Camera Meter vs Hand held Meter

What metering mode were you using in the camera.
Also, is your Polaris meter a spot meter, a reflective meter, or average meter.
Just because you were standing in one place doesn't mean your meters were reading the same areas.

René Damkot wrote in post #5953690
First off: I agree with PacAce: The second shot is a bit blown, the first is way overexposed (the flower that is).

Second: From f/11 to f/16 is one full f stop. So 1/125; f/11 at ISO 200 equals 1/60s f/16 at ISO 200.

I'm not sure what you are trying to calculate here, but as said, f/11 is one stop faster then f/16.
Each aperture stop is a factor of SQR 2 (1.41 and a bit) bigger then the next:
f/1; f/1.4; f/2; f/2.8; f/4; f/5.6; f/8; f/11; f/16; f/22 and so on.

I think everyone can see based on my calculations that I know how f stops work.

I simply took the core concept or calculating the difference by numbers (you take your focal length i.e.: 85mm and divide by your f-stop).

I did this for both 85/16 and 85/11 then I obtained the difference.

You can open both pics in ACR and adjust the overall exposure (which may affect teh red) channel and find that +.75 is what it will show.

I did the math and came up with +.65 as the difference.

Unless I missed something an f stop is a ratio @ a focal length.

So f/16 on an 85mm lens means 85/16 = 5.31
85/11 = 7.73....
....you can refer to my previous post in this thread since I've already calculated the numbers there.

Leo- One can argue that messing around too much with recovery, fill and the other things in ACR will affect the color of the subject.

Taking red for example and lowering the exposure, etc. can result in a color that is quite a bit off from red and is starting to show hints of burgundy.

Alexajlex wrote in post #5954298
So f/16 on an 85mm lens means 85/16 = 5.31
85/11 = 7.73....
....you can refer to my previous post in this thread since I've already calculated the numbers there.

And what does this have to do with the price of soybeans on the Chicago options trading market?! Care to elaborate as to why you think this calculation has a bearing in the discussion?

Alexajlex wrote in post #5941654
That is true and of course some variations would be needed.

Something like .64 would probably work better on paper.

In reality +.64 exposure will clip red a bit so .60 would work a bit better.

I'll use 85 as an example focal length.

85mm/16 = 5.31
85mm/11 = 7.73

7.73 /5.31 =1.46

Verify
5.31 * .46 = 2.44 + 5.31 = 7.75

100-46 = 64

+.64 stop light.

Are these paper gamma factors?

Alexajlex wrote in post #5941569
Well if it was bright daylight then "Sunny 16" would be ISO 100 f16 @ 1/125

@ ISO 200 the Sunny 16 would be 1/200

Given the fact that you were already @ f16 ISO200 1/125 and it was nearly 1 stop under exposed I'd say that it was a bit less light than the regular Sunny 16 (more like the slight overcast).

Strictly talking about the flower exposure the Polaris handheld meter shot @ f16 needed +.75 stop increase to bring it to proper exposure.

To me the 40D did a outstanding job.
The Polaris underexposed by nearly 1 stop (~.75).

For f16 to work @ ISO200 you'd need 1/80~ shutter speed.

No, Sunny 16 would have ISO 100 = 1/100 f/16, not 1/125. Yes at ISO 200 Sunny 16 is 1/200 f/16.

PacAce wrote in post #5954073
Neither. In the first picture, the red flower is overexposed so it'll look much lighter and a different shade of red than what the eyes would see. In the 2nd picture, the red is truer to life but, because of the exposure, darker than what the eyes would see. The eyes has a wider dynamic range so both the flower and the green foliage would look brighter than what's recorded in the 2nd image. But, having said that, the 2nd image can be post processed to come closer to looking like what the eyes would see than the 1st image can be.

That was kind of the direction I was going. I agree, neither look Exactly like what you see, but which one is closer and would require less PP. I am not a foliage shooter, so I dont have any experience in exposure or metering of flowers, but based on others and what I have read here before, they are extremely difficult to meter and often times require PP.

Wilt wrote in post #5954371
No, Sunny 16 would have ISO 100 = 1/100 f/16, not 1/125. Yes at ISO 200 Sunny 16 is 1/200 f/16.

1/100 1/125 ~ it all varies (that is why most places show 1/100 and 1/125 in parenthesis). Some place like luminouslandscape uses 1/250 @ f16 @ ISO 200 so at at ISO100 it would be 1/125.

http://www.luminous-landscape.com …/understandexpo​sure.shtml

Slight variations occur in light every second. That is why I believe most resources will say 1/100 or 1/125).

Wilt wrote in post #5954324
And what does this have to do with the price of soybeans on the Chicago options trading market?! Care to elaborate as to why you think this calculation has a bearing in the discussion?

The calculations I put there are what I was able to get when I took the pic posted @ f16 in ACR.

I simply moved the exposure slider up until I noticed the red overexposure lights start to show up on the flower.

I also looked at the reds in the histogram.

It appeared that +.75 was when clipping was occurring (the histogram indicated the same by having the reds push up against the right side and starting to form the upward straight line that occurs when clipping shows).

I then did some math to see what the "numbers" would say.

