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Analytics of Stop Difference And Diameter (Radius) Of Exit P
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mjkzz



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PostPosted: Mon Dec 02, 2019 4:50 pm    Post subject: Analytics of Stop Difference And Diameter (Radius) Of Exit P Reply with quote

Recent discussion made me think about relationship between stop difference and diameter (radius) of exit pupil (thanks Lou) and found it is so beautiful and easy to think about. Since I have learned so much here, I will contribute back Very Happy

A = PI*R*R, where A is the area of exit pupil, R is the radius, taking first order of derivative gives

dA = 2*PI*R*dR, and this gives

dA/A = (2*PI*R*dR / PI*R*R), cancelling out the PI and one R gives

dA/A = 2*dR/R ==> dR/R = 0.5*dA/A (1)

Note, the term dA/A is THE stop difference, ie, when people say one lens is dA/A stops faster, and dR/R is the relative change in radius (diameter).

This relevation shows how elegant and simple to estimate dR/R for a given (small) stop difference without using calculator.

The formual (1) is true for infinitely small dA, however, it is still close enough to estimate dR/R, take the following:

1/10 stop --> 0.05 or 5%
1/5 stop --> 0.1 or 10%
1/4 stop --> 0.125 or 12.5%
1/3 stop --> 0.167 or 16.7%

Numbers corresponding to 1/4 and 1/3 is probably high, maybe 2nd order derivative should be factored in. I have left school so long, so I tend to use numerical analysis to analyze anything higher than 2nd order. As matter of fact, it is not difficult to write a Python program to produce ACTUAL result. But that loses the elegance of estimating dR/R quickly in mind.

The underlying assumptions are: everything else about the two lenses are identical Or maybe the same lens under different stops.
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rjlittlefield
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PostPosted: Mon Dec 02, 2019 6:14 pm    Post subject: Reply with quote

Hhmm...

I tried doing the calculation a different way and got a different answer.

In the calculation that I did, we note that a change of one stop is a factor of 2 in area and thus a factor of sqrt(2) in radius. The radial change for 1/N stop must then be a factor of sqrt(2)^(1/N), so that applying it N times in succession produces the required total factor of sqrt(2).

But then the problem is that sqrt(2)^(1/10) ~= 1.035265, not 1.05.

Thinking that the discrepancy might lie in not having taken the limit far enough, I tried N = 1000. But then sqrt(2)^(1/1000) ~= 1.0003466, not 1.0005, so the discrepancy is not much different and has gotten slightly bigger, not smaller. So I think the limit is not the problem.

I then looked harder at your analysis to see why the two methods are giving different answers.

My conclusion is that it lies in the words, not the math. In your method, you're treating "1/N stop" as it if were a factor of 1+1/N in area, where in mine, it is a factor of 2^(1/N). In the limit, these are different by a factor of 1/ln(2) ~= 1.443 , which is why I think all your numbers are too large by about that much.

The numbers that I get are:

1/10 stop --> 3.5%
1/5 stop --> 7.2%
1/4 stop --> 9.1%
1/3 stop --> 12.2%

Let me know if I've messed this up somehow.

--Rik
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Lou Jost



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PostPosted: Mon Dec 02, 2019 6:32 pm    Post subject: Reply with quote

Rik, yours are the numbers I get as well. Your number for 1/4 stop is 9%, which is what I calculated for the post where mjkzz and I were discussing this issue. A quarter of a stop is a factor of [Sqrt(2)]^(1/4). Like you said, there is nothing wrong with mjkzz's math, it is a valid approximation when the stop fraction is very close to zero.
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mjkzz



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PostPosted: Mon Dec 02, 2019 7:02 pm    Post subject: Reply with quote

rjlittlefield wrote:
Hhmm...

I tried doing the calculation a different way and got a different answer.

