## collision detection – voxel crossing for constant gravity parabolic projectile arc

If your gravity vector is purely vertical (let's call it the Y axis) and there is no side wind to take into account, the XZ position of the projectile follows a straight line. If you do not have resistance to air, then it moves along this line at a steady pace, just like our friendly linear ray emissions!

We can project a "shadow" of the projectile directly on the XZ plane. We can determine the squares of the grid on this plane through which this line passes using your ordinary 2D scatter / scroll algorithm (or some other online rasterization algorithm like Bresenham). I show an example of this type of raymarching in this answer.

With this we can calculate flight time when shadow intersects each new box along its way. This tells us when the 3D projectile enters itself into a new column of voxels.

By calculating the height at each column entry timestamp, using:

$$vec p (t) = vec h_0 + vc v_0 cdot t + frac { vec a} 2 cdot t ^ 2$$

… then we can get the height at which the projectile enters and leaves each column.

The other height we need is the height at the top of the parable. This is the only point where the projectile can touch voxels in a column outside the range from its entrance to the exit height. We can find this point with:

$$t_ * = frac {-v_y} {a_y}$$

We can now review each of these key timestamps in sequence, since entering a new column of voxels and moving up / down this column to the next timestamp (either the vertex or the entry in the next column). This ensures that we visit every voxel touched by the dish, with only a small amount of work beyond the basic 2D raycast.

## number of frames per second – Constant or variable frame rate in GoPro Hero 3?

The recordings of a GoPro Hero 3 camera at 60 fps with a constant or variable frame rate? How can I check the corresponding mp4 file for a constant or variable scroll speed?

Here is the release of `ffmpeg -i` order:

``````Input # 0, mov, mp4, m4a, 3gp, 3g2, mj2, from /media/102GOPRO/GOPR0333.MP4&#39 ;:
major_brand: avc1
minor version: 0
compatible_brands: avc1isom
creation_time: 2019-04-16 13:18:14
Duration: 00: 19: 53.73, start: 0.000000, speed: 15122 kb / s
Stream # 0: 0 (eng): Video: h264 (Main) (avc1 / 0x31637661), yuvj420p (pc, bt709), 1280x720 [SAR 1:1 DAR 16:9], 14982 kb / s, 59.94 fps, 59.94 tbr, 60k tbn, 119.88 tbc (default)
creation_time: 2019-04-16 13:18:14
encoder: GoPro AVC encoder
timecode: 13: 17: 26: 44
Stream 0: 1 (eng): Audio: aac (LC) (mp4a / 0x6134706D), 48000 Hz, stereo, fltp, 127 kb / s (default)
creation_time: 2019-04-16 13:18:14
timecode: 13: 17: 26: 44
Stream # 0: 2 (eng): Data: none (tmcd / 0x64636D74) (default)
creation_time: 2019-04-16 13:18:14
timecode: 13: 17: 26: 44
``````

## Complex analysis – Integer function such that \$ f (z) = f (z + 1) \$ for all \$ z \$, then f is constant?

Yes $$f$$ is whole function as $$f (z) = f (z + 1)$$ for everyone $$z$$ is not that enough to consider that $$f$$ is a constant. Why? Because we have $$f (0) = f (1) = f (2) = cdots$$ so what $$f (z) = f (z + 1) = f (z + 2) = cdots$$ so the function $$f$$ is determined by the image of the dots on the desk of the unit but $$f$$ is delimited in the desk of the unit without poles because it is so whole according to Liouville's theorem $$f$$ is a constant?
Something is missing! Am I right

In fact, the initial question poses another condition $$f (z) = f (z + 1) = f (z + a)$$ for an irrational $$a$$ to understand that $$f$$ is a constant.

## enumerative combinatorial – Enumerative functions that are at most \$ c \$ -to- \$ 1 \$ for a constant \$ c \$ 2 \$

rating: $$[m] : = {1, 2, dots, m }$$.

How many functions are there $$f: [a] at [b]$$? It is easy to see that the answer is $$b ^ a$$.

How many injective the functions are there $$f: [a] at [b]$$? Again, the answer is well known and sometimes called the factorial down:
$$b (b-1) dots (b-a + 1).$$

And if, more generally, we limit the size of the fibers $$f ^ {- 1} (x)$$but the limit could be more than $$1$$?

