AG-Schnozz asked:
"If these are the "standard" multiplication factors regarding
general "magnification" of 35mm cameras, then why is 3x 135mm
and not 150mm?"
Magnification is not linear.
3x is not half way between 2x and 4x.
It's not at all like asking "what point on a meter stick is mid-way between
100cms and 200cms?"
You initially asked what focal length is "half way between". My
interpretation of the question was "what focal length produces a
magnification that is the median magnification between 2x and 4x.
Here's a visual metaphor. Imagine you're at a camera show. You've lined up
exactly 24 OM-2n bodies, side-by-side. Mount a 50mm lens on an OM-4 and set
the camera on a tripod , at a distance so you can just barely see all 24
bodies.
When you switch to a 100mm lens, you'd see exactly 12 OM-2n bodies across
the field of view, right? When you double your magnification, you reduce
the field of view by a _factor_ of two.
Would you agree that if we swapped in a 200mm (4x) lens, we'd see exactly 6
bodies?
The 'half-way' question is "What focal length will show me exactly 9 OM-2n
bodies across the field of view?" [9 is midway between 6 and 12, right?]
By my math,
1x = 50
2x = 50 x2
4x = 50 x 2 x 2
8x = 50 x 2 x 2 x 2 = 400
16x = 50 x 2 x 2 x 2 x 2 = 800
Look closely at the sequence.
The sequence is 50 x 2 ^n
(where n is the power)
50 = 50 x (2 ^0)
(two to the zero power = one.) 50 x 1 = 50
100 = twice the focal length is 50 x (2^1)
(two to the power of one is just two.)
50 x 2 = 100
200 = twice the focal length again is 50 x (2^2)
(two to the power of two is four.)
50 x 4 = 200
400 = twice the focal length again is 50 x (2^3)
(two to the power of three is eight.)
50 x 8 = 400
800 = twice the focal length again is 50 x (2^4)
(two to the power of four is sixteen.)
50 x 16 = 800
The sequence is about _doubling_. It's about multiplying by two. It takes
a multiplication of 2 to get to the next step. To get half way there, then
the other half of the way, you use "the square root of two" first, then "the
square root of two" to get the rest of the way.
Half way between 2x and 4x is = 50 x 2 x (square root of 2)
Half way = 50 x 2 x (1.4142136)
Half way = 50 x (2.8)
Half way = 141 mm
Still don't believe me? Let's check it. If we start at 141 mm, and
multiply it by "the square root of two" again, we should get to 200, right?
My method to get 'half way' was to multiply by 1.4
Let's see if that works:
141 x (1.4142136) = 200mm Check
Or think about it this way. You already know it's not a linear
relationship. The "focal length difference" between 1x and 2x is 50mm. So
how come the "focal length difference" between 2x to 4x isn't also 50mm?
How come "focal length difference" between 4x and 8x isn't also 50mm?
Because the "focal length differences" are not linear.
Lama
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