Center of gravity of an Octagon

CENTER OF GRAVITY OF OCTAGON

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 Octagon 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 an Octagon which is an 8 sided regular polygon. Let ‘a’ be the side length and let ‘h’ be the height from the center to the base. This regular Octagon can be divided equally into 8 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 Octagon as depicted in Fig .1

Fig .1 Octagon

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

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

Each angle (θ) of a Octagon 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 135/2 = 67.5°

Consider the triangle as shown in fig .2, 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 POA

Consider ∆POA, ⦤PAO = ϕ = 45°, ⦤POA = 90°

By simple trigonometry we can calculate distance OA as follows

OA = IA*cos45° = a* cos45° = a/√2

Triangle AXC

Consider ∆AXC, C₁ is the center of gravity and the x and y distances are as follows;

X₁ = OA + AB = a/√2 + a/2

Y₁ = h/3

Triangle EXG

Consider ∆EXG, C₃ is the center of gravity and the x and y distances are as follows;

X₃ = OA + AB + 2h/3 = a/√2 + a/2 + 2h/3

Y₃ = h

Triangle IXK

Consider ∆IXK, C₅ is the center of gravity and the x and y distances are as follows;

X₅ = OA + AB = a/√2 + a/2

Y₅ = h + 2h/3

Triangle MXP

Consider ∆MXP, C₇ is the center of gravity and the x and y distances are as follows;

X₇ = h/3

Y₇ = h

Triangle PXA

Consider ∆XB₁C₈ and ∆PXA, ⦤AXB = ϕ/2 = 45/2 = 22.5°

Now ⦤C₈XB₁ = 2*⦤AXB = 45°

Point C₈ represents the center of gravity of ∆PXA, length C₈X = 2h/3 and ⦤ XB₁C₈ = 90°

Now using simple trigonometry we can determine length C₈B₁ and XB₁,

C₈B₁ = C₈X*sin45° = 2h/3√2

XB₁ = C₈X*cos45° = 2h/3√2

Now X₈ = OB – C₈B₁ = a/√2 + a/2 – 2h/3√2

Y₈ = XB – XB₁ = h – 2h/3√2

Triangle CXE

Consider ∆ XB₁C₂, Point C₂ is the center of gravity of ∆CXE

C₂X = C₈X = 2h/3 and ⦤ C₂XB₁ = ⦤ C₈XB₁, hence we can determine the coordinates as;

C₂B₁ = C₂X*sin45° = 2h/3√2

XB₁ = C₈X*cos45° = 2h/3√2

X₂ = OA + AB + C₂B₁ = a/√2 + a/2 + 2h/3√2

Y₂ = XB – XB₁ = h – 2h/3√2

Triangle KXM

Consider ∆C₆XB₃ and ∆KXM, ⦤C₆XK = ϕ/2 = 45/2 = 22.5°

Now ⦤ C₆XB₃ = 2*⦤ C₆XK = 45°

Point C₆ represents the center of gravity of ∆KXM, length C₆X = 2h/3 and ⦤ C₆B₃X = 90°

Now using simple trigonometry we can determine length C₆B₃ and XB₃,

C₆B₃ = C₆X*sin45° = 2h/3√2

XB₃ = C₆X*cos45° = 2h/3√2

Now X₆ = OB – C₆B₃ = a/√2 + a/2 – 2h/3√2

Y₆ = XB + XB₃ = h + 2h/3√2

Triangle GXI

Consider ∆ XB₃C₄, Point C₄ is the center of gravity of ∆GXI

C₄X = C₆X = 2h/3 and ⦤ C₄XB₃ = ⦤ C₆XB₃, hence we can determine the coordinates as;

C₄B₃ = C₄X*sin45° = 2h/3√2

XB₃ = C₄X*cos45° = 2h/3√2

X₄ = OA + AB + C₄B₃ = a/√2 + a/2 + 2h/3√2

Y₄ = XB + XB₃ = h + 2h/3√2

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

In above equations, areas a₁, a₂… represent area of triangles inscribed in the Octagon. 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 Octagon is located at (h, h)

Center of gravity of Heptagon

CENTER OF GRAVITY OF HEPTAGON

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 Heptagon 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 Heptagon which is a 7 sided regular polygon. Let ‘a’ be the side length and let ‘h’ be the height from the center. This regular Heptagon can be divided equally into 7 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 Heptagon as depicted in Fig .1

Fig.1 Heptagon

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

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

Each angle (θ) of a Heptagon 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 128.57/2 = 64.285°