In the calculation that I did, we note that a change of one stop is a factor of 2 in area and thus a factor of sqrt(2) in radius. The radial change for 1/N stop must then be a factor of sqrt(2)^(1/N), so that applying it N times in succession produces the required total factor of sqrt(2).

But then the problem is that sqrt(2)^(1/10) ~= 1.035265, not 1.05.

Thinking that the discrepancy might lie in not having taken the limit far enough, I tried N = 1000. But then sqrt(2)^(1/1000) ~= 1.0003466, not 1.0005, so the discrepancy is not much different and has gotten slightly bigger, not smaller. So I think the limit is not the problem.

I then looked harder at your analysis to see why the two methods are giving different answers.

My conclusion is that it lies in the words, not the math. In your method, you're treating "1/N stop" as it if were a factor of 1+1/N in area, where in mine, it is a factor of 2^(1/N). In the limit, these are different by a factor of 1/ln(2) ~= 1.443 , which is why I think all your numbers are too large by about that much.

The numbers that I get are:

1/10 stop --> 3.5%
1/5 stop --> 7.2%
1/4 stop --> 9.1%
1/3 stop --> 12.2%

Let me know if I've messed this up somehow.

--Rik


The bolded part is the difference between you and me. 1/N stop difference does not mean if you apply it N times, it would give you one full stop. Think about it, 1/N stop differences means 1/N difference in area and if you apply N time, it gives you (1+1/N)^N difference in area.

Using your interpretation, this means 1/10 difference, applying 10 times give you (1+1/10)^10 = 2.594 difference in area, more than 2. (1+1/3)^3 = 2.37, again more than 2 in terms of area, meaning doing that in succession gives you more than one stop difference.

Or maybe I should used the term EV . . .
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mjkzz



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PostPosted: Mon Dec 02, 2019 7:24 pm    Post subject: Reply with quote

It all comes down to the definition of a lens B is 1/N faster than lens A. What it means is that lens B is letting in 1/N times more (or 1+1/N times) light in, or its area is (1+1/N) larger.

If you use successive method, apply it N times, you end up a lens with more than 1 stop.

If you calculate the area for all lenses, you will see what I mean.
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PostPosted: Mon Dec 02, 2019 7:37 pm    Post subject: Reply with quote

mjkzz wrote:
1/N stop difference does not mean if you apply it N times, it would give you one full stop.

With respect, that is exactly what it means. If your camera adjusts in 1/3 stop increments, then you click the button 3 times to make a one stop change. Click it 6 times, you get a two stop change, and so on.

Quoting from https://en.wikipedia.org/wiki/F-number,
Quote:
Photographers sometimes express other exposure ratios in terms of 'stops'. Ignoring the f-number markings, the f-stops make a logarithmic scale of exposure intensity. Given this interpretation, one can then think of taking a half-step along this scale, to make an exposure difference of "half a stop".

If you want to imagine that "1/N stop differences means 1/N difference in area", then of course you're free to do that, but you won't be following photographic tradition.

--Rik
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Lou Jost



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PostPosted: Mon Dec 02, 2019 7:43 pm    Post subject: Reply with quote

Quote:
if you apply N time, it gives you (1+1/N)^N difference in area.


That's where we differ. We agree that an aperture at f/5.6 has twice the area of an aperture at f/8. Each stop gets MULTIPLIED or DIVIDED by the factor that Rik and I are talking about, to produce that relationship between areas.

One stop down from 5.6 is [Sqrt(2)]*5.6=8. Two stops down is {[Sqrt(2)]^2}*5.6 = 11. N stops down from 5.6 is {[Sqrt(2)]^N}*5.6. This is valid for any N including fractions.

Edit: I didn't see Rik's post until I finsihed mine, but our results agree.
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mjkzz



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PostPosted: Mon Dec 02, 2019 8:02 pm    Post subject: Reply with quote

Thanks Rik and Lou. Here is some exercise:

Lens B is 1/N stops faster than lens A, Lens C is 2/N stops faster than lens A. Does that make lens C (2/N - 1/N) = 1/N stop faster than B?

