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Barcelona (Spain) has the coordinates (approx): $\theta = 2^\circ$, $\phi = 41^\circ$, and New York has the coordinates: $\theta = −74^\circ$, $\phi = 41^\circ$. Notice that both cities lie on the same latitude, which makes the calculations easier. For your calculations, use Earth’s radius $R = 6378$ km, and give all your result in kilometers.

  1. On most maps, the naive ‘straight line’ between Barcelona and New York lies on the line of latitude $\phi = 41^\circ$. What is the distance between the two cities along this line?

  2. What is the actual shortest distance between those two cities?

    Hint: Notice that it is determined by the angle Barcelona–Earth’s center–New York. Use the fact that the angle $\psi$ between two unit vectors $nA$ and $nB$ satisfies $\cos \psi = nA \cdot nB$.

    Hint: Observe that both curves are parts of circles. By using this, you can avoid integration altogether.

For the first part, I'm trying to take the Euclidean distance. \begin{eqnarray} x &=& R\cos\theta\cos\phi \\ y &=& R\cos\theta\sin\phi \\ z &=& R\sin\theta, \end{eqnarray} Using this, I get coordinates for NY and Barcelona.

New York: $(x, y, z) = (2620.58, 421.01, 5799)$

Barcelona: $(x, y, z) = (-1081.34, -173.72, 6283.26)$

If I try to find the distance by the equation

\begin{eqnarray} \sqrt{x^2 + y^2 + z^2} \end{eqnarray}

I get 3780 km.

Am I doing this right? I feel a bit lost because this "naive" distance differs so greatly from the real one (ie 6157 km)

The Euler-Lagrange equation is this: \begin{eqnarray}ϑ˙^2 cos ϕ sin ϕ/ \sqrt{ϕ˙^2 + ϑ˙^2 cos^2 ϕ} + d/dt ϕ˙^2 \sqrt{ϕ˙^2 + ϑ˙^2 cos^2 ϕ} = 0 \end{eqnarray}

Henry
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mathinthecity
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4 Answers4

3

Be careful. All angles are measured in radians. In accord with the question $\phi$ means the latitude ($=\frac{\pi}{2}-\text{polar angle}$) and $\theta$ means the longitude (azimuth angle).

If one considers the Earth being a ball, the shortest path is: $$ d=R\arccos(\sin\phi_1\sin\phi_2+\cos\phi_1\cos\phi_2\cos(\theta_2-\theta_1)), $$ the argument of $\arccos$ being the scalar product of unit vectors directed from the Earth center to the points on the surface.

The "naive path" for $\phi_1=\phi_2\equiv\phi$ has length: $$ d^*=R(\theta_2-\theta_1)\cos\phi. $$

In the last expression $0\le\theta_2-\theta_1\le\pi$ is assumed. Generally $\arccos\cos(\theta_2-\theta_1)$ can be used instead.

user
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  • For the naive path, won't d = 0 since \begin{eqnarray} \phi_2=\phi_1=41

