troubleshooting – Is there a focusing issue with the Sigma 17-50mm f/2.8 on a Canon 70D?

The first sample image in the question is focused well in front of the central pillar. The second is focused well behind the flowers.

You have told us that you are using single point AF but you haven’t told us which AF mode you are using: One Shot, AI Servo, or AI Focus? If you’re trying to focus and recompose using AI Servo the camera will refocus when you move the camera to point at a different spot. If you use AI Focus the camera will initially hold focus as it would in One Shot mode, but if you recompose and hold the camera too long in the new position the camera will sense that your selected AF point is no longer in focus and will switch over to AI Servo.

There appear to be other issues at work that might also be contributing to your results:

Diffraction The 70D is a Canon APS-C camera with 4.1µm pixel pitch. The Diffraction Limited Aperture (DLA) of the 70D is f/6.6. This is the point at which the effects of diffraction begin when viewed at the pixel level. As apertures are narrowed beyond the DLA the results get more and more noticeable at normal viewing sizes. The best way to avoid this is to shoot at around f/8 or wider and at f/6.3 or wider if possible.

Camera movement Not everyone can hold a camera steady enough to use the 1/focal length rule-of-thumb, even when viewing at the standard 8×10 sizes for which it applies. You may get useable results for viewing at smaller sizes, but nowhere near the equivalent viewing size of looking at part of an image at 100% on your monitor. If you have an HD (1920×1080 pixels) monitor that measures 23″ diagonally you are viewing images at 96 ppi. That means an 18MP image viewed at 100% is being magnified at the equivalent of 54×36 inches! That’s 5X the magnification of the standard 8×10 print.

The optical limits of your lens I’d like to know where you read excellent reviews of this lens. I’ve never seen any critical reviews from reputable reviewers written about it that impressed me very much. Before you can blame AF you need to be sure that something else isn’t causing your images to be blurred. To do that you need to eliminate as many of the other possible causes as you can.

  • Mount your camera on a stable tripod, turn off optical image stabilisation, and use a cable release or the self timer to release the shutter. This will help eliminate camera movement as the source of your problem.
  • Shoot under bright enough constant lighting that your shutter speed at ISO 100 can be 1/100 second or faster. Use the fullest spectrum lights available to you. This will further help to eliminate camera movement including vibrations caused by the movement of the camera’s mirror. Properly exposing using low ISO will also help eliminate poor image quality caused by a low signal-to-noise ratio and the resulting noise reduction.
  • Use a flat target that is lined up parallel with your camera’s image sensor and perpendicular to the optical axis of the lens. An easy way to do this is to aim your camera at a flat, stable mirror. Center the viewfinder on the center of reflection of the lens in the mirror. Then tape your focus target onto the mirror being careful not to move the mirror.
  • Use careful manual focus with magnified Live View. Take several samples while refocusing manually between each sample.
  • Repeat the test shots using One Shot AF mode with the single center focus point selected. Move the lens to infinity or minimum focus between each test shot. Use a half shutter press with your cable release to allow the AF to confirm focus before taking the photo.
  • Compare the best of the manually focused shots to the best of the AF shots.

If there is a significant difference then you have an AF issue. If there is not a significant difference then your problem lies elsewhere.

Comment from the OP:

I still doubt that there is a focus issue. I have been using Single shot AF in almost all pictures. And shooting at 4-5 times of 1/focal length at ISO less than 800 I doubt the hardware. Of course its difficult to doubt on ones own abilities! 🙂

Look at the examples you posted carefully at 100% magnification. You can tell by the cobblestones in the first image that focus was missed (based on what you said you attempted to focus). The second image is focused well behind the flowers in the foreground. The clock tower in the background is the most in focus area ofthe image. There are areas in both images that are in focus, they’re just not where you wanted to focus.

sigma – How do you add lens profiles to Adobe Lightroom?

There is a readme.txt at both



C:Program FilesAdobeAdobe Lightroom ClassicResourcesLensProfiles1.0ThirdParty

that reads

Install third-party (non-Adobe) lens profiles here.

By default, .lcp files are saved to C:UsersbensoAppDataRoamingAdobeCameraRawLensProfiles1.0

I left my original file in AppData and then copied that file to both these locations. I didn’t investigate which location Lightroom is dependant upon, but Adobe keeps identical lens profile directories on my computer; adding to their file clutter worked for me.

ct.category theory – In the category of sigma algebras, are all epimorphisms surjective?

