This theory had been proposed by H.C. Rajpoot @ his college presently M.M.M. University of Technology, GKP-273010 in Oct, 2013, for finding out the solid angle subtended by any polygonal plane at any point in the space. It gives the simplest, easiest & the most versatile methods for calculating the mathematically correct value of solid angle subtended by any plane bounded by the straight lines like triangle, quadrilateral like rectangle, square, rhombus, trapezium etc., any regular or irregular polygon like pentagon, hexagon, heptagon, octagon etc.) at any point in the space. It is the unified theory applied for any polygon by using one standard formula only. This can derive expression for solid angle subtended by any plane bounded by the straight lines. This theory is equally applicable for atomic distances & stellar distances in the Universe.
Solid angle subtended by a rectangular plane at any point in the space Harish Chandra Rajpoot
This is the most general case for any location of given point in the space which is derived by using basic formula taken from the book "Advanced Geometry by H.C. Rajpoot". Its derivation & detailed explanation has been given in author's book of research articles of 3-D Geometry.
1) The document discusses the calculation of bending and shear stresses in a beam with a rectangular cross-section.
2) Shear stress is derived to be proportional to the first moment (Q) of the cross-sectional area above the level of interest.
3) For a rectangular cross-section, the maximum shear stress occurs at the neutral axis and is 50% larger than the average shear stress.
This document discusses calculating the moment of inertia for composite cross-sections made up of multiple simple geometric shapes. It introduces the parallel axis theorem, which allows calculating the moment of inertia of each individual shape about a common reference axis so that the individual values can be added to determine the total moment of inertia of the composite cross-section. Several examples are provided to demonstrate calculating moments of inertia for composite areas using this approach.
The document derives the Law of Sines and Law of Cosines, which relate the angles and sides of triangles. It discusses using these laws to solve oblique triangles given certain information like two angles and a side, two sides and the angle opposite one of the sides, two sides and the angle between them, or all three sides. It also covers finding the area of a triangle given two sides and the included angle, or using Heron's Formula with all three sides. Application problems demonstrate using these concepts to solve real-world geometry problems.
Este documento presenta varios ejercicios sobre mecánica estadística. El primer ejercicio describe la distribución de Maxwell-Boltzmann y calcula la probabilidad de que la energía de una partícula esté fuera del promedio. El segundo ejercicio introduce el modelo de Ising unidimensional y calcula el número de configuraciones posibles y la entropía. El tercer ejercicio considera una cadena polimérica retorcida y relaciona la longitud total con el número de segmentos orientados hacia la derecha.
Equilibrium is the state of rest of a particle or body where the net force and net torque are zero. For a body to be in static equilibrium, the sum of the forces in the x and y directions and the sum of moments must equal zero. There are various types of loads that can act on structures including concentrated loads, uniformly distributed loads, and uniformly varying loads. Equilibrium problems of concurrent coplanar forces can be solved using the principle of two forces or three forces. Examples are provided to demonstrate solving for tensions, reactions, and other unknown forces in equilibrium problems.
Solutions manual for statics and mechanics of materials 5th edition by hibbel...zaezo
Solutions manual for statics and mechanics of materials 5th edition by hibbeler ibsn 9780134301006
download at: https://goo.gl/rTKRQ1
people also search:
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statics and mechanics of materials 5th edition ebook
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Here are the key steps to solve this problem:
* Future sum (F) = Rs. 1,00,000
* Interest rate (i) = 15%
* Number of periods (n) = 10 years
* Interest is compounded annually
* To find the present worth (P):
* P = F / (1 + i)^n
* P = Rs. 1,00,000 / (1 + 0.15)^10
* P = Rs. 1,00,000 / 2.472
* P = Rs. Rs. 40,320
Therefore, the single payment that should be made now to receive Rs. 1,00,000 after 10 years at
Solid angle subtended by a rectangular plane at any point in the space Harish Chandra Rajpoot
This is the most general case for any location of given point in the space which is derived by using basic formula taken from the book "Advanced Geometry by H.C. Rajpoot". Its derivation & detailed explanation has been given in author's book of research articles of 3-D Geometry.
1) The document discusses the calculation of bending and shear stresses in a beam with a rectangular cross-section.
2) Shear stress is derived to be proportional to the first moment (Q) of the cross-sectional area above the level of interest.
3) For a rectangular cross-section, the maximum shear stress occurs at the neutral axis and is 50% larger than the average shear stress.
This document discusses calculating the moment of inertia for composite cross-sections made up of multiple simple geometric shapes. It introduces the parallel axis theorem, which allows calculating the moment of inertia of each individual shape about a common reference axis so that the individual values can be added to determine the total moment of inertia of the composite cross-section. Several examples are provided to demonstrate calculating moments of inertia for composite areas using this approach.
The document derives the Law of Sines and Law of Cosines, which relate the angles and sides of triangles. It discusses using these laws to solve oblique triangles given certain information like two angles and a side, two sides and the angle opposite one of the sides, two sides and the angle between them, or all three sides. It also covers finding the area of a triangle given two sides and the included angle, or using Heron's Formula with all three sides. Application problems demonstrate using these concepts to solve real-world geometry problems.
Este documento presenta varios ejercicios sobre mecánica estadística. El primer ejercicio describe la distribución de Maxwell-Boltzmann y calcula la probabilidad de que la energía de una partícula esté fuera del promedio. El segundo ejercicio introduce el modelo de Ising unidimensional y calcula el número de configuraciones posibles y la entropía. El tercer ejercicio considera una cadena polimérica retorcida y relaciona la longitud total con el número de segmentos orientados hacia la derecha.
Equilibrium is the state of rest of a particle or body where the net force and net torque are zero. For a body to be in static equilibrium, the sum of the forces in the x and y directions and the sum of moments must equal zero. There are various types of loads that can act on structures including concentrated loads, uniformly distributed loads, and uniformly varying loads. Equilibrium problems of concurrent coplanar forces can be solved using the principle of two forces or three forces. Examples are provided to demonstrate solving for tensions, reactions, and other unknown forces in equilibrium problems.
Solutions manual for statics and mechanics of materials 5th edition by hibbel...zaezo
Solutions manual for statics and mechanics of materials 5th edition by hibbeler ibsn 9780134301006
download at: https://goo.gl/rTKRQ1
people also search:
hibbeler statics and mechanics of materials; 5th edition pdf
statics and mechanics of materials 5th edition pdf download
statics and mechanics of materials 5th edition ebook
statics and mechanics of materials 5th edition solutions pdf
statics and mechanics of materials hibbeler pdf download
statics and mechanics of materials solutions
hibbeler statics and mechanics of materials 4th edition pdf
statics and mechanics of materials 5th edition pdf free
Here are the key steps to solve this problem:
* Future sum (F) = Rs. 1,00,000
* Interest rate (i) = 15%
* Number of periods (n) = 10 years
* Interest is compounded annually
* To find the present worth (P):
* P = F / (1 + i)^n
* P = Rs. 1,00,000 / (1 + 0.15)^10
* P = Rs. 1,00,000 / 2.472
* P = Rs. Rs. 40,320
Therefore, the single payment that should be made now to receive Rs. 1,00,000 after 10 years at
Mathematical analysis of sphere resting in the vertex of polyhedron, filletin...Harish Chandra Rajpoot
The generalized formula derived here by the author are applicable to locate any sphere, with a certain radius, resting in a vertex (corner) at which n no. of edges meet together at angle α between any two consecutive of them such as the vertex of platonic solids, any of two identical & diagonally opposite vertices of uniform polyhedrons with congruent right kite faces & the vertex of right pyramid with regular n-gonal base. These are also useful for filleting the faces meeting at the vertex of the polyhedron to best fit the sphere in that vertex. These are used to determine the distance of sphere from the vertex, distance of sphere from the edges, fillet radius of the faces etc. The formula have been generalized for packing the spheres in the vertices of right pyramids & all five platonic solids.
Mathematical Analysis of Tetrahedron (dihedral angles between the consecutive...Harish Chandra Rajpoot
All the articles have been derived by the author Mr H.C. Rajpoot by using HCR's Inverse cosine formula & HCR's Theory of Polygon. These formula are very practical & simple to apply in case of any tetrahedron to calculate the internal (dihedral) angles between the consecutive lateral faces meeting at any of four vertices & the solid angle subtended by it (tetrahedron) at the vertex when the angles between the consecutive edges meeting at the same vertex are known. These are the generalized formula which can also be applied in case of three faces meeting at the vertex of various regular & uniform polyhedrons to calculate the solid angle subtended by polyhedron at its vertex.
Mathematical Analysis of Regular Spherical Polygons (Application of HCR's The...Harish Chandra Rajpoot
This document presents a mathematical analysis of regular spherical polygons. It derives a characteristic equation relating the radius of the sphere, length of each side, interior angle, and number of sides. This equation can be used to calculate any parameter when the other three are known. The document also discusses calculating the solid angle, area, and other properties of both regular spherical polygons and their corresponding planar polygons. It provides examples of applying the analysis to polygons on both general spheres and the unit sphere.
Mathematical Analysis of Spherical Triangle (Spherical Trigonometry by H.C. R...Harish Chandra Rajpoot
All the important parameters of a spherical triangle have been derived by Mr H.C. Rajpoot by using simple geometry & trigonometry. All the articles (formula) are very practical & simple to apply in case of a spherical triangle to calculate all its important parameters such as solid angle, covered surface area, interior angles etc. & also useful for calculating all the parameters of the corresponding plane triangle obtained by joining all the vertices of a spherical triangle by straight lines. These formula can also be used to calculate all the parameters of the right pyramid obtained by joining all the vertices of a spherical triangle to the center of sphere such as normal height, angle between the consecutive lateral edges, area of base etc.
HCR's method of concentric cones (solid angle subtended by a torus at any poi...Harish Chandra Rajpoot
HCR's Method of concentric cones is the simplest & most versatile method to find out the solid angle subtended by a torus at any point lying on its geometrical axis.
HCR's Formula for Regular n-Polyhedrons (Mathematical analysis of regular n-p...Harish Chandra Rajpoot
This formula was derived by H.C. Rajpoot to calculate all the important parameters of a regular n-polyhedron such as inner radius, outer radius, mean radius, surface area & volume. This formula is a generalized dimensional formula which can be applied on any existing n-polyhedron since it depends on two parameters of any regular polyhedron as the no. of faces & the no. of edges in one face only. It can be used in analysis, designing & modelling of regular n-polyhedrons. It's named as "H. Rajpoot's Formula" by the author.
All the standard formula from 'Advanced Geometry' by the author Mr H.C. Rajpoot have been included in this book. These formula are related to the solid geometry dealing with the 2-D & the figures in 3-D space & miscellaneous articles in Trigonometry & Geometry. These are useful the standard formula for case studies & practical applications.
