Why doesn’t horizon distance move exactly proportional to the height of the observer?

For instance if someone is 8 inches above the surface of the Earth, they can see approximately 1 mile to the horizon. However, if someone is viewing the horizon at an eye level of 5’5 they can only see about 3 miles out. If the height of the observer increases by a factor of about 8, from 8 inches to 65 inches, why does the distance they can see only increase by a factor of 3?(from 1 mile to 3 miles)

• Trivially, the Earth is a finite size whereas the height can be infinite, so there can't be a linear relationship. Commented Apr 25, 2022 at 8:45

Consider the following image showing the earth (with radius $$R$$) and an observer at height $$h$$ above the ground. The distance from the observer to the horizon is $$s$$.

The theorem of Pythagoras applied to the right triangle gives $$R^2+s^2=(R+h)^2$$ With a little bit of algebra we get $$R^2+s^2=R^2+2Rh+h^2$$ $$s^2=2Rh+h^2$$

Because $$h$$ is much smaller than $$R$$ we can neglect the $$h^2$$ on the right side. $$s^2 \approx 2Rh$$ $$s \approx \sqrt{2Rh}$$

Now you see that $$s$$ is not proportional $$h$$, but proportional to $$\sqrt{h}$$.

On earth, the distance to the horizon, say $$d_h$$ and the height of an observer, say $$h_o$$ cannot have a linear relationship $$d_h=\text{constant}\cdot h_o$$ or proportional relationship you speak of, since this would assume the earth has some geometry other than spherical.

Instead, in reality, since the earth has curvature and is a sphere, you can show with a little trigonometry, $$d_h\propto \sqrt h_o$$

The exact relationship is $$d_h=r_e\cos^{-1}(\frac{r_e}{r_e+h_o})$$ where $$r_e$$ is the radius of earth, and this equation is usually simplified to the approximate relationship $$d_h\approx\text{1.22}\sqrt h_o$$ where $$d_h$$ is in miles and $$h_o$$ is in feet.

• @DJohnM Yes. The distance to the horizon is in miles and the heigh of the observer is in feet. Commented Apr 25, 2022 at 8:00