## what tolerance level/parameters/targets should be considered while framing cyber security readiness system?

I think the title itself defines my question. Can anyone help me with identifying tolerance level/parameters/targets that should be considered while framing security readiness system?

## framing – How to straighten the Black Stare print

I have printed 11×17 print on Black Styrene (metallic print) couple of years back. It was put on my office wall, now the office is closing so I brought it back home. I have noticed that over time it has bent a little. Is there a way to straighten it without damaging it. Someone from my office really liked the picture so she is asking if she can have it. So, I want to make sure I give a straighten frame than a bent one. Please help.

## dg.differential geometry – Obstruction of “Framing = trivialization of normal bundle”

Is seems that “Framing” is a synonym as “trivialization of normal bundle” in many contexts I saw.

Is this always true? What if there is no way or an obstruction to “trivialize normal bundle” — what does it imply? Does it have any thing to do with the anomaly of a quantum field theory? Which again relates to
the anomaly in physics is the determinant line bundle in math.

Can we put this story into a streamline context (is this correc?t):

Obstruction of “Framing” = Obstruction of “trivialization of normal bundle” = anomaly in physics = determinant line bundle in math?

Ref: Anomaly in QFT physics v.s. determinant line bundle

Merci!

## Sensor – Why is FF sharper than the crop for the same framing of the same object?

It is primarily the difference in the different subject distances and magnification ratios and how they interact with the resolution limits of the objective.

The effects of diffraction and sensor bloom, although measurable under laboratory conditions, are more subtle. If an image is taken below the diffraction-limited aperture for a particular sensor and there are no fully saturated pixels, these effects will be much less shown in the resulting image than at an aperture above the DLA and with a substantial number of fully saturated pixels.

In order to fill the frame with the same flat test chart, the harvesting body must be shot at a 1.6-fold distance, as with the full frame. If you use 10 feet for the entire frame, you must shoot the crop at 16 feet. However, what you do not increase is the number of lines per inch (projected on the sensor) that the lens can resolve. The image of the object cast by the lens is smaller at 16 feet than at 10 feet. Therefore, the resolution limit of the lens is greater with respect to the surface features of the object and the size of each pixel (assuming APS-C) and FF sensor have the same number of pixels).

In order to obtain the same display size, the image must be enlarged from a cut-out body by 1.6 times compared to a picture from a full screen body. For a 4×6 print, the full image only has to be increased by a factor of 4.23, compared to 6.77 for the body part image.

With the longer shooting distance (1.6x) and the larger magnification (1.6x), you'll stretch the resolution limits of the lens more (2.56x). To put it another way, to get the same sharpness as the crop housing, you'll need a lens capable of resolving 1,800 lines per inch to match the full-frame camera with a lens capable of resolving 700 lines per inch !

Even if you have both an 80mm FF camera lens and a 50mm crop box lens so you can shoot at the same distance, you still need the 50mm lens attached to the APS The 700-inch lens used for the FF package is equivalent to about 700 lines per inch, as the result is still 1.6 times higher is increased to get the same display size.

For the sake of simplicity, the following theoretical representation assumes a 1.5 times smaller APS-C sensor than the FF sensor (even if the original question concerns a camera with a 1.6-fold crop factor sensor). ,

Imagine, you have a lens with a theoretical resolution limit of 1000 line pairs per mm. With a 24 mm wide sensor, 24,000 line pairs could be projected. With a 36 mm wide sensor, 36,000 line pairs could be projected. Now take a test chart with 36,000 line pairs, which fills the frame of the FF camera to ten feet. If you move back down to 15 feet to fill the grain camera frame with the same test chart, the 36,000 line pairs in the test chart will exceed the resolution of the lens, as 36,000 line pairs try to fit on it, a 24mm wide sensor.

They do not secure, because the lens is more enlarged when it is attached to a fruiting body. The lens projects the same image in both directions. The reason for the fuse is that the smaller sensor can detect the same framing. This reduces the angle size of the subject in the virtual image actually projected by the lens by a factor of f / 1.5. However, they do not reduce the angle size of the lens' resolution limit by 1 / 1.5-fold by performing a backup,

At a distance of 15 feet from the chart, the angle difference between each line pair is 1 / 1.5 times the angle size when the camera was 10 feet from the chart. However, the lens still has the same resolution limit, which is ultimately based on the angular size of the line pairs in the test chart. Line pairs per mm can only make sense if the distance from the entrance pupil of the lens to the sensor remains constant and if the magnification factor from the virtual image projected onto the sensor remains constant at a certain display size.

You then zoom in on the APS-C image 1.5 times more than the FF image to display both images in the same display size. This means that with the image from the APS-C sensor, we can perceive blur circles (measured on the sensor before enlarging the display) which are 1/1 / 1.5 times the blur circles at the boundary of our perception on the FF -Image. Slightly blurred edges that would look sharp in the FF image may appear blurry due to the larger magnification of the APS-C image.

When the 1.5-fold image of the trimming body of a 24-K line chart taken from 15 & phgr; is printed with 4×6 and that at 10 & # 39; If the captured FF image of a 36k line chart is printed at 6×9, the sharpness should be equal because of the line pairs would be the same width on both prints. However, if you print the 1.5x section image at 6×9, the line pairs (which are at the resolution limit of your lens) will now be magnified 1.5 times. You will not get any additional subject details by enlarging more, because the lens can not resolve these details, which are smaller than the width of the line pairs. At this point, you only reveal blur.

The two effects are multiplied: Retracting the same frame reduces the angle size of the subject detail 1.5-fold. Increasing the size of the picture by 1.5 times at the same size reduces the acceptable circle of confusion by a factor of 1.5.

Here's another way to look at it: If the 1.5-fold body part of a 15K 24K line graph captured is printed with 4×6 and the FF image of a 36k line graph is printed with 10 & # 39; printed with 6×9, the line pairs would be the same width on both prints. Note that the FF image resolves 36K line pairs with a resolution of 6×9 inches, while the 1.5X section resolves only 24X line pairs with a resolution of 4×6 inches. However, if you zoom in on the 1.5-frame image to 6×9, the line pairs (which are at the resolution limit of your lens) will now be magnified 1.5 times.