Response to “Comment on ‘Decluttering our thinking with the life-changing magic of twiddle’” [Am. J. Phys. 86, 325 (2018)] Free

I agree with Rod Cross's first two comments¹ on my “Decluttering our thinking with the life-changing magic of twiddle,”² including the short time window within which the two balls must be dropped (a point also noted by Feinberg³). In Cross's third comment, he states that using the proposed dimensionless ratio

to estimate the height difference is physically not even “approximately right,” as it neglects air's fluid nature and thereby incorrectly assumes that the air in front of the ball is accelerated to the ball's speed. Cross states, with some justification, that I did not elaborate. In support of his argument, he provides a momentum analysis.

In this response, I follow Cross in considering momentum but show that the R_drag approach is fine. Then, by approximating early and often to avoid power-series expansions, I refine the approach to explain the correction factors that convert the unvarnished R_drag estimate of a 10-m height difference into the actual height difference of 80 cm. (The following analysis assumes that drag modifies free fall only slightly.)

So first, by considering momentum, I twiddle estimate the drag force F on a ball with cross-sectional area A moving at speed v through a fluid of density ρ—assuming that the ball accelerates all the swept-out air to a speed ∼v and leaves the rest of the air undisturbed. Over a time t, the ball travels a distance vt, sweeping out a volume vtA and a mass of fluid $mfl∼ρvtA$⁠. This fluid gains, and the ball loses, a momentum $ΔP∼mflv∼ρv2tA$⁠. The momentum-loss rate $ΔP/t$ is the drag force F, and so, $F∼ρv2A.$ Thus, even neglecting the fluid nature of the air, the assumption that the ball accelerates the swept-out fluid to the ball's speed gives a drag force that depends correctly on v, ρ, and A.

But air is indeed a fluid and flows around the ball. The result is that some of the air behind the ball, which the ball had just accelerated to a speed ∼v, pushes the ball on its backside and hurries it along—that is, some of the air behind the ball returns momentum to the ball. All this complicated physics is captured by the drag coefficient c_d. Including c_d and the conventional factor of 1/2 that rides with it makes the drag force exact:

Thus, for the only moderately streamlined shape of a sphere (⁠$cd≈0.5$⁠), the returned momentum contributes a physical factor of 1/4. (The twiddle estimate of F that ignored the drag coefficient implicitly used c_d = 2, which is reasonable for bluff objects like a flat plate oriented with its large surface perpendicular to the flow.)

This drag force reduces the ball's momentum and speed and therefore its fall distance (imagine the ball racing against a drag-free, freely falling ball, with the race ending when the freely falling ball lands). The drag-induced fractional momentum loss in an unvarnished twiddle analysis is

where $tff$ is the free-fall time and v is the impact speed. The ball's fractional momentum loss is also its fractional speed loss Δv/v; furthermore, the fall time is $tff∼h/v$⁠. With these substitutions, the fractional loss in speed is $∼Rdrag$⁠:

Thus, the returned momentum contributes to ΔP/P and Δv/v a physical factor of c_d/2—the same factor as it contributed to F.

The changing fall speed contributes several further factors that I call geometric. The first geometric factor is hidden in the fall time t_ff, which is actually 2h/v (because the average speed over the fall is v/2). So, correcting the fall time contributes a factor of 2 to Δv/v.

The second geometric factor arises because F varies. Thus, the momentum change $ΔP∼Ftff$⁠, which is the numerator of Eq. (3), should really be $⟨Ft⟩tff$⁠, where $⟨·⟩$ denotes the time average over the journey and $⟨Ft⟩$ is the time average of the time-varying drag force F_t. Because $Ft∝vt2$⁠, where v_t is the fall speed after a time t and $vt∝t$ (in the small-drag approximation), $Ft∝t2$⁠. Thus, $⟨Ft⟩=F/3$⁠, where F is the maximum drag force, at the end of the race. So, the second geometric factor, and the final one for Δv/v, is 1/3. Including the physical factor of c_d/2 and the first geometric factor of 2 gives

The fractional speed loss is now correct.⁴ 

Converting it to the fractional height loss Δh/h contributes the third and final geometric factor. The height loss Δh is

where Δv_t is the drag-induced speed loss due after the ball has fallen for a time t. Converting the fractional speed loss Δv/v in Eq. (5) to the absolute speed loss Δv_t gives

where $mtair$ is the mass of air swept out after a time t. In the small-drag approximation, $vt∝t$⁠. Furthermore, $mtair$ is proportional to y_t, the fall distance after a time t; this distance is itself proportional to t² (in the small-drag approximation). Thus, $Δvt∝t3$⁠, its time average is 1/4th of its final value Δv (at t = t_ff, when the drag-free ball lands).

The fractional height loss picks up this factor of 1/4:

Because h/t_ff is the average speed v/2 and it is in the denominator, a factor of 2 appears; thus

In summary, the complicated physics of fluids contributes a factor of c_d/2 ≈ 1/4; the geometric factors combine to contribute a factor of 1/3. The overall result is a factor of ≈1/12. It converts the unvarnished twiddle estimate of 10 m into the actual height difference of 80 cm.

A twiddle analysis omits geometric factors. However, even before I calculated these factors, they seemed unlikely to conspire enough to save Galileo's 5-cm claim, whose rescue requires a factor of 1/200. Alas, all these factors physical and geometrical could not put Galileo's estimate together again.

As a final, philosophical point, the line separating physical from geometric factors is more porous than I have pretended because these geometric factors arise from a physical effect (the varying fall speed). The porosity is probably inevitable. Here, Cross and I, and our readers, can all perhaps agree with Galileo: The book of Nature is indeed “written in the language of mathematics.”⁵