Or if lens B is 1/N faster than A, lens C is 1/N faster than B, does it make lens C 2/N faster than A?

I think if we stick to the difference between two apertures, forget about succession . . .
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mjkzz



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PostPosted: Mon Dec 02, 2019 8:09 pm    Post subject: Reply with quote

Another way to think about it, we are talking about the DIFFERENCE between two apertures, not f/stop themselves. This is why I think I should have used the term EV instead of stops.
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mjkzz



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PostPosted: Mon Dec 02, 2019 8:51 pm    Post subject: Reply with quote

I think I know where I got it WRONG, thanks Rik and Lou . . . I am stubborn but I do admit it that I am wrong Very Happy
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PostPosted: Mon Dec 02, 2019 8:59 pm    Post subject: Reply with quote

mjkzz wrote:
I think I know where I got it WRONG

As a teacher of sorts, I always want to know: can you explain the process by which you got it wrong?

For example did you find some reference with incorrect information, or misinterpret some reference with correct information, or did you just get things scrambled a little in your own head?

--Rik
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mjkzz



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PostPosted: Tue Dec 03, 2019 5:48 am    Post subject: Reply with quote

rjlittlefield wrote:
mjkzz wrote:
I think I know where I got it WRONG

As a teacher of sorts, I always want to know: can you explain the process by which you got it wrong?

For example did you find some reference with incorrect information, or misinterpret some reference with correct information, or did you just get things scrambled a little in your own head?

--Rik


I was hoping somebody else point out the error I made, but since you asked, I guess no body wants to do it now. The error is the part that I assumed dA/A is the stop difference, note I actually capitalized the THE, that is glaringly WRONG.


It is actually more elegant than what I had before.

Let log2(X) denotes log of X base 2

I remember that EV = log2(C*A) where C is some constant and A is the area of aperture. This makes sense -- the larger the A, the larger the EV given all other things equal. If A' is twice as large as A, then EV' = log2(2A*C) = 1 + log2(A*C) = 1 + EV or one stop up.

Now log2(X) = ln(X)/ ln(2) where ln denotes natural log

so now EV = ln(C*A) / ln(2)

EV = (ln(C) + ln(A)) / ln(2)

note here ln(C) / ln(2) is constant, and remember derivative of ln(x) is 1/x

so we have

dEV =((0 + 1/A)*dA) / ln(2) ==> dEV = (dA / A) / ln(2)

this means ln(2) * dEV = dA / A, we also know dA/A = 2*dR / R

so we have dR/R = (ln(2)/2)*dEA, ln(2) = 0.69315, so we have

dR/R = 0.346575 * dEV

Note, dEV is what we are talking about, the stop difference.

So, it is so elegant that we can simply multiply the stop difference by 0.346575 to get percentage change in R for an infinitely small dR

so for 1/10 stop, it is 0.0346575 or 3.46575%, etc, etc. They should match very closely to how you calculated your results.

Hope this makes sense.

And BTW, I am not trolling, the first post was an honest mistake.
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mjkzz



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PostPosted: Tue Dec 03, 2019 2:20 pm    Post subject: Reply with quote

OK, it is 5 AM and I am doing this . . .

Anyways, for those who have doubts about EV = log2(C*A) where C is a constant and log2 denotes log base 2, here is another way to look at it.

According to EV formula, we have

EV = log2(N^2 / t) where t is exposure duration and N is relative aperture where N = 2R / F and F is the focal length of the lens and is a constant. So this gives us the following:

EV = log2(4*R^2 / F^2 / t) ==> EV = log2(4/F^2 /t) + log2(R^2) = log2(4/F^2/t) + log2(R) + log2(R)

with log2(R) = ln(R) / ln(2) so here we got:

dEV = (0 + 1/R + 1/R)*dR/ln(2)

The derivative of first term, log2(4/F^2/t) with respect to R is zero, not only just for a given t. This is true if you think of it as part of multi variable system and taking partial derivative of it with respect to variable R

so, this give us:

ln(2) * dEV = (1/R + 1/R) * dR ==> dR/R = ln(2)/2*dEV

where dEV is the stop difference and dR/R is the percentage change

I should have used the EV approach at the very first post. The result in the very first post is off by a multiplicative factor of ln(2), so even if I am really stubborn, the estimate is still off by ln(2) regardless how small dEV is. You can verify this using dEV = 0.01 (1/100 stop), dEV = 0.02, etc.

