**Tips and Tricks #24Drift in Holding Patterns**

Keith Thomassen, PhD, CFII

One nice feature found in many modern GPS navigators is the capability to create holding patterns at arbitrary waypoints. It would seem easy to properly track such a holding pattern: simply follow the magenta line. But whether it's a standard pattern with a one-minute inbound leg or the four-mile-leg pattern now common on course reversals on an approach, winds come into play. With winds affecting your track, and observing standard rate turns, the patterns become distorted. Proper patterns are no longer simple race tracks, but are elongated or foreshortened on the outbound leg for parallel winds, and pear shaped for crosswinds. So, let's examine what happens in both of these cases and determine the proper way to fly them.

First, a standard pattern 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.

If there is a parallel wind as shown in the figure above, the outbound leg is stretched. While flying the outbound half circle the wind will cause the aircraft to drift a distance h in one min. On the outbound arc you will clearly fall short by h of reaching the standard point opposite the hold. To start the inbound arc you have to fly past the opposite point by the distance h since you will be blown inbound by that amount on that turn, and must therefore begin late.

So the outbound leg is longer; length = L + 2h. If you convert the winds in mph to miles/min (divide by 60) then that number is the distance h in miles. A 15 mph wind is ¼ mi/min so h = ¼ mile. To keep the standard inbound leg to 1 min, and to fly standard rate turns, you have to increase time on the outbound leg. If your airspeed is S and the wind speed is W, your ground speed is S + W inbound, and S - W outbound. The ratio of these two [(S + W)/(S - W) min] is the time to fly the L portion of the outbound leg. This must be corrected by the factor 1 + 2h/L for the extra distance. Since inbound takes 1 min, you can calculate L from that time and the speed S + W.

If you write the Wind as a fraction p of the airspeed, the outbound time in minutes is simply (1 + 3p)/(1 - p). For a headwind inbound just reverse each sign, (1 - 3p)/1 + p). Note that for 10% winds (p = 0.1) the outbound times are 1.44 min or 0.64 min. These times in seconds are plotted in the graphs here, so you could look it up quickly if you keep this in the plane.

Now consider the crosswind component, which we take to be blowing across the pattern towards the outbound side as below. The ground track will now be pear shaped with a large bulge on the outbound arc and a smaller bulge on the inbound arc. Note that this is our track around the course, not our heading.

The inbound crab angle to track the given inbound course is α, and you will determine that by reading your heading while holding course inbound. A good way to do that is let your autopilot track the GPS inbound leg. Having a magnetometer with a good (slaved) heading reading also helps.

Now, with airspeed S the inbound crosswind is S tan α, and accumulates a drift over one min of S/60 tan α. With a 90 knot airspeed, and for α = 5° (tan 5° = 0.0875) we accumulate a cross-drift of 0.14 nm. In general, whatever the cross-drift correction is on the inbound leg, the total cross drift is 4 times that because the pattern is 4 min (4 times as long as the inbound leg). On the other 3 legs you have to make up the other 3 parts of that.

Let's look at drift on the arcs. On the outbound arc there is a crosswind from the left up to the 90° point, while we turned α° + 90°. On the rest of the arc there is a crosswind from the right and we turn 90° + β°. Flying the inbound arc, turning 90° - β° there is a crosswind from the right, and on the remainder, turning 90° - α° there is a wind from the left. So adding up the arc lengths having a right crosswind, the β° portions cancel out, as do the α° portions on the other arcs, so there is no net drift. [Note that the outbound arc is 180° + α° + β°, while the inbound arc is 180° - α° - β°, so the times are in that ratio].

Consequently, we're left with correcting the other 3 parts of the drift on the outbound leg, which we also assume to be about 1 min. The question is, what is the outbound crab angle β° to do that? Since the drift on that leg is s tan β° we require tan β = 3 tan α° I call this the tangent formula for drift correction. Here, s is airspeed in mi/min.

It is often stated (incorrectly) that the outbound crab angle should be 3 times that of the inbound angle. In fact, it's the tangent that must be 3 times larger than that of the inbound tangent. While the rule is nearly true for small angles (the tangent of small angles is equal to the angle itself, in radians), it is not true for larger angles. Plotted below is that factor as a function of the inbound crab angle. As the crab angles increase, the multiplying factor decreases. At 20° inbound crab, the factor is down to less than 2.4.

Finally, to fly a hold with wind at an arbitrary angle, break the problem in two parts --- for the parallel and the cross winds. Then apply this combination of corrections. Lengthen or shorten the outbound leg for tailwind or headwind inbounds, and apply the tangent formula for the outbound crab. There will be an error in the outbound crab leg since it will now be flown either more or less than 1 min. When the time is more than 1 min, you have longer to correct the drift so you should reduce β so that tan β is reduced by that factor. As a first cut, reduce β by that factor. This is certainly more than you could do while flying a hold, but you understand the spirit of this last correction. Keeping these two graphs in the aircraft would help you a lot in hitting that 1 min mark for the inbound leg.