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Arc and Chord Relationship

Every circle arc has a close relationship with the chord that connects its two endpoints. A chord is a straight line connecting two endpoints of an arc, while an arc is a curved path along the circle's circumference. Imagine it like a bow and arrow, where the string is the straight line and the bow is the curved wood.

This relationship is very important in various engineering and architectural applications. The longer the arc, the longer the chord that connects it, but this relationship is not linear.

Arc and Chord Relationship
Comparison of arc with chord at various central angles.

Chord Length Formula

The chord length can be calculated using a trigonometric formula involving the central angle and circle radius:

c=2rsin(θ2)c = 2r \sin\left(\frac{\theta}{2}\right)

Where:

  • cc = chord length
  • rr = circle radius
  • θ\theta = central angle in radians

This formula is very useful in engineering calculations, especially in curved structure design and material strength analysis.

Arc Height and Sagitta

Arc height or sagitta is the perpendicular distance from the chord midpoint to the highest point of the arc. This concept is very important in arch bridge design and architectural structures.

h=rrcos(θ2)=r(1cos(θ2))h = r - r\cos\left(\frac{\theta}{2}\right) = r\left(1 - \cos\left(\frac{\theta}{2}\right)\right)

Where:

  • hh = arc height (sagitta)
  • rr = circle radius
  • θ\theta = central angle in radians

Observe the following visualization:

Arc Height Visualization
Arc height (sagitta) at various central angles.

Arc in Coordinate System

In the Cartesian coordinate system, an arc can be represented using parametric equations:

x=rcos(t)x = r\cos(t)
y=rsin(t)y = r\sin(t)

Where tt is a parameter that varies from the initial angle to the final angle of the arc.

Arc in Cartesian Coordinates
Arc representation using parametric equations.

Engineering Calculation Example

Let's apply this concept in engineering calculations. Suppose we design an arch bridge with a radius of 25 meters and a central angle of 120°.

Calculating chord length:

c=2rsin(θ2)c = 2r \sin\left(\frac{\theta}{2}\right)
c=2×25×sin(120°2)c = 2 \times 25 \times \sin\left(\frac{120°}{2}\right)
c=50×sin(60°)c = 50 \times \sin(60°)
c=50×32=253 metersc = 50 \times \frac{\sqrt{3}}{2} = 25\sqrt{3} \text{ meters}

Calculating arc height:

h=r(1cos(θ2))h = r\left(1 - \cos\left(\frac{\theta}{2}\right)\right)
h=25(1cos(60°))h = 25\left(1 - \cos(60°)\right)
h=25(112)=12.5 metersh = 25\left(1 - \frac{1}{2}\right) = 12.5 \text{ meters}

Exercises

  1. An arch bridge has a radius of 30 meters and a central angle of 90°. Calculate the chord length and arc height of the bridge.

  2. In a mosque dome design, the arc height is 8 meters and the circle radius is 15 meters. Determine the central angle of the arc.

  3. A circle arc has a chord length of 24 meters and a radius of 15 meters. Calculate the central angle and arc height.

  4. In a coordinate system, an arc starts from point (4, 0) and ends at point (0, 4) on a circle centered at the origin. Determine the parametric equations of the arc.

Answer Key

  1. Solution:

    Given: r=30 mr = 30 \text{ m} and θ=90°\theta = 90°

    Chord length:

    c=2rsin(θ2)=2×30×sin(45°)c = 2r \sin\left(\frac{\theta}{2}\right) = 2 \times 30 \times \sin(45°)
    c=60×22=302 m42.43 mc = 60 \times \frac{\sqrt{2}}{2} = 30\sqrt{2} \text{ m} \approx 42.43 \text{ m}

    Arc height:

    h=r(1cos(θ2))=30(1cos(45°))h = r\left(1 - \cos\left(\frac{\theta}{2}\right)\right) = 30\left(1 - \cos(45°)\right)
    h=30(122)8.79 mh = 30\left(1 - \frac{\sqrt{2}}{2}\right) \approx 8.79 \text{ m}
  2. Solution:

    Given: h=8 mh = 8 \text{ m} and r=15 mr = 15 \text{ m}

    Step 1: Use the arc height formula

    8=15(1cos(θ2))8 = 15\left(1 - \cos\left(\frac{\theta}{2}\right)\right)

    Step 2: Isolate cos

    815=1cos(θ2)\frac{8}{15} = 1 - \cos\left(\frac{\theta}{2}\right)
    cos(θ2)=1815=715\cos\left(\frac{\theta}{2}\right) = 1 - \frac{8}{15} = \frac{7}{15}

    Step 3: Calculate angle

    θ2=arccos(715)62.18°\frac{\theta}{2} = \arccos\left(\frac{7}{15}\right) \approx 62.18°
    θ124.36°\theta \approx 124.36°
  3. Solution:

    Given: c=24 mc = 24 \text{ m} and r=15 mr = 15 \text{ m}

    Step 1: Use the chord formula

    24=2×15×sin(θ2)24 = 2 \times 15 \times \sin\left(\frac{\theta}{2}\right)
    sin(θ2)=2430=0.8\sin\left(\frac{\theta}{2}\right) = \frac{24}{30} = 0.8

    Step 2: Calculate central angle

    θ2=arcsin(0.8)53.13°\frac{\theta}{2} = \arcsin(0.8) \approx 53.13°
    θ106.26°\theta \approx 106.26°

    Step 3: Calculate arc height

    h=15(1cos(53.13°))=15(10.6)=6 mh = 15\left(1 - \cos(53.13°)\right) = 15(1 - 0.6) = 6 \text{ m}
  4. Solution:

    • Starting point: (4, 0) → t1=0°t_1 = 0°
    • Ending point: (0, 4) → t2=90°t_2 = 90°
    • Radius: r=4r = 4

    Parametric equations:

    x=4cos(t)x = 4\cos(t)
    y=4sin(t)y = 4\sin(t)
    0°t90°0° \leq t \leq 90°