According to the math +.64 was a good value to work with.

I plugged this into ACR and it turns out that it is a good value (naturally since it is less than .75).

Alexajlex wrote in post #5954590
1/100 1/125 ~ it all varies (that is why most places show 1/100 and 1/125 in parenthesis). Some place like luminouslandscape uses 1/250 @ f16 @ ISO 200 so at at ISO100 it would be 1/125.

http://www.luminous-landscape.com …/understandexpo​sure.shtml

Slight variations occur in light every second. That is why I believe most resources will say 1/100 or 1/125).

Easy to explain...film cameras before electronic shutters never had shutter speeds at 1/3 EV increments! So the convention was to merely round to the closest available shutter speed...1/125, 1/250, etc.

Alexajlex wrote in post #5954661
The calculations I put there are what I was able to get when I took the pic posted @ f16 in ACR.

I simply moved the exposure slider up until I noticed the red overexposure lights start to show up on the flower.

I also looked at the reds in the histogram.

It appeared that +.75 was when clipping was occurring (the histogram indicated the same by having the reds push up against the right side and starting to form the upward straight line that occurs when clipping shows).

I then did some math to see what the "numbers" would say.

According to the math +.64 was a good value to work with.

I plugged this into ACR and it turns out that it is a good value (naturally since it is less than .75).

But the question was where these values (0.75, 0.64, etc) have relevance...you are dividing FL by f/stop, so those values are the physical size of the aperture in millimeters, and are no different in their relevance than the native f/stop value (f/11, f/16). They (millimeters) do not relate directly to density values resultant from exposure to light.

If I were discussing Miles Per Gallon, a similar irrelavant computation might be to calculate Gallons / Deciliters of Oxygen consumed in the combustion process. Yes, it may be a valid computation, but what does the computation have to do with how far you go on a gallon of gas?! So I am trying to understand why you think 85/11 vs 85/16 is significant??

Wilt wrote in post #5954915
But the question was where these values (0.75, 0.64, etc) have relevance...you are dividing FL by f/stop, so those values are the physical size of the aperture in millimeters, and are no different in their relevance than the native f/stop value (f/11, f/16). They (millimeters) do not relate directly to density values resultant from exposure to light.

If I were discussing Miles Per Gallon, a similar irrelavant computation might be to calculate Gallons / Deciliters of Oxygen consumed in the combustion process. Yes, it may be a valid computation, but what does the computation have to do with how far you go on a gallon of gas?! So I am trying to understand why you think 85/11 vs 85/16 is significant??

Irrelevant.

You know what man if you are going to call the fact that an f-stop is a ratio (focal length divided by the f-stop) then we got our answer.

I'd really like to hear this new f-stop theory you may have since you seem to challenge the current one that everyone accepts.

You can't really seriously expect us to believe that if I divide the focal length by the f stop I don't have the essence of what an f-stop means.
I just took the issue at its core.

F stop is a ratio.

According with what I know you divide the focal length (I chose 85mm) by the f-stop.

I Simply divided 85/16 and 85/11

You can call it irrelevant of whatever you want.

The fact still stands that I came within .10 of determining when clipping would occur in ACR.

If you are disputing that then all you have to do is to load the image in ACR.

Unless you are somehow saying that ACR is wrong.

Alexaja, the computation you did was to find the aperture diameter, rather than its normal expression as a ratio. So if you used a different lens in the exact same lighting condition (e.g. 200mm f/11 vs. 200mm f/16) would that computation be anywhere close to the number you found (which is by coincidence alone!) the number in ACR -- and what is the number in ACR supposed to represent?!

Wilt wrote in post #5955103
Alexaja, the computation you did was to find the aperture diameter, rather than its normal expression as a ratio. So if you used a different lens in the exact same lighting condition (e.g. 200mm f/11 vs. 200mm f/16) would that computation be anywhere close to the number you found (which is by coincidence alone!) the number in ACR -- and what is the number in ACR supposed to represent?!

Yes the number would be the same

200/16 = 12.5

200/11 = 18.1

18.1/12.5 = 1.45

100-45 = 65

.65 (came within .01 since the other number was .64).

I don't really see why you are arguing this part since we are dealing with ratios the final numbers will be the same as long as the ratios are constant.

I can't really invest any more time in this.

I've proven the fact that I can get within .01 of the initial numbers with the new numbers you provided.

You asked what the numbers represent.

The numbers represent the light difference between these 2 apertures (f11 and f16) at a give focal length (85mm).

They formula works (why wouldn't it is just a ratio) even with the 200mm focal length you provided.

Alexajlex, I am not 'arguing'. I am trying to understand the logic of thinking and the scientific basis behind it. Yes, the f/stop is a ratio. So if I arbitrarily choose a 30 lens and f/8 vs. f/11,
30/8 = 3.75 and
30/11 = 2.72 and
3.75/2.72 = 1.37 and
100-37 = 63
So now what what? What does 0.63 represent, I do not use ACR and am trying to understand what relationship there supposed is being proven.


It is a given that every whole f/stop lets in twice (or half) as much light as the others, which makes the aperture either 1.414x or 0.707x as large. So what is the relationship to these ACR values (which units are those?)