text:
# Ubuntu – sudo: unable to resolve host ubuntu-server

sudo

Every time I try to use sudo I get this:

sudo: unable to resolve host ubuntu-server

My /etc/hosts file:

127.0.0.1       localhost
127.0.1.1       ubuntu-server

# The following lines are desirable for IPv6 capable hosts

::1     ip6-localhost ip6-loopback
fe00::0 ip6-localnet
ff00::0 ip6-mcastprefix
ff02::1 ip6-allnodes
FF02::2 ip6-allrouters

/etc/hostname has:

ubuntu-server

I think there is some other setting that causes the unable to resole host error, but I cannot find these.

• This may mean your DNS lookups are broken.

Your /etc/hosts file looks fine. You want it too look like this:

127.0.0.1       localhost
127.0.1.1       ubuntu-server

But there are several legitimate variants.

Make sure you have a matching hostname in /etc/hostname .

Check /etc/resolv.conf to see if you have working name servers defined.

See if you can resolve any hostnames. Does nslookup google.com work ? If not, your DNS lookup is failing, either due to configuration or a network or firewall problem (possibly at a point beyond what you have control of with your server).

In my experience, sudo will eventually ask you for you password, and grant you elevated privileges, after domain lookup times out.

Try manually adding the correct DNS name servers to /etc/resolv.conf.

In a pinch, you can use public servers. Here's what the file looks like using Google DNS servers:

/etc# cat resolv.conf
nameserver 8.8.8.8
nameserver 8.8.4.4
nameserver 2001:4860:4860::8888
nameserver 2001:4860:4860::8844

format: plain


text:
# GetDP

GetDP is an open source finite element solver using mixed elements to discretize de Rham-type complexes in one, two and three dimensions. GetDP is developed by the ACE group from the Montefiore Institute at the University of Liège, and is released under the GNU GPL.

## Getting started

ONELAB allows to use GetDP as a black-box solver: you don't need to know anything about finite elements or de Rham complexes in order to run your first simulations:

2. Launch the app
3. Open a GetDP model:
• Desktop version: go to the File/Open menu and select a GetDP .pro file, e.g. models/magnetometer/magnetometer.pro
• Mobile version: select one of the preloaded models
4. Press Run.

## GetDP models

### Basic templates

These are basic physical templates, that can either be used interactively to define new problems from scratch, or be included in other problem definition files.

### Featured physical models

These are complete, parametric application examples, ready to be solved and modified.

Acoustics Electromagnetism Heat transfer Multi-physics

## How does it work?

GetDP input files (.pro files) can be instrumented to share parameters with the ONELAB server, through the same syntax as the one used in Gmsh.

## Building GetDP from the source code

If you want to recompile GetDP directly from the source code, you should

tar zxvf petsc-3.7.4.tar.gz

• Configure and build PETSc. The configuration options depend on the calculations you want to perform (complex- or real-valued), as well as your compiler setup. For a sequential build (without MPI), run (remove --with-scalar-type=complex to build in real arithmetic):
cd petsc-3.7.4
export PETSC_DIR=\$PWD
export PETSC_ARCH=complex_mumps_seq
make
cd ..

• Configure, compile and install a minimal Gmsh library (it will be used by GetDP):
cd gmsh-xxx
mkdir lib
cd lib
cmake -DDEFAULT=0 -DENABLE_PARSER=1 -DENABLE_POST=1 -DENABLE_BUILD_LIB=1 ..
make lib
sudo make install/fast
cd ../..

• Configure and compile GetDP:
cd getdp-xxx
mkdir bin
cd bin
cmake ..
make
cd ../..

Instructions for building the parallel (MPI) version of GetDP are available here


text:
# 27-5 Multithreading a simple stencil calculation

Computational science is replete with algorithms that require the entries of an array to be filled in with values that depend on the values of certain already computed neighboring entries, along with other information that does not change over the course of the computation. The pattern of neighboring entries does not change during the computation and is called a stencil. For example, Section 15.4 presents a stencil algorithm to compute a longest common subsequence, where the value in entry $c[i, j]$ depends only on the values in $c[i - 1, j]$, $c[i, j - 1]$, and $c[i - 1, j - 1]$, as well as the elements $x_i$ and $y_j$ within the two sequences given as inputs. The input sequences are fixed, but the algorithm fills in the two-dimensional array $c$ so that it computes entry $c[i, j]$ after computing all three entries $c[i - 1, j]$, $c[i, j - 1]$, and $c[i - 1, j - 1]$.

In this problem, we examine how to use nested parallelism to multithread a simple stencil calculation on an $n \times n$ array $A$ in which, of the values in $A$, the value placed into entry $A[i, j]$ depends only on values in $A[i' , j']$, where $i' \le i$ and $j' \le j$ (and of course, $i' \ne i$ or $j' \ne j$). In other words, the value in an entry depends only on values in entries that are above it and/or to its left, along with static information outside of the array. Furthermore, we assume throughout this problem that once we have filled in the entries upon which $A[i, j]$ depends, we can fill in $A[i, j]$ in $\Theta(1)$ time (as in the $\text{LCS-LENGTH}$ procedure of Section 15.4).

We can partition the $n \times n$ array $A$ into four $n / 2 \times n / 2$ subarrays as follows:

$$A = \begin{pmatrix} A_{11} & A_{12} \\ A_{21} & A_{22} \tag{27.11} \end{pmatrix} .$$

Observe now that we can fill in subarray $A_{11}$ recursively, since it does not depend on the entries of the other three subarrays. Once $A_{11}$ is complete, we can continue to fill in $A_{12}$ and $A_{21}$ recursively in parallel, because although they both depend on $A_{11}$ , they do not depend on each other. Finally, we can fill in $A_{22}$ recursively.

a. Give multithreaded pseudocode that performs this simple stencil calculation using a divide-and-conquer algorithm $\text{SIMPLE-STENCIL}$ based on the decomposition $\text{(27.11)}$ and the discussion above. (Don't worry about the details of the base case, which depends on the specific stencil.) Give and solve recurrences for the work and span of this algorithm in terms of $n$. What is the parallelism?

b. Modify your solution to part (a) to divide an $n \times n$ array into nine $n / 3 \times n / 3$ subarrays, again recursing with as much parallelism as possible. Analyze this algorithm. How much more or less parallelism does this algorithm have compared with the algorithm from part (a)?

c. Generalize your solutions to parts (a) and (b) as follows. Choose an integer $b \ge 2$. Divide an $n \times n$ array into $b^2$ subarrays, each of size $n / b \times n / b$, recursing with as much parallelism as possible. In terms of $n$ and $b$, what are the work, span, and parallelism of your algorithm? Argue that, using this approach, the parallelism must be $o(n)$ for any choice of $b \ge 2$. ($\textit{Hint:}$ For this last argument, show that the exponent of $n$ in the parallelism is strictly less than $1$ for any choice of $b \ge 2$.)

d. Give pseudocode for a multithreaded algorithm for this simple stencil calculation that achieves $\Theta(n\lg n)$ parallelism. Argue using notions of work and span that the problem, in fact, has $\Theta(n)$ inherent parallelism. As it turns out, the divide-and-conquer nature of our multithreaded pseudocode does not let us achieve this maximal parallelism.

a. In this part of the problem, we will assume that $n$ is an exact power of $2$, so that in a recursive step, when we divide the $n \times n$ matrix $A$ into four $n / 2 \times n / 2$ matrices, we will be guaranteed that $n / 2$ is an integer, for all $n \ge 2$. We make this assumption simply to avoid introducing $\lfloor n / 2 \rfloor$ and $\lceil n / 2 \rceil$ terms in the pseudocode and the analysis that follow. In the pseudocode below, we assume that we have a procedure $\text{BASE-CASE}$ available to us, which calculates the base case of the stencil.

  1 2 3 4 5 6 7 8 9 10 11 SIMPLE-STENCIL(A, i, j, n) if n == 1 A[i, j] = BASE-CASE(A, i, j) else // Calculate submatrix A[1, 1]. SIMPLE-STENCIL(A, i, j, n / 2) // Calculate submatrices A[1, 2] and A[2, 1] in parallel. spawn SIMPLE-STENCIL(A, i, j + n / 2, n / 2) SIMPLE-STENCIL(A, i + n / 2, j, n / 2) sync // Calculate submatrix A[2, 2]. SIMPLE-STENCIL(A, i + n / 2, j + n / 2, n / 2)

To perform a simple stencil calculation on an $n \times n$ matrix $A$, we call $\text{SIMPLE-STENCIL}(A, 1, 1, n)$. The recurrence for the work is $T_1(n) = 4T_1(n / 2) + \Theta(1) = \Theta(n^2)$. Of the four recursive calls in the algorithm above, only two run in parallel. Therefore, the recurrence for the span is $T_\infty(n) = 3T_\infty(n / 2) + \Theta(1) = \Theta(n^{\lg 3})$, and the parallelism is $\Theta(n^{2 - \lg 3}) \approx \Theta(n^{0.415})$.

b. Similar to $\text{SIMPLE-STENCIL}$ of the previous part, we present $\text{P-STENCIL-3}$, which divides $A$ into nine submatrices, each of size $n / 3 \times n / 3$, and solves them recursively. To perform a stencil calculation on an $n \times n$ matrix $A$, we call $\text{P-STENCIL-3}(A, 1, 1, n)$.

  1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 P-STENCIL-3(A, i, j, n) if n == 1 A[i, j] = BASE-CASE(A, i, j) else // Group 1: compute submatrix A[1, 1]. P-STENCIL-3(A, i, j, n / 3) // Group 2: compute submatrices A[1, 2] and A[2, 1]. spawn P-STENCIL-3(A, i, j + n / 3, n / 3) P-STENCIL-3(A, i + n / 3, j, n / 3) sync // Group 3: compute submatrices A[1, 3], A[2, 2], and A[3, 1]. spawn P-STENCIL-3(A, i, j + 2n / 3, n / 3) spawn P-STENCIL-3(A, i + n / 3, j + n / 3, n / 3) P-STENCIL-3(A, i + 2n / 3, j, n / 3) sync // Group 4: compute submatrices A[2, 3] and A[3, 2]. spawn P-STENCIL-3(A, i + n / 3, j + 2n / 3, n / 3) P-STENCIL-3(A, i + 2n / 3, j + n / 3, n / 3) sync // Group 5: compute submatrix A[3, 3]. P-STENCIL-3(A, i + 2n / 3, j + 2n / 3, n / 3)

From the pseudocode, we can informally say that we can solve the nine subproblems in five groups, as shown in the following matrix:

$$\begin{pmatrix} 1 & 2 & 3 \\ 2 & 3 & 4 \\ 3 & 4 & 5 \end{pmatrix} .$$

Each entry in the above matrix specifies the group of the corresponding $n / 3 \times n / 3$ submatrix of $A$; we can compute in parallel the entries of all submatrices that fall in the same group. In general, for $i = 2, 3, 4, 5$, we can calculate group $i$ after completing the computation of group $i - 1$.

