calculation – What is the rate of change of the average distance?

Whenever I have encountered the expression "rate of change of distance", it simply means the derivative of distance from the independent variable. If I remember correctly, the independent variable has always been time, as is the case in your particular question, so the sentence refers to speed, that is, $ s = frac {d ( text {distance})} {dt} $.

$ epsilon $ -net under Hausdorff distance – MathOverflow

Consider the linear subspaces of $ mathbb {R} ^ n $. For two subspaces $ X $ and $ Y $, we define their Hausdorff distance as
{ displaystyle d _ { mathrm {H}} (X, Y) = max left {, sup _ {x in X, | x | _2 = 1} inf _ {y in Y, | y | _2 = 1} d (x, y), , sup _ {y in Y, | y | _2 = 1} inf _ {x in X, | x | _2 = 1} d (x, y) , right }. !}

For $ epsilon> 0 $, all of the subspaces $ X_1, dots, X_m $ is called a $ epsilon $-net under Hausdorff distance if for any subspace $ Y $, there are $ X_i $ such as
d _ { mathrm {H}} (X_i, Y) < epsilon.

The question is to provide upper and lower bounds of $ m $.

html – Decrease distance from H3 to Slider

Hi, my website is in non semantic HTML, I am trying to reduce the distance of the cursor H3 pro, I have created my own DIV does not work, I have changed everything I didn & # 39; I haven't found the problem if someone can:



latest projects


21.Project Page
.project-page-sec .sec-title {
    text-align: center;
    padding-bottom: 20px;
    padding-top: 20px;
    margin: 0 200px 50px 200px;
.single-project-inner {
    margin-bottom: 20px;
.project-thumb {
    margin-bottom: 20px;
    position: relative;
    overflow: hidden;
.single-project-inner:hover .project-thumb-overlay {
    transition: all 0.4s ease 0s;
.project-thumb-overlay {
    position: absolute;
    bottom: 0;
    height: 100%;
    width: 100%;
    background-color: rgba(15, 25, 52, 0.90);
    transition: all 0.4s ease 0s;
    left: -100%;
.project-thumb-overlay::before {
    position: absolute;
    content: "";
    width: 85%;
    height: 85%;
    border: 1px solid #006b23;
    left: 0;
    right: 0;
    margin: 0px auto;
    transform: translateY(-50%);
    top: 50%;
.project-icon a i {
    color: #006b23;
    font-size: 32px;
.project-icon {
    position: absolute;
    top: 50%;
    transform: translateY(-50%);
    margin: 0px auto;
    text-align: center;
    width: 100%;
.project-inner-desc h2 {
    font-size: 20px;
    text-transform: capitalize;
    font-weight: 600;
    line-height: 25px;
.project-inner-desc h2 {
    font-size: 20px;
    text-transform: capitalize;
    font-weight: 600;
.single-project-inner:hover .project-inner-desc h2 a{
22.Project Details
.sec-title {
    text-align: center;
    margin: 0 250px 50px 250px;

In this section of the code, if I increase the distance, it increases, but if I decrease it, it doesn't work.

.project-page-sec .sec-title {
    text-align: center;
    **padding-bottom: 20px;
    padding-top: 20px;**
    margin: 0 200px 50px 200px;

Link to my website:
insert description of image here

I tried to create a new div didn't work,

Distance in graph theory

Prove it for everyone $ u, v in V (G), $ if $ f: G rightarrow H $ is homomorphism, then $ d_ {G} (u, v) geq d_ {H} (f (u), f (v)). $

powershell – problem with robocopy on copying files from source to long distance destination

I am using robocopy $ source $ destination / MIR, the fact is that it copies the files in synchronization with the source but ignores the files and directories of the source which have long path names.

i wish i had a fix for that, i tried several ways but it didn't work.

Distance from camera so that the photo prints in its actual size

How far do I need to be to take a full size photo of a 6.25 by 6.25 inch object or, if you prefer; 158.25 mm X 158.75 mm?

Is the aperture really important if the distance from the subject is large?

This observation is true.

The closer the lens, the smaller the depth of field.

If you use an online DoF calculator like eg. you can easily see this.

