**Flying the Perfect Hold**

Keith Thomassen, PhD, CFII

How do you fly the perfect hold? Tracking inbound to the hold, having determined the correct crab angle, roll into a standard rate turn at the hold point, whether right or left. Roll out on the correct heading (what is it?) and fly the outbound leg for the correct amount of time (when?). Finally, roll into a standard rate turn from outbound and roll out at the inbound cab angle. If you’re on the inbound course and it takes 1 min to get to the hold, you did it perfectly.

But note that you needed two pieces of information about the outbound leg; its heading and the time to fly it. That’s all you need. It turns out you can calculate both of these if you know two things; the wind speed as a fraction of your true airspeed, and the direction the wind is blowing.

The fact that just these two numbers are all you need to fly the hold perfectly is the good news. The bad news is that the calculations are not simple, even sitting at home with pen and paper, or calculator and computer. So later on we’ll discuss what to do about that situation. For now, lets talk about the solutions and how to get them.

We’re not talking about what that perfect pattern looks like when drawn on paper, although you could figure out that trajectory. With a computer program and trajectory equations you could calculate where a very small step along course will take you in a very short time (from current position). You use your ground speed vector to determine the change in coordinates at that point. Now take more steps, each from the new position, and with a very large number of very small steps you’ll eventually return to the hold. This same procedure is used to find trajectories as used to write equations for each leg as later described.

In calculus, this small step procedure is called integration (sum up all the steps) over the orbit or along the trajectory. On the straight legs you’re following your ground track, with a velocity vector the sum of the wind and speed vectors. The two integral limits are the start and the end of those legs. On the turns you are drifting along the wind vector and your speed vector is increasing at a constant angular rate, so add both steps (drift and turn) starting at the initial heading and ending at the heading for the next leg.

But in this process you need to know when to start and stop turning, or equivalently, what heading to take on the outbound leg and how long to fly it. So let’s get on that procedure for each of the 4 legs.

**The General Solution**

I was recently made aware of these general solutions in a paper by its author, Leslie Glatt who has graciously shared the paper with me. The solutions given here are his, from that paper. In this article I describe his steps (essentially those above) but leave out many of the details in order to get to the solution (but for a standard pattern; his is non-standard).

Let’s define terms and coordinate systems for right and left patterns. The inbound hold leg is the x-axis and is the reference for all angles in the computations. Those angle solutions are readily converted to magnetic, using the magnetic course for the hold. A standard right pattern measures angles increasing clockwise, while those for a left pattern increase counterclockwise. So there will be a different conversion rule for these two in order to convert computed angles to actual magnetic headings (a compass rose has a clockwise coordinate system).

The solution here is for a standard pattern with the wind coming from the holding side (our reference). It’s easy to use these answers for winds from the non-holding side, and for each of these cases in a left pattern. So lets discuss our reference situation).

The first number needed for the equations later on is the wind parameter w, the ratio of the wind speed W and (true) airspeed S, so that w = W/S. We also need the wind angle ⍺ to the x-axis. With these two parameters everything else can be calculated. But the wind angle in these equations is the angle the wind is coming from, as used in aviation. To be clear, when ⍺ = 0 there is a pure headwind inbound and when ⍺ = +90° there is a pure crosswind from the right (the holding pattern side). For an angle of – 90° it comes from the left.

Let’s describe how the equations are created. Flying around the pattern from the hold point, with coordinates taken as x = 0 and y = 0, the aircraft heading angle ø changes by 360° (all done in the two turns). The outbound leg heading is defined as 𝛳 (which we don’t know) and T is the time you need to fly it (the second unknown). Our inbound heading angle (crab) is σ° and computed from the wind triangle inbound; sin σ = w sin ⍺. The wind angle is positive in the reference case but negative with wind coming from the non-holding side.

The equations we seek come from adding all the accumulated x,y coordinate changes on the legs and then separately forcing the x and y changes to sum to zero (no net change) when you get back to the hold point. That gives us the two equations, one from each of the coordinates, resulting in the solutions for 𝛳 and T.

