Help with an exposure problem.
Hi I have an exposure problem that I think I have figured out but would like to run past everyone.
I am using 5x4 and I need to photograph at F/16. With my bellows extension and Infrared filter I need to factor in 10 extra stops in total. My subject is delicate and prone to imperceptible movements of which I cannot control so I need to try and capture the image as fast as possible to eliminate movement.
My flash heads will fire at F/128 giving me 6 extra stops (from F/16) in one flash. If I flash once, then reduce the power on the flash heads to F/64 giving me 4 extra stops (from F/16) and flash again will this give me the 10 extra stops I need in 2 flash pops?
thank you for your advice
Jon
Re: Help with an exposure problem.
Each stop means that you double the quantity of light. If with one pop you are at 6 stops, with two, you are doubling the light, you are at 7. Four will leave you at 8. Eight at 9. Sixteen at 10.
Best,
Pau
Re: Help with an exposure problem.
Sorry, posted this before seeing the first reply, but here goes.
It's easiest to think of it in terms of exposure times. Suppose the exposure time for F/16 is 1 second. You need 10 more stops, which is 2^10 = 1024x more exposure, or 1024 seconds. Your F/128 setting is giving you 6 more stops, or 2^6 = 64s total. To get to 1024s exposure, you 1024/64 = 16 exposures at your F/128 setting to get to 1024s equivalent exposure.
HTH
Ed Freniere
Re: Help with an exposure problem.
Quote:
Originally Posted by
pau3
Each stop means that you double the quantity of light. If with one pop you are at 6 stops, with two, you are doubling the light, you are at 7. Four will leave you at 8. Eight at 9. Sixteen at 10.
Best,
Pau
Thank you for your replies. I understand that each flash pop doubles the amount of light but if with one flash pop I am 6 stops over the intended exposure (and I need to be 10 stops over) then with 2 flash pops will I not be 12 stops over? I cant see how with one pop I will gain 6 stops but with 2 I will only gain one more stop making 7?
Sorry if its obvious, I just need to be sure.
2 Attachment(s)
Re: Help with an exposure problem.
Quote:
Originally Posted by
bakerbang
Thank you for your replies. I understand that each flash pop doubles the amount of light but if with one flash pop I am 6 stops over the intended exposure (and I need to be 10 stops over) then with 2 flash pops will I not be 12 stops over? I cant see how with one pop I will gain 6 stops but with 2 I will only gain one more stop making 7?
Read Ed's post again, or consider the exposure as the number of flashes to give ten greater f-stops (ignoring reciprocity failure of using multiple pops instead on one huge flash from an impossible source.)
Begin at 1, then add 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024
That gives you ten stops.
A smaller number, real life example. A flash gun that adds one stop.
Attachment 146463
These guys knew what they were doing. O. Winston Link with lights and assistant.
Attachment 146464
Re: Help with an exposure problem.
Your second 6-stop-over exposure adds 64s to the exposure in my example above, giving a total of 128s. You need to add another 14 of them to get to 1024s.
Re: Help with an exposure problem.
Ok I got it! thank you everyone for your help.
I would need 1024 pops of the flash to achieve F/16 (if I need an extra 10 stops) but I already have 6 extra stops with the power of my lights so If originally I needed
1024 then going backwards 512 256 128 64 32 16. I need 16 pops of the flash to achieve 10 extra stops if my lights are giving me an extra 6 stops.
Thank you everyone I appreciate your time
JB
Re: Help with an exposure problem.
Quote:
Originally Posted by bakerbang
My flash heads will fire at F/128 giving me 6 extra stops (from F/16) in one flash.
Three stops, not six.
Re: Help with an exposure problem.
Jac,
He's right, six extra stops.
Re: Help with an exposure problem.
Quote:
Originally Posted by
photog_ed
Jac,
He's right, six extra stops.
Thanks, Ed. I must be mucking up powers of two.