Consider the triangle as shown in fig .2, 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 MOA

Consider ∆MOA, ⦤MAO = ϕ = 51.43° and ⦤MOA = 90°  

By simple trigonometry we can calculate distance OA as follows

OA = MA*cos51.43° = a* 0.623° = 0.623a

Triangle AXC

Consider ∆AXC, C₁ is its center of gravity, the x and y distances are as follows;

X₁ = OA + AB = 0.623a + a/2 = 1.123a

Y₁ = h/3

Triangle MXA

Consider ∆C₇XWand ∆MXA, ⦤AXB = ϕ/2 = 51.43/2 = 25.715°

Now ⦤C₇XW = 2*⦤AXB = 51.43°

Point C₇ represents the center of gravity of ∆MXA, length C₇X = 2h/3 and ⦤ C₇WX = 90°

Now using simple trigonometry we can determine length C₇W and XW,

C₇W = C₇X*sin51.43° = 0.7818(2h/3) = 0.5212h

XW = C₇X*cos51.43° = 0.6234(2h/3) = 0.4156h

Now X₇ = OA + AB – C₇W = 1.123a – 0.5212h

Y₇ = XB – XW = h – 0.4156h

Triangle CXE

Consider ∆C₂XW where Point C₂ is the center of gravity of ∆CXE and ⦤C₂WX = 90°

⦤C₂XW = ⦤C₇XW = 51.43° and C₂X = 2h/3, hence we can determine the coordinates as,

C₂W = C₂X*sin51.43° = 0.7818(2h/3) = 0.5212h

XW = C₇X*cos51.43° = 0.6234(2h/3) = 0.4156h

X₂ = OA + AB + C₂W = 1.123a + 0.5212h

Y₂ = XB – XW = h – 0.4156h

Triangle IXK

Consider ∆C₅PX and ∆IXK, ⦤C₅XP = ϕ/2 = 51.43/2 = 25.715°

Point C₅ represents the center of gravity of ∆IXK, length C₅X = 2h/3 and ⦤C₅PX = 90°

Now using simple trigonometry we can determine length C₅P and XP,

C₅P = C₅X*sin25.715° = 0.4338(2h/3) = 0.289h

XP = C₅X*cos25.715° = 0.9(2h/3) = 0.6h

Now X₅ = OA + AB – C₅P = 1.123a – 0.289h

Y₅ = XB + XP = h + 0.6h = 1.6h

Triangle GXI

Consider ∆C₄PX where Point C₄ is the center of gravity of ∆GXI and ⦤C₄PX = 90°

⦤C₄XP = ⦤C₅XP = 25.715° and C₄X = 2h/3, hence we can determine the coordinates as,

C₄P = C₄X*sin25.715° = 0.4338(2h/3) = 0.289h

XP = C₄X*cos25.715° = 0.9(2h/3) = 0.6h

X₄ = OA + AB + C₅P = 1.123a + 0.289h

Y₄ = XB + XP = h + 0.6h = 1.6h

Triangle KXM

Consider ∆C₆QX and ∆KXM, ⦤C₆XQ = ⦤KXI + ⦤LXK = ϕ + ϕ/2 = 3ϕ/2 = 77.145°

Point C₆ represents the center of gravity of ∆KXM, length C₆X = 2h/3 and ⦤C₆QX = 90°

Now using simple trigonometry we can determine length C₆Q and XQ,

C₆Q = C₆X*sin77.145° = 0.974(2h/3) = 0.649h

XQ = C₆X*cos77.145° = 0.222(2h/3) = 0.148h

Now X₆ = OA + AB – C₆Q = 1.123a – 0.649h

Y₆ = XB + XQ = h + 0.148h = 1.148h

Triangle EXG

Consider ∆C₃QX where Point C₃ is the center of gravity of ∆EXG and ⦤C₃QX = 90°

⦤C₃XQ = ⦤C₆XQ = 77.145° and C₃X = 2h/3, hence we can determine the coordinates as,

C₃Q = C₃X*sin77.145° = 0.974(2h/3) = 0.649h

XQ = C₃X*cos77.145° = 0.222(2h/3) = 0.148h

X₃ = OA + AB + C₃Q = 1.123a + 0.649h

Y₃ = XB + XQ = h + 0.148h = 1.148h

Center of gravity

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

In above equations, areas a₁, a₂… represent area of triangles inscribed in the Heptagon. 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 Heptagon is located at (1.123a, h)