The recurrence for the work is $$T_1(n) = 9T_1(n / 3) + \Theta(1) = \Theta(n^2).$$

The recurrence for the span is

$$T_\infty(n) = 5T_\infty(n / 3) + \Theta(1) = \Theta(n^{\log_3 5}).$$

Therefore, the parallelism is

$$\Theta(n^{2 - \log_3 5}) \approx \Theta(n^{0.535}).$$

c. Similar to the previous part, we can solve the $b^2$ subproblems in $2b - 1$ groups:

$$\begin{pmatrix} 1 & 2 & 3 & \cdots & b - 2 & b - 1 & b \\ 2 & 3 & 4 & \cdots & b - 1 & b & b + 1 \\ 3 & 4 & 5 & \cdots & b & b + 1 & b + 2 \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots & \vdots \\ b - 2 & b - 1 & b & \cdots & 2b - 5 & 2b - 4 & 2b - 3 \\ b - 1 & b & b + 1 & \cdots & 2b - 4 & 2b - 3 & 2b - 2 \\ b & b + 1 & b + 2 & \cdots & 2b - 3 & 2b - 2 & 2b - 1 \end{pmatrix} .$$

The recurrence for the work is

$$T_1(n) = b^2 T_1(n / b) + \Theta(1) = \Theta(n^2).$$

The recurrence for the span is

$$T_\infty(n) = (2b - 1) T_\infty(n / b) + \Theta(1) = \Theta(n^{\log_b (2b - 1)}).$$

The parallelism is

$$\Theta(n^{2 - \log_b (2b - 1)}).$$

As the hint suggests, in order to show that the parallelism must be $o(n)$ for any choice of $b \ge 2$, we need to show that $2 - \log_b (2b - 1)$, which is the exponent of $n$ in the parallelism, is strictly less than $1$ for any choice of $b \ge 2$. Since $b \ge 2$, we know that $2b - 1 > b$, which implies that $\log_b (2b - 1) > \log_b b = 1$. Hence, $2 - \log_b (2b - 1) < 2 - 1 = 1$.

d. The idea behind achieving $\Theta(n / \lg n)$ parallelism is similar to that presented in the previous part, except without recursive division. We will compute $A[1, 1]$ serially, which will enable us to compute entries $A[1, 2]$ and $A[2, 1]$ in parallel, after which we can compute entries $A[1, 3]$, $A[2, 2]$ and $A[3, 1]$ in parallel, and so on. Here is the pseudocode:

  1 2 3 4 5 6 7 8 9 10 P-STENCIL(A) n = A.rows // Calculate all entries on the antidiagonal and above it. for i = 1 to n parallel for j = 1 to i A[i - j + 1, j] = BASE-CASE(A, i - j + 1, j) // Calculate all entries below the antidiagonal. for i = 2 to n parallel for j = i to n A[n + i - j, j] = BASE-CASE(A, n + i - j, j)

For each value of index $i$ of the first serial for loop, the inner loop iterates $i$ times, doing constant work in each iteration. Because index $i$ ranges from $1$ to $n$ in the first for loop, we require $\Theta(1 + 2 + \cdots + n) = \Theta(n^2)$ work to calculate all entries on the antidiagonal and above it. For each value of index $i$ of the second serial for loop, the inner loop iterates $n - i + 1$ times, doing constant work in each iteration. Because index $i$ ranges from $2$ to $n$ in the second for loop, we require $\Theta((n - 1) + (n - 2) + \cdots + 1) = \Theta(n^2)$ work to calculate all entries on the antidiagonal and above it. Therefore, the work of $\text{P-STENCIL}$ is $T_1(n) = \Theta(n^2)$.

Note that both for loops in $\text{P-STENCIL}$, which execute parallel for loops within, are serial. Therefore, in order to calculate the span of $\text{P-STENCIL}$, we must add the spans of all the parallel for loops. Given that any parallel for loop in $\text{P-STENCIL}$ does constant work in each iteration, the span of a parallel for loop with $n'$ iterations is $\Theta(\lg n')$. Hence,

\begin{aligned} T_\infty(n) & = \Theta((\lg 1 + \lg 2 + \cdots + \lg n) + (\lg(n - 1) + \cdots + 1)) \\ & = \Theta(\lg(n!) + \lg(n - 1)!) \\ & = \Theta(n\lg n), \end{aligned}

giving us $\Theta(n / \lg n)$ parallelism.


text:
# How to prove the Hahn-Banach constructively

I am just wondering, how to prove the Hahn-Banach theorem constructively for a finite dimensional normed vector space.

Thanks in advance for any helpful answers.

-
I am not sure this question reflects the level of this site. I am a bit puzzled about the attention it gets. –  András Bátkai Sep 27 '11 at 21:55
@Andras: IMHO, the question might be borderline, but it is definitely not trivial (or at least, it is as non-trivial as the usual proof of the Hahn-Banach theorem, where the inductive step is actually a bit delicate). –  Yemon Choi Sep 28 '11 at 1:00
I believe that constructive versions of the Hahn-Banach theorem have been studied in several papers by Thierry Coquand. You might look at, e.g., his paper "Geometric Hahn-Banach Theorem" in Math. Proc. Camb. Phil. Soc. (2006), 140, 313. –  Michael A Warren Sep 28 '11 at 2:40
I was confused because this is usually optional homework on my beginning FA course. –  András Bátkai Sep 28 '11 at 7:42

## 2 Answers

Same way as for the infinite dimensional case, except you avoid Zorn's lemma by counting dimensions.

-
Isn't the whole thing trivial as soon as you know that your subspace is complemented? –  András Bátkai Sep 27 '11 at 20:52
Preserving norms isn't trivial. –  Benjamin Hayes Sep 27 '11 at 20:54

The idea is to show that one can extend a linear functional from an $n$-dimensional space to a space of dimension $n+1$ without increasing its norm. See, for instance, my notes (Lemma E.2)

In fact, by doing so, you can prove THBT constructively for any separable space.

-


text:
# listing with labeled frame

I'm trying to get 'labeled' frames on an lstlisting environment that looks like this:

The top and bottom frames are easy are fairly easy to achieve:

\begin{lstlisting}[frame=tb]
Code
Code
Code
\end{lstlisting}

But after perusing through the listings documentation, I'm unsure how to achieve the "Some Text" labels on the bottom and top frames (it seems that kind of thing is not supported).

I've looked to alternative packages such as mdframed and fancyvrb. fancyvrb comes close to what I'm looking for, but from what I can by the options only allows the "Some Text" labels to be aligned in the center, i.e.:

Also, one consideration: I already have a nice (and working) lst style definition for my language, so It'd be nice if any solution to this could still using lstlisting (and hence my language definition).

• Check the tcolorbox documentation. – Ulrike Fischer Apr 9 '17 at 21:08

A version with tcolorbox and its listings library (which was my first idea to realize this)

The appearance can be changed by styles or by changing the overlay portion of the code.

\documentclass{article}

\usepackage[most]{tcolorbox}

\makeatletter
\newtcblisting[auto counter]{somecode}[2][]{%
enhanced jigsaw,
boxrule=0pt,
toprule=1pt,
bottomrule=1pt,
listing options={language={C}},
listing only,
colback=white,
sharp corners,
overlay={\node[inner sep=2pt,xshift=-2cm,fill=tcbcol@back] (A) at (frame.south east) {#2};
\node[inner sep=2pt,xshift=+2cm,fill=tcbcol@back] (B) at (frame.north west) {#2};
},
#1,
}
\makeatother

\begin{document}

\begin{somecode}{Some Text}
int main
\end{somecode}

\begin{somecode}[colback=yellow!10!white]{Some Text}
int main
\end{somecode}

\end{document}


text:
MORE IN Applied Mathematics - 3
MU Electronics and Telecom Engineering (Semester 3)
Applied Mathematics - 3
May 2015
Total marks: --
Total time: --
INSTRUCTIONS
(1) Assume appropriate data and state your reasons
(2) Marks are given to the right of every question
(3) Draw neat diagrams wherever necessary

1 (a) Determine the constants a,b,c,d if f(z)=x2+2axy+by2+i(dx2+2cky+y2) is analytic.
5 M
1 (b) Find a cosine series of period 2π to represent sin x in 0≤x≤π.
5 M
1 (c) Evaluate by using Laplace Transformation $\int^\infty_0 e^{-3x} t \cos t \ dt.$
5 M
1 (d) A vector field is given by $\overline {F} = (x^2 + xy^2)i + (y^2 + x^2 y)j. Show that F is irrotational and find its potential. Such that F=∇ϕ. 5 M 2 (a) Solve by using Laplace Transform. (D2+2D+5)y=e-t sin t, when y(0)=0, y(0)=1. 6 M 2 (b) Find the total work done in moving a particle in the force field. F=3xy i-5z j+10x k along x=t2+1, y=2t2, z=t3 from t=1 and t=2. 6 M 2 (c) Find the Fourier series of the function f(x)=e-x. 0 8 M 3 (a) Prove that \[ J_{1/2}(x) = \sqrt{ \dfrac {2} {\pi x } }\cdot \sin x$
6 M
3 (b) Verify Green's theorem in the plane for ∮(x2-y)dx+(2y2+x)dy Around the boundary of region defined by y=x2 and y=4.
6 M
3 (c) Find the Laplace transforms of the following. $i) \ e^{-t} \int^t_0 \dfrac {\sin u} {u} du \\ ii) \ t \sqrt{1+\sin t}$
8 M

4 (a) f f(x)=C1Q1(x) + C2Q2(x) + C3Q3(x)t where C1, C2, C3 constants and Q1, Q2, Q3 are orthonormal sets on (a,b), show that. $\int^b_a [f(x)]^2 dx = c^2_1 + c^2_2 + c^2_3.$
6 M
4 (b) If v=ex sin y, prove that v is a Harmonic function. Also find the corresponding harmonic conjugate function and analytic function.
6 M
4 (c) Find inverse Laplace transform of the following: $i) \ \dfrac {S^2} {(S^2 + a^2) (S^2+b^2)} \\ ii) \ \dfrac {S+2}{S^2 -4S+13}$
8 M

5 (a) Find the Fourier series if f(x)=|x|, -k
6 M
5 (b) Define solenoidal vector. Hence prove that $$\overline {F} = \dfrac{\overline{a}\times\overline{r}}{r^n}$$ is a solenoidal vector.
6 M
5 (c) Find the bilinear transformation under which 1, i, -1 from the z-plane are mapped onto 0, 1, ∞ of w-plane. Further show that under this transformation the unit circle in w-plane is mapped onto a straight line in the z-plane. Write the name of this line.
8 M

6 (a) Using Gauss's Divergence theorem $$\iint_s \ \overline {F} .d\overline {s}$$ where F2x2yi-y2j+4xz2 k and s is the region bounded by y2+z2=9 and x=2 in the first octant.
6 M
6 (b) Define billinear transformation, And prove that in a general, a bilinear transformation maps a circle into a circle.
6 M
6 (c) Prove that $\int xJ_{2/3} (x^{3/2})dx = - \dfrac {2}{3} x^{-1/2}J_{-1/3}(x^{3/2}) .$
8 M

More question papers from Applied Mathematics - 3


text:
year-month-weekday-boundary {clock} R Documentation

## Boundaries: year-month-weekday

### Description

This is a year-month-weekday method for the calendar_start() and calendar_end() generics. They adjust components of a calendar to the start or end of a specified precision.

