Christiane Roh wrote:
>
> If one were doing the math (anyone ?)
Yes?
OK, for simplicity we'll make the following asumptions:
1) film records objects in 2D, so all we need to consider is the
displacement as seen on the image, not absolute differences. In other
words we will be considering the angle that a certain distance subtends
with the lens.
2) we'll assume ideal lens (no aberrations), film (no grain), and
development (no adjacency effects).
3) we'll assume that the distances are measured from the nodal point of
the lens, and that both nodes are coincident.
4) we'll also assume that complaints that these assumptions somehow
invalidate the mathematics, or unduelt influence the result will be
proved rigourously ;-)
5) we'll consider only those rays of ligh which remain undeviated by the
lens (ignoring the effect of the nodal space)
Lets assume that the first object (A) is 1000 metres away, the second
(B) 1100 metres, and the third (C) 1200 metres away A straight line
could be drawn connecting the tops of these objects. Also assume that
these objects are 2 metres in height and the top of object B is aligned
directly along the axis of the lens. Furthermore, the angle subtended
by the chord connecting the tops of objects A and B as measured from the
lens' nodal point is 0.01 radian. The objects are appropximatly
cylindrical in shape and are parallel.
(If that doesn't make sense, think of three people standing on a field
in a line, such that you're 1 km from the first one, 1100m form the
second, etc.)
Film is assumed to be 35mm (not that it matters) and lens focal lengths
are 28mm and 500mm
Call the lens nodal point N
The distance from the nodal point to the film plane (along the lens
axix) is approx:
28mm 500mm
1/f = 1/u + 1/v 1/f = 1/u + 1/v
v = 1/(1/f - 1/u) v = 1/(1/f - 1/u)
= 1/(1/28 - 1/1 100 000) = 1/(1/500 - 1/1 100 000)
= 28.0007127 mm = 500.2273761 mm
The distance between the objects A and B:
AB^2 = 1000^2 + 1100^2 - 2.1000 x 1100 x Cos(0.01) (cosine rule)
= 10110.00
AB = 100.55 m
The angle at object A between the camera and B:
Sin A = (Sin(0.01) x 1100) / 100.55 (sin rule)
= 0.109396
A = 3.0319768
Therefore angle at B between camera and C is:
B = 0.01 + 3.0319768 (NBC supplimentary to ABN)
= 3.0419768
The angle at object C between the camera and B:
Sin C = (1100 x Sin 3.0419768) / 1200 (Sin rule)
= 0.091163600
C = 0.09129
Thus the angle at the lens between onjects B and C is:
BNC = pi - (3.0419768 + 0.09129)
= 0.00833
And the distance BC is:
BC^2 = 1100^2 + 1200^2 - 2 x 1100 x 1200 x Cos 0.0833
= 19154.04
BC = 138.40 m
Now, knowing certain facts about nodal points, we know that the angle
subtended by a chord between 2 points on the object plane will equal the
angle subtended by those same 2 points on the image plane.
Note that this does not hold for fisheye lenses, or other lenses where
barrel or pincushion distortion are significant.
the distance between pts A and B on the film plane are:
28mm 500mm
AB' = 28.0007127 Sin 0.01 AB' = 500.2273761 sin 0.01
= 0.280002 mm = 5.002190 mm
and between B and C are:
BC' = 28.0007127 Sin 0.00833 BC' = 500.2273761 sin 0.00833
= 0.233243 mm = 4.166846 mm
This is actually an indication of perspective. Remember that the
distance AB is actually 100.55m and the distance BC is actually
138.40m. But on the film, the distance BC' is actually shorter than
AB'. This is because A is closer to us than C.
Look at the ratio between AB' values, BC' values for both lenses:
28mm / 500mm
AB'/AB'
= 0.280002 / 5.002190
= 0.055976
BC'/BC'
= 0.233243 / 4.166846
= 0.055976
Aha! It appears that the ratio of the sizes is the same (0.055976:1),
even though the actual distances and angles subtended are different!
Now this should be proof enough, but I know that many of you are far
more stubborn than I, and will want to see more proof.
So I shall continue.
The next step will be to look at the relative heights of the objects.
We need to know the angle subtended by the objectds A, B, and C
for A: sin a = 2/1000
= 0.002
a = 0.00200000133
for B: sin b = 2/1100
= 0.00181818181818
b = 0.00181818282
for C: sin c = 2/1200
= 0.001666666666
c = 0.00166666744
The height of B is:
28mm 500mm
B = 28.0007127 x 0.00181818181818 B = 500.2273761 x 0.00181818181818
= 0.0509104 mm = 0.909504
(note the ratio is once again 0.055976:1)
The height of A is:
A = sqrt(AB'^2 + 28.0007127^2) x 0.002
= 0.0560042 A = sqrt(AB'^2 + 500.2273761^2) x
0.002
= 1.0005048
(And the ratio is again 0.055976:1)
And finally, the height of C is:
C = sqrt(BC'^2 + 28.0007127^2) x 0.001666666666
= 0.0466695 C = sqrt(BC'^2 + 500.2273761^2) x
0.001666666666
= .833741
(And you would not be surprised to learn that the ratio is again
0.055976:1)
So what have we shown?
1) that the *only* difference between the image produced from such a
system with lenses of 2 different focal lengths is one of scale.
2) that the scale remains constant whether the objects are the same
height, but different distances, or indeed different distances from the
same point.
3) that greater displacement in the object space may show up as smaller
distances on the image but that the ratio again remains the same.
Steve
