How to Calculate Distance Using Bearings

Calculating distance using bearings is a practical skill often used in navigation, surveying, and even in certain sports like orienteering. To understand this concept, let’s break it down into clear, manageable steps.

Understanding Bearings

Bearings are a way of describing direction. They are usually given in degrees, with North being 0° or 360°, East being 90°, South being 180°, and West being 270°. Bearings help you understand the direction from one point to another.

Tools You Need

To calculate distance using bearings, you’ll typically need:

  1. A Compass or a GPS device to determine bearings.
  2. A Map or Chart to visualize the points.
  3. A Calculator for the final calculations.

Step-by-Step Guide

  1. Identify Your Points
    Let’s say you have two points, Point A and Point B. You need to know the coordinates (latitude and longitude) of these points. For example:

    • Point A: (Lat1, Lon1)
    • Point B: (Lat2, Lon2)

  1. Calculate the Bearing
    The bearing from Point A to Point B can be calculated using the following formula:
    $theta = text{atan2}(sin(Delta text{Lon}) cdot cos(text{Lat2}), cos(text{Lat1}) cdot sin(text{Lat2}) – sin(text{Lat1}) cdot cos(text{Lat2}) cdot cos(Delta text{Lon}))$
    where:

    • $Delta text{Lon}$ is the difference in longitude between Point A and Point B.
    • $theta$ is the bearing angle in radians.

  1. Convert Bearing to Degrees
    Bearings are typically given in degrees, so you’ll need to convert the angle from radians to degrees:
    $text{Bearing} = theta times frac{180}{pi}$

  1. Calculate the Distance
    To calculate the distance between two points on the Earth’s surface, you can use the Haversine formula:
    $a = sin^2left(frac{Delta text{Lat}}{2}right) + cos(text{Lat1}) cdot cos(text{Lat2}) cdot sin^2left(frac{Delta text{Lon}}{2}right)$
    $c = 2 cdot text{atan2}(sqrt{a}, sqrt{1-a})$
    $d = R cdot c$
    where:

    • $Delta text{Lat}$ and $Delta text{Lon}$ are the differences in latitude and longitude, respectively.
    • $R$ is the Earth’s radius (mean radius = 6,371 km).
    • $d$ is the distance between the two points.

Example Calculation

Let’s go through an example. Suppose Point A is at (40.748817, -73.985428) and Point B is at (34.052235, -118.243683).

  1. Identify Your Points
    • Point A: (40.748817, -73.985428)
    • Point B: (34.052235, -118.243683)

  1. Calculate the Bearing
    $Delta text{Lon} = -118.243683 – (-73.985428) = -44.258255$
    $theta = text{atan2}(sin(-44.258255) cdot cos(34.052235), cos(40.748817) cdot sin(34.052235) – sin(40.748817) cdot cos(34.052235) cdot cos(-44.258255))$

  1. Convert Bearing to Degrees
    Convert $theta$ from radians to degrees.

  1. Calculate the Distance
    Using the Haversine formula:
    $Delta text{Lat} = 34.052235 – 40.748817 = -6.696582$
    $a = sin^2left(frac{-6.696582}{2}right) + cos(40.748817) cdot cos(34.052235) cdot sin^2left(frac{-44.258255}{2}right)$
    $c = 2 cdot text{atan2}(sqrt{a}, sqrt{1-a})$
    $d = 6,371 cdot c$

Conclusion

Calculating distance using bearings involves understanding both the direction and the actual distance between two points on the Earth’s surface. By using the formulas and steps outlined, you can accurately determine these values, which is invaluable in navigation and other practical applications.

1. Wikipedia – Bearing (navigation)2. NOAA – Haversine Formula

Citations

  1. 3. Math is Fun – Bearings

Related

(2) O3 + H → O2 + OH k2 = 1.78×10^-11 cm^3 s^-1 (3) O + OH → O2 + H k3 = 4.40×10^-11 cm^3 s^-1 (5) O + HO2 → O2 + OH k5 = 3.50×10^-11 cm^3 s^-1 (6) H + HO2 → O2 + H2 k6 = 5.40×10^-12 cm^3 s^-1 (9) OH + HO2 → O2 + H2O2 k9 = 4.00×10^-11 cm^3 s^-1 (10) HO2 + HO2 → O2 + H2O2 k10 = 2.50×10^-12 cm s^-1 (11) O + O2 + M → O3 + M k11 = 1.05×10^-34 cm^6 s^-1 (14) H + O2 + M → HO2 + M k14 = 8.08×10^-32 cm^6 s^-1 (15) H + H + M → H2O + M k15 = 3.31×10^-27 cm^6 s^-1 (16) O2 + hv → 2 O k16 = (1.26×10^-8 s^-1) φ (17) H2O + hv → H + OH k17 = (3.4×10^-6 s^-1) φ (18) O3 + hv → O2 + O k18 = (7.10×10^-5 s^-1) φ

