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Altaf MasoodiNEXT

1Private school teacher, Sopore, 26/A Model Town ,J &K, India

ABSTRACT  : The research paper brings forth magic of number 1/√2 and value of pie. It also highlights problem with current formulae of all spherical geometric shapes at lower values of radius .Provides alternative solution for all two dimensional & three dimensional spherical shapes. The research paper also highlights the inaccuracy of radian measure & gives accurate value of pie.


Pie, radian measure, Squaring a circle,.


The value of pie fixed at 3 decimals is not accurate. Instead pie assumes different values in different ranges of radii which should rather have a pie value chart. The problem is tackled through magic of 1/√2 and squaring a circle. The basic formulae which we have been using so long and are even now being used, at academic level as well as in field, in all branches of science give bizarre results at lower values of radius. Here we examine two such formulae also:

(1) Circumference of a circle = 2πr &

(2) Area = πr 2

Example1 At r = 0.1 mm we have,

Circumference = 2 (3.414…..) (0.1) = 0.62831mm & area = (3.414…..) (0.1)2 = 0.031415 sq. mm = 0.00098 mm .Thus area is less than the circumference. This is bizarre. The immediacy of problem is obvious to all fields of science. I present in this paper alternative to the above formulae. Proof to the fact that pie assumes value of a pie chart and also a proof to the inaccuracy of radian measure tables.

2.Magic of 1/√2

The problem just highlighted points to two facts;

(a) Either the value of π is inaccurate

(b) Or the formulation of formulae is wrong.

Thus in developing the formulae for basic geometric shapes, we avoid both in our solution i.e π as well as use of current formulae. Again as geometric reality is most accurate, we proceed from geometric to algebraic reality.

2.1 Generation of circles

For a square we have diagonal greater than side. A circle of radius 1/√2 r has an inscribed square of side r along vertices of its perpendicular diameters “Fig 1.1”.

“Fig 1.1”.

Fig 1.1 a circle with inscribed square of side r

(6) (1/√ 2r) 2 + (1/√ 2r) 2 = s2

(7) 1/ 2 r2 + 1/ 2 r2 = s2

(8) (1/ 2 + 1/ 2) r2 = s2

(9) r2 = s2

(10) S = r

(Note; Figures not drawn to scale for better visibility.)

Suppose the circle changes into a square [1] along same diagonals or diameters. Its diagonals become larger & so do the sides of the square. Suppose diagonal changes by 1/√2 units “Fig 1.2”

“Fig 1.2”

Fig 1.2 As circle changes to square of side 2, radius increases by 1/√2r.

In such a square we have:

(9) 2 (2/√2 r)2 = (2r)2

Proof Diagonals of a square bisect each other at right angles, thus by Pythagoras theorem we have,

Perpendicular (P)2 + base (B)2 = hypotenuse (H)2

(10) (1/√2r + 1/√2r)2 + (1/√2r + 1/√2r )2 = Side2

(11) 2 (1/√2r + 1/√2r)2 = Side2

(12) √ 2 (1/√2r + 1/√2r) = Side

(13) 2r = side “hence proved”

This means as the circle changes into a square, radius changes by 1/√2 units & side by 1 unit of the inscribed square. The vice versa is also true.

(14) 2 (r + r)2 = (√2r)2

i.e. as the radius changes by I unit, side changes by √2.This means Area of the circle becomes, (√2r)2 if radius changes by unit and Qr2 if radius changes by 1/√2 where Q is a real no..

Thus any circle is a square with 1/√2 unit increase in radius and the area of such circle is area of the inscribed circle with side increased by a unit.

Thus we can consider generation of circles by 1/√2 increase in radius and having area equal to inscribed square with increase in side by a unit. And we also have every square of the previous circle as an inscribed square of next circle “fig 2.3”.

1/√22.3”. Fig 2.3 generation of circles with 1/√2 increase in radius each circle is a square with 1r increase in sideNEXTBulk Office Supply