Table of Contents

ASA's CX-2 Pathfinder is an electronic flight computer and can be used in place of the E6-B. This aviation computer can solve all flight planning problems, as well as perform standard mathematical calculations.

Finding True Course, Time, Rate, Distance, and Fuel

True course is expressed as an angle between the course line and true (geographic) north. Lines of longitude (meridians) designate the direction of north-south at any point on the surface of the Earth and converge at the poles. Because meridians converge toward the poles, true course measurement should be taken at a meridian near the midpoint of the course rather than at the point of departure