In other words, how many functions are there $$f: [a] at [b]$$ who are no more than $$c$$-at-$$1$$?

I do not necessarily hope that there is an exact formula and I am more interested in asymptotics. For example, can we give upper and lower limits "reasonable", in the case where $$c 2$$ and $$| A | / | B |$$ are fixed, and $$| A | to infty$$?

For a concrete example, how many functions are there around $$[5n] at [n]$$ who are at most $$8$$-at-$$1$$? Call this function $$g (n)$$.

Clearly we have
$$frac {(5n)!} {n! ^ 5} the g (n) nn {5n}.$$
Function $${(5n)!} / {N! ^ 5}$$ counts functions that are exactly 5-to-1, and function n ^ {5n} counts all functions.

The application of Stirling's approximation to the first function gives something like
$$5 ^ {5n} f (n) n ^ {5n}.$$

These limits are not far enough apart and do not even fix themselves if the true order of growth is exponential or "factorial". (We could say that the growth rate of a function $$f (n)$$ is factorial if $$log f (n) = Theta ( log n)$$.)

## macos – What could be the cause? I / O completely dead. Constant rotation, even the connection takes forever

Owner of a 5K iMac, model late 2014 here. The weirdest thing happened yesterday. This machine is usually very powerful – 8 GB of RAM, 3 TB Fusion. I double-boot Windows on this one using Boot Camp.

Until now, I still have 980 GB free – this problem can not be due to insufficient storage.

The Mac worked perfectly until I used it `Photos.app` to browse photos (the application has performed a 'repair permissions library'). Shortly after, I started experiencing frequent crashes from the macOS UI (all applications, beachball spinning) for sometimes a few minutes.

The activity monitor showed very low RAM pressure and virtually no strain on the part of the processor. Because I could not isolate the problem, I restarted the Mac.

Now even by clicking on the connection image suspends the computer for about 4-5 minutes. Yes, it takes so long to display the password field. Once the password is entered, the wallpaper appears after 2 minutes, the dock appears in 3 minutes and the widgets in the menu bar take about 2 minutes. Opening the Launchpad depends on loading applications for several minutes, and each application icon loads at the correct time.

1. Opening no matter what – System Preferences, About this Mac, Spotlight, Finder takes ageand freezes the entire interface in the meantime. In this frozen state, only the cursor moves and nothing else responds in any way.
2. If you do not click anything, the entire interface freezes periodically for 2-3 minutes.
3. For whatever reason, open applications work perfectly – you can drag them and select items from the list. It seems that everything that requires I / O, such as switching folders in the Finder or System preference pane, takes forever.
4. When I am able to open Activity Monitor, there is 0 culprit. No application using the disk, the RAM pressure is extremely low – although, perhaps normally, it is used at 0 bytes – and the processor is virtually inactive. Strange.
5. SSHing in the Mac from another box presents these same symptoms, suggesting that this is not `loginwindow`. SSH takes about a minute to connect and the execution of each command takes forever.
• I have run Apple Hardware Diagnostics (cmd+re while booting). No hardware problem found.
• I have run First Aid on the APFS volume in Recovery OS (cmd+R while booting). No major problem found, and it seemed OK.
• I removed it `/var/db/.applesetupdone` in single-user mode to force macOS to create a new administrator account. The administrator account suffers from the same symptoms.

I use macOS Mojave 10.14.6 Beta. Yes, I know it's a beta software, but it has worked well for ages until yesterday. I do not want to reinstall macOS and it is impossible to perform a backup because I do not have an external hard drive of the Mac's capacity. In fact, the kind of Mac was my hard drive backup one way.

What can I do? Do these symptoms tell you something? It seems to me that I am / I am, but honestly I am utterly lost.

## oracle – Declare constant variables in advance or when I need them?

I have a PL / SQL script that requires a bunch of `CONSTANT VARCHAR2 (4)` variables to declare. If I declare them all, then the script will have about 200 lines more than if I just use their string value when I need it.

