During instrument flight training, pilots are usually taught to perform teardrop entries to holds or course reversals by flying a leg angled 30° away from the inbound course for 1 minute before turning inbound. Is this combination of time (1 minute) and angle (30°) correct to arrive exactly on the inbound course when the turn is complete? ![[AIM hold teardrop modified 2.png]] ## Definitions and Abbreviations Throughout this page, the following definitions and abbreviations are used: $v$ - true airspeed, in knots $t$ - time, in seconds Feet will be used for all measurements of distance. Conversion of knots to feet/second will use the following formula: $ ft/s = \frac{6076.115}{3600}v=1.6878v $ # Modeling the Teardrop We will begin by constructing a mathematical model of the teardrop maneuver. To simplify things, we will make a few assumptions: there is **no wind**, the aircraft crosses the fix already established on a track to maintain the desired angle during the teardrop, and instantaneously sets the correct bank angle for a standard rate turn when beginning the turn inbound. ![[teardrop original 1.png]] - $x$ represents the angle selected for the teardrop maneuver - $d$ represents the distance flown prior to beginning the turn inbound - $a$ represents the lateral distance from the inbound course upon beginning the turn inbound - $r$ represents the radius of turn during the turn inbound To arrive exactly on the inbound course when the turn is complete, $a$ must be equal to the lateral distance covered during the **entire** turn inbound. The amount of turn required will be 180° greater than the original teardrop angle, or $x+180°$. Note that $a$ will be slightly **less** than $2r$, as the aircraft is still moving further from the inbound course during the first $x$ degrees of turn: ![[teardrop original 2.png]] - $b$ represents the lateral distance flown away from the inbound course during the first $x$ degrees of the turn inbound The following equation describes a teardrop maneuver where the aircraft arrives exactly on the inbound course upon completing the turn: $ a = 2r-b $ ## Equation for $a$ Since $a$ and $d$ form the legs of a right triangle with the angle $x$ opposite of $a$: $ \frac{a}{d}=\sin(x) $ or $ a=d\sin(x) $ With $1.6878$ used to convert knots to ft/sec, and $d=vt$ $ a=1.6878vt\sin(x) $ ## Equation for $b$ ![[teardrop original 3.png]] To solve for $b$, the first $x$ degrees of the turn inbound can be represented by an isosceles triangle with side lengths $r$ and $c$, and the angle $x$ opposite of $c$. Using the law of cosines (isosceles case): $ c=\sqrt{2r^2(1-\cos(x))} $ The angle between $r$ and $c$ must be $(180°-x)/2$. $b$ is parallel to $r$; $b$ and $c$ form the adjacent and hypotenuse legs of a right triangle (shown in red), so: $ \frac{b}{c}=\cos(\frac{180°-x}{2}) $ $ b=\cos(\frac{180°-x}{2})\sqrt{2r^2(1-\cos(x))} $ ## Solving for $t$ and $x$ The original equation $ a = 2r-b $ can be filled in for $a$ and $b$: $ 1.6878vt\sin(x)=2r-\cos(\frac{180°-x}{2})\sqrt{2r^2(1-\cos(x))} $ Radius can be substituted as a function of velocity using the formula for [[Radius of Turn at Standard Rate]]: $ r = 32.297v $ $ 1.6878vt\sin(x)=64.594v-\cos(\frac{180°-x}{2})\sqrt{2(32.297v)^2(1-\cos(x))} $ Additionally, using the half-angle formulae allows the following substitution: $ \cos(\frac{180°-x}{2})=\sin(\frac{x}{2})=\sqrt{\frac{1-\cos(x)}{2}} $ $ 1.6878vt\sin(x)=64.594v-\sqrt{\frac{1-\cos(x)}{2}}*\sqrt{2(32.297v)^2(1-\cos(x))} $ The following reductions can simplify the equation: $ 1.6878vt\sin(x)=64.594v-32.297v\sqrt{\frac{1-\cos(x)}{2}*2(1-\cos(x))} $ $ 1.6878vt\sin(x)=64.594v-32.297v(1-\cos(x)) $ Both sides of the equation can be divided by $v$. This means that **the relationship between teardrop angle and time is independent of velocity!** $ 1.6878t\sin(x)=64.594-32.297(1-\cos(x)) $ Continuing the reductions: $ 1.6878t\sin(x)=64.594-32.297+32.297\cos(x) $ $ 1.6878t\sin(x)=32.297+32.297\cos(x) $ $ t=\frac{32.297+32.297\cos(x)}{1.6878\sin(x)} $ $ t=\frac{1+\cos(x)}{\sin(x)}*19.13556 $ or $ t=-\tan(\frac{x}{2}+90°)*19.13556 $ rearranging the equation to solve for $x$: $ x=-2\tan^-¹(\frac{t}{19.13556})+180° $ ## Findings The "normal" method for performing a teardrop (30°/1 minute) is close, but not an exact fit for the model described above: - if a 30° angle is used, 71.4 seconds is the correct value for time - if timed for 60 seconds, 35.4° is the correct value for angle # Conclusion While the 30°/1 minute method for performing a teardrop may not be mathematically perfect, it is fairly close. Variables omitted from the model for simplicity tend to favor timing for a slightly shorter period; most pilots reach the predetermined time value, *then* begin to roll into the inbound turn, possibly adding several seconds to establish the appropriate bank angle. Wind, of course, complicates matters further. As with most things in flying, understanding the basic model for the teardrop, and how additional variables (winds, turn roll-in/roll-out, etc) affect the aircraft's performance, will help the pilot make informed decisions and adjustments that work in the real world. ### By Kevin Sakson