📚 Bearings and Scale Drawings | 方位角与比例绘图
Bearings and scale drawings are practical tools in mathematics that connect classroom geometry with real-world navigation, architecture and map reading. This revision guide explains how to measure bearings, use map scales, convert between drawing lengths and actual distances, apply back bearings and triangulation, and avoid common errors in KS3 Cambridge-style questions.
方位角和比例绘图是数学中非常实用的工具,把课堂几何与真实世界的导航、建筑和地图阅读联系起来。本复习指南将讲解如何测量方位角、使用地图比例尺、在图上长度与实际距离之间进行换算、运用后方位角与三角定位,并避开剑桥 KS3 风格题目中的常见错误。
1. What is a Bearing? | 什么是方位角?
A bearing is a direction measured from north. In mathematics we always measure bearings clockwise from the north line, and we write them as three-figure angles such as 045° or 125°. A bearing tells you which way to turn and how far round to turn, which is why it is used constantly in navigation, aviation, surveying and map reading.
方位角是从正北开始测量的方向。数学中我们总是从正北线沿顺时针方向测量方位角,并写成三位数的角度,例如 045° 或 125°。方位角告诉你朝哪个方向转、转多少度,因此在导航、航空、测量和地图阅读中被频繁使用。
There are three key rules: bearings are measured from north, measured clockwise, and must be written with three digits. If the angle is less than 100°, add leading zeros.
有三个关键规则:从正北测量、顺时针测量、必须写成三位数。如果角度小于 100°,需要在前面补零。
A bearing is always described as the bearing of B from A, where the point you start at comes second. The bearing of A from B is usually different from the bearing of B from A, and confusing the two is one of the most frequent mistakes in exam papers.
方位角总以 B 相对于 A 的方位角这种形式表述,起始点放在后面。A 相对于 B 的方位角通常与 B 相对于 A 的方位角不同,把这两者搞混是考卷中最常见的错误之一。
2. Measuring Bearings with a Protractor | 用量角器测量方位角
To measure the bearing of point B from point A, place your protractor with 0° pointing north at A. Join A to B with a line, then read the clockwise angle from the north line. Make sure the centre cross of the protractor sits exactly on A, and that the 0° mark lies along the north line rather than the edge of the paper.
要测量点 B 相对于点 A 的方位角,将量角器的 0° 对准 A 点的正北方向。连接 A 与 B 成一条直线,然后从正北线顺时针读取角度。务必让量角器的中心十字正好落在 A 点上,并且 0° 刻度沿着正北线,而不是沿着纸边。
If the line points left, you may measure an anticlockwise angle and subtract it from 360°. Always double-check that your angle is measured clockwise from the north line at the correct point.
如果线指向左侧,你可以先测量逆时针角,再用 360° 减去它。始终检查角度是否从正确点的正北线顺时针测量。
Drawings should be done in pencil with a sharp ruler and a clear north arrow at every point where a bearing is measured. A small diagram error of 2° can shift a position by tens of metres in a real survey, so neatness matters.
作图应使用铅笔、直尺,并在每一个测量方位角的点上画出清晰的北向箭头。在真实测量中,2° 的小误差就可能让位置偏移几十米,所以整洁规范非常重要。
3. Three-Figure Bearings and Compass Points | 三位数方位角与罗盘方向
Bearings are always written as three figures. This means 5° is written as 005°, 45° as 045°, and 105° remains 105°. An angle of exactly 90° is written 090°, and a full turn back to north is 360° or 000°.
方位角始终写成三位数。这意味着 5° 写作 005°,45° 写作 045°,105° 仍写作 105°。正好 90° 写作 090°,转一圈回到正北则是 360° 或 000°。
The eight principal compass directions link directly to standard bearings, so it is worth memorising them before moving on to scale work.
八个主要罗盘方向与标准方位角直接对应,在进入比例绘图之前值得先记住它们。
| Compass direction | 罗盘方向 | Bearing | 方位角 |
|---|---|
| North | 北 | 000° |
| Northeast | 东北 | 045° |
| East | 东 | 090° |
| Southeast | 东南 | 135° |
| South | 南 | 180° |
| Southwest | 西南 | 225° |
| West | 西 | 270° |
| Northwest | 西北 | 315° |
Bearing of east is 090°. | 正东的方位角是 090°。
Bearing of south is 180°. | 正南的方位角是 180°。
Bearing of west is 270°. | 正西的方位角是 270°。
A direction such as 022° is between north and northeast. | 像 022° 这样的方向位于正北与东北之间。
4. Back Bearings and Reverse Directions | 后方位角与反向方位角
A back bearing is the bearing you would travel on to return along the same straight line, but in the opposite direction. The rule is simple: add 180° if the original bearing is less than 180°, and subtract 180° if it is 180° or more.