Mathematical analysis of n-gonal trapezohedron with 2n congruent right kite f...Harish Chandra Rajpoot
The generalized formula are applicable on any n-gonal trapezohedronhaving 2n congruent right kite faces, 4n edges & 2n+2 vertices lying on a spherical surface with a certain radius. These formula have been derived by the author Mr H.C. Rajpoot to analyse infinite no. of the trapezohedrons having congruent right kite faces simply by setting n=3,4,5,6,7,………………upto infinity, to calculate all the important parameters such as ratio of unequal edges, outer radius, inner radius, mean radius, surface area, volume, solid angles subtended by the polyhedron at its vertices, dihedral angles between the adjacent right kite faces etc. These formula are very useful for the analysis, modeling & designing of various n-gonal trapezoherons/deltohedrons.
Mathematical Analysis of Rhombicuboctahedron (Application of HCR's Theory)Harish Chandra Rajpoot
The author H C Rajpoot has derived the radius of circumscribed sphere passing through all 24 identical vertices of a rhombicuboctahedron with given edge length by applying ‘HCR’s Theory of Polygon’ & subsequently derived various formula to analytically compute the normal distances of equilateral triangular & square faces from the centre of rhombicuboctahedron, radius of mid-sphere, surface area, volume, solid angles subtended by each equilateral triangular face & each square face at the centre by using ‘HCR’s Theory of Polygon’, dihedral angle between each two faces meeting at any of 24 identical vertices (i.e. truncated rhombic dodecahedron), solid angle subtended by truncated rhombic dodecahedron at any of its 24 identical vertices.
All the standard formula from 'Advanced Geometry' by the author Mr H.C. Rajpoot have been included in this book. These formula are related to the solid geometry dealing with the 2-D & 3-D figures in the space & miscellaneous articles in Trigonometry & Geometry. These are useful the standard formula to be remembered for case studies & practical applications. Although, all the formula for the plane figures (i.e. planes bounded by the straight lines only) can be derived by using standard formula of right triangle that has been explained in details in "HCR's Theory of Polygon" published with International Journal of Mathematics & Physical Sciences Research in Oct, 2014.
And the analysis of oblique frustum of right circular cone has been explained in his research paper 'HCR's Infinite-series' published with IJMPSR
Volume & surface area of right circular cone cut by a plane parallel to its s...Harish Chandra Rajpoot
All the articles have been derived by the author by using simple geometry, trigonometry & calculus. All the formula are the most generalized expressions which can be used for computing the volume & surface area of minor & major parts usually each with hyperbolic section obtained by cutting a right circular cone with a plane parallel to its symmetrical (longitudinal) axis.
HCR’s Inverse Cosine Formula derived by Mr H.C. Rajpoot is a trigonometric relation of four variables/angles. It is applicable for any three straight lines or planes, either co-planar or non-coplanar, intersecting each other at a single point in the space. It directly co-relates the internal angles (i.e. angles between the consecutive lateral faces) & the face angles (i.e. angles between the consecutive lateral edges) of any tetrahedron. This formula is very useful to find out all the unknown internal angles if all the face angles of any tetrahedron are known & vice versa.
This formula holds good for all the regular spherical polygons. It is a very important formula (mathematical relation) applicable on any regular spherical polygon having each of its sides as an arc of the great circle on a spherical surface. It is of crucial importance to find out any of the four important parameters (i.e. radius of sphere, no. of sides, length of side, interior angle of polygon) if other three are given (known) for any regular spherical polygon. It also concludes that any three of the four parameters are self-sufficient to exactly represent a unique regular spherical polygon. A regular spherical polygon having three known parameters can be created or drawn only on a unique spherical surface of a radius which is easily found out by HCR's formula.
Mathematical Analysis of Spherical Rectangle (Application of HCR's Theory of ...Harish Chandra Rajpoot
All the articles have been derived by Mr H.C. Rajpoot by using simple geometry & trigonometry. All the formula are very practical & simple to apply in case of any spherical rectangle to calculate all its important parameters such as solid angle, surface area covered, interior angles etc. & also useful for calculating all the parameters of the corresponding plane rectangle obtained by joining all the vertices of a spherical rectangle by the straight lines. These formula can also be used to calculate all the parameters of the right pyramid obtained by joining all the vertices of a spherical rectangle to the center of the sphere such as normal height, angle between the consecutive lateral edges, area of plane rectangular base etc.
1. The document discusses various coordinate systems used to represent points in 2D and 3D space, including rectangular, polar, cylindrical and spherical coordinates.
2. It also provides equations for lines, circles, ellipses, parabolas, and hyperbolas in rectangular coordinate systems. These include the slope-intercept form for lines, the standard form for circles, and equations involving eccentricity for conic sections.
3. The equations are illustrated with diagrams showing their geometric significance in terms of distances from foci and directrixes or lengths of axes.
Mathematical analysis of a small rhombicuboctahedron (Archimedean solid) by HCRHarish Chandra Rajpoot
All the important parameters of a small rhombicuboctahedron (an Archimedean solid having 8 congruent equilateral triangular & 18 congruent square faces each of equal edge length) such as normal distances & solid angles subtended by the faces, inner radius, outer radius, mean radius, surface area & volume have been calculated by using HCR's formula for regular polyhedrons. This formula is a generalized dimensional formula which is applied on any of the five platonic solids i.e. reguler tetrahedron, regular hexahedron (cube), regular octahedron, regular dodecahedron & regular icosahedron to calculate their important parameters. It can also be used in analysis, designing & modelling of truncated polyhedrons.
This document discusses different types of projections used to transform 3D coordinates into 2D. There are two main types: parallel projection and perspective projection. Parallel projection involves projecting points along parallel lines, preserving properties like parallel lines. Perspective projection involves projecting points that converge on a center of projection, making distances appear smaller with depth. The document provides formulas and examples for performing orthographic, oblique, and perspective projections of 3D points onto a 2D plane.
This document discusses different types of projections used to transform 3D coordinates into 2D. There are two main types: parallel projection and perspective projection. Parallel projection involves projecting points along parallel lines, preserving properties like parallel lines. Perspective projection involves projecting points that converge on a center of projection, making distances appear smaller with depth. The document provides formulas for orthographic, oblique, and perspective projections, working through examples of projecting individual points from 3D to 2D coordinates under each projection type.
Mathematical analysis of identical circles touching one another on the whole ...Harish Chandra Rajpoot
All the articles discussed & analysed here are related to all five platonic solids. A certain no. of the identical circles are touching one another on a whole (entire) spherical surface having certain radius then all the important parameters such as flat radius & arc radius of each circle, total surface area & its percentage covered by all the circles on the sphere have been easily calculated by using simple geometry & table for the important parameters of all five platonic solids by the author Mr H.C. Rajpoot. These parameters are very useful for drawing the identical circles on a spherical surface & for designing & modeling all five platonic solids having identical flat circular faces.
1. The document introduces analytic geometry and its use of Cartesian coordinate systems to determine properties of geometric figures algebraically.
2. It defines key concepts like directed lines and rectangular coordinates, and explains how to find the distance between two points and the area of polygons using their coordinates.
3. Formulas are provided to calculate distances between horizontal, vertical and slanted line segments, as well as the area of triangles and general polygons from the coordinates of their vertices. Sample problems demonstrate applying these formulas.
Mathematical analysis of a non-uniform tetradecahedron having 2 congruent reg...Harish Chandra Rajpoot
All the important parameters of a non-uniform tetradecahedron, having 2 congruent regular hexagonal faces, 12 congruent trapezoidal faces & 18 vertices lying on a spherical surface with a certain radius, have been derived by the author by applying "HCR's Theory of Polygon" to calculate solid angle subtended by each regular hexagonal & trapezoidal face & their normal distances from the center of non-uniform tetradecahedron, inscribed radius, circumscribed radius, mean radius, surface area & volume. These formulas are very useful in the analysis, designing & modeling of various non-uniform polyhedra.
Mathematical analysis of non-uniform polyhedra having 2 congruent regular n-g...Harish Chandra Rajpoot
All the important formulas have been generalized which are applicable to calculate the important parameters, of any non-uniform polyhedron having 2 congruent regular n-gonal faces, 2n congruent trapezoidal faces each with three equal sides, 5n edges & 3n vertices lying on a spherical surface, such as solid angle subtended by each face at the centre, normal distance of each face from the centre, inner radius, outer radius, mean radius, surface area & volume. These are useful for the analysis, designing & modeling of different non-uniform polyhedra.
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Similar to Hcr's Theory of Polygon (Proposed by Mr Harish Chandra Rajpoot)
Mathematical analysis of sphere resting in the vertex of polyhedron, filletin...Harish Chandra Rajpoot
The generalized formula derived here by the author are applicable to locate any sphere, with a certain radius, resting in a vertex (corner) at which n no. of edges meet together at angle α between any two consecutive of them such as the vertex of platonic solids, any of two identical & diagonally opposite vertices of uniform polyhedrons with congruent right kite faces & the vertex of right pyramid with regular n-gonal base. These are also useful for filleting the faces meeting at the vertex of the polyhedron to best fit the sphere in that vertex. These are used to determine the distance of sphere from the vertex, distance of sphere from the edges, fillet radius of the faces etc. The formula have been generalized for packing the spheres in the vertices of right pyramids & all five platonic solids.
Mathematical Analysis of Tetrahedron (dihedral angles between the consecutive...Harish Chandra Rajpoot
All the articles have been derived by the author Mr H.C. Rajpoot by using HCR's Inverse cosine formula & HCR's Theory of Polygon. These formula are very practical & simple to apply in case of any tetrahedron to calculate the internal (dihedral) angles between the consecutive lateral faces meeting at any of four vertices & the solid angle subtended by it (tetrahedron) at the vertex when the angles between the consecutive edges meeting at the same vertex are known. These are the generalized formula which can also be applied in case of three faces meeting at the vertex of various regular & uniform polyhedrons to calculate the solid angle subtended by polyhedron at its vertex.
Mathematical Analysis of Regular Spherical Polygons (Application of HCR's The...Harish Chandra Rajpoot
This document presents a mathematical analysis of regular spherical polygons. It derives a characteristic equation relating the radius of the sphere, length of each side, interior angle, and number of sides. This equation can be used to calculate any parameter when the other three are known. The document also discusses calculating the solid angle, area, and other properties of both regular spherical polygons and their corresponding planar polygons. It provides examples of applying the analysis to polygons on both general spheres and the unit sphere.
Mathematical Analysis of Spherical Triangle (Spherical Trigonometry by H.C. R...Harish Chandra Rajpoot
All the important parameters of a spherical triangle have been derived by Mr H.C. Rajpoot by using simple geometry & trigonometry. All the articles (formula) are very practical & simple to apply in case of a spherical triangle to calculate all its important parameters such as solid angle, covered surface area, interior angles etc. & also useful for calculating all the parameters of the corresponding plane triangle obtained by joining all the vertices of a spherical triangle by straight lines. These formula can also be used to calculate all the parameters of the right pyramid obtained by joining all the vertices of a spherical triangle to the center of sphere such as normal height, angle between the consecutive lateral edges, area of base etc.
