shown by
the formula Steinmetz cited.
(2) AOV can also depend on magnification. The
formula Steinmetz gave applies only when the focus
is at infinity. A formula allowing for
magnification is
2 * tan^-1(L/2(F(1+m))),
where m is magnification -- the film dimension
divided by the object dimension.
....
From Hasselblad Mailing List:
Date: 26 Dec 99
From: Patrick Bartek bartek@access1.com
Subject: Re: C Lenses and Personal Preferences
Klaus Knuth wrote:
> I'm a - so far - 35mm person, trying to become acquainted with a used 500 > c/m and el/m setup I bought a few weeks back and I'm interested in member's > opinions about their favorite, best and most used lenses. What to go for > and the pitfalls to avoid. Like: Is the Schneider zoom something to start > saving for? Is the 40mm really useful or just for rare occasions? Are any > of the older lenses lacking in quality? When and with what kind of lenses > do you use a tripod?
One bit of advice I can give you: DON'T get the equivalent focal
lengths in 2 1/4 that match to your 35mm lenses; since 35mm is very
rectangular and the 'Blad is very square, there is not an equal
relationship. For example, my normal lens for 35mm is the 35 with a
54 degree horizontal angle of view. The equivalent 'Blad lens would
be the 50 at 58 degrees, but the 50 is just too wide for a normal.
But the 60 (50 degrees) is just about perfect.
Most people buy the 'Blad with an 80, then get the 50 and 150 to round
out the system. I had the experience of owning a Mamiya TLR with
an 80, before getting the Hasselblad and the 80 was either too long or
not wide enough. So, when I got the Hasselblad, the 60 was the
perfect normal lens for me. I don't even own an 80. Over the past 15
years of shooting professionally with it (mostly commercial work), the
60 is used about 2/3 of the time.
For head and shoulder portraits and magazine covers, I preferred the
135 from the Mamiya days, but I didn't want to mess with the 'Blad's
135 and bellows. The 120 was too short (plus both were too much
money even used). So, I compromised and got the 150.
For a wide angle, the only choice was a SWC. The 50 was too close to
the 60 (only 10 degrees wider), and the 40 of the day was that monster
C 40 with the 104mm filter size.
So, there you have it: SWC, 60, 150. I can do about 99% of all the
medium format work I get with just those 3 lenses. There is only one
more lens I'd like to have: the black 100 C T*. It would be perfect
for product work or group head and shoulder shots like for CD covers,
but I get by with the 150 and sometimes a #1 Closeup lens.
PS. I purchased all my equipment used. So, just because it's old,
doesn't mean it's bad.
PPS. I don't think the Schneider zoom is worth the money. Too big.
Too heavy. Too small a range. Wrong range. Like a 90 to 180 on
35mm.
Oh, and by the way: Welcome to the club.
--
Patrick Bartek
NoLife Polymath Group
bartek@access1.com
From hasselblad Mailing List:
Date: Sun, 23 Jul 2000
From: InfinityDT@aol.com
Subject: Re: HUG: Advice please; 40mm C vs. 50mm CF
jeffcoatphoto@sumter.net writes:
BTW it's about the same as a 28 on 35 mm
In diagonal coverage, yes. But where it counts, left-to-right horizontal
coverage, the 50 translates to a 33mm in 35mm format, so it's between a 28
and a 35, closer to the 35. The 40 is about a 26mm so it comes right
between a 24 and a 28. The 60 is approximately 39mm, a tad longer than a
35 but well short of a 50.
From Hasselblad Mailing List;
Date: Tue, 16 Jan 2001
From: "Q.G. de Bakker" qnu@worldonline.nl
To: hasselblad@kelvin.net
Subject: Re: Close-up Portraits
Paul Clark wrote:
> I don't even pretend to understand the use of extension tubes, but this > thread has me wondering. > > Can I use a 56mm extension tube on my 80mm CB lens on my 501CM, effectively > creating a 136 mm lens? Wouldn't the extension tube be a whole lot cheaper > than a 120mm macro planar or 150 mm Sonnar lens?
Just an addition to what has been answered already:
you can determine the field of view your lens will give by dividing its
focal length by the amount of extension used, and then multiplying the
result by the size of the frame.
For instance: a 150 mm lens with 16 mm extension on a Hasselblad having a
56.5 mm format
150/16 = 9.375 * 56.5 = 529.6875 mm = almost 53 cm.
Keep in mind that a lens has a built-in extension of its own. The 150 mm
lens has approximately 21 mm maximum.
From rollei mailing list:
Date: Fri, 9 Mar 2001
From: John Kufrovich jkufrovich@ev1.net
Subject: [Rollei] Zeiss
FYI, Zeiss has added a new feature to their site. It gives you a view
angle of an image using different lenses.
http://www.zeiss.de/de/photo/home_e.nsf/allBySubject/Launch+-+Zeiss-engl+JavaNavigator
Just click on the "Changing Lenses - Interactive"
jk
From Rollei Mailing List:
Date: Sat, 24 Mar 2001
From: DKFletcher@aol.com
Subject: Re: RE: [Rollei] Angle of view
Say you photograph a person with a 90 mm lens on 4x5 and shoot the same
person (from the same distance with the same background) with your 90mm
lens on your G2 you can lay the 35mm slide on top of the 4x5 sheet of film
and the two should be identical. the only thing you should see is the
sprocket holes of the roll film. The size of the head and the
relationship between the person and the background will be identical.
The only difference will be the 4x5 will have a LOT more image around it.
....
From Rollei Mailing List:
Date: Sat, 26 May 2001
From: Mark Rabiner mark@rabiner.cncoffice.com
Subject: [Rollei] Lens angle of view table Long dimension
I'm made a color A3 PDF 31k chart and uploaded it for people.
It goes from half frame to 8x10 inch formats.
Take a look if you have Acrobat Reader. It should print well.
Lens angle of view table Long dimension
http://www.rabiner.cncoffice.com/pdfs/LensangleofviewtableLong.pdf
If there are focal lengths that you use which are missing please fill me
in.
Remember. These are not diagonal angles if view but horizontal or long
side as the cinematographers know are more valuable. I'm made sure to
include the focal lengths for Leica M and R and Hasselblad but i also
looked up all the sites of the usual suspects to get the focal lengths of
interest.
although i see i forgot 40mm for my Rollei 35. I shoot with a Rolleiflex
by the way.
