What is the derivation of 15 and 11.3 used for estimating pivotal altitude?
A possibly surprising result is even airplanes as different as a humble Cessna 152 and a Boeing 747, if able to fly the same speed in the same conditions, would use the same pivotal altitude. The airplane’s velocity and acceleration due to gravity are the only factors in determining pivotal altitude. Detailed derivations by John S. Denker and by ERAUSpecialVFR (13:57 YouTube) are included below.
The exact formula for height above the pylon is
$$h = \frac{v_{air}\cdot v_{gnd}}{g}$$
where $v_{air}$ and $v_{gnd}$ are velocities relative to the air and ground, and $g$ is acceleration due to Earth’s gravity. This shows that the common approximation of squaring groundspeed is the calm-day special case.
Assuming we want to know $h$ in feet, we need to connect the building blocks, namely the units of measure, appropriately. Given that $g$ is 32.17405 ft/s² (that is speeding up by 32-ish feet per second every second), compatible velocity will also be denominated in ft/s. To see why, you can think of the units as canceling out, as in
$$\frac{\frac{\textrm{ft}^2}{\textrm{s}^2}\equiv v^2}{\frac{\textrm{ft}}{\textrm{s}^2}\equiv g} \Rightarrow \frac{\textrm{ft}^2}{\textrm{s}^2} \cdot \frac{\textrm{s}^2}{\textrm{ft}} \Rightarrow \textrm{ft}$$
At least in the airplanes I fly, the airspeed indicator displays knots or miles per hour. The conversion factor from knots to feet per second is $\frac{6{,}076}{3{,}600}$ because there are 6,076 feet in a nautical mile and 3,600 seconds per hour. For statue miles to feet per second, the factor is $\frac{5{,}280}{3{,}600}$. Remember that the formula for pivotal altitude has two velocity factors, so we must square the conversion factor.
$$d_{mph} = 32.17405\ \textrm{ft}/\textrm{s}^2 \cdot \Biggl(\frac{3{,}600\ \textrm{s/hr}}{5{,}280\ \textrm{ft/SM}}\Biggr)^2 \approx 14.9569\ \textrm{mph}^2/\textrm{ft}$$
and
$$d_{kts} = 32.17405\ \textrm{ft}/\textrm{s}^2 \cdot \Biggl(\frac{3{,}600\ \textrm{s/hr}}{6{,}076\ \textrm{ft/NM}}\Biggr)^2 \approx 11.2947\ \textrm{knots}^2/\textrm{ft}$$
Then performing the dimensional analysis to see that the units of measure fit in the knots case, for example:
$$\frac{\textrm{NM}^2}{\textrm{hr}^2} \cdot \frac{\frac{\textrm{ft}^2}{\textrm{NM}^2}}{\frac{\textrm{s}^2}{\textrm{hr}^2}} \cdot \frac{\textrm{s}^2}{\textrm{ft}} \Rightarrow \textrm{ft}$$
That is, the squared airspeed converted to $\frac{\textrm{ft}^2}{\textrm{s}^2}$ divided by the acceleration due to gravity yields a quantity in feet.
Thinking of the magic numbers in the denominator may show more clearly what’s happening. In the case of knots, $\frac{1}{11.3}$ is around 0.0885 feet per knots squared. This means approximately the same as 9 feet per knots gained or lost, per 100 knots (because 9 ≈ 0.0885 × 100). Likewise for statute miles, the pivotal altitude changes by about 7 feet per mph airspeed change, per 100 mph the airplane is traveling. In either case, given two airplanes where one is flying twice as fast as another, a unit of airspeed gained for the faster will have twice the impact on its pivotal altitude as compared with its slower counterpart.


5. Derivation of the Nifty Formula


We wish to derive the nifty formula for the altitude required during a turn on pylon in the presence of wind. We will be using the following quantities:



$R$, horizontal position vector, from pylon to aircraft
$h$, height above the base of the pylon
$V_{Gnd}$, velocity relative to the ground
$V_{Air}$, velocity relative to the air
$W$, wind vector = constant, independent of time
$a$, acceleration vector = derivative of $V_{Gnd}$

or (equivalently) = derivative of $V_{Air}$


$g$, acceleration of gravity



Assumption: We assume constant $\lvert V_{Air}\rvert$ i.e. constant airspeed.



