Spherical trigometry for Sailing Problems

Question

A man travel from Pekin (ΦP = 39◦54′N, ΛP = 116◦23′E) to Singapur (ΦS = 1◦17′N, ΛS = 103◦51′E). Suppose the trajectory is a great circle:

a) find the distance in nmi betwen Pekin and Singapur

b) find the final rhumb in Singapur

c) find the distance from Singapur to the point X ( intersection betwten the trajectory Pekin-Singapur and the equator.

Answers

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a)

$\text{N}=\Lambda_P-\Lambda_S=116º23'-103º51'=12º32'$

$p=90º-\Phi_S=90º-1º17'= 88º43'$

$s=90º-\phi_P =90º-39º54'=50º6'$

Using the cosine rule:

$$\cos n = \cos s \cos p + \sin s \sin p \cos \text{N}$$

$$n=\arccos (\cos s \cos p + \sin s \sin p \cos \text{N})$$

$ n = \arccos( \cos (50º6')\cos (88º43') + \sin (50º6') \sin (88º43')\cos (12º32')$ \

$$\boldsymbol{n}= 40º15'54.43'' = \boldsymbol{2415.90717}\, \textbf{nmi}$$

Is this correct?

For b) I don't know excatly how solve it, but I did this:

The final rhumb is $180º+S$. $$\frac{\sin N}{\sin n} = \frac{\sin S}{\sin s}\Rightarrow \sin S= \frac{\sin N\sin s}{\sin n}\Rightarrow S = \arcsin\left( \frac{\sin N\sin s}{\sin n}\right)$$

$$ S= \arcsin\left( \frac{\sin (12º32')\sin(50º6') }{\sin (40º15'54.43'')}\right) = 14º55'35.51''$$

$$ \textbf{Final rhumb}=180º+14º55'35.51''= \boldsymbol{ 194º 55'35.51''} =\textbf{S}\boldsymbol{14.9265303º}\, \textbf{W}$$

For c what I did is:

First made a new spherical triangle with points $N,X,S'$. We know that

$p = 88º43', s'=90º, S'$ =180-S=$ 165º 4'24.49''$

$$\cos s' =0= \cos n' \cos p + \sin n' \sin p \cos S' $$ $$ \cos n'\cos p = -\sin n'\sin p \cos S'$$ $$ \cos n' = -\sin n'\tan p \cos S '$$ $$ \cot n' = -\tan p \cos S'$$ $$n'=\arctan\left( (-\tan p \cos S'\right)^{-1})$$ $$\boldsymbol{n'}=1º19'41.28''=\boldsymbol{79.68800445}\,\textbf{nmi}$$

Can you help me please? I don't know if my answers are correct

I did a draw with Geogebra but I don't know if it is right ...

Answer 1

First off you have massive errors in calculation starting with getting $N$ wrong, then rounding the value of $\cos n$ before finding $n$ and then either converting from degrees to radians incorrectly or multiplying by the wrong radius of the earth to get the distance.

In the second part, I don't know what is the convention for the bearing, but your drawing says $\text{rhumb}=180°+S$, inconistent with your equation. You have to decide which is correct. Then you subtracted incorrectly from $360°$.

In the third part I would draw a spherical triangle with vertices at $\chi$, the north pole, and Beijing. Then we can get the angle $P$ at Beijing from the law of sines for the first triangle: $$\sin P=\frac{\sin N}{\sin n}\sin p$$ Then we can get $S'$, the angle at $\chi$ in the second triangle from the law of sines $$\sin S'=\frac{\sin P}{\sin 09°}\sin s$$ And then we can solve the law of cosines $$\cos s=\cos90°\cos a+\sin90°\sin a\cos S'$$ To get the arc from $\chi$ to Beijing $$\sin a=\frac{\cos s}{\cos S'}$$ Then the arc from $\chi$ to Singapore is $a-n$ and the distance is $R(a-n)$.

