The Physics Computer – Physicspupil

November 8, 2022November 8, 2022

202211081129 Python Formula 002

Coulomb’s law states that $F$ the magnitude of the electrostatic force of attraction or repulsion between two point charges $Q_1$ and $Q_2$ is directly proportional to $Q_1Q_2$ the product of the magnitudes of charges and inversely proportional to $r^2$ the square of the distance between them:

$\displaystyle{F=\frac{Q_1Q_2}{4\pi\epsilon_0r^2}}$

_{Wikipedia on Coulomb’s law}

Roughwork.

pi = 3.1415926535
epsilon_0 = 8.8541878128e-12
charge_1 = float(input("Charge of Q_1: "))
charge_2 = float(input("Charge of Q_2: "))
distance = float(input("Separating distance: "))
force = charge_1 * charge_2 / (4 * pi * epsilon_0 * distance ** 2)
print("Electrostatic force (in Newton) is:", force)

pi = 3.1415926535

epsilon_0 = 8.8541878128e-12

charge_1 = float(input("Charge of Q_1: "))

charge_2 = float(input("Charge of Q_2: "))

distance = float(input("Separating distance: "))

force = charge_1 * charge_2 / (4 * pi * epsilon_0 * distance ** 2)

print("Electrostatic force (in Newton) is:", force)

The nucleus of a helium atom contains two protons $\sim 3.8\times 10^{-15}\,\mathrm{m}$ apart each of electric charge $+e=+1.6\times 10^{-19}\,\mathrm{C}$ . The electrostatic force of repulsion between them is

Charge of Q_1: 1.6e-19
Charge of Q_2: 1.6e-19
Separating distance: 3.8e-15
Electrostatic force (in Newton) is: 15.933609826070786

Charge of Q_1: 1.6e-19

Charge of Q_2: 1.6e-19

Separating distance: 3.8e-15

Electrostatic force (in Newton) is: 15.933609826070786

November 8, 2022November 8, 2022

202211081044 Python Formula 001

Converting between temperature scales.

$\begin{aligned} & x\,^\circ\mathrm{F}\triangleq (x-32)\times \frac{5}{9}\,^\circ\mathrm{C} & \qquad & x\,^\circ\mathrm{C}\triangleq (x\times \frac{9}{5}+32)\,^\circ\mathrm{F} \\ & x\,^\circ\mathrm{F}\triangleq (x+459.67)\times \frac{5}{9}\,\mathrm{K} & \qquad & x\,\mathrm{K}\triangleq (x\times \frac{9}{5}-459.67)\,^\circ\mathrm{F} \\ & x\,^\circ\mathrm{F} \triangleq (x+459.67) ^\circ\mathrm{R} & \qquad & x\,^\circ\mathrm{R}\triangleq (x-459.67)\,^\circ\mathrm{F} \\ \end{aligned}$

Roughwork.

degree_Celsius = float(input("Input degrees Celsius: "))
degree_Fahrenheit = degree_Celsius * (9 / 5) + 32
print(degree_Fahrenheit)

degree_Celsius = float(input("Input degrees Celsius: "))

degree_Fahrenheit = degree_Celsius * (9 / 5) + 32

print(degree_Fahrenheit)

One has a slight fever when one’s body temperature is $37.8\,^\circ\mathrm{C}$ , or in degrees Fahrenheit,

Input degrees Celsius: 37.8
100.03999999999999

1 2	Input degrees Celsius: 37.8 100.03999999999999

The rest of conversion of temperature scales, in Celsius, Fahrenheit, Kelvin, Rankine, etc., are left for the reader to code.

May 26, 2022July 24, 2022

202205261415 Arithmetic in Lisp

Given two operands $a$ and $b$ which are real numbers, define the operators,

$a+b$ , $a-b$ , $a\times b$ , $a\div b$ , $a^{b}$ , $\sqrt[b]{a}$ , $a\bmod b$ ,

with respect to arithmetic operations such as addition, subtraction, multiplication, division, exponentiation, rooting, and Modulo.

+------+------------+-----+-----+-------------------------+
|  ID  |  operator  |  a  |  b  |  instance for example   |
+------+------------+-----+-----+-------------------------+
|  1   |  +         |  2  |  3  |  a + b  = 2 + 3    = 5  |
|  2   |  -         |  5  |  2  |  a - b  = 5 - 2    = 3  |
|  3   |  ×         |  3  |  2  |  a * b  = 3 * 2    = 6  |
|  4   |  ÷         |  9  |  3  |  a / b  = 9 / 3    = 3  |
|  5   |  ^         |  3  |  2  |  a ^ b  = 3 ^ 2    = 9  |
|  6   |  √         |  3  |  8  |  a √ b  = 3 √ 8    = 2  |
|  7   |  mod       |  7  |  4  |  a mod b = 7 mod 4 = 3  |
+------+------ -----+-----+-----+-------------------------+

+------+------------+-----+-----+-------------------------+

| ID | operator | a | b | instance for example |

+------+------------+-----+-----+-------------------------+

| 1 | + | 2 | 3 | a + b = 2 + 3 = 5 |

| 2 | - | 5 | 2 | a - b = 5 - 2 = 3 |

| 3 | × | 3 | 2 | a * b = 3 * 2 = 6 |

| 4 | ÷ | 9 | 3 | a / b = 9 / 3 = 3 |

| 5 | ^ | 3 | 2 | a ^ b = 3 ^ 2 = 9 |

| 6 | √ | 3 | 8 | a √ b = 3 √ 8 = 2 |

| 7 | mod | 7 | 4 | a mod b = 7 mod 4 = 3 |

+------+------ -----+-----+-----+-------------------------+

Not so accustomed to these, Lisp cannot interpret or compile them until we have defined those seven primitive functions, better known as symbolic computations; as were newborn infants taught human intelligence by grownup adults between homo sapiens.

To Lisp, arithmetic operations are usually $\textrm{\scriptsize{NOT}}$ predefined. We have in our toolkit $\textrm{\scriptsize{ONLY}}$ three mathematical functors, i.e., list, iteration, and recursion.

The roots of quadratic equation are given by the formula:

$\displaystyle{x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}}$

the expected outcome of which is calculated by parts, first the discriminant ‘delta’:

(define (delta a b c) (- (square b) (* 4 a c)))

1	(define (delta a b c) (- (square b) (* 4 a c)))

$\Delta = b^2-4ac$ ;

then the plus-minus ‘pm’:

(define (pm u v)
	(values (cons (+ u v) (- u v))))

1 2	(define (pm u v) (values (cons (+ u v) (- u v))))

$\pm (x)=(+x,-x)$ ;

and hence

(/ (pm (- b) (sqrt (delta a b c))) (* a 2))

1	(/ (pm (- b) (sqrt (delta a b c))) (* a 2))

will solve for real roots, ignoring the complex though.

$\therefore$ Predefining the seven primitive functions is in order for solving any computational problems by Lisp.

(to be continued)

May 25, 2022May 26, 2022

202205251335 SQL Table 005

CREATE TABLE SQL_Table_005
    (ID int,
    Function varchar(255),
    Derivative varchar(255));
INSERT INTO SQL_Table_005
    (Function, Derivative)
    VALUES
    ('position', 'velocity')
    ('velocity', 'acceleration')
    ('momentum', 'force')
    ('work', 'power')
    ('angular momentum', 'torque')
    ('electric charge', 'electric current')
    ('magnetic flux', 'induced voltage')

CREATE TABLE SQL_Table_005

(ID int,

Function varchar(255),

Derivative varchar(255));

INSERT INTO SQL_Table_005

(Function, Derivative)

VALUES

('position', 'velocity')

('velocity', 'acceleration')

('momentum', 'force')

('work', 'power')

('angular momentum', 'torque')

('electric charge', 'electric current')

('magnetic flux', 'induced voltage')

Output.
+------+--------------------+--------------------+
|  ID  |  Function          |  Derivative        |
+------+--------------------+--------------------+
|  1   |  position          |  velocity          |
|  2   |  velocity          |  acceleration      |
|  3   |  momentum          |  force             |
|  4   |  work              |  power             |
|  5   |  angular momentum  |  torque            |
|  6   |  electric charge   |  electric current  |
|  7   |  magnetic flux     |  induced voltage   |
+------+--------------------+--------------------+

Output.

+------+--------------------+--------------------+

| ID | Function | Derivative |

+------+--------------------+--------------------+

| 1 | position | velocity |

| 2 | velocity | acceleration |

| 3 | momentum | force |

| 4 | work | power |

| 5 | angular momentum | torque |

| 6 | electric charge | electric current |

| 7 | magnetic flux | induced voltage |

+------+--------------------+--------------------+

December 2, 2021October 24, 2022

202112021138 SQL Table 004

Background. (Equilibrium)

Equilibrium is established by the exchange of energy, volume, or particle number between different systems or subsystems.

_{Section 2.2. Arovas. (2019). Thermodynamics and Statistical Mechanics}

CREATE TABLE SQL_Table_004
    (ID int,
    Object_of_exchange varchar(255),
    Constancy varchar(255)
    Subject_of_equilibrium varchar(255)
    );
INSERT INTO SQL_Table_004
    (Object_of_exchange, Constancy, Subject_of_equilibrium)
    VALUES
    ('energy', 'temperature (T)', 'thermal')
    ('volume', 'momentum/temperature (p/T)', 'mechanical')
    ('particle', 'chemical potential/temperature (\mu /T)', 'chemical')

CREATE TABLE SQL_Table_004

(ID int,

Object_of_exchange varchar(255),

Constancy varchar(255)

Subject_of_equilibrium varchar(255)

);

INSERT INTO SQL_Table_004

(Object_of_exchange, Constancy, Subject_of_equilibrium)

VALUES

('energy', 'temperature (T)', 'thermal')

('volume', 'momentum/temperature (p/T)', 'mechanical')

('particle', 'chemical potential/temperature (\mu /T)', 'chemical')

Output.
+------+---------------------+-------------------------------+--------------------------+
|  ID  | Object_of_exchange  |  Constancy                    |  Subject_of_equilibrium  |
+------+---------------------+-------------------------------+--------------------------+
|  1   |  energy             |  temperature (T)              |  thermal                 |
|  2   |  volume             |  momentum/temperature (p/T)   |  mechanical              |
|  3   |  particle           |  chemical potential/          |  chemical                |
|      |                     |    temperature (\mu /T)       |                          |
+------+---------------------+-------------------------------+--------------------------+

Output.

+------+---------------------+-------------------------------+--------------------------+

+------+---------------------+-------------------------------+--------------------------+

+------+---------------------+-------------------------------+--------------------------+

November 25, 2021October 24, 2022

202111251402 Exercise 2.59 in Lisp

Implement the union-set operation for the unordered-list representation of sets.

_{Extracted from Structure and Interpretation of Computer Programs, SICP}

Attempts.

Define a function element-of-set?by

(define (element-of-set? x set)
(cond ((null? set) false)
((equal? x (car set)) true)
(else (element-of-set? x (cdr set)))))

(define (element-of-set? x set)

(cond ((null? set) false)

((equal? x (car set)) true)

(else (element-of-set? x (cdr set)))))

to determine whether an object xis an element of a set set.

Define a function intersection-set by

(define (intersection-set set1 set2)
(cond ((or (null? set1) (null? set2)) ’())
((element-of-set? (car set1) set2)
(cons (car set1)
(intersection-set (cdr set1) set2)))
(else (intersection-set (cdr set1) set2))))

(define (intersection-set set1 set2)

(cond ((or (null? set1) (null? set2)) ’())

((element-of-set? (car set1) set2)

(cons (car set1)

(intersection-set (cdr set1) set2)))

(else (intersection-set (cdr set1) set2))))

to draw the intersection intersection-set of two sets set1 and set2.

Define a function union-set by

(define (union-set set1 set2)
(cond ((and (null? set1) (null? set2)) ’())
((null? set1) set2)
((null? set2) set1)
((element-of-set? (car set1) set2)
(cons (car set1)
(union-set (cdr set1) set2)))
(else (union-set (cdr set1) set2))))

(define (union-set set1 set2)

(cond ((and (null? set1) (null? set2)) ’())

((null? set1) set2)

((null? set2) set1)

((element-of-set? (car set1) set2)

(cons (car set1)

(union-set (cdr set1) set2)))

(else (union-set (cdr set1) set2))))

to draw the union union-set of two sets set1 and set2.

September 29, 2021August 11, 2022

202109291137 SQL Table 003

CREATE TABLE SQL_Table_003
	(ID int,
	Formulations_of_quantum_mechanics varchar(255),
	Proposer varchar(255),
	Time_(Yr) varchar(255)
	);
INSERT INTO SQL_Table_003
	(Formulations_of_quantum_mechanics, Proposer, Time_(Yr))
	VALUES
	('transformation theory', 'Paul Dirac', '~1927 ')
	('matrix mechanics', 'Werner Heisenberg, Max Born, Pascual Jordan', '~1925')
	('wave mechanics', 'Erwin Schrödinger', '~1925')

CREATE TABLE SQL_Table_003

(ID int,

Formulations_of_quantum_mechanics varchar(255),

Proposer varchar(255),

Time_(Yr) varchar(255)

);

INSERT INTO SQL_Table_003

(Formulations_of_quantum_mechanics, Proposer, Time_(Yr))

VALUES

('transformation theory', 'Paul Dirac', '~1927 ')

('matrix mechanics', 'Werner Heisenberg, Max Born, Pascual Jordan', '~1925')

('wave mechanics', 'Erwin Schrödinger', '~1925')

Output.
+------+-------------------------------------+--------------------------------+-------------+
|  ID  |  Formulations_of_quantum_mechanics  |  Proposer                      |  Time_(Yr)  |
+------+-------------------------------------+--------------------------------+-------------+
|  1   |  transformation theory              |  Paul Dirac                    |  ~1927      |
|  2   |  matrix mechanics                   |  Werner Heisenberg, Max Born,  |  ~1925      |
|      |                                     |    Pascual Jordan              |             |
|  3   |  wave mechanics                     |  Erwin Schrödinger             |  ~1925      |
|  4   |  path integral formulation          |  Richard Feynman               |  ~1948      |
+------+-------------------------------------+--------------------------------+-------------+

Output.

+------+-------------------------------------+--------------------------------+-------------+

+------+-------------------------------------+--------------------------------+-------------+

| 1 | transformation theory | Paul Dirac | ~1927 |

| 2 | matrix mechanics | Werner Heisenberg, Max Born, | ~1925 |

| 3 | wave mechanics | Erwin Schrödinger | ~1925 |

| 4 | path integral formulation | Richard Feynman | ~1948 |

+------+-------------------------------------+--------------------------------+-------------+

September 28, 2021November 29, 2021

202109281720 SQL Table 002

CREATE TABLE SQL_Table_002
    (ID int,
    Velocity varchar(255),
    Nonquantum varchar(255),
    Quantum varchar(255)
    );
INSERT INTO SQL_Table_002
    (Velocity, Nonquantum, Quantum)
    VALUES
    ('v << c', 'Classical Mechanics (Newton)', 'Quantum Mechanics (Bohr, Heisenberg, Schrödinger, et al.)')
    ('0 < |v-c| << 1', 'Special Relativity (Einstein)', 'Quantum Field Theory (Dirac, Pauli, Feynman, Schwinger, et al.)')

CREATE TABLE SQL_Table_002

(ID int,

Velocity varchar(255),

Nonquantum varchar(255),

Quantum varchar(255)

);

INSERT INTO SQL_Table_002

(Velocity, Nonquantum, Quantum)

VALUES

('v << c', 'Classical Mechanics (Newton)', 'Quantum Mechanics (Bohr, Heisenberg, Schrödinger, et al.)')

('0 < |v-c| << 1', 'Special Relativity (Einstein)', 'Quantum Field Theory (Dirac, Pauli, Feynman, Schwinger, et al.)')

Output.
+------+------------------+-----------------------+----------------------------+
|  ID  |  Velocity        |  Nonquantum           |  Quantum                   |
+------+------------------+-----------------------+----------------------------+
|  1   |  v << c          |  Classical Mechanics  |  Quantum Mechanics         |
|      |                  |   (Newton)            |   (Bohr, Heisenberg,       |
|      |                  |                       |   Schrödinger, et al.)     |
|  2   |  0 < |v-c| << 1  |  Special Relativity   |  Quantum Field Theory      |
|      |                  |   (Einstein)          |   (Dirac, Pauli, Feynman,  |
|      |                  |                       |    Schwinger, et al.)      |
+------+------------------+-----------------------+----------------------------+
(D. J. Griffiths on Introduction to Electrodynamics (1999), pg. xi)

Output.

+------+------------------+-----------------------+----------------------------+

+------+------------------+-----------------------+----------------------------+

| 1 | v << c | Classical Mechanics | Quantum Mechanics |

| 2 | 0 < |v-c| << 1 | Special Relativity | Quantum Field Theory |

+------+------------------+-----------------------+----------------------------+

(D. J. Griffiths on Introduction to Electrodynamics (1999), pg. xi)

September 28, 2021November 29, 2021

202109281636 SQL Table 001

CREATE TABLE SQL_Table_001
    (ID int,
    Subject varchar(255),
    Object varchar(255),
    Category varchar(255)
    );
INSERT INTO SQL_Table_001
    (Subject, Object, Category)
    VALUES
    ('algebraic geometry', 'regular (polynomial) functions', 'algebraic varieties')
    ('topology', 'continuous functions', 'topological spaces')
    ('differential topology', 'differentiable functions', 'differentiable manifolds')
    ('complex analysis', 'analytic (power series) functions', 'complex manifolds');

CREATE TABLE SQL_Table_001

(ID int,

Subject varchar(255),

Object varchar(255),

Category varchar(255)

);

INSERT INTO SQL_Table_001

(Subject, Object, Category)

VALUES

('algebraic geometry', 'regular (polynomial) functions', 'algebraic varieties')

('topology', 'continuous functions', 'topological spaces')

('differential topology', 'differentiable functions', 'differentiable manifolds')

('complex analysis', 'analytic (power series) functions', 'complex manifolds');

Output.
+------+-------------------------+-------------------------------------+----------------------------+
|  ID  |  Subject                |  Object                             |  Category                  |
+------+-------------------------+-------------------------------------+----------------------------+
|  1   |  algebraic geometry     |  regular (polynomial) functions     |  algebraic varieties       |
|  2   |  topology               |  continuous functions               |  topological spaces        |
|  3   |  differential topology  |  differentiable functions           |  differentiable manifolds  |
|  4   |  complex analysis       |  analytic (power series) functions  |  complex manifolds         |
+------+-------------------------+-------------------------------------+----------------------------+
(J.S. Milne on Algebraic Geometry (2017), pg. 7)

Output.

+------+-------------------------+-------------------------------------+----------------------------+

+------+-------------------------+-------------------------------------+----------------------------+

+------+-------------------------+-------------------------------------+----------------------------+

(J.S. Milne on Algebraic Geometry (2017), pg. 7)

August 5, 2021July 25, 2022

202108050932 Project Euler Problem 1 solved by C

Problem 1 of Project Euler

If we list all the natural numbers below $10$ that are multiples of $3$ or $5$ , we get $3$ , $5$ , $6$ and $9$ . The sum of these multiples is $23$ .

/* running a check on the first two statements by a direct computation */
#include <stdio.h>;
#include <stdlib.h>;
int main()
    {
    printf("The sum of all the natural numbers below 10 that are multiples of 3 and 5 is %d.", 3 + 5 + 6 + 9);
    system("pause");
    return 0;
    }

/* running a check on the first two statements by a direct computation */

#include <stdio.h>;

#include <stdlib.h>;

int main()

{

printf("The sum of all the natural numbers below 10 that are multiples of 3 and 5 is %d.", 3 + 5 + 6 + 9);

system("pause");

return 0;

}

Find the sum of all the multiples of $3$ or $5$ below $1000$ .

Ans. $233168$

(step-by-step approach, slow but sure logic)

I wish to find all the multiples of $3$ (excluding multiples of $5$ ) below $1000$ :

/* finding all the multiples of 3 (excluding multiples of 5) below 1000*/
#include <stdio.h>;
#include <stdlib.h>;
int main()
    {
    int sum = 0;
    for (int i = 0; i < 1000; i += 3)
        {
        if (i % 5 != 0)
            {
            sum += i;
            }
        }
    printf("The sum of multiples of three (excluding multiples of five) below one thousand is %d.", sum);
    }

/* finding all the multiples of 3 (excluding multiples of 5) below 1000*/

#include <stdio.h>;

#include <stdlib.h>;

int main()

{

int sum = 0;

for (int i = 0; i < 1000; i += 3)

{

if (i % 5 != 0)

{

sum += i;

}

printf("The sum of multiples of three (excluding multiples of five) below one thousand is %d.", sum);

}

Output.
The sum of multiples of three (excluding multiples of five) below one thousand is 133668.

1 2	Output. The sum of multiples of three (excluding multiples of five) below one thousand is 133668.

Then I wish to find all the multiples of $5$ (excluding multiples of $3$ ) below $1000$ :

/* finding all the multiples of 5 (excluding multiples of 3) below 1000:*/
#include <stdio.h>;
#include <stdlib.h>;
int main()
    {
    int sum = 0;
    for (int i = 0; i < 1000; i += 5)
        {
        if (i % 3 != 0)
            {
            sum += i;
            }
        }
    printf("The sum of multiples of five (excluding multiples of three) below one thousand is %d.", sum);
    }

/* finding all the multiples of 5 (excluding multiples of 3) below 1000:*/

#include <stdio.h>;

#include <stdlib.h>;

int main()

{

int sum = 0;

for (int i = 0; i < 1000; i += 5)

{

if (i % 3 != 0)

{

sum += i;

}

printf("The sum of multiples of five (excluding multiples of three) below one thousand is %d.", sum);

}

Output.
The sum of multiples of five (excluding multiples of three) below one thousand is 66335.

1 2	Output. The sum of multiples of five (excluding multiples of three) below one thousand is 66335.

Lastly I wish to find all the multiples of $15=3\times 5$ below $1000$ :

/* finding all the multiples of 15 below 1000:*/
#include <stdio.h>;
#include <stdlib.h>;
int main()
    {
    int sum = 0;
    for (int i = 0; i < 1000; i += 15)
        {
        sum += i;
        }
    printf("The sum of multiples of fifteen below one thousand is %d.", sum);
    }

/* finding all the multiples of 15 below 1000:*/

#include <stdio.h>;

#include <stdlib.h>;

int main()

{

int sum = 0;

for (int i = 0; i < 1000; i += 15)

{

sum += i;

}

printf("The sum of multiples of fifteen below one thousand is %d.", sum);

}

Output.
The sum of multiples of fifteen below one thousand is 33165.

1 2	Output. The sum of multiples of fifteen below one thousand is 33165.

In sum,

$\begin{aligned} & \quad \textrm{The sum of all multiples of 3 or 5 below 1000} \\ & = 133\small,668 + 66\small,335 + 33\small,165 \\ & = 233\small,168 \end{aligned}$

Ans. $233168 \quad\checkmark$

(to be continued)

Going off at a tangent, the sum of all numbers from $1$ to $1000$ is $500500$ . See:

#include <stdio.h>;
#include <stdlib.h>;
int main()
    {
    int i,sum;
    i = 1;
    sum = 0;
    do
    {
    sum += i++;
    }
    while (i <= 1000);
    printf("sum = %d", sum);
    system("pause");
    return 0;
    }

#include <stdio.h>;

#include <stdlib.h>;

int main()

{

int i,sum;

i = 1;

sum = 0;

{

sum += i++;

}

while (i <= 1000);

printf("sum = %d", sum);

system("pause");

return 0;

}

Output.
sum = 500500

1 2	Output. sum = 500500

(proof)

$\begin{aligned} & \quad 1+2+3+\cdots +1000 \\ & = \sum_{k=1}^{1000}k \\ & = \frac{(1000)(1000+1)}{2} \\ & = 500500 \\ \end{aligned}$

(continue)

Solution. (fast and furious attempt)

/* The sum of all the multiples of 3 or 5 below 1000 */
#include <stdio.h>;
#include <stdlib.h>;
int main()
    {
    int sum = 0;
    int i = 0;
    while(i < 1000)
    {
    if(i % 3 == 0 || i % 5 == 0)
    sum += i;
    i++;
    }
    printf("The sum of all the multiples of 3 or 5 below 1000 is %d.", sum);
    }

/* The sum of all the multiples of 3 or 5 below 1000 */

#include <stdio.h>;

#include <stdlib.h>;

int main()

{

int sum = 0;

int i = 0;

while(i < 1000)

{

if(i % 3 == 0 || i % 5 == 0)

sum += i;

i++;

}

printf("The sum of all the multiples of 3 or 5 below 1000 is %d.", sum);

}

Output.
The sum of all the multiples of 3 or 5 below 1000 is 233168.

1 2	Output. The sum of all the multiples of 3 or 5 below 1000 is 233168.