This method is restricted to only "year" and "month" precisions, and x can't be more precise than month precision. Computing the "start" of a day precision year-month-weekday object isn't defined because a year-month-weekday with day = 1, index = 1 doesn't necessarily occur earlier (chronologically) than day = 2, index = 1. Because of these restrictions, this method isn't particularly useful, but is included for completeness.

### Usage

## S3 method for class 'clock_year_month_weekday'
calendar_start(x, precision)

## S3 method for class 'clock_year_month_weekday'
calendar_end(x, precision)

### Arguments

 x [clock_year_month_weekday] A year-month-weekday vector. precision [character(1)] One of: "year" "month"

### Value

x at the same precision, but with some components altered to be at the boundary value.

### Examples

# Month precision
x <- year_month_weekday(2019, 1:5)
x

# Compute the last month of the year
calendar_end(x, "year")

[Package clock version 0.6.1 Index]


text:
# Automatically add category tag(s) to questions, or at least suggest them

It seems almost all requests to add tag category/hierarchy/organization/etc have been shot down.

I'm suggesting a different variety: automatically add the appropriate "category" tag(s) when the question is posted.

Here's an example for my personal interest, very similar to a previous answer: On SO there's swing tag, but many people (especially n00bs) use specific tags like jtable, jlabel, etc. without using the swing tag. It would be a huge pain, if not impossible, to add all JComponent subclasses to my interested tags and RSS feed.

Per ThePony's answer, moderators would try to merge/retag, but it obviously hasn't happened (at least not completed) with those JComponent tags.

So instead of imposing it on our dearly-beloved mods, why not let the engine do it automatically, so that whenever a question with only jtable is posted, add swing to it? Or at least it can suggest the category tag(s) to the user.

The tag categorization/hierarchy should be maintained by mods or with high rep users on the category tag (the swing tag in my case).

• Duplicate of Could we make tags imply other tags? and even more of Auto assign language tags to questions based on other tags.. The problem I see is, "child" tags don't necessarily imply any relevance to their parent: plenty of jQuery questions aren't about JavaScript, plenty of C# questions aren't about the .NET libraries. This isn't a technical problem... See also: On guidelines to tagging and avoiding unnecessary tags. – Shog9 Sep 27 '10 at 16:55
• No and no. In my example, 1) jtable tag stays as is. It does NOT imply swing. Instead swing is automatically added (or at least suggested) to the question. 2) I did NOT suggest adding java. swing is a large and self-sufficient enough tag to stand on itself. Only St. Skeet can follow language tags. – Geoffrey Zheng Sep 27 '10 at 18:47


text:
# God Rays

## Recommended Posts

Asheh    354
I have been trying to impliment god rays from gpu gems 3 i have the pipeline setup where each light is rendered as a sphere (circle) in white then the geometry in black. Using this texture i perform a post process, taken from nvidia:
float4 PostProcessGodRaysPS( float2 OriginalUV : TEXCOORD0 ) : COLOR
{
float4 lightPos = mul(g_LightSources[0].Position, g_CurrWorldViewProj);
float2 screenLightPos = lightPos.xy / lightPos.w * 0.5f - 0.5f;
screenLightPos.y = 1.0f - screenLightPos.y;

// Calculate vector from pixel to light source in screen space.
half2 deltaTexCoord = (OriginalUV - screenLightPos.xy);

// Divide by number of samples and scale by control factor.
deltaTexCoord *= 1.0f / NUM_SAMPLES * Density;

// Store initial sample.
half3 color = tex2D(GodRaysSamp, OriginalUV);

// Set up illumination decay factor.
half illuminationDecay = 1.0f;

// Evaluate summation from Equation 3 NUM_SAMPLES iterations.
for (int i = 0; i < NUM_SAMPLES; i++)
{
// Step sample location along ray.
OriginalUV -= deltaTexCoord;
// Retrieve sample at new location.
half3 sample = tex2D(GodRaysSamp, OriginalUV);
// Apply sample attenuation scale/decay factors.
sample *= illuminationDecay * Weight;
// Accumulate combined color.
color += sample;
// Update exponential decay factor.
illuminationDecay *= Decay;
}
// Output final color with a further scale control factor.
return float4( color * Exposure, 1);
}

The outcome is not light streaks but mearly a repeating sphere from center to top right of the screen. Im unsure what im doing wrong. Ash.

Asheh    354

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lefthandman    138
Perhaps you could try several smaller "god rays" because the code you posted seems to be dead on with the article (unless I'm missing something). You could even try combining rays to make a ray that big. And if you want to make it look extra "god-like" increase the decay, and increase the illumination so that it fades out even more and has a more visible starting place.

##### Share on other sites
ankharlyn    122
I'm attempting to implement this for work right now as well, and I have the *exact* same results as Asheh. It seems to me that it's just a bug of some sort, since the results are definitely incorrect (regardless of tweaking the other parameters, it shouldn't be rendering a single solid shaft like that, but instead should spread them in a radial fashion).

The GPU Gems 3 article mentions rendering the quad for this shader using an ortho projection, though I honestly can't see how that would make a difference, since the shader uses the texture coordinates of the pixel, not the vertex positions at all.

@Asheh, if you'd like, perhaps we could hook up in IM and attempt to help eachother fix this bug. Just PM me if you want to try that :)

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ankharlyn    122
(I'm making a separate post since it's been a bit since my last, and the content is different.)

So, I managed to get the effect working. The problem that I had (and you're most likely having Asheh) is that the range for the texture coordinates and screen space light position in the fragment/pixel shader is different between OGL/DX. My issue was caused by having the sun position in the -1 to 1 range, and the texture coords in the 0 to 1 range.

To get more specific information about what's going wrong in your case, go ahead and just use the backbuffer for the occluded texture (instead of writing occluders as black, just use the normal backbuffer). As you move your camera around you'll be able to see where the center of the blur actually is in screen space, and more easily pinpoint what's going wrong.

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nicmenz    169
hello,

I implemented this effect three weeks ago, just have a look at

Volumetric Lighting with XNA

Just send me an PM with your email and I send you the code or any explanations.

greetings,
nicolas

[Edited by - nicmenz on May 6, 2009 4:43:31 AM]

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phresnel    953
Quote:
 Original post by nicmenz******@**********.***********.**

Better not post such on public fora.

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nicmenz    169
thanks changed :-)

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Krokhin    218
Quote:
 Original post by AshehI have been trying to impliment god rays from gpu gems 3 i have the pipeline setup where each light is rendered as a sphere (circle) in white then the geometry in black.Im unsure what im doing wrong.

Why you think so? It works pretty nice.
demo -volume.zip file(press F1 for help)
This demo shows "true" volumetric light,i.e god rays is visible when the light behind you.

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nicmenz    169
krokhin running your demo seems to be corrupted on my computer. I run a 8800 gtx, I guess you have an ATI card? Neither do I see volumetric lighting.

What technique do you use? I currently (desperately) try to perform a ray casting from the viewing perspective, but it doesn't work :-(

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Krokhin    218
I render depth and colors from transparent objects in cubemaps from light position, and draw spheres with blend around light using this info.Technique based at HLSL shaders,version 1.1 .This demo itself is pretty old (was writen ~4 years ago) and some bugs can take place of couse.


text:
Using the Intermediate Value Theorem on two functions?

Homework Statement

Given that f and g are continuous on [a, b], that f(a) < g(a), and g(b) < f(b), show that there exists at least one number c in (a, b) such that f(c) = g(c). HINT: Consider f(x) - g(x).

Homework Equations

If f is continuous on [a, b] and K is a number between f(a) and f(b), then there is at lease one number c between a and b for which f(c) = K.

The Attempt at a Solution

I know the proof for the Intermediate Value Theorem (IVT) is as follows:

Suppose F(a) < K < f(b)
g(x) = f(x) - K is continuous on [a, b], since
g(a) = f(a) - K < 0 and g(b) = f(b) - K > 0
We know there is a number c between a and b for which g(c) = 0
Then, f(c) = K

From the hint it seems to say to make a function p(x) = f(x) - g(x) like how g(x) = f(x) - K in the proof. But I don't know where to go from there... it gets confusing with so many functions!

Put h(x)=f(x)-g(x). Then h(a)<0 and h(b)>0. Thus $$0\in [h(a),h(b)]$$, which implies that h(c)=0 for some c.


text:
The transverse ballistic impact on a two-dimensional (2D) membrane causes a truncated deformation cone to develop in the wake of tensile implosion waves. Here, the cone wave reflected from the finite boundaries of the elastic membrane has been studied analytically. A first-order linear nonhomogeneous differential equation for the ratio of the reflected cone wave front velocity to the speed of tensile waves is derived, which is further used to calculate the traveling time taken by the reflected cone wave to reach to the projectile surface. Since the reflected wave starts when the membrane is already in a deformed configuration, the speed of the reflected cone wave is a function of radius r in the cylindrical coordinates as opposed to almost constant speed of the incoming cone wave studied in the literature. The analytical results are validated with molecular dynamics (MD) simulations of the ballistic impact of projectiles onto a single layer of coarse-grained (CG) graphene. In the second part of the paper, we analyze the membrane impact problem for linear isotropic viscoelastic materials and find that the tensile wave speed for stresses and displacements is the same as that obtained in the case of a linear isotropic elastic material. We also show that only under special conditions, self-similar solutions for the cone wave are possible in viscoelastic materials modeled by Maxwell, Kelvin–Voigt, or a combination of similar models. Our findings lay some grounds on which further studies on the ballistic response of viscoelastic materials can be performed.

## Introduction

The research in effective body armors with low cost, lightweight, and increased ballistic protective performance has a long history [1]. Protective systems against ballistic impact have several other applications too, mainly in the composite parts of aerodynamic vehicles, boat hulls, electronic parts, and coated fabrics. Several analytical, numerical, and experimental studies have been performed to assess ballistic impacts on high performance yarns, fabrics, metals, and other materials [2]. They involve deriving constitutive relations, modeling overall fabric behavior, increasing perforative performance, studying the effect of layers, size, temperature, friction, heat generation, etc. [3].

