10.01c - Scale drawings and bearings

Reading a map scale

Two paths measure $4\,\mathrm{cm}$ and $7\,\mathrm{cm}$ on the same map. How much longer is the second in real life? The centimetres alone cannot tell us: we need the map's scale.

For example, $1:25000$ means $1\,\mathrm{cm}$ on the map represents $25000\,\mathrm{cm}$, which is $250\,\mathrm{m}$. It does not mean $25000\,\mathrm{m}$. Recall that $100\,\mathrm{cm}=1\,\mathrm{m}$ and $1000\,\mathrm{m}=1\,\mathrm{km}$; we will examine unit conversions more fully next lesson.

With scale $1\,\mathrm{cm}$ to $k\,\mathrm{m}$, multiply the drawing length in centimetres by $k$ to get metres. Divide the actual length in metres by $k$ to get drawing centimetres. Match the units before calculating.

Keep the scale and change only the direction of conversion. An actual length must be made smaller to fit the map.

These scales apply to lengths. If a model has lengths divided by ten, a square face has both dimensions divided by ten: its area is divided by one hundred. Do not apply a length factor once to an area or volume.

The opening paths differ by $3\,\mathrm{cm}$. On a $1:25000$ map that represents $3\times250=750\,\mathrm{m}$.

Measuring bearings from north

A bearing gives a direction from a named starting point. “The bearing of $B$ from $A$” means stand at $A$, face north, then turn towards $B$.

Put the centre of the protractor at the starting point and align its zero with north. Read clockwise to the direction line. For a bearing greater than $180^\circ$, either use a full-circle protractor or measure the smaller anticlockwise angle from north and subtract it from $360^\circ$. “$040^\circ$” means forty degrees, not four.

[DIAGRAM: asset_name: Clockwise and return bearings; asset_slug: clockwise-return-bearings; caption: The bearing of B from A is 065 degrees. The reverse direction from B is 245 degrees.; description: A lies southwest of B. Parallel north arrows rise from A and B. A clockwise arc from north at A to AB is labelled 065 degrees. At B a clockwise reflex arc from north to BA is labelled 245 degrees. The arrows on the arcs show the clockwise sense.]

A return bearing reverses the direction, so it differs by $180^\circ$. If the original bearing is below $180^\circ$, add $180^\circ$. If it is at least $180^\circ$, subtract $180^\circ$. This keeps the answer between $000^\circ$ and just below $360^\circ$. The north lines at different places are treated as parallel on an ordinary local scale drawing.

Change only the original bearing so it is now above $180^\circ$. Subtract the half-turn to keep the answer in range.