HCR's method of concentric cones (solid angle subtended by a torus at any poi...Harish Chandra Rajpoot
HCR's Method of concentric cones is the simplest & most versatile method to find out the solid angle subtended by a torus at any point lying on its geometrical axis.
HCR's Formula for Regular n-Polyhedrons (Mathematical analysis of regular n-p...Harish Chandra Rajpoot
This formula was derived by H.C. Rajpoot to calculate all the important parameters of a regular n-polyhedron such as inner radius, outer radius, mean radius, surface area & volume. This formula is a generalized dimensional formula which can be applied on any existing n-polyhedron since it depends on two parameters of any regular polyhedron as the no. of faces & the no. of edges in one face only. It can be used in analysis, designing & modelling of regular n-polyhedrons. It's named as "H. Rajpoot's Formula" by the author.
All the standard formula from 'Advanced Geometry' by the author Mr H.C. Rajpoot have been included in this book. These formula are related to the solid geometry dealing with the 2-D & the figures in 3-D space & miscellaneous articles in Trigonometry & Geometry. These are useful the standard formula for case studies & practical applications.
Mathematical analysis of n-gonal trapezohedron with 2n congruent right kite f...Harish Chandra Rajpoot
The generalized formula are applicable on any n-gonal trapezohedronhaving 2n congruent right kite faces, 4n edges & 2n+2 vertices lying on a spherical surface with a certain radius. These formula have been derived by the author Mr H.C. Rajpoot to analyse infinite no. of the trapezohedrons having congruent right kite faces simply by setting n=3,4,5,6,7,………………upto infinity, to calculate all the important parameters such as ratio of unequal edges, outer radius, inner radius, mean radius, surface area, volume, solid angles subtended by the polyhedron at its vertices, dihedral angles between the adjacent right kite faces etc. These formula are very useful for the analysis, modeling & designing of various n-gonal trapezoherons/deltohedrons.
Mathematical Analysis of Rhombicuboctahedron (Application of HCR's Theory)Harish Chandra Rajpoot
The author H C Rajpoot has derived the radius of circumscribed sphere passing through all 24 identical vertices of a rhombicuboctahedron with given edge length by applying ‘HCR’s Theory of Polygon’ & subsequently derived various formula to analytically compute the normal distances of equilateral triangular & square faces from the centre of rhombicuboctahedron, radius of mid-sphere, surface area, volume, solid angles subtended by each equilateral triangular face & each square face at the centre by using ‘HCR’s Theory of Polygon’, dihedral angle between each two faces meeting at any of 24 identical vertices (i.e. truncated rhombic dodecahedron), solid angle subtended by truncated rhombic dodecahedron at any of its 24 identical vertices.
All the standard formula from 'Advanced Geometry' by the author Mr H.C. Rajpoot have been included in this book. These formula are related to the solid geometry dealing with the 2-D & 3-D figures in the space & miscellaneous articles in Trigonometry & Geometry. These are useful the standard formula to be remembered for case studies & practical applications. Although, all the formula for the plane figures (i.e. planes bounded by the straight lines only) can be derived by using standard formula of right triangle that has been explained in details in "HCR's Theory of Polygon" published with International Journal of Mathematics & Physical Sciences Research in Oct, 2014.
And the analysis of oblique frustum of right circular cone has been explained in his research paper 'HCR's Infinite-series' published with IJMPSR
Volume & surface area of right circular cone cut by a plane parallel to its s...Harish Chandra Rajpoot
All the articles have been derived by the author by using simple geometry, trigonometry & calculus. All the formula are the most generalized expressions which can be used for computing the volume & surface area of minor & major parts usually each with hyperbolic section obtained by cutting a right circular cone with a plane parallel to its symmetrical (longitudinal) axis.
HCR’s Inverse Cosine Formula derived by Mr H.C. Rajpoot is a trigonometric relation of four variables/angles. It is applicable for any three straight lines or planes, either co-planar or non-coplanar, intersecting each other at a single point in the space. It directly co-relates the internal angles (i.e. angles between the consecutive lateral faces) & the face angles (i.e. angles between the consecutive lateral edges) of any tetrahedron. This formula is very useful to find out all the unknown internal angles if all the face angles of any tetrahedron are known & vice versa.
This formula holds good for all the regular spherical polygons. It is a very important formula (mathematical relation) applicable on any regular spherical polygon having each of its sides as an arc of the great circle on a spherical surface. It is of crucial importance to find out any of the four important parameters (i.e. radius of sphere, no. of sides, length of side, interior angle of polygon) if other three are given (known) for any regular spherical polygon. It also concludes that any three of the four parameters are self-sufficient to exactly represent a unique regular spherical polygon. A regular spherical polygon having three known parameters can be created or drawn only on a unique spherical surface of a radius which is easily found out by HCR's formula.
Mathematical Analysis of Spherical Rectangle (Application of HCR's Theory of ...Harish Chandra Rajpoot
All the articles have been derived by Mr H.C. Rajpoot by using simple geometry & trigonometry. All the formula are very practical & simple to apply in case of any spherical rectangle to calculate all its important parameters such as solid angle, surface area covered, interior angles etc. & also useful for calculating all the parameters of the corresponding plane rectangle obtained by joining all the vertices of a spherical rectangle by the straight lines. These formula can also be used to calculate all the parameters of the right pyramid obtained by joining all the vertices of a spherical rectangle to the center of the sphere such as normal height, angle between the consecutive lateral edges, area of plane rectangular base etc.
1. The document discusses various coordinate systems used to represent points in 2D and 3D space, including rectangular, polar, cylindrical and spherical coordinates.
2. It also provides equations for lines, circles, ellipses, parabolas, and hyperbolas in rectangular coordinate systems. These include the slope-intercept form for lines, the standard form for circles, and equations involving eccentricity for conic sections.
3. The equations are illustrated with diagrams showing their geometric significance in terms of distances from foci and directrixes or lengths of axes.
Mathematical analysis of a small rhombicuboctahedron (Archimedean solid) by HCRHarish Chandra Rajpoot
All the important parameters of a small rhombicuboctahedron (an Archimedean solid having 8 congruent equilateral triangular & 18 congruent square faces each of equal edge length) such as normal distances & solid angles subtended by the faces, inner radius, outer radius, mean radius, surface area & volume have been calculated by using HCR's formula for regular polyhedrons. This formula is a generalized dimensional formula which is applied on any of the five platonic solids i.e. reguler tetrahedron, regular hexahedron (cube), regular octahedron, regular dodecahedron & regular icosahedron to calculate their important parameters. It can also be used in analysis, designing & modelling of truncated polyhedrons.
This document discusses different types of projections used to transform 3D coordinates into 2D. There are two main types: parallel projection and perspective projection. Parallel projection involves projecting points along parallel lines, preserving properties like parallel lines. Perspective projection involves projecting points that converge on a center of projection, making distances appear smaller with depth. The document provides formulas and examples for performing orthographic, oblique, and perspective projections of 3D points onto a 2D plane.
This document discusses different types of projections used to transform 3D coordinates into 2D. There are two main types: parallel projection and perspective projection. Parallel projection involves projecting points along parallel lines, preserving properties like parallel lines. Perspective projection involves projecting points that converge on a center of projection, making distances appear smaller with depth. The document provides formulas for orthographic, oblique, and perspective projections, working through examples of projecting individual points from 3D to 2D coordinates under each projection type.
Mathematical analysis of identical circles touching one another on the whole ...Harish Chandra Rajpoot
All the articles discussed & analysed here are related to all five platonic solids. A certain no. of the identical circles are touching one another on a whole (entire) spherical surface having certain radius then all the important parameters such as flat radius & arc radius of each circle, total surface area & its percentage covered by all the circles on the sphere have been easily calculated by using simple geometry & table for the important parameters of all five platonic solids by the author Mr H.C. Rajpoot. These parameters are very useful for drawing the identical circles on a spherical surface & for designing & modeling all five platonic solids having identical flat circular faces.
1. The document introduces analytic geometry and its use of Cartesian coordinate systems to determine properties of geometric figures algebraically.
2. It defines key concepts like directed lines and rectangular coordinates, and explains how to find the distance between two points and the area of polygons using their coordinates.
3. Formulas are provided to calculate distances between horizontal, vertical and slanted line segments, as well as the area of triangles and general polygons from the coordinates of their vertices. Sample problems demonstrate applying these formulas.
Similar to Hcr's Theory of Polygon (Proposed by Mr Harish Chandra Rajpoot) (20)
Mathematical analysis of a non-uniform tetradecahedron having 2 congruent reg...Harish Chandra Rajpoot
All the important parameters of a non-uniform tetradecahedron, having 2 congruent regular hexagonal faces, 12 congruent trapezoidal faces & 18 vertices lying on a spherical surface with a certain radius, have been derived by the author by applying "HCR's Theory of Polygon" to calculate solid angle subtended by each regular hexagonal & trapezoidal face & their normal distances from the center of non-uniform tetradecahedron, inscribed radius, circumscribed radius, mean radius, surface area & volume. These formulas are very useful in the analysis, designing & modeling of various non-uniform polyhedra.
Mathematical analysis of non-uniform polyhedra having 2 congruent regular n-g...Harish Chandra Rajpoot
All the important formulas have been generalized which are applicable to calculate the important parameters, of any non-uniform polyhedron having 2 congruent regular n-gonal faces, 2n congruent trapezoidal faces each with three equal sides, 5n edges & 3n vertices lying on a spherical surface, such as solid angle subtended by each face at the centre, normal distance of each face from the centre, inner radius, outer radius, mean radius, surface area & volume. These are useful for the analysis, designing & modeling of different non-uniform polyhedra.
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A regular n-gonal right antiprism is a semiregular convex polyhedron which has 2n identical vertices all lying on a sphere, 4n edges, and (2n+2) faces out of which 2 are congruent regular n-sided polygons, and 2n are congruent equilateral triangles such that all the faces have equal side. The equilateral triangular faces meet the regular polygonal faces at the common edges and vertices alternatively such that three equilateral triangular faces meet at each of 2n vertices. This paper presents, in details, the mathematical derivations of the generalized and analytic formula which are used to determine the different important parameters in terms of edge length, such as normal distances of faces, normal height, radius of circumscribed sphere, surface area, volume, dihedral angles between adjacent faces, solid angle subtended by each face at the centre, and solid angle subtended by polygonal antiprism at each of its 2n vertices using HCR’s Theory of Polygon. All the generalized formulae have been derived using simple trigonometry, and 2D geometry which are difficult to derive using any other methods.