Mark Rabiner
Portland, Oregon
USA
http://www.rabiner.cncoffice.com/
Date: 6 Jun 2001
From: Drew W Saunders dru@leland.stanford.edu
Newsgroups: rec.photo.equipment.large-format
Subject: Re: focal lengths
christine, christine@napc.com writes:
>Does anyone know of a chart someplace online that list short, long, and >normal focal lengths for lenses for both 4x5 and 8x10? Or am I asking >for way too much in one page? :>)
I created an Excel spreadsheet (actually, it started out in Claris
Works, then I converted it) that allows you to compare based on
the diagonal or on the horizontal angle of view. Being a
spreadsheet, you can modify any value you want, so if you always
crop 35mm to 8x10, then you could consider it to be 24x32mm and
modify the chart accordingly. It's at:
http://www.stanford.edu/~dru/Lens_Comparison2.xls
It's an older version so should be compatable with older versions if you
don't have the latest.
I dug up actual measured dimensions of 95.3x120.7mm for 4x5 and
interpolated to 5x7 and 8x10, so if someone has the correct
numbers for those formats, I'd love to know them and I'll update
the chart accordingly.
Drew W.Saunders
Stanford Network Consulting
mailto:Drew.Saunders@Stanford.EDU
http://www.stanford.edu/~dru
Date: 4 Jul 2001
From: john@stafford.net (John Stafford)
Newsgroups: rec.photo.equipment.medium-format
Subject: Re: usage vs. coverage Re: comparing 35mm to med. format lens
rmonagha@smu.edu (Robert Monaghan) wrote in
True! I think 35mm photographers might be confused by the metric, so
to clarify in other words: In practice (usage), the SWC's 38mm Bigon
over 6x6 is SIGNIFICANTLY wider than a 24mm over 35mm format. It
covers more of the same scene. Your degree-of-angle figure compares
the long side of 35mm to the 58mm side of 6x6. In that regard, the
angle is almost the same, however, the 'vertical' dimension of the 6x6
is far greater than the short side vertical of 35mm. The SWC is simply
'wider' in application for this case.
> the usage is also different; I have a hasselblad "superwide" with 38mm > biogon lens, this is about as wide as it gets in 6x6cm; that said, the > superwide lens only covers about 73 degrees horizontally (and vertically) > which is about the same as my 24mm nikkor on 35mm SLR; I don't consider > 24mm "superwide" on 35mm, but I do consider 73 degrees of coverage > superwide on 6x6cm. So you really have to go by usage versus coverage...
From Bronica Mailing List:
Date: Fri, 22 Jun 2001
From: Robert Monaghan rmonagha@post.smu.edu
Subject: re: 40/50mm dilemma
yep, just been there; ended up with a hassy SWC/M as 38mm biogon was tops
for architecture and cityscapes I like to shoot, near zero distortion, and
high contrast (true non-retrofocus lens). The biogon is similar to my 24mm
on 35mm, but quite a bit less wide than my Veriwide 100 (18mm equiv on
35mm) I figure I can always crop a 50mm angle of view photo out of the
38mm shot; but I can't do the reverse, and I can't move back in a cramped
room for the 50mm, but if I can't get it in the 38/40mm, the only option
is fisheyes
see http://www.rabiner.cncoffice.com/pdfs/LensangleofviewtableLong.pdf
the horiz. long axis of the 40mm on 6x6cm ~ to 69 degrees, or circa 26mm
horiz. long length coverage on 35mm SLR; 50mm ~ 58 degrees, about 33mm on
a 35mm SLR lens horizontally (35mm = 54 degrees); personally, I find a
35mm a "wide normal" so a 28mm is fairly nominal as a wide angle on 35mm
SLRs; in other words, a 50 is not very wide on 6x6cm ;-)
another trick is to take a 6x6cm slide mount and position it 40mm and then
50mm from your eye, using a tape measure under the eyeball on the eye
socket. The view is approx. what a 40mm and 50mm lens will see (and ditto
to see 150mm vs 250mm etc.). To make it more accurate, double the
dimensions and cut out of black cardboard (112x112mm hole at 80mm and
100mm) Use this as a composition tool and to help you decide what the 40mm
gets you in general use shots that the 50mm doesn't etc.
a third trick is to buy one of the front of lens adapters; the 0.6x gives
a view of what a 50mm will do, while the 0.5x will give view of 40mm when
used on 75-80mm normal lenses ;-) the 0.42x also give a nice fisheye
effect for low $$ versus 35mm semi-fisheyes ;-) (see fisheye adapter pages
http://www.smu.edu/~rmonagha/bronfe.html for examples etc.) Compare the
two for utility and how often you need the extra coverage of the 40mm etc.