Example: full frame, 50mm f1.4
at  1m -> DoF appr  0.02m
at  3m -> DoF appr  0.19m
at 10m -> DoF appr  2.16m
at 50m -> DoF appr 75.40m 
at 95m -> DoF infinite (behind)

So, once you have reached a certain distance, the aperture will not give you more DoF behind the focal point, but it could influence the near limit where the DoF begins. You can use it to your advantage to calculate the best DoF for landscape photography.

relationship between focal length, perspective projection and camera distance

One thing and only one thing determines perspective: the distance from the subject. Period.

For more information, please see:
Is there a difference between a distant shot on a 50mm lens and a close-up shot on a 35mm lens?
Is the wide angle equivalent in the image of the crop sensor?
Can a telephoto lens have a wide field of vision?
How does the focal length change perspective?
Why is the background bigger and more blurry in one of these images?
What does it really mean that telephoto lenses "flatten" the scenes?

I think your fundamental confusion expressed in the question is due to the different ways in which the different sources you are looking at use the word perspective. They don't all mean the same thing when they use the same word.

The term perspective projection does not describe the same thing as "the perspective of the image". The term geometric projection, in my opinion, is a more precise way of saying what is meant by perspective projection. In other words, it describes how the lens projects an image of the 3D world onto a 2D medium such as a film or a sensor.

The difference is covered in more detail on this question here at Photography SE:
What is the difference between perspective distortion and barrel or pincushion distortion?

Perspective describes the distance relationships between the camera and various elements of the lens' field of vision. As long as the camera is in the same place and everything in front of the camera is in the same place, the perspective will be the same regardless of the focal length lens used. The focal length will determine the angle of view (as long as the film or the size of the sensor does not change), but this will not affect the perspective at all. If you take a picture with a very wide angle lens and crop the middle of the resulting image, it will look like a picture taken in the same place with a longer focal length lens providing a narrower field of view. your cropping. You would lose resolution by throwing away most of the pixels from the camera, but the perspective would be the same.

From the description of the (Perspective) tag here at Photography SE:

Perspective is the spatial relationship between a camera and the things the camera photographs. If two objects of the same size are in the scene and one is much closer to the camera than the other, the closest object will appear to be much larger than the distant object. If the two objects remain at the same distance from each other but the distance from the camera increases by two, then the difference in the apparent size of the two objects will decrease.

You ask:

If the change in focal length does not affect the perspective of the image, what affects the perspective distortion of the image?

The only thing that affects perspective is the shooting distance and the relative distances of various objects from the scene to the camera. To change the perspective, you have to change the relative distances between the camera and various elements of the scene. Here is an example of an answer to this question:

Suppose you are 10 feet from your friend Joe and take his portrait orientation photo with a 50mm lens. Let's say there is a building 100 feet behind Joe. The building is 10 times farther from the camera than Joe, so if Joe is 6 feet and the building is 60 feet, they will appear to be the same height in your photo, as the two would occupy approximately 33º of the image. 39º angle of view of a 50mm lens on the longest dimension.

Now save 30 feet and use a 200mm lens. Your total distance from Joe is now 40 feet. Since you are using a focal length 4 times greater than the original 50 mm (50 mm X 4 = 200 mm), it will appear on the second photo at the same height as on the first. The building, however, is now 130 feet from the camera. It's only 1.3 times higher than when you first shot (100 feet X 1.3 = 130 feet), but you've increased the focal length by 4 times. Now the 60 foot tall building will appear to be about 3X the height of Joe in the image (100 feet / 130 feet = 0.77; 0.77 X 4 = 3.08). At least, that would be the case if all of this could get into the picture, but that is not the case.

Another way of looking at it is that in the first photo with the 50mm lens, the building was 10X further away than Joe (100 feet / 10 feet = 10). In the second photo with the 200mm objective, the building was only 3.25 times further away than Joe (130 feet / 40 feet = 3.25), even if the distance between Joe and the building was the same. What has changed is the relationship between the distance between the camera and Joe and the distance between the camera and the building. This is what defines the perspective: The ratio of the distances between the camera and various elements of a scene.

On the other hand, even two lenses with the same focal length can have different geometric projections. This will affect the shapes of the elements of the image, but it will not change what the camera can and cannot see from the same shooting position. If the "A" box is in front of the "B" box and hides half of the "B" box from the camera, switching from a straight lens to a fisheye lens will not change the amount of box "B" visible by the camera. . That's what the prospect is. Different geometric projections can cause straight lines to appear curved or elements at the edge of the field of view to be "stretched", but that does not change the perspective.