The document presents mathematical derivations of parameters for a regular pentagonal right antiprism. It derives analytic formulas for the antiprism's normal heights, normal distances from faces to the center, radius of the circumscribed sphere, surface area, volume, and other values in terms of the antiprism's edge length. All formulas are derived using trigonometry and 2D geometry applied to the transformation of a regular icosahedron into the antiprism shape.
Derivation of great-circle distance formula using of HCR's Inverse cosine for...Harish Chandra Rajpoot
The author derives the great-circle distance formula using hcr's inverse cosine formula. An analytic and the most generalized formula has been derived to accurately compute the minimum distance or great circle distance between any two arbitrary points on a sphere of finite radius which is equally applicable in the geometry of sphere. This formula is extremely useful to calculate the geographical distance between any two points on the globe for the given latitudes & longitudes. This formula is the most power tool which is applicable for all the distances on the tiny sphere as well as the large sphere like giant planet assuming them the perfect spheres.
The circumscribed and the inscribed polygons are well known and mathematically well defined in the context of 2D-Geometry. The term ‘Circum-inscribed Polygon’ has been proposed by the author and used as a new definition of the polygon which satisfies the conditions of a circumscribed polygon and an inscribed polygon together. In other words, the circum-inscribed polygon is a polygon which has both the inscribed and circumscribed circles. The newly defined circum-inscribed polygon has each of its sides touching a circle and each of its vertices lying on another circle. The most common examples of circum-inscribed polygon are triangle, regular polygon, trapezium with each of its non-parallel sides equal to the Arithmetic Mean (AM) of its parallel sides (called circum-inscribed trapezium) and right kite. This paper describes the mathematical derivations of the analytic formula to find out the different parameters in terms of AM and GM of known sides such as radii of circumscribed & inscribed circles, unknown sides, interior angles, diagonals, angle between diagonals, ratio of intersecting diagonals, perimeter, area, and distance between circum-centre and in-centre of circum-inscribed trapezium. Like an inscribed polygon, a circum-inscribed polygon always has all of its vertices lying on infinite number of spherical surfaces. All the analytic formulae have been derived using simple trigonometry and 2-dimensional geometry which can be used to analyse the complex 2D and 3D geometric figures such as cyclic quadrilateral and trapezohedron, and other polyhedrons.
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3) The analysis involves deriving the truncated polyhedron from a rhombic dodecahedron and establishing relationships between their geometric properties.
Mathematical Analysis of Rhombic Dodecahedron by applying HCR's Theory of Pol...Harish Chandra Rajpoot
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HCR's theorem (Rotation of two co-planar planes, meeting at angle bisector, a...Harish Chandra Rajpoot
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This theorem is very important for creating pyramidal flat containers with polygonal (regular or irregular) base, closed right pyramids & polyhedrons having two regular n-gonal & 2n congruent trapezoidal faces.
The author has also presented some paper models of pyramidal flat containers with regular pentagonal, heptagonal and octagonal bases
How to compute area of spherical triangle given the aperture angles subtended...Harish Chandra Rajpoot
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Reflection of a point about a line & a plane in 2-D & 3-D co-ordinate systems...Harish Chandra Rajpoot
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LAND USE LAND COVER AND NDVI OF MIRZAPUR DISTRICT, UPRAHUL
This Dissertation explores the particular circumstances of Mirzapur, a region located in the
core of India. Mirzapur, with its varied terrains and abundant biodiversity, offers an optimal
environment for investigating the changes in vegetation cover dynamics. Our study utilizes
advanced technologies such as GIS (Geographic Information Systems) and Remote sensing to
analyze the transformations that have taken place over the course of a decade.
The complex relationship between human activities and the environment has been the focus
of extensive research and worry. As the global community grapples with swift urbanization,
population expansion, and economic progress, the effects on natural ecosystems are becoming
more evident. A crucial element of this impact is the alteration of vegetation cover, which plays a
significant role in maintaining the ecological equilibrium of our planet.Land serves as the foundation for all human activities and provides the necessary materials for
these activities. As the most crucial natural resource, its utilization by humans results in different
'Land uses,' which are determined by both human activities and the physical characteristics of the
land.
The utilization of land is impacted by human needs and environmental factors. In countries
like India, rapid population growth and the emphasis on extensive resource exploitation can lead
to significant land degradation, adversely affecting the region's land cover.
Therefore, human intervention has significantly influenced land use patterns over many
centuries, evolving its structure over time and space. In the present era, these changes have
accelerated due to factors such as agriculture and urbanization. Information regarding land use and
cover is essential for various planning and management tasks related to the Earth's surface,
providing crucial environmental data for scientific, resource management, policy purposes, and
diverse human activities.
Accurate understanding of land use and cover is imperative for the development planning
of any area. Consequently, a wide range of professionals, including earth system scientists, land
and water managers, and urban planners, are interested in obtaining data on land use and cover
changes, conversion trends, and other related patterns. The spatial dimensions of land use and
cover support policymakers and scientists in making well-informed decisions, as alterations in
these patterns indicate shifts in economic and social conditions. Monitoring such changes with the
help of Advanced technologies like Remote Sensing and Geographic Information Systems is
crucial for coordinated efforts across different administrative levels. Advanced technologies like
Remote Sensing and Geographic Information Systems
9
Changes in vegetation cover refer to variations in the distribution, composition, and overall
structure of plant communities across different temporal and spatial scales. These changes can
occur natural.
A review of the growth of the Israel Genealogy Research Association Database Collection for the last 12 months. Our collection is now passed the 3 million mark and still growing. See which archives have contributed the most. See the different types of records we have, and which years have had records added. You can also see what we have for the future.
Hcr's Theory of Polygon (Proposed by Mr Harish Chandra Rajpoot)
1. “HCR’s Theory of Polygon”
“Solid angle subtended by any polygonal plane at any point in the space”
Madan Mohan Malaviya University of Technology, Gorakhpur-273010 (UP) India
Abstract: This is the unifying principle proposed by the author to calculate mathematically correct value of solid angle subtended by any polygonal plane at any point in the space. According to this theory, if the location of foot of perpendicular drawn from the given point in the space to the plane of polygon is specified then the polygonal plane can be internally or externally or both divided into certain number of triangles by joining all the vertices of polygon to the foot of perpendicular. Further each of the triangles can be internally or externally sub-divided in two right triangles having common vertex at the foot of perpendicular. Thus, the solid angle subtended by the polygonal plane at any point is the algebraic sum of solid angles subtended by the triangles at the same point such that the algebraic sum of areas of all these triangles is equal to the area of polygon. This theory requires only one standard formula for the solid angle subtended by a right triangular plane for finding out the solid angle subtended by any polygonal plane. The applications of solid angle subtended by the planes & plates are wider in the field of Radiometry for analysis of radiation energy emitted by point-sources. This field requires precise values of radiation energy emitted by uniform point-sources like radioactive elements. Also, this theory is extremely useful in case studies & practical computations.
Keywords: Theory of Polygon, HCR’s Standard Formula-1, Solid angle, Polygonal plane, F.O.P., Element Method, Algebraic sum
I. Introduction
The graphical method overcomes the limitations of all other analytical methods provided the location of foot of perpendicular drawn from the given point to the plane of polygon is known. It involves theoretically zero error if the calculations are done correctly. A polygon is the plane bounded by the straight lines. But, if the given polygon is divided into certain number of elementary triangles then the solid angle subtended by polygon at given point can be determined by summing up the solid angles subtended by all the elementary triangles at the same point. This concept derived a Theory. According to it “for a given configuration of plane & location of point in the space, if a perpendicular, from any given point in the space, is drawn to the plane of given polygon then the polygonal plane (polygon) can be internally or externally or both divided into a certain number of elementary triangles by joining all the vertices of polygon to the foot of perpendicular (F.O.P.) Further each of these triangles can be internally or externally sub-divided in two right triangles having common vertex at the foot of perpendicular. Thus the solid angle subtended by the given polygonal plane at the given point is the algebraic sum of solid angles subtended by all the elementary triangles at the same point such that algebraic sum of the areas of all these triangles is equal to the area of given polygonal plane”
Let’s study in an order to easily understand the Theory of Polygon in a simple way
II. HCR’s Standard Formula-1 (Solid angle subtended by a right triangular plane at any point lying at a normal height h from any of the acute angled vertices)
Using Fundamental Theorem of Solid Angle:
Let there be a right triangular plane ONM having perpendicular ON = p & the base MN = b and a given point say P(0, 0, h) at a height ‘h’ lying on the axis (i.e. Z-axis) passing through the acute angled vertex say ‘O’
(As shown in the figure1 below)
In right ⇒ ⇒
2. ⇒
Now, the equation of the straight line OM passing through the origin ‘O’ ⇒ ( ) ⇒
Now, consider a point Q(x, y) on the straight line OM
⇒
In right ⇒ √ . / √. / √
In right
⇒ √ √
Let’s consider an imaginary spherical surface having radius ‘R’ & centre at the given point ‘P’ such that the area of projection of the given plane ONM on the spherical surface w.r.t. the given point ‘P’ is ‘A’
Now, consider an elementary area of projection ‘dA’ of the plane ONM on the spherical surface in the first quadrant YOX
(As shown in the figure 2)
Now, elementary area of projection in the first quadrant
⇒ ( )( ) ( )( )
Hence, area of projection of the plane ONM on the spherical surface in the first quadrant is obtained by integrating the above expression in the first quadrant &
Applying the limits of ( ⁄ ) ⁄ & Fig 1: Right Triangular Plane ONM, ON=p & MN=b Fig 2: Elementary area dA on spherical surface
3. ⇒ ∫[∫ ] ∫[∫ ] ∫ , - ∫ ∫ √ ∫ √ ( ) ∫ √ ∫ √ √ √ ⇒ 0 2 31 [ 2 3 { }] [ { } 2 3]
Now, on setting the values of ‘K’ & , we get ⇒ [ { √ } { √ √ } ] 6 8 √ 9 84 √ 54 √ 597
Now, the solid angle subtended by the right triangular plane ONM at the given point ‘P’
= solid angle subtended by area of projection of triangle ONM on the spherical surface at the same point ‘P’
Using Fundamental Theorem ⇒ ∫ ∫ ∫
( ) ⇒ 6 8 √ 9 84 √ 54 √ 597 8 √ 9 84 √ 54 √ 59 ⇒ 6 8 √ 9 84 √ 54 √ 597 ( )
4. If the given point P (0, 0, h) is lying on the axis normal to the plane & passing through the acute angled vertex ‘M’ then the solid angle subtended by the right triangular plane ONM is obtained by replacing ‘b’ by ‘p’ & ‘p’ by ‘b’ in the equation(1), we have ⇒ 6 8 √ 9 84 √ 5( √ )97
Note: Eq(1) is named as HCR’s Standard Formula-1 which is extremely useful to find out the solid angle subtended by any polygon at any point in the space thus all the formulae can be derived by using eq(1).