Consider:

``````TO DECLARE
- Cds
lvsSIGN CONSTANT VARCHAR2 (4): = & # 39; SIGN & # 39 ;;
lvsOWNS CONSTANT VARCHAR2 (4): = & # 39; OWNS & # 39 ;;
lvsPIPE CONSTANT VARCHAR2 (4): = & # 39; PIPE & # 39 ;;
lvsABCD CONSTANT VARCHAR2 (4): = & # 39; ABCD & # 39 ;;
- 200 more
TO START
INSERT INTO Some_Table (Column1)
VALUES (lvsSIGN);
- Repeat this insertion for the next 200 CDs.
END;
``````

Against:

``````TO DECLARE
- Some other declarations
TO START
INSERT INTO Some_Table (Column1)
VALUES ("SIGN");
- Repeat this insertion for the next 200 CDs.
END;
``````

Is there a different approach that will save the LOC so that the script is not filled with statements and `INSERT`s?

## Learn How to Make Constant Forex Profits – Discussions & Help

Very interesting information

Have you got been able to get coherent perofit after In progress these advice? But I think, no matter strategy we use if do not disciplined then everything will be useless.

Well thought guys …

These are the rules of consistency in trade:

Repeat your path when you get the profit

Repeat your path when you get the profit

Repeat your path when you get the profit

Repeat your path when you get the profit

Repeat your path when you get the profit

I've negotiated with many brokers such as Liteforex, Insta, Oanda, etc. But it is so difficult to trade regularly. I come to exchange, exchange, exchange and exchange and exchange with small, huge, lose, small, huge, lose profits. In NDD lite forex, I'm pretty consistent there, because the very low spread there …

## Manipulation of the list – Why does a table with a constant defined in its index compute 10X slower?

The problem lies mainly in the interior `Table`:

``````Timing[Total[Table[Total[Table[data[[i]], {i, j, 10 + j}

], {j, 1, length[data] - 5 * 10}]]]m = 10;
Timing[Total[Table[Total[Table[data[[i]], {i, j, m + j}

], {j, 1, length[data] - 5 * 10}]]]
``````

{0.366407, 5.50276 * 10 ^ 6}

{8.01738, 5.50276 * 10 ^ 6}

I think the reason is:
Because the global variable `m` could theoretically change its value during calculations, the body of the external table can not be compiled (without calling MainEvaluate). At least the JIT compiler does not analyze the body of the outer loop enough to decide that `m` will not change.

You can help the JIT compiler by using `With`:

``````With[{m = 10},
Timing[Total[Table[Total[Table[data[[i]], {i, j, m + j}

], {j, 1, length[data] - 5 * m}]]]]
``````

{0.369601, 5.5049 * 10 ^ 6}

## lens – To what extent do manual zooms with constant aperture have a mechanical influence on aperture when zooming?

I've seen some examples of constant aperture zoom (for example, a Ricoh of 28-100mm f / 4) in second-hand markets where the aperture does not seem to be fully open (visibly no hole). circular) with certain zoom settings. Have any designs actually done this intentionally to force the opening constant, or is it only mechanical defects in which unrelated mechanical parts interfere with one another due to a rubbing or poor rendering?

## Functional analysis – Asymptotically constant sequence of the eigenfunctions of the Laplacian

I am not an expert in this kind of thing, so forgive me if I am naive.

I wonder if there is a sequence of functions $$f_t: B _ { mathbb {R} ^ 2} (0, t) to [0,infty]$$ (right here $$B (0, t)$$ means radius ball $$t$$ focused on $$0$$) such as, if $$triangle$$ is the Laplacian dish

1) $$triangle f_t = 4f_t$$ (or more generally $$triangle f_t = cf_t$$ for a fixed positive $$c$$)

2) $$f_t to 0$$ punctually as $$t to infty$$

3) $$f_t (x, y) / f_t (0) to 1$$ punctually as $$t to infty$$

In other words, I would like to know if there exists a family of eigenfunctions of the Laplacian, defined on a growing series of marbles, converging towards $$0$$ and such as the convergence rate, at least $$infty$$, does not depend on the point in question.

Here is a vague idea of ​​why, in my opinion, such a sequence may exist: I can imagine a proper function of the Laplacian on the unit disk whose norm goes up and down slightly from a small amount in small disks . The domain stretch forces the oscillation to go down and finally converge towards $$0$$. Maybe I'm looking for a Bessel function.

Thank you.