resell the adapters, or keep them for other projects ;-)
HTH bobm
From: "Leonard Evens" len@math.northwestern.edu> Newsgroups: rec.photo.equipment.medium-format Subject: Re: Normal lens for 6x9? Date: Tue, 15 Jan 2002 "Leonard Evens" len@math.northwestern.edu> wrote: > "Eldritch" > TiredofSpa_m@hotmail.com> wrote: > >> Could some kind soul tell me what a normal lens is for a 6x9cm camera? >> Would it be 105cm? I bought an old Zeiss Ikon Ikonta 521/2 and that's >> the lens it came with. >> >> E >> >> > As others have commented, the normal focal length is supposed to be the > diagonal of negative. Also, one might want to distinguish the nominal > dimensions from the actual dimensions of the image area, which will be > slightly smaller. Roll film and sheet film holders may produce > different results for the film area. However, a few mm either way > shouldn't make an enormous difference. > > But I've always thought that taking the diagonal without reference to > the aspect ratio didn't make much sense. 6 x 9 is pretty close to 35 > mm with an aspect ratio of close to 1.5. 6 x 7 is usually 56 x 70 mm, > with an aspect ratio of 1.25, which is the same as a lot of photographic > paper, e.g., 8 x 10, 16 x 20. 6 x 7 and 6 x 9 would have roughly the > same angle of view in the short dimension but different angles of view > in the long direction. If one printed on standard paper, one would > have to crop the 6 x 9 image anyway. > > Keep in mind also that the diagonal of a 24 x 36 mm frame is about 43 > mm, while the normal focal length for such a camera is usually taken to > be 50 mm or even 55 mm. If one crops to 8 x 10 format, that makes such > a lens look definitely long rather than normal. The "normal lens" for > 6 x 6 format is usually taken to be about 75 or 80 mm. The latter is > pretty close to the diagonal. But 6 x 6 images are usually cropped to > an 8 x 10 aspect ratio, which again makes this appear to be a bit long. > It might be added that the rationale behind a so-called normal focal length is something like the following. If you view an 8 x 10 print at about 12 inches, its diagonal subtends about 53 degrees. If your lens also has the same angle of view, then when you view that print at that distance, you recover the perspective of the camera when you view the print. One could argue that for a print to look like the real thing, one should always view it with the eye placed at the appropriate center of perspective when taking into account the difference in scale between the print and the actual scene. However, as Pirenne has pointed out, the print is also an object in its own right, and our visual system responds differently to it because of that. That means you need not view the print from the center of perspective nor even at any special distance in order to interpret it properly. Moreover, it is often necessary to have some sort of perspective distortion to compensate for properties of the visual system like size constancy. -- Leonard Evens len@math.northwestern.edu 847-491-5537 Dept. of Mathematics, Northwestern Univ., Evanston, IL 60208
Newsgroups: rec.photo.equipment.medium-format Subject: Re: Normal lens for 6x9? From: josh@WOLFENET.COM (Joshua_Putnam) Date: Wed, 16 Jan 2002 Leonard Evens len@math.northwestern.edu> wrote: >But I've always thought that taking the diagonal without reference to the >aspect ratio didn't make much sense. 6 x 9 is pretty close to 35 mm >with an aspect ratio of close to 1.5. 6 x 7 is usually 56 x 70 mm, with >an aspect ratio of 1.25, which is the same as a lot of photographic >paper, e.g., 8 x 10, 16 x 20. 6 x 7 and 6 x 9 would have roughly the >same angle of view in the short dimension but different angles of view in >the long direction. If one printed on standard paper, one would have to >crop the 6 x 9 image anyway. I find it much more sensible to use sqrt (2 * long_side^2), in essence treating the rectangular negative as having been cropped from a square negative, since almost nobody really prints squares. In 35mm, that means the normal lens would be the square root of 2*36^2, or 50.9mm As for printing on standard paper, I cut the next larger size in half, so that 1 sheet of 16x20 becomes two sheets of 10x16. You get that same 1.6 aspect ratio from cutting down other standard sizes based on 8x10, so really there's no shortage of paper suited for rectangular formats, the manufacturers just haven't finished cutting it yet. -- josh@phred.org is Joshua Putnam http://www.phred.org/~josh/ Updated Infrared Photography Books List: http://www.phred.org/~josh/photo/irbooks.html
From: Tony Polson tony.polson@btinternet.com Newsgroups: rec.photo.equipment.medium-format Subject: Re: Naturalism/ was: Depth of Field - 35mm vs MF Date: Fri, 03 Aug 2001 Phil Stripling phil_stripling@cieux.zzn.com> wrote: > I read somewhere years ago that the 135mm lens (in 35mm cameras) was 135mm > because it represented what the human eye saw when it was focused on a > particular object. (Not that the eye zooms in, but that it focuses on > something to the exclusion of other things in a way similar to the 135mm > lens.) Hi Phil, A survey on this precise same topic was conducted by a UK magazine over 20 years ago. The conclusion drawn was that most people questioned by the magazine felt the focal length which gave a field of view and perspective similar to that of the human eye was ... ... around 70mm. -- Best regards, Tony Polson
Date: Tue, 18 Dec 2001 From: Robert Monaghan rmonagha@post.cis.smu.edu Subject: Re: Equivalent lens focal lengths for different formats To: Brian Swale bj@caverock.net.nz yep, you will find such charts in Roger Hicks Medium Format Handbook IIRC I also like his The Lens Book for 35mm stuff, lots of good insights. Bob Shell of shutterbug editor fame noted hicks was one of the top guys he knew into lenses for photography, and so I've tended to like his stuff too but the usage is quite a bit different than the diagonal (esp.) or even the horiz. coverages would suggest. partly, you crop 6x6 and even 35mm a bit but not 645 or 6x7 as much, so it isn't exactly the same except for 35mm and 6x9cm. And a 50mm is pretty wide in 6x6cm while an equiv 35mm SLR might be math-wise a 33mm, it is used like a 28mm on 35mm SLR usually. this may partly be due to most 35mm finders chopping off much of the edges while 6x6 shows you the whole thing? You see about the same thing, you just get more on edges of film and slide with 35mm than 6x6cm... just my theory ;-) bobm
