# How can I correct the heading for this navigation problem?

I'm a pilot trainee in Hungary and just encountered a navigation question problem which is from the PPL Training Questions book. Would be glad if someone can briefly explain how to do this typical question which I come across a lot.

An aeroplane has a heading of 090°. The distance which has to be flown is 90 NM. After 45 NM the aeroplane is 4.5 NM north of the planned flight path. What is the corrected heading to reach the arrival aerodrome directly?

a) 6° to the right,

b) 9° to the right,

c) 18° to the right

d) 12° to the right

• You must understand the "1 in 60 rule". It will be in your exams and is very useful if you are flying without a GPS.
– Ben
May 15, 2019 at 2:43

## 3 Answers

The correct answer is 12 degrees. The problem is as follows:

To determine wind correction use an E6-b or similar.

Set distance off course over distance already flown. In this case, $$\frac{4.5}{45}$$. The degrees to eliminate wind drift will be under the rate arrow, in this case: 6.

Then, determine how much additional correction is needed to rejoin. In this case, $$\frac{4.5}{45}$$ again. You must add 6 more degrees to the first 6 to rejoin.

The total (therefore the answer) is 12.

• your answer is excellent, but we may wish to clarify "heading" as the direction the nose and directional indicator (little airplane on instrument panel) is pointing. With cross wind, "heading" of 090 will produce ground track of 084. The pilot must "point" 096 to aim at destination plus 6 degrees to the right (102) to track in that direction. Little nitpick, your thoughts on this are appreciated. +1 May 14, 2019 at 23:14
• It's a good comment especially in this day and age where handheld GPS users are used to seeing "heading" used to indicate the direction of ground track if the device has no internal compass. May 15, 2019 at 16:49

Not familiar with the aviation way, but geometrically: After 45 NM, wind pushed you 4.5 NM aside. Over the next 45 NM, you need to be 9 NM to the other side. (4.5 to compensate the side wind to come, 4.5 to go back over what you already lost).

Arctangent of 9/45 is quite close to 12 degrees.

This only holds if the perpendicular component is negligible (at 1/10 it is).

I did arc sin 4.5/45 and 6 degrees to the right was the closest answer. This is assuming no cross wind, just navigational drift. Arc sin was made from a triangle of 4.5 to correct and remaining distance of around 45 NM.

However, you may ask to specify whether or not this is a cross wind issue. If it is, the effect of wind on the remaining track would have to be calculated from the point you were at and the new heading to your destination.

If the cause of the drift was unknown, 12 degrees right would also make sense. But in real life this is where you would start looking for landmarks such as roads, terrain, lakes etc. to help find your destination.

EDIT: correction without wind: point nose to 096 degrees (this is the direction to your destination) correction with wind: point nose to 102 degrees

We can see how this is a confusing question, as the CORRECTION in both cases is 6 degrees to the right. Because it is highly doubtful anyone would fly 084 the whole time before noticing the error, the most probable cause is wind drift. The correction is 12 degrees right from ORIGINAL heading and/or 6 degrees right from the NEW heading.

Even if I got this "wrong" on a test, it would be something remembered and learned from.

• This is why it is a good idea to have checkpoints along the way to stay on course. May 14, 2019 at 19:33
• In the aviation context I think there is a presumption here that the deviation was caused by a wind which is assumed to continue for the rest of the flight, but you do have point. May 15, 2019 at 16:51