magento2 – Changing the minimum length of the administrator password

I put my minimum password to 6 from the backend via

Store> Configuration> Clients> Client Configuration> Minimum Password Length

But I can not find a parameter to manage the length of the main administrator password for each configuration. I want to change the length of the main administrator password to 8.

focal length – How to calculate the pixel height of the sensor?

You can not use this information because "pixel pitch" is: sensor length divided by the number of pixels along this dimension. You do not have the size of the sensor – the 2/3 "designation is usually around 11mm diagonal, but may vary in detail.

It is possible that "pixel size" in this specification also means "pixel pitch", but it is also possible that it has been measured in a different way. The number shown aligns quite well (with a 1920 × 1200 sensor, this would be a diagonal of about 10.9 mm).

Depending on your needs, this could be pretty close. Do you have the exact dimensions of the imaging area from which the 1920 × 1200 display is produced? If you need more precision, you will need it.

From your comment:

Is pixel size and height the same?

Not necessarily: the pixel pitch is measured from one center to the other and it is possible that there is a small gap between each pixel, so it would be different from the pixel size. However, if you know this number to get a general idea of ​​the light gathering area of ​​each pixel when comparing camera specifications, you are close enough so that it does not matter.

If your goal is to measure accurately, you will probably need to take measurements in the field based on your current total system in practice rather than making theoretical assumptions. This is especially the case when the tool you buy is manufactured and you market it as a tool for creating images, but rather you want to use it as a camera. measured.

Is it possible to bring a rod of length $ sqrt {3} l $ into a side cube of length l?

A cube of side length l may be suitable for a rod of maximum length $ sqrt {3} l $. But can such a rod be brought into a cube to be fixed in this position?

(Not even sure it's a mathematical question 🙂 and any help with the tags would be appreciated.)

Optical – What should be the width of the front of the lens, given the focal length and aperture?

Objectives with very narrow viewing angles require frontal elements roughly equivalent to the size of the entrance pupil. A main telephoto lens will have a front element less than 10% larger than the entrance pupil at the maximum aperture of the lens. Indeed, the light rays captured by the lens are almost perpendicular to the imaging plane and the entrance pupil will not be much larger than the diameter of the front element.

But with wider viewing angles and closer subject distances, the entrance pupil may be much larger than the element before:

A single objective single element:

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A goal composed of several elements:

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If the element before a wider-angle lens was large enough for the entrance pupil to be fully visible subjects focused on the optical axis of the lens, the Purpose would have the effect of seriously vignetting the light from off-axis parts of the mount. Thus, wide-angle lenses tend to have much larger front-end elements than the size of the entrance pupil, so that a larger portion of the entrance pupil is visible from the most peripheral parts of the field of vision.

When parts of the field of view of the lens are obstructed by a full view of the entrance pupil, dark corners and unusually fuzzy shape reflections may result. Consider opening goals with even a Ordinary field of view:

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Such an objective would have a bokeh "cat's eye":

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Even when there is no mechanical vignetting caused by the lens barrel, at a wider angle, the entrance pupil appears to be an oblong shape, rather than an eyebrow. a circle.

enter the description of the image here

Compare these examples, all for full frame cameras:

  • Canon EF 300mm f / 4 has 77mm filter wires. 300mm / 4 is 75mm
  • Canon EF 100 mm f / 2 with 58 mm filter thread. 100mm / 2 is 50mm
  • Canon EF 85mm f / 1.8 has 58mm filter wires. 85 / 1.8 is 47mm
  • Canon EF 50mm f / 1.4 has 58mm filter wires. 50mm / 1.4 is 36mm
  • Canon EF 35mm f / 2 with 67mm filter thread. 35/2 is 17 mm
  • Canon EF 24mm f / 1.4 has 77mm filter wires. 24mm / 1.4 is 17mm

Your 24-240mm f / 3.5-6.3 lens with a front end of nearly 72mm is probably more focused on reducing vignetting to 24mm and f / 3.5 than on the entrance pupil needed for 240mm and f / 6.3.

Does an APS-C lens self-label with actual or actual focal length?

I know that the effective focal length of the 50mm FX lens is actually 50 x 1.5 = 75mm because I shoot with an APS-C sensor.

Not exactly. Not "incorrect", but you have to understand what it means. The 75mm will not be a useful number on DX. The 50mm lens is NOT 75mm, and there is no real reality called effective focal length. It is a purely hypothetical concept. No matter what lens (at any zoom) has only one focal length, where it actually focuses the light to infinity.

The 50mm lens is still only 50mm, whether it's a DX or FX sensor. It's 50mm, end point. That's why it is marked 50mm. 50mm is the only focal length it has.

Now, you may want to compare his field of view to another lens on another sensor. And it's true that the small size of the DX sensor reduces its field of view, so that (if with a 50mm lens) its reduced field of view compares to that of a 75mm lens on a film frame 35 mm see (or an FX frame has the same size as a 35 mm film). But if your lens is marked 50mm, there is still a 50mm lens on any sensor.