From: Lassi lahippel@ieee.org Newsgroups: rec.photo.equipment.medium-format Subject: Re: Normal lens for 6x9? Date: Thu, 17 Jan 2002 Leonard Evens wrote: ...> > Metric size papers all yield the same aspect ratios when cut in half, > namely sqrt(2) = 1.4.14... Personally, I find the 8 x 10 aspect ratio > of 1.25 most pleasing, but that may just be from years of printing on > 8 x 10 paper in the darkroom. I would also like to see the "normal lens" as the "diagonal of the largest rectangle with 1.4142... aspect ratio that fits in the frame". In most film sizes that rectangle covers more than 90% of the frame. The exceptions are the extremes, panoramic cameras and squares. I'd like to go even further. For other lenses the ratio (focal length)/(normal lens) can be called "lens magnification". Using magnification rather than focal length would eliminate the confusion of different film and detector sizes. Lenses could be classified according to their magnification value, for example as: - superwide: less than 0.5 - wide: 0.5 to 0.8 - normal: 0.8 to 1.5 - tele: 1.5 to 5 - supertele: more than 5 The numeric values are negotiable, of course. Square formatters will be allowed use mm :-) They are all 6x6 cameras, and their focal lenghts can be compared as such. Anyway, as digitalisation and bubble jets win ground (even in postproduction of MF photos) standard paper sizes of office automation will take over from traditional photographic sizes. The metric A-series is very handy and rules most of the world. You can still crop it to 1.25 if you want. Loses only 12% of the sheet. -- Lassi
From minolta mailing list: Date: Thu, 08 Nov 2001 From: tonyfleres@home.com Subject: Re: Formula for calculation the width of a pic A quick approximation that doesn't involve trig functions is: distance / focal length x 36 = width of scene, and distance / focal length x 24 = height of scene. HTH, Tony Fleres
From minolta mailing list: Date: Wed, 07 Nov 2001 From: ctgardener@yahoo.com Subject: Re: Formula for calculation the width of a pic --- In Minolta@y..., Christian Knigge christian.knigge@g... wrote: > hello all, > if I know the focal length and I know the distance I assume I could > calculate the width (of the scenery/house whatever I shot) fitting > on the film. Does > anybody knows the formula? A reasonable approximation should be 2*distance*tan(angle-of-view/2) where angle-of-view is dependent on the focal length. So if you were shooting straight at a house centered in the frame 50 feet away with a lens giving you 60 degree coverage, 2 * 50 * tan(60/2) = 57.7 feet. - Dennis
From: bg174@FreeNet.Carleton.CA (Michael Gudzinowicz) Newsgroups: rec.photo.equipment.medium-format Subject: Re: Best Head and shoulder focal length for 6x9 Date: 5 Feb 2002 "Joe Lacy" jmlacy1@attbi.com wrote: > I need to get tight on my subject without "the big nose" syndrome at 6x9. > What focal length length lens will accomplish this in the 6x9 format. I'm > shooting a Galvin. subject_lens_distance = (1 + 1/M) x focal_length focal_length = subject_lens_distance / (1 + 1/M) If you want to reduce a 10x15" area to 2x3" (6x9), the magnification is 2/10 = 0.2 X For a 6 foot shooting distance, focal_length = 6' / (1 + 1 / 0.2) = 6' / 6 = 1' = 12" = 305 mm extension_required = (1 + M) x focal_length = 1.2 x 305 = 366 mm = 14.4"
Date: Sat, 26 May 2001 From: Richard Knoppow dickburk@ix.netcom.com To: rollei@mejac.palo-alto.ca.us Subject: Re: [Rollei] Lens angle of view table Long dimension you wrote: > Mark Rabiner at mark@rabiner.cncoffice.com wrote: > >> If there are focal lengths that you use which are missing please fill me in. > >Mark... > >Thanks for sharing the chart. I'd personally like to see the following focal >lengths added: > >90mm 6x6 >72mm 4x5 > >Doug > >-- >Doug Brightwell >doug@dougbrightwell.com Its easy enough to calculate angle of view. The calculator built into Windows (and also Mac's)will do it. The angle is as follows: Angle of view = 2* arctan (FL / D)/2 Where arctan is inverse tangent FL is focal length of the lens D is the length of the dimention the angle is to be calculated for (width, length, diagonal, whatever). The reason for deviding and multiplying is that we must calculate the half-angle and double it for the full angle of view. Of course the inverse of this formula can be used for determining the focal length of lens needed for a particular angle of view. Here the formula is: FL = 2* tan(A/2)*D Where A is the desired angle. ---- Richard Knoppow Los Angeles,Ca. dickburk@ix.netcom.com
From rollei mailing list: Date: Sun, 27 May 2001 From: Richard Urmonas rurmonas@senet.com.au Subject: Re: [Rollei] Lens angle of view table Long dimension Richard, There is an error in your formula. The angle of view is: Angle of view = 2 * arctan(D / (2 * FL)) or 2 * arctan(D / FL / 2) if you prefer. And so FL = 2 * D / tan(A/2) Richard.
From: Drew Saunders dru@stanford.edu Newsgroups: rec.photo.equipment.large-format Subject: Re: Equivalent focal lengths from medium or 35 mm to large format? Date: Tue, 14 May 2002 "Leonard Evens" len@math.northwestern.edu wrote: > I am beginning to toy with the idea of getting a 4 x 5 system and I've > been looking at the recommendations for appropriate lenses. I now have > a Horseman Technical camera with limited movements. In the past, when > comparing with 35 mm, I've calculated a certain way and I wonder if anyone has > any comments on it. > > Assuming that I will usually print with a 4:5 > aspect ratio, I take the 35 mm format as being 24 x 30 with "movement" > possible in the long dimension, in that one can crop where desired along the > 36 mm length. > > With that assumption, the diagonal is about 38.4 mm. > Comparing that to 6 x 7 (which has about a 90 mm diagonal), I find a 28 > mm lens for 35 mm is about equivalent to a 65 mm lens for 6 x 7. Taking > the diagonal of 4 x 5 film to be about 150 mm, that means the same 35 mm > lens is equivalent to about 110 mm. Of course, I would be even better > off with a 90 mm lens, but I wondered if there was anything wrong with my > analysis. (I had a reference for the actual frame dimensions for 4 x 5 > film somewhere, but I can't seem to find it.) I have an excel spreadsheet that you might like, mainly because it lets you easily change the dimensions for any of the film sizes, so you could re-do all the calculations for both diagonal and horizontal angle of view for 135 by changing the "36" to "30" and everything will fall into line. It's at http://www.stanford.edu/~dru/Lens_Comparison2.xls. I wanted a lens that approximated the angle of view of a 35mm on 35mm and I ended up with a 120mm lens and have been happy with it. Drew -- Drew W. Saunders dru (at) stanford (dot) eee dee you