The analytical literature on transverse impact of two-dimensional (2D) membranes has come a long way from the early works of Grigoryan [4] and Rakhmatulin and Dem'͡ianov [5] to the recent works of Walker [6], Phoenix and Porwal [2], Naik et al. [7], and several others. All of these works take into account different stress waves originating in the system after the impact traveling at different speeds. The transverse impact on 2D membrane also causes truncated cone to develop in the wake of tensile implosion waves. Normally, the membrane radius is infinite and reflections of the waves are not accounted for. But Freeston et al. [8] developed a numerical model to describe the strain build-up in yarns due to multiple strain wave reflections from yarn crossover interactions in a woven fabric under the ballistic impact. Ha-Minh et al. [9] have included the reflection of the longitudinal tensile waves from the adjacent yarn in their analytical model for the ballistic impact behavior of 2D plain-weave fabrics. However, the reflection of the cone wave due to finite size of the target has not been considered in the ballistic impact literature. The study of these reflections is increasingly becoming possible after some recent advances in experiments and computer simulations of graphene membranes subject to the ballistic impact. The laser-induced projectile impact tests on finite length (85 μm) multilayer graphene (MLG) have shown that the MLG membranes can be excellent ballistic barrier materials [10]. Inspired by this work, atomistic molecular dynamics (MD) simulations over graphene membranes of nanometer sizes have been carried out which are able to capture stress wave propagation speed, formation of the cone wave, specific penetration energy, etc. [11,12]. In another work [13], with the help of MD simulations, we showed the existence of a critical membrane size below which the cone wave reflections from the boundaries induce perforation resulting into reduction of ballistic proof capability. The present work focuses on a detailed analytical study of the reflection cone wave.

In the first part of the paper, Sec. 2, we study the reflected cone wave occurring due to the finiteness of the specimen size in the analytical framework of the membrane model of Phoenix and Porwal [2]. The model was developed to study the stress and cone waves caused by the ballistic impact on fibrous systems. We derive a first-order linear nonhomogeneous differential equation for the ratio of the reflected cone wave front velocity to the speed of tensile waves. This is used to calculate the time taken by the reflected cone wave from the boundaries of the membrane to reach to the projectile surface. Then, we validate our analytical results with MD simulations of the ballistic impact of projectiles onto a single layer of coarse-grained (CG) graphene. The CG model of graphene maps every four atoms of graphene onto a coarse-grained bead using a mapping that preserves hexagonal symmetry [14]. Our model captures the elasticity, strength, and fracture properties of the monolayer graphene and MLG very well as reported in the previous studies [14] and has the potential to capture nano- and mesoscale failure mechanisms and size effects in the MLG systems. Details regarding the CG model have been provided in the Supplemental material which is available under the “Supplemental Data” tab for this paper on the ASME Digital Collection. The projectiles are made of CG beads forming a diamond cubic lattice structure with a lattice constant of 0.72 nm, which is double the actual lattice constant of diamond, in consistent with the CG mapping scheme of graphene. The default projectile bead mass is 96 g/mole, which gives rise to a density of 3.42 g/cm3, close to the actual diamond density of 3.5 g/cm3. More detailed information about the simulation settings can be found in Supplemental material which is available under the “Supplemental Data” tab for this paper on the ASME Digital Collection .

In the second part of the paper, Sec. 3, we analyze the 2D membrane impact problem for the linear isotropic viscoelastic materials. The deformation of a thin viscoelastic membrane of infinite size under the ballistic impact with constant projectile velocity is considered. We study the problem with two most general viscoelastic models, Maxwell and Kelvin–Voigt, and find that the tensile wave speed for stresses and displacements is the same as that for the linear isotropic elastic material. We also show that self-similar solutions for the cone wave generated in the wake of the tensile wave are possible only under special conditions. We conclude in Sec. 4 with a summary and suggestions for future work.

## Reflected Cone Wave Analysis

We first consider the deformation of a thin linear isotropic elastic membrane with fixed boundaries due to the ballistic impact by a right circular cylinder (projectile) of radius rp. The membrane of radius a is stress-free before the projectile hits at the center of the membrane. Both tensile and cone wave develop in the wake of the impact. Following the incoming cone wave hitting the fixed boundaries of the membrane, the cone wave reflects from the boundary, travels toward the center of the membrane, and starts to deform it further along its traveling path. This has been shown in Fig. 1 with the help of an MD simulation where the membrane radius is a = 1000 Å and the projectile radius is rp = 115 Å. The circular membrane dimensions are along the X- and Z-axes of the MD box and the projectile is traveling along the Y-axis. The front views in Fig. 1 are shown in such a way that the Z-axis goes into the plane and axisymmetric deformation of the membrane becomes visible. Figure 1(a) shows the membrane at the time of impact, t = −tp, where tp is the time taken by the cone wave to reach from the projectile surface r = rp to the outermost edge r = a of the membrane. We take our time origin t = 0 exactly at the point when the incoming wave reaches the membrane boundary and the reflected cone wave starts propagating backward toward the projectile surface as shown in Fig. 1(b). Figure 1(c) shows the propagation of the reflected cone wave at time t = t1 > 0. Here, we locate a membrane element at a distance r from the center under the deformation due to the reflected cone wave. We produce the illustration of the cylindrical coordinate system (r, ϕ, y) and displacements (u, 0, v) of the material element with center coordinates (r, ϕ, 0) in Fig. 1(d) inspired by Fig. 2(b) of Phoenix and Porwal [2], where u and v are displacements along radial r and downward y directions, respectively. Assuming axisymmetric deformation of the membrane, the black curve in Fig. 1(d) shows the curvature of the cone shape at time t = 0. We analyze this reflected cone wave with two basic assumptions: (a) For time t > 0, the projectile has already achieved near zero velocity, i.e., Vp(t = 0) ≈ 0. This makes sure that the deformation caused by the projectile after time t > 0 is negligible. (b) The additional strain caused by the fixed boundary is negligible. Following the MD simulations, we observe that both of these assumptions are justified. The impact velocity of the projectile at time t = −tp is Vp(t = −tp) = 500 m/s, where tp is approximately equal to 59 ps. The projectile velocity at time t = 0 is Vp(t = 0) ≈ 50 m/s. For our case, the incoming cone wave speed and the tensile wave speed are c ≈ 1.5 km/s and a0 ≈ 20 km/s, respectively. This means $Vp(t=0). So, the projectile keeps going down with a low velocity after the incoming cone wave reaches the membrane edge but this low velocity does not cause much deformation at the projectile surface. From our simulations, we note that the local in-plane strain defined in Eq. (3) at the projectile surface r = rp and time t = 0 is εt ≈ 0.035, whereas at time $t=trp−$, just before the reflected cone wave hits the projectile surface r = rp, we obtain εt ≈ 0.045. We neglect this additional strain during the time period t = 0 to t = trp in which we limit our analysis. This assumption helps simplify our analysis and we leave it upon future study to relax this assumption. Figure 1(e) shows the membrane shape when the reflected cone wave reaches the projectile surface and Fig. 1(f) shows the immediate membrane failure.

We note that all equations up to Eq. (16) in the work done by Phoenix and Porwal [2] remain the same for the displacements (u, v) caused by the reflected cone wave. Assuming that u always remains negative, the displacements are given by
$1a02∂2u∂t2=1r∂(εtr cos γ)∂r$
(1)

$1a02∂2v∂t2=−1r∂(εtr sin γ)∂r$
(2)
where $a0≈E/ρ$ is the speed of tensile waves with tensile stiffness E and density ρ of the membrane. Here, εt is the local in-plane strain given by
$εt=(1+∂u∂r)2+(∂v∂r)2−1$
(3)

and γ is the angle between the local tangential surface to the membrane element and the ground plane.

Before analyzing the reflecting cone wave in detail, we first intend to find in-plane strain εt at time t = 0 as a function of radial distance r from the center. This can be obtained from Eq. (80) of Phoenix and Porwal [2] as
$εt(r;t=0−)≈εc0ar$
(4)
where $εc0$ is the local in-plane strain at r = a and time t = 0, i.e., $εc0=εt(r=a;t=0−)$. Further, it can be seen from Eq. (92) of Phoenix and Porwal [2] that the ratio of the speed of the incoming cone wave c to that of the tensile wave a0 is related with $εc0$ as
$c/a0≈εc0$
(5)

Now we intend to verify Eq. (4) with the help of the MD simulation data of the projectile impact on the graphene membrane. A typical case has been chosen with the membrane radius a = 1000 Å and the projectile radius rp = 115 Å. The plot of εt computed with the help of the MD data at time t = 0 for r = rp to r = a is shown in Fig. 2. We try to fit the analytical expression of εt(r; t = 0) given by Eq. (4) over the MD data and obtain an excellent fit. The nonlinear least square solver “lsqcurvefit” of matlab is used for this purpose with the only fitting coefficient $εc0$. With the help of $εc0$ obtained from this fitting procedure and Eq. (5), we find that c/a0 = 0.069±0.001 with 95% confidence interval. We also calculate c and a0 independently from the MD data assuming that c remains constant for the incoming cone wave, and as noted above in the beginning of this section, we find that c ≈ 1.5 km/s, a0 ≈ 20 km/s, and c/a0 ≈ 0.075. This is quite close to the analytical fit result, and therefore, validates the analytical relation expressed in Eq. (4).

We seek self-similar solutions for (u, v) with fixed boundaries at r = a, i.e., $u(r=a,t)=v(r=a,t)=0$, based on the reflected cone wave front velocity, cr, which is a function of time t as the reflected wave propagates. We define two functions αr(t): = cr(t)/a0, which is the ratio of the reflected cone wave speed to the tensile wave speed, and $z(t):=a−cr(t)t=a−αr(t)a0t$, where $∫0tαr(t)a0dt$ is the distance traveled by the reflected cone wave from the boundary r = a in time t. The function z(t) is analogous to the variable z introduced in the analysis of the incoming cone wave (see Eq. (32)) and helps obtain self-similar solutions for the reflected cone wave.

