Question 970283: The exposure volume EV of a photograph can be calculated from the time, t, in seconds that the shutter was open and the value of the f-stop,f, the picture was taken at using by using the logarithmic equation:
EV=logbase2(f^2/t)
use the properties of logarithms to expand the formula for EV:
For this part I did: 2 Logbase2 f - logbase2 t
b. Keith takes a picture of a waterfall with a shutter speed of 1/60 second at f/5.657. he likes the exposure of his picture, but want to leave the shutter open four times as long to better capture the motion of the water. What f-stop should he use on his next picture.
This is where I am having some trouble. I started by doing 4(1/60)=1/15 and that is my t. I know I am solving for f, but is 5.657 my EV?
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! The exposure volume EV of a photograph can be calculated from the time, t, in seconds that the shutter was open and the value of the f-stop,f, the picture was taken at using by using the logarithmic equation:
EV=logbase2(f^2/t)
use the properties of logarithms to expand the formula for EV:
The correct equation is 2*log2(f)-log2(t)
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b. Keith takes a picture of a waterfall with a shutter speed of 1/60 second at f/5.657. he likes the exposure of his picture, but want to leave the shutter open four times as long to better capture the motion of the water. What f-stop should he use on his next picture.
This is where I am having some trouble. I started by doing 4(1/60)=1/15 and that is my t. I know I am solving for f, but is 5.657 my EV?
No, your EV is what it was on the original photo.
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Find the EV on the original picture::
EV = 2log2(5.657) - log2(1/60) = 5 - (-5.91) = 10.91
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Solve for "f"::
10.91 = 2*log(f) - log2(1/15)
10.91 = 2*log(f) - 3.91
log2(f) = (1/2)14.82
log2(f) = 7.41
f = 2^7.41 = 170.07
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Cheers,
Stan H.
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