III. Specifying the location of given point & foot of perpendicular in the Plane of Polygon
Let there be any point say point P & any polygonal plane say plane ABCDEF in the space. Draw a perpendicular PO from the point ‘P’ to the plane of polygon which passes through the point ‘O’ i.e. foot of perpendicular (F.O.P.) (See front & top views in the figure 3 below showing actual location of point ‘P’ & F.O.P. ‘O’)
Fig 3: Actual location of point ‘P’ & F.O.P. ‘O’ Fig 4: Location of ‘P’ & ‘O’ in the plane of paper
Now, for simplification of specifying the location of given point ‘P’ & foot of perpendicular ‘Q’ we assume
a. Foot of perpendicular ‘O’ as the origin &
b. Point P is lying on the Z-axis
If the length of perpendicular PO is h then the location of given point ‘P’ & foot of perpendicular ‘O’ in the plane of polygon (i.e. plane of paper) is denoted by ( ) similar to the 3-D co-ordinate system
(See the figure 4 above)
IV. Element Method (Method of dividing the polygon into elementary triangles)
It is the method of Joining all the vertices A, B, C, D, E & F of a given polygon to the foot of perpendicular ‘O’ drawn from the given point P in the space (as shown by the dotted lines in figure 4 above) Thus, are the elementary triangles which have common vertex at the foot of perpendicular ‘O’.
Note: are not taken as the elementary triangles since they don’t have common vertex at the foot of perpendicular ‘O’
The area ( ) of polygon ABCDEF is given as follows (from the figure 4)
5. Since the location of given point & the configuration of polygonal plane is not changed hence the solid angle ( ) subtended by polygonal plane at the given point in the space is the algebraic sum of solid angles subtended by the elementary triangles obtained by joining all the vertices of polygon to the F.O.P.
Now, replacing the areas of elementary triangles by their respective values of solid angles in the above expression as follows
While, the values of solid angles subtended by elementary triangles are determined by using axiom of triangle i.e. dividing each elementary triangle into two right triangles & using standard formula-1 of right triangle.
V. Axiom of Triangle
“If the perpendicular, drawn from a given point in the space to the plane of a given triangle, passes through one of the vertices then that triangle can be divided internally or externally (w.r.t. F.O.P.) into two right triangles having common vertex at the foot of perpendicular, simply by drawing a normal from the common vertex (i.e. F.O.P.) to the opposite side of given triangle.”
An acute angled triangle is internally divided w.r.t. F.O.P. (i.e. common vertex)
A right angled triangle is internally divided w.r.t. F.O.P. (i.e. common vertex)
An obtuse angled triangle is divided
Internally if and only if the angle of common vertex (F.O.P.) is obtuse
Externally if and only if the angle of common vertex (F.O.P.) is acute
Now, consider a given point ( ) (located perpendicular to the plane of paper) lying at a height h on the normal axes passing through the vertices A, B & C where the points ( ) ( ) ( ) are the different locations of given point P on the perpendiculars passing through the vertices A, B & C respectively at the same height h.
(See the different cases in the figures (5), (6) & (7) below)
Acute Angled Triangle:
Consider the given point ( ) at a normal height h from the vertex ‘A’ (i.e. foot of perpendicular drawn from the point to the plane of )
Now, draw the perpendicular AM from F.O.P. ‘A’ to the opposite side BC to divide the into two sub-elementary right triangles
In this case, the area ( ) of is given by
( )
Hence replacing the areas by the corresponding values of solid angles, we get Fig 5: Acute Angled Triangle
6. ( ) ( )
The values of subtended by the right triangles respectively are calculated by using standard formula-1 of right triangle (from eq(1) or eq(2)) as follows 8 √ 9 84 √ 54 √ 59
Measure the dimensions from the drawing & set
( ) ( ) in above expression
We get, the solid angle subtended by the right at the given point ( ) 8 √( ) ( ) 9 84 √( ) ( ) 54 √( ) ( ) 59
Similarly, solid angle subtended by the right at the the same point ( ) 8 √( ) ( ) 9 84 √( ) ( ) 54 √( ) ( ) 59
Hence, the solid angle subtended by the given at the the given point ( ) is calculated as ⇒
Similarly, for the location of given point ( ) lying at a normal height h from the vertex ‘B’ (i.e. foot of perpendicular drawn from the point ( ) to the plane of ) (See figure 5 above)
By following the above procedure, we get the solid angle subtended by the given at the the given point ( ) as follows ⇒ 8 √( ) ( ) 9 84 √( ) ( ) 54 √( ) ( ) 59 8 √( ) ( ) 9 84 √( ) ( ) 54 √( ) ( ) 59
Similarly, for the location of given point ( ) lying at a normal height h from the vertex ‘C’ (i.e. foot of perpendicular drawn from the point ( ) to the plane of ) (See figure 5 above)
By following the above procedure, we get the solid angle subtended by the given at the the given point ( ) as follows ⇒
We can calculate the corresponding values of using standard formula-1 as follows
7. 8 √( ) ( ) 9 84 √( ) ( ) 54 √( ) ( ) 59 8 √( ) ( ) 9 84 √( ) ( ) 54 √( ) ( ) 59
Right Angled Triangle:
Consider the given point ( ) at a normal height h from the vertex ‘A’ (i.e. foot of perpendicular drawn from the point ( ) to the plane of right ) (See the figure 6)
Now, draw the perpendicular AM from F.O.P. ‘A’ to the opposite side BC to divide the right into two sub-elementary right triangles
In this case, the area ( ) of right is given by
( )
Hence replacing the areas by the corresponding values of solid angles, we get ( ) ( )
The values of subtended by the right triangles respectively are calculated by using standard formula-1 of right triangle (from eq(1) or eq(2)) as follows 8 √ 9 84 √ 54 √ 59
Measure the dimensions from the drawing & set
( ) ( ) in above expression
We get, the solid angle subtended by the right at the given point ( ) 8 √( ) ( ) 9 84 √( ) ( ) 54 √( ) ( ) 59
Similarly, solid angle subtended by the right at the the same point ( ) 8 √( ) ( ) 9 84 √( ) ( ) 54 √( ) ( ) 59
Hence, the solid angle subtended by right at the the given point ( ) is calculated as ⇒ Fig 6: Right Angled Triangle
8. Now, for the location of given point ( ) lying at a normal height h from the vertex ‘B’ (i.e. foot of perpendicular drawn from the point ( ) to the plane of right ) (See figure 6 above)
Using standard formula-1 & setting
( ) ( )
We get the solid angle subtended by the right at the the given point ( ) as follows 8 √( ) ( ) 9 84 √( ) ( ) 54 √( ) ( ) 59
Similarly, for the location of given point ( ) lying at a normal height h from the vertex ‘C’ (i.e. foot of perpendicular drawn from the point ( ) to the plane of right ) (See figure 6 above)
Using standard formula-1 & setting
( ) ( )
We get the solid angle subtended by the right at the the given point ( ) as follows 8 √( ) ( ) 9 84 √( ) ( ) 54 √( ) ( ) 59
Obtuse Angled Triangle:
Consider the given point ( ) at a normal height h from the vertex ‘A’ (i.e. foot of perpendicular drawn from the point ( ) to the plane of ) (See the figure 7)
Now, draw the perpendicular AM from F.O.P. ‘A’ to the opposite side BC to divide the obtuse into two sub- elementary right triangles
In this case, the area ( ) of obtuse is given by
( )
Hence replacing the areas by the corresponding values of solid angles, we get ( ) ( )
The values of subtended by the right triangles respectively are calculated by using standard formula-1 of right triangle (from eq(1) or eq(2)) as follows 8 √ 9 84 √ 54 √ 59 Fig 7: Obtuse Angled Triangle
9. Measure the dimensions from the drawing & set
( ) ( ) in above expression
We get, the solid angle subtended by the right at the given point ( ) 8 √( ) ( ) 9 84 √( ) ( ) 54 √( ) ( ) 59
Similarly, solid angle subtended by the right at the the same point ( ) 8 √( ) ( ) 9 84 √( ) ( ) 54 √( ) ( ) 59
Hence, the solid angle subtended by obtuse at the the given point ( ) is calculated as ⇒
Now, for the location of given point ( ) lying at a normal height h from the vertex ‘B’ (i.e. foot of perpendicular drawn from the point ( ) to the plane of ) (See figure 7 above)
Draw the perpendicular BQ from F.O.P. ‘B’ to the opposite side AC to divide the obtuse into two sub-elementary right triangles
In this case, the area ( ) of obtuse is given by ( )
Hence replacing the areas by the corresponding values of solid angles, we get ( ) ( )
The values of subtended by the right triangles respectively are calculated by using standard formula-1 of right triangle (from eq(1) or eq(2)) as follows 8 √ 9 84 √ 54 √ 59
Measure the dimensions from the drawing & set
( ) ( ) in above expression
We get, the solid angle subtended by the right at the given point ( ) 8 √( ) ( ) 9 84 √( ) ( ) 54 √( ) ( ) 59
Similarly, solid angle subtended by the right at the the same point ( ) 8 √( ) ( ) 9 84 √( ) ( ) 54 √( ) ( ) 59
10. Hence, the solid angle subtended by obtuse at the the given point ( ) is calculated as ⇒
Similarly, for the location of given point ( ) lying at a normal height h from the vertex ‘C’ (i.e. foot of perpendicular drawn from the point ( ) to the plane of ) (See figure 7 above)
Following the above procedure, solid angle subtended by obtuse at the the given point ( ) is calculated as ⇒ ( ) 8 √( ) ( ) 9 84 √( ) ( ) 54 √( ) ( ) 59 ⇒ 8 √( ) ( ) 9 84 √( ) ( ) 54 √( ) ( ) 59
Now, we will apply the same procedure in case of any polygonal plane by diving it into the elementary triangles with common vertex at the F.O.P. & then each elementary triangle into two sub-elementary right triangles with common vertex at the F.O.P. Now, it’s easy to understand the theory of polygon
VI. Axiom of polygon
“For a given point in the space, each of the polygons can be divided internally or externally or both w.r.t. foot of perpendicular (F.O.P.) drawn from the given point to the plane of polygon into a certain number of the elementary triangles, having a common vertex at the foot of perpendicular, by joining all the vertices of polygon to the F.O.P. by straight lines (generally extended).”
Polygon is externally divided if the F.O.P. lies outside the boundary
Polygon is divided internally or externally or both if the F.O.P. lies inside or on the boundary depending on the geometrical shape of polygon (i.e. angles & sides)
(See the different cases in the figures (8), (9), (10) & (11) below as explained in Theory of Polygon)
Elementary Triangle: Any of the elementary triangles obtained by joining all the vertices of polygon to the foot of perpendicular drawn from the given point to the plane of polygon such that all the elementary triangles have common vertex at the foot of perpendicular (F.O.P.) is called elementary triangle.
