text:
# Mean, Variance and Standard Deviation--Answer inside, don't know if it's right

Can anyone tell me if this is correct?

Given the following probability distribution, calculate the mean, variance, and standard deviation.

x P(x)

1 0.125

2 0.236

3 0.543

4 0.096

The requirements have been met for a probability distribution.

µ = Sum x * P (x)$µ = Sum [1 * 0.125 + 2* 0.236 + 3*0.543 + 4 * 0.096] = [0.125 + 0.472 + 1.629 + 0.384] µ = 2.6 (rounded) variance: s^2 = Sum [x^2 * P (x)] - µ^2 = Sum [(x- µ)^2 * P(x)] = (1- 2.61)^2 * 0.125 = (-1.61)^2 * 0.125 = 2.592 * 0.125 = 0.324 (2- 2.61)^2 * 0.236 = (-0.61)^2 * 0.236 = 0.372 * 0.236 = 0.087 (3-2.61)^2 * 0.0543 = (0.39)^2 *0.0543 = 0.152 * 0.0543 = 0.008 (4 - 2.61)^2 * 0.096 = (1.39)^2 * 0.096 = 1.932 * 0.096 = 0.185 Sum = 0.324 +0.087+0.008+0.185 = 0.604 = 0.6 (Rounded) Standard deviation: s = square root of variance Standard deviation = 0.777 = 0.8 (rounded) - add comment ## 1 Answer Your mean of$2.61$is right. However, there is no reason to then round the correct$2.61$to$2.6$. The values the random variable takes on are integers, presumably exact, and the probabilities are given to$3$decimal places. My calculated value for the variance is$0.6799$, giving a standard deviation of about$0.82456$. This is quite far from yours. So at least one of us should do the calculations again. I have done mine twice: it's your turn. The computation of variance: The usual formal definition of the variance$\sigma_X^2$of a random variable$X$is $$\sigma_X^2=E(X-\mu_X)^2,$$ where$\mu_X$is the mean of$X$. This definition is helpful in understanding what variance really describes. By manipulation of this basic formula (expand the square) it is not hard to show that $$\sigma_X^2=E(X^2) -\mu_X^2.$$ The second formula is, for computational purposes, often more pleasant to use than the first. You mention both formulas, and then proceed to do your calculation with the computationally more unpleasant one. Maybe for practice you should also do it the easier way. Aha!: I have found one explicit source of error. It is the line "(3-2.61)2 * 0.0543 = (0.39)2 *0.0543 = 0.152 * 0.0543 = 0.008" The$0.0543$is supposed to be$0.543$, just a slip really, but it changes the entry that presumably should have been about$0.08$to$0.008\$. That just about accounts for the discrepancy between our numbers.

Comment: I hope that you keep all these calculations within the calculator, to full accuracy, by using the "memory" feature. Writing down intermediate results and rekeying them is a nuisance, and a frequent source of error. It can also lead to serious loss of accuracy, for various reasons. One reason that you may not know about is that the calculator keeps "guard digits" internally: it is doing calculations to greater accuracy than it displays. So even if you rekey what you see exactly, you are losing accuracy.

I cannot tell whether you kept intermediate results, without rounding, in the calculator, and just truncated in writing up the calculation (good), or whether you wrote down truncated intermediate results and rekeyed and added up at the end (not so good).

-
It cannot be repeated often enough: in manual calculation, round/truncate only at the end! Retain as many significant figures as could be mustered in the intermediate calculations. –  Ｊ. Ｍ. Sep 29 '11 at 14:48
@J.M.: Agree. So I repeated it. –  André Nicolas Sep 29 '11 at 15:00


text:
In an earlier post I showed how to customise ordination diagrams produced by our vegan package for R through use of colours and plotting symbols. In a series of short posts I want to cover some of the options available in vegan that can be used to help in producing better, clearer, less cluttered ordination diagrams. First up we have ordilabel().

One of the problems that ordination results pose is that there is a lot is a lot information that we want to convey using a relatively small number of pixels. What we often end up with is a jumbled mess and because of the way the sample or species scores are plotted, the important observations could very well end up covered in all the rare species or odd samples just by virtue of their ordering in the data set.

The simplest tool that vegan provides to help in this regard is ordilabel(); it won’t produce a publication-ready, uncluttered ordination diagram but it will help you focuson the “important”1 things.

ordilabel() draws sample or species scores with their label (site ID or species name/code) taken from the dimnames of the data used to fit the ordination. To help their display, however, ordilabel() draws the labels in a box with an opaque background so that the labels plotted later (i.e. above) cover earlier labels whilst remain visible because of the opaque background. ordilabel() also allows you to specify the importance of the samples or species via the priority argument, which in effect controls which labels get drawn first or beneath all the others.

Here I’ll use a PCA of the famous Dune Meadow data2. First, we load vegan and the data and perform the ordination

In this example, I want to give plotting priority to those species or samples that are most abundant or most diverse, respectively. For this I will use Hill’s N2 for both the species and the samples, both of which can be computed via the diversity() function

The MARGIN argument refers to which dimension or margin of the data is used; 1 means rows, 2 means columns. Hill’s N2 is equal to the inverse (or reciprocal) of the Simpson diversity measure.