We now aim to obtain αr(t) in the analytical framework of Sec. 3.2 in Phoenix and Porwal [2]. Self-similar solutions mean that functions $u′$ and $v′$ can be defined as $u′(r,z)≡u(r,t)$ and $v′(r,z)≡v(r,t)$, respectively. Without loss of generality, we drop the primes over the displacement functions $(u′,v′)$ and treat them as (u, v). Equations (1) and (2) in terms of z can be written as
$(dzdt)2∂2u∂z2+d2zdt2∂u∂z=a02r∂(εtr cos γ)∂r$
(6)

$(dzdt)2∂2v∂z2+d2zdt2∂v∂z=−a02r∂(εtr sin γ)∂r$
(7)
As validated by MD simulations, $((d2z/dt2)(∂u/∂z),(d2z/dt2)(∂v/∂z))$ are negligible in comparison to the other two terms in each of Eqs. (6) and (7). Now substituting $(dz/dt)=a0(αr+tαr˙)$, we have
$(αr+tαr˙)2∂2u∂z2≈1r∂(εtr cos γ)∂r$
(8)

$(αr+tαr˙)2∂2v∂z2≈−1r∂(εtr sin γ)∂r$
(9)
where $αr˙:=(dαr/dt)$ is the first derivative of the function αr(t). We note that the forms of Eqs. (8) and (9) are exactly similar to Eqs. (57) and (58) in Phoenix and Porwal [2], except for the fact that α becomes $αr+tαr˙$. Taking $α:=αr+tαr˙$, we recover exactly same equations from Eq. (57) to Eq. (72) of Phoenix and Porwal [2] with only change in α getting replaced by $αr+tαr˙$. Defining the variables ξ = r/z and ξa = ξ(r = a; z) so that ξ becomes only a function of r for a fixed z, we note that ξ varies from ξa to the outer cone edge ξ = 1. In this sense, α becomes a parameter for Eq. (76) of Phoenix and Porwal [2] and we further recover Eq. (76) to Eq. (91) of the same paper. Defining εrcεt(ξ = 1; z) as the in-plane strain at the outer cone edge of the reflected wave, we rewrite Eq. (92) of Phoenix and Porwal [2] as
$αr+tαr˙≈εrc/(1+εrc)=εrc.$
(10)
Since εrc ≪ 1 is a function of time t, we can rewrite Eq. (10) as $αr+tαr˙=g(t)$, where $g=εrc$, and then we solve Eq. (10) as first-order linear nonhomogeneous equation. The complementary homogeneous solution to this ordinary differential equation (ODE) is αr = C/t, where C is a constant. The particular solution to this ODE is $αr=(1/t)∫0tg(t)dt$. Therefore, the actual solution is $αr=C/t+(1/t)∫0tg(t)dt$. To avoid any singularity at t = 0, we can have C = 0, and therefore, $αr=(1/t)∫0tg(t)dt$. This is time-average of g(t), so this should be the same as the average over the distance $|r−a|$ traveled by the reflected cone wave in time t. This means $αr̃(r):=αr(t)$ can be expressed as
$αr̃(r)=1(r−a)∫arεrcdr$
(11)
where εrc can be treated as a function of r. We can safely assume that the in-plane strain at the outer cone edge εrc is the sum of two strains: (i) the existing strain εt(r; t = 0) caused by the incoming wave at t = 0, and (ii) the strain caused by the reflected wave, which should be approximately the same as the strain $εc0$ at the outer cone edge of the incoming cone wave. So using Eq. (4), εrc (r) can be written as
$εrc(r)≈εc0(1+a/r)$
(12)
The speed of the reflected wave $cr̃(r):=cr(t)=a0αr̃(r)$ for time 0 < ttrp becomes
$cr̃(r)≈a0εc0(r−a)∫ar(1+a/r)dr$
(13)
Using Eq. (5) and the table of basic integrals, we have the nondimensionalized version of Eq. (13) in terms of x = r/a
$cr̃(ax)≈c[1x−1∫1x(x+1)/xdx]=cx−1[x(x+1)+ln(x+x+1)−2−ln(1+2)]$
(14)
Therefore, the time taken by the reflected cone wave to reach the surface of the projectile r = rp is
$trp=−a∫1rp/a dx/cr̃=ac∫1rp/a(1−x)dx/[x(x+1)+ln(x+x+1)−2−ln(1+2)]$
(15)
On the other hand, the time taken by the incoming cone wave to reach r = a from r = rp is
$tp=ac(1−rp/a)$
(16)
With the help of Eqs. (15) and (16), we obtain a ratio trp/tp as a function of ζ = rp/a as
$trptp(ζ)=11−ζ∫1ζ(1−x)dx/[x(x+1)+ln(x+x+1)−2−ln(1+2)]$
(17)

For our MD simulation case described earlier, we have rp/a = 0.115. From Eqs. (15) and (16), we obtain trp = 0.575a/c and tp = 0.885a/c. This means trp = 0.65tp, analytically. We also calculate trp and tp from the MD data and find that trp ≈ 0.69tp. Thus, MD results are very close to the analytical results and the small differences can be attributed to the assumptions made in the analysis. We also plot time ratio $(trp/tp)$ versus ζ = rp/a in Fig. 3 and find that the time ratio $(trp/tp)$ is a monotonically increasing function of ζ with minimum around 0.63 and maximum around 0.71. This completes our analysis of the reflected wave for the present purpose.

## Ballistic Impact Response for Viscoelastic Materials

### Governing Equations Under Constant Projectile Velocity.

Here, we extend the analysis of the 2D membrane impact problem for the linear isotropic viscoelastic materials. We consider the deformation of a thin linear isotropic viscoelastic membrane of infinite size over time t > 0 due to the ballistic impact by a cylindrical projectile of radius rp at t = 0. Working with cylindrical coordinate system (r, ϕ, y), the membrane element with center coordinates (r, ϕ, y = 0) at t = 0 undergoes the displacement (u, 0, v) at time t, where u and v are the in-plane and out-of-plane components of the membrane displacement vector, respectively. It can be shown that these displacements are governed by the following partial differential equations (PDEs) [2,5]:
$ρ∂2u∂t2=1r∂∂r(σtr cos γ)−σϕr$
(18)

$ρ∂2v∂t2=−1r∂∂r(σtr sin γ)$
(19)
where ρ is the membrane initial density, σt and σϕ are the local engineering normal stress components along the deformation curve and along the circumferential direction (hoop stress), respectively, and
$cos γ=(1+∂u∂r)/(1+∂u∂r)2+(∂v∂r)2$
(20)

$sin γ=−(∂v∂r)/(1+∂u∂r)2+(∂v∂r)2$
(21)
For the 2D isotropic elastic membrane, the stresses are related to their corresponding strains in the following way:
$σt=E1−ν2(εt+νεϕ)$
(22)

$σϕ=E1−ν2(εϕ+νεt)$
(23)
where E and ν are Young's modulus and Poisson's ratio, respectively, and the strains εt and εϕ for any linear isotropic material are given by Eq. (3) and
$εϕ=u/r$
(24)
respectively. Now, we use the elastic-viscoelastic correspondence principle, see Findley et al. [15], to find the solutions of linear isotropic viscoelastic problem based on the analogy between the Laplace transformation of the governing equations for stresses, strains, and displacements and the corresponding governing equations for linear isotropic elastic problem itself in the real time domain, provided the boundary surfaces are time independent. The Laplace transforms of Eqs. (18) and (19), transforming any function f(t) in time space t to the complex frequency space s such that $L{f(t)}:=f̂(s)=∫0∞f(t)e−stdt$, yield the following results:
$ρs2û(r,s)=1r∂∂r(rσt cos γ〪)$
(25)

$ρs2v̂(r,s)=−1r∂∂r(rσt sin γ〪)$
(26)
Here, we have neglected the hoop stresses thereby making the equations independent of the circumferential direction ϕ and used the zero initial conditions at t = 0, i.e., $u(r,t=0)=(∂u/∂t)(r,t=0)=0$ and $v(r,t=0)=(∂u/∂t)(r,t=0)=0$. In order to come up with stress–strain relations analogous to Eqs. (22) and (23) in the complex frequency domain, we need to replace E and ν with their viscoelastic analogies $sEv̂$ and $νv̂$, which are functions of the corresponding transformed operators of time-dependent deviatoric P1, Q1 and dilatational P2, Q2 linear differential operators for the stress–strain constitutive equations [15]
$sEv̂(s)=3Q1̂Q2̂P2̂Q1̂+2P1̂Q2̂,νv̂(s)=P1̂Q2̂−P2̂Q1̂P2̂Q1̂+2P1̂Q2̂$
(27)
Likewise, Phoenix and Porwal [2] who neglect the effect of ν for elastic membranes, we also assume that the effect of operator $νv̂$ is negligible in comparison to other operators. This means $Q2̂≈P2̂Q1̂/P1̂$. We also assume the hydrostatic pressure independence of the elastic response, which yields $P2̂=1$, implying $Q2̂≈Q1̂/P1̂$. Using Eq. (27), we obtain the transformed modulus $sEv̂(s)$ as
$sEv̂(s)=Q1̂P1̂$
(28)
For the viscoelastic materials whose deviatoric behavior follows either the Maxwell or Kelvin–Voigt (K–V) model, we have the values of $Q1̂,P1̂$ given in Table 1, where E and η refer to Young's modulus and viscosity, respectively. Based on Eq. (22) and all the assumptions made earlier, Table 1 also provides the stress–strain relations in the transformed space. Since $σt cos γ〪=sEv̂(s)εt cos γ〪$, Eqs. (25) and (26) become
$Ω(s)û(r,s)=1r∂∂r(rεt cos γ〪)$
(29)

$Ω(s)v̂(r,s)=−1r∂∂r(rεt sin γ〪)$
(30)
where
$Ω(s)=ρsEv̂(s)={s2+EηsE/ρ,Maxwell,s2E/ρ(1+ηEs),K−V$
(31)

From these equations, it becomes clear that the stress and discontinuity waves propagate with the speed $a0=(E/ρ)$ for the Maxwell and the K–V model, respectively.2 Unlike the pure elastic waves, these stress and discontinuity waves for the Maxwell and the K–V model attenuate and possess a relaxation time τ = η/E so that for time tτ the behavior differs little from that of the elastic membrane [17], but for time t ≫ τ, the stress and discontinuity waves relax and get dissipated.

### Cone Wave Analysis.

Following the ballistic impact, a truncated cone develops in the wake of the tensile wave. Here, our purpose is to seek a self-similar solution based on the cone wave front velocity c with the assumption that c is constant. We let α = c/a0 be the ratio of cone wave speed to tensile wave speed and hope to find a self-similar solution in terms of the variable z
$z=ct+rp=αa0(t+tp)$
(32)

where tp = rp/c is the time taken by the cone wave to travel the distance rp. We show below that self-similar solutions for the cone wave can only be obtained under special conditions for both Maxwell and K–V models, and therefore, we need to solve full PDEs for complete solutions. For this purpose, we first start with the Maxwell model.

