quiz: if you increase by 1/3 stop, how many % of total exposure are you increasing?

Jack I get what you are saying. Just for discussion sake and let's not get agitated over this, let me rephrase again:

 

We AGREED that exposure increases in this geometric progression, for sunsequent stops: T, 2T,4T,8T,16T...

 

If you are saying that stops changes should be linear with exposure changes, then the above is wrong! Because then you should get the below:

 

..for subsequent stop change, t,2t,3t,4t,5t,6t...etc, an additional progression which is not how light exposure works through a lens.

 

 

So if like you said Jack, f stop changes should be linear with exposure changes, then when you increase 2 stops, you should by your definition get 3T and not 4T!

 

The exponential curve for light exposure vs f stop is there for you to plot. It can never be a straight line.

and you don't need to bring the diameter changes of the aperture into play here because that's not in question.

<i>The exponential curve for light exposure vs f stop is there for you to plot. It can never be a straight line.</i><P>It can, if you do a funky mathematical thing to one of the axes... ;-)

"The exponential curve for light exposure vs f stop is there for you to plot. It can never be a straight line.

It can, if you do a funky mathematical thing to one of the axes... ;-)

"

 

Oh yeah... I really have to get funky to do this: Plot the relative percentages of light transmission for each f-stop increase against the corresponding shutter speed required to maintain equivalent exposure... In other words, assume your meter reads 1/1000th at f2.0 and you are using your 50mm Summicron. You have 100% transmission at f2, 50% at f2.8, 25% at 4, etc. My graph would look like this: 100%@1/1000th; 50%@1/500th; 25%@1/250th... Really tough math and a straight-line graph too ;)

 

Now if you persist in plotting shutter speed against f-stop, then you will get a curve -- But that is simply the wrong way to look at it because light TRANSMISSION is what we are talking about, and transmission is dependant on the AREA of the circle the aperture inscribes (and shutter open time) -- NOT the absolute f-number!

 

Cheers,

If f stop changes is linear with exposure changes, why do you get T,2T,4T,8T,16T etc. where T=units of light, as the F stop increases in subsequent full stops?

 

 

f1,f1.4,f2,f2.8,f4<=> 16T, 8T, 4T,2T,T. Anyone disagree?

 

How can one agree on this and then say the relationship is not geometrical? And that the f Stop number is in linear relationship with amount of light units?

 

If you plot the above : T vs f stop, you get an exponential curve. Agree? y=2^x

 

SO if you choose any 1/3 increase in x starting from any f stop number, the y increase is 2^1/3=26%. Isn't this what the question was asking? What has aperture size got to do with the question in this case?

 

When you click on your aperture ring, every click follows this exponential curve. Every lens does this.

 

Really, what Jean said.

Hello,

 

My comment about the inverse square law was to show that the change in light hitting the film plotted against fstop/exposure/shutter speed is exponential. Hence the r squared.

 

Certainly you may convert exponential curves to a straight line by using a log scale. Jack is "proving" his point by describing his logic using a log scale. Problem is, it only represents a straight line. Hence the thought connecting the dots between 100% and 50% and 25% is a straight line. Yes it is--when graphed on log paper. The physics however, as pointed out by many, is really curved lines (not straight) mathematically symbolized by squares or square root's of the power of 2.

 

No matter how Jack plots it, Jean Christophe is still correct.

In Electrical Engineering; the cube root of 2 is used alot in motor design; and wire sizing.<BR><BR>Each wire guage increases in cross sectional areas by 1.26 ; Three guages vary by a factor of 2 in area. 36 guage wire is exactly 0.005 inch diameter; 0000 wire is exactly 0.4600 inches in diameter. ALL OTHER GUAGES ARE DETERNINED BY THESE TWO DEFINITIONS. <br><br>WIRE diameters vary almost as the 1/3 root of two for the difference in two guages; ie 30 guage is about 0.010; and 28 guage is about 0.0126 <BR><BR>In Photography; a 1/3 stop is 1.26; 2/3rds varyis 1.59 ; 3/3 stop is 2.00 <br><BR>The 26% number is correct for a 1/3 stop; and the 33% a grave error; at least if this was a math test!.

"In Electrical Engineering; the cube root of 2 is used alot in motor design; and wire sizing. Each wire guage increases in cross sectional areas by 1.26 ; Three guages vary by a factor of 2 in area. 36 guage wire is exactly 0.005 inch diameter; 0000 wire is exactly 0.4600 inches in diameter. ALL OTHER GUAGES ARE DETERNINED BY THESE TWO DEFINITIONS. WIRE diameters vary almost as the 1/3 root of two for the difference in two guages; ie 30 guage is about 0.010; and 28 guage is about 0.0126 In Photography; a 1/3 stop is 1.26; 2/3rds varyis 1.59 ; 3/3 stop is 2.00 The 26% number is correct for a 1/3 stop; and the 33% a grave error; at least if this was a math test!."