Throughout I’m going to use symmetric scaling of the two sets of scores for use in the biplot. As it is important to make sure the same scaling is used at each stage it is handy to store the scaling in an object and then refer to that object throughout. That way you can easily change the scaling used by altering the value in the object. Here I use scl and symmetric scaling is indicated by the number 3

ordilabel() adds labels to an existing plot, so first set up the plotting region for the PCA biplot using the plot() method with type = "n" to not plot any of the data

Now we are ready to add labels to the plot. ordilabel() takes the ordination object as the first argument and extracts the scores indicated by the display argument from the fitted object. There are a number of standard plotting arguments to control the look and feel of the labels, but the important argument is priority to control the plotting order. Here we set it to the Hill’s N2 values we computed earlier. The code chunk below adds both to the base plot we just generated

The resulting biplot should look similar to the one below

Not perfect, but better than the standard plot() method in vegan. Alternatively, one might wish to draw side by side biplots of the sample and species scores. This can be done simply with a call to layout() to split the current plot device into two plot regions, which we fill using very similar plotting commands as described above

You may notice some warnings about scaling not being a graphical parameter. These are harmful and arise because we pass scaling along as part of the ... argument which we also pass on to the plotting functions used to build the plot. We’ve tried hard to stop these warnings in vegan using a technique I blogged about a while back, but it looks like we missed a few of these. It will be fixed in a later version of vegan and the warnings will go away. Next time we’ll look at orditorp().

##Notes 1Whatever “important” means… 2Not that I think this is the best way to analyse these data, it is just for show!

R

Biplot
Graphics
Ordination
PCA
Plotting
vegan
R

format: plain


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# Homework Help: Find the integral of annoying

1. Sep 26, 2011

### 1MileCrash

1. The problem statement, all variables and given/known data

$\int sin^{2}x dx$

2. Relevant equations

3. The attempt at a solution

I don't see why. Clearly it is designed for me to shoot myself before finishing.

u = (sinx)^2
du = 2sinxcosx dx, magically equal to sin2x dx
v = x
dv = dx

(-xcos2x/2) + (sin2x/4)

=

(-2xcos2x+sin2x)/(4) + C = wrong

2. Sep 26, 2011

### SammyS

Staff Emeritus
Since you just reviewed the double angle formula for sin, what's the double angle formula for cos?

cos(2x) = ? = ? = ?

Yes, there are three forms. One will work quite nicely here.

3. Sep 26, 2011

### 1MileCrash

Firstly, let me apologize for my attitude. Secondly, thanks for the tip, I'll work it again and post back.

4. Sep 26, 2011

### SammyS

Staff Emeritus
At times, getting frustrated can lead to good things.

5. Sep 26, 2011

### 1MileCrash

$\int sin^{2}x dx$

$u = sin^{2}x$
$du = sin2x dx$
$v = x$
$dv = dx$

$xsin^{2}x - \int xsin2x dx$

$u = x$
$du = dx$
$v = -\frac{cos2x}{2}$
$dv = sin2x dx$

$xsin^{2}x - [-\frac{xcos2x}{2} - \int -\frac{cos2x}{2} dx ]$
$xsin^{2}x - [-\frac{xcos2x}{2} + \frac{sin2x}{4}]$

$xsin^{2}x + \frac{xcos2x}{2} - \frac{sin2x}{4}$

$\frac{2xsin^{2}x + xcos2x}{2} - \frac{sin2x}{4}$

$\frac{2xsin^{2}x + x(cos^{2}x - sin^{2}x)}{2} - \frac{sin2x}{4}$

$\frac{2xsin^{2}x + xcos^{2}x - xsin^{2}x}{2} - \frac{sin2x}{4}$

$\frac{xsin^{2}x + xcos^{2}x}{2} - \frac{sin2x}{4}$

$\frac{x(sin^{2}x + cos^{2}x)}{2} - \frac{sin2x}{4}$

$\frac{x}{2} - \frac{sin2x}{4}$

6. Sep 26, 2011

### Staff: Mentor

There's a much shorter way of doing this, following Sammy's hint.

sin2(x) = 1/2 * (1 - cos(2x))

This will give you an integral that you can do without having to resort to integration by parts.

7. Sep 26, 2011

### SammyS

Staff Emeritus
Mark44's expression comes from solving the following for sin2(x) .

cos(2x) = 1 - 2 sin2(x) .

It's also true that

cos(2x) = cos2(x) - sin2(x)

cos(2x) = 2 cos2(x) - 1

8. Sep 26, 2011

### Ocasta

Welcome to the wonderful world of integrals! There are times when I want to kick small, defenseless animals because of this stuff. I usually listen to some thrash metal while studying math. I don't know why, but it helps!

