| General use | Close-up Photography | Miscellaneous |
 |
Click on square if you need the calculator ready.. |
 |
where:
BE = Bellows extension FL = Focal length BEF = Bellows exposure factor . |
| Example
1 |
Bellows extension, measured from front
panel to filmplane = 450mm.Focal length used 300 mm.
We want to know the exposure factor to multiply with the shutter speed measured.... Calculation|
2025:900= 2.25 .This is the exposure
factor to use.
|
 |
where:
BE = Bellows extension FL = Focal length BEF = Bellows exposure factor . |
| Example
2 |
Bellows extension, measured from front
panel to filmplane = 450mm
Focal length used 300 mm. We want to know the exposure compensation in stops. Calculation|
450:300 = 1.5 stops |
| When you are in a
hurry...
hence, you should not... Take your time........ |
Rough method:
For every 25% Focal length, ( More Bellows extension) increase Exp. time by 1/2 stop or.. multiply Exposure time by factor 1.5 |
Note:If you know the degree of magnification
in relation to
Circle of confusion: 0.092 for 4x5 inch films
Near limit measured from camera lens: = Far limit measured from camera lens: = Hyperfocal distance (near limit depth of field when lens is set at infinity)
Example picture not supplied yet.. I've been over this quite a few times - I think I've got it right,
but if you disagree I'd like to hear from you.
With a very high quality sighting compass, you
would be very surprised at the accuracy that can be achieved
with this system of measuring distance.
d = (Tan (90 - (A -B))) x Ref
1.25
+ 1/3
1.5
+ 2/3
2
+ 1
2.5
+ 1 1/3
3
+ 1 2/3
4
+ 2
5
+ 2 1/3
6
+ 2 2/3
8
+ 3
10
+ 3 1/3
12
+ 3 1/2
40
+ 5 1/3
100
+ 6 2/3
1000
+10
1/4 : 1
1.5
1/2
1/2 : 1
2
1
3/4 : 1
3
1 1/2
1 : 1
4
2
2 : 1
9
3
3 : 1
16
4
4 : 1
25
4 1/2
5 : 1
36
5
6 : 1
50
5 1/2
7 : 1
64
6
8 : 1
80
6 1/4
9 : 1
100
6 1/2
10 : 1
120
7
1/ 100,000
1 Stop more
none
20% increase
1/ 10,000
1/2 Stop more
none
15% increase
1/ 1,000
No correction
No correction
Normal
1 /10
No correction
No correction
Normal
1
1 Stop less
2 seconds
10% decrease
10
2 Stops less
50 seconds
20% decrease
100
3 Stops less
1200 seconds
30% decrease
Schematic
view of Hyperfocal Distance principle
where:
f = focal length
F-N = f-number
C = acceptable circle of confusion
Example
1Focal length to be used= 65mm
F-Number used 32.
Circle of confusion:
0.124 for5x7 inch
Circle of confusion:
0.182 for 8x10 inch
Calculation |
square 65 mm = 4225mm
32 x 0.092 = 2.944 x 4225 results in 124384
Hyperfocal distance = 124 cm
Images will be sharp from 62cm till infinity...
where:
F = focal length of lens
f = f - number of relative
aperture
H = Hyperfocal distance
u = distance for which camera is focused.
d = diameter of circle of confusion
Circle of confusion: 4x5 & larger:
F/1720 for critical use
F/1000 for liberal use
Circle of confusion: 4x5 & larger:
F/1720 for critical use
F/1000 for liberal use
Circle of confusion: 4x5 & larger:
F/1720 for critical use
F/1000 for liberal use
Format
(inches)Format (mm) Coversion to other formats.
Vert. Horiz. Vert. Horiz. Focal length & Diagonal (mm)
Circle of Confusion at film plane (mm) f-
stop Hyper-
focal distance
(meters)
Equiv. f-stop
24 36 43.27
0.024724 8 9.46 8
60 70 92.20
0.052683 8 20.17 17
4 5 102 127 162.64
0.092937 8 35.58 30
5 7 127 177.8 218.50
0.124857 8 47.80 40.5
8 10 203 254 325.28
0.185874 8 71.15 60
We have made a picture focused at infinity in an open landscape.
Camera used: 4x5 inch, Lens mounted 65mm wide-angle, height of mounted
camera 1.75
Angle of view: 105 o
The picture came out very well, but I wonder as this was made on the
puszta in Hungary...
You could see tremendous space all around.
What is the distance of the horizon showing on the print. (whole negative
printed)
We would like to know if there's somebody in Cyberspace who can give us
the formula to
give us the answer on this question.
Mail your solution to: info@johndesq.com
I have constructed the
following files, to state what the situation was. Finally
I will give the solution the way I think it is..
Known factors :
Unkown factor :
Step # 1
Click on square if you need the calculator now..
Step # 2
Final note by author:
In practice due through dispersion probably about
10 Km (Kilometers)
Finding a fixed point at the horizon, we could sight 2 bearings.
However to get accurate results we must choose a base reference of 1000m
The first bearing would have been 210 degrees. Now
walk or drive :-) exactly 1000 meters perpendicular in relation to the horizon
(As perpendicular as possible).For shorter distances up to 500 meter a reference of 10meter will do fine.
The second bearing will read 222 degrees.
Now calculate 90 - (222-210). You should get 78. Now find the
Tangent of 78. This will give you 4.704 Multiply this value
by the perpendicular reference (Ref) distance (1000 meters).
This would be (4.704 x 1000)=4704. The distance (d) to the horizon
is 4.7 km.
d = Distance (to be calculated)
Tan = Tangent value of the resultant angle
A = Greater value of the two measured bearing angles
B = Lower value of the two measured bearing angles
Ref = Measured reference distance
Site Design by
JohnD (+31(0)343 513927)
Content is Copyright
© John D de Vries 1997 - 2001 All rights reserved.
email
the Webmaster
Page modified: May 21st 2001
http://www.johndesq.com/wfpa