I set up everything in a spreadsheet as you should too because then you don't have to do it all over again when you make a mistake early on. $$\begin{array}{|l|r|r|}\hline R &3958.76 &\text{mi}\\\hline \Lambda_P &116.3833333 &116°23'0''\\\hline \Phi_P &39.9 &39°54'00''\\\hline \Lambda_S &103.85 &103°51'00''\\\hline \Phi_S &1.283333333 &1°17'00''\\\hline N &12.53333333 &12°32'00''\\\hline p &88.71666667 &88°43'00''\\\hline s &50.1 &50°06'00''\\\hline \cos n &0.763061941 &\\\hline n &40.2651195 &40°15'54''\\\hline n &2782.053859 &\text{mi}\\\hline \sin S &0.257580237& \\\hline S &14.92653028 &14°55'36''\\\hline \text{Rhumb} &345.0734697 &345°04'24''\\\hline \sin P &0.335671696& \\\hline P &160.3866114 &160°23'12''\\\hline \sin S' &0.257515627 &\\\hline S' &14.92269918 &14°55'22''\\\hline \sin a &0.663838148 &\\\hline a-n &41.59325291 &41°35'36''\\\hline a-n &1.328133406 &1°19'41''\\\hline a-n &91.76524774 &\text{mi}\\\hline \end{array}$$ EDIT: This maybe should be a comment but it's gonna be too long so...

When we see 'miles' in the US we usually think $5280\text{ ft}$ particularly in our beautiful square state. Thus if nautical miles is intended one should say 'nautical miles' rather than the unqualified 'miles'. Maybe in a naval context it's the other way around, but it's quite the ship that can sail in a straight line from Beijing to Singapore.

When you crunch out $\cos n$ it's clear that you don't know how to use your calculator. On Windows calculator in scientific, degrees mode, if you punch in
50.06 i m o * 88.43 i m o + 50.06 i m s * 88.43 i m s * 12.32 i m o =
You get $0.763062$. Then
i o m
It's $40.1554$ which means $40°15'54''$. It seems that you are copying numbers down from your calculator then punching them in again, with rounding. Don't do that! Punch everything in in one big expression like I did just now, or use your calculator's memories to save intermediate results. This not only leads to more accurate results but guards against gross errors made when copying and re-entering numbers. Also your calculator should have a key that converts between degrees and decimals and degrees, minutes, and seconds. Use that.

Note that you got a new value of $n$ in part (a) but then re-used the old, crappy value in part (b). Thus your answer in part (b) is wrong. That's why I encourage you to set everything up on a spreadsheet. Fixing the value of $n$ in part (a) would then propagate to part (b) with no problems. Not to mention that you would never be tempted to copy and re-enter intermediate results by hand with all the extra effort and potential for error that entails.

Now in part (c) I thought about it a little more, recalling that there are $10$ spherical trigonometric identities valid for right spherical triangles and similar identities hold for triangles with a $90°$ side like the triangle with vertices at $\chi$, Singapore, and the north pole. In this case you can use the law of cosines in the form $$\cos s^{\prime}=0=\cos n^{\prime}\cos p+\sin n^{\prime}\sin p\cos S^{\prime}$$ And solve to get $$\begin{align}\cot n^{\prime}&=-\tan p\cos S^{\prime}=-\tan p\cos(180°-S)=\tan p\cos S\\ &=\tan88°43'00''\cos14°55'36''=43.1323\end{align}$$ So $n^{\prime}=1°19'41''=91.77\text{ mi}$ again. Your approach to part (c) was way too complicated. Review what you know about right spherical triangles and their duals and you should be able to see this one step answer. Your answer to part (c) is again wrong at least through extra rounding and also because you used wrong intermediate results to start the calculation.

Answer 2

Your answer to a) looks correct, but GeoGebra gives a slightly different result for angle $n$. I checked b) with GeoGebra and found $S=14.92653°=14°55'35.5''$, again a value slightly different from yours.

For c) I would find unit vector $\hat n$ perpendicular to the plane $\alpha$ of the great circle as: $$ \hat n=\hat P\times\hat S \\ =(\sin p\cos s\sin\Lambda_P-\sin s\cos p\sin\Lambda_S, \cos p\sin s\cos\Lambda_S-\cos s\sin p\cos\Lambda_P, -\sin{p}\sin s\sin N) $$ so that the equation of plane $\alpha$ can be written as $n_x x+n_y y+n_z z=0$. I used $s=90°-\Phi_S$ and $p=90°-\Phi_P$, because your definitions look confusing.

Letting $z=0$ into that equation leads to $y/x=-n_x/n_y$ and longitude $\Lambda_X$ of point $X$ satisfies then $$ \tan\Lambda_X=-{n_x\over n_y}= {\sin p\cos s\sin\Lambda_P-\sin s\cos p\sin\Lambda_S \over -\cos p\sin s\cos\Lambda_S+\cos s\sin p\cos\Lambda_P}. $$