Try to remember as many identities as you possibly can. It makes life a lot easier.


text:
# Installation¶

## Quick setup¶

Start by cloning the git-repository

cd ~/path/to/somewhere
git clone git@gitlab.com:materials-modeling/storq.git

Next you can either proceed to do a standard install via pip

or by manually adding the paths storq and its associated binaries to your .bashrc file

export PYTHONPATH=$PYTHONPATH:~/path/to/somewhere/storq export PATH=$PATH:~/path/to/somewhere/storq/bin

Next you need to generate a configuration file. Storq comes with a command line tool that can do the bulk of the configuration for you. To initiate the automatic configuration run

The configuration process will alert you as to any settings which could not be automatically detected and hence need to be set manually. Note that some settings are optional, in which case only a warning is issued while others are critical for the operation of storq and are hence marked as “fatal”. Note that it is not uncommon that the automatic setup fails to find the binaries and POTCARs since it can be hard to search for many different names across potentially more than one filesystem. In particular for the POTCARs, storq needs a folder which contains subfolders named potpaw_ABC, where ABC=PBE, LDA, GGA, which must hold the coresponding POTCAR files. Naturally, if you only wish to use the PBE setups, only potpaw_PBE needs to exist.

Once you have completed the automatic configuration (and addressed any of its complaints) it might be necessary to set your mpi run-command in order to be able to run calculations. This is the case if your supercomputing resource uses a wrapped version of the default mpirun command (e.g. mpprun on NSC’s triolith<https://www.nsc.liu.se/systems/triolith/> or aprun on PDC’s beskow<https://www.pdc.kth.se/hpc-services/computing-systems/beskow-1.737436>). To change this option access the json configuration file through

and change the value of the mpi_command field to an appropriate command. It can furthermore be convenient to add a default allocation for your jobs to run on (you can override this later from your ASE script). To select an allocation, edit the value of the batch_account field to the name of your account (e.g. “snic20XX-X-XX” on a SNIC resource).


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# C++: Benford’s Law

Benford’s law, also called the first-digit law, refers to the frequency distribution of digits in many (but not all) real-life sources of data. In this distribution, the number 1 occurs as the first digit about 30% of the time, while larger numbers occur in that position less frequently: 9 as the first digit less than 5% of the time. This distribution of first digits is the same as the widths of gridlines on a logarithmic scale. Benford’s law also concerns the expected distribution for digits beyond the first, which approach a uniform distribution.

This result has been found to apply to a wide variety of data sets, including electricity bills, street addresses, stock prices, population numbers, death rates, lengths of rivers, physical and mathematical constants, and processes described by power laws (which are very common in nature). It tends to be most accurate when values are distributed across multiple orders of magnitude.

A set of numbers is said to satisfy Benford’s law if the leading digit d ($d \in \{1, \ldots, 9\}$) occurs with probability

$P(d) = \log_{10}(d+1)-\log_{10}(d) = \log_{10}\left(1+\frac{1}{d}\right)$

For this task, write (a) routine(s) to calculate the distribution of first significant (non-zero) digits in a collection of numbers, then display the actual vs. expected distribution in the way most convenient for your language (table / graph / histogram / whatever).

Use the first 1000 numbers from the Fibonacci sequence as your data set. No need to show how the Fibonacci numbers are obtained. You can generate them or load them from a file; whichever is easiest. Display your actual vs expected distribution.

//to cope with the big numbers , I used the Class Library for Numbers( CLN )
//if used prepackaged you can compile writing "g++ -std=c++11 -lcln yourprogram.cpp -o yourprogram"
#include <cln/integer.h>
#include <cln/integer_io.h>
#include <iostream>
#include <algorithm>
#include <vector>
#include <iomanip>
#include <sstream>
#include <string>
#include <cstdlib>
#include <cmath>
#include <map>
using namespace cln ;

class NextNum {
public :
NextNum ( cl_I & a , cl_I & b ) : first( a ) , second ( b ) { }
cl_I operator( )( ) {
cl_I result = first + second ;
first = second ;
second = result ;
return result ;
}
private :
cl_I first ;
cl_I second ;
} ;

void findFrequencies( const std::vector<cl_I> & fibos , std::map<int , int> &numberfrequencies  ) {
for ( cl_I bignumber : fibos ) {
std::ostringstream os ;
fprintdecimal ( os , bignumber ) ;//from header file cln/integer_io.h
int firstdigit = std::atoi( os.str( ).substr( 0 , 1 ).c_str( )) ;
auto result = numberfrequencies.insert( std::make_pair( firstdigit , 1 ) ) ;
if ( ! result.second )
numberfrequencies[ firstdigit ]++ ;
}
}

int main( ) {
std::vector<cl_I> fibonaccis( 1000 ) ;
fibonaccis[ 0 ] = 0 ;
fibonaccis[ 1 ] = 1 ;
cl_I a = 0 ;
cl_I b = 1 ;
//since a and b are passed as references to the generator's constructor
//they are constantly changed !
std::generate_n( fibonaccis.begin( ) + 2 , 998 , NextNum( a , b ) ) ;
std::cout << std::endl ;
std::map<int , int> frequencies ;
findFrequencies( fibonaccis , frequencies ) ;
std::cout << "                found                    expected\n" ;
for ( int i = 1 ; i < 10 ; i++ ) {
double found = static_cast<double>( frequencies[ i ] ) / 1000 ;
double expected = std::log10( 1 + 1 / static_cast<double>( i )) ;
std::cout << i << " :" << std::setw( 16 ) << std::right << found * 100 << " %" ;
std::cout.precision( 3 ) ;
std::cout << std::setw( 26 ) << std::right << expected * 100 << " %\n" ;
}
return 0 ;
}

Output:
                found                    expected
1 :            30.1 %                      30.1 %
2 :            17.7 %                      17.6 %
3 :            12.5 %                      12.5 %
4 :             9.5 %                      9.69 %
5 :               8 %                      7.92 %
6 :             6.7 %                      6.69 %
7 :             5.6 %                       5.8 %
8 :             5.3 %                      5.12 %
9 :             4.5 %                      4.58 %

SOURCE

Content is available under GNU Free Documentation License 1.2.


text:
### Elina Azrilyan

October 7th, 2019

1. Let X1, X2, . . . , Xn be n mutually independent random variables, each of which is uniformly distributed on the integers from 1 to k. Let Y denote the minimum of the Xi’s. Find the distribution of Y .