Sub-elementary Right Triangles: These are the right triangles which are obtained by drawing a perpendicular from F.O.P. to the opposite side in any of the elementary triangles. Thus one elementary triangle is internally or externally divided into two sub-elementary right triangles. These sub-elementary right triangles always have common vertex at the foot of perpendicular (F.O.P.)
11. VII. HCR’s Theory of Polygon
(Proposed by the Author-2014)
This theory is applicable for any polygonal plane (i.e. plane bounded by the straight lines only) & any point in the space if the following parameters are already known
1. Geometrical shape & dimensions of the polygonal plane
2. Normal distance (h) of the given point from the plane of polygon
3. Location of foot of perpendicular (F.O.P.) drawn from given point to the plane of polygon
According to this theory “Solid angle subtended by any polygonal plane at any point in the space is the algebraic sum of solid angles subtended at the same point by all the elementary triangles (obtained by joining all the vertices of polygon to the foot of perpendicular) having common vertex at the foot of perpendicular drawn from the given point to the plane of polygon such that algebraic sum of areas of all these triangles is equal to the area of given polygon.” It has no mathematical proof.
Mathematically, solid angle ( ) subtended by any polygonal plane with ‘n’ number of sides & area ‘A’ at any point in the space is given by ⇒ 0Σ 1 0Σ 1
are the areas of elementary triangles subtending the solid angles at the given point which are determined by using standard formula-1 (as given from eq(1)) along with the necessary dimensions which are found out analytically or by tracing the diagram which is easier.
Element Method: It the method of diving a polygon internally or externally into sub-elementary right triangles having common vertex at the F.O.P. drawn from the given point to the plane of polygon such that algebraic sum of areas of all these triangles is equal to the area of given polygon.”
Working Steps:
STEP 1: Trace/draw the diagram of the given polygon (plane) with known geometrical dimensions.
STEP 2: Specify the location of foot of perpendicular drawn from a given point to the plane of polygon.
STEP 3: Join all the vertices of polygon to the foot of perpendicular by the extended straight lines. Thus the polygon is divided into a number of elementary triangles having a common vertex at the foot of perpendicular.
STEP 4: Further, consider each elementary triangle & divide it internally or externally into two sub-elementary right triangles simply by drawing a perpendicular from F.O.P. (i.e. common vertex) to the opposite side of that elementary triangle.
STEP 5: Now, find out the area of given polygon as the algebraic sum of areas of all these sub-elementary right triangles i.e. area of each of right triangles must be taken with proper sign whose sum gives area of polygon.
12. Remember: All the elementary triangles & sub-elementary right triangles must have their one vertex common at the foot of perpendicular (F.O.P.).
STEP 6: Replace areas of all these sub-elementary right triangles by their respective values of solid angle subtended at the given point in the space.
STEP 7: Calculate solid angle subtended by each of individual sub-elementary right triangles by using the standard formula-1 of right triangle (from eq(1) as derived above) as follows 8 √ 9 84 √ 54 √ 59
STEP 8: Now, solid angle subtended by polygonal plane at the given point will be the algebraic sum of all its individual sub-elementary right triangles as given (By Element Method)
* Finally, we get the value of as the algebraic sum of solid angles subtended by right triangles only.
VIII. Special Cases for a Polygonal Plane
Let’s us consider a polygonal plane (with vertices) & a given point say ( ) at a normal height from the given plane in the space.
Now, specify the location of foot of perpendicular say ‘O’ on the plane of polygon which may lie
1. Outside the boundary
2. Inside the boundary
3. On the boundary
a. On one of the sides or b. At one of the vertices
Let’s consider the above cases one by one as follows
1. F.O.P. outside the boundary:
Let the foot of perpendicular ‘O’ lie outside the boundary of polygon. Join all the vertices of polygon (plane) 123456 to the foot of perpendicular ‘O’ by the extended straight lines
(As shown in the figure 8) Thus the polygon is divided into elementary triangles (obtained by the extension lines), all having common vertex at the foot of perpendicular ‘O’. Now the solid angle subtended by the polygonal plane at the given point ‘P’ in the space is given
By Element-Method ( ) Fig 8: F.O.P. lying outside the boundary
13. Thus, the value of solid angle subtended by the polygon at the given point is obtained by setting these values in the eq. (I) as follows ⇒ ( ) ( ) ( ) ( )
Further, each of the individual elementary triangles is divided into two sub-elementary right triangles for which the values of solid angle are determined by using standard formula-1 of right triangle.
2. F.O.P. inside the boundary:
Let the foot of perpendicular ‘O’ lie inside the boundary of polygon. Join all the vertices of polygon (plane) 123456 to the foot of perpendicular ‘O’ by the extended straight lines (As shown in the figure 9)
Thus the polygon is divided into elementary triangles (obtained by the extension lines), all having common vertex at the foot of perpendicular ‘O’. Now the solid angle subtended by the polygonal plane at the given point P in the space is given
By Element-Method
( )
Where,
Thus, the value of solid angle subtended by the polygon at the given point is obtained by setting these values in the eq. (II) as follows ⇒ ( )
Further, each of the individual elementary triangles is divided into two sub-elementary right triangles for which the values of solid angle are determined by using standard formula-1 of right triangle.
3. F.O.P. on the boundary: Further two cases are possible
a. F.O.P. lying on one of the sides:
Let the foot of perpendicular ‘O’ lie on one of the sides say ‘12’ of polygon. Join all the vertices of polygon (plane) 123456 to the foot of perpendicular ‘O’ by the straight lines
(As shown in the figure 10)
Thus the polygon is divided into elementary triangles (obtained by the straight lines), all having common vertex at the foot of perpendicular ‘O’. Now the solid angle subtended by the polygonal plane at the given point ‘P’ in the space is given By Figure 9: F.O.P. lying inside the boundary Figure 10: F.O.P. lying at one of the sides
14. Element-Method
( )
Further, each of the individual elementary triangles is divided into two sub-elementary right triangles for which the values of solid angle are determined by using standard formula-1 of right triangle.
b. F.O.P. lying at one of the vertices:
Let the foot of perpendicular lie on one of the vertices say ‘5’ of polygon. Join all the vertices of polygon (plane) 123456 to the foot of perpendicular (i.e. common vertex ‘5’) by the straight lines
(As shown in the figure 11)
Thus the polygon is divided into elementary triangles (obtained by the straight lines), all having common vertex at the foot of perpendicular ‘5’. Now the solid angle subtended by the polygonal plane at the given point ‘P’ in the space is given
By Element-Method ( )
Further, each of the individual elementary triangles is divided into two sub-elementary right triangles for which the values of solid angle are determined by using standard formula-1 of right triangle.
IX. Analytical Applications of Theory of Polygon
Let’s take some examples of polygonal plane to find out the solid angle at different locations of a given point in the space. For ease of understanding, we will take some particular examples where division of polygons into right triangles is easier & standard formula can directly be applied by analytical measurements of necessary dimensions i.e. drawing is not required. Although, random location of point may cause complex calculations
Right Triangular Plane
F.O.P. lying on the right angled vertex: Let there be a right triangular plane ABC having orthogonal sides AB = a & BC = b & a given point say P(0, 0, h) at a normal height h from the right angled vertex ‘B’
(As shown in the figure 12 below)
Now, draw a perpendicular BN from the vertex ‘B’ to the hypotenuse AC to divide the right into elementary right triangles . By element method, solid angle subtended by the right at the given point P ⇒
Where, the values of are calculated by standard Fig 11: F.O.P. lying on one of the vertices Fig 12: Point P lying on the normal axis passing through the right angled vertex B
15. formula-1 as follows √ √ √ 8 √( ) ( ) 9 84 √( ) ( ) 54 √( ) ( ) 59 { √ √( √ ) ( √ ) } { ( √ √( √ ) ( √ ) ) ( √( ) ( √ ) ) } ⇒ { √ } 8 √ ( ) 9
Similarly, we can calculate solid angle subtended by the at the given point ( ) 8 √( ) ( ) 9 84 √( ) ( ) 54 √( ) ( ) 59 { √ √( √ ) ( √ ) } { ( √ √( √ ) ( √ ) ) ( √( ) ( √ ) ) } ⇒ { √ } 8 √ ( ) 9
Hence, the solid angle subtended by given right at the given point ( ) lying at a normal height h from the right angled vertex ‘B’ is calculated as follows ⇒ { √ } 8 √ ( ) 9 { √ } 8 √ ( ) 9 [ { √ } { √ }] 6 8 √ ( ) 9 8 √ ( ) 97
Using, ( √ √ ) ( ( ) ) & simplifying, we get ⇒ 8 9 8 ( √ √ ) ( ) 9 8 ( √ √ ) ( ) 9
16. 8 ( √ √ ) ( ) 9 ( )
Note: This is the standard formula to find out the value of solid angle subtended by a right triangular plane, with orthogonal sides at any point lying at a normal height h from the right angled vertex.
Rectangular Plane
F.O.P. lying on one of the vertices of rectangular plane: Let there be a rectangular plane ABCD having length & width & a given point say P(0, 0, h) at a normal height h from any of the vertices say vertex ‘A’
Now, draw a perpendicular PA from the given point ‘P’ to the plane of rectangle ABCD passing through the vertex A (i.e. foot of perpendicular). Join the vertex ‘C’ to the F.O.P. ‘A’ to divide the plane into elementary triangles I.e. right triangles
It is clear from the figure (13) that the area of rectangle ABCD
Hence, using Element Method by replacing areas by corresponding values of solid angles, the solid angle subtended by the rectangular plane ABCD at the given point P
Now, using standard formula-1, solid angle subtended by the right at the point ( ) 8 √( ) ( ) 9 84 √( ) ( ) 54 √( ) ( ) 59 { √ } {( √ )( √ )} 8 √( ) ( ) 9 84 √( ) ( ) 54 √( ) ( ) 59 { √ } {( √ )( √ )}
Hence, the solid angle subtended by rectangular plane ABCD at the given point ( )lying at a normal height h from vertex A { √ } {( √ )( √ )} { √ } {( √ )( √ )} [ { √ } { √ }] [ {( √ )( √ )} {( √ )( √ )}]
Using, ( √ √ ) ( ( ) ) & simplifying, we get Fig 13: Point P lying on the normal axis passing through one of the vertices of rectangular plane
17. ⇒ 8 9 8 ( )√ ( )√ √ 9 8 √ √ √ 9 8 √ √ √ 9 { √ √ } 8 √( )( ) 9 ( )
Note: This is the standard formula to find out the value of solid angle subtended by a rectangular plane of size at any point lying at a normal height h from any of the vertices.
F.O.P. lying on the centre of rectangular plane: Let the given point P(0, 0, h) be lying at a normal height h from the centre ‘O’ (i.e. F.O.P.) of rectangular plane ABCD. Join all the vertices A, B, C, D to the F.O.P. ‘O’.
