With an approximate solution in hand, Euler could then derive formulas for the the balloon’s maximum altitude, total ascent time, and maximum speed. First, he noted that the maximum altitude \(h\) is reached when the vertical velocity vanishes: \[ 0 \;=\; \frac{4\lambda gb}{k} \left((b+f)(1-e^{-h/b}) - h\right).

\]

Since \(h\) will be considerably larger than \(b\) an approximation of \(e^{-h/b} \approx 0\) is enough to approximate the maximum altitude, as \(h = b+f\). Euler could have (and likely had) approximated \(h\) by appealing directly to his differential equation: if we use a linear approximation for \(e^{-x/k}\), the equation (1) becomes

\[

2v \partial v + \left(\frac{1}{b} \cdot vv\left(1-\frac{x}{k}\right)\right) \partial x \;=\; 4g \left(\lambda \left(1-\frac{x}{k}\right)-1\right)\partial x.

\]

Since the maximum altitude \(h\) is reached when velocity and acceleration *both* vanish, the differential equation may be reduced to \(0 = 4g \left(\lambda \left(1-\frac{h}{k}\right)-1\right)\), in which case \(h = \big(1-\frac{1}{\lambda}\big)k = f\).

Regardless of the approximation used for maximum altitude, the total ascent time is now easily found. Since \(v\) is velocity, Euler used \(v \partial t = \partial x\) to integrate solution \(v\) as a separable differential equation. After integrating, Euler set \(x = h\) and again used the approximation \(e^{-h/b} \approx 0\) to get \(\sqrt{\frac{kh}{\lambda gb}}\).

Maximum speed is not much more difficult. Euler needed the differential of \(v^2\) to vanish, i.e.,

\[

\partial \left((b+f)(1-e^{-x/b})-x\right) \;=\; 0.

\]

Some elementary differential calculus shows that the maximum speed is attained at an altitude of \(x = b\cdot \ln\left(\frac{b+f}{b}\right)\). From here, the speed itself can be seen to equal \(\sqrt{\frac{4\lambda gb}{k}\left(f-b\ln \left(\frac{b+f}{b}\right)\right)}\). Euler again used an approximation, this time \(\frac{b}{f}\approx 0\), to simplify his speed formula to \(2b\sqrt{\frac{\lambda gf}{bk}}\).

##### An Example

As we now see, the notes from Euler’s blackboard gave a set of explicit formulas to calculate (at least approximately) a hot-air balloon’s maximum altitude, ascent time, and maximum speed. One only needs to know the balloon’s radius \(a\) and buoyancy ratio \(\lambda\). From there, the intermediate parameter \(b = \frac{8a}{3\lambda}\) and air column of height \(k = 24,\!000\) feet provide the key to understanding the formulas:

\[ \begin{cases} \text{Maximum altitude} & \;=\; h \;=\; b+f \;=\; b+\Big(1-\frac{1}{\lambda}\Big)k, \\ \text{Ascent time} & \;=\; \sqrt{\frac{kh}{\lambda gb}}, \\ \text{Altitude of maximum speed} & \;=\; b\cdot \ln\left(\frac{b+f}{b}\right) \;=\; b\cdot \ln\left(1+\Big(1-\frac{1}{\lambda}\Big)\frac{k}{b}\right), \\ \text{Maximum speed} & \;=\; 2b\sqrt{\frac{\lambda gf}{bk}} \;=\; 2\sqrt{ gb(\lambda-1)}. \end{cases} \] The text of the article ends with a brief example, in which \(a = 30\) feet and \(\lambda = 5\) (thus making \(b = 16\)). Euler reported a maximum altitude as 19,200 feet, a maximum velocity of 64 feet per second (at an altitude of 112 feet), and an ascent time of 10 minutes and 32 seconds. As Gillispie notes, Euler’s son Johann Albrecht Euler “copied off the equations and sent them to the Academy of Science in Paris. Its *Mémoires* thus had the privilege of inaugurating mathematical analysis of the flight of aircraft with the publication of Euler’s valedictory” [Gillispie 1983, p. 32].

In this reading of Euler’s valedictory, we have generally assumed that his calculation was accurately transcribed, and that the work was correct. However, a closer analysis of Euler’s example reveals some discrepancies that deserve further attention. First of all, the formula he stated for maximum altitude should give 19,216 feet, not 19,000 feet. This is easily explained by the alternative calculation for \(h\) described above, which does not require a solution to the differential equation; so Euler could justifiably have used \(h = \big(1-\frac{1}{\lambda}\big)k = 19,\!200\) feet, as reported here. Second, the maximum speed was reported to occur at an altitude of 112 feet, even though the formula above gives an altitude of 113.45 feet. While the formula will clearly require \(\ln (1201) \approx 7.0909\), approximating the logarithm as 7 would have been entirely reasonable for the time. Of course, this approximation aligns with the maximum altitude of 112 feet recorded in the paper.

A greater difficulty lies with the reported ascent time of 10 minutes and 32 seconds, since his formula appears to give an answer of 10 minutes and 0.25 seconds. What happened to the missing time? Using the alternative value for \(h\) does not help, since this reduces the ascent time to exactly 10 minutes. A partial explanation may lie in the choice of units Euler used in his work. Since his calculations relied on the observations of flights across France, it is likely that when Euler measured distance he was using French units. While we have used the English word *foot* until now, this is somewhat misleading: the French unit (*pied*) was not the same as the English Imperial unit of the same name. Specifically, a pied was defined as 1/6 of the *Toise de l’Académie*, making it about 1.07 English feet. Crucially, this would make Euler’s gravitational constant \(g\) smaller than the typical value for English Imperial feet—and thus would make the ascent time greater. For pieds, \(g\) is very near 15, which gives an ascent time of 10 minutes and 20 seconds—much closer to the reported value. While this value of \(g\) would also cause the maximum speed to change (to about 62 pieds per second), overall this more closely approximates Euler’s own calculations.

In spite of the discrepancies, in a roundabout way these equations allow us to view a sort of poetic coincidence. Let us return to the flight of 19 September 1783, in which a sheep, a duck, and a rooster lifted off from Versailles the day after Euler’s death. Piroux reported that their balloon had a volume of 40,000 cubic feet, which corresponds to a radius of approximately 21.216 feet. At dinner the night before, Euler remarked to Lexell that a value of \(\lambda = 2\) should correspond to a maximum speed of 41 feet per second. If we understand this balloon’s radius to be measured in pieds, and use a value of \(g = 15\) for gravitational acceleration, Euler’s formula produces a maximum velocity of . . . 41.2 pieds per second! So while Euler’s example suffers from some inaccuracies, there is little doubt that his analysis was close to the mark.