Consider the category of abstract $sigma$-algebras ${mathcal B} = (0, 1, vee, wedge, bigvee_{n=1}^infty, bigwedge_{n=1}^infty, overline{cdot})$ (Boolean algebras in which all countable joins and meets exist), with the morphisms being the $sigma$-complete Boolean homomorphisms (homomorphisms of Boolean algebras which preserve countable joins and meets). If a morphism $phi: {mathcal A} to {mathcal B}$ between two $sigma$-algebras is surjective, then it is certainly an epimorphism: if $psi_1, psi_2: {mathcal B} to {mathcal C}$ are such that $psi_1 circ phi = psi_2 circ phi$, then $phi_1 = phi_2$. But is the converse true: is every epimorphism $phi: {mathcal A} to {mathcal B}$ surjective?

Setting ${mathcal B}_0 := phi({mathcal A})$, the question can be phrased as follows. If ${mathcal B}_0$ is a proper sub-$sigma$-algebra of ${mathcal B}$, does there exist two $sigma$-algebra homomorphisms $phi_1, phi_2: {mathcal B} to {mathcal C}$ into another $sigma$-algebra ${mathcal C}$ that agree on ${mathcal B}_0$ but are not identically equal on ${mathcal B}$?

In the case that ${mathcal B}$ is generated from ${mathcal B}_0$ and one additional element $E in {mathcal B} backslash {mathcal B}_0$, then all elements of ${mathcal B}$ are of the form $(A wedge E) vee (B wedge overline{E})$ for $A, B in {mathcal B}_0$, and I can construct such homomorphisms by hand, by setting ${mathcal C} := {mathcal B}_0/{mathcal I}$ where ${mathcal I}$ is the proper ideal
$$ {mathcal I} := { A in {mathcal B}_0: A wedge E, A wedgeoverline{E} in {mathcal B}_0 }$$
and $phi_1, phi_2: {mathcal B} to {mathcal C}$ are defined by setting
$$ phi_1( (A wedge E) vee (B wedge overline{E}) ) := (A)$$
$$ phi_2( (A wedge E) vee (B wedge overline{E}) ) := (B)$$
for $A,B in {mathcal B}_0$, where $(A)$ denotes the equivalence class of $A$ in ${mathcal C}$, noting that $phi_1(E) = 1 neq 0 = phi_2(E)$. However I was not able to then obtain the general case; the usual Zorn’s lemma type arguments don’t seem to be available in the $sigma$-algebra setting. I also played around with using the Loomis-Sikorski theorem but was not able to get enough control on the various null ideals to settle the question. (However, Stone duality seems to settle the corresponding question for Boolean algebras.)

Canon – EOS-M4 / 3 adapter glued to sigma lens?

I just bought a new Canon camera and found that my Sigma 24mm 2.8 Superwide lens is compatible with it because it is an EF mount. The lens itself was given to me second hand and I had just used it for M4 / 3rd cameras, so I never had to remove the adapter. I'm trying now, but it's not moving at all. Am I just an idiot and is it all part of the lens? This & # 39; adapter & # 39; acts as an aperture, since it has a dial to adjust. I have attached pictures to see if anyone can help / give advice !?

PS: Is there a small hole in which the adapter release button may have been?

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The Canon EOS 6D Mark II with Sigma lens looks strange on the LCD monitor

Based on your comment, you have a Sigma 17-70mm f / 2.8-4 DC Macro OS HSM lens, an APS-C lens, on a full frame camera. An APS-C sensor is smaller than a full-frame sensor, so the lenses developed for these cameras project a smaller image circle. APS-C specific lenses are not designed for use with full frame cameras. (For more information on sensor sizes, see this post.)

Canon's APS-C-specific lenses (with EF-S in the model number instead of EF) have a different mounting flange to prevent you from attaching them to full-frame cameras like your 6D, but almost all APS-C lenses third-party mount in the EF mount is mounted on Canon FF cameras.

Product descriptions of the lenses usually (ideally) always) State fairly clearly whether it is a crop sensor or full-frame camera. You can use this information to determine if a lens is suitable for your camera. If the description of a lens indicates that it is suitable for APS-C, Crop-Senor or only for digital, it is an APS-C-specific lens that was not developed for your camera. If it is full screen, this is the case (see note 1 below).