I am still stunned by how these "natural" numbers play a role in our daily analytical life, if not mathematically analyzed, I won't believe it.

Anyways, I think I have rectified it
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PostPosted: Tue Dec 03, 2019 5:48 pm    Post subject: Reply with quote

mjkzz wrote:
Anyways, I think I have rectified it.

Close, but not quite.

Quote:
According to EV formula, we have

EV = log2(N^2 / t) where t is exposure duration and N is relative aperture where N = 2R / F and F is the focal length of the lens and is a constant.

Sorry, I beg to differ. That is not what your reference at Wikipedia says. You have added all the part about "N = 2R / F", and it looks to me like you've gotten it backwards.

A symbol N -- maybe the same one, but we need to check that -- is defined at https://en.wikipedia.org/wiki/F-number . There, it is written that
Quote:
The f-number N is given by
N = f/D
where f is the focal length, and D is the diameter of the entrance pupil

We can check whether this is the meaning of N that is intended by the EV article by seeing what makes their formula match their table of values.

For example in your reference's Table 1, we see that EV=5 is associated with t=1 and f-number 5.6 . The crosscheck is log2(5.6^2/1)=4.97 . At t=1 and f-number 16, the table gives EV=8 while the calculation gives log2(16^2/1)=8. This is just a couple of spot-checks; you can run others if you like. So, the table agrees with the formula when using N = f/D. But if you try it the other way around, with N=2R/F, the table does not match the formula. This tells us that N=f/D is what the formula is referring to.

In other words, making the aperture physically smaller makes the EV number bigger, using the definition of EV provided by Wikipedia.

So, to be formally correct, you would have to rework your entire analysis starting with EV = log2(F^2 / (2R)^2 / t), that is, reverse the F and the 2R.

Fortunately, I think in the end the only effect would be to flip the sign of your result, and that particular correction would probably be done almost without thinking, by anyone trying to apply the method.

The person's thinking would go something like this: "The EV difference is 0.2, so the radius change must be 0.2*0.346575 ~= 0.07 = 7%, and, ah, let's see, lens A gives a darker image so it must have a smaller aperture."

But anyway, yes, it is definitely elegant that we can accurately estimate small percentage change in R/F as simply a fixed multiple of the difference in EV. I had not noticed that relationship.

--Rik
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mjkzz



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PostPosted: Wed Dec 04, 2019 8:38 pm    Post subject: Reply with quote

Hmmm, OK, I thought I checked about definition of N . . . in the link I provided, EV = log2(N^2/t) means the larger the N, the larger the EV and it makes sense -- meaning under the same lighting conditions, EV measured by camera (or other instruments) will be larger if N is larger. I think the N in that formula is 1/(f-number which is F/2R -- the smaller f-number, the larger N, f 2 vs f 5.6 -- the same lens set at f 2 or N=1/2 vs set at f 5.6 or N = 1/5.6)

[edit] there is another form of EV expressed in ISO, Luminance, and some calibration constant K (varies from camera to camera or light sensor) and that formula is what I used to build a light (color) meter.

But in any event, a multiplicative of constant factor of N in relation to R does NOT play any role, it just shift the log curve a bit. For example, we can say N = C*R/F, as long as C is constant, log2(C/F) is zero when taking derivative.

Anyways, I agree that everything comes out so elegantly, if not proven mathematically, I really would not believe this relationship.
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