# How are intermediate f/stops calculated?

Discussion in 'Nikon' started by robert_lai, Jun 5, 2005.

1. ### Robert_Lai

Does anyone have a simple equation to allow me to tell how far apart
in stops, two different f/stops are? That is, an equation that will
tell me that f/1.2 is 1/2 stop faster than f/1.4? Or, if I wanted to
know what 2/3 stops slower than f/1.4 are, that f/1.8 is the answer?
<p>There is also a second related question: how much more light do
you obtain if you are 1/2 stop faster? Tediously working out
diameters and cacluating areas, I get about 36% more light. Is that
correct? <p>For light transmission ratios, I get:<p>
Transmission(Aperture1 - Aperture 2) = [(1/aperture1)/(1/aperture2)]
squared.<p>That is, f/1.2 vs f/1.4 gives me 36% more light coming
in.<p>Thanks, in advance, to all you optical experts out there.

2. ### Robert_Lai

A moment's reflection on paper made me realize that the formula I gave above simplifies to:<p>
Transmission (Aperture1/Aperture2) = (Aperture2/Aperture1)Squared.<p>
Using this formula, if I use aperture2 = 1.4, then I get the following for various values of Aperture1:<p>
f/1.4 (no change) = 1<p>
f/1.3 (1/3 stop faster) = 1.28 X more light<p>
f/1.2 (1/2 stop faster) = 1.36<p>
f/1.1 (2/3 stop faster) = 1.65<p>
f/1 (1 stop faster) = 2 x more light, which is 1 stop.<p>Alas, this is only fully available to Leitz Noctilux users!

3. ### Robert_Lai

If you want the results in percentages, subtract 1 from the Transmission result above, and multiply by 100.<p>
Percentage light transmission change = [Trans (A1/A2) -1] x 100 = [(A2/A1)squared - 1] x 100.<p>
Example: A2 = 1.8, A1 = 1.2<p>
Percentage light transmission change = [(1.8/1.2)*2 - 1]X100 = 125%<p>That is, the Nikkor 50mm f/1.2 lets in 125% more light wide open than the Nikkor 50mm f/1.8 wide open. This works out in stops to 7/6 stops.<p>Now, does anyone have that intermediate stop calculation formula?

4. ### richard_meyers

Robert -

IIRC, the f stops are the square roots of a series of numbers -- the number two taken to successive powers: 2 (raised to 1), 2 (raised to 2, or 2 squared), 2 (raised to 3), etc. Using this formula, the intermediate 1/2 stops would be the square root of 2 (raised to 1.5), 2 (raised to 2.5), and so on. The 1/3 stops would be the square root of 2 (raised to 1.25) etc. and the 2/3 stops would be the square root of 2 (raised to 1.75) etc.

Hope this helps.

5. ### richard_meyers

Ops!

"The 1/3 stops would be the square root of 2 (raised to 1.25) etc. and
the 2/3 stops would be the square root of 2 (raised to 1.75) etc."

Should read: "The 1/3 stops would be the square root of 2 (raised to
1.33) etc. and the 2/3 stops would be the square root of
2 (raised to 1.66) etc."

Sorry.

6. ### Robert_Lai

I did find a good basic explanation of the f/stop in Matthew Cole's Tedious Explanation. He also has a good pdf chart listing all of the intermediate f/stops. I guess I'll have to contact him regarding the equation that he used.

7. ### kelly_flanigan|1

If F1.0 passes 4 gallons of mollases per hour then; <BR><BR> F1.414 passes 2 gallons of mollases per hour; then<BR><BR> F2.0 passes 1 gallon of mollases per hour<BR><BR><BR>The 1/3 stop sequence is:<BR><BR>1.000 ; 1.12246 ; 1.25992 ; 1.4142<BR><BR>another stop<BR><BR>1.4142; 1.5874; 1.7818; 2.00<BR><BR>1.12246 is about the sixth root of two. ie what number mulitiplied by itslf six times equals 2.0000<BR><BR><BR>The practical sequence is 1.0; 1.12; 1.26; 1.4 ; 1.6; 1.8; 2.0

8. ### bjørn rørslett

If N0 is your reference f-number and t0 your exposure time, the corresponding values for another f-number N1 is as follows,

difference in stops dN = log(N1/N0)/log(sqrt(2))

exposure factor = 2^dN

% light transmitted vs reference = 2 ^(-dN)*100

So, f/1.2 vs 1.4 (I've set dN to 0.5 in this case, since we're talking about a halfstop, and the traditional f-numbers are rounded a little bit weird), will give you 41 % more light, leading to an exposure time 0.71 times that of the smaller aperture.