Thus, rectangle ABCD is divided into four elementary triangles . Hence, Area of rectangle ABCD
⇒ ( )
Now, draw perpendiculars OE & OF from the F.O.P. ‘O’ to the opposite sides AB in & AD in to divide them into sub- elementary right triangles
⇒ ( ) ( )
Hence, using Element Method by replacing areas by corresponding values of solid angles, the solid angle subtended by the rectangular plane ABCD at the given point P
( )
Now, using standard formula-1, solid angle subtended by the right at the point ( ) 8 √( ) ( ) 9 84 √( ) ( ) 54 √( ) ( ) 59 { . / √. / . / } { ( . / √. / . / ) ( √ . / ) } { √ } {( √ )( √ )} 8 √( ) ( ) 9 84 √( ) ( ) 54 √( ) ( ) 59 { . / √. / . / } { ( . / √. / . / ) ( √ . / ) } Fig 14: Point P lying on the normal axis passing through the centre of rectangular plane
18. { √ } {( √ )( √ )}
Hence, the solid angle subtended by rectangular plane ABCD at the given point ( )lying at a normal height h from vertex A ( ) [ { √ } {( √ )( √ )} { √ } {( √ )( √ )}] [ { √ } { √ }] [ {( √ )( √ )} {( √ )( √ )}]
Using, ( √ √ ) ( ( ) ) & simplifying, we get ⇒ 6 8 9 8 ( )√ ( )√ √ 97 6 8 √ √ √ 97 8 √ √ √ 9 { √ √ } 8 √( )( ) 9 ( )
Note: This is the standard formula to find out the value of solid angle subtended by a rectangular plane of size at any point lying at a normal height h from the centre.
Rhombus-like Plane
Let there be a rhombus-like plane ABCD having diagonals bisecting each other at right angle at the centre ‘O’ and a given point say P(0, 0, h) at a height ‘h’ lying on the normal axis passing through the centre ‘O’ (i.e. foot of perpendicular) (See the figure 15)
Join all the vertices A, B, C & D to the F.O.P. to divide the rhombus ABCD into elementary right triangles Now, area of rhombus ABCD
⇒
( )
Hence, using Element Method by replacing area by corresponding value of solid angle, the solid angle subtended by the rhombus-like plane ABCD at the given point P.
From eq(3), we know that solid angle subtended by a right triangular plane at any point lying on the vertical passing though right angled vertex is given as Fig 15: Point P lying on the normal axis passing through the centre of rhombus
19. 8 ( √ √ ) ( ) 9
On setting, in the above equation, we get { 4 √ √ 5 ( ) }
Hence, the solid angle subtended by rhombus-like plane ABCD at the given point ( )lying at a normal height h from centre ‘O’ ⇒ { 4 √ √ 5 ( ) } ( )
Note: This is the standard formula to find out the value of solid angle subtended by a rhombus-like plane having diagonals at any point lying at a normal height h from the centre.
Regular Polygonal Plane
Let there be a regular polygonal plane having ‘n’ no. of the sides each equal to the length ‘ ’ & a given point say P(0, 0, h) at a height ‘h’ lying on the normal height h from the centre ‘O’. (i.e. foot of perpendicular) (See the figure 16)
Join all the vertices to the F.O.P. ‘O’ to divide the polygon into elementary triangles which are congruent hence the area of polygon
( ) ( )
Now, draw a perpendicular OM from F.O.P. ‘O’ to opposite side to divide into two sub- elementary right triangles which are congruent. Hence area of polygon ( ) ( )
Hence, using Element Method by replacing area by corresponding value of solid angle, the solid angle subtended by the regular polygonal plane at the given point P. ( )
Now, using standard formula-1, solid angle subtended by the right at the point ( ) 8 √( ) ( ) 9 84 √( ) ( ) 54 √( ) ( ) 59 Fig 16: Point P lying on the normal axis passing through the centre of regular polygonal
20. Now, setting the corresponding values in above expression as follows { √. / . / } { ( √. / . / ) ( √ . / ) } { } { : ; ( √ ) } 2 3 { √ } { √ }
Hence, the solid angle subtended by regular polygonal plane at the given point ( ) lying at a normal height h from centre ‘O’ [ { √ } ] ⇒ { √ } ( )
Note: This is the standard formula to find out the value of solid angle subtended by a regular polygonal plane, having number of sides each of length at any point lying at a normal height h from the centre.
Regular Pentagonal Plane
Let there be a regular pentagonal plane ABCDE having each side of length and a given point say P(0, 0, h) lying at a normal height ‘h’ from the centre ‘O’ (i.e. foot of perpendicular) (See figure 17)
Join all the vertices A, B, C, D & E to the F.O.P. ‘O’ to divide the pentagon ABCDE into elementary triangles
Now, draw perpendiculars AN, AQ & AM to the opposite sides BC, CD & DE in respectively to divide each elementary triangle into two right triangles. Hence the area of regular pentagon
( )
( ) ( ) Fig 17: Point P lying on the normal axis passing through the vertex A of regular pentagon
21. Hence, using Element Method by replacing areas by corresponding values of solid angle, the solid angle subtended by the regular pentagonal plane ABCDE at the given point P. ( ) ( )
Necessary dimensions can be calculated by the figure as follows
Now, using standard formula-1, solid angle subtended by the right at the point ( ) 8 √( ) ( ) 9 84 √( ) ( ) 54 √( ) ( ) 59 8 √( ) ( ) 9 84 √( ) ( ) 54 √ ( ) 59 8 √ ( ) 9 84 √ ( ) 5( √ )9 {√ } {:√ ;( √ )} ( √ ) : √ ;
Similarly, we get solid angle subtended by right at the given point ‘P’ 8 √( ) ( ) 9 84 √( ) ( ) 54 √( ) ( ) 59 8 √( ) ( ) 9 84 √( ) ( ) 54 √ ( ) 59 * + {( )( √ )} ( √ )
Similarly, we get solid angle subtended by right at the given point ‘P’ 8 √( ) ( ) 9 84 √( ) ( ) 54 √( ) ( ) 59 { √. / . / } { ( √. / . / ) ( √ . / ) } { } {( )( √ )}
22. ( √ )
Hence on setting the corresponding values, solid angle subtended by the regular pentagonal plane ABCDE at the given point P is given as [ ] [ ( √ ) ( √ ) ( √ )] [ ( √ ) ( √ ) ( √ )]
Note: This is the standard formula to find out the value of solid angle subtended by a regular pentagonal plane, having each side of length at any point lying at a normal height h from any of the vertices.
Regular Hexagonal Plane
By following the same procedure as in case of a regular pentagon, we can divide the hexagon into sub-elementary right triangles.
(As shown in the figure 18)
By using Element Method, solid angle subtended by given regular hexagonal plane ABCDEF at the given point P lying at a normal height h from vertex ‘A’ (i.e. foot of perpendicular) ⇒ ( )
Necessary dimensions can be calculated by the figure as follows √ √
Now, using standard formula-1, solid angle subtended by the right at the point ( ) 8 √( ) ( ) 9 84 √( ) ( ) 54 √( ) ( ) 59 { √. / 4 √ 5 } { ( √. / 4 √ 5 ) ( √ 4 √ 5 ) } 8√ 9 84√ 5( √ )9 4 √ √ 5
Similarly, we get solid angle subtended by right at the given point ( ) 8 √( ) ( ) 9 84 √( ) ( ) 54 √( ) ( ) 59 Fig 18: Point P lying on the normal axis passing through the vertex A of regular hexagon
23. { √. / 4 √ 5 } { ( √. / 4 √ 5 ) ( √ 4 √ 5 ) } { } {( )( √ )} ( √ )
Similarly, we get solid angle subtended by right at the given point ( ) 8 √( ) ( ) 9 84 √( ) ( ) 54 √( ) ( ) 59 { √ ( √ ) } { ( √ ( √ ) ) ( √ ( √ ) ) } { } {( )( √ )} ( √ )
Hence on setting the corresponding values, solid angle subtended by the regular hexagonal plane ABCDEF at the given point P is given as , - 6 4 √ √ 5 ( √ ) ( √ )7 6 ( √ ) 4 √ √ 5 ( √ )7
Note: This is the standard formula to find out the value of solid angle subtended by a regular hexagonal plane, having each side of length at any point lying at a normal height h from any of the vertices.
Thus, all above standard results are obtained by analytical method of HCR’s Theory using single standard formula-1 of right triangular plane only. It is obvious that this theory can be applied to find out the solid angle subtended by any polygonal plane (i.e. plane bounded by the straight lines only) provided the location of foot of perpendicular (F.O.P.) is known.
Now, we are interested to calculate solid angle subtended by different polygonal planes at different points in the space by tracing the diagram, specifying the F.O.P. & measuring the necessary dimensions & calculating.
X. Graphical Applications of Theory of Polygon
Graphical Method:
This method is similar to the analytical method which is applicable for some particular configurations of polygon & locations of given point in the space. But graphical method is applicable for any configuration of polygonal plane & location of the point in the space. This is the method of tracing, measurements & mathematical calculations which requires the following parameters to be already known
24. 1. Geometrical shape & dimensions of the polygonal plane
2. Normal distance (h) of the given point from the plane of polygon
3. Location of foot of perpendicular (F.O.P.) drawn from given point to the plane of polygon
First let’s know the working steps of the graphical method as follows
Step 1: Trace the diagram of the given polygon with the help of known sides & angles.
Step 2: Draw a perpendicular to the plane of polygon & specify the location of F.O.P.
Step 3: Divide the polygon into elementary triangles then each elementary triangle into two sub-elementary right triangles all having common vertex at the F.O.P.
Step 4: Find the area of the polygon as the algebraic sum of areas of sub-elementary right triangles i.e. area of each of the right triangles must be taken with proper sign (positive or negative depending on the area is inside or outside the boundary of polygon)
Step 5: Replace each area of sub-elementary right triangle by the solid angle subtended by that right triangle at the given point in the space.
Step 6: Measure the necessary dimensions (i.e. distances) & set them into standard formula-1 to calculate the solid angle subtended by each of the sub-elementary right triangles.
Step 7: Thus, find out the value of solid angle subtended by given polygonal plane at the given point by taking the algebraic sum of solid angles subtended by the sub-elementary right triangles at the same point in the space.
We are interested to directly apply the above steps without mentioning them in the following numerical examples
Numerical Examples:
Example 1: Let’s find out the value of solid angle subtended by a triangular ABC having sides at a point P lying at a normal height 3cm from the vertex ‘A’.
Sol. Draw the triangle ABC with known values of the sides & specify the location of given point P by ( ) perpendicularly outwards to the plane of paper & F.O.P. (i.e. vertex ‘A’) (as shown in the figure 19 below)
Divide into two right triangles by drawing a perpendicular AN to the opposite side BC (extended line). All must have common vertex at F.O.P.
