RE: [Leica] OT: Yet another f-stop post

["B. D. Colen" <bdcolen@earthlink.net>]

Thanks to one and all for these "wonderful" posts - they've served once
again to remind me why it took me two tries to get through math during my
junior year in high school...Oh well, ask a simple question, get a bunch of
equations..:-)

B. D.

- -----Original Message-----
From: owner-leica-users@mejac.palo-alto.ca.us
[mailto:owner-leica-users@mejac.palo-alto.ca.us]On Behalf Of Richard
Edwards
Sent: Tuesday, March 07, 2000 5:23 PM
To: 'leica-users@mejac.palo-alto.ca.us'
Subject: [Leica] OT: Yet another f-stop post


>
> Marc - I'm curious...How does one calculate these "stop fractions?"
>
> B. D.
>

Let r represent the change in stops; r can be positive, negative, or
fractional.

Then a change by r stops multiplies the initial ratio number by a factor of
2^(r/2) --- in English, this
is 2 raised to the power (0.5)r. (By 'ratio numbers' I mean the figures on
the lens ring, which are not
literally f-stops but ratios of opening divided by focal length.)

So:
A change of 1 stop multiplies the ratio number by 2^(1/2) = 1.41421356237...
A change of 1/3 stop multiplies the ratio number by 2^(1/6) =
1.12246204831...
A change of -1/2 stop multiplies the ratio number by 2^(-1/4) =
0.840896415254...
A change of 3.5 stops multiplies the ratio number by 2^(1.75) =
3.36358566101...

Moving up from f/1.0 in 1/6-stop increments gives the table:

1.0
1.06
1.12   <- 1/3 stop
1.19
1.26   <- 2/3 stop
1.33
1.41   <- 1 stop

You can continue this exciting [SARCASM] process as long as it pleases you.

Cheers,

- -Al