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In the figure's sphere does:

$\widehat{AB} = \widehat{A'B'}$ or Not?

I mean the angles represented by arcs. (not the lengths).

In a book it said that they're not equal and they are: $\widehat{AB} = \widehat{A'B'} * \cos{\widehat{AA'}}$.

Is it right? If yes, How it is possible?

enter image description here

Example

In the above picture we know $AA' = BB' = 21^{o}$ (I don't write the degree in the other ones). and also A'B' (something like difference in longitudes of A and B) is: $A'B' = APB = A'PB' = \Delta\lambda = 66$ The book where I said the question solved like this:

The formula I said above say that the Arc $ADB = 66 * \cos (21) = 61.61630814881532$ which is obviously not $66$.

If we think the green arc ($ACB$) is part of great circle then we have this formula: $\cos (ADB) = \cos (PB) \cos(PA) + \sin(PB)\sin(PA)cos(APB)$

$\cos (ADB) = \cos (90-21) \cos(90-21) + \sin(90-21)\sin(90-21)cos(66)$

And from that we get $ADB = 61.123188817554769358$

so $66\neq 61.123188817554769358 \neq 61.61630814881532 $

So what does each of formulas actually measure?

titansarus
  • 749

1 Answers1

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On the assumptions that

  • $A'$ and $B'$ lie on the equator,

  • $AA'$ and $BB'$ are longitudes,

  • $A$ and $B$ lie on the same latitude,

the respective angles $\angle A'B'$ and $\angle AB$ subtended from the centers of the corresponding latitude lines are equal, but the angles subtended from the center of the sphere (i.e., the spherical distances) are not equal.

The formula $\widehat{AB} = \widehat{A'B'} \cos\widehat{AA'}$ appears to refer to a third quantity: length measured along the latitude segment $D$. This also clearly depends on the latitude.

Spherical arcs and angles

Quantitatively, assume $A$ and $A'$ have longitude $0$, that $B$ and $B'$ have longitude $\theta = \angle A'B'$, that $A$ and $B$ have latitude $\phi = \angle AA'$, and that all four points lie on the unit sphere centered at the origin. In terms of these spherical coordinates, the four points have Cartesian coordinates: \begin{align*} A &= (\cos\phi, 0, \sin\phi) & A' &= (1, 0, 0) \\ B &= (\cos\theta \cos\phi, \sin\theta \cos\phi, \sin\phi) & B' &= (\cos\theta, \sin\theta, 0). \end{align*} The dot product gives $$ \angle AB = \arccos[\cos\theta \cos^{2}\phi + \sin^{2}\phi] = \arccos[\cos\theta + \sin^{2}\phi(1 - \cos\theta)] $$ for the arc length of the spherical segment $C$.

The latitude $AB$, viewed as a circle in Euclidean $3$-space, has radius $\cos\phi$, and the formula $\widehat{AB} = \widehat{A'B'} \cos\widehat{AA'}$ gives length of the latitude segment $D$.

Andrew D. Hwang
  • 78,195