    and

    \phi_2-\phi_1 = 0 \end{eqnarray}

     – mathinthecity Apr 25 '18 at 04:24
  • @mathinthecity NO. By mistake I interchanged the meaning of $\theta $ and $\phi $ compared with your question, so that in my expressions $\theta $ means latitude and $\phi $ means longitude. This is however in accord with the spherical coordinate system introduced by you (so that you also have interchanged the meaning). If you insist I can change the names back to original notation. – user Apr 25 '18 at 05:25
  • According to this website, I'm not getting the correct answer. It's not even close.. https://www.movable-type.co.uk/scripts/latlong.html – mathinthecity Apr 25 '18 at 05:35
  • The website reports the distance 5856 km. My expression with R=6371 km gives 6157 km. I would not say "it's not even close". The deviation is probably caused by the fact that Earth is not a ball. – user Apr 25 '18 at 05:54
  • Which expression are you referring to, the 1st or 2nd? – mathinthecity Apr 25 '18 at 05:57
  • The first one. The second gives 6378 km. – user Apr 25 '18 at 05:59
  • d* = 6371cos(4176) = 5715 The website gives 6157KM – mathinthecity Apr 25 '18 at 05:59
  • Am I inserting the values wrong? – mathinthecity Apr 25 '18 at 06:02
  • Now I see your problem with understanding. $(\phi_2-\phi_1)$ is not in argument of $\cos$. It is a multiplicative factor equal to $\pi*76/180$. – user Apr 25 '18 at 06:04
  • Oh i see! Why do we convert it to radians? – mathinthecity Apr 25 '18 at 06:05
  • Do we not convert the 41 to radian? – mathinthecity Apr 25 '18 at 06:05
  • @mathinthecity I have rewritten the formula to avoid misunderstanding. – user Apr 25 '18 at 06:05
  • We should convert it to radians to compute the length. $\arccos$ in my first expression gives results also in radians. – user Apr 25 '18 at 06:07
  • I see! But still for the first formula, I set my calculator to degrees and input the following values:

    Rarccos(sin^2(41) +cos^2(41)cos(76))

    Still don't get the same answer as you

     – mathinthecity Apr 25 '18 at 06:19
  • @mathinthecity Check if the calculator gives the result of $\arccos$ in radians. If not divide it by 180 and multiply by π. I have rewritten the answer using the angle definitions from the question. – user Apr 25 '18 at 07:42
  • @mathinthecity I have checked the distance from New-York to Barcelona. It appears to be 6167 km, practically coinciding with the result of the first formula, whereas its approximation on the website you linked is in a heavy error. – user Apr 25 '18 at 07:55
  • +1 Because the great circle distance formula is certainly relevant. But, the question seems to indicated that a more elementary technique should be used. – Mark McClure Apr 26 '18 at 15:43
  • @MarkMcClure Thank you very much for the vote. But I disagree with suggestion about "a more elementary technique". Quite contrary, the sentence "Use the fact that the angle ψ between two unit vectors nA and nB satisfies cosψ=nA⋅nB." (which I did not notice while answering the question) implies that exactly the "great circle distance formula" was expected. – user Apr 26 '18 at 15:56
2

Use that chord length and the Earth's radius to determine the arc length around the curved path.

Phil H
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1

Using the distance equation you will get the "straight line distance" if you bored a hole through the earth.

The radius of the $41^\circ$ parallel is $R\cos 41^\circ$

And the distance a person would travel if they followed this arc would be this radius times the longitude traveled in radians: $(R\cos 41)(\frac {76 \pi}{180})$

The great circle path...

$a\cdot b = \|a\|\|b\| \cos \psi\\ (R\cos 74^\circ\cos 41^\circ, R\cos 74^\circ\sin 41^\circ, R\sin 41^\circ)\cdot(R\cos -2^\circ\cos 41^\circ, R\sin -2^\circ\cos 41^\circ, R\sin 41^\circ) = R^2(\cos 76^\circ\cos^2 41^\circ + \sin^2 41^\circ)\\ \psi = \arccos (\cos 76\sin^2 41 + \cos^2 41)\\ D = R\psi$

Doug M
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0

Cut the Earth along its center $O$ and the two cities $N$ and $B$ to get a circle with the radius $R$ and the angle $\measuredangle BON=76^\circ$.

a) The length of the chord $BN$ is: $$BN=\sqrt{NO^2+BO^2-2NO\cdot BO\cdot \cos{\measuredangle BON}}=\\ \sqrt{2\cdot 6378^2-2\cdot 6378^2\cdot \cos{76^\circ}}\approx 7853.38 \ \text{km}.$$ Note: The Cosine Theorem is applied in the triangle $\Delta BON$.

b) The length of the arc $BN$ is: $$l_{BN}=2\pi R\cdot \frac{\alpha^\circ}{360^\circ}=2\pi\cdot 6378\cdot \frac{76^\circ}{360^\circ}\approx 8460.1 \ \text{km}.$$

farruhota
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