new lens doesn't stop at the biggest F

[edenl]

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hello all 

I recently picked up a brand new 50/1.4 asph black chrome edition. I noticed that when I turn the f ring to wide open (f2) it doesn't mechanically stop at f2 like all my other M lens do, it can move a bit pass f2(that is to say very little force applied) Was wondering if that is something I need to worry about or it is just normal for a remake like this 😂 It does feel kinda annoying having to dial it back a tad bit every time I over turn it.

[jaapv]

It should click in at the widest aperture, but it is quite normal to be able to turn the aperture ring a few degrees further. That will not increase the aperture, though.

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[frame-it]

2 minutes ago, jaapv said:

but it is quite normal to be able to turn the aperture ring a few degrees further.

and sometimes in the stop down direction as well! though the OP's post is confusing  is it a Max f1.4 or f2 lens?

[jaapv]

Well, if it is the Summilux (which the asph suggests) it is indeed not surprising that it will turn past f 2.0  

[edenl]

Much thanks for the quick reply! while it won't increase the aperture further, it is my observation that the blades do go few degrees hidden from just a bit protruding at the widest stop. Will it result in a slight change of imaging since the blades are not fully concealed at f 1.4? 

[edenl]

6 minutes ago, frame-it said:

and sometimes in the stop down direction as well! though the OP's post is confusing  is it a Max f1.4 or f2 lens?

it stops perfectly at the smallest stop and can mechanically go no further; it is a sumilux

[Jean-Michel]

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Most of my lenses go past the open f stop detent. Zero effect on imaging. 

[jaapv]

46 minutes ago, edenl said:

Much thanks for the quick reply! while it won't increase the aperture further, it is my observation that the blades do go few degrees hidden from just a bit protruding at the widest stop. Will it result in a slight change of imaging since the blades are not fully concealed at f 1.4? 

No. The diameter of the lens is restricted throughout. 

[Dzongkha]

It is the same on my Summicron 35 v2 at f2 and f16. At the beginning i was wondering a bit but it doesn't effect the image.

[adan]

On 8/14/2022 at 9:47 AM, edenl said:

it is my observation that the blades do go few degrees hidden from just a bit protruding at the widest stop. Will it result in a slight change of imaging since the blades are not fully concealed at f 1.4

1) Leica often slightly overdesigns the diameter of their glass, and then adjusts the aperture blades to accurately produce "f/1.4" or whatever, in each individual lens made.

I suspect this is because there are always slight variations in true focal length (51mm, 51.2mm, 51.5mm, 51.6mm, etc. - see the tiny sideways numbers next to the  m  symbol on your focusing ring).

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For which Leica then corrects the aperture diameter (51mm/1.4 = 36.428mm opening, 51.6mm/1.4 = 36.857mm opening), and also the focusing cam. So that is normal.

2) The protruding blades may change the shape of out-of-focus blur-circles ("bokeh balls"), very slightly, to polygons. That will depend on a lot of things - contrast of the blur edge (intense bright pin-point lights out-of-focus vs. low-contrast midtone blurs or large lights), focused distance, background distance.

[Siriusone59]

4 hours ago, adan said:

1) Leica often slightly overdesigns the diameter of their glass, and then adjusts the aperture blades to accurately produce "f/1.4" or whatever, in each individual lens made.

I suspect this is because there are always slight variations in true focal length (51mm, 51.2mm, 51.5mm, 51.6mm, etc. - see the tiny sideways numbers next to the  m  symbol on your focusing ring).

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For which Leica then corrects the aperture diameter (51mm/1.4 = 36.428mm opening, 51.6mm/1.4 = 36.857mm opening), and also the focusing cam. So that is normal.

2) The protruding blades may change the shape of out-of-focus blur-circles ("bokeh balls"), very slightly, to polygons. That will depend on a lot of things - contrast of the blur edge (intense bright pin-point lights out-of-focus vs. low-contrast midtone blurs or large lights), focused distance, background distance.

Thanks for posting this info, adan.  I've wondered why my 50 cron had a sideways 22 past the m.

[Jean-Michel]

As soon as you stop down a lens with aperture blades that do not make a perfect circle the equation for calculating the open area of the lens goes out the window, if one is insisting on a perfect geometrical result. With fewer than an infinite number blades the opening is no longer a circle and therefore does not have a diameter or radius. For all photographic uses the close approximation of the f stops is perfectly fine. All really a tempest in a teacup. 

For a more precise calculation of the amount of light transmitted one would use the T stop system whereby actual transmission is measured, cine lenses usually use this system.

[adan]

42 minutes ago, Jean-Michel said:

As soon as you stop down a lens with aperture blades that do not make a perfect circle the equation for calculating the open area of the lens goes out the window, if one is insisting on a perfect geometrical result.

Really?

https://www.cuemath.com/geometry/area-of-polygons/

Quote

In order to determine the area of a regular polygon, if the number of its sides are known, is given by:

• Area of regular polygon = (number of sides × length of one side × apothem)/2, where the length of apothem is given as the L/(2 tan(180/n)),

• where L is the side length and n is the number of sides of the regular polygon.
• In terms of the perimeter of a regular polygon, the area of a regular polygon is given as, Area = (Perimeter × apothem)/2, in which perimeter = number of sides × length of one side

Example: Find the area of a regular pentagon whose side is 7 inches long.
Solution: Given the length of one side = 7 inches.
Hence, the area of the regular pentagon is given as A = 1/4×√5(5+2√5)×(side)2

⇒ A = 1/4×√5(5+2√5)×(7)2

⇒ A = 84.3 square inches

Thus, the area of the regular pentagon is 84.3 square inches.

.........

Once you have the area of the opening (which, as you say, is what determines the amount of light getting through), then simply take the old "pi" equation for area of a circle... Area = π*radius-squared,

and solve it for equivalent radius = square-root of Area/π

https://virtualnerd.com/pre-algebra/perimeter-area-volume/circles/circle-sector-area-examples/circle-radius-from-area

If the aperture blades are curved, then adjusting ± for the "area under a curve" is just a bit of basic calculus.

Although admittedly solving for some of Leica's star-shaped apertures is best left to Leica's computers.

https://www.kenrockwell.com/leica/90mm-f28-tele.htm

[Jean-Michel]

Hi,

Well, it proves that I did not do so well in higher math 😀.

However, I still maintain that all of our camera setting are just excellent approximations, be it aperture or shutter times. Last time I had my M3 shutter adjusted (years ago and I still have the hand-written cardboard note from the technician !), the actual shutter times varied quite a bit from the ones on the shutter dial, the 1/1000 sec was 1/850; the only one that was bang on was the 1/125 sec. Who checks the shutter times on their digital Ms, probably no one, and it really does not matter, we figure out how to best expose for the results we look for. And is 'exposure compensation' sometimes just compensating for inaccurate shutter times? Essentially, the markings (f/ or shutter times) are more than good enough.