后方位角是指沿同一条直线原路返回时所行的方位角,方向正好相反。规则很简单:如果原方位角小于 180°,加 180°;如果大于或等于 180°,减 180°。
Back bearing = Bearing + 180° (if Bearing < 180°) | 后方位角 = 方位角 + 180°(当方位角 < 180°)
Back bearing = Bearing − 180° (if Bearing ≥ 180°) | 后方位角 = 方位角 − 180°(当方位角 ≥ 180°)
°. If the bearing of C from D is 295°, then the bearing of D from C is 295° − 180° = 115°.
例如,如果 B 相对于 A 的方位角是 062°,那么 A 相对于 B 的方位角就是 062° + 180° = 242°。如果 C 相对于 D 的方位角是 295°,那么 D 相对于 C 的方位角就是 295° − 180° = 115°。
You can also see why this works by drawing parallel north lines through both points. Alternate angles between the two north lines are equal, so turning round at the second point and measuring clockwise gives an angle exactly 180° different from the first.
通过在这两个点上画出互相平行的正北线,就能看出其中的道理。两条正北线之间的内错角相等,因此在第二点转身再顺时针测量,得到的角度恰好比第一个角相差 180°。
5. Map Scales and Scale Drawings | 地图比例尺与比例绘图
A scale tells you how many times smaller the drawing is than the real object. A scale of 1 : 25 000 means that 1 unit on the map represents 25 000 of the same unit in real life. If the map unit is a centimetre, then 1 cm on the map is 25 000 cm on the ground, which is 250 m or 0.25 km.
比例尺告诉你图形比实物缩小了多少倍。比例尺 1 : 25 000 表示图上 1 个单位代表实际的 25 000 个相同单位。如果图上的单位是厘米,那么图上 1 cm 代表地面上的 25 000 cm,也就是 250 m 或 0.25 km。
Scales appear in three common forms on maps and in exam questions. A ratio scale is written as 1 : 50 000. A written or verbal scale says something like 1 cm represents 2 km. A linear scale is drawn as a bar marked with real distances, which allows distances to be read off directly with a strip of paper.
在地图和考题中,比例尺有三种常见形式。比例式比例尺写作 1 : 50 000。文字式比例尺写成类似 1 cm 代表 2 km 的说法。线段比例尺则画成一条标有实际距离的横条,可以用纸条直接比着读出距离。
1 : 10 000 means 1 cm = 100 m. | 1 : 10 000 表示图上 1 cm = 实际 100 m。
1 : 25 000 means 1 cm = 250 m. | 1 : 25 000 表示图上 1 cm = 实际 250 m。
1 : 50 000 means 1 cm = 500 m. | 1 : 50 000 表示图上 1 cm = 实际 500 m。
1 : 100 000 means 1 cm = 1 km. | 1 : 100 000 表示图上 1 cm = 实际 1 km。
6. Converting Between Drawing Lengths and Actual Distances | 图上长度与实际距离的换算
Conversion is a two-step process: multiply or divide by the scale factor, then convert the units. Because there are 100 cm in a metre and 1000 m in a kilometre, there are 100 000 cm in a kilometre. Getting this unit step right is worth more marks than any other part of the topic.
换算是两步过程:先乘或除以比例尺分母,再换算单位。由于 1 m = 100 cm、1 km = 1000 m,因此 1 km = 100 000 cm。把这个单位步骤做对,比本专题其他任何环节都更能拿分。
Actual distance = Drawing length × Scale factor | 实际距离 = 图上长度 × 比例尺分母
Drawing length = Actual distance ÷ Scale factor | 图上长度 = 实际距离 ÷ 比例尺分母
villages are 6 cm apart. The actual distance is 6 × 25 000 = 150 000 cm = 1500 m = 1.5 km.
示例:在比例尺为 1 : 25 000 的地图上,两个村庄相距 6 cm。实际距离为 6 × 25 000 = 150 000 cm = 1500 m = 1.5 km。
ong in reality. Its length on the map is 4.5 km = 450 000 cm, and 450 000 ÷ 50 000 = 9 cm.
反向示例:在比例尺为 1 : 50 000 的地图上,一条道路实际长 4.5 km。它的图上长度为 4.5 km = 450 000 cm,450 000 ÷ 50 000 = 9 cm。
7. Drawing and Solving Bearing Problems | 方位角作图的步骤与求解
Most bearing questions can be solved by following the same six steps in the same order. Skipping steps is the fastest way to lose method marks, even when the final answer is nearly right.
大多数方位角题目都可以按同样顺序的六个步骤来解决。跳步是最快丢方法分的方式,即使最终答案接近正确也一样。
Choose a sensible scale, for example 1 cm = 1 km, and write it on the diagram. | 选择合理的比例尺,例如 1 cm = 1 km,并标注在图上。
Draw a north line at the starting point. | 在起点画一条正北线。
Measure the clockwise angle from north with a protractor. | 用量角器从正北顺时针量出角度。
Draw the line of travel in the correct direction. | 按正确方向画出运动路线。
Mark the distance along that line using your scale. | 按比例尺在该线上标出距离。
Draw a north line at the new point before measuring any second bearing. | 在测量第二个方位角之前,先在新位置画出正北线。
At the end, measure the required length or angle on the diagram and convert it back to real units. If the question asks for a bearing, remember to give the answer as three figures.
最后,在图上量出所需的长度或角度,再换算回实际单位。如果题目要求方位角,记得用三位数作答。
8. Triangulation and Locating Positions | 三角定位与确定位置
A single bearing gives a direction but not a fixed position. To pinpoint a location you need two pieces of information, such as two bearings taken from two known points, or one bearing plus one measured distance.
单独一个方位角只能给出方向,无法确定唯一位置。要精确定位,需要两条信息,例如从两个已知点测得的两条方位角,或者一条方位角加一段测得的距离。
The classic method is this: draw a ray from point A along the given bearing, draw a second ray from point B along its given bearing, and the object lies at the intersection of the two rays. This process is called triangulation and it is the basis of traditional map-and-compass navigation as well as modern GPS positioning.
经典方法是:从 A 点沿给定方位角画一条射线,再从 B 点沿给定方位角画第二条射线,目标就位于两条射线的交点上。这个过程称为三角定位,它既是传统地图与罗盘导航的基础,也是现代 GPS 定位的基础。
A scale drawing is essential here, because without it you cannot convert the measured map distance back into a real distance such as kilometres. Always show the scale, the north arrows and the intersecting rays clearly.
此时比例绘图必不可少,因为不按比例作图就无法把图上量得的距离换回公里等实际长度。务必清楚标出比例尺、北向箭头和相交的射线。
9. Multi-Stage Journeys and Direction Changes | 多段行程与方向改变
Many KS3 questions describe a journey with two or three legs, for example a ship sailing 12 km on a bearing of 070°, then 9 km on a bearing of 160°. Draw each leg separately, always starting the new north line at the new position, and never simply adding the two bearings together.
很多 KS3 题目描述的是由两段或三段组成的行程,例如一艘船先沿 070° 方位角航行 12 km,再沿 160° 方位角航行 9 km。要分别画出每一段,每到一个新位置都要重新画正北线,绝不能简单地把两个方位角相加。
Once the whole route is drawn to scale, measure the straight-line distance from the start point to the finish point, then measure the bearing of that line from the start point. This gives the overall displacement, which is the net effect of the whole journey.
整条路线按比例画好后,先量出起点到终点的直线距离,再量出这条直线相对于起点的方位角。这样就得到了总位移,也就是整段行程的净效果。
at 15 km/h for 40 minutes covers 15 × ⅔ = 10 km, and this 10 km is what you draw to scale.
如果题目给出了速度,先用 距离 = 速度 × 时间 换算成实际距离再作图。例如,一艘船以 15 km/h 航行 40 分钟,行驶距离为 15 × ⅔ = 10 km,按比例画出的就应是这 10 km。
10. Worked Examples with Full Reasoning | 完整解题示例
Example 1: the bearing of B from A is 062°. Find the bearing of A from B. Since 062° is less than 180°, add 180°: the answer is 242°. Check that the answer is a three-figure bearing and that it points in roughly the opposite direction on a sketch.
示例 1:B 相对于 A 的方位角是 062°,求 A 相对于 B 的方位角。因为 062° 小于 180°,所以加 180°:答案是 242°。检查答案是否为三位数,并在草图上确认它大致指向相反方向。
nd then 3 km due north. Find the direct distance and bearing back to the start. Distance = √4² + 3² = √25 km. The angle east of north satisfies tan θ = 4 ÷ 3, so θ ≈ 53°, giving a bearing of 053°.
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