F-stops Explained

ClickAddict · Feb 8, 2013

JayJeep said:
ClickAddict said:

Both Sparky and O|||||||O have the same math. Just one is longer form.
Sparky's math Breaking it down:

focal lens = m
diameter = d
aperture = f
He states m/d = f........................... (His 200mm lense to 25 diameter = f8 example)
which means m=df ...........................(multiply both sides by d)
and d=m/f .....................................(divide both sides by f)

opening area = A
A = pr^2 .............................(This is a known calculation for area of circle)
p=3.14 ..................................(rounded)
r = radius

r= 1/2 d .....................................(By definition radius = 1/2 diameter)

r= 1/2X(m/f)................................ (replacing d with m/f as per above)
r= m/2f........................................ (result of 1/2 X m/f)

A1 ...................area of lens at a given aperture
A2 .................area of same lense at a different aperture
r1............... radius of lense to get Area = A1
r2.................... radius of lense to get Area = A2

if A1 = 1/2 of A2 then ............................(looking for next stop which is half the area)

A1 = A2/2
pr1^2=(pr2^2)/2................................... (A1 = pr1^2 as per area formula, same for A2))
p(m1/2f1)^2 = p(m2/2f2)^2/2 ............................(replacing r with m/2f)
(m1/2f1)^2 = (m2/2f2)^2/2 ...........................divide both sides by p
2(m1/2f1)^2 = (m2/2f2)^2 ........................multiply both sides by2
2 = ((m2/2f2)^2)/(m1/2f1)^2...................... divedi both sides by
2 = (m2^2/(2f2)^2)/(m1^2)/(2f1)^2 ......................(square the fractions)
2= (m2^2)(2f1)^2)/((2f2)^2)(m1^2) ......................(a/b)/(c/d) = a/b X d/c = ad/bc
2= (m1^2)(2f1)^2)/((2f2)^2)(m1^2) .....................since were keeping m the same and only changing f, m1=m2)
2=(2f1)^2)/((2f2)^2) ...................................(xy/zx = y/x)
2= 4f1^2 / 4f2^2 ........................................(ab)^2 = (a^2)(b^2)
2=f1^2 / f2^2................................ (4A/4B = A/B)
sqrt(2) = f1/f2 ..................................sqrt = square root)
1.4 = f1/f2

so to get the half amount of light as per sparkys math AND O||||||O you need to have that ration

Essentially many of the variables eliminate themselves to get to the 1.4 value

Click to expand...

Except O||||||O is wrong. He says to multiply 1.4*(integer) to get 2, 2.8, 4, 5.6, 8, 11. You don't multiply you take it to the power of the integers.

sqrt(2)^0=1, sqrt(2)^1=1.41, sqrt(2)^2=2, sqrt(2)^3=2.82, sprt(2)^4=4

If I used his math of rounding 7.07 to 8 and 8.48 to 11 in my line of work I would most likely be fired and endangering lives.

The problem I see is that he insisted that he was right and the OP was wrong to post a complicated explanation for something I know has been asked.

"where do they get f/5.6 as double the light of f/8" and "where do those numbers come from"

You are correct. I thought he had said it was 1.4 X previous number. (approximately)

Helen B · Feb 8, 2013

ClickAddict said:
Both Sparky and O|||||||O have the same math. Just one is longer form.
Sparky's math Breaking it down:

focal lens = m
diameter = d
aperture = f
He states m/d = f........................... (His 200mm lense to 25 diameter = f8 example)
which means m=df ...........................(multiply both sides by d)
and d=m/f .....................................(divide both sides by f)

opening area = A
A = pr^2 .............................(This is a known calculation for area of circle)
p=3.14 ..................................(rounded)
r = radius

r= 1/2 d .....................................(By definition radius = 1/2 diameter)

r= 1/2X(m/f)................................ (replacing d with m/f as per above)
r= m/2f........................................ (result of 1/2 X m/f)

A1 ...................area of lens at a given aperture
A2 .................area of same lense at a different aperture
r1............... radius of lense to get Area = A1
r2.................... radius of lense to get Area = A2

if A1 = 1/2 of A2 then ............................(looking for next stop which is half the area)