#### Maxwell Model.

By taking the inverse Laplace transform of Eqs. (29) and (30) with the help of the Maxwell model in Eq. (31), we obtain
$1a02[∂2u∂t2+1τ∂u∂t]=1r∂∂r(εtr cos γ)$
(33)

$1a02[∂2v∂t2+1τ∂v∂t]=−1r∂∂r(εtr sin γ)$
(34)
where $a0=E/ρ$ is the speed of the tensile wave and τ = η/E is the relaxation time. In terms of z, these PDEs become
$α2[∂2u∂z2+1τc∂u∂z]=1r∂∂r(εtr cos γ)$
(35)

$α2[∂2v∂z2+1τc∂v∂z]=−1r∂∂r(εtr sin γ)$
(36)
Introducing a new variable ξ = r/z, we define two new displacement functions $X(ξ):=(u+r)/z=(u/r)ξ+ξ$ and $Y(ξ):=v/z=(v/r)ξ$ in terms of which we have
$∂u∂z=(X−ξX′), ∂v∂z=(Y−ξY′), ∂2u∂z2=ξ2zX″ and ∂2v∂z2=ξ2zY″$
where the notation $(•)′≡d(•)/dξ$ means the derivative with respect to the dependent variable ξ. Substituting these into Eqs. (35) and (36), we obtain
$α2[ξ3X″+rτc(X−ξX′)]=(ξεt cos γ)′$
(37)

$α2[ξ3Y″+rτc(Y−ξY′)]=−(ξεt sin γ)′$
(38)
where Eqs. (20), (21), and (3) give
$cos γ=X′(ξ)/X′(ξ)2+Y′(ξ)2$
(39)

$sin γ=−Y′(ξ)/X′(ξ)2+Y′(ξ)2$
(40)

$εt=X′(ξ)2+Y′(ξ)2−1$
(41)
In order to have self-similar solutions, we must have Eqs. (37) and (38) turning into ODEs in ξ. Since r is an independent variable, hence self-similarity transformation is possible only when either (a) τ, i.e., material is viscous, or (b) $X−ξX′=Y−ξY′=0$, which means
$X=C1ξ, Y=C2ξ$
(42)
where C1 > 0 and C2 > 0 are constants. The conditions in Eq. (42) are purely geometrical as they can be satisfied for any Maxwell material under proper conditions. Since $X(ξ=r/z)=(u+r)/z$ and $Y(ξ=r/z)=v/z$, the conditions in Eq. (42) also imply $u=(C1−1)r$ and v = C2r, which means that we obtain simple shear motion in the rz plane. Under these conditions, the PDEs of elastodynamics turn into similar equations as obtained for the linear isotropic elastic membrane for which it has been shown that [2]
$α=c/a0=εc/(1+εc)$
(43)

where $εc≡εt|ξ=1$ is the local membrane strain along the deformation curve at the cone wave front ξ = 1 and remains constant as the cone wave travels [2]. Using Eq. (41) and the relations in Eq. (42), the constants C1 and C2 must satisfy $C12+C22=(εc+1)2$.

#### Kelvin–Voigt Model.

This time we take the inverse Laplace transform of Eqs. (29) and (30) with the help of the K–V model in Eq. (31) and obtain
$1a02∂2u∂t2=1r∂∂r[(εt+τ∂εt∂t)r cos γ]$
(44)

$1a02∂2v∂t2=−1r∂∂r[(εt+τ∂εt∂t)r sin γ]$
(45)
where a0 and τ have the same meaning as above. In terms of z, these PDEs become
$α2∂2u∂z2=1r∂∂r[(εt+αa0τ∂εt∂z)r cos γ]$
(46)

$α2∂2v∂z2=−1r∂∂r[(εt+αa0τ∂εt∂z)r sin γ]$
(47)
Analogous to Eqs. (37) and (38), we have
$α2ξ3X″=(ξεt cos γ)′−αa0τz(ξ2εt′ cos γ)′$
(48)

$α2ξ3Y″=−(ξεt sin γ)′+αa0τz(ξ2εt′ sin γ)′$
(49)
where $cos γ, sin γ$, and εt are given by Eqs. (39)(41). Here too, we obtain self-similar solutions only for the special cases for which Eqs. (48) and (49) turn into ODEs in ξ. Since z is an independent variable, this is possible only under the conditions when either (a) τ → 0, i.e., material is elastic, or (b) $(ξ2εt′ cos γ)′=(ξ2εt′ sin γ)′=0$, which means
$ξ2εt′ cos γ=C3, ξ2εt′ sin γ=C4$
(50)

where C3 and C4 are constants. We again treat the conditions in Eq. (50) as purely geometrical since any Kelvin–Voigt material can satisfy them under proper conditions. As noted for the Maxwell model, the analysis follows the elastic membrane hereafter for which the cone wave speed is given by the relation in Eq. (43). Using Eq. (41) and the relations in Eq. (50), the constants C3 and C4 must satisfy $C32+C42=εt′2|ξ=1$.

When these special conditions are not available, we fail to obtain self-similar solutions and then the governing PDEs for u and v in Eqs. (35) and (36) for the Maxwell model and Eqs. (46) and (47) for the K–V model need to be solved for the purpose of obtaining the cone wave speed. For any general constitutive model [16] of similar kinds, by extension, it can be said that the self-similar solution for cone wave is possible only for special cases.

## Conclusion

In this paper, we first study the cone wave reflected from the finite boundaries of the elastic membrane in the framework of Phoenix and Porwal [2]. We are able to derive a first-order linear nonhomogeneous differential equation for the ratio of the reflected cone wave front velocity to the speed of tensile waves, which is used to calculate the time taken by the reflected cone wave from the boundaries of the membrane to reach to the projectile surface. The analytical results are verified with the MD simulations of the ballistic impact of projectiles onto a single layer of coarse-grained graphene. Then, we analyze the 2D membrane impact problem for the linear isotropic viscoelastic materials and find that the tensile wave speed for stresses and displacements is the same as that obtained for the linear isotropic elastic material. Then, we also show that self-similar solutions for the cone wave are possible only under special conditions in the absence of which full governing PDEs must be solved. For the present purpose, only the cone wave analysis has been done for the viscoelastic materials. We also plan to analyze the deceleration of projectile due to reactive forces of the viscoelastic membrane. The analysis can also be extended to the impact response of a viscoelastic balanced fabric. Another interesting thing would be to study the widening and thickening of the crazes in polymers under the ballistic impact.

## Acknowledgment

The authors acknowledge the support from the Departments of Civil and Environmental Engineering and Mechanical Engineering at Northwestern University, as well as the Northwestern University High Performance Computing Center for a supercomputing grant.

## Funding Data

• Army Research Office (Award No. W911NF-13-1-0241).

2

A detailed discussion on the wave speeds in the frequency domain can be found in Eringen [16].

## References

References
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,
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,
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An Analysis of the System Effects in Woven Fabrics Under Ballistic Impact
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, and
Porwal
,
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,
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,”
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Ballistic Impact of Dry Woven Fabric Composites: A Review
,”
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,
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,”
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284
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Rakhmatulin
,
K. A.
, and Dem'͡ianov, I. A.,
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,
,
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6.
Walker
,
J. D.
,
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, “
Ballistic Limit of Fabrics With Resin
,”
19th International Symposium on Ballistics
, Interlaken, Switzerland, May 7–11.
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Naik
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,
Shrirao
,
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, and
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B.
,
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Ballistic Impact Behaviour of Woven Fabric Composites: Formulation
,”
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,
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), pp.
1521
1552
.
8.
Freeston
,
J. R.
,
William
, D.
, and
Claus
,
J. R.
,
1973
, “
Strain-Wave Reflections During Ballistic Impact of Fabric Panels
,”
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,
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(
6
), pp.
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351
.
9.
Ha-Minh
,
C.
,
,
A.
,
Boussu
,
F.
, and
Kanit
,
T.
,
2013
, “
On Analytical Modelling to Predict of the Ballistic Impact Behaviour of Textile Multi-Layer Woven Fabric
,”
Compos. Struct.
,
99
, pp.
462
476
.
10.
Lee
,
J.-H.
,
Loya
,
P. E.
,
Lou
,
J.
, and
Thomas
,
E. L.
,
2014
, “
Dynamic Mechanical Behavior of Multilayer Graphene Via Supersonic Projectile Penetration
,”
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,
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(
6213
), pp.
1092
1096
.
11.
Haque
,
B. Z. G.
,
Chowdhury
,
S. C.
, and
Gillespie
,
J. W.
,
2016
, “
Molecular Simulations of Stress Wave Propagation and Perforation of Graphene Sheets Under Transverse Impact
,”
Carbon
,
102
, pp.
126
140
.
12.
Yoon
,
K.
,
,
A.
, and
van Duin
,
A. C.
,
2016
, “
Atomistic-Scale Simulations of the Chemomechanical Behavior of Graphene Under Nanoprojectile Impact
,”
Carbon
,
99
, pp.
58
64
.
13.
Meng
,
Z.
,
Singh
,
A.
,
Qin
,
X.
, and
Keten
,
S.
,
2017
, “
Reduced Ballistic Limit Velocity of Graphene Membranes Due to Cone Wave Reflection
,”
EML
,
15
, pp.
70
77
.
14.
Ruiz
,
L.
,
Xia
,
W.
,
Meng
,
Z.
, and
Keten
,
S.
,
2015
, “
A Coarse-Grained Model for the Mechanical Behavior of Multi-Layer Graphene
,”
Carbon
,
82
, pp.
103
115
.
15.
Findley
,
W. N.
,
Lai
,
J. S.
, and
Onaran
,
K.
,
1989
,
Creep and Relaxation of Nonlinear Viscoelastic Materials: With an Introduction to Linear Viscoelasticity
, Dover Publications, Mineola, NY.
16.
Eringen
,
A. C.
,
1980
,
Mechanics of Continua
,
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,
Robert E. Krieger Publishing Co
, Huntington, NY.
17.
Lee
,
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, and
Kanter
,
I.
,
1953
, “
Wave Propagation in Finite Rods of Viscoelastic Material
,”
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,
24
(
9
), pp.
1115
1122
.


text:
calculate the average of 545 and 777?

• Calculate The Average Of 545 And 777

• Average =

(545 + 777) / (2)

(1322) / (2)


text:
# Can you see yourself in a mirror when you are riding on top of a light stream?

What happens if you would ride on top of a light stream and you would look into a mirror that is in front of you, could you actually see your own face? I am asking this because I heard that nothing can be faster than light--shouldn't this be the case here as well?

-
–  Qmechanic Apr 6 '11 at 20:04
You won't do anything at lightspeed. Literally. Time dilation is infinite, so absolutely nothing happens. Physicists consider any internal change to indicate that a particle is traveling slower than light (there was something going on about neutrinos changing type a while back, and that would indicate that they weren't massless particles at light speed). –  David Thornley Apr 7 '11 at 2:04
What about an external observer on earth, will he sees the image of the person riding the beam of light? –  user5411 Sep 25 '11 at 21:45
@user5411 You should have asked that as a new question, but seems like you're gone... –  Tobias Kienzler Jun 11 '13 at 13:18

## migrated from skeptics.stackexchange.comApr 6 '11 at 19:55

This question came from our site for scientific skepticism.

This question has a great pedigree! Supposedly it's what Einstein asked himself when he was first thinking about relativity.

The answer is that you can't "ride on top of a light stream" -- that is, you can't go as fast as the speed of light.

The speed of light is invariant -- same for all observers. So at any sub-light speed you can see yourself in a mirror just fine. No matter how fast you're going, a beam of light going past you still moves at the speed of light, relative to you. No matter how hard you fire your rocket engines, you never catch up with it.

