I was trying to derive by myself the distance on a great circle.

To confirm my calculations are correct I was comparing against the formula (4) in this link, which is something like

$$ \cos(\delta_1)\cos(\delta_2) \cos(\lambda_1 - \lambda_2) + \sin(\delta_1) \sin(\delta_2) $$

According to wikipedia instead the formula looks more:

$$ hav(\Theta) = hav(\varphi_2 - \varphi_1) + \cos(\varphi_1) \cos(\varphi_2) hav(\lambda_2 - \lambda_1) $$


$$ hav(\theta) = \sin^2\left( \frac{\theta}{2}\right) = \frac{1 - \cos(\theta)}{2} $$

They don't seem equivalent to me, I've tried to use some trig to prove they're an identity, but unless I'm making mistakes it doesn't seem the case.

My derivation leads exactly to the first formula I've shown. The question is... which one is correct, are they the same?

  • $\begingroup$ Please use MathJax to enter mathematical expressions in your posts instead of images. Images can't be searched and give some readers difficulties. You can find a MathJax tutorial here: math.meta.stackexchange.com/questions/5020/… $\endgroup$
    – awkward
    May 29, 2019 at 20:52
  • $\begingroup$ As a first sanity check, try special values, such as $\lambda_1-\lambda_2=0$ or $\frac\pi2$. $\endgroup$
    – user65203
    May 30, 2019 at 7:25

1 Answer 1


Note that $$ \operatorname{hav}\Theta = \operatorname{hav}(\varphi_1-\varphi_2)+\cos(\varphi_1)\cos(\varphi_2)\operatorname{hav}(\lambda_1-\lambda_2) $$ iff $$ 1-\cos\Theta = 1-\cos(\varphi_1-\varphi_2)+\cos(\varphi_1)\cos(\varphi_2)[1-\cos(\lambda_1-\lambda_2)] $$ iff $$ \cos\Theta = \cos(\varphi_1-\varphi_2)+\cos(\varphi_1)\cos(\varphi_2)[\cos(\lambda_1-\lambda_2)-1], $$ and recall $\cos(\varphi_1-\varphi_2)=\cos\varphi_1\cos\varphi_2-\sin\varphi_1\sin\varphi_2$.

  • $\begingroup$ What's your conclusion then? $\endgroup$ May 30, 2019 at 12:17
  • $\begingroup$ I think I have given away enough for you to reach a conclusion. $\endgroup$ May 30, 2019 at 14:00
  • $\begingroup$ It doesn't seem the same to me. $\endgroup$ May 30, 2019 at 14:17

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