New Nokton 50mm f/1.1 coming from Cosina

[luigi bertolotti]

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This is getting interesting !

 

My numbers came from these equations:

 

π * R^2 = (4/3) * π * r^2

R = r * 2√3

≈ 1,1547 r

 

π * R^2 = (3/2) * π * r^2

R = r * √(3/2)

≈ 1,2247 r

 

π * R^2 = (5/3) * π * r^2

R = r * √(5/3)

≈ 1,29099 r

 

where R is the bigger radius, r the smaller radius.

 

I don't know how correct this is, but it seems to work with the powers of 2, which in turn give the familiar series 1,4 - 2 - 2,8 ---...

 

The formulas I used also bring to the usual series... for me the real problem of how to master this math is that I haven't ever known what is, in MATH terms, the strict definition of the "measurement unit" known as "STOP", and in particular what exactly means A FRACTION of it.... just to explain :

a) "closing one stop" means "halves the area through which the light enters the camera", i.e., "divide by SQRT(2) the diameter of the iris"

"closing half a stop" means ... what exactly ?

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To state the obvious: the f-stop is the ratio of the aperture diameter to the focal length. Hence, a lens with an aperture of 1:1 and a focal length of 5cm has an aperture (diameter) of 5cm. The diaphragm makers arranged the stops on their scales such that each succeeding stop halved or doubled the amount of light passing through the lens.

 

Now as to the question of the half-stop.

 

Keep in mind that for some time, the sensitivity of the film was reckoned in logarithmic scales where increasing the sensitivity by three units doubled the speed of the film.

 

Hence, one f-stop corresponds to three DINs or Schreiners. It wouldn't be completely inconsistent to call a difference of one and a half DIN "half an f-stop", I think.

[kenneth]

No

 

You'll have to spell it out to me. Wilson's avatar is of him in a helmet in a car

 

Now Andy, you are either a moderator or you are chief whip, you cannot be both. I think you know exactly what I mean

[andybarton]

Now Andy, you are either a moderator or you are chief whip, you cannot be both. I think you know exactly what I mean

 

Kenneth

 

I don't. Honestly. I am not in a position to play games here...

 

Please, tell me what you have in mind.

[kenneth]

Fast lenses are like fast cars, unnecessary but good for male ego, but I guess it gives bragging rights.

 

Andy. As you can see the comment I made above the post received an instant reply from no less than a member with some sort of racing car on his avatar to which I said, I rest my case.

 

With regard to the comment I made, I stand by it and yes money can buy you the fastest most expensive glass in the world but you cannot buy talent and good taste. I remember, in the olden days if one did not have a 50mm 1.4 Nikkor lens bolted onto an F2 one was not cool event though the 1.8 produced superior results for general use

[marknorton]

Certainly food for thought that this lens is 1/10 of the cost of the new Noctilux, roughly.

 

Just waiting for the "My CV 50/1.1 blows the new Noctilux away" threads...

[SJP]

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The formulas I used also bring to the usual series... for me the real problem of how to master this math is that I haven't ever known what is, in MATH terms, the strict definition of the "measurement unit" known as "STOP", and in particular what exactly means A FRACTION of it.... just to explain :

a) "closing one stop" means "halves the area through which the light enters the camera", i.e., "divide by SQRT(2) the diameter of the iris"

"closing half a stop" means ... what exactly ?

It really is very simple - the shutter times are traditionally in a series of 2 (roughly): 4, 2, 1, 1/2, 1/4, 1/8, 1/16, 1/30. 1/60, 1/125, 1/250, 1/500, 1/1000 etc. Similarly the diafraghm values are in a series of (roughly) √2 = 1.4ish, so 1, 1.4, 2, 2.8, 4, 5.6, 8, 11, 16, 22, 32 etc. Each increment is called a stop. The reason for the √2 series with the aperture is that the aperture indicates the relative diameter of the diafraghm, and the surface area is proportional to the square of the diameter ((√2)² = 2 obviously). So changing the aperture by one stop, from 1.4 to 2 for example, reduces the light intensity by a factor 2.

 

For the shutter speeds we therefore get a series like 2^n where n is the nr. of stops, and for the aperture we get (√2)^n. This also then defines what is meant by a fractional stop value. A ½ stop for the shutter speed, say between 1/30 and 1/60, would be a factor 2^½ = √2 = 1.414... so 1/42.426... sec = 1/40 roughly, 1/3 stop would give 2^1/3 and 2^2/3 repectively at 1/38 and 1/48.

 

So how much is f/1.1 away from f/1 in stops?

A factor 1.1 in aperture is a factor 1.1² = 1.21 in intensity, we need to calculate the ²log of that ratio which yields 0.2750070474998698 stop, slightly more than 1/4 stop.

 

Note ²log(x) = ln(x)/ln(2) or log(x)/log(2), so ²log(1.21) = ln(1.21)/ln(2) = 0.191/0.693 = 0.275.

[luigi bertolotti]

It really is very simple - the shutter times are traditionally in a series of 2 (roughly): 4, 2, 1, 1/2, 1/4, 1/8, 1/16, 1/30. 1/60, 1/125, 1/250, 1/500, 1/1000 etc. Similarly the diafraghm values are in a series of (roughly) √2 = 1.4ish, so 1, 1.4, 2, 2.8, 4, 5.6, 8, 11, 16, 22, 32 etc. Each increment is called a stop. The reason for the √2 series with the aperture is that the aperture indicates the relative diameter of the diafraghm, and the surface area is proportional to the square of the diameter ((√2)² = 2 obviously). So changing the aperture by one stop, from 1.4 to 2 for example, reduces the light intensity by a factor 2.

 

For the shutter speeds we therefore get a series like 2^n where n is the nr. of stops, and for the aperture we get (√2)^n. This also then defines what is meant by a fractional stop value. A ½ stop for the shutter speed, say between 1/30 and 1/60, would be a factor 2^½ = √2 = 1.414... so 1/42.426... sec = 1/40 roughly, 1/3 stop would give 2^1/3 and 2^2/3 repectively at 1/38 and 1/48.

 

So how much is f/1.1 away from f/1 in stops?

A factor 1.1 in aperture is a factor 1.1² = 1.21 in intensity, we need to calculate the ²log of that ratio which yields 0.2750070474998698 stop, slightly more than 1/4 stop.

 

Note ²log(x) = ln(x)/ln(2) or log(x)/log(2), so ²log(1.21) = ln(1.21)/ln(2) = 0.191/0.693 = 0.275.

 

Thanks... I was really stupid not to catch the math analogy with music / scales / octaves / cord lengths etc... ... so the correct computation of the scale between

1 and 1,4 was the one posted by Masjah :

 

1/3 stop = 1,1225

1/2 stop = 1,1892

2/3 stop = 1,2599

 

and 1/4 stop is 1,0905 ... 1,1 is, as you say, slightly more than 1/4...and very next to 1/3... pity CV doesn't make the 1,1 in LTM... "old syle times" of IIIc/IIIf could be useful

BTW... from 1 to 0,7 1/5 of stop gives 0,933... the aperture of the new Noctilux is really made just for "beating the wall"...

[wlaidlaw]

... pity CV doesn't make the 1,1 in LTM... "old syle times" of IIIc/IIIf could be useful

BTW... from 1 to 0,7 1/5 of stop gives 0,933... the aperture of the new Noctilux is really made just for "beating the wall"...

 

Luigi,

 

I agree about LTM. It would mean I could use it on my IIF as well. It also makes coding easier with a screw on JM adapter, with the pits already milled.

 

I look forward to Sean's test of a production version.

 

Wilson