-
"So at any sub-light speed you can see yourself in a mirror just fine." - ignoring relativistic aberration and doppler shift, which would make a whole lot of difference near $c$ :) –  dbrane Apr 6 '11 at 20:13
@dbrane I think it is assumed the mirror is not moving relative to you (ie. you are holding the mirror, not travelling at a large relative velocity towards the mirror). So aberration and doppler shift will not cause any effect for the person holding the mirror, no matter how fast they are moving according to an arbitrary choice of coordinate system. –  Edward Apr 6 '11 at 20:21
@Edward Sure, agreed. –  dbrane Apr 6 '11 at 20:22


text:
# Chapter 6 - Work and Energy - Problems: 6

$6.0 \times 10^3$ J.

#### Work Step by Step

The lawn’s area is 200 square meters, and the lawnmower’s width is 0.5 m. That means that the mower must be pushed a distance of (200/0.5) = 400 meters. The force and the displacement are both forward. The angle between them is 0 degrees. Use equation 6–1. $$W=(15 N)(400 m)(cos 0^{\circ}) = 6.0 \times 10^3 J$$

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.


text:
# How to make an (open) object behave like rubber when penetrated by another

So uh i've got this problem with a penetrating object and another object that serves as something like a condom if you will

So what I want, is for the penetrating object to not deform while moving into this "rubber cylinder", while the rubber cylinder does deform and reshape itself to the outer form of the penetrating object. That is, if the pen. object is bigger than the entrance hole ofc.

Also, when the object leaves the "rubber cylinder" I want the outer edges to follow (mid object in picture) and then jump back (right object in picture)

Parenting the size and position of the "rubber cylinder"-entrance to the position of the penetrating object would work i guess, but I think it'd create problems when the pen. object enters from another angle (not perpendicular to the entrance). I just read another post where someone mentions Vertex Weight Proximity modifier in a combination with a cast modifier. Don't know if thats gonna work yet. Couldn't find the time to test it.

• Are you looking for a quick effect like example in answer ?
– Ratt
Dec 8, 2018 at 6:25
• Hey, thanks! Yea that looks great, too! I'll have to check that one out later :) Dec 8, 2018 at 10:48

You could use cloth physics for the deformation part.

Pin the corners or edges of the rubber object so it keeps its shape by adding it to a Vertex Group. Let the physics do the rest.

Not sure how one could do the holes. Weight Proximity could help fake it, I don't think there is any elegant solution with Blender. You can also assign each hole size to a shape key and manually control its diameter as necessary by keyframing the influence.

Another option is perhaps a Mask modifier that would hide those parts of the mesh.

This technique might work too. Punch holes in your mesh, Add a vertex group and set only the rim of the hole with a weight of 1. Also add creases to the edges when necessary.

Add a Subdiv modifier to form the circle shape of the hole, and a solidify to add thickness to the object.

Now add a cast modifier and set it first in order. Select only the Z Axis, set the Factor to 1 and the Radius to 2 (Might need tweaking depending on the scale of your scene).

Add a an empty object and set that to be the target of the cast modifier. Also set the vertex group to the one you just created.

Already we can see the effect of the cast modifier. Add your "penetrating" object at Z = 2. Right click into the Z field of its location and choose "Copy as new driver". Select the empty, right click into the Z field of its location and choose "Paste Driver".

The field should become purple. Right-click again in this field, choose "Edit Driver". Change the type from "Average Value" to "Scripted Expression" and in the expression field, type 2 - location. This should ensure the empty affectes the hole accordingly.

Move the penetrating object along the Z axis to see the effect :


text:
Basic College Mathematics (10th Edition)

Since the prefix M means one million, 1Mm is one million meters. A unit fraction is a fraction with a value of 1. 1Mm=1,000,000m, so $\frac{1,000,000m}{1Mm}$=1. This fraction can be used to convert Mm to m. $3.5Mm\times\frac{1,000,000m}{1Mm}= 3,500,000m$


text:
At Sunrise Corporation, direct materials are added at the beginning of the process and conversion...

At Sunrise Corporation, direct materials are added at the beginning of the process and conversions costs are uniformly applied. Other details include: WIP beginning (50% for conversion) 24, 200 units Units started 125, 500 units Units completed and transferred out 106, 700 units WIP ending (60% for conversion) 43,000 units Beginning WIP direct materials $37,000 Beginning WIP conversion costs$23, 500 Costs of materials added $384, 100 Costs of conversion added$267, 125 What is the total cost of units remaining in ending WIP? $177, 332$171, 045 $165, 475$120, 957


text:
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• 1.
YKI – Ytkemiska institutet.
Template-stripped gold surfaces with 0.4-nm rms roughness suitable for force measurements: Application to the Casimir force in the 20-100 nm range2000Ingår i: Physical Review A. Atomic, Molecular, and Optical Physics, ISSN 1050-2947, E-ISSN 1094-1622, Vol. 62, s. 0621041-8Artikel i tidskrift (Refereegranskat)

Using a template-stripping method, macroscopic gold surfaces with root-mean-square (rms) roughness less than 0.4 nm have been prepared, making them useful for studies of surface interactions in the nanometer range. The utility of such substrates is demonstrated by measurements of the Casimir force at surface separations between 20 and 100 nm, resulting in good agreement with theory. The significance and quantification of this agreement is addressed, as well as some methodological aspects regarding the measurement of the Casimir force with high accuracy

• 2. Ninham, BW
YKI – Ytkemiska institutet.
Lifshitz theory of Casimir forces at finite temperature1998Ingår i: Physical Review A. Atomic, Molecular, and Optical Physics, ISSN 1050-2947, E-ISSN 1094-1622, Vol. 57, s. 1870-1880Artikel i tidskrift (Refereegranskat)

We reexamine the Casimir and Lifshitz theories of intermolecular forces at nonzero temperature. For dilute media and atomic interactions, the limits of validity of the London-van der Waals potential between atoms at finite temperature are established by a detailed asymptotic analysis. In the retarded limit, the Casimir-Polder interaction potential is shown to be rigorously correct only in the limit of zero temperature. At any non-zero temperature a different analytic form obtains and is derived. We then consider Casimir forces between perfectly conducting plates. Existing results for the case of intervening vacuum are recovered by a novel method. Moreover, we show that the Mellin transform technique and theory of generalized zeta functions allows a detailed asymptotic treatment of a system of perfectly conducting plates with an intervening electron plasma, useful in the modeling of forces between metal plates, where the finite metallic skin-depth is an important consideration.

• 3. Wennerström, H
YKI – Ytkemiska institutet.
Temperature dependence of atom-atom interactions1999Ingår i: Physical Review A. Atomic, Molecular, and Optical Physics, ISSN 1050-2947, E-ISSN 1094-1622, Vol. 60, s. 2581-2584Artikel i tidskrift (Refereegranskat)

The interaction potential at finite temperature between two atoms in the asymptotic limit of large separations $R\gg\hbar c/k_{B}T$ is shown to be $V\left(R\right) = -3k_{B}T\alpha^{2}\left(0\right)/R^{6}$, identical to the expression for two classical polarizable dipoles. This is in contrast to the zero-temperature, large-separation Casimir-Polder retarded potential $V\left(R\right) = -23\hbar c\alpha^{2}\left(0\right)/4\pi R^{7}$. The result is demonstrated by taking the low-density limit in the Lifshitz theory of interacting macroscopic bodies, and independently through a quantum-electrodynamical fourth-order perturbation theory calculation with averaging over a thermally excited blackbody radiation field. An interpretation is offered in terms of the nature of the thermal excitations of the field, leading to a manifestation of the correspondence principle at sufficiently large interatomic distances.

• 4.
Lund University.
Lund University. Lund University. RISE - Research Institutes of Sweden, Material och produktion, IVF. Lund University. Lund University.
Wall-collision line broadening of molecular oxygen within nanoporous materials2011Ingår i: Physical Review A. Atomic, Molecular, and Optical Physics, ISSN 1050-2947, E-ISSN 1094-1622, Vol. 84, nr 4Artikel i tidskrift (Refereegranskat)

Wall-collision broadening of near-infrared absorption lines of molecular oxygen confined in nanoporous zirconia is studied by employing high-resolution diode-laser spectroscopy. The broadening is studied for pores of different sizes under a range of pressures, providing new insights on how wall collisions and intermolecular collisions influence the total spectroscopic line profile. The pressure series show that wall-collision broadening is relatively more prominent under reduced pressures, enabling sensitive means to probe pore sizes of porous materials. In addition, we show that the total wall-collision-broadened profile strongly deviates from a Voigt profile and that wall-collision broadening exhibits an additive-like behavior to the pressure and Doppler broadening. © 2011 American Physical Society.

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v. 2.35.9


text:
# Accounting for Zero-Coupon Bonds

A zero-coupon bond is a bond which does not pay any periodic interest but whose total return results from the difference between its issuance price and maturity value.

For example, if Company Z issues 1 million bonds of $1000 face value bonds due to maturity in 5 years but which do not pay any interest, it is a zero-coupon bond. The company must price the bond significantly below par to be able to sell it. ## Issuance of a zero-coupon bond In order to determine the amount at which to recognize the issuance of a zero-coupon bond, we need to be able to value it. The value of a zero-coupon bond equals the present value of its maturity value determined as follows: $$\text{PV} = \frac{\text{FV}}{(\text{1}+\text{r})^{\text{n}}}$$ Where PV is the present value of the bond, FV is the maturity value, r is the periodic discount rate and FV is the maturity value. Considering the example above, if the required return is 10%, the present value of a$1,000 bond due in 5 years would be $680.58. $$\text{PV} = \frac{\text{\1,000}}{(\text{1}+\text{8%})^{\text{5}}}=\text{\680.58}$$ This would result in total proceeds of$680.58 million which are recognized as follows:

 Bank $680.58 M Bond discount$319.42 M Bonds payable $1,000 M On the statement of financial position, the bonds are presented at an amount equal to the face value of the bonds minus the discount:  Bonds payable$1,000 M Bond discount ($319.42 M) Carrying amount$680.58 M

## Recognition of interest expense

Under the effective interest method, interest expense equals the product of carrying amount of the bonds and the effective interest rate:

Interest Expense = (Bonds Payable - Discount) × Effective Interest Rate

In the example above, interest expense equals $54.45 million (=$680.58 million × 8%). This is recognized using the following journal entry:

 Interest Expense $54.45 M Bond discount$54.45 M

The carrying amount of the bonds after the first interest payment is as follows:

 Bonds payable $1,000 M Bond discount ($319.42 million - $54.45 million) ($264.97 M) Carrying amount \$735.03 M


text:
# Tangents of sums

Geometry Level 3

Let $$e_k$$ (for $$k = 0, 1, 2, 3, ...$$) be the $$k$$th-degree elementary symmetric polynomial in the variables

$x_i= \tan \theta_i$

for $$i = 0, 1, 2, \ldots$$ i.e.

$e_0 = 1$

$e_1 = \displaystyle \sum_i \tan \theta_i$

$e_2 =\displaystyle \sum_{i<j} \tan \theta_i \tan \theta_j$

and so forth.

