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F/stop question

  • 29-07-2009 6:04am
    #1
    TomCo
    Registered Users Posts: 2,528 ✭✭✭


    I've been up all night trying to figure this out.
    I'm writing a piece of code to normalise exposure times from images with different f-stops.

    If I have an image with an exposure of 1/500sec and f/3.6, and an image with 1/500sec and f/6.3. I want to know what the second image's exposure time would be if it had an f/stop of f3.6. Still with me?

    Right, so here's the formula.
    Shorter time - 
new exposure time  = original time / 2^(f/stop difference)

Longer time -
new exposure time  = original time * 2^(f/stop difference)

    The problem here is, I dont know what the f/stop difference is.
    How do I know there are exactly 1 and 2/3 stops between f/3.6 and f/6.3 (I think its 5/3 anyway).

    How do I know there are 2/3 stops between f/1.4 and f/1.8.

    How do I calculate that f-stop difference?
    Something to do with logarithms?

    Thanks.


Comments

  • #2
    quilmore
    Registered Users, Registered Users 2 Posts: 1,037 ✭✭✭quilmore


    here you go in 1/3rds:
    http://www.riversidecardiology.com/fstop/f-stop.htm#4

    mind you, 1.8 it's not a "pure" F-stop
    1.4 or 2.0 are


  • #3
    chezzer
    Closed Accounts Posts: 184 ✭✭chezzer


    its in increments of sqrt(2) isn't it ?


  • #4
    charybdis
    Registered Users, Registered Users 2 Posts: 802 ✭✭✭charybdis


    f-number = sqrt(2)^(stops)

    stops = log[sqrt(2)](f-number)

    difference in stops = log[sqrt(2)](6.3) - log[sqrt(2)](3.6) = ~1.6


  • #5
    eas
    Registered Users, Registered Users 2 Posts: 1,435 ✭✭✭eas


    TomCo wrote: »
    Still with me?


    err.....no. You lost me at "normalise exposure times" :)


  • #6
    TomCo
    Registered Users Posts: 2,528 ✭✭✭TomCo


    charybdis wrote: »
    f-number = sqrt(2)^(stops)

    stops = log[sqrt(2)](f-number)

    difference in stops = log[sqrt(2)](6.3) - log[sqrt(2)](3.6) = ~1.6

    Nice, this is exactly what I was looking for. Actually discovered the formula this morning but its nice to get verification.

    I'm using,

    log(f1) - log(f2)
    log(sqrt(2))

    Same thing though, I think.


    Anyone interested in a free and incredibly easy to use, command line based composite HDR image generator?
    Might stick it up when Im finished (if there's interest), image alignment is proving slightly tricky at the moment.


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  • #7
    charybdis
    Registered Users, Registered Users 2 Posts: 802 ✭✭✭charybdis


    TomCo wrote: »
    Nice, this is exactly what I was looking for. Actually discovered the formula this morning but its nice to get verification.

    I'm using,

    log(f1) - log(f2)
    log(sqrt(2))

    Same thing though, I think.

    Yeah, just expressed differently using logarithmic identities, which you'd want to do to solve it computationally.
    TomCo wrote: »
    Anyone interested in a free and incredibly easy to use, command line based composite HDR image generator?
    Might stick it up when Im finished (if there's interest), image alignment is proving slightly tricky at the moment.

    You can already use Qtpfsgui via command line (despite the name), but by all means feel free to add your own software.


  • #8
    chezzer
    Closed Accounts Posts: 184 ✭✭chezzer


    Now im confused ... I thought it was sqrt(2) ... thats why 2 stops was always exactly double ....

    and sqrt(2) is ~1.4 .. not 1.6 ...

    anyway ... I'm sure I'm wrong...


  • #9
    charybdis
    Registered Users, Registered Users 2 Posts: 802 ✭✭✭charybdis


    chezzer wrote: »
    Now im confused ... I thought it was sqrt(2) ... thats why 2 stops was always exactly double ....

    and sqrt(2) is ~1.4 .. not 1.6 ...

    anyway ... I'm sure I'm wrong...

    f-numbers are powers of sqrt(2), i.e.:

    sqrt(2)^0 = 1
    sqrt(2)^1 = ~1.4
    sqrt(2)^2 = 2
    sqrt(2)^3 = ~2.8
    sqrt(2)^4 = 4
    sqrt(2)^5 = ~5.6
    sqrt(2)^6 = 8
    sqrt(2)^7 = ~11
    sqrt(2)^8 = 16

    1.6 is the difference, in stops, between the two f-numbers posted in the original question.


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