From: wiltw@aol.com (Wilt W) Newsgroups: rec.photo.equipment.medium-format Date: 31 May 2002 Subject: Re: 35mm vs 645 focal length roland.rashleigh-berry wrote: > >> Standard focal length is roughly the length of the diagonal of the film >> frame. Because the 35mm frame is overly long, compared to other formats, a better way to convert equivalent focal lengths is to compare the lens with the short dimension of the film rather than the diagonal. So a 24mm lens in 35mm format is like a 40mm lens in 645 is like a 90mm lens in 4x5. Similarly a 100mm lens in 35mm format is 4x the short dimension, so the equivalent 645 length is 160mm and the 4x5 format is 360mm. A 50mm lens would be 2x the short dimension in 35mm format, so you choose a 85mm lens in 645 and 180mm in 4x5. Note that this illustrates the fact that a 50mm lens on the 35mm format is longer than the 'diagonal' dimesion dictates, and that is an abberation caused by the fact that a 50mm lens allowed the mirror to clear the back of the lens easily without resorting to more expensive optics design in the mass produced 'normal' lenses that were provided with every SLR body! --Wilt
rec.photo.equipment.large-format From: "Martin Trucco" martintrucco@tutopia.com [1] Re: Lens Chart Date: Fri Jun 07 2002 I do have an Excel chart with three equivalence comparisons: Long side, short side and diagonal/hypotenuse. You can find them at my web site www.martintrucco.com.ar Rgds
From: sog@amaterasu.scd.ucar.edu (Steve Gombosi) Newsgroups: rec.photo.equipment.medium-format Subject: Re: Hasselblad 150mm vs. 180mm portrait lens comparison Date: 15 Jul 2002 Q.G. de Bakker qnu@worldonline.nl wrote: >Stik wrote: > >> 2. Head and shoulder shot - 150mm requires 8mm or 16mm extension tube. >> 180mm does not require an extension tube. >It does. >Both 150 mm and 180 mm lens have about the same minimum field of view >(roughly 35 cm / 14"). So if you need extension tubes to get tighter framing >using the 150 mm lens, you do too when using the 180 mm lens (16 or 32 mm >tube). Huh? How do you get that? I figure that field of view = 2*distance*(tan(t/2)), where t = angle of view. For 150mm CFi (t=30 degrees, minimum distance - 1.4 meters), diagonal field of view = 2.8tan(15 degrees) = 0.75 meter (= 29.5 inches) For 180mm CFi (t=24 degrees, minimum distance = 1.55 meters), diagonal field of view = 3.1*tan(12 degrees) = 0.64 meter (= 25.2 inches) Vertical or horizontal field of view ought to be diagaonal/sqrt(2) (assuming square format), which gives: 0.53 meters for the 150 (= 20.9 inches) 0.47 meters for the 180 (= 18.5 inches) 2.4 inches is a lot of difference in a portrait. Look, the thing to do is go down to a dealer, put a 150 on a body, focus as close as you can, then do the same with a 180. Look at the difference in the viewfinder. See if you find the working distance difficult, or if you think the lens is too bulky. One of the distinct advantages of Hasselblads is that you can find them pretty much anywhere. You're the only person who can actually determine whether a given piece of gear is suitable for you. >However, 35 cm is already pretty tight for a head & shoulder shot, so you >will probably not need a tube at all with either lens for H&S.; > >> 3. Small studio - 180mm focal length maybe too long to use in a small >room. > >Not that much difference between the two really (approx. 15 cm / 6" less >working distance, at most). But if you're really struggling for space, every >bit might help. ell, there's a 20% increase in focal length so you can expect a corresponding increase in working distance for the same magnification. > >> 5. Lens flare - Someone mentioned that 180mm has flare problems, and >> another poster mentioned 150mm having flare problems. With proper lens >> hood, can I assume that this is a none issue? > >Yes. I've never heard of flare problems with the 180. I'd be really surprised if this were the case (and, as Q.G. notes, a hood will take care of it). The 150 and the 180 have the same number of elements, so I wouldn't expect any significant difference in flare between them. >> 6. Ease of use - 150mm lens is smaller, and easier to handle. 180mm is >> heavier and more difficult to handle. Has anyone have specific problems >> with the weight of 180mm lens on a tripod, e.g. tipping over? Someone It's not a 350 or 500. Both the 150 and 180 are quite compact. Neither affects the balance of the camera to any significant degree. They're both easy to use hand-help. >> mentioned that 180mm is clunkier (can someone please confirm this?) Define "clunkier". Steve
From: "Paul Saunders pvs1@wildwales.fsnet.co.uk Newsgroups: rec.photo.equipment.medium-format Subject: Re: MF lens basics. Date: Thu, 1 Aug 2002 "Bo Wrangborg" bo@visicon.se wrote > I think this is sad in a way - the "mm" for a lense! It tells nothing! I agree. > Why not use the "lense opening angle" as I and some others do. > Then you can easily compare 35 mm MF and LF. Much better. > (Ps "opening angle" is direct transation from Swedish - but you know what I > mean and what you might call it!) It's usually called "Field of View" or "Angle of View" Here's a field of view calculator; http://www.dudak.baka.com/fovcalc.html But be warned. When they list the angle, it's the diagonal angle, not the horizontal angle, so when you compare films with different ratios (like 6x6) the result is misleading. Two lenses may have the same diagonal field of view, but different horizontal fields of view. Horizontal field of view makes much more sense, especially if you're a landscape photographer studying maps. Here's another site that calculates horizontal and vertical fields of view; http://www.mat.uc.pt/~rps/photos/angles.html Paul --