 

 

Except we ARE NOT talking about DIAMETER when we are discussing TRANSMISSION capability of a given f-stop -- we are talking AREA of the circle that f-stop covers!!! Read that last sentence again! Which is PRECISELY why wire is "guaged", because as you increase the cross-sectional AREA you increase the "flow". BUT, you only need to increase the DIAMETER of the wire by root2 to DOUBLE the area, and hence DOUBLE the transmission capability. And it's the same with lens apertures. F-stops only represent the aperture's DIAMETER, while the light TRANSMISSION of any given f-stop is a function of the aperture's AREA --

 

AREA, NOT DIAMETER!!! RELATIVE LIGHT TRANSMISSION QUANTITIES, NOT JUST F-NUMBERs!

 

Cheers,

Exactly, it is a function of area which is why 1/3 stop more is 26% more (1.26x) in area of light transmission and 1/2 stop more is 41% more (1.41x) in area of light transmission. Do you still not see why, Jack? You really need to sit down and think about it more instead of making physics conform to your own wrong assumptions. First off, your assumption that 1/2 stop more corresponds to 50% increase in area or 1/3 stop more corresponding to 33% more in area is wrong. Disabuse yourself of that wrong assumption and it will be clear to you.

 

As a function of area, exposures multiply, not add. F-stops add up because they are the exponents of base 2.

 

Your mistake lies in adding. If we follow your line of reasoning i.e. one 1/3 stop = 1.33 (33% more), then 1/3 stop + 1/3 stop + 1/3 stop = 1 stop = 1.33 x 1.33 x 1.33 = 2.35x! And 1/2 stop + 1/2 stop = 1 stop = 1.50% x 1.50% = 2.25x!! You are going to insist that 1/4 stop is 25% more so let us do the maths your way: 1/4 stop + 1/4 stop + 1/4 stop + 1/4 stop = 1 stop = 1.25 x 1.25 x 1.25 x 1.25 = 2.44.

 

Sp, according to your mathematics, 1 stop can be 2x, 2.25x, 2.35x or 2.44x. It simply cannot be.

 

Now, as what we have said in this thread all along:

1/3 stop + 1/3 stop + 1/3 stop = 1 stop = 1.26 x 1.26 x 1.26 = 2.

And 1/2 stop + 1/2 stop = 1 stop = 1.41 x 1.41 = 2.

And 1/4 stop + 1/4 stop + 1/4 stop + 1/4 stop = 1 stop = 1.189 x 1.189 x .189 x 1.189 = 2!

 

That is exactly what 1/3 stop more means: 1.26x more in area than the base and so on for the rest, no matter if it is 1/2 stop or 1/24th of a stop. The area of light transmission which you have maintained increases or decreases in this fashion.

 

I wonder how you are going to spin this and still insist that we are all mistaken.

 

Yes, you can get 33% and 50% more in area of light transmisson or exposure in other words but they are not 1/3 stop more and 1/2 stop more respectively.

 

Work it out for yourself, Jack. Start with the formula for area of a disc and then do the mathematics.

Jack ; <h3>1/3 stop is 1 26% increase; do this again and it is a 59% total increase; do this again and you are at 1 full stop; or a 100% increase. </h3> 1/3 reduction in stop requires a 26% more exposure; a 2/3 stop down requires a 59% more exposure; a full stop closure requires a 100% more exposure. These concepts have been around for about 150 years. Also; the legal definition of wire gauges is by diameter; in larger sizes circular mils is used. <BR><BR>The basic flaw many people have in this thread; is they dont know what a cube root is. This is quite sad; we learned this in 6th grade in Indiana; my teacher probably made 1k per year; and wasnt even a college graduate. The school was poor; and built in 1909. They used a paddle; and ran a tight ship. People today are chicken to give teachers any power; the lawyers have ruined education; by pure fear and lawsuits.<BR><BR>The 3 geometric increments from going from 1 to 2 by 3 parts are roughly 1.26 ; 1.59; 2.00. This was known probably several thousand years ago; but seems to confuse some here. The progression is the cube root of 2; which is the number; multiplied by itself three times equals 2.00. This number is about 1.259921050; squaring this number is 1.587401051; cubing it is about 2.00000.<BR><BR>A 1/3 stop increase is a 26% increase; a 2/3 stop increase is a 59% increase; a full stop is a 100% increase; this is old stuff.<BR><BR>A doubleing in area in wire cross section is 3 full wire guages; a doubling in area of a lens's opening is only 1 stop . 30 guage is about 0.010 diameter; 27 guage is about 0.0142 diameter; this 3 guage jump has double the cross section; since area varies as the square of the diameter.<BR><BR> 1/3 stop Differences in exposure; are like 1 guage differences in wire guage.<BR><BR> 1/3 stop difference is a 26% difference in exposure; a 1 guage difference in wire guage; is about a 13% difference in diameter; or a 26% difference in area; the same as the 1/3 stop in exposure. <BR><BR>Being a registered Electrical Engineer; these analogies are second nature; the cube root of 3 has been a part of electical Engineering for well over 100 years; the jumps of 1; 1.26, 1.59., 2.0 are etched in most electrical Engineers who have designed solenoids; and dc motors.