The velocity relative to the air, $V_{Air}$, is perpendicular to $R$. That is, $$V_{Air} \cdot R = 0 \tag{4}\label{eq4}$$ This is required by the rules of the game. The heading is perpendicular to $R$ because the wing is pointing at the pylon, and the airspeed vector is parallel with the heading because we require coordinated flight (zero slip).




The acceleration $a$ is antiparallel to $R$. This is required by the rules of the game, since the wing is pointing at the pylon. That means the horizontal component of lift is pointing at the pylon. Meanwhile, all other forces add up to zero in accordance with the constant-airspeed assumption. Because $a$ is antiparallel to $R$, we have:



$$a \cdot R = −\lvert a\rvert \lvert R\rvert \tag{5}\label{eq5}$$



We will be interested in what happens to all these quantities after some time $\Delta t$ has passed. Time-stepping the equations of motion gives us: $$
\begin{align}
\textrm{new}\ V_{Air} &amp; = \textrm{old}\ V_{Air} + a\Delta t \\
\textrm{new}\ R &amp; = \textrm{old}\ R + V_{Gnd} \Delta t
\end{align} \tag{6}\label{eq6}
$$




Combining equation $\eqref{eq6}$ with equation $\eqref{eq4}$ gives us the time-delayed version of equation $\eqref{eq4}$:$$(V_{Air} + a \Delta t) \cdot (R + V_{Gnd} \Delta t) = 0 \tag{7}\label{eq7}$$




Multiplying out equation $\eqref{eq7}$ gives us: $$0 = V_{Air}\cdot R + a\cdot R\Delta t + V_{Gnd}\cdot V_{Air}\Delta t + a \cdot V_{Gnd}(\Delta t)^2 \tag{8}\label{eq8}$$ The first term vanishes in accordance with equation $\eqref{eq4}$. The second term reduces to $\lvert a\rvert \lvert R\rvert \Delta t$ because of equation $\eqref{eq5}$. The fourth term is negligible in the limit of small $\Delta t$.




Rearranging the remaining parts of equation $\eqref{eq8}$ gives $$\lvert a\rvert = \frac{V_{Air} \cdot V_{Gnd}}{\lvert R\rvert} \tag{9}\label{eq9}$$ This tells us the magnitude of the acceleration. The direction is antiparallel to $R$ as mentioned [elsewhere]. So now $a$ is fully determined, since we know its direction and magnitude.




As usual, when the aircraft is properly banked toward the pylon, the geometry of the situation is shown in figure 3 [shown below]. In this geometry, the law of similar triangles tells us $$\lvert a\rvert = \frac{h}{\lvert R\rvert} g \tag{10}\label{eq10}$$ 




Combining equation $\eqref{eq9}$ and equation $\eqref{eq10}$ gives us the nifty expression for the required height at any point during a turn on a pylon:$$h = \frac{V_{Air} \cdot V_{Gnd}}{g}\tag{11}\label{eq11}$$ You can readily verify that this reduces to the conventional expression for the pivotal altitude in the no-wind case.




Physics of Flight - Pivotal Altitude - Lesson 6 by ERAUSpecialVFR



Figure created with Khan Academy


$\sum F_y = 0 \Rightarrow L \cos\theta = mg$


$\sum F_x = ma_c \Rightarrow L \sin\theta = \frac{mv^2}{R}$ (centripetal force)


$\frac{L\sin\theta}{L\cos\theta} = \frac{\frac{mv^2}{R}}{mg} \Rightarrow \tan\theta = \frac{v^2}{Rg}$


$\tan\theta = \frac{h}{R} = \frac{v^2}{Rg} \Rightarrow h = \frac{v^2}{g}$


$\therefore PA = \frac{GS^2}{g}$


$\begin{aligned}
  PA = \frac{GS^2}{g} &amp;\approx GS^2\cdot\frac{\left[\left(1\frac{\textrm{nm}}{\textrm{hr}}\right)\left(\frac{1.15\ \textrm{sm}}{1\ \textrm{nm}}\right)\left(\frac{5{,}280\ \textrm{ft}}{1\ \textrm{sm}}\right)\left(\frac{1\ \textrm{hr}}{3{,}600\ \textrm{s}}\right)\right]^2}{32.17\ \textrm{ft}/\textrm{s}^2}{} \\
                      &amp;\approx GS^2\cdot\frac{(1.687)^2}{32.17} \approx GS^2\cdot\frac{2.845}{32.17} \approx \frac{GS^2}{11.3}
  \end{aligned}$