A1 = A2/2
pr1^2=(pr2^2)/2................................... (A1 = pr1^2 as per area formula, same for A2))
p(m1/2f1)^2 = p(m2/2f2)^2/2 ............................(replacing r with m/2f)
(m1/2f1)^2 = (m2/2f2)^2/2 ...........................divide both sides by p
2(m1/2f1)^2 = (m2/2f2)^2 ........................multiply both sides by2
2 = ((m2/2f2)^2)/(m1/2f1)^2...................... divedi both sides by
2 = (m2^2/(2f2)^2)/(m1^2)/(2f1)^2 ......................(square the fractions)
2= (m2^2)(2f1)^2)/((2f2)^2)(m1^2) ......................(a/b)/(c/d) = a/b X d/c = ad/bc
2= (m1^2)(2f1)^2)/((2f2)^2)(m1^2) .....................since were keeping m the same and only changing f, m1=m2)
2=(2f1)^2)/((2f2)^2) ...................................(xy/zx = y/x)
2= 4f1^2 / 4f2^2 ........................................(ab)^2 = (a^2)(b^2)
2=f1^2 / f2^2................................ (4A/4B = A/B)
sqrt(2) = f1/f2 ..................................sqrt = square root)
1.4 = f1/f2

so to get the half amount of light as per sparkys math AND O||||||O you need to have that ration

Essentially many of the variables eliminate themselves to get to the 1.4 value

You can leave the unneccessary stuff out at the beginning if you want cleaner workings.

N=f/D

(which means f-number = focal length divided by entrance pupil diameter)

therefore

N∝1/D (1)

A = (pi D^2)/4

therefore

A∝D^2 and thus D∝A^0.5

Substituting in (1)

N∝1/A^0.5

which leads to

N2/N1 = (A1/A2)^0.5

thunderkyss · Feb 8, 2013

This is turning into a very informative thread.

bunny99123 · Feb 8, 2013

OMG. I am going to have nightmares from all those physic courses and teaching it for so many years. You guys have triggered my PTSD caused from Math. I am too old for this intelligent thinking, so I am just going to turn my zoom and let the light flow in or shut some out. Seriously, you guys have done an excellent job on explaining it simply to college level

IByte · Feb 8, 2013

Helen B said:
They are mostly rounded to two significant figures. The 0.5 - 1 - 2 - 4 - 8 series do not need to be rounded because they are exact numbers, 0.5 being the lowest f-number theoretically possible. The others usually get rounded to a nominal f-number with no more than two significant figures, such as 1.4, 2.8. 5.6, 11 etc.

The sequence is (√2)^x, where x is a positive or negative integer equal or greater than -2, or zero.

F/4 on a 50 mm lens is equivalent to f/4 on a 300 mm lens in terms of image brightness. As I mentioned earlier, the 300 mm lets more light in from an object, but it forms a larger image, so the brightness is the same. How much detail do you want on this? How good are you at maths?

It's the weekend and I have nothing else but crunching the numbers

IByte · Feb 8, 2013

JacaRanda · Feb 8, 2013

My wife would surely bash me in the face with her camera if I even hinted at trying to teach/show/explain any of this to her. It's dangerous enough telling her the magic trackpad is easier to deal with than that stupid mouse she has to pick up and put down every couple of seconds :lol:

kundalini · Feb 8, 2013

While it's interesting to see how passionate some of you are with the minute (pronounced my-noot) intricacies for the math involved, it seems a bit over the top. With time behind the camera, the results of your settings becomes intuitive and the mathematical formulas become impractical in the field. The substance is retained, but the details are of no consequence, other than in a classroom environment.

Helen B · Feb 8, 2013

kundalini said:
While it's interesting to see how passionate some of you are with the minute (pronounced my-noot) intricacies for the math involved, it seems a bit over the top. With time behind the camera, the results of your settings becomes intuitive and the mathematical formulas become impractical in the field. The substance is retained, but the details are of no consequence, other than in a classroom environment.

I have no idea who in particular you are dissing here, but I do wonder why you make such assumptions about other people's practice? Do you really think we worry about this stuff when shooting? It is possible to have decades behind the camera and also to understand the trivial physics being discussed here. If someone is curious, should we just ignore them?

shefjr · Feb 8, 2013

I just memorized the f-stops. Is it bad that I don't know the precise why it is what it is? I mean before this post, because I have an understanding now.

Helen B · Feb 8, 2013

shefjr said:
I just memorized the f-stops. Is it bad that I don't know the precise why it is what it is? I mean before this post, because I have an understanding now.

I'm pretty sure that anyone could be an excellent photographer without knowing this sort of stuff, but it doesn't seem to hurt too much to understand it either. For some of us it is much easier to understand than to try to remember by rote.

480sparky · Feb 8, 2013

I may have a terrible memory, but I'm fairly certain I never stated that it's required to know this in order to operate a camera. There's not going to be a pop quiz tomorrow. No test at the end of the month. No Finals in order to 'graduate'.

shefjr · Feb 8, 2013

480sparky said:
I may have a terrible memory, but I'm fairly certain I never stated that it's required to know this in order to operate a camera. There's not going to be a pop quiz tomorrow. No test at the end of the month. No Finals in order to 'graduate'.

I meant no offense. I just honestly wondered if there was something that I was missing that would eventually lead me to understand further the importance of the f-stops. I have found at times I think I know it all until someone points something out that corrects me.