Here is some information from Michael C's answer to this question and from a community wiki for decoding lens model numbers to help you find the right lens types in the future:

Look for it

  • Sigma lenses: model numbers labeled DG (Not DC)
  • Canon lenses: model numbers starting with EF (Not EF-S or EF-M)
  • Tamron lenses: model numbers that denote the Di (Not Di II or Di III)
  • Tokina lenses: model numbers that contain the identifier FX (Not DX)
  • Samyang lenses: model numbers that do not contain the identifier CS

If you shop on a website like B&H Photo, you can only use the filter function to display lenses that were developed for your full-screen camera:

Screenshot of the check boxes for the lens cover filter from B & H.

Note 1: You may see a product description for a full frame lens that describes the effective or equivalent focal length of the lens on a crop sensor camera. This is because reversing your situation – using a full frame lens with a crop sensor camera – is perfectly fine and fairly common.

So what now?

Due to the fact that you linked to B & H in your comment, I assume that you bought your lens from B & H Photo. You have a pretty good return policy, so you may be able to return this lens and use the credit for a full-frame lens. Hopefully, if you bought both your lens and camera from there, they will understand

If for some reason you can't return the lens, you can still use it as it seems to at least work properly on your camera.

  • You can use it to take photos and then use software such as Photoshop, GIMP, etc. to cut out the unexposed portions of the photos. In fact, you would turn your full-frame camera into a crop sensor camera. You will lose some image size, but it is not a total loss.
  • You can use this as an artistic effect and take photos where the circular view is part of the art.
  • Use it to take pictures of crop circles. Like meta.

Large double onion ring in the middle of the shoots when I use a Sigma 35mm f1.4 DG HSM Art lens on my Canon 6D Mark II

I bought a Sigma 35mm f1.4 DG HSM Art lens, it works well on my APS-C Camara, but doesn't work with My Canon 6D Mark II (large double onion ring in the middle of the shots). Is this lens meant? work on both? Please let me know.

Thanks a lot

Limitation of $ | Sigma ^ {- 1/2} (X- mu) | _2 ^ 3 $ for two-dimensional Bernoulli

To let $ X in {0,1 } ^ 2 $ have in common $ mu = left ( begin {smallmatrix} p_1 \ p_2 end {smallmatrix} right) $ and $ Pr (X_1 = X_2 = 1) = p $. We can then calculate the covariance matrix $ Sigma = E ((X- mu) (X- mu) ^ T) = left ( begin {smallmatrix} p_1 (1-p_1) & p-p_1p_2 \ p-p_1p_2 & p_2 (1- p_2) end {smallmatrix} right) $.

I want to use the berry-food limit, and for that we have to limit the amount $ gamma = | Sigma ^ {- 1/2} (X- mu) | _2 ^ 3 $.

I think you should be able to show it
$$ gamma le C left ( tfrac1 { sqrt {p_1 (1-p_1)}} + tfrac1 { sqrt {p_2 (1-p_2)}} + tfrac1 { sqrt { min { p_1, p_2 } – p}} right)

for a universal constant $ C> 0 $.

The symbolic calculation of $ Sigma ^ {- 1/2} $ is a bit unwieldy, however, and so I wonder if there are some tricks that can help me achieve this result better?

If not, any evidence would be welcome.

Statistics – Find the variance of the residuals for $ y_i = beta_ {i} x_i + sigma epsilon_i $

Given the simple linear regression $ y_i = beta_ {i} x_i + sigma epsilon_i $I have to find out the variance of the residuals.

I found out $ has { beta} = frac { sum_ {i = 1} ^ {n} x_iy_i} { sum_ {i = 1} ^ {n} x_i ^ 2} $ and $ E (r_i) = 0 $. I tried to apply the definition of residuals $ r_i = y_i – has {y_i} $ and $ hat {y_i} = x_i hat { beta} $. This is $ Var (r_i) = Var (y_i – hat {y}) = Var (y_i) + Var ( hat {y}) – 2Cov (y_i, hat {y}) $. But I'm having trouble figuring it out $ Cov (y_i, hat {y}) $

ag.algebraic geometry – Cremona transformation $ sigma: mathbb {P} ^ 2 dashrightarrow mathbb {P} ^ 2 $ and push forward of the divisor

Let us consider the birational Cremona transformation $ sigma: mathbb {P} ^ 2 dashrightarrow mathbb {P} ^ 2 $ defined by $$ (X: Y: Z) mapsto (X ^ {- 1}: X ^ {- 1}: Z ^ {- 1}) = (YZ: XZ: XY) $$.

