Let's get back to the classic astronomical problem of calculating horizontal dip angle using small angle approximations. Let, our observer is a point object situated at an height $h$ on the earth surface with radius $r$. And his/her horizon is situated at a distance $D$. So I tried calculating $\theta$ in two different ways: one route was forming a Right Angled Triangle with $(h+r)$ as the Hypotanuse, $D$ and $r$ as the two other sides, cancelling $h^2$ term for being negligible - and simplifying for $\theta$ using $tan\theta$ approximation.

The other method just creates an expression in terms of $D,\theta$ and $r$ for the length $(h+r)$, and then another expression showing the length of perpendicular drawn from the horizon upon the segment $(h+r)$ being subtracted from itself. Finally, the two expressions are simplified with small angle approximations of sin and cos, and combined and simplified to get a formula for $\theta$.

And to my knowledge, both ways are correct. Then why are we getting two different results?

picture of math

Pardon for somewhat untidy drawing, I'm still pretty much new at this.

Assumption used: Considering $\theta$ as small angle, we know $sin\theta \approx tan\theta \approx \theta $ and $cos\theta \approx 1$

Result obtained in the first method: $ \theta = \sqrt\frac{2h}{r}$
And the second method yields: $ \theta = \sqrt\frac{h}{r}$


2 Answers 2


This edit is a doozy. The first equation on the right side, "rcos(theta) + Dsin(theta)= h+R" is correct. It's saying that the portion of the r+h line above the yellow chord plus the portion below is equal to r+h. That's obviously so.

The problem is in the application of the small angle approximations. The second two equations make the r terms disappear because r is very close to rcos(theta). It's close but not close enough: r is a very large number so multiplying it by (1-cos(theta)) doesn't leave a neglibible product. It's approximately equal to h, so the equation becomes D(theta) = 2h. The small angle approximations have to be used carefully.

Incidentally, answer to your possible next question: the thing that threw me, years ago, was that the dip table in the Nautical Almanac doesn't agree with the geometry. I eventually found that the table has a factor for refraction as well as the geometrical factor. The refraction table on the same page only has a correction for the observed height of the object seen. The dip table puts in the correction for the height of the observer.

  • $\begingroup$ Isn't r/cos(theta) equal to h+r? $\endgroup$
    – JohnHoltz
    Mar 2, 2022 at 17:06

If you want $D$ to go away so that you have a relationship between $r,h,\theta$, the exact relationship is

$$ \frac{r}{r+h} = \cos\theta = 1 - \frac{\theta^2}{2!} + \frac{\theta^4}{4!} - \cdots $$

In the small-angle approximation you can throw away "most" of the terms on the right-hand side, and use the additional approximation $(1+\epsilon)^n \approx 1 + n\epsilon$ to invert both sides:

\begin{align} \frac{r + h}{r} & \approx \left(1 - \frac{\theta^2}{2}\right)^{-1} \\ 1 + \frac{h}{r} & \approx 1 + \frac{\theta^2}{2} \end{align}

This is the result you get from the Pythagorean approach, $\theta \approx \sqrt{2h/r}$. This is sometimes called a "second-order" approximation, because the small parameter $\theta$ shows up squared.

In your "plane geometry" approach you have used the first-order approximation $\sin\theta \approx \tan\theta \approx \theta$ twice, but when you set $r\cos\theta = r$ you threw away an additional second-order term. To get the correct second-order result from your plane-geometry approach, you would write

\begin{align} h + r &= r\cos\theta + D\sin\theta \\ h& = D\sin\theta + r(\cos\theta-1) \\ &= r\tan\theta\sin\theta + r(\cos\theta - 1) & \text{because } \tan\theta &= D/r \\ \frac{h}{r} & \approx \theta^2 + \left( 1 - \frac{\theta^2}{2} - 1 \right) = \frac{\theta^2}{2} \end{align}

As a general rule, if your approximation involves subtracting things that are "approximately" equal, you're in trouble unless you can keep track of the difference. In your "Pythagorean" treatment you subtracted two things that were exactly the same, $r^2 = r^2$. But in your plane geometry approach you subtracted two things that were approximately the same, $r\cos\theta \approx r$, and the error you introduced was as big as the approximation you were making.


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