Center of gravity of Hexagon

CENTER OF GRAVITY OF HEXAGON

INTRODUCTION

The center of gravity COG of an object is the point of action of the gravitational force. It is also known as the balancing point since all objects [simple or complex] have their COG within the object’s sphere of influence and this ensures stability. There are two values to be noted which is the geometrical center of gravity and the actual center of gravity. The geometrical COG is the exact center of the object. It can be calculated by various available methods like summation, moment of inertia, etc. The geometrical COG is valid as long as there is uniform distribution of mass and or uniform gravitational field. In case of uneven mass distribution or use of composite or heterogeneous materials, the actual COG will no longer coincide with the geometrical COG. This is because mass is distributed unevenly and the COG will shift where there is more mass. In this article we intend to determine the COG of Hexagon and we assume uniform mass distribution for simplicity.

ASSUMPTIONS

1. Mass of the object is evenly distributed

2. The gravitational field is uniform

CALCULATION

Consider a Hexagon which is a 6 sided regular polygon. Let ‘a’ be the side length and let ‘h’ be the height from the center. A regular hexagon can be divided equally into 6 equilateral triangles. This is a very special case and is only possible on a hexagon due to its internal angle. All other regular polygons can be divided only in to equal isosceles triangles. The equilateral triangle inside the hexagon has all three equal sides ‘a’. Two sides of the triangle other than the base bisect the angle between two sides of the Hexagon as depicted in Fig .1

Fig .1 Hexagon

The angle between each side of the Hexagon can be determined by using the angle formula,

The sum of all internal angles of a Hexagon is given by,

Each angle (θ) of a Hexagon can be obtained by dividing the above answer by the number of sides

Now the angle bisector (θ/2) which also represents the angle of triangle opposite to side ‘a’ is 120/2 = 60°

Consider the triangle as shown in fig .1, we can determine the third angle ϕ by using the equation,

We will determine the center of gravity using the area moment of inertia method. We need to follow three steps:

1    1. Determine the distance between the x coordinate of C.O.G. of triangle and the reference Y axis.

2    2. Determine the distance between the y coordinate of C.O.G. of triangle and the reference X axis.

3   3. Repeat the above two steps for the other triangles. Use the Summation equation to determine the C.O.G. coordinates. There is no need to determine the area of triangles since they will cancel out as it will be evident from the calculations.

The real advantage with equilateral triangles inside Hexagon is that all medians inside equilateral triangle bisect equal angles. This implies that we can just drop a perpendicular from the center of gravity of the triangle directly to the base and measure the distance between the center and the x or y axis.

Triangle KOA

Consider ∆KOA from fig .1, by simple trigonometry we can calculate distance OA as follows

OA = KA*cos60° = a* cos60° = a/2

Triangle AXC

Consider ∆AXC, Point A₁ is the center of gravity of the triangle, x and y distances from the center of triangle are as follows;

X₁ = OA + AB = a/2 + a/2 = a

Y₁ = h/3

Triangle CXE

Consider ∆CXE, Point B₁ is the center of gravity of the triangle, x and y distances from the center of triangle are as follows;

X₂ = OA + AC = a/2 + a = (3/2)*a

Y₂ = 2h/3

Triangle GXE

Consider ∆GXE, Point C₁ is the center of gravity of the triangle, x and y distances from the center of triangle are as follows;

X₃ = OA + AC = a/2 + a = (3/2)*a

Y₃ = h + h/3 = 4h/3

Triangle GXI

Consider ∆GXI, Point D₁ is the center of gravity of the triangle, x and y distances from the center of triangle are as follows;

X₄ = OA + AB = a/2 + a/2

Y₄ = h + 2h/3 = 5h/3

Triangle IXK

Consider ∆IXK, Point E₁ is the center of gravity of the triangle, x and y distances from the center of triangle are as follows;

X₅ = OA/2 = a/2

Y₅ = h + h/3 = 4h/3

Triangle KXA

Consider ∆KXA, Point F₁ is the center of gravity of the triangle, x and y distances from the center of triangle are as follows;

X₆ = OA/2 = a/2

Y₆ = 2h/3

Center of gravity

The center of gravity coordinates x and y of Hexagon can be computed by using the formula,

In above equations, areas a₁, a₂… represent area of triangles inscribed in the Hexagon. Since all triangles are equal to each other, they have equal areas. Thus the area term can be eliminated from the above equations.