It is clear from the diagram, the solid angle subtended by at the point ‘P’ is given by Element Method as follows
Now, measure the necessary dimensions & perform the following calculations, by using standard formula-1
( )
Fig 19: Point P is lying perpendicularly outwards to the plane of paper. All the dimensions are in cm.
25. Now, solid angle subtended by right at the given point ‘P’
On setting the corresponding values in formula of right triangle ⇒ 8 √( ) ( ) 9 84 √( ) ( ) 54 √( ) ( ) 59 { √ } {( √ )( √ )}
Similarly, solid angle subtended by right at the given point ‘P’ ⇒ 8 √( ) ( ) 9 84 √( ) ( ) 54 √( ) ( ) 59 { √ } {( √ )( √ )}
Hence, solid angle subtended by at the point ‘P’ (by Element Method) ⇒
Example 2: Let’s find out solid angle subtended by a quadrilateral ABCD having sides at a point lying at a normal height 2cm from the vertex ‘A’ & calculate the total luminous flux intercepted by the plane ABCD if a uniform point-source of 1400 lm is located at the point ‘P’
Sol. Draw the quadrilateral ABCD with known values of the sides & angle & specify the location of given point P by ( ) perpendicularly outwards to the plane of paper & F.O.P. (i.e. vertex ‘A’) (as shown in the fig 20)
Divide the quadrilateral ABCD into two triangles by joining the vertex C to the F.O.P. ‘A’. Further divide into two right triangles and respectively simply by drawing perpendicular to the opposite side in having common vertex at F.O.P.
It is clear from the diagram, the solid angle subtended by quadrilateral ABCD at the point ‘P’ is given by Element Method ( ) ( )
Now, measure the necessary dimensions then perform the following calculations
( )
⇒ Fig 20: Point P is lying perpendicularly outwards to the plane of paper. All the dimensions are in cm.
26. Now, solid angle subtended by right at the given point ‘P’
On setting the corresponding values in formula of right triangle ⇒ 8 √( ) ( ) 9 84 √( ) ( ) 54 √( ) ( ) 59 { √ } {( √ )( √ )}
Similarly, solid angle subtended by right at the given point ‘P’ ⇒ 8 √( ) ( ) 9 84 √( ) ( ) 54 √( ) ( ) 59 { √ } {( √ )( √ )}
Similarly, solid angle subtended by right at the given point ‘P’ ⇒ 8 √( ) ( ) 9 84 √( ) ( ) 54 √( ) ( ) 59 { √ } {( √ )( √ )}
Similarly, solid angle subtended by right at the given point ‘P’ ⇒ 8 √( ) ( ) 9 84 √( ) ( ) 54 √( ) ( ) 59 { √ } {( √ )( √ )}
Hence, solid angle subtended by at the point ‘P’ (by Element Method) ⇒ ( ) ( )
Calculation of Luminous Flux: If a uniform point-source of 1400 lm is located at the given point ‘P’ then the total luminous flux intercepted by the quadrilateral plane ABCD ( )
27. It means that only 142.9628753 lm out of 1400 lm flux is striking the quadrilateral plane ABCD & rest of the flux is escaping to the surrounding space. This result can be experimentally verified. (H.C. Rajpoot)
Example 3: Let’s find out solid angle subtended by a pentagonal plane ABCDE having sides at a point ‘P’ lying at a normal height 6cm from a point ‘O’ internally dividing the side AB such that & calculate the total luminous flux intercepted by the plane ABCDE if a uniform point-source of 1400 lm is located at the point ‘P’
Sol: Draw the pentagon ABCDE with known values of the sides & angles & specify the location of given point P by ( ) perpendicularly outwards to the plane of paper & F.O.P. ‘O’ (as shown in the figure 21)
Divide the pentagon ABCDE into elementary triangles by joining all the vertices of pentagon ABCDE to the F.O.P. ‘O’. Further divide each of the triangles in two right triangles simply by drawing a perpendicular to the opposite side in the respective triangle. (See the diagram)
It is clear from the diagram, the solid angle subtended by pentagonal plane ABCDE at the point ‘P’ is given by Element Method as follows
( )
From the diagram, it’s obvious that the solid angle subtended by the pentagon ABCDE is expressed as the algebraic sum of solid angles of sub-elementary right triangles only as follows
Now, setting the values in eq(I), we get ( ) ( ) ( ) ( ) ( )
Now, measure the necessary dimensions & set them into standard formula-1 to find out above values of solid angle subtended by the sub-elementary right triangles at the given point ‘P’ as follows ⇒ 8 √( ) ( ) 9 84 √( ) ( ) 54 √( ) ( ) 59 8 √( ) ( ) 9 84 √( ) ( ) 54 √( ) ( ) 59 ⇒ 8 √( ) ( ) 9 84 √( ) ( ) 54 √( ) ( ) 59 8 √( ) ( ) 9 84 √( ) ( ) 54 √( ) ( ) 59 Fig 21: Point P is lying perpendicularly outwards to the plane of paper. All the dimensions are in cm.
29. ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )
Calculation of Luminous Flux: If a uniform point-source of 1400 lm is located at the given point ‘P’ then the total luminous flux intercepted by the pentagonal plane ABCDE ( )
It means that only 86.63434241 lm out of 1400 lm flux is striking the pentagonal plane ABCDE & rest of the flux is escaping to the surrounding space. This result can be experimentally verified. (H.C. Rajpoot)
Example 4: Let’s find out solid angle subtended by a quadrilateral plane ABCD having sides at a point ‘P’ lying at a normal height 4cm from a point ‘O’ outside the quadrilateral ABCD such that & calculate the total luminous flux intercepted by the plane ABCD if a uniform point-source of 1400 lm is located at the point ‘P’
Sol: Draw the quadrilateral ABCD with known values of the sides & angle & specify the location of given point P by ( ) perpendicularly outwards to the plane of paper & F.O.P. ‘O’ (See the figure 22)
Divide quadrilateral ABCD into elementary triangles by joining all the vertices of quadrilateral ABCD to the F.O.P. ‘O’. Further divide each of the triangles in two right triangles simply by drawing perpendiculars OE, OG & OF to the opposite sides AB, AD & CD in the respective triangles. (See the diagram)
It is clear from the diagram, the solid angle subtended by quadrilateral plane ABCD at the given point ‘P’ is given by Element Method as follows
( ) ( ) ( )
Now, replacing areas by corresponding values of solid angle, we get ( ) ( ) ( ) ( )
Now, draw a perpendicular OH from F.O.P. to the side BC to divide into right triangles & express the above values of solid angle as the algebraic sum of solid angles subtended by the right triangles only as follows
Now, setting the above values in eq(I), we get Fig 22: Point P is lying perpendicularly outwards to the plane of paper. All the dimensions are in cm.
31. ⇒ 8 √( ) ( ) 9 84 √( ) ( ) 54 √( ) ( ) 59 8 √( ) ( ) 9 84 √( ) ( ) 54 √( ) ( ) 59 ⇒ 8 √( ) ( ) 9 84 √( ) ( ) 54 √( ) ( ) 59 8 √( ) ( ) 9 84 √( ) ( ) 54 √( ) ( ) 59
Hence, by setting the corresponding values in eq(II), solid angle subtended by the pentagonal plane at the given point ‘P’ is calculated as follows
Calculation of Luminous Flux: If a uniform point-source of 1400 lm is located at the given point ‘P’ then the total luminous flux intercepted by the quadrilateral plane ABCD ( )
It means that only 34.56302412 lm out of 1400 lm flux is striking the quadrilateral plane ABCD & rest of the flux is escaping to the surrounding space. This result can be experimentally verified.
Thus, all the mathematical results obtained above can be verified by the experimental results. Although, there had not been any unifying principle to be applied on any polygonal plane for any configuration & location of the point in the space. The symbols & names used above are arbitrary given by the author Mr H.C. Rajpoot.
XI. Conclusion
It is obvious from results obtained above that this theory is a Unifying Principle which is easy to apply for any configuration of a given polygon & any location of a point (i.e. observer) in the space by using a simple & systematic procedure & a standard formula. Necessary dimensions can be measured by analytical method or by tracing the diagram of polygon & specifying the location of F.O.P.
Though, it is a little lengthy for random configuration of polygon & location of observer still it can be applied to find the solid angle subtended by polygon in the easier way as compared to any other methods existing so far in the field of 3-D Geometry. Theory of Polygon can be concluded as follows
Applicability: It is easily applied to find out the solid angle subtended by any polygonal plane (i.e. plane bounded by the straight lines only) at any point (i.e. observer) in the space.
Conditions of Application: This theory is applicable for any polygonal plane & any point in the space if the following parameters are already known
32. 1. Geometrical shape & dimensions of the polygonal plane
2. Normal distance (h) of the given point from the plane of polygon
3. Location of foot of perpendicular (F.O.P.) drawn from given point to the plane of polygon
While, the necessary dimensions (values) used in standard formula-1 are calculated either by analytical method or by graphical method i.e. by tracing the diagram & measuring the dimensions depending on which is easier. Analytical method is limited for some particular location of the point while Graphical method is applicable for all the locations & configuration of polygon w.r.t. the observer in the space. This method can never fail but a little complexity may be there in case of random locations & polygon with higher number of sides.
Steps to be followed:
1. Trace the diagram of the given polygon with the help of known sides & angles.
2. Draw a perpendicular to the plane of polygon & specify the location of F.O.P.
3. Divide the polygon into right triangles having common vertex at the F.O.P. & find the solid angle subtended by the polygon as the algebraic sum of solid angles subtended by right triangles such that algebraic sum of areas of right triangles is equal to the area of polygon.
4. Measure the necessary dimensions & set them into standard formula-1 to calculate solid angle subtended by each of the right triangles & hence solid angle subtended by the polygon at the given point.
Ultimate aim is to find out solid angle, subtended by a polygon at a given point, as the algebraic sum of solid angles subtended by the right triangles, measuring the dimensions, applying standard formula-1 on each of the right triangle & calculating the required result.
Future Scope: This theory can be easily applied for finding out the solid angle subtended by 3-D objects which have surface bounded by the planes only Ex. Cube, Cuboid, Prism, Pyramid, Tetrahedron etc. in 3-D modelling & analysis by tracing the profile of surface of the solid as a polygon in 2-D & specifying the location of a given point & F.O.P. in the plane of profile-polygon as the projection of such solids in 2-D is always a polygon for any configuration of surface of solid in the space.
Acknowledgements:
Author is extremely thankful to all his teachers & professors who inspired & guided him for the research work in the field of Applied Mathematics/Geometry. He also thanks to his parents who have been showering their blessing on him. Finally, he thanks to Almighty God who kept him physically fit during the completion of his research work.