📚 5D3 The Earth: Spherical Distance and Navigation | 5D3 地球：球面距离与导航
In A Level Further Mathematics, the topic ‘The Earth’ applies spherical geometry to model our planet for navigation and distance calculations. You will learn how to use latitude and longitude, compute distances along meridians and parallels, find great-circle distances, and understand nautical miles and time zones. This article covers all the key ideas, formulas, and worked examples you need for Paper 5D3.
在A Level进阶数学中，“地球”这一考点运用球面几何为地球建模，解决导航和距离计算问题。你将学习如何利用纬度和经度，计算沿子午线和纬线的距离，求解大圆距离，并理解海里和时区。本文涵盖5D3所需的全部核心概念、公式和详细例题。
1. Spherical Earth and Key Terminology | 球体地球与关键术语
For most navigation problems, we treat the Earth as a perfect sphere with a mean radius R. The commonly used value is R ≈ 6370 km (or 3960 miles). A great circle is any circle on the sphere whose centre coincides with the Earth’s centre; the Equator is a great circle. A small circle is a circle whose plane does not pass through the centre, e.g. all parallels of latitude except the Equator. The North and South Poles are the endpoints of the axis of rotation.
在大多数导航问题中，我们把地球视为一个完美的球体，平均半径 R 常用 R ≈ 6370 km（或3960英里）。大圆是球面上以球心为圆心的圆；赤道就是一个大圆。小圆是平面不经过球心的圆，例如除赤道外的所有纬线圈。北极和南极是自转轴的两个端点。
2. Latitude and Longitude | 纬度和经度
Latitude φ (phi) is the angle measured north or south of the Equator, ranging from 0° at the Equator to 90° at the poles. North is taken as positive, south as negative. Longitude λ (lambda) is the angle measured east or west of the Prime Meridian (0°), ranging from 0° to 180° east or west. By convention, eastern longitudes are positive and western longitudes are negative.
纬度 φ 是从赤道向北或向南测量的角度，赤道为0°，极点为90°。北纬取正值，南纬取负值。经度 λ 是从本初子午线（0°）向东或向西测量的角度，范围0°到180°。习惯上东经为正，西经为负。
3. Angular Distance and Arc Length | 角距离与弧长
On any circle of radius R, the arc length s corresponding to a central angle θ measured in radians is: s = R θ. Angles given in degrees must first be converted: θ (rad) = θ° × π/180. This fundamental relationship is the basis of all distance calculations on the Earth’s surface.
在任何半径为 R 的圆上，圆心角 θ（弧度制）所对的弧长 s 为：s = R θ。以度为单位的角度需先转换：θ (弧度) = θ° × π/180。这一基本关系是所有地球表面距离计算的基础。
4. Great Circles and Small Circles | 大圆与小圆
A great circle gives the shortest path between two points on a sphere. All meridians (lines of longitude) are halves of great circles. The Equator is a great circle. Any other parallel of latitude is a small circle, with radius r = R cos φ, where φ is the latitude. Only travel along a great circle or a meridian yields the shortest distance; along a parallel of latitude the path is not the shortest unless it is the Equator.
大圆是球面上两点之间的最短路径。所有经线（子午线）都是大圆的一半。赤道是一个大圆。其他任何纬线圈都是小圆，其半径 r = R cos φ，其中 φ 为纬度。只有沿大圆或经线的路径才是最短距离；沿纬线（除赤道外）行走并不是最短路径。
5. Distance Along a Meridian | 沿经线的距离
For two points on the same meridian (same longitude), the angular difference is simply Δφ = φ₂ − φ₁ (in degrees). Converting to radians and multiplying by R gives the distance: d = R × (|Δφ| × π/180). For example, moving from 30°N to 45°N along a meridian gives Δφ = 15°, so d ≈ 6370 × (15π/180) ≈ 6370 × 0.2618 ≈ 1668 km.