Date: Mon, 07 Oct 2002 From: Manu Schnetzler marsu@earthling.net Reply to: hasselblad@kelvin.net To: hasselblad@kelvin.net Subject: Re: [HUG] 35 vs. 6x6 I think one issue is exactly how you would compare the two formats since one is square and the other rectangular. I see at least three ways: 1) diagonal against diagonal - this seems a bit unnatural since they have different aspect ratios 2) cropped 35mm to square against full frame 6x6 3) full frame 35mm agains cropped 6x6 to the same aspect ratio A little while I set up a little excel spreadsheet to compare the formats since I also started in 35mm and was used to the focal lengths. Here's how it goes: Diagonal (full) Square 24x36 ratio 35mm 6x6 35mm 6x6 35mm 6x6 16.8 30 13.2 30 19.8 30 21.3 38 16.7 38 25.1 38 28.0 50 22.0 50 33.0 50 33.6 60 26.4 60 39.6 60 44.8 80 35.2 80 52.7 80 61.6 110 48.3 110 72.5 110 84.0 150 65.9 150 98.9 150 140.1 250 109.9 250 164.8 250 280.1 500 219.7 500 329.6 500 14 25 14 32 14 21 20 36 20 46 20 30 24 43 24 55 24 36 28 50 28 64 28 42 35 62 35 80 35 53 50 89 50 114 50 76 85 152 85 193 85 129 105 187 105 239 105 159 135 241 135 307 135 205 200 357 200 455 200 303 300 535 300 683 300 455 400 714 400 910 400 607 500 892 500 1138 500 758 600 1071 600 1365 600 910 You can also find the spreadsheet with the calculations (let me know if I screwed up the maths...): http://www.schnetzler.com/Technical/focal.xls My conclusion: forget about equivalents, and learn to "see" your 6x6 lenses coverages. I'm still working on that advice... Manu Bill Pearce wrote: >>Regards field of view, the 80mm on 6x6cm is close to 35mm on 35 format. >>60mm on 6x6 is closer to 28mm on 35 format, 50mm on 6x6 is close to >>24mm on 35 format, 38-40mm on 6x6 is close to 21mm on 35 format. > > >>Godfrey > > > Godfrey, > > Are you sure about this? I have an old chart from the seventies, source long > forgotten, that gives horizontal angles of coverage as thus: > 69 degrees = 40mm in 6x6, 26mm in 35; 57.5 degrees = 50mm in 6x6, 33mm in > 35mm; and 35 degrees 36 minutes = 50mm in 35, and 76 in 6x6. These numbers > seem close to what I perceive in actual use. > > I have noted that angle of coverage is sometimes specified as horizontal, > and sometimes diagonal. Like the size of TV picture tubes, I find the > diagonal measurement of little use. > > I find it interesting that most of us think of lenses coverage in 35mm > equivelents, as we most likely started with 35. Is there anyone that > struggles to compare 35 lenses to 6x6 lenses because they started with a > Rollei or Yashikamat? > > Bill Pearce
From: "Chuck Moulton" chuckmoulton@cox.net To: cameramakers@rosebud.opusis.com Subject: Re: [Cameramakers] field of view question Date: Sun, 22 Dec 2002 Angle of View The "angle of view" is the measure of a given lens' angle of view for a given film format. For example, a 50mm lens can see a 47=B0 view on a 35mm camera, and an 81=B0 view on a 6=D76 camera. Several problems are expressed in terms of angle of view: 1.. What focal length lens on a 4=D75 view camera would be equivalent to a 35mm camera's 50mm lens? (Answer: About 180mm.) 2.. Given that the full moon is half a degree wide as seen from the Earth, how long a lens will I have to use to capture an image of the moon that fills half the width of the frame, using a 35mm camera? (Answer: About 2000mm.) 3.. If I place my 6=D77 medium format camera in the corner of a room, how wide a lens will I need to capture the entire room wall-to-wall, in landscape orientation? (Answer: At least 30mm.) Formula Used The angle of view is calculated as: http://tangentsoft.net/fcalc/help/angle-of-view.gif where F is the length of the frame in the direction for which you want = the angle measurement, and f is the focal length of the lens. Both = values must be in the same unit, such as millimeters. Also try this links: http://www.photo-software.com/dataguide.htm#camera And a Ho Ho Ho back to you, Chuck Moulton - Phoenix, AZ
From: Stefan Patric tootek2@yahoo.com Subject: Re: What focal length.... Newsgroups: rec.photo.equipment.medium-format Date: Mon, 06 Jan 2003 Patrick L. wrote: > on my Mamiya TLR (6x6) would equate to 85mm or 100mm in a 35mm > format? What is the rule on this? Equating the long dimension of the 35mm frame to the horizontal dimension of the 6x6 frame, and not the diagonals, one gets these equivalences: 35mm 6x6 35 55 45 65 50 80 65 100 90 135 100 150 135 180 180 250 -- Stefan Patric tootek2@yahoo.com
Date: Mon, 06 Jan 2003 From: Leonard Evens len@math.northwestern.edu Newsgroups: rec.photo.equipment.medium-format Subject: Re: What focal length.... This calcuation makes perfect sense if you think that the 24 x 36 frame of 35 mm has the ideal aspect ratio and you want to replicate it in everything you do. But that aspect ratio 24:36 = 2:3 = 1:1.5 is really an historical accident. 35 mm film was originally introduced for motion pictures. The frame size was 18 x 24 mm with an aspect ratio of 3:4. That aspect ratio is still commonly used for most TV sets, for computer monitors, and for many digital cameras. When someone thought of using 35 mm film for still photography, it was decided to double the size of the image by using two frames, which resulted in 36 x 24 mm. Before this, the aspect ratio 4:5 as in 4 x 5, 8 x 10, 16 x 20, etc. was in common use. I believe that most medium and large format photographers, who aren't predisposed to judge by 35 mm biases, still prefer the 4:5 aspect ratio. That was certainly true in my case. I started off with a TLR, which of course produced a square image. When I started doing my own printing, I soon gravitated to cropping to 4:5 aspect ratio to fill an 8 x 10 print. I used the square format as a way to shift the image much as one would do by shifting the lens in large format photography. When I later got a 35 mm camera, I treated it as basically a 24 x 30 format with the ability to shift 3 mm either way in the long dimension. There are some that argue that the golden mean which is about 1:62 is the ideal aspect ratio, but I think most photographers would find that the length was much too long in relation to the height (in landscape orientation). -- Leonard Evens len@math.northwestern.edu
From: "Nick Payne" njpayne@pcugDOTorg.au Newsgroups: rec.photo.equipment.medium-format Subject: Re: "New" Medium Format 36x72mm using zeiss biogon lens... Date: Tue, 28 Jan 2003 Make the calculation if you're capable...the field angle for the corner of 35mm format with a 21mm lens is 46 degrees. The fourth power of the cosine of 46 degrees is 0.23. So the corner illumination is 23%. Two stops down would be 25%. Why do you think Zeiss sell a graduated 2-stop ND filter for their G-series Hologon lens? See http://www.schneideroptics.com/info/white_papers/quality_criteria_of_lenses. pdf. It will tell you the same. ...