Find the value of

$\tan \left( \displaystyle \sum_i \theta_i \right) .$

×


text:
# What are the effects of ports?

Status
Not open for further replies.

#### suyan

##### Newbie level 6
Can i know what is the effect of ports? why do my ciruit need ports to analyse my project? why ports need to be grounded? pls help.. thanks

#### armsgroup

What are ports?

it's for simplfy the analysis using network th. commonly used is s-parameters


text:
# Consequences on a system of ODEs after performing operations

• I
Hi, I have derived a matrix from a system of ODE, and the matrix looked pretty bad at first. Then recently, I tried the Gauss elimination, followed by the exponential application on the matrix (e^[A]) and after another Gauss elimination, it turned "down" to the Identity matrix. This is awfully simple, and I may wonder if the system at hand is either completely redundant, or if the procedure of using Gauss elimination on a matrix for a system of ODEs, followed by e^A and then another Gauss elimination is a reasonable and valid approach to solve ODE systems, and does not "make critical information of the ODE vanish along the way"?

Thanks!

Homework Helper
Not clear what you are doing. What is in the first matrix ? What does Gauss elimination give you when you do that ? Can you give an example ?

Not clear what you are doing. What is in the first matrix ? What does Gauss elimination give you when you do that ? Can you give an example ?

The first matrix is:

A = \begin{bmatrix}
1.0545718E-68 & 1.1271866E-(61i) \\
1 & -1\\
\end{bmatrix}

the first operation in gauss elimination with rref[A] in matlab

that gives:

B = \begin{bmatrix}
1 & -1 \\
0 & 0\\
\end{bmatrix}

the exponential of B gives then:

C= \begin{bmatrix}
2.7183 & 0.3679 \\
1.0000 & 1.0000 \\
\end{bmatrix}

finally gauss elim of C with rref[C] gives the identity matrix

Does that make sense? I suspect MATLAB is rounding exponentially small values to 0, and this is not a correct result.

Thanks

Mentor
2022 Award
Due to the very small first row, the matrix has been interpreted as ##A=\begin{bmatrix}0&0\\1&-1\end{bmatrix}## by the program, which is a singular matrix. It results in ##\exp(A)=\begin{bmatrix}1&0\\1+e^{-1}&e^{-1}\end{bmatrix}##.

Now that you exchanged the rows of ##A## for whatever reason, you have ##B=\begin{bmatrix}1&-1\\0&0\end{bmatrix}## instead, which leads to ##\exp(B)=\begin{bmatrix}e&e^{-1}\\0&1\end{bmatrix}##, if I made no mistakes by exponentiation.

So what do you want to do or say? As it is a numerical error, which led to that result, almost zero instead of equal to zero, we have algorithmic stability as one topic, or the general case of a Lie Group and its tangent space: ##\mathfrak{g} \stackrel{\exp}{\longrightarrow} G## as another.

SeM
Thanks, I prefer to rectify the numerical error! Cheers


text:
Series/parallel combination of resistance/capacitance in an RC circuit (wrt LCR meter readings)

I have this equivalent circuit I am working on. I have to find out individual values of all components in it. I can measure resistance, capacitance and impedance of the whole circuit. I am not sure resistance and capacitance values I measure with LCR meter are equivalent values of capacitance and resistance in this circuit? If so, how Ra, Rs and Rc are connected in terms of series and parallel to find equivalent resistance? Similarly, for capacitance?

In this particular circuit, if frequency goes towards zero, the measured resistance is equal to Ra + Rc + Rs. This is due to the impedance of the capacitors becoming infinite. On the opposite, if frequency goes towards infinity (i.e. is large enough), the measured resistance is equal to Rs.

It is harder to obtain CPE (and the exact value of Ra and Rc). What I would do is :

1. measure the impedance as a function of frequency, and save the results in a file ;
2. create a little program to calculate the same impedance for whatever values of the parameters (CPE, Ra and Rc). This can be done for example with Matlab (or its free clone Octave), or SciPy;
3. plot the measured impedance along with the calculated one, and adjust the parameters till there is a good correlation between the two curves

You will get a good set of parameters.

Note that you never will be able to distinguish Rc from Ra: these two resistance could be swapped, and you would get exactly the same impedance.

EDIT: typical LCR meter operation

Usually, you can choose the way an LCR meter gives you information about your circuit. This can be:

• Z (impedance) and $$\\theta\$$ (phase)
• Cs (equivalent series capacitance) and Rs (equivalent series resistance)
• Cp (equivalent parallel capacitance) and Rp (equivalent parallel resistance)
• ... (same for the inductance)

Note that, whatever the kind of data you choose, you will get a pair of values. Also: each of these two values is a function of frequency. For example, at 1kHz, a given circuit can be equivalent to a 1$$\\Omega\$$ resistor in series with a 1 µF capacitor, but at 10kHz it will be equivalent to a 2$$\\Omega\$$ resistor in series with a 0.2 µF capacitor.

In other words: LCR meters (except some high-end ones, and only for some basic circuits) won't give you an equivalent circuit, with FIXED values for a large range of frequencies. They will give you a very basic circuit with values (capacitance, resistance) accurate for ONE frequency only.

On the contrary, you want to have a (rather) complicated equivalent circuit with FIXED values for the different components. Provided the circuit is a good model for the actual component (it is not always true), you can try to fit for example the pair (Z(f), $$\\theta\$$(f)) calculated with a set of parameters (Ra, Rc, CPE...) with the experimental values. When the set of parameter gives you a good fit, you can consider that your equivalent circuit with your set of parameters is a good description of your actual circuit.

• Thank you for the answer, its very useful. However, I am stuck at how LCR meter is calculating equivalent resistance and capacitance values at different frequencies? – NHB Nov 27 '18 at 20:58
• Usually LCR meters give the L and C value that FOR A GIVEN FREQUENCY would have the same the same impedance as your circuit. You can usually choose between a series or a parallel combination. I will try to add a paragraph to my answer – Charles JOUBERT Nov 27 '18 at 21:14
• I am sorry if I am not getting my point through, let me elaborate. For a similar circuit with different values of resistances and capacitances, I am measuring C and R of whole circuit with LCR meter at different frequencies. I am getting resistance values of 6.78k, 6.41k, 0.83k 0.15k and 36.8 ohm at 0.1, 0.12, 1k, 10k and 100k frequencies. Is there a way around to find how it is calculating these values using series or parallel combination? Or LCR meter is just calculating these values with the help of impedance and phase? – NHB Nov 27 '18 at 21:29
• Yes, basically LCR meters measure impedance and phase, and give you the values of R and C that match the impedance for each particular frequency. – Charles JOUBERT Nov 27 '18 at 21:47


text:
# Right angled triangle to all acute angled triangles

What is the minimum number of cuts needed to dissect a right angled triangle into acute-angled triangles ?

• you may want to explain more. Such as What do we cut actually? A4?
– Oray
Aug 26 '17 at 7:43
• :-) @Oray, probably you can cut it diagonally and take one half and proceed with the problem ! Aug 26 '17 at 8:52

## 2 Answers

It depends very much on what you consider to be a cut. I have assumed every cut is like a straight line segment, and that an endpoint of a cut is allowed to be in the interior of the figure.

7 cuts as follows: Cut off two acute triangles so that you are left with a pentagon with four obtuse angles and one right angle. Then make 5 cuts from the corners to the centre of that pentagon.

• Yes, @Jaap Sch...s, you got it correct. Good illustration too. Aug 26 '17 at 8:39
• Can you prove this is optimal, or is that really difficult? Sep 4 '17 at 13:10
• @greenturtle3141 See boboquacks answer for why there must be at least one interior point where some cuts meet. At such a point there must be at least 5 cuts meeting there, otherwise the average angle size at that point is not acute. So that gives you 5 radial cuts. Where cuts meet an edge of the main triangle you get an obtuse angle unless two cuts meet there. I think that means you need at least two more cuts, because I think only one of the radial cuts can go to a corner of the main triangle. Sep 4 '17 at 13:45

The minimum number of cuts is:

You can't :P

Since:

Suppose the minimum number of cuts is $m$ with $\triangle ABC$, which has its non-acute angle at $A$. Then some cut $K$ must pass through $\angle A$ else it would be part of a non-acute-angled triangle. Let that cut meet $BC$ at $D$. WLOG $\angle ADB\geq\angle ADC$, then, since they sum to $180^\circ$, $\angle ADB\geq 90^\circ$. So then $\triangle ADB$ must be partitioned into acute-angled triangles in $\leq m-1$ cuts (because $K$ won't help cut up the triangle), contradicting the minimality of $m$. So such a dissection is impossible.

Note:

I am assuming cuts go all the way through the triangle.

• well that is the first impression. However, if you try, you can get the solution. Aug 26 '17 at 6:52
• Well, I agree that it depends upon how a cut is defined / interpreted as. It need not go through entirely (from a vertex to an opposite side altogether). It can stop in between ! Aug 26 '17 at 9:09
• @MeaCulpaNay To be fair, that should have been included in the question.
– Rubio
Aug 26 '17 at 21:01


text:
Bessel function

A function that is the solution to the differential equation —

\begin{align} \label{eq:11101a} x^2 \, \frac{d^2y}{dx^2} + x \, \frac{dy}{dx} + (x^2 - p^2) \, y = 0 \end{align}

where —

 p = a complex number (the order of the Bessel function)

Two common Bessel functions are shown in the following graph. Note that $$J_0(x)$$ resembles a decaying cosine while $$J_1(x)$$ resembles a decaying sine.

 Figure 1. Bessel functions - J0(x) & J1(x)

Modified Bessel function

A function that is the solution to the differential equation —

\begin{align} \label{eq:11102a} x^2 \, \frac{d^2y}{dx^2} + x \, \frac{dy}{dx} - (x^2 + p^2) \, y = 0 \end{align}

Two common modified Bessel functions are shown in the following graph. Note that $$I_0(x)$$ resembles a hyperbolic cosine while $$I_1(x)$$ resembles a hyperbolic sine.

 Figure 2. Modified Bessel functions - I0(x) & I1(x)

Bessel functions and modified Bessel functions describe the flexural and radial vibration of disks and the axial amplitude of a pressure wave in an infinitely long cylinder (and many other physical phenomena).

References:
Abramowitz, pp. 355 - 434.
Kinsler, pp. 449 - 452 — includes tables for Bessel functions $$J_0(x)$$, $$J_1(x)$$, and $$J_2(x)$$ and modified Bessel functions $$I_0(x)$$, $$I_1(x)$$, and $$I_2(x)$$
Wikipedia


text:
# Chapter 4, Exponential and Logarithmic Functions - Chapter 4 Review - Concept Check: 23

$10^{y} = x$

#### Work Step by Step

RECALL: $\log{x} = y \longrightarrow 10^y=x$ Use the rule above to obtain: $\log{x}=y \longrightarrow 10^{y} = x$

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.