AngelicBeaver
Leading Member
I know there's a full stop between F/2.0 and F/1.4, and between F/1.4 and F/1.0, but what's 1 stop brighter than F/1.8? Is it F/1.2?
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F/1.2 one stop faster than F/1.8?
- Thread starter AngelicBeaver
- Start date
T
Tony Collins
Guest
Always divide or multiply by 1.41 (Then round to 1 or 0 decimal places)
So one stop brighter than 1.8 would be about 1.3 (1.27)
So one stop brighter than 1.8 would be about 1.3 (1.27)
tkbslc
Forum Pro
there is a factor of 1.4 in between stops, so just do the math. 1.8/1.4 = 1.28, so roughly f1.3 is your answer.
Precise math is as follows:
Area of a circle is r^2*pi, r being radius. One stop equals doubling the amount of light, ie doubling the area of the circle. The radius we know in this case is the f/1.8
Thus equation is:
2 = ((f/x)^2 * pi) / ((f/1.8)^2 * pi)
= ((f^2)/(x^2)) / ((f^2)/(1.8^2))
= 1.8^2 / x^2
We end up with
1.8^2 / x^2 = 2
x^2 = (1.8^2)/2
x = sqrt((1.8^2)/2) = 1.272792
Thus it's about f/1.3. F-numbers in general are rounded anyway: Going from f/2.8 to f/2 is actually only +96% light as opposed to +100% doubling the light.
Area of a circle is r^2*pi, r being radius. One stop equals doubling the amount of light, ie doubling the area of the circle. The radius we know in this case is the f/1.8
Thus equation is:
2 = ((f/x)^2 * pi) / ((f/1.8)^2 * pi)
= ((f^2)/(x^2)) / ((f^2)/(1.8^2))
= 1.8^2 / x^2
We end up with
1.8^2 / x^2 = 2
x^2 = (1.8^2)/2
x = sqrt((1.8^2)/2) = 1.272792
Thus it's about f/1.3. F-numbers in general are rounded anyway: Going from f/2.8 to f/2 is actually only +96% light as opposed to +100% doubling the light.
Solution
f/1.8 is actually f/1.782…, only rounded for convenience.
log(1.782, sqrt(2)) = 1.667
1.667 – 1 = 0.667
sqrt(2)^0.667 = 1.26
1.26 ≈ 1.3
So it's f/1.3.
f/1.2 (more precisely f/1.189) is 1/6 stop faster.
tkbslc
Forum Pro
Be careful with your implied precision. Real-world lenses certainly are not designed so strictly.
Dutch Newchurch
Veteran Member
Yes.
(For all practical purposes, near enough.)
(For all practical purposes, near enough.)
Of course not. I'm discussing idealised cases.
T
Tommi K1
Guest
f-number - Wikipedia
Great explanation!
Just to take it one tiny step further:
x = 1.8 / sqrt(2) = 1.272792
The 1.4 or 1.41 other posters mentioned is just sqrt(2) which equals roughly to 1,4142135623730950488016887242097
tkbslc
Forum Pro
Those are just half stops. f/1.8 is on the third stop scale. Nothing more or less proper about them, and they are often rounded quite a bit, too.
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Astrotripper
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Funny thing about f/1.2 is that it can be both 1/2 of a stop faster than f/1.4, or 1/3 of a stop faster, depending on which scale you use. That's because f-numbers are rounded up for convenience, and it gets a bit too crowded between f/1 and f/1.4.
So you can have two f/1.2 lenses and they can actually be different in terms of actual f-number
So you can have two f/1.2 lenses and they can actually be different in terms of actual f-number
So how can there be two 1/2 stops = 1 stop between f/1.2 and f1/1.7, and at the same time three 1/3 stops = 1 stop between f/1.2 and f/1.8. That can only be true if f/1.7 = f/1.8.
Here you go.
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Mark
Astrotripper
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I'm sure you've heard of rounding numbers.
You can say that for each and every f-stop.Funny thing about f/1.2 is that it can be both 1/2 of a stop faster than f/1.4, or 1/3 of a stop faster, depending on which scale you use. That's because f-numbers are rounded up for convenience, and it gets a bit too crowded between f/1 and f/1.4.
So you can have two f/1.2 lenses and they can actually be different in terms of actual f-number
tkbslc
Forum Pro
1/2 stop up from f1.0 is f1.19. 2/3 stop up from f1.0 is f1.26. Many camera makers decided it was easier to round both to f1.2 and not bother with an f1.3 setting in camera. And besides, the lens design won't hit the exact setting anyway.So how can there be two 1/2 stops = 1 stop between f/1.2 and f1/1.7, and at the same time three 1/3 stops = 1 stop between f/1.2 and f/1.8. That can only be true if f/1.7 = f/1.8.
Here you go.
As a particularly annoying example, what if you design a 52.1mm f1.249 and call it a 50mm f1.2? Isn't f1.249 closer to f1.26, than f1.19? But it rounds mathmatically to f1.2. So is it f1.2 and half stop, or f1.2 1/3 stop or since f1.26 rounds to f1.3, is it an f1.3 1/3 stop?
tkbslc
Forum Pro
Not really. The difference between a sixth stop at f1.2 is 0.07. So rounding can totally throw things off. When you are at f2.8, that sixth stop is 0.2 difference, meaning you aren't going to round your way to the next stop by losing a significant digit.You can say that for each and every f-stop.Funny thing about f/1.2 is that it can be both 1/2 of a stop faster than f/1.4, or 1/3 of a stop faster, depending on which scale you use. That's because f-numbers are rounded up for convenience, and it gets a bit too crowded between f/1 and f/1.4.
So you can have two f/1.2 lenses and they can actually be different in terms of actual f-number
Photo Pete
Veteran Member
They're all f1
- tkbslc wrote:
1/2 stop up from f1.0 is f1.19. 2/3 stop up from f1.0 is f1.26. Many camera makers decided it was easier to round both to f1.2 and not bother with an f1.3 setting in camera. And besides, the lens design won't hit the exact setting anyway.So how can there be two 1/2 stops = 1 stop between f/1.2 and f1/1.7, and at the same time three 1/3 stops = 1 stop between f/1.2 and f/1.8. That can only be true if f/1.7 = f/1.8.
Here you go.
As a particularly annoying example, what if you design a 52.1mm f1.249 and call it a 50mm f1.2? Isn't f1.249 closer to f1.26, than f1.19? But it rounds mathmatically to f1.2. So is it f1.2 and half stop, or f1.2 1/3 stop or since f1.26 rounds to f1.3, is it an f1.3 1/3 stop?
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