In the real world, there is no lens with an infinite focal length. It is a figment of the imagination that can be used in CGI. To get an orthographic projection with a real camera, you need a telecentric lens.

To get a view where the objects at the end of a three-dimensional
the subject appears the same size as the objects on the side near the
subject we have to use a telecentric type lens which will give us a
orthographic view of our subject. One of the basic requirements
of a telecentric lens is that the lens must be at least as large
diameter as subject. This tends to make them very expensive.

terminology – How to explain the focal distance to someone who is not a photography lover?

Without going into the formulas, I think the easiest way to visually explain the focal length is to use an empty 35mm slide as a framing guide. (Note that over time, less and less people know what a 35mm film slide looks like, so the visual guide is less apt …)

First of all, you need to explain that focal length is a property of the lens. Just like a milk jug can hold 1 or 1/2 gallon or 1 liter, or a certain bottle of water can hold 1/2 liter, so any particular lens has a particular focal length. (In this analogy, zoom lenses are like collapsible water bottles, which have a certain minimum volume when folded down and a maximum volume when folded down). Just like volume is a property of this particular bottle, the focal length is therefore a property of this particular lens.

(Note: I didn't have to use a bottle volume for analogy. I could have used the height of the bottle as easily as the property. It doesn’t matter – it’s just an analogy)

Extending the analogy, it doesn't matter if the bottle is full, half full or empty – the capacity of the bottle is fixed. Just like with a lens: it doesn't matter if it's focused far or near – the focal distance of the lens is unchanged.

Related: What is the focal length and how does it affect my photos?

Now back to the cameras. Different focal lenses modify field of view when mounted on a certain camera. Conversely, when mounting different cameras (with different film or sensor sizes) on a particular lens, the field of vision is also affected.

Here is where the 35mm slide comes in when explaining to people: for a lens with a focal length ƒ (say, 50mm), if it was mounted on a film camera 35 mm (those that most people using film cameras know), then you will get the same field of vision just as you hold a 35mm film slide at a distance of ƒ (50mm, or about 2 inches, in this case) in front of your eye.

Another example: early in the evening of a full moon night, when the moon is low on the horizon and it looks impressive, if you want to capture it in full glory, imagine holding a empty 35 mm slide at arm's length (about 3 feet or or about 900 mm) to frame the moon. When framed with a slide holder at this distance, the moon will fill about 1/3 the height of the frame. So that gives you an idea of ​​the viewing angle of a 900mm lens on a 35mm film camera (or a 35mm full frame DSLR).

Related: What Focal Length Lens Do I Need To Photograph The Moon?

Now if you are talking about a camera with a smaller sensor, such as a 1.5 or 1.6 APS-C crop sensor on entry level DSLRs and mid-range, a 35mm film slide holder no longer works. The framing tool should be 1.5 times smaller. In this case, it would be 24 x 16 mm. Using the smaller "1.5 APS-C slider holder" as a framing guide, you can place it at the focal distance of the lens ƒ from your eye to judge the size of the field of view .

Related: Does My Crop Sensor Camera Really Turn My Lenses Into A Longer Focal Length?

This is the simplest way I have found to explain and visualize the focal distance, without delving into math with the formula of the thin lens and the formula of the angle pinhole vision.

probable probability – $ 2 $ -Wasserstein distance between mixtures

I am stuck on the following problem. I have a discrete distribution $ mu_0 $ (this is actually an empirical distribution). i have some $ mu_i $ (again discrete, a certain empirical distribution). I have some on the distance from Wasserstein $ W_2 ( mu_0, mu_i). $ I now want to consider a simple mixture of $ mu_i, $ C & # 39; is, $ nu = sum limits_ {i = 1} ^ {m} lambda_i mu_i $ or $ sum lambda_i = 1, lambda_i> 0. $

My goal is to link $ W_2 ^ 2 ( mu_0, nu). $ I thought it would be easy to $ W_2 ^ 2 ( mu_0, nu) $ in terms of $ W_2 ^ 2 ( mu_0, mu_i), $ but I can't prove anything. I want something like $$ W_2 ^ 2 ( mu_0, nu) le sum lambda_i ^ 2 W_2 ^ 2 ( mu_0, mu_i). $$

It doesn't sound terribly difficult, but I'm stuck. Can anyone tell if it is true or not? If someone can just demonstrate why this is true, it would be great.