**The Equation**

Let’s begin with the first turn, flown at constant angular rate (dø/dt = 3° per sec), so that the integration is over the heading angles ø = σ° to 𝛳 °. This determines the end coordinates x_{1} and y_{1}. Now fly the outbound leg for T seconds at its speed along its course at angle 𝛳 giving the changes in x and y from start to end. Increment the starting coordinates of the leg by those changes to get the coordinates x_{2} and y_{2} at the end of the leg. Use the same method on the inbound turn as used for the outbound turn, but now with integration limits ø = 𝛳° to (360 + σ)° to get the changes on this turn. Add them to get x_{3} and y_{3}, the start of the inbound leg.

Now, our equations for x_{3} and y_{3} involve the outbound heading and time, in terms of the wind fraction and angle. But we know that adding x_{3} to the inbound distance L gives zero, the x coordinate at the hold point. Here, L is the distance flown in 1 minute at inbound speed.

This gives us our first equation. The second comes from setting y_{3} = 0 since we must be on the inbound leg there. From the y_{3} = 0 equation, Glatt finds T as a function of sin 𝛳

T = 120 sin σ /[sin ⍬ – sin σ]

He next finds a quadratic equation in the variable cos 𝛳, with coefficients a, b, and c, by also using x_{3} + L = 0. The equation is

(a

^{2}+ b^{2}) cos^{2}𝛳 + 2 bc cos 𝛳 + (c^{2}– a^{2}) = 0a = cos σ – 3w cos ⍺ b = 2 sin σ c = – [cos σ – w cos ⍺] sin σ

You can solve for cos 𝛳 by the quadratic formula, and get

cos 𝛳 = – [ bc + aX]/[a

^{2}+ b^{2}] with X^{2}= a^{2}+ b^{2}– c^{2}

Now you find the inverse (arc cosine) of this to get 𝛳 itself (there are both plus and minus options; here we chose plus). Solving for cos 𝛳 lets you calcuate sin 𝛳 using simple trig, then you can compute 𝛳 and T. Finally, knowing 𝛳 you can determine the outbound crab angle β as the difference from 180°.

Some publications suggest using 3 times the inbound crab for the outbound crab, but that is only an approximation for very small wind percentages. It comes from assuming the same drift on the turns as there is on the inbound (each of these legs around 1 min), so you need to make up all three drifts on the outbound. For small angles, a factor 3 does that. The more general angle ratio R = β/σ is found from our solutions by first solving for 𝛳. The difference between 𝛳 and 180° is β. Now divide by σ. It turns out that R can both increase and decrease from 3 as the wind increases, for different cases as will be seen.

This is our reference solution, but if the crosswind is coming from the non-holding side (⍺ is negative) it’s easy to find out what happens. First, the wind angle is negative, therefore so is inbound crab angle. So the all sine terms in the coefficients above change sign but all cosine terms do not. Consequently the coefficient a does not change sign but b and c both do. Therefore cos 𝛳 is the same, since bc is unchanged, as are a and all squared coefficients in the quadratic. But 𝛳 must change sign since the outbound crab is now into the pattern (the minus sign choice for our cos 𝛳 equation).

Before considering left patterns lets look at reference solutions for special values of wind speed and angle. After that we’ll talk about the left pattern solutions, and about converting all 4 cases to magnetic headings.

**Solution for winds Parallel to Inbound leg**

First, a standard pattern (flown at standard rate of 3 degrees per sec) has 4 one-minute legs, two straight and two half circles. Since it takes the same time to fly the half circle as a straight leg, the path lengths L and πd/2 are the same. So the separation between straight legs is d = 2L/π. With winds along the inbound leg you can easily find an exact formula for T without the above equations. For tailwinds inbound (see below) the ratio of inbound to outbound speeds is simply (1 + w)/(1 – w), and there is a drift h = W/60 in one minute during the outbound turn, which adds to the length of the outbound leg at each end. The inbound length is L= (S + W)/60 and the outbound is L + 2h, giving the pattern shape in Figure 1.