CONCLUSION

Thus the center of gravity of a regular Hexagon is located at (a, h)

Center of gravity of Pentagon

CENTER OF GRAVITY OF PENTAGON

INTRODUCTION

The center of gravity COG of an object is the point of action of the gravitational force. It is also known as the balancing point since all objects [simple or complex] have their COG within the object’s sphere of influence and this ensures stability. There are two values to be noted which is the geometrical center of gravity and the actual center of gravity. The geometrical COG is the exact center of the object. It can be calculated by various available methods like summation, moment of inertia, etc. The geometrical COG is valid as long as there is uniform distribution of mass and or uniform gravitational field. In case of uneven mass distribution or use of composite or heterogeneous materials, the actual COG will no longer coincide with the geometrical COG. This is because mass is distributed unevenly and the COG will shift where there is more mass. In this article we intend to determine the COG of Pentagon and we assume uniform mass distribution for simplicity.

ASSUMPTIONS

1. Mass of the object is evenly distributed

2. The gravitational field is uniform

CALCULATION

Consider a Pentagon which is a 5 sided regular polygon. Let ‘a’ be the side length and let ‘h’ be the height from the center. This regular pentagon can be divided equally into 5 isosceles triangles. Each triangle has base of length ‘a’ and two other sides of length ‘b’. The side of length ‘b’ bisects the angle between two sides of the Pentagon as depicted in Fig .1

Fig .1 Pentagon

The angle between each side of the pentagon can be determined by using the angle formula,

The sum of all internal angles of a pentagon is given by,

Each angle (θ) of a Pentagon can be obtained by dividing the above answer by the number of sides

Now the angle bisector (θ/2) which also represents the angle of triangle opposite to side ‘b’ is 108/2 = 54°

Consider the triangle as shown in fig .1, we can determine the third angle ϕ by using the equation,

We will determine the center of gravity using the area moment of inertia method. We need to follow three steps:

1    1.  Determine the distance between the x coordinate of C.O.G. of triangle and the reference Y axis.

2    2. Determine the distance between the y coordinate of C.O.G. of triangle and the reference X axis.

3   3. Repeat the above two steps for the other triangles. Use the Summation equation to determine the C.O.G. coordinates. There is no need to determine the area of triangles since they will cancel out as it will be evident from the calculations.

Triangle IOA

Consider ∆IOA from fig .1, by simple trigonometry we can calculate distance OA as follows

OA = IA*cos72° = a* cos72° = 0.309a

Triangle AXC

Consider ∆AXC, the x and y distances are as follows;

X₁ = OA + AB = 0.309a + a/2 = 0.809a

Y₁ = h/3

Triangle AXI

Consider ∆A₁XS and ∆AXC, ⦤AXS = ϕ/2 = 72/2 = 36°

Now ⦤A₁XS = 2*⦤AXS = 72°

Point A₁ represents the center of gravity of ∆AXI, length A₁X = 2h/3 and ⦤XSA₁ = 90°

Now using simple trigonometry we can determine length A₁S and XS,

A₁S = A₁X*sin72° = 0.951(2h/3) = 0.634h

XS = A₁X*cos72° = 0.309(2h/3) = 0.206h

Now X₅ = OA + AB – A₁S = 0.809a – 0.951(2h/3) = 0.809a – 0.634h

Y₅ = XB – XS = h – 0.206h = 0.794h

Triangle CXE

Consider ∆C₁XS, Point C₁ is the center of gravity of ∆CXE

C₁X = A₁X = 2h/3 and ⦤ C₁XS = ⦤A₁XS, hence we can determine the coordinates as;

X₂ = OA + AB + C₁S = 0.809a + 0.634h

Y₂ = XB – XS = 0.794h

Triangle GXI

Consider ∆E₁GX where Point E₁ is the center of gravity of ∆GXI and ⦤ E₁GX = 90°

⦤ E₁XG = ϕ/2 = 72/2 = 36° and E₁X = 2h/3

Now using simple trigonometry we can determine length E₁G and XG,

E₁G = E₁X*sin36° = (2h/3)*0.587 = 0.391h

XG = E₁X*cos36° = (2h/3)*0.809 = 0.539h

X₄ = OA + AB - E₁G = 0.809a - 0.391h

Y₄ = XB + XG = h + 0.539h = 1.539h

Triangle GXE

Consider ∆D₁GX where Point D₁ is the center of gravity of ∆GXE and ⦤ D₁GX = 90°

⦤ E₁XG = ⦤D₁XG = 36° and D₁X = 2h/3, hence we can determine the coordinates as,

X₃ = OA + AB + D₁G = 0.809a + 0.391h

Y₃ = XB + XG = h + 0.539h = 1.539h

Center of gravity

The center of gravity coordinates x and y of Pentagon can be computed by using the formula,

In above equations, areas a₁, a₂… represent area of triangles inscribed in the pentagon. Since all triangles are equal to each other, they have equal areas. Thus the area term can be eliminated from the above equations.

CONCLUSION

Thus the center of gravity of a regular pentagon is located at (0.809a, h)