Okay Travis and Anatole, here is the math with a Lieca example to keep this ON topic ;)

 

The area of a circle is pi x r^2. A Noctilux�s aperture is 50mm in diameter at f1 (50mm/50mm = f1). Its radius is 50mm/2 or 25mm, so its area is 25^2 x pi or roughly 1963 sq. mm. At f1.4 its diameter is 50/1.414 or 35.36mm. Its radius is now 35.36/2 or 17.68mm. The new area of the circle is 17.68mm^2 x 3.141 or roughly 982sq mm �Which is essentially EXACTLY 1/2 the original 1963sq mm area. One down...

 

Now, 1/3 of a stop over f1.4 has to add in 1/3 more area. 982sq mm x 0.333 = 327 sq mm. 327sq mm + 982sq mm = 1309sq mm. The radius^2 of that circle equals 1309sq mm/3.141 or 416.746sq mm. The sqrt of that number is (416.746sq mm)^1/2 = 20.41mm. SO the new diameter of the aperture for f1.4 =1/3 stop is 20.41mm x 2 or 40.82mm. (BTW, this numerical value of this new aperture would be 50/40.82 or f1.23.) Two down.

 

The difference between f1.4+1/3 at 40.82mm and f 1.4 at 35.36mm is 1 � (35.36mm/40.82) = .134 or 13.4%. NOTE THAT THIS CHANGE IS ESSENTIALLY EQUAL TO HARRY�S EARLIER CALCULATION FOR 1/3 STOP CHANGES AT (sqrt(2) ^ 1/3) -- I obviously suffered from rounding error by to only using 2 decimal points in my above arithmetic. Three down.

 

So, I did the math, and it proves what I have been saying all along. I stand by my claim that you increase the AREA of the aperture by 33% when you increase an exposure by 1/3 stop!

 

Cheers,

13.4%...how does this correlate with your 33% that you claimed?

Like Kelly said, 13% change in diameter<=> 26% change in exposure.<=>1/3 stop.

1309/982=1.33=33% <=> areas of aperture at (f1.4+1/3 ) area of aperture at(f1.4).

 

But you are ASSUMING THAT for a 1/3 stop increase, the area increases by 1/3!! from the start.

 

"Now, 1/3 of a stop over f1.4 has to add in 1/3 more area..."--- Jack.

 

 

How can you assume that to be true when you are supposed to proved it in the first place?!

 

cheers.

Kelly, you posted as I was writing. Yes I understand that wire sizes go up by the cube root. And I do understand that if you want to multiply by some number three times to get to the next one you have to use cube roots.

 

And, a 1/3 increase in the aperture's area only requires a 13% increase in the aperture's diameter. And yes, the f-NUMBER changes by 13% too when you increase the exposure by 33%, but this is because it is a number derived using the aperture's diameter. In either case the area has increased by 33%. See the math above.

 

You do not increase the area by 26% and gain 33% more transmission, and you do not increase the area by 13% and gain 33% more transmission -- aperture areas to transmission percentages are a direct linear relationship; double the area, double the light; quadruple the aperture area, quadruple the light; increase the aperture area by 33%, increase light flow by 33%.

 

As I said from the start, f-stops are based on the aperture's diameter, while light tranmission is based on the aperture's area.

 

You guys keep arguing percents the DIAMETER changes or the percent the f-number changes when the percent the AREA is changing is what is relevant to the original question: "Subject: quiz: if you increase by 1/3 stop, how many % of total exposure are you increasing?" Answer: A 1/3 stop increase is a 33% increase in the exposure value, and also a 33% increase in aperture area. And yes, the 33% exposure increase only requires a 13% increase in the aperture's diameter, and the f-number only changes by 13%, but these numbers have no direct relation to the increase in the exposure VALUE -- they only represent the new diameter and the new f-number.