The effective focal length only concerns this other lens on this other sensor (35mm film), only because this other 75mm lens seems to have the same field of view on a 35mm film (or the same FX format) as the 50mm lens of the DX sensor. The focal length on the DX sensor is 50 mm. Only this other lens measures 75mm and on this other larger sensor, it has the same field of view as the 50mm on DX. We are talking about two different lenses and two different sensors.

The fact is that many people have used 35mm film for years, even decades. They are very used to what a 50 or 75mm lens will see and do on a 35mm film. Their experience just knows.

Today's smaller digital sensors are changing things (smaller field of view of smaller sensors). This smaller sensor requires shorter lenses now, to see the "same width of view" as that of larger 35mm films. So, their experience only knows (still) about their new camera. The goal of this "effective focal length" is to compare, to tell users familiar with 35mm film what a lens will do on their new cropped sensor. If we say that this 50mm lens works on DX, just as we are used to 75mm on 35mm film (in regards to the field of view), so it makes sense to them, they know what to do with it. ;expect. However, if you are not familiar with the use of a 35mm film, the effective focal length of a 35mm film will probably not be helpful.

Effective focal length published with objectives for smaller sensors always compare to a 35mm film size (same as FX size, called Full Frame). However, we can compare the field of view of any two sensor sizes. For example, imagine 1/2 inch and 2 inch sensors (movie maybe). The largest is 4x larger than the smallest, so the framing factor is 4x, and (with the same lens), the larger one will have a field of view 4x wider than the smallest one and will need to. an effective focal 4x longer to see. the same reduced field of view as the smaller one. Different sensors can have different shapes (3: 2, 4: 3, 16: 9), so the crop factor actually compares the diagonal of the images.

For FX and DX, this ratio is 1.5.

A similar report was still true for different film sizes, but it was only up to FX and DX Digital that we were able to use the same lens on different size sensors. This is therefore becoming a topic of discussion today.

Does the expansion of the universe actually lose the contraction and has it a limit to reach the value of the total length?

Your question contains many invalid statements.

You do not apply special relativity to the expansion of the universe. You use general relativity which, under certain circumstances, allows expansion by gravity. The notion of "total length" does not exist in general relativity.

If you want to understand how the extension works, just ask it. Do not put invalid stuff in a question.

.

c # – How can I make fun of the length of a string without typing a physical string

I have a service layer that connects to a repository layer. The repository layer invokes a stored process with a parameter with a restricted string length of 200.

I have written the basic service that attaches to a validation service. The validation service checks the length of the string and throws an exception if the string is greater than 200, thus preventing the repository layer from being called. I am writing my unit tests for my validation service and I want to pass a string below 200 and above 200. I do not really want to write.

var teststring = "aa …." for more than 200 characters. there must be a cleaner way to set it up for testing purposes. I use Moq with MSTest.

links or ideas appreciated.

python – How can the length of a string be O (log (n))?

I have just started to learn the complexity of time and I am currently reading the logarithmic complexity. Here is an example code that is O (log (n)).

def intToStr (i):
& # 39; & # 39; Assume that i is a non-negative int Returns the string representation of i & # 39; & # 39; & # 39;

digits = 0123456789 & # 39;
if i == 0:
return "0"
result = & # 39; & # 39;
while I> 0:
result = figures[i % 10] + result
i = i // 10
returned result

So, that gives is the str of an integer. intToStr (1243) results & # 39; 1243 & # 39;. I fully understand that the above function is of logarithmic complexity.

In continuation,

def addDigits (n):

& # 39; & # 39; & Nbsp; Assume n is a non-negative integer
Returns the sum of the digits in ## EQU3 ## & # 39;
stringRep = intToStr (n)
val = 0
for c in stringRep:
val + = int (c)
return val

My book says that:

The complexity of converting n to string is O (log (n)) and intToStr returns a string of length O (log (n)).

Can any one please explain how intToStr returns a string of length O (log (n))? Does not it just return a string of length n where n is the length of the integer we spend !?

algorithms – Count paths of $ n $ length on an infinite 2D grid

I write a video game and I try to find an effective way to calculate it. The goal is to count the number of paths of length $ n $ on an infinitely large 2D grid from a starting node. Both warnings are

(a) the value y of a path can not decrease (you can never go down, only laterally or upwards)

(b) you can not revisit a node in a given path

I've come up with a mediocre brute force method for counting paths, but ideas on an effective way to calculate it?

Calculations – Why does measuring the width and length of the same fish object of a photo give different distance results?

I'm using a Canon EOS Kiss X7 Canon DSLR camera with a Canon EF-S 18-55mm f / 3.5-5.6 IS STM lens to take two shots.

The standard length of the fish should be 62 cm and the height of 14 cm.

The formula (extracted from this discussion of Matt Grum) gives different results of distance from the same photo:
Part 1: IMG_9228-62.jpg (with a ruler of 62 cm).
Part 2: IMG_9230-001.jpg (with a ruler of 14 cm).

IMG_9228.jpg
IMG_9230-001.jpg

Please help me overcome this problem.

Thank you very much in advance.

Warmest regards,
Suryadi

Exif Information for IMG_9228-62.jpg enter the description of the image here
Exif Information for IMG_9230-001.jpg Exif Information for IMG_9230-001.jpg