Table 1 Reactions, rate constants and activation energies used in the model* No. Reaction kopt (M⁻¹ s⁻¹) 1 OH + H₂ → H + H₂O 3.74 x 10⁷ 2 OH + HO₂ → HO₂ + OH⁻ 5 x 10⁹ 3 OH + H₂O₂ → HO₂ + H₂O 3.8 x 10⁷ 4 OH + O₂ → O₂ + OH 9.96 x 10⁹ 5 OH + HO₂ → O₂ + H₂O 7.1 x 10⁹ 6 OH + OH → H₂O₂ 5.3 x 10⁹ 7 OH + e⁻aq → OH⁻ 3 x 10¹⁰ 8 H + O₂ → HO₂ 2.0 x 10¹⁰ 9 H + HO₂ → H₂O₂ 2.0 x 10¹⁰ 10 H + H₂O₂ → OH + H₂O 3.44 x 10⁷ 11 H + OH → H₂O 1.4 x 10¹⁰ 12 H + H → H₂ 1.94 x 10¹⁰ 13 e⁻aq + O₂ → O₂⁻ 1.9 x 10¹⁰ 14 e⁻aq + O₂ → HO₂⁻ + OH⁻ 1.3 x 10¹⁰ 15 e⁻aq + HO₂ 2.0 x 10¹⁰ 16 e⁻aq + H₂O₂ 1.1 x 10¹⁰ 17 e⁻aq + HO₂ → OH + OH⁻ 1.3 x 10¹⁰ 18 e⁻aq + H⁺ → H 2.3 x 10¹⁰ 19 e⁻aq + e⁻aq → H₂ + OH⁻ + OH⁻ 2.5 x 10⁹ 20 HO₂ + O₂ → O₂ + HO₂ 1.3 x 10⁹ 21 HO₂ + HO₂ → O₂ + H₂O₂ 8.3 x 10⁵ 22 HO₂ + HO₂ → O₂ + OH + H₂O 3.7 23 HO₂ + HO₂ → O₂ + O₂ + OH + H₂O 7 x 10⁵ s⁻¹ 24 H⁺ + O₂⁻ → HO₂ 4.5 x 10¹⁰ 25 H⁺ + O₂⁻ → O₂ 2.0 x 10¹⁰ 26 H⁺ + OH⁻ 1.4 x 10¹¹ 27 H⁺ + HO₂⁻ 2 x 10¹⁰ 28 H₂O₂ → HO₂ + H⁺ + OH⁻ 2.5 x 10⁻⁵ s⁻¹ 29 H₂O₂ → H⁺ + OH⁻ 1.4 x 10⁻⁷ s⁻¹ 30 O₂ + O₂ → O₂ + HO₂ + OH⁻ 0.3 31 O₂ + H₂O₂ → O₂ + OH + OH 16 32

(2) O3 + H → O2 + OH k2 = 1.78×10^-11 cm^3 s^-1 (3) O + OH → O2 + H k3 = 4.40×10^-11 cm^3 s^-1 (5) O + HO2 → O2 + OH k5 = 3.50×10^-11 cm^3 s^-1 (6) H2O + O → 2 OH k6 = 5.40×10^-12 cm^3 s^-1 (9) OH + HO2 → O2 + H2O k9 = 4.00×10^-11 cm^3 s^-1 (10) HO2 + HO2 → O2 + H2O2 k10 = 2.50×10^-12 cm s^-1 (11) O + O2 + M → O3 + M k11 = 1.05×10^-34 cm^6 s^-1 (14) H + O2 + M → HO2 + M k14 = 8.08×10^-32 cm^6 s^-1 (15) OH + H + M → H2O + M k15 = 3.31×10^-27 cm^6 s^-1 (16) O2 + hv → 2 O k16 = (1.26×10^-8 s^-1) φ (17) H2O + hv → H + OH k17 = (3.4×10^-6 s^-1) φ (18) O3 + hv → O2 + O k18 = (7.10×10^-8 s^-1) φ