*Figure 1. Holding pattern distortion for pure tailwinds inbound to the hold.*

From these it’s easy to find the outbound time,

T (sec) = 60 (1 + 3w)/(1 - w)

For headwinds inbound, just reverse the signs in the numerator and denominator. The second solution for 𝛳 = 180 is obvious. These times are plotted in Figure 2 for winds with w = 0 to 0.33.

*Figure 2. Outbound leg times for winds parallel to the inbound course.*

These same solutions can be found from the full equations above, but you need to resolve a difficulty in the formula for T since in this parallel wind case the ratio of sin 𝛳 and sin σ are both are zero, so T is indeterminate. The resolution by Glatt is explained in his article, but for here we only need to know his result is the same as above.

Note here that with inbound headwinds there is no outbound straight section (T = 0) when w = 1/3. This you simply start turning at the hold and continue through 360° to the start of the inbound leg. So for winds that strong, to have an outbound leg you need an inbound time more than 1 min, or you need to fly faster so that w < 1/3.

**Solution for winds Perpendicular to Inbound leg**

Another special case is drift perpendicular to the inbound leg, with winds from ⍺ = 90° (so ⍬ should be < 180°). Now, sin ⍺ = 1, and cos ⍺ = 0, so that the coefficients become

a = cos σ b = 2 sin σ c = – cos σ sin σ

and the inbound crab angle is found from sin σ = w, in which case cos^{2} σ = 1 – w^{2}. The solutions can be found by substituting these coefficients into the general formula. But Glatt has found direct analytic forms in this case (for cos 𝛳 and T) instead of substituting these a, b, and c coefficients into the quadratic for T and cos 𝛳; he finds

T = 60(1 + 3w

^{2})/(1 – w^{2}) and cos 𝛳 = – (1 – w^{2})^{3/2}(1 +3w^{2})

These are easily solved in an Excel spreadsheet and graphed. Here, the answers for a range of w from 0 to 0.35 in steps of 0.05 are shown in Figure 3. Time T goes from 60 s to 93 s and the heading goes from 180° down to 127°. If winds are from the opposite direct (-⍺) the changes are simply that σ and 𝛳 change sign. Whereas the outbound heading was less than 180°, it is now the negative of that, which means there is now more than a 180° turn outbound. You have the same size crab angle β, now in a different direction, and still into the wind.

*Figure 3. Outbound headings and times for increasing cross winds. Subtract this heading from 180° for the outbound crab angle β (0 to 53°).*

We can compute the R factor from the outbound crab using the heading answers, then divide by inbound crab which goes from 0 to 20.5° over this ranges. We see in Figure 4 that the factor decreases from 3 to 2.59 as in the figure below.

*Figure 4. Ratio of outbound crab angle to inbound crab angle for increasing cross winds.*

The pear shaped pattern for crosswinds generally looks like the one in Figure 5, for a right hand pattern. If this were a left hand pattern it would be correct for a wind in the reverse direction. The larger radius turn is always on the “downwind turn”, and of course the shorter radius is upwind.

*Figure 5. Holding pattern distortion for perpendicular winds.*

**Solution for mixed winds **

To solve for arbitrary winds and angles you can use our reference formula, but let’s just solve and illustrate a case of winds coming from ⍺ = 45° (quartering winds with inbound headwind). We again expect that 𝛳 < 180°. Both cos ⍺ and sin ⍺ are 1/√2, therefore sin σ = w/√2 and cos σ = √(1 – w^{2}/2). Solving for 𝛳 and T using these values gives the heading and time over the range of fractional winds in Figure 6.