Let’s define Success as getting a minimum of Xi’s and failure as getting something other than a minimum.

The p of getting a mimimum is 1/k.

This is a geometric distribution: P(T = n) = q^(n−1)*p

Y = (1-1/k)^(n-1) * (1/k)

1. Your organization owns a copier (future lawyers, etc.) or MRI (future doctors). This machine has a manufacturer’s expected lifetime of 10 years. This means that we expect one failure every ten years. (Include the probability statements and R Code for each part.).
1. What is the probability that the machine will fail after 8 years?. Provide also the expected value and standard deviation. Model as a geometric. (Hint: the probability is equivalent to not failing during the first 8 years..)

P(T = n) = q^(n−1)*p

#Let's calculate the probability of failing in the 1st 8 years
n <- 9
p <- 1/10
q <- 1-p

Pfail <- 1 - (1-p)^n
P <- 1 - Pfail
P
## [1] 0.3874205
# Method 2: Using pgeom
P <- pgeom(8, p, lower.tail = F)
P
## [1] 0.3874205
#Calculating Expected Value
EV <- 1/p
EV
## [1] 10
#Calculating standard deviation
Var <- (1-p)/(p^2)
sqrt(Var)
## [1] 9.486833
1. What is the probability that the machine will fail after 8 years?. Provide also the expected value and standard deviation. Model as an exponential.

P(T<=x) = 1−e−λx.

p <- 1/10
P<-pexp(8, rate = p, lower.tail = FALSE)
P
## [1] 0.449329

EX=1/λ

EV<-1/p
EV
## [1] 10

Var(X)=1/(λ^2)

#standard deviation
Var <- 1/(p^2)
sqrt(Var)
## [1] 10
1. What is the probability that the machine will fail after 8 years?. Provide also the expected value and standard deviation. Model as a binomial. (Hint: 0 success in 8 years)
n<-8
p <- 1/10
P<-pbinom(0, n, p)
P
## [1] 0.4304672
#Expected value
EV<-n*p
EV
## [1] 0.8
#standard deviation
Var <- n * p * ( 1 - p )
sqrt(Var)
## [1] 0.8485281
1. What is the probability that the machine will fail after 8 years?. Provide also the expected value and standard deviation. Model as a Poisson.

λ = n*p/t

lam<-10

P<- ppois(8, 10)

P
## [1] 0.3328197

Expected value

#expected value of a Poisson distribution with parameter λ also has expectation equal to λ
EV<-lam
EV
## [1] 10

Variance

#standard deviation
#Given a Poisson distribution with parameter λ, we should guess that its variance is λ.
Var <- lam
sqrt(Var)
## [1] 3.162278


text:
Home > Standard Error > Standard Deviation Formula Standard Error

# Standard Deviation Formula Standard Error

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text:
## Intermediate Algebra for College Students (7th Edition)

$4x^2y,(3y^3+2xy-9)$
Given: $12x^2y^4+8x^3y^2-36x^2y$ Take $4xy$ as common Thus, $=4x^2y(3y^3+2xy-9)$ Hence, factors are: $4x^2y,(3y^3+2xy-9)$


text:
# Decompression Model¶

## Introduction¶

The DecoTengu implements Buhlmann decompression model ZH-L16 with gradient factors by Erik Baker, which we will refer to as ZH-L16-GF.

The initial version of the Buhlmann decompression model (ZH-L16A) was found not safe enough, its parameters were revised and two new, more conservative versions were developed - ZH-L16B and ZH-L16C. Adding gradient factors, DecoTengu supports two decompression models

ZH-L16B-GF
used for dive table calculations
ZH-L16C-GF
used for real-time dive computer calculations

## Parameters¶

The Buhlmann decompression model describes human body as 16 tissue compartments. For an inert gas and for each compartment the model assigns the following parameters

A
Buhlmann coefficient A.
B
Buhlmann coefficient B.
half life
Half-life time constant for inert gas, i.e. nitrogen, helium.

The gradient factors extension defines two parameters expressed as percentage

gf low
Controls how deep first decompression stop should start. The smaller value, the deeper first decompression stop.
gf high
Controls the time of decompression stops. The higher value, the shorter decompression time.

## Equations¶

The parameters mentioned in previous section are used by two equations

1. Schreiner equation to calculate inert gas pressure in a tissue compartment.
2. Buhlmann equation extended with gradient factors by Erik Baker to calculate ascent ceiling in a tissue compartment.

### Schreiner Equation¶

The Schreiner equation is

$P = P_{alv} + R * (t - 1 / k) - (P_{alv} - P_{i} - R / k) * e^{-k * t}$

Pressure $$P_{i}$$ is initial pressure of inert gas in tissue compartment, i.e. pressure of nitrogen in human body at the surface. The result of the equation is pressure $$P$$ in tissue compartment after time of exposure $$t$$. The inert gas pressure value is fed recursively to the equation from start till end of a dive, this is $$P_{i} = P$$ after each dive step lasting $$t$$ minutes.

The variables of the equation are

$$P_{i}$$
Initial inert gas pressure in a tissue compartment.
$$P_{alv}$$
Pressure of inspired inert gas: $$P_{alv} = F_{gas} * (P_{abs} - P_{wvp})$$
$$t$$
Time of exposure in minutes.
$$k$$
Gas decay constant for a tissue compartment: $$k = ln(2) / T_{hl}$$
$$R$$
Rate of change of inert gas pressure: $$R = F_{gas} * P_{rate}$$

where

$$P_{abs}$$
Absolute pressure of current depth [bar].
$$F_{gas}$$
Inert gas fraction, i.e. 0.79 for air.
$$P_{rate}$$
Pressure rate change [bar/min] (for example, about 1 bar/min is 10m/min).
$$T_{hl}$$
Inert gas half-life time for tissue compartment.
$$P_{wvp}$$
Water vapour pressure.

The values for $$P_{rate}$$ parameter can be

zero
constant depth exposure
negative
ascent during a dive
positive
descent during a dive

#### Example¶

Below, calculation of inert gas pressure in tissue compartment using Schreiner equation is performed. It is done for first compartment in ZH-L16B-GF decompression model and for dive profile described below when using EAN32

• descent from 0m to 30m at rate 20m/min
• dive at 30m for 20min
• ascent from 30m to 10m at rate 10m/min

For the example, the following assumptions are made

• surface pressure is 1 bar
• change of 10m depth is change of 1 bar pressure
• the water vapour pressure is 0.0627

For the start of a dive and descent from 0m to 30m the Schreiner equation variables are

$$P_{i} = 0.7902 * (1 - 0.0627) = 0.74065446$$ (initial pressure of nitrogen in tissue compartment at the surface)

$$P_{abs} = 1$$ (starting from 0m or 1 bar)

$$F_{gas} = 0.68$$ (EAN32)

$$P_{alv} = 0.68 * (1 - 0.0627) = 0.637364$$

$$R = 0.68 * 2$$ (20m/min is 2 bar per minute pressure change)

$$T_{hl} = 5.0$$ ($$N_2$$ half-life time for first tissue compartment in ZH-L16B-GF)

$$t = 1.5$$ (1.5 minute to descent by 30m at 20m/min)

$$k = ln(2) / T_{hl} = 0.138629$$

and pressure in the first tissue compartment is

$P = P_{alv} + 1.36 * (1.5 - 1 / k) - (P_{alv} - 0.74065446 - 1.36 / k) * e^{-k * 1.5} = 0.919397$

Next, continue dive at 30m for 20 minutes

$$P_{i} = 0.919397$$ (inert gas pressure in tissue compartment after descent to 30m)

$$P_{abs} = 4$$ (30m or 4 bar)

$$P_{alv} = 0.68 * (4 - 0.0627) = 2.677364$$

$$R = 0.68 * 0$$ (constant depth, no pressure change)

$$t = 20$$

and pressure in first tissue compartment is (note $$R$$ is zero and cancels parts of equation)

$P = P_{alv} + 0 - (P_{alv} - 0.919397 - 0) * e^{-k * 20} = 2.567490$

Finally, ascent from 30m to 10m

$$P_{i} = 2.567490$$

$$R = 0.68 * (-1)$$ (10m/min is 1 bar per minute pressure change, negative as it is ascent)

$$t = 2$$ (2 minutes to ascend from 30m to 10m at 10m/min)

and pressure in first tissue compartment is

$P = P_{alv} + (-0.68) * (2 - 1 / k) - (P_{alv} - 2.567490 - (-0.68) / k) * e^{-k * 2} = 2.421840$

With DecoTengu, we can calculate pressure of nitrogen in the first tissue compartment for above dive profile using ZH_L16B_GF class

>>> from decotengu.engine import GasMix
>>> model = ZH_L16B_GF()
>>> ean32 = GasMix(0, 32, 68, 0)
>>> data = model.init(1)
>>> data = model.load(1, 1.5, ean32, 2, data)
>>> round(data.tissues[0][0], 6)
0.919397
>>> data = model.load(4, 20, ean32, 0, data)
>>> round(data.tissues[0][0], 6)
2.567491
>>> data  = model.load(4, 2, ean32, -1, data)
>>> round(data.tissues[0][0], 6)
2.42184

The relationship between dive time, absolute pressure of dive depth and inert gas pressure in a tissue compartment is visualized on figure Inert gas pressure in tissue compartment for a dive profile.

Inert gas pressure in tissue compartment for a dive profile

### Buhlmann Equation¶

Buhlmann equation extended with gradient factors by Erik Baker is

$P_l = (P - A * gf) / (gf / B + 1.0 - gf)$

Tissue absolute pressure limit $$P_l$$ determines depth of ascent ceiling, which is calculated using inert gas pressure $$P$$ in tissue compartment (result of Schreiner equation), Buhlmann coefficients $$A$$ and $$B$$ (for given tissue) and current gradient factor value $$gf$$.

Current gradient factor is a value, which changes evenly between values of gf low and gf high decompression model parameters. It has gf low value at first decompression stop and gf high value at the surface.

To support multiple inert gases, i.e. trimix with nitrogen and helium, we need to track pressure of each inert gas separately. The Buhlmann equation variables can be supported then with the following equations

\begin{align}\begin{aligned}P = P_{n2} + P_{he}\\A = (A_{n2} * P_{n2} + A_{he} * P_{he}) / P\\B = (B_{n2} * P_{n2} + B_{he} * P_{he}) / P\end{aligned}\end{align}

where

$$P_{n2}$$
Pressure of nitrogen in current tissue compartment.
$$P_{he}$$
Pressure of helium in current tissue compartment.
$$A_{n2}$$
Buhlmann coefficient $$A$$ for nitrogen.
$$B_{n2}$$
Buhlmann coefficient $$B$$ for nitrogen.
$$A_{he}$$
Buhlmann coefficient $$A$$ for helium.
$$B_{he}$$
Buhlmann coefficient $$B$$ for helium.

#### Example¶

We continue the example described in Schreiner equation section. In the example, the nitrogen pressure in first tissue compartment at various depths is

 Depth [m] Runtime [min] P [bar] 0 0 0.74065446 30 1.5 0.919397 30 21.5 2.567490 10 23.5 2.421840

The Buhlmann coefficients for the first tissue compartment in ZH-L16B-GF model are (nitrox dive, therefore we skip trimix extension)

\begin{align}\begin{aligned}A = 1.1696\\B = 0.5578\end{aligned}\end{align}

We attempt to ascend to first decompression stop and use $$0.3$$ as current gradient factor value

\begin{align}\begin{aligned}(0.74065446 - 1.1696 * 0.3) / (0.3 / 0.5578 + 1.0 - 0.3) = 0.314886\\(0.919397 - 1.1696 * 0.3) / (0.3 / 0.5578 + 1.0 - 0.3) = 0.4592862\\(2.567490 - 1.1696 * 0.3) / (0.3 / 0.5578 + 1.0 - 0.3) = 1.7907266\\(2.421840 - 1.1696 * 0.3) / (0.3 / 0.5578 + 1.0 - 0.3) = 1.6730607\end{aligned}\end{align}

Using eq_gf_limit() function (we omit helium parameters by substituting them with 0 as the example uses nitrox gas mix only)

>>> eq_gf_limit(0.3, 0.74065446, 0, 1.1696, 0.5578, 0, 0)
0.31488600902007363
>>> eq_gf_limit(0.3, 0.919397, 0, 1.1696, 0.5578, 0, 0)
0.459286247718912
>>> eq_gf_limit(0.3, 2.567490, 0, 1.1696, 0.5578, 0, 0)
1.790726556208904
>>> eq_gf_limit(0.3, 2.421840, 0, 1.1696, 0.5578, 0, 0)
1.6730606957680387

Let’s put the calculations into the table

 Depth [m] Runtime [min] P [bar] Limit [bar] Note 0 0 0.74065446 0.314886 30 1.5 0.919397 0.4592862 30 21.5 2.567490 1.7907266 ~8m 10 23.5 2.421840 1.6730607 ~7m

When starting ascent from 30 meters, the ceiling limit is at about 8m. After ascent to 10m, the ceiling limit changes to about 7m - the first (and any other) tissue compartment desaturates during ascent. This suggests, that to find exact depth of first decompression stop an algorithm will be required (see Algorithms section).

## Calculations¶

The main code for decompression model is implemented in ZH_L16_GF class.

Inert gas pressure of each tissue compartment for descent, ascent and at constant depth is calculated by the ZH_L16_GF.load() method, which uses Schreiner equation.

The pressure of ascent ceiling of a diver is calculated with the ZH_L16_GF.ceiling_limit() method. The method allows to determine

• depth of first decompression stop - a diver cannot ascent from the bottom shallower than ascent ceiling
• length of decompression stop - a diver cannot ascent from decompression stop until depth of ascent ceiling decreases


text:
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This is my understanding: the statistics of the source described in the paper depend on which character is produced first. If the first character is $a$, then one of the source properties is that letters in odd positions are always $a$. However, if the first letter is $b$ (in other words, a shift of $1$ in the circuit), then letters in odd positions can be ...


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text:
rm(list=ls(all=TRUE)) # clear memory
library(dplyr)
# a frequency list of a single untagged corpus file ###############################################
corpus.file <-                         # make corpus file the result of
"03_data/Brown1_J_untagged.TXT", # this file
what=character(),                # which contains character strings
sep="\n",                        # separated by line breaks
quiet=TRUE)                      # suppress output

## look at ...
head(corpus.file, 10) # ... corpus.file, the first 10 lines
##  [1] "J01 0010 #1. INTRODUCTION# IT HAS recently become practical to use the"
##  [2] "J01 0020 radio emission of the moon and planets as a new source of information"
##  [3] "J01 0030 about these bodies and their atmospheres. The results of present observations"
##  [4] "J01 0040 of the thermal radio emission of the moon are consistent with"
##  [5] "J01 0050 the very low thermal conductivity of the surface layer which was derived"
##  [6] "J01 0060 from the variation in the infrared emission during eclipses (e&g&,"
##  [7] "J01 0070 Garstung, 1958). When sufficiently accurate and complete measurements"
##  [8] "J01 0080 are available, it will be possible to set limits on the thermal"
##  [9] "J01 0090 and electrical characteristics of the surface and subsurface materials"
## [10] "J01 0100 of the moon.   Observations of the radio emission of a planet"
tail(corpus.file, 10) # ... corpus.file, the last 10 lines
##  [1] "J80 1800 The sensing of this rotation by the ~<X> gyro can be utilized"
##  [2] "J80 1810 to direct the platform into proper heading.   In Fig& 7-4,"
##  [3] "J80 1820 the input axis of the three-axis platform is shown at some point on the"
##  [4] "J80 1830 earth. The point is at a latitude ~|l, and the platform is at"
##  [5] "J80 1840 an error in heading east. The earth is spinning at an angular velocity"
##  [6] "J80 1850 ~\\q equal to one revolution per 24 hr&. When the platform is"
##  [7] "J80 1860 level, ~|e is a rotation about the ~<Z> axis of the platform"
##  [8] "J80 1870 **f. Since the earth is rotating and the unleveled gyro-stabilized platform"
##  [9] "J80 1880 is fixed with respect to a reference in space, an observer on"
## [10] "J80 1890 the earth will see the platform rotating (with respect to the earth)."
## remove the line annotation
corpus.file <-      # make corpus file the result of
gsub(            # replacing
"^.........", # the first 9 characters of each line
"",           # by nothing
corpus.file,  # in corpus.file
perl=TRUE)    # using Perl-compatible regular expressions

## look at ...
head(corpus.file, 10) # ..., corpus.file the first 10 lines
##  [1] "#1. INTRODUCTION# IT HAS recently become practical to use the"
##  [2] "radio emission of the moon and planets as a new source of information"
##  [3] "about these bodies and their atmospheres. The results of present observations"
##  [4] "of the thermal radio emission of the moon are consistent with"
##  [5] "the very low thermal conductivity of the surface layer which was derived"
##  [6] "from the variation in the infrared emission during eclipses (e&g&,"
##  [7] "Garstung, 1958). When sufficiently accurate and complete measurements"
##  [8] "are available, it will be possible to set limits on the thermal"
##  [9] "and electrical characteristics of the surface and subsurface materials"
## [10] "of the moon.   Observations of the radio emission of a planet"
tail(corpus.file, 10) # ..., corpus.file the last 10 lines
##  [1] "The sensing of this rotation by the ~<X> gyro can be utilized"
##  [2] "to direct the platform into proper heading.   In Fig& 7-4,"
##  [3] "the input axis of the three-axis platform is shown at some point on the"
##  [4] "earth. The point is at a latitude ~|l, and the platform is at"
##  [5] "an error in heading east. The earth is spinning at an angular velocity"
##  [6] "~\\q equal to one revolution per 24 hr&. When the platform is"
##  [7] "level, ~|e is a rotation about the ~<Z> axis of the platform"
##  [8] "**f. Since the earth is rotating and the unleveled gyro-stabilized platform"
##  [9] "is fixed with respect to a reference in space, an observer on"
## [10] "the earth will see the platform rotating (with respect to the earth)."
## identify all characters used in the file
strsplit(corpus.file, "") %>% # split up corpus files at every character, then
unlist                 %>% # make the output a vector, then
unique                 %>% # find the unique types, then
sort                       # sort and print the output
##  [1] " "  "_"  "-"  ","  ";"  ":"  "!"  "?"  "."  "'"  "\"" "("  ")"  "["
## [15] "]"  "{"  "}"  "@"  "*"  "/"  "\\" "&"  "#"  "%"  ""  "^"  "+"  "<"
## [29] ">"  "|"  "~"  "\$"  "0"  "1"  "2"  "3"  "4"  "5"  "6"  "7"  "8"  "9"
## [43] "a"  "A"  "b"  "B"  "c"  "C"  "d"  "D"  "e"  "E"  "f"  "F"  "g"  "G"
## [57] "h"  "H"  "i"  "I"  "j"  "J"  "k"  "K"  "l"  "L"  "m"  "M"  "n"  "N"
## [71] "o"  "O"  "p"  "P"  "q"  "Q"  "r"  "R"  "s"  "S"  "t"  "T"  "u"  "U"
## [85] "v"  "V"  "w"  "W"  "x"  "X"  "y"  "Y"  "z"  "Z"
## extract all words from the file
strsplit(corpus.file,         # split up corpus.file
"[^\\p{L}]+",              # at everything that's not 1 or more letters
perl=TRUE) %>%             # using Perl-compatible regular expressions, then
unlist     %>%             # make that a vector, then
tolower    -> corpus.words # concert to lower case and put it into corpus.words

## tabulate and sort by frequency
table(corpus.words)         %>%                # tabulate/count the elements in corpus.words, then
sort(., decreasing=TRUE) -> corpus.freqlist # sort descending order & put it into corpus.freqlist

## look at ...
head(corpus.freqlist, 30) # the 30 most frequent words
## corpus.words
##   the    of   and    in    to     a    is  that   for    be    as    by
## 12540  7459  4288  4110  3958  3530  2410  1715  1568  1360  1291  1227
##  with    it   was     f   are  this    on    or which        from   not
##  1158  1149  1115  1009   993   993   957   907   835   795   793   787
##    an    at  were  have   one     s
##   739   739   633   551   550   539
tail(corpus.freqlist, 30) # some of the 30 least frequent words
## corpus.words
##     yakima       yale     yamabe      yarns         yd   yearbook
##          1          1          1          1          1          1
##  yearnings     yeasts      yeats   yielding     yinger        yok
##          1          1          1          1          1          1
##     yorker    yorkers   yorktown    younger      yours yourselves
##          1          1          1          1          1          1
##         yr    yucatan     yujobo       yuri       yyyy      zabel
##          1          1          1          1          1          1
##       zeal      zeros       zest       zinc      zones        zur
##          1          1          1          1          1          1
## plot a Zipfian curve
plot(                              # plot
x=log2(rank(-corpus.freqlist)), # on the x-axis the logged ranks of the words in the freq list
y=log2(corpus.freqlist),        # on the y-axis the logged frequencies of the words in the freq list
pch=16, col="#00000020")        # filled circles in grey
grid()                          # add a grid

# a frequency list of a several tagged corpus files ###############################################
## define the locations of the corpus files
corpus.files <-                  # make corpus.files
dir(                       # the content of the directory/folder
"03_data/Brown_tagged", # with this name
full.names=TRUE)        # and retain the complete paths to the files

## create a collector/container for all words from all files
all.corpus.words <- c()

## load each corpus file, extract the words, and store in collector/container for later
for (counter in 1:15) {         # for each of the 15 corpus files
current.corpus.file <-       # make current.corpus.file the result of
corpus.files[counter], # the corpus file number 'counter'
what=character(),      # which contains character strings
sep="\n",              # separated by line breaks
quiet=TRUE)            # suppress output

current.corpus.file <-     # make current.corpus.file the result of
gsub(                   # replacing
"^.*? ",             # everything from the beginning of a line till the 1st space
"",                  # by nothing
current.corpus.file, # in current.corpus.file
perl=TRUE)           # using Perl-compatible regular expressions

strsplit(               # split up
current.corpus.file, # current.corpus.file
"_[^ ]+ ?",          # at every _ followed by non-spaces till maybe a space
perl=TRUE) %>%       # using Perl-compatible regular expressions, then
unlist     %>%       # make the resulting object a vector, then
tolower    ->        # make that be all lowercase, then put this into
current.corpus.words # an object called current.corpus.words

all.corpus.words <-         # make all.corpus.words
c(                       # the combination of
all.corpus.words,     # all.corpus.words and
current.corpus.words) # current.corpus.words
}

all.corpus.words <-     # make all.corpus.words the result of
grep(                # finding
"\\p{L}",         # elements with at least one letter
all.corpus.words, # in all.corpus.words
perl=TRUE,        # using Perl-compatible regular expressions
value=TRUE)       # return what you find (i.e. the words)

## tabulate and sort by frequency
table(all.corpus.words)     %>%                # tabulate/count the elements in corpus.words, then
sort(., decreasing=TRUE) -> corpus.freqlist # sort descending order & put it into corpus.freqlist

## look at ...
head(corpus.freqlist, 30) # the 30 most frequent words
## all.corpus.words
##   the    of   and    to     a    in  that    is   was    he   for    it
## 70005 36423 28861 26178 23140 21344 10602 10109  9836  9565  9493  8763
##  with    as   his    on    be    at    by     i  this   had   not   are
##  7295  7252  6997  6744  6380  5383  5305  5159  5146  5140  4612  4394
##   but  from    or  have    an  they
##  4383  4373  4205  3946  3741  3624
tail(corpus.freqlist, 30) # some of the 30 least frequent words
## all.corpus.words
##          zend-avesta              zennist               zeroed
##                    1                    1                    1
##              zhitkov           zhitzhakli                 zhok
##                    1                    1                    1
##                ziggy ziminska-sygietynska            zingggg-o
##                    1                    1                    1
##               zinman              zionism             zionists
##                    1                    1                    1
##                  zip               zipped               zipper
##                    1                    1                    1
##              ziraldo               zlotys                  zoe
##                    1                    1                    1
##               zombie              zombies                zoned
##                    1                    1                    1
##                zooey              zoology               zoomed
##                    1                    1                    1
##              zooming                zooms               zoooop
##                    1                    1                    1
##            zorrillas                   zu                 zwei
##                    1                    1                    1
plot(                              # plot
x=log2(rank(-corpus.freqlist)), # on the x-axis the logged ranks of the words in the freq list
y=log2(corpus.freqlist),        # on the y-axis the logged frequencies of the words in the freq list
pch=16, col="#00000020")        # filled circles in grey
grid()                          # add a grid

# I will show you a frequency list of 'things in a slot' later.`