If the numbers 1.2 and 1.4 are plugged into the equations, you do get 136% transmission (36% increase). Just to show that there is a little leeway in the numbers we calculate, since the scale of f-numbers is "nice" not accurate in steps of sqrt(2).

f/4 vs f/1 will give dN = 4 (4 stops), exposure increase 16 times, and 6.25% transmitted light of the reference. And so on.

9. ### kelly_flanigan|1

Many lenses are rounded down; to make the numbers better for marketing. this is like the girl who weighs 134 pounds; and round down to 130Lbs for the Bikini contest.<BR><BR> A typical Nikkor 50mm F1.4 might be usually F 1.44 ; rounded down to F1.4. A 1.7 Minolta might be rally a 1.74 lens. <BR><BR>In optical engineering; a 35mm wideangle F2 lens might be this to marketing; but really the nominal design target is to us engineers really 36.5mm; at say F 2.04. Alot of numbers are weaseled around; this makes the design easier. With a super fast lens; a tiny change in f stop can help a design alot; many abberations vary with the CUBE of the aperture<bR><BR>with film asa/iso's; the numbers are rounded to the nearest 1/3 F-stop; ie standard FTC numbers; 100; 125; 160; 200; 250; 320; 400; 500; 640; 800. <BR><BR>In lenses; many lenses are slower due to tranmission losses. In the movie industry folks use T stops; many older zooms have a horrid loss before multicoating came out.<BR><BR>F stops are used for DOF tables in film work; T stops used for exposure. <BR><BR>Many times folks chase and buy old FAST 1950's lens designs with still cameras; that really T-stop less in speed/performance than one would guess.

10. ### todd peach (seattle, washington)

I was curious about this some time back, Robert. When I found the 'relationship', I did a quicky excel table, added in the traditional 'breaks' in the math formula, and published it as a web page:
http://www.thepeaches.com/photography/fractional_stops.htm
Now I refer to it when I'm trying to remember how they go....

11. ### Robert_Lai

Thank you all for your help. However, the basic equation still evades me. Why do you use the 1/6th root of 2 for 1/3 stops, and 14th root of 2 for 1/2 stops?

12. ### Robert_Lai

Looking at Todd's tables, the basic equation seems to be:<p>
f/stop = 2exp([stops from f/1]/2)<p>
Thus, 1/2 stop slower than f/1 is:<p>
f/stop = 2^(1/4) = 1.189207
This is usually rounded off to f/1.2.<p>By Bjorn's equations, exposure time has to be increased 1.414X, and there is a 30% reduction in light transmission at f/1.2 compared to f/1.0

13. ### kelly_flanigan|1

It has to do with areas. The iris is round. Light density on the film. One stop is twice as much light. A half stop and third stops are fairly common. If a pizza is 12 inches around; and it feeds 1 person; how big around is one to feed 2 people? If a farmer uses on bag of seeds for a 100 foot diameter plot; how big is it around for 2 bags. What if he only has 1 1/3 bags; or 1 2/3 bags? Think areas

14. ### bjørn rørslett

I cannot see any discrepancy here? I used the general equation, Todd the special case with reference to f/1. Both are identical.

Stops from a reference N0=1 are given by 2^(0.5*dN), so +0.5 stop is 2^(0.5*0.5)=2^(1/4) and +1/3 stop is 2^(0.5*0.33)=2^(1/6). Hence the powers of 4 and 6.

Using the exact value 1.189 for the +0.5 stop over 1.0, transmission of light is 70.7% and correspondingly, exposure time must be increased by a factor of 1.41.

If you start with 1.4(14...) and open up 0.5 stop, transmission is 141% (41% increase) and exposure factor is 0,707. (if the traditional, rounded numbers of f/1.2 and f/1.4 are used, transmission increase is 36% and exposure factor is 0,73).

Of course the calculations are mostly superfluous today with all the electronic wizardy inside our cameras. However, I have programmed them for use with my Multiphot photomacrographic camera, which lacks an exposure meter. Here, I need to find the exact exposure time given the lens magnification (computed from lens characteristics and bellows draw), and the exposure reading given by my spot meter of the subject (but without metering through the lens). Since I use 4x5" sheets for this kind of photography, I certainly don't want to bracket excessively to get a good exposure, and accordingly wrote a small computer program to do the tedious calculations. It works perfectly, so "my" equations seem to be correct.

15. ### david_h._hartman

Here is an Excel workbook I made sometime ago. It&#146;s 4:45
AM so what you see is what you get. It&#146;s about 100KB so it&#146;s
a quick download. I just scanned it with Norton Antivirus (up to
date at the time of posting.<br>
<br>
Best,<br>
<br>
Dave Hartman.

16. ### briany

The actual equation is pretty clean. The distance between two f-stops is the Log(StopA/StopB), where the base of the log is the square root of two (~1.4). In Excel, you can write this as "=LOG(Ref1/Ref2, 2^0.5)"
If StopA > StopB, the answer will be a negative number of stops.

You've already gotten the transmission equation. f-stop is inversely proportional to the relative diameter of the aperture. By extension, it's proportional to the inverse radius of the aperture, since radius = diameter/2. Twice the f-stop means half the aperture radius. Since light entering is proportional to size of the hole, and the size of a circle is PI*r^2, if you substitute (1/fstop) for r, you get relative light gathering = PI*(1/fstop)^2. Comparing two fstops, you get PI*(1/fstopA)^2 / PI*(1/fstopB)^2. This falls out to become your equation above, (fstopB/fstopA)^2

PS - for my first equation above, if you don't have Excel and you only have a calculator that does log base 10, there is a change of base law for logarithms: Log base A of B = (Log base X of A) / (Log base X of B). So the equation on a calculator would be LOG(Ref1/Ref2) / LOG(2^0.5)

17. ### Robert_Lai

Thank you, Brian! Yours was the explanation that I was looking for.

18. ### briany

Glad it works for you. I gave you the difference between two stops above. If you want to know what stop is X stops faster than another, the equation is:

ResultingStop = (2^0.5)^numberOfStops * OriginalStop

The resulting stop is the square root of two raised to the number of stops you want to add (also works with negative numbers when decreasing stops) multiplied by your starting fstop. Now lets go take some photos... ;-)

19. ### briany

PS - Bjorn did give all of those equations above, though I admit I didn't see them the first time around.

20. ### jovan

Wrong, as others are already pointed out it is log(f-base)X equation. <br>
For example if f1.2 is 'B' and f1.8 is 'C' then <br>
1.EV of B = log(base=1.4)(B) or it can be developed to log(B)/log(1.4) = 0.54 EV <br>
<br>
That mean F1.2 is 0.56EV or 56% more light coming in then f1.4. <br>
2. EV of C is log(C)/log(1.4) = 1.75 the is +0.75EV more then 1EV <br>
Difference is now linear and we can say f1.8 is 1.75-0.54 = 1.21 EV dimer then F1.2. That is 150% less light energy.<br>

21. ### pnance

Jovan, I believe you round too much. If you had used the square root of 2 instead of 1.4 your answers would have been:

F1.2 is 0.45EV or 45% more light coming in then f1.4 (not 56%)

F1.2 compared to F1.8 is 1.17 (not 1.21)

"Difference is now linear and we can say f1.8 is 1.75-0.54 = 1.21 EV dimer then F1.2. That is 150% less light energy" 1.21 does not equal 150! Less than 1 and a quarter stops.

22. ### jovan

Ok I do round but, can you be 100% shure that 1.2 is 1.2 and not 1.187776 od 1.2412.
Second using 1.4 for base is 2^(1/2) rounded on 1 decimal for standardization.
But in any case I do some mistake 1.2->1.8 is 0.526EV (unrounded) to 1.414.. + 0.696EV to 1.8 => 1.222EV
or 1+1/3EV if ve use standardized ISO numeric sequence.

23. ### pnance

Jovan, there's is really nothing wrong with rounding, its just that its the last thing you do, not the first!