对于同一经线上两点，角度差仅为 Δφ = φ₂ − φ₁（度）。化为弧度后乘以 R 即得距离：d = R × (|Δφ| × π/180)。例如，沿经线从30°N到45°N，Δφ = 15°，因此 d ≈ 6370 × (15π/180) ≈ 6370 × 0.2618 ≈ 1668 km。
6. Distance Along a Parallel of Latitude | 沿纬线的距离
Along a parallel of latitude φ, the circle radius is r = R cos φ. The angular difference in longitude is Δλ (degrees). The distance along this small circle is d = r × (|Δλ| × π/180) = R cos φ × (|Δλ| × π/180). Note: this path is not the shortest route between the two points unless both lie on the Equator. For example, at latitude 60°N, cos 60° = 0.5, so a 10° longitude difference gives d ≈ 6370 × 0.5 × (10π/180) ≈ 556 km.
在纬度为 φ 的纬线上，小圆半径为 r = R cos φ。经度差为 Δλ（度）。沿此小圆的距离为 d = r × (|Δλ| × π/180) = R cos φ × (|Δλ| × π/180)。注意：这条路径并非两点间的最短路线（除非两点都在赤道上）。例如，在纬度60°N，cos 60° = 0.5，经度差10°，则 d ≈ 6370 × 0.5 × (10π/180) ≈ 556 km。
7. Great Circle Distance Between Two Points | 两点间的大圆距离
The shortest distance between two points A (φ₁, λ₁) and B (φ₂, λ₂) on the Earth’s surface lies along the great circle passing through them. The central angle c (in radians) is found using the spherical cosine rule:
地球表面两点 A (φ₁, λ₁) 和 B (φ₂, λ₂) 之间的最短距离在通过这两点的大圆上。球心角 c（弧度）由球面余弦定理求得：
cos c = sin φ₁ sin φ₂ + cos φ₁ cos φ₂ cos(λ₂ − λ₁)
Here latitudes φ₁, φ₂ and the longitude difference Δλ = λ₂ − λ₁ must all be in degrees or all in radians when evaluating the trigonometric functions. Once c is obtained (by taking arccos), the great-circle distance is d = R c. This formula works for any two points on the sphere.
这里纬度 φ₁、φ₂ 和经度差 Δλ = λ₂ − λ₁ 在计算三角函数时可同为度制或同为弧度制。求得 c（通过反余弦）后，大圆距离为 d = R c。该公式适用于球面上任意两点。
Worked Example: Find the great-circle distance between London (51.5°N, 0°) and New York (40.7°N, 74°W). Use R = 6370 km.
例题：求伦敦 (51.5°N, 0°) 与纽约 (40.7°N, 74°W) 之间的大圆距离，取 R = 6370 km。
- London: φ₁ = 51.5°, λ₁ = 0°
- New York: φ₂ = 40.7°, λ₂ = −74° (west is negative)
- Δλ = |0 − (−74)| = 74°
- Compute: sin 51.5° ≈ 0.7826, sin 40.7° ≈ 0.6521, cos 51.5° ≈ 0.6225, cos 40.7° ≈ 0.7581, cos 74° ≈ 0.2756
- cos c ≈ (0.7826 × 0.6521) + (0.6225 × 0.7581 × 0.2756) ≈ 0.5102 + 0.1300 = 0.6402
- c ≈ arccos(0.6402) ≈ 50.18° = 50.18 × π/180 ≈ 0.8758 rad
- Distance d = 6370 × 0.8758 ≈ 5579 km
Thus the great-circle distance is approximately 5580 km, close to the actual air distance. This method gives the minimum surface distance between the two cities.
因此大圆距离约为 5580 km，与实际航线距离接近。该方法给出了两城市之间的最短地表距离。
8. Nautical Miles and Knots | 海里与节
A nautical mile (nmi) is defined as the length of one minute of arc along a meridian. Since 1° = 60 minutes, and 1° of arc along a meridian corresponds to a distance of (π/180)× R, one minute of arc equals (π/10800)× R. Using R = 6370 km = 6,370,000 m, we get 1 nmi ≈ 1852 m. The knot (kn) is a unit of speed equal to one nautical mile per hour.
海里 (nmi) 定义为沿经线上1弧分对应的长度。1° = 60弧分，沿经线1°弧长对应 (π/180)× R，因此1弧分 = (π/10800)× R。取地球半径 R = 6370 km = 6,370,000 m，计算得 1 nmi ≈ 1852 m。节 (kn) 是速度单位，等于每小时1海里。
9. Time Zones and Longitude | 时区与经度
The Earth rotates 360° in 24 hours, so in 1 hour it rotates 15° of longitude. The world is divided into 24 time zones, each roughly 15° wide. The Prime Meridian (0°) defines Greenwich Mean Time (GMT). Moving eastwards, time advances by 1 hour for every 15°; moving westwards, time goes back by 1 hour. When solving problems, you can relate the difference in local time (in hours) to the difference in longitude using: Δλ = 15° × (time difference in hours).
地球每隔24小时自转360°，因此每小时转过 15° 经度。全球分为24个时区，每个大致宽15°。本初子午线（0°）定义的格林威治标准时间（GMT）为基准。向东每15°时间增加1小时；向西每15°时间减少1小时。解题时，可用关系式 Δλ = 15° × (时差，小时) 将地方时差与经度差联系起来。
10. Combining Latitude and Time Zone Problems | 经纬度与时区综合应用
Examination questions often ask you to calculate both the distance and the time difference between two cities. For example, if Town X is at (φ, λ₁) and Town Y is at (φ, λ₂) on the same parallel, first compute the distance along the parallel as shown in Section 6. Then, using the longitude difference Δλ, find the time difference: time difference (hours) = |Δλ| / 15. If additional route details are given along a meridian and a parallel, you can work out the total journey distance by summing the respective arc lengths.
考题常要求同时计算两城市间的距离与时差。例如，城镇 X 位于 (φ, λ₁)，城镇 Y 位于 (φ, λ₂) 同纬度，首先按第6节计算沿纬线距离。再利用经度差 Δλ 求时差：时差（小时）= |Δλ| / 15。如果还涉及沿经线和纬线的组合航线，你可将各段弧长相加得到总路程。
11. Common Mistakes and Tips | 常见错误与提示
- Angle units: Always ensure trigonometric functions are evaluated in the same unit (degrees or radians). Convert Δφ and Δλ to radians before multiplying by R for distance. Use π/180 to convert.
- 单位混淆：三角函数计算时确保角度单位一致。计算距离前务必将 Δφ 和 Δλ 转换为弧度。用 π/180 进行转换。
- Sign of latitude and longitude: Use positive values for north and east, negative for south and west. The great-circle cosine rule only needs the absolute longitude difference, but be careful when finding Δλ manually.
- 经纬符号：北纬和东经取正值，南纬和西经取负值。大圆余弦定理仅需经度差的绝对值，但手动计算 Δλ 时要小心。
- Small circle radius: The radius of a parallel is R cos φ, not R. This is often forgotten, leading to overestimating the distance along a latitude line.
- 小圆半径：纬线圈半径是 R cos φ，而非 R。这一点常被忽略，导致高估纬线距离。
- Nautical miles: Remember that 1 nautical mile corresponds to 1 minute of latitude along a meridian. This provides a quick conversion when given distances in minutes or degrees.
- 海里换算：记住1海里对应于沿经线纬度1弧分的长度。当题目给出台里或度分单位的距离时，可快速换算。
12. Summary and Key Formulas | 总结与关键公式
Mastering ‘The Earth’ topic means you can confidently convert between angular differences and distances on a sphere. The core formulas to memorise are:
掌握“地球”这一考点，意味着你能自信地在球面上进行角差与距离的转换。需要记住的核心公式有：
Meridian distance: d = R × |Δφ| × π/180
Parallel distance: d = R cos φ × |Δλ| × π/180
Great-circle distance: cos c = sin φ₁ sin φ₂ + cos φ₁ cos φ
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