From: Patrick Bartek [bartek@lvcm.com] Sent: Fri 4/25/2003 To: hasselblad@kelvin.net Subject: Re: [HUG] What 120 or 150 lens should I get? you wrote: > Patrick, > > >However, I find the 60 on 6x6 more equivalent to a 35 on > >35mm than the 50 > > I agree. > > >I never found any great need for the 180 (135) lenses. > > [....] > > >In my case, with Hasselblad, the 60 is my normal. > > See it all depends on your shooting style. I've newer been able to > make the ultra-wide (<28mm for 35mm format) lenses work for me. To > me a 100 is "normal" in 6x6. Thus, I find the 180 highly useful. > Personal preference. The main reason I need focal length equivalents among systems is that in my professional work, it is not uncommon to be shooting the same things with two different formats for two or more different purposes for one or more clients. Like with architectural interiors: the primary shooting is with 4x5; but the client needs 35mm slides, too. Since my 35mm lenses are equivalent to the 4x5 ones, I just replace the 4x5 with the 35 with the proper lens and shoot the slides. Don't have to change the lighting or the point of view. Just raise the tripod head a few inches. Saves a lot of time, produces better slides than duping the 4x5 chromes, and it's cheaper than duping. -- Patrick Bartek NoLife Polymath Group bartek@lvcm.com
From: Patrick Bartek [bartek@lvcm.com] Sent: Thu 4/24/2003 To: hasselblad@kelvin.net Subject: Re: [HUG] What 120 or 150 lens should I get? you wrote: > >Seems a strange selection, though. 40 to 80: 30 degree > > difference; big gap, like a 50 and a 24 on 35mm. 120 to 180: 8 > > degrees; very small, almost nothing spacing like owning both an > > 85 and a 105. 80 to 120: 15 degrees; okay, any less and it would > > be too little. I wonder what Hasselblad's thinking was? > > I believe 50mm on 6x6 is roughly 28mm in 24x36. 80mm is 50, 120 > roughly 70, 180 about 135. A 50 on 6x6 is a little wider (58 degrees to 54, hardly noticeable) than the 35 on 35mm. A 28 on 35 is 65 degrees, 7 degrees wider than the 50. However, I find the 60 on 6x6 more equivalent to a 35 on 35mm than the 50, even though the 60 is 4 or 5 degrees narrower than the 35, and 9 degrees narrower than the 50. Yes, an 80 is almost dead-on equal to a 50 on 35mm using AOV along the long dimension of the 35mm frame. The 120 is more equal to an 80 than a 70: 120 = 25 or 26 degrees; 85 = 24 degrees. (I don't have the exact AOV for an 80. Interpolated (85 at 24, 75 at 27), it's around 25.5 degrees. 180 at 17 degrees, 135 at 15. Close enough for government work. I never found any great need for the 180 (135) lenses. Before upgrading to Hasselblad, I used professionally for about 10 years the Mamiya C-220. During my initial lens choosing, I looked at its 180, but didn't like it, just like the 135 on 35mm. For my shooting situations, it was either too long or not long enough. I opted for the 135 on the Mamiya, which gave the working distances I desired. I shot mostly 3/4 length portraits for corporate reports and magazine covers with it. (When I went to the 'Blad, I had to go with the 150, since their 135 was a macro lens without a focusing mount.) > Your 20 degree rule only works for moderately wide angle lenses. > Take an example (using 24x36 lenses): The angle of view for a 200mm > lens is 12 degrees. A 1200mm lens gives you a 2-degree field of > view. "But that's only 10 degrees difference", I hear you cry. > Thus, the 1200mm lens isn't worth getting (ask any bird > photographer and (s)he will tell you different)... The 20 degree "rule" is not a rule, but only a guideline, and, yes, it only applies to lenses shorter than your personal normal. For lenses longer than "your" normal, it's double the previous focal length or, really, half the angle of view. Same-O. Same-O. Double is easier to figure. In my case, with Hasselblad, the 60 is my normal. The next, optimum shorter lens should be the 40 (49 + 20 = 69 degrees), but I opted for the SWC. At the time, 20 years ago, the 40 was a monster hunk of glass. I was going for a minimal, compact and lightweight system. For the first tele, the guide says double the focal length of your personal normal. In my case, that's would be a 120, but I didn't want or need a macro. My only real choice for a general purpose telephoto was the 150. Like I've said previously, my Lens Selecting Algorithm is ONLY a guide, not law. Also, it doesn't apply to zooms. > When trying to choose between two focal lengths, I find that the > approach which works for me is to rent both lenses for a weekend. > Granted, you spend $50 on the rental, but I'd rather be out that 50 > than spend $3k on a lens I end up not liking. That would be the best way, but most people don't -- I surely didn't, at the time -- have local rental houses in their cities. And out of town places require a credit check or "collateral" for the retail price of the lens before they'll even consider you. And, if you check out, the minimum rental is usually 7 days, of which 2 or 3 is shipping time. Too much expense. Better to befriend someone, who has the lenses you're considering, and invite them on a field trip: "Oh, I haven't used that lens before. Mind if I shoot a few shots with it?" -- Patrick Bartek NoLife Polymath Group bartek@lvcm.com
From: David Littlewood david@nospam.demon.co.uk Newsgroups: rec.photo.equipment.35mm Subject: Re: Questions on angle of view of wide angle lenses Date: Wed, 7 May 2003 wazza cameraspam@cameraspwm.?.invalid writes >Hi >If my fisheye lens has 180 degrees field of view and while my 24mm lens has >a view of 84 degrees. Is the view diagonal or horizontal. I have read the >specs sheets and the 17-28mm fisheye zoom and it has a field of view of >180-90 degrees while the sigma 15-30mm lens has an angle of view of 110-71 >degrees. On a dlsr this with the 1.5 magnifier would be 120-60 (13mm - 36mm) >and 74 - 47 degrees (effectively a 27mm to 50mm) am I on the right track or >a little off. >Or is the field of view for the fisheye going to be more like 20-35mm on a >dslr and the sigma 22.5mm -45mm. >thanks The 84 degree field of view for a 24mm lens on 35mm film is the diagonal view (you can usually trust manufacturers to quote the most impressive figures). The equivalent horizontal and vertical angles are approximately 104 and 81 degrees respectively. Translating this into field of view for a DSLR is not a simple ratio; you need to do a little maths. For 35mm film, the diagonal dimension of the image is (24^2+36^2)^0.5 = 43.3mm (Pythagoras' theorem). Half of this divided by the lens to film distance is equal to the tangent of half the angle of view. Taking the lens to film distance as equal to the focal length of the lens* (24mm), this gives the angle of view as: 2*tan^-1(21.6/24) = 2*tan^-1(0.9) = 2*42 = 84 degrees Now, if you know (or can calculate) the diagonal of the sensor of your DSLR, you can do the same sum. Say it is 22.5 x 15 mm (i.e. smaller than the 35mm film by a factor of 1.6. The diagonal is then 27.06mm and the angle of view becomes: 2*tan^-1(13.53/24) = 2*tan^-1(0.564) = 2*29.4 = 58.8 degrees Note that 58.8 times 1.6 does not equal 84; the tangent of an angle is not linearly related to the size of the angle (except at very small angles). The 35mm lens which gives the same field of view on 35mm film as your 24mm lens does on the smaller sensor of the DSLR is 38.3 mm - which is (allowing for rounding) exactly what you get by multiplying 24 by 1.6 (the tangent relationship does not appear in this calculation). IOW, you can calculate the equivalent focal length for your DSLR by simply multiplying by the sensor size ratio, but the angle of view is not so simply related, and has to be calculated using trigonometry. However, unless you bear in mind exactly what you are doing, this can sometimes lead to errors (for example in DOF calculations). Fisheye lenses are a bit of a law unto themselves, and I am not sure the same formula will always work. You need to know a little more about the lens itself - for example, is it a full-frame fisheye or one giving a reduced circular image on 35mm film? *Depending on the internal design of the lens, this may not be strictly correct, but is close enough for most purposes. -- David Littlewood
From: "T Rittenhouse" gray_wolf@charter.net Newsgroups: rec.photo.equipment.medium-format Subject: Re: 24X36mm equivalent to 6X7 50 mm ? Date: Fri, 8 Aug 2003 Well, comparing the diagonals is fine if you want to know the coverage of the lenses. But if you want to know what the pictures will look like you have to compare similar aspect ratios. The easiest way to do that is to pick the side of the actual negative area, not the film size, that is common to that aspect ratio for both formats. For instance to make an 8x10 (4x5 aspect ratio) from 35mm and 6x7, you will use 24x30mm of the 24x36 35mm negative, and 56x70 of the 56x72 6x7 negative. So, you use the ration of 56:24 which is 2.33. That means to compare the two focal lengths you multiply by 2.3 to get the equivalent 6x7 focal length, and divide by 2.3 to get the equivalent 35mm focal length. If you were comparing to the 2x3 aspect ration of 35mm you would use the long sides 72:36 which would give you a multiplier of 2.0. These comparisons are all a simple matter of ratios. -- Ciao, Graywolf http://pages.prodigy.net/graywolfphoto "Q.G. de Bakker" qnu@tiscali.nl wrote > Victor Bazarov wrote: > > > How do you calculate those, BTW? > > > > I used to think that ratio was determined by the ratio of > > the diagonals. If I use that rule, 50 mm on 6x7 gives about > > 24mm on 24x36 (considering that the frame size of 6x7 is > > about 56x70mm, which gives 90mm diagonal, which is more than > > twice the size of the 24x36mm diagonal). I can be easily > > mistaken, of course. It's very easy with 6x9, the shape of > > the frame is basically the same as 24x36mm (2:3 ratio), and > > 2.5 times greater... > > I did the same (though i took 6x7 to be 56 x 72 mm). > You forgot to add in that the focal length of what in 35 mm photography is > deemed to be the "normal lens" is actually 15% longer (!) than the diagonal. > This must be considered when giving "24x36 mm equivalent" focal lengths.
