or
Valuable formulae you will have to master..
when working with the Large Format Camera.

General Use..

General use Close-up Photography Miscellaneous

 
Click on square if you need the calculator ready..

# 1 Bellows extension/Exposure factor
This is the most important formula to get
properly exposed negatives or chromes.
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.
i.e: measured time is 1 second, multiply by the factor: 1 x 2.25
Result corrected exposure time is 2.25 seconds

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 
square  1.5 = 2.25 will give factor to multiply by exposure time
i.e exposure time 1 second, corrected exposure time is 2.25 seconds.

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
 

 

Conversion of filter factors to increase in Exposure stops
FILTER FACTOR
++ STOPS
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
FILTER FACTOR
+STOPS

Note: When using long exposures, check the reciprocity effect
of the film being used..
 
 
Magnification
Amount of
Exposure
increase
Number of
stops needed
for required
Exposure increase
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

Note:If you know the degree of magnification in relation to

the subject  ..Use this table
Reciprocity corrections for most B&W films
 
Indicated Exposure times
(in seconds)
Aperture Correction
Time Correction
Development
Correction
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 

Hyperfocal distance (1)
 
 where: 
 f    = focal length 
F-N = f-number 
C     = acceptable circle of confusion
Example 
1		 Focal length to be used= 65mm 
F-Number used 32.  

Circle of confusion: 0.092 for 4x5 inch films  
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...


Hyperfocal distance (2)

 

 
 
   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
H
 
 Circle of confusion: 4x5 & larger: 
F/1720  for critical use 
F/1000  for liberal use 

Near limit measured from camera lens: =
H
 
 Circle of confusion: 4x5 & larger: 
F/1720  for critical use 
F/1000  for liberal use 

Far limit measured from camera lens: =
H 
 
 
 Circle of confusion: 4x5 & larger: 
F/1720  for critical use 
F/1000  for liberal use 

Hyperfocal distance (near limit depth of field when lens is set at infinity)


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
243643.27 0.02472489.468
607092.20 0.052683820.1717
45102127162.64 0.092937835.5830
57127177.8218.50 0.124857847.8040.5
810203254325.28 0.185874871.1560

*

Question
Answer?
This picture was made in an open landscape. (International Falconry meeting) 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 actual width of the horizon showing on the print. (whole negative printed)

Highslide JS

The author changing film during a report from the puszta in Hungary.
Image made by a local newspaper photographer with a Technika.
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: Johndesq
 

Example

 
"Just for my peace of mind"
I have constructed the following files, to explain the situation .  I will give the solution the way I think it is.. 

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. 

My Solution :
The approach
 
Known factors :
  • Height of camera, axial placement of lens
  • h = 1.75 cm or 1.75 x 10-3 Km 
  • Angle of view 105o 
Unkown factor :
  • Distance to horizon d 
  • Width of horizon D [ Our question ] 
 
 
Step # 1
Middle radius of earth
 
 
 
 
 
Click on square if you need the calculator now..
Step # 2
 
  Final note by author:  
  Sine/tangent without dispersion
In practice due through dispersion probably about
10 Km (Kilometers) 

If we had carried a compass, we could have known d (Distance to horizon)

 


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.

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.

Measuring distance with a compass

 

d = (Tan (90 - (A -B))) x Ref

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

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