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Pivotal Altitude Explained: Teaching eights on pylons the easy way at AOPA repeats a common formula for pivotal altitude.

To estimate pivotal altitude, square the groundspeed and then divide by 15 if you use mph, or 11.3 if you prefer knots.

Where do the 15 and 11.3 magic numbers come from?

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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.

We are ultimately chasing the denominator, so use the reciprocals of the above conversion factors to get

$$d_{mph} = 32.17405\ \textrm{ft}/\textrm{s}^2 \cdot \Biggl(\frac{3{,}600\ \textrm{s/hr}}{5{,}280\ \textrm{ft/SM}}\Biggr)^2 \approx 14.9569\ \textrm{mph}^2/\textrm{ft}$$

and

$$d_{kts} = 32.17405\ \textrm{ft}/\textrm{s}^2 \cdot \Biggl(\frac{3{,}600\ \textrm{s/hr}}{6{,}076\ \textrm{ft/NM}}\Biggr)^2 \approx 11.2947\ \textrm{knots}^2/\textrm{ft}$$

Take reciprocals to gain clearer insight on 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.

  1. 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).

  2. 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:

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

  1. 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: $$ \begin{align} \textrm{new}\ V_{Air} & = \textrm{old}\ V_{Air} + a\Delta t \\ \textrm{new}\ R & = \textrm{old}\ R + V_{Gnd} \Delta t \end{align} \tag{6}\label{eq6} $$

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

  3. 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 $\Delta t$.

  4. 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.

  5. 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 $$\lvert a\rvert = \frac{h}{\lvert R\rvert} g \tag{10}\label{eq10}$$ Figure 3

  6. 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

Pivotal altitude diagram
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)

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

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

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

$\begin{aligned} PA = \frac{GS^2}{g} &= \frac{\left[\left(1\frac{\textrm{nm}}{\textrm{hr}}\right)\left(\frac{1.15\ \textrm{sm}}{1\ \textrm{nm}}\right)\left(\frac{5{,}280\ \textrm{ft}}{1\ \textrm{sm}}\right)\left(\frac{1\ \textrm{hr}}{3{,}600\ \textrm{s}}\right)\right]^2}{32.2\ \textrm{ft}/\textrm{s}^2}{} \\ &= \frac{(1.68)^2}{32.2} = \frac{2.845}{32.2} = \frac{1}{11.3} \\ &\Rightarrow PA = \frac{GS^2}{11.3} \end{aligned}$

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  • $\begingroup$ If the Vair is constant and there is significant wind at altitude, Vground will vary throughout the turn. So... is the approved solution to hold bank constant & vary altitude throughout the turn, or to hold altitude constant & vary the bank angle to adjust for wind? $\endgroup$ – Ralph J Dec 29 '18 at 18:20
  • $\begingroup$ @RalphJ At any given moment, the airplane’s $V_{air}$ and $V_{gnd}$ will combine to yield a particular pivotal altitude, so the answer is to hold constant bank and vary altitude because $h$ varies as a function of the velocities, not bank. $\endgroup$ – Greg Bacon Dec 29 '18 at 18:57
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After you have an understanding of the base formula to calculate your pivotal altitude and the units used, which is:

PA = V^2/g

That is...

V: tangential velocity in knots (AKA NM/HR)

g: the acceleration of gravity in feet per second^2

PA (or could just be h): pivotal altitude height in feet


It is a given that the output unit of distance must be feet because feet is the unit used on the altimeter (or at least every one I've ever used) which makes any other unit more or less useless to a pilot wanting to fly pivotal altitude based on the altimeter in feet.

The problem that the 11.3 conversion is solving is correcting, V in NM/HR to FT/S without having to write out the entire conversion each time and it bundles that conversion with division of V^2 by acceleration of gravity.

I wrote this answer despite the age of the question and thorough first answer because I feel that there is a missed point:

11.3 is the coversion of NM/HR to FT/S twice multiplied by the acceleration of gravity in FT/S.

The "long" way of writing this formula (after cancelling out tangent theta and mass leaving PA = V^2/g) is:

* Take V^2 and split it to V * V.

PA = V * V / g

I need to convert each V to FT/S since we will use NM/HR from the airplane's readings, so the expanded version is to write the conversions twice.

((V NM/HR * (6076.12 FT/NM / 3600 S/HR)) * (V NM/HR * (6076.12 FT/NM / 3600 S/HR))) / g

Well, that's a mouth full, so it can be re-written as:

(V NM/HR * (6076.12 FT/NM / 3600 S/HR))^2 / g

If it can be grouped by squaring, then you can split it squared, so we would have:

V^2 * (6076.12/3600)^2 / g

Since the conversion factor of 6076.12ft/3600S is a stand-alone fraction, you can take the reciprocal to the denominator (AKA flip it and move it to the bottom).

V^2 / ((3600/6076.12)^2 * g)

3600/6076.12 = 0.59248... which squares to 0.35104... (S^2/FT^2)

Since the decimal result of the squared conversion factor is now in the denominator with g, it can be multiplied against g to get a single number.

0.35104 * 32.2 Ft/S^2 (AKA g) = 11.30338...

The end result is a number that effectively bundles the division of V^2 by gravity with the double conversion factor needed of "one per velocity unit" making the output of KTS^2/11.3 be pivotal altitude in feet.

V^2/11.3 = PA (in feet)


To understand V^2/15 for MPH simply replace 6076.12 ft with 5280 ft in order to convert in statute miles instead of nautical miles. Otherwise, the math and logic is identical.

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