For the past day, I have been trying to calculate the arc length of a sphere from a pole to a longitudinal cross-section with a specified circumference.

I have diagrams and equations belo w. However, I am getting answers that seem incorrect. Are equations correct? Any improvements to it?

enter image description here

  • $\begingroup$ Why the down vote? I have shown my work and how I got to the equation $\endgroup$ May 14 '15 at 21:33
  • $\begingroup$ See math notation guide. $\endgroup$
    – user147263
    May 15 '15 at 0:47

Computing the distance $d$ using $\theta$ in degrees is...ill-advised. :) Radians are defined so that $d = r_{0}\theta$; mathematics seldom gets simpler than that. Further, if you're using a calculator to test values, there's a non-negligible chance the outputs are coming to you in radians.

That aside, you're essentially there. If you measure $\theta$ in radians, then \begin{gather*} C = 2\pi r_{i},\quad\text{or}\quad r_{i} = C/(2\pi); \tag{1} \\ \sin\theta = r_{i}/r_{0},\quad\text{or}\quad \theta = \arcsin(r_{i}/r_{0}); \tag{2} \\ d = r_{0}\theta = r_{0} \arcsin\bigl[C/(2\pi r_{0})\bigr]. \tag{3} \end{gather*} For example, if $r_{0} = 1$, the equator has $C = 2\pi$, and the formula gives $d = \pi/2$ as expected. If $C = \pi$, then $d = \pi/6$ or $5\pi/6$, again as expected.

  • $\begingroup$ if $r_0 = 1km$, then $d =( \pi / 2 )km$ ? $\endgroup$ May 14 '15 at 22:54
  • $\begingroup$ Indeed, the calculator was set to radians when I was using degrees $\endgroup$ May 14 '15 at 23:08
  • $\begingroup$ If $r_{0} = 1$ km, you still need a circumference to find $d$. :) (In case it matters, trigonometric formulas in calculus hold only when angles are measured in radians; nature prefers radians as the unit of angle.) $\endgroup$ May 14 '15 at 23:29
  • $\begingroup$ Yes, I got it to work with degress, but I will convert it to use radians instead. Also, my km question was in reference to your example ..... if $r_0=1$km, the equator has $C=2π$km, and the formula gives $d=π/2$km as expected. I just wanted to make sure i didn't have to do anything else with my answer - I don't have to I think $\endgroup$ May 14 '15 at 23:34
  • $\begingroup$ Ah...yes, $\pi/2$ km. :) $\endgroup$ May 14 '15 at 23:39

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