Reviewing my trigonometry for an exam next week, I got stuck on this question:
A ship at latitude 36 deg North and longitude 120 deg East is traveling at 16 knots. How long will it take the ship to cross the equator at longitude 170 deg East ?
Reviewing my trigonometry for an exam next week, I got stuck on this question:
A ship at latitude 36 deg North and longitude 120 deg East is traveling at 16 knots. How long will it take the ship to cross the equator at longitude 170 deg East ?
1. I don't know which trigonometry you use to do this problem. Therefore I'll use vectors to show you how to solve it.
2. Define a coordinate system with the origin at the center of the Earth, the x-axis pointing to E(E90°; N0°) and the y-axis pointing N(E180°; N0°), the z-axis pointing to the Northpole. Then every place on the surface of the Earth is determined by unit stationary vector.
3. The town Qingdao (E120°; N36°) is determined by:
$\displaystyle \vec q=\left(\begin{array}{c}\cos(36^\circ) \cdot \cos(120^\circ-90^\circ) \\ \cos(36^\circ) \cdot \sin(120^\circ-90^\circ) \\ \sin(36^\circ)\end{array}\right)$ with $\displaystyle |\vec q|= 1 (\text{radius of the earth})$
4. The point F is determined by
$\displaystyle \vec f=\left(\begin{array}{c} \cos(170^\circ-90^\circ) \\ \sin(170^\circ-90^\circ) \\ 0\end{array}\right)$
5. The angle $\displaystyle \alpha$ which is included by $\displaystyle \vec q$ and $\displaystyle \vec f$ is calculated by:
$\displaystyle \cos(\alpha)=\dfrac{\vec q \cdot \vec f}{1 \cdot 1}= \cos(36^\circ) \cdot\cos(30^\circ) \cdot\cos(80^\circ) +\cos(36^\circ) \cdot\sin(30^\circ) \cdot\sin(80^\circ)$
You'll get $\displaystyle \alpha \approx 58.666^\circ$
6. Since 1° represents a distance of 60 nautical miles (nmi) you'll get:
$\displaystyle |\widehat{QF}|=3520\ nmi$
At a speed of $\displaystyle 16\ \frac{nmi}h = 16\ kts$ the elapsed time will be 220 h.
7. Be sure that you pass the south Korean islands and the south japanese islands at a safe distance