# Calculating the distance on a cone or a sphere and the implications for understanding the geometry of space

I am reading Eddington’s book on space and time

https://www.gutenberg.org/files/29782/29782-pdf.pdf

And on page 71 he gives the formula for the distance, $$ds$$, between two points on a spherical surface.

$$ds^2 =d Lat^2 + (\cos Lat)^2 \cdot d Long^2$$

I interpret this as the Pythagorean formula where the distance, $$ds$$, is the hypotenuse of a triangle whose bases are $$dLong$$ and $$dLat$$. Latitude is an evenly spaced vertical measurement. And since $$d Long$$ gets smaller and smaller as latitude increases, the $$\cos Lat$$ adjusts for this.

To really see if I understood it, I tried to derive a similar formula for a planet shaped like a cone, but then I got the same result. How could these be the same? Clearly confused, I searched how to calculate distance along a geodesic on a sphere. I discovered Haversine’s formula-Where $$\phi$$ is latitude and $$\lambda$$ is longitude.

$$\sin^2(Δφ/2) + \cos \phi_1 ⋅ \cos \phi_2 ⋅ \sin^2(\Delta \lambda/2)$$

For small angles we assume $$\phi_1 \approx \phi_2$$ so we can approximate with

$$\sin^2(\Delta\phi/2) + \cos^2 \phi_1 ⋅ \sin^2(\Delta\lambda/2)$$

Why has Eddington omitted the $$\sin$$ function? Is he using an approximation for small angles like $$\sin x \approx x$$ ?

If that is the case, then can we not also use the approximation of $$\cos x \approx 1$$ for small angles?

In which case the formula becomes the distance calculation for Euclidean space. So what is the reason that Eddington would use his formula? What am I missing?

And of course this formula goes straight to the heart of Minkowski’s and Einstein’s calculations as well in calculations of spherical geometry.

Thank you for your help in clearing up my misunderstanding.

• $\Delta\phi$ and $\Delta\lambda$ are "small", but $\phi_1$ and $\phi_2$ are not, in general. Hence you cannot set $\cos\phi_1\approx1$. Apr 12 '20 at 12:59
• thank you! I knew my brain was overcooked somewhere and just was not seeing things. Enter it as a response so I can mark it answered. Apr 12 '20 at 14:39

$$Δ𝜙$$ and $$Δ𝜆$$ are "small", but $$𝜙_1$$ and $$𝜙_2$$ are not, in general. Hence you cannot set $$\cos \phi_1\approx 1$$.