CONTENTS

Working out - Endurance of the fight

WORKING OUT:

1)- We construct the true route of 252º with the upper side of the page as true north.  On the line of the true route we put two arrows in the flight direction.

2)- The target is located somewhere on the route and is indicated by means of a little square (see drawing: point D).

We put this little square on the most acceptable spot on our drawing.  We now know that the airplane is located somewhere above this route, but that it does not fly in the direction of the target.

Indeed, imagine that the airplane is at a distance of 1 hour from the target.  If it continues to fly in that direction, it will locate 19 km wind downward from that target after 1 hour, since an imaginary cloud above the target would be located there under influence of the wind.

Consequently, we must find a point that will be located above the target after 1 hour.

That point is situated at 1 hour windward (scale: 1mm = 1 km).

3)- We will construct the wind direction in that target and we put an imaginary cloud, indicated by means of a little triangle, somewhere on the line.

This cloud will reach our target after 1 hour.  We know that the wind speed equals 19 km/h.  Therefore we will draw our imaginary cloud 19 mm wind upward on the wind line.

A- Now we know that the position of the airplane, one hour before the arrival, should be located at a point on the route and that the airplane must be located above that line at every moment of the flight.

B- Our target should be located at a distance of 120 km vis-a-vis the cloud, since we will have reached this point after one hour.

4)- We draw a circle with the cloud on the wind line as center.

The circle’s radius is 120 mm.  According to our scale, 120 mm = 120 km, which is the airplane’s airspeed.  The circle’s intersection with the route is the point that is situated at one hour from our target.  This point shows the airplane’s position one hour before arrival at the target.

5)-The following calculation is the connection of the cloud on the wind line and with the intersection on the route.  This will result in a new line in the speed triangle.  This line is the course to be followed by the airplane in order to reach the assumed target.  This course is 244°.

6)- If we now measure the length between the point of arrival and the intersection on the route, we obtain a distance of 113 mm, or 113 km, since 1mm = 1 km.  This means that the groundspeed is 113 km/h (see drawing).

Although we have created the speed triangle, the problem is not solved yet.  We will only now see what is really required.  Indeed, no true course, but a compass course was required.

The transition from true course to compass course requires VARIATION and DEVIATION.

Variation = -5°.  Result: 244° + 5° = 249° MAGNETIC COURSE.

So the MAGNETIC COURSE is 249°.  Now we will check the DEVIAGRAM and see that the DEVIATION with a MAGNETIC COURSE of 249° equals +7°.

We determine the direction of the COMPASS NORTH (+ 7° to the right of the MAGNETIC NORTH) and measure the angle between the COMPASS NORTH and the COURSE.

Result: 249° - 7° = 242°.  This means that the compass course for the outward flight is 242°.

ENDURANCE OF THE FLIGHT:

The GROUNDSPEED that resulted from our speed triangle is not requested directly, but we need it to calculate the endurance of the flight between Antwerp and Ghent.

This GROUNDSPEED was 113 km/h.

If the distance between Antwerp and Ghent was exactly 113 km, the endurance of the flight would be exactly 1 hours.

If the distance between Antwerp and Ghent was exactly 1 km, the endurance would be 60 :113.

However, since the distance between Antwerp and Ghent is 56 km, we will use the famous rule of three.

RESULT:

The fourth question was the compass course for the return flight.

This requires another speed triangle.

The airplane’s airspeed, as well as the wind direction and speed keep the same value as for the outward trip.

The true route will be 180° more or less than the true route.  Therefore we will combine the new speed triangle with the existing one.

We extend the existing ROUTE in the direction of the OUTWARD FLIGHT.  From the existing cloud we make a new circle on the route, resulting in the true ROUTE and the OUTWARD GROUNDSPEED.  We then convert the data obtained in the usual way, i.e. with VARIATION, DEVIATION, etc.

The fifth and last question was the “POINT OF NO-RETURN”.

We will first calculate the TIME POINT OF NO-RETURN via the following formula:

• The ENDURANCE is 4 hours or 240 minutes.
• The OUTWARD GROUNDSPEED is 113 km/h (first speed triangle)
• The RETURN GROUNDSPEED is 126 km/h (second speed triangle)

We will reach the POINT OF NO-RETURN after 126 minutes.

We can locate this point by using the rule of three.

If we reached the P.N.R. after 60 minutes, the airplane would have flown 113 km.

If we reached the P.N.R. after 1 minute, the airplane would have flown 113 km : 60.

However, we reached the P.N.R. after 126, which means that the airplane will be at a distance of: