# Happy Holi! : The Village Tour

Holi, the festival of colors, was celebrated this year on 27th and 28th of March all over India. I decided to move to my own village, Kasturwa and then to Surya’s Village, Shiv Patti, on this occasion. Here are some images from the events  taken with my Nokia device, which I thought were worth sharing.

Rangoli is one of the most beautiful arts made on festivals in Uttar Pradesh. The below one isn’t that good, but fair enough to be shared.

This is a photo taken at nearby market of my village. Left to right we are, Vishal Shahi, Suryakant Tripathi, Me, Raman Pandey, Atul Pandey & Deepak Kumar.

LOCATION: Chaura Khas, Kasturwa

Now, here are the Tripathi brothers: Shivakant, Surya and Chandrakant. Nice pic, I’d say.

# Analysis of Meteorological Data of Pantnagar Weather Station

About: This post is actually a summary of a research project I took under INSPIRE-SHE Scholarship Program by Dept. of Science and Technology, Govt. of India. My plan was to make the content open-source on the web that faults could be corrected by time. The language is simple and very easy to understand and the ease of understanding is focused to A-level (10+2) students and beyond.

#### Abstract

The present project is based on 20 years of meteorological data provided by IMD approved weather station at Govind Ballabh Pant University of Agriculture and Technology, Pantnagar. The observations are taken twice a day and therefore fourteen times a week. My analysis is based on the weekly reports provided by the agro-meteorological department. The main aim of this project is to develop my analytic skills in meteorology and to provide some better ideas on the meteorological studies. The meteorological observatory at Pantnagar is located at 290 N latitude, 79.30 E longitude and 243.84 m altitude under N. E. B. Crop Research Center. Pantnagar region lies in the Tarai belt of Uttarakhand state of India. It is comparatively hot and wet place to nearby places like famous hill station Nainital. The annual rainfall is about 145 cm, which has a lot of variations throughout the year. The rainy season starts at the end of June and end in September. Maximum rain is received from south-west monsoon during four months rainy season from June to September. The soil of this region is good for agriculture and holds enough moisture to produce good crops. The average pH value of the soil is 7.2 – 7.4. The temperature variation is very large, as summer holding temperature maxima of around 42-45 degree Celsius while in winter season it falls heavily to 2-4 degree Celsius.

This report was aimed to study and analyze the collected weekly data within the limited time period of two months, as proposed by DST, Govt. Of India under INSPIRE-SHE summer project.

### Introduction

Meteorology is a word made by a combination of two Greek words ‘Meteors‘ which means ‘atmospheric‘ or ‘lofty‘ and ‘logos‘ which means discourse or science. Therefore we can define Meteorology as the Science of Atmosphere. Meteorology is actually the study of atmospheric processes using extensive use of applied physics. The science of atmosphere involves the static and dynamic components of atmosphere, so they when combined called weather.

#### Constitution of Atmosphere

The atmosphere of earth is a relatively thin sheet of gas firmly attracted to the surface by gravitational force of earth. The atmosphere is mainly treated to be made of something special that is invisible and odorless substance called Air. Air is the most important component of earth atmosphere. When we say ‘Atmosphere’, in behavior we mean ‘Air’. Air itself has two constituents: Distinct and Variable!

Distinct gaseous constituents are major and cover almost 100% of atmosphere. Main gases present in atmosphere are Nitrogen (78%) and Oxygen (21%) and minor constituents (~1%) are Argon types of  inactive gases. Main Variable constituents are water-vapour, ozone, carbon dioxide and dust. The availability of these constituents depend upon the location at the earth.

The standard state of atmosphere has been well explored with the help of meteorological and satellite observations. The paragraph below will feature the layers of atmosphere. The temperature and physical standards of every atmospheric layer has different values. Hence , it is highly recommended to study about them before we proceed. The lowest layer contains about three fourth of the mass and almost all the moisture and dust of the atmosphere. It is called the troposphere. All the meteorological phenomena which are called Weather and thus highly related to this project are confined to the troposphere. The top of Troposphere is called tropopause. The height of tropopause varies from 8km (at the poles) to 16km (at equator).

In troposphere the temperature decreases with the elevation at an average rate of 6.5 degree Celsius per km. This is known as the lapse rate. Above the tropopause is the stratosphere which is free of daily and annual heating of earth’s surface and contains very little dust and moisture. Stratosphere contains a major portion of lifesaving Ozone. Stratopause is located at 30 to 50km above the earth surface as the upper limit of stratosphere. Above stratosphere, there are the mesosphere and mesopause, a warm layer at 80km above the earth surface. The fourth major layer above the mesopause is ionosphere which contains a very little pressure of about 0.01milibars at 90 km. This layer contains ultraviolet radiation, satellites etc. Ionosphere merges gradually into the outermost  shell called the exosphere. In exosphere the mean free path is very large and the atmosphere has lost the property of continuum.

As already mentioned troposphere is the domain of study of  meteorology with an extension to stratosphere involving ozone.

#### Meteorological observations

Using applied physics, engineers do periodical measurements on meteorological parameters to understand long term or short term atmospheric phenomena. The meteorological parameters are like coordinates to explain atmospheric observations and these are dependent of conditions (time and location). There are a number of parameters depending on the meteorological observations one uses, out of which some parameters are most important and shall be used in this project are:

1. Atmospheric pressure

2. Wind Velocity

3. Wind direction

4. Temperature

5. Humidity

7. Sunshine

8. Precipitation (rain, snowfall, hail)

9. Evaporation

Meteorological instruments are used to collect the data of these parameters at meteorological observatory or the weather station. In India there are a number of weather stations which are regulated by Indian Meteorological Department (IMD). All weather stations work on specific standards formulated by IMD for the location. The meteorological data can be obtained from the weather stations. I have collected the data from weather station Pantnagar, Uttarakhand. Generally the observations are taken twice a day, i.e., at 07:12 & at 14:12 hours.

The minimal understanding of physics of meteorological parameters is required before we can do statistical observations.

##### Atmospheric Pressure

All in One Equipments

This is the force exerted per unit surface area of earth at a location. Atmospheric pressure is usually measured by Fortins barometer consisting of an inverted U-tube filled with mercury kept in a cistern, or an aneroid barometer. The instrument which records the variation of the atmospheric pressure with respect to time is called a barograph.

The standard atmospheric pressure is $1.02 times 10^5 N/m^2$ .

##### Wind Direction and Wind Velocity

Wind Gauge

Air in the motion is called the wind. The horizontal component of the air in movement parallel to the earth’s surface is generally referred to as wind while the vertical components are referred to as the air currents. Measurement of wind direction is extremely easy and is measured by wind vane, while measurement of wind velocity is a little tricky and is measured using the cup anemometer.

##### Temperature

Thermometers

The most studied parameter of meteorology is temperature and it can be measured simply using mercury thermometer in degree Celsius. A continuous record of temperature with time can be obtained by an automatic recording instrument called the thermograph. The thermograph measures the temperature using the principle that a bimetallic strip changes its shape under the influence of the change in temperature. The maximum and minimum temperatures at a station are measured by maximum thermometer and minimum thermometer respectively. The mean daily temperature is computed as the arithmetic average of the maximum and minimum temperatures recorded on  that day. The daily range is the temperature difference between maximum and minimum temperatures for the particular day. The mean monthly temperature is computed as the arithmetic average of the mean daily temperatures of all days in a month. The mean annual temperature is the arithmetic average of the mean temperatures of all days in an year. The normal daily temperature is the average of the daily mean temperatures for a period of 30 years. Similarly, the normal monthly temperature is the average of the mean monthly temperatures for a period of 30 years. The normal annual temperature is the average of the mean annual temperatures for a period of 30 years.

##### Humidity

Meteorologically known as relative humidity, humidity is measured by the psychrometer. The continuous recording of humidity with time is done by automatic recording instrument called hydrograph. Humidity is relative quantity and is measured in unit per cent and is defined by the percentage measured by the ratio of actual and saturation vapor pressures at a given temperature.

Radiation is a process in which energetic particles or energetic waves travel through a medium or space. In meteorological observations, thermal radiation causing increase in temperature and solar radiation causing effect on ozone layers etc. are studied.

##### Sunshine

Sunshine meter

Sunshine is actually the measure of hours Sun emits light at a weather station a day. It is measured by an instrument called Sunshine recorder.

##### Precipitation

In meteorology, precipitation means the gathering of water from the atmosphere to earth in any physical form. Rainfall, Snowfall, Hails, Fog etc. are the ways water reaches to earth. Rainfall is measured using rain gauge. The average rainfall over a significant area is approximated by three methods:

1. Arithmetic Mean Method: The result is obtained by dividing the sum of the rainfall amounts recorded at all the rain gauge stations which are located within the area under consideration by the number of stations. i.e.,
$P=dfrac{P_1+P_2+ldots+P_n}{n}=dfrac{1}{n} sum_{k=1}^n P_k$

This method is also known as the unweighted mean method within the area.

2. Thiessen Polygon Method: Suggested by Theissen in 1911, this method allows irregularities in gauge locations by measurement of each gauge in proportion to the area which is closer to that gauge than to any other gauge. The average depth of rainfall by this method is given by
$P=dfrac{A_1 P_1+A_2 P_2+ldots+A_n P_n}{A_1+A_2+ldots+A_n}$
where $P_1, P_2, ldots, P_n$ are the rainfalls recorded at raingauge stations with polygonal areas around them.

3. Isohyetal Method: A weakly converging method but perhaps the most accurate method for rainfall measurement is Isohyetal method. The accuracy actually depends upon the skill of analyst. An isohyet is defined by a line joining points with equal rainfall.

For the ease of calculations, I have used first method in the report.

##### Evaporation

As usual water evaporates into vapor at every temperature. As the temperature is increased, evaporation occurs readily. As a side-effect of temperature, water-resources convert into vapor and it is very essential to know how much water was evaporated in order to predict the future precipitation predictions. A tank of water is used to determine the evaporation for a day, called Pan Evaporimeter.

## Meteorological Analysis

Many meteorological phenomena are physical processes and in every physical process there should be a relation between the cause and the effect. Once the relation is executed, the output of the process can be easily and precisely predicted. But in some cases there is an element of uncertainty regarding the outcome of the process. The first type of processes are called deterministic processes, while the second type of processes are undeterministic. Prediction of sunset and sunrise are deterministic and that of rainfall and wind velocity are undeterministic. The most of the meteorological parameters are connected to such processes which are random, and where arises randomness there comes the Statistics and Probability. So before we proceed, let we have a quick review on some statistical topics which are closely related to the the calculations I shall use in this project.

• Random Process / Probabilistic Process / Random Phenomenon is a process/phenomenon in which no certain relation between the cause and the effect is observed. A random process always has an element of unpredictability.
• If an experiment / event is conducted N times, or if the outcome of a process is observed N times, and if a particular attribute A occurs n times, then the limit of n/N as N becomes large, is defined as probability.

Therefore, probability $P=displaystyle{lim_{N to infty}} frac{n}{N}$ .

• The maximum value of probability is 1 and the minimum value is 0. In first case event is said to occur and in later case it is said not to occur.
• Probability is a numerical quantity and hence its application is implemented to meteorological processes by defining some variables and functions.
• A variable associated with a random process which can’t be predicted uniquely, is called a random variable.
• Random variable is classified into two ways: discrete random variable and continuous randim variable.

# The Area of a Disk

[This post is under review.]

If you are aware of elementary facts of geometry, then you might know that the area of a disk with radius $R$ is $\pi R^2$ .

The radius is actually the measure(length) of a line joining the center of disk and any point on the circumference of the disk or any other circular lamina. Radius for a disk is always same, irrespective of the location of point at circumference to which you are joining the center of disk. The area of disk is defined as the ‘measure of surface‘ surrounded by the round edge (circumference) of the disk.

The area of a disk can be derived by breaking it into a number of identical parts of disk as units — calculating their areas and summing them up till disk is reformed. There are many ways to imagine a unit of disk. We can imagine the disk to be made up of several concentric very thin rings increasing in radius from zero to the radius of disc. In this method we can take an arbitrary ring, calculate its area and then in similar manner, induce areas of other rings -sum them till whole disk is obtained.

Rings and Sections

Mathematically, we can imagine a ring of with radius $x$ and thickness $dx$ , any where in the disk having the same center as disk, calculate its area and then sum up (integrate) it from $x=0$ to $x=R$ . Area of a thin ring is since $\pi x dx$ . And after integrating we get, area of disk $A=2 \int_0^R \pi x \ dx$ or $A=\pi R^2$ .

An Inscribed Triangle

There is another approach to achieve the area of a disk, A.

Imagine a disk is made up of a number equal sections or arcs. If there are $n$ number of arcs then interior angle of an arc is exactly $\frac{2\pi}{n}$ , since $2 \pi$ is the total angle at the center of disk and we are dividing this angle into $n$ -equal parts. If we join two ends of each sections –we can get $n$ -identical triangles in which an angle with vertex O is $\frac{2 \pi}{n}$ . Now, if we can calculate the area of one such section, we can approach to the area of the disk intuitively. This approach is called the method of exhaustion.

Let, we draw two lines joining center O of the disk and points A & B at circumference. It is clear that OB and OA are the radius of the disk. We joined points A and B in order to form a triangle OAB. Now consider that the disk is made up of n-number of such triangles. We see that there is some area remaining outside the line AB and inside the circumference. If we had this triangle thinner, the remaining area must be lesser.

Area remaining after the Triangle

So, if we increase the number of triangles in disk —-we decrease the remaining areas. We can achieve to a point where we can accurately calculate the area of disk when there are infinitely many such triangles or in other words area of one such triangle is very small. So our plan is to find the area of one triangle —sum it upto n — make $n$ tending to infinty to get the area of disk. It is clear that the sum of areas of all identical triangles like OAB must be either less than or equal to area of the disk. We can call triangles like OAB as inscribed triangles.

Now, if we draw a radius-line OT’, perpendicular to AB at point T and intersecting the circumference at point T’, we can easily draw another triangle OA’B’ as shown in figure. AOB and A’OB’ are inscribed and superscribed triangles of disk with same angle at vertex O. So, it is clear that the angle A’OB’ is equal to the angle AOB. Triangle A’OB’ is larger than the circular arc OAB and circular arc OAB is larger than the inscribed triangle AOB. Also, the sum of areas of triangles identical to OA’B’ is either greater than or equal to area of the disk.

# My Five Favs in Math Webcomics

Cartoons and Comics are very useful in the process of explaining complicated topics, in a very light and humorous way. Like:

Source of These Cartoons
Here my five most favorite math-webcomics sites. (Click on images to visit them.)

# Welcome 2012 – The National Mathematical Year in India

Srinivasa Ramanujan (Photo credit: Wikipedia)

I was very pleased on reading this news that Government of India has decided to celebrate the upcoming year 2012 as the National Mathematical Year. This is 125th birth anniversary of math-wizard Srinivasa Ramanujan (1887-1920). He is one of the greatest mathematicians India ever produced. Well this is ‘not’ the main reason for appointing 2012 as National Mathematical Year as it is only a tribute to him. Main reason is the emptiness of mathematical awareness in Indian Students. First of all there are only a few graduating with Mathematics and second, many not choosing mathematics as a primary subject at primary levels. As mathematics is not a very earning stream, most students want to go for professional courses such as Engineering, Medicine, Business and Management. Remaining graduates who enjoy science, skip through either physical or chemical sciences. Engineering craze has developed the field of Computer Science but not so much in theoretical Computer Science, which is one of the most recommended branches in mathematics. Statistics and Combinatorics are almost ‘died’ in many of Indian Universities and Colleges. No one wants to deal with those brain cracking math-problems: neither students nor professors. Institutes where mathematics is being taught are struggling with the lack of talented lecturers. Talented mathematicians don’t want to teach here since they aren’t getting much money and ordinary lecturers can’t do much more. India is almost ‘zero’ in Mathematics and some people including critics still roar that we discovered ‘zero’, ‘pi’ and we had Ramanujan.

Image by Terriko via Flickr

Indian education divided into three categories: —-Primary, Secondary and Higher Education, looks like a mountain climbing after a smooth beginning.

In primary classes, a student is taught about elementary topics (like Elementary Operations, Introductory Algebra and Geometry in Mathematics). Primary classes take about 8 to 10 years. When promoted to Secondary Classes, which are of 4 years exactly, students are distributed among two categories: one studying science subjects (Math, Physics, Chem, Bio etc.) and others not studying science subjects (History, Politics etc.). Only 2 out of 10 students opt for Mathematics. Reason? The students are introduced with some very complicated topics, which should not be there if primary classes were very easy. Mathematics suddenly becomes extremely tough to handle and the mathematician dies within him/her. The total eliminators of mathematics are IIT-JEE and AIEEE type exams. These lead students to study in a fast forward way and they start their chase to marks, money and machines. Students who succeed in these exams, go for engineering in premier IITs and NITs while others pursue so with private institutes. And here starts the higher education. Out of 10 students passed with Math as subject in secondary classes, 3 students choose to pursue under-graduation with mathematics. There are two ways to pursue under-graduation with Math. First is doing honours course in mathematics and second is regular bachelors degree of math with two other science subjects like Physics or Mathematics of Psychology or Geography or Economics. This also divides the mathematical interest. And believe me, 90% of these math-undergrads are those who tried for engineering exams but could never succeed. They have no heart with math. They are trained to write in exams. Only 2% of undergrads go for graduation with mathematics. IITs and NITs also operate graduate courses in mathematical sciences, but there are only a few which go for it.

# The Cattle Problem

This is a famous problem of intermediate analysis, also known as ‘Archimedes’ Cattle Problem Puzzle’, sent by Archimedes to Eratosthenes as a challenge to Alexandrian scholars. In it one is required to find the number of bulls and cows of each of four colors, the eight unknown quantities being connected by nine conditions. These conditions ultimately form a Pell equation which solution is necessary in case of finding the answer of the puzzle. The Greek puzzle is stated below with a little deviation. I have just tried to make the language simpler than the original, hope you’ll be able to grasp the puzzle easily.

O Stranger! If you are intelligent and wise, find the number of cattle of the Sun, who once upon a time grazed on the fields of an Island, divided into four groups (herds) of different colors, one white, another a black, a third yellow and the last dappled color.In each herd were bulls, mighty in number according to these proportions:

• White bulls were equal to a half and a third of the black together with the whole of the yellow.
• The black bulls were equal to the fourth part of the dappled and a fifth, together with, once more, the whole of the yellow.
• The dappled bulls, were equal to a sixth part of the white and a seventh, together with all of the yellow.

So, these were the proportions of bulls, now the proportions of the cows were as following:

• White cows were equal to the third part and a fourth of the whole herd of the black.
• Black cows were equal to the fourth part once more of the dappled and with it a fifth part, when all cattle, including the bulls, went to pasture together. Now the dappled in four parts were equal in number to a fifth part and a sixth of the yellow herd.
• Yellow cows were in number equal to a sixth part and a seventh of the white herd.

Keeping above conditions in focus, find the number of cattle of the Sun, giving separately the number of well-fed bulls and again the number of females according to each color. But come, this solution is not complete unless you understand  all these conditions regarding the cattle of the Sun:

• When the white bulls mingled their number with the black, they stood firm, equal in depth and breadth. Number of bulls in a row were equal to the number of columns.
• When the yellow and the dappled bulls were gathered into one herd they stood in such a manner that their number, beginning from one, grew slowly greater till it completed a triangular figure, there being no bulls of other colors in their midst nor none of them lacking.

Find the number of cows and bulls of each color separately.

Solution:

#### $W$

= number of white bulls
$B$ = number of black bulls
$Y$ = number of yellow bulls
$D$ = number of dappled bulls
$w$ = number of white cows
$b$ = number of black cows
$y$ = number of yellow cows
$d$ = number of dappled cows

The relations come as:

•   $W = (\frac{1}{2} + \frac{1}{3})B + Y$ The white bulls were equal to a half and a third of the black bulls together with the whole of the yellow bulls.
• $B = (\frac{1}{4} + \frac{1}{5})D + Y$ The black [bulls] were equal to the fourth part of the dappled bulls and a fifth, together with, once more, the whole of the yellow bulls
•   $D = (\frac{1}{6} + \frac{1}{7})W + Y$ The remaining bulls, the dappled, were equal to a sixth part of the white bulls and a seventh, together with all of the yellow bulls

# A Yes No Puzzle

This is not just math, but a very good test for linguistic reasoning. If you are serious about this test and think that you’ve a sharp [at least average] brain then read the statement (only) below –summarize it –find the conclusion and then answer that whether summary of the statement is Yes or No.
[And if you're not serious about the test ...then read the whole post to know what the stupid author was trying to tell you. ]

# Blog of the Month Awards – October 2011

Reader’s brain is variable. It changes according to what it read. I have changed the pattern of selection and style of writing about BLOG OF THE MONTH. At the beginning of August, I planned that I will select some blogs from the education blog-o-sphere and will award to appreciate them for their excellent work. I know these awards will probably never make a difference but hope too that they’ll keep their good works on. So, here is the list of my 10 most favorite blogs, one of which, Gowers’s Weblog, is my Blog of The Month, for October 2011.

# The problem of the Hundred Fowls

This is a popular Chinese problem, on Linear Diophantine equations, which in wording seems as a puzzle or riddle. However, when used algebraic notations, it looks obvious. The problems states :

 If a cock is worth 5 coins, a hen 3 coins, and three chickens together 1 coin, how many cocks, hens and chickens, totaling 100 in number, can be bought for 100 coins?

This puzzle in terms of algebraic equations can be written as $5x+3y+\frac{1}{3}z=100$ and $x+y+z=100$
where $x, y, z$ being the number of cocks, hens and chicks respectively.
We find that there are two equations with three unknown quantities. So eliminating one of the unknowns, by putting $z=100-x-y$ from second equation into first one such that $5x+3y+\frac{1}{3} (100-x-y)=100$
or, $15x+9y+100-x-y=300$
or, $14x+8y=200$
or, $7x+4y=100$ .
Which is a linear Diophantine equation (with only two unknown quantities).