From: Severi Salminen severi.salminen@NOT_THISsiba.fi Newsgroups: rec.photo.equipment.medium-format Subject: Re: another 645 format question.. Date: Tue, 18 Nov 2003 Victor Bazarov wrote: > "Tom Thackrey" use.signature@nospam.com wrote... >>"Victor Bazarov" v.Abazarov@comAcast.net wrote: >>>"Tom Thackrey" use.signature@nospam.com wrote... >>>> aschuh@cogeco.ca (Adalbert) wrote: >>>> >>>>>my favoutite 35mm format lens is 35 mm focal length. Slightly wide >>>>>angle effect without being obvious. What is the equivelent focal >>>>>length in the 645 format? Thanks. >>>> >>>>65mm is about the same FOV >>> >>>How does one measure the FOV of a lens? Thanks. >> >>I look it up in the Kodak Professional Photoguide. > > > Do _they_ explain how it's measured? Thanks. If you want to know what is the FOV along the diagonal of certain film/lens combo: FOV = 2 * (INVtan(d/2f)) FOV is the field of view (in degrees for example), d = film diagonal (mm) and f = focal length (mm). INVtan is the inverse of tan function, or arcustan - or whatever it is called :) And if you know the FOC, you can calculate f: d/(2xtan(FOV/2)) = f Could someone please double check these... Severi
From: thomandpam@yahoo.com.au (Thom) Newsgroups: rec.photo.equipment.medium-format Subject: Re: Why does 80mm seem like a wide angle in 6x6? Date: Fri, 09 Jul 2004 Mxsmanic mxsmanic@hotmail.com wrote: >Is this just an illusion of some sort or what? Technically they >_should_ be similar, but they just don't seem to look that way when I >actually sight through the viewfinder. I'm always surprised by the >angle covered by the 80mm. > >I know that my 80mm lens is supposedly a "standard" focal length for >6x6, but why does it seem like a wide angle when I compose with it? It >seems wider than a 50mm lens in 35mm, for some reason. Often I realize >that a shot I might have composed with a 50mm in 35mm seems to demand a >150mm in 6x6. 50mm is not standard for a 24x36mm format. If you use the whole image standard is 43mm, if you use the "ideal" portion it 38mm which means for an 8x10 inch print 50mm is actually 1.32 power or equal to a 94mm on a 56x44mm image (ideal part of a 6x6. THOM
From: thomandpam@yahoo.com.au (Thom) Newsgroups: rec.photo.equipment.medium-format Subject: Re: Why does 80mm seem like a wide angle in 6x6? Date: Thu, 08 Jul 2004 Mxsmanic mxsmanic@hotmail.com wrote: >I know that my 80mm lens is supposedly a "standard" focal length for >6x6, but why does it seem like a wide angle when I compose with it? It >seems wider than a 50mm lens in 35mm, for some reason. Often I realize >that a shot I might have composed with a 50mm in 35mm seems to demand a >150mm in 6x6. 80mm isn't normal for 120 square negs. The so called "Ideal portion" of the negative is 56x44.8mm which means an 80mm has a magnifcation factor of 1,11X. 50mm on a 35mm format has a magnification factor of 1.316X. That's why it looks "wide angle to you. 50mm is not a "normal" lens for the 35mm format. THOM ...
From: Bob Salomon bob_salomon@mindspring.com Newsgroups: rec.photo.equipment.large-format Subject: Re: Figuring out coverage for non square formats? Date: Sat, 22 May 2004 Nick Zentena zentena@hophead.dyndns.org wrote: > I understand the formula > > X^2 + y^2 =z^2 > > With X one side of the negative. Y the other. Z the diagonal and > therefore the coverage needed. > > Now I can understand this working just fine for formats like 4x5 > or 8x10. Both are fairly square. For a format like 6x12 won't it give a > number that is too small? The image coming out of the lens is round. So the > coverage should be a circle big enough to fit the long side of the > negative. No? > > So something like > > X^2+X^2 = z^2 > > With X being the long side. > > Nick Linhof Technorama 612 with 56 x 120mm image size. 56x56 = 3136, 120 x 120 = 14,400. 3136 + 14,400 = 17,536 Square root of 17,536 = 132.42 Normal lens on Linhof Technorama 612 is the 135mm 5.6. so X^2 + y^2 =z^2 works fine. Linhof Technorama 617 SIII with 56 x 172mm image area is 56x56 = 3136 + 172x172 = 29,584 giving a square root of 180.9. The normal lens for the 617 SIII is the 180mm 5.6. The formula seems to work just fine. Using your formula results in what Linhof sells as a telephoto for the 617 SIII. Not the normal lens. -- To reply no_ HPMarketing Corp
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