Travis wrote: "Like Kelly said, 13% change in diameter<=> 26% change in exposure.<=>1/3 stop."

 

I'll correct you slightly and say a 13% change in diameter increases area by 26%. BUT this is only true when you are going UP in number...

 

To INCREASE light flow through an F-stop we have to use a larger aperture, which is a SMALLER f-number... (This is not an assumption, this is standard photographic knowledge ;) ) And guess what happens next? We DIVIDE the LARGER number by 1.13 to get the new SMALLER f-number. And now if you do the math, this 13% smaller f-number will generate an aperture diameter that is 13% smaller and an aperture area that is 33% LARGER than your original -- NOT 26% larger!

 

Cheers,

My last sentence should read: "And now if you do the math, this 13% smaller f-number will generate an aperture diameter that is 13% LARGER and an aperture area that is 33% <b> LARGER </b> than your original -- NOT 26% larger!"

...and I bolded the wrong LARGER -- sighhhhh. I'm tired and I'm going to bed. Hopefully I now have you all straightened out on the math :)

area=pie (radius^2)

area= pie (Diameter/2)^2

 

Therefore,

area proportional to Diameter^2

F stop proportional to Diameter

Therefore,

area proportional to (F stop)^2

 

 

Light transmission proportional to area

therefore, Light transmission proportional to F stop^2

 

Increase/decrease in F stop will not equal Linear decrease/increase in area/light transmission.

 

AM I right so far?

 

So by your LINEAR definition Jack, 1/3 stop increase=33%, 1/2 stop=50%, 1 stop=100%. Isn't this Linear?

 

How can you say Light transmission is only based on area when in fact area is based on diameter^2 and diameter is proportional to f stop ?

 

cheers.

Area is directly proportional to the change in radius. A=PI.R^2. Everything else is a constant. So, how does area change by 33% with a 13% change in radius or diameter? Area changes by 26% with a 13% change in radius or diameter.

 

"Now, 1/3 of a stop over f1.4 has to add in 1/3 more area." This is flawed. You have made an assertion without proving it. No, it does not follow that if 1/2 the area of a disc equals 1 stop change then 1/3 stop equals 33% change. You have proved nothing. So, nothing down. And this is what we've been trying to prove: that 1/3 stop change is NOT 1/3 more area but 26% more area.

 

Start with variables like X and Y and prove them for all cases. If you do not, then that is sloppy mathematics, sloppy physics and sloppy logic.

 

Remember that mathematics has to be consistent and hold for all cases.

You cannot pull figures out from the air and say that that is the truth.

 

I have proved it for myself and as a mathematician, I damn well should as this is basic mathematics.

Area=pi(R^2)

If R increases by 13%, new r=13/100R +R=113/100R.

 

New area therefore = pi (113/100R)^2=pi(1.2769R^2)

 

Change in area= 1.2769-1= 27.7%

 

Am I right?

Travis:

 

Yes, from a math point ov view, you're right. 1/3 stop increases the diameter of the aperture by (approximately) 12.2% (which is the sixth root of 2). And mathematicians prefer to approximate to the closest value, and to be explicit whenever they approximate.

 

In physics, though, (and hence in optics) it's usual to factor in an implicit approximation and to say that the implicit precision interval of any expressed result is a variation of plus or minus one on the last digit. i.e. 13% is a valid approximation of the sixth root of 2 for a physicist, so are 12%, 12.2% and 12.3%, but not 13.0%.

Ok, where did this 13% thing first come about in this thread? I believe Kelly was the first to mention this 13% increase in radius/diameter<=> 26% increase in area/exposure for a 1/3 stop

 

My question is, isn't this working backwards? From 26% increase in area, you derive 13% increase in diameter/radius?

 

Kelly, how did you come out with the 13% thingy? ;)

Ah, Jean-Baptiste; your sly dig at mathematicians is duly noted and not forgotten :p Just using the numbers provided by the respondents so as not to confuse the issue further, significant decimal places be damned.

 

It is clear that you can drag a horse to the edge of the water but you cannot make it drink. Oh well, more power to Jack and his logic then. Let's hope that he is better at photography than mathematics.

 

Before you try to prove an erroneous point, Jack, please consult a mathematics teacher or textbook. Your argument and assumption are obviously unsound, trust me on this one. Do not make yourself look anymore foolish by arguing.

Anatole,

 

I don't suppose you'd be willing to share the proof you have. Not that i doubt you, i'm just interested. If you think it will bore the living crap out of everyone here then you can email it to me.

 

Thanks,

 

Moiz