The sections in the language of birational geometry $ YZ, XZ, XY in H ^ 0 ( mathbb {P} ^ 2, O _ { mathbb {P} ^ 2} (2)) $ induce it as

$$ sigma: mathbb {P} ^ 2 backslash B = mathbb {P} ^ 2_ {YZ} cup mathbb {P} ^ 2_ {XZ} cup mathbb {P} ^ 2_ {XY} to mathbb {P} ^ 2 $$.

with the Base location $ B = operatorname {Supp} V (YZ) cap operatorname {Supp} V (XZ) cap operatorname {Supp} V (XY) $ Where $ sigma $ is not defined and $ mathbb {P} ^ 2_ {YZ}: = {p in mathbb {P} ^ 2 vert (YZ) _p neq 0 } $ the non-vanishing place. The same applies to $ XZ $ and $ XY $.

By construction $ mathbb {P} ^ 2_ {YZ} = sigma ^ {- 1} (D_ + (X)), mathbb {P} ^ 2_ {XZ} = sigma ^ {- 1} (D_ + (Y .)), mathbb {P} ^ 2_ {XY} = sigma ^ {- 1} (D_ + (Z)) $. Obviously $ sigma $ is not defined at points $ (0: 0: 1), (0: ​​1: 0), (1: 0: 0) in B $.

So far we have introduced some notations; now the essential part of my question::

To let $ V (F) = C subset mathbb {P} ^ 2 $ be a curve defined by a homogeneous polynomial $ F (X, Y, Z) = sum_ {i + j + k = n} a_ {ijk} X ^ iY ^ jZ ^ k $ With $ deg (F) = n $.

We assume that $ C $ got to the point $ (0: 0: 1) $ Variety $ d $, $ (0: 1: 0) $ Variety $ e $ and $ (1: 0: 0) $ Variety $ f $.

Remember how multiplicity is defined: $ (0: 0: 1) in D _ + (Z) $ and if we dehomogenize $ F $ With respect $ Z $ we get a new polynomial $ f ^ Z $ in variables $ x = X / Z, y = Y / Z $ that has structure

$$ f ^ Z (x, y) = sum_ {i + j + k = n} a_ {ijk} x ^ iy ^ j = \ sum_ {i + j + k = n, i + j = d } a_ {ijk} x ^ iy ^ j + sum_ {i + j + k = n, i + j> d} a_ {ijk} x ^ iy ^ j =: f ^ Z_d + f ^ Z _ {> d } $$

Remember again that by definition of multiplicity $ d $ is a minimal natural number with this property: for all $ a_ {ijk} neq 0 $ we have $ i + j ge d $. That’s the same thing $ d $ is at most with the property that the $ d $-th power $ (x, y) ^ d $ of the maxiaml ideal $ (x, y) subset k (x, y) $ contains $ f ^ Z (x, y) $.

Analogously for multiplicities $ e $ and $ f $ in memory of $ (0: 1: 0) $ and $ (1: 0: 0) $.

QUESTION: Why the push forward from $ C $ by $ sigma $ is the disappearing set

$$ sigma_ * C = (X ^ {- f} Y ^ {- e} Z ^ {- d} F (YZ, XZ, XY) = 0) ? $$

For reasons of symmetry, it is clear that it is sufficient to show locally that this applies to the restriction to an affine open subschema $ D _ + (Z) = operatorname {Spec} k (X / Z, Y / Z) subset mathbb {P} ^ 2 $.

We know that from construction

$$ sigma_ * C (D _ + (Z)) = C ( sigma ^ {- 1} D _ + (Z)) = C ( mathbb {P} ^ 2_ {XY}) = C (D _ + (XY))) = C vert _ {D _ + (X)} (D (y))
= k (y, z) _y / f ^ X (y, z). $$

Recall, $ k (y, z) _y = k (y ^ { pm 1}, z) $ is the localization
by $ y $. If we designate now

$ H: = X ^ {- f} Y ^ {- e} Z ^ {- d} F (YZ, XZ, XY) $ and $ S: = V (H) $, then $ H $ localized
at the $ Z $ is $ h ^ Z (x, y): = x ^ {- f} y ^ {- e} F (y, x, xy) $ and thus

$$ S (D _ + (Z)) = k (x, y) / h ^ Z (x, y). $$

And we have to check $ k (y, z) _y / f ^ X (y, z) = k (x, y) / h ^ Z (x, y) $.

Although I've only used definitions (right?), The equation I need to check doesn't seem to make sense. I think I've made a serious mistake somewhere, but I can't find it. Can someone help? Is there an easier way to review the claim on the question?

Thanks for your help.