*Figure 6. Heading and Time for a quartering wind. Figure 7. Crab angle ratio increases. *

Heading drops from 180° to 108° (crab angle from 0 to 72°) with times dropping from 60s to 36s at w = ¼, then rising to 39 s. We also find that R now increases from 3 to 5.3 over this range, as in Figure 7. Alarming indeed!

From the general solution this R factor can be calculated for any combination of w and ⍺, as in Figure 8 taken with permission from the paper by Glatt. For any wind fraction you can see what wind angle gives the largest crab. You can also see why you should avoid wind speeds over ¼, where peak crab is already about 50°. In this figure, his vw is my parameter w, and the Outbound Wind Correction Angle (OWCA) is my angle β. The peaks in the curves are shifted towards wind angles that are mostly headwinds inbound. As wind increases, the wind angle for the peak shifts closer to zero (closer to the inbound direction).

From a graph of the R factor in Glatt’s paper, at about 80° he finds R = 3 for all winds, and it decreases below that as ⍺ increases. But, below 80° the R factor increases, to values above 10 near very low wind angles (but bear in mind that the actual inbound crab angles get very small there).

*Figure 8. Crab angle **β vs wind angle **a for various wind fractions w courtesy of Leslie Glatt*

**Left patterns and conversion to magnetic.**

When the wind is coming into the hold the solutions are the same for right and left patterns. But you convert those to magnetic headings differently. When the winds reverse directions in the standard hold you simply change all the angle signs. The same is true for left patterns.

To convert the angle solutions to magnetic, there is one rule for right turns and another for left, since the angle coordinates go in opposite directions. They go in the same direction as a compass rose for right patterns, so you add the inbound hold course angle to the solutions for crabs and headings to get their magnetic directions. This may in some cases give you an angle greater than 360°, but then you just subtract 360° to get answers in the conventional range between 0 and 360°. For left turns you subtract the angle of the holding course from the results to get their magnetic direction. Now you may need to add 360° to put it in the conventional range.

**Now What? **

How does this help us the next time we fly a holding pattern? First, just looking at the graphs and values gives some sense of the corrections you can have from a no-wind pattern, and at what wind fractions do these become large. Curves like those in figure 8 are very useful to keep in mind. Maybe you want to increase your holding speed for bigger winds, and at what wind speed would you start doing that? What kinds of corrections are there for winds mostly parallel or mostly in the perpendicular direction to the hold. Since you have a lot to do while you’re flying the plane, it’s the knowledge you bring to the flight that will help you, and not any calculations you would do on the spot.

Glatt has given many more examples, and plotted patterns for them. He also allows for different bank angles and for a longer inbound time (90 sec is required above 14,000 ft), or for arbitrary inbound distances. All these results, plus a full derivation of the general solutions are contained in his paper previously cited in this article. He has even discovered a new type of hold for higher winds than plotted here. They are characterized by ⍬ < 90° (yes, that’s possible!). You fly outbound roughly perpendicular to the inbound for values of ⍬ near 90°. Picture a more than 270° inbound turn! These patterns require wind fractions greater than 1/3 and can be avoided if you keep your airspeed up. To avoid this, and lessen major effects keep w < ¼ (or stay home).

Ideally, having an iPad app for this could be workable. You would put in the inbound heading, left or right pattern, then the wind speed and direction and your airspeed. It would compute the inbound and outbound crab angles translated to actual aircraft headings to fly, and give the outbound time.

Come to think about it, some GPS navigators today have all of that information if it gets a wind input. Wind comes from air data, which also gives your airspeed. It knows your heading, and knows the inbound for any hold in the database and the turn direction. So you wouldn’t have to provide any input. If software were written for this it could give an announcement in the outbound turn of the next (outbound) heading, and give a countdown time on the outbound leg to the turn point. From there you simply fly standard rate until intercepting the inbound course. Your GPS likely makes announcements now for turns, holding type and entry, procedure turns, etc.

Dream on, but this may not be that difficult. Whether it will someday come into your GPS or be available as an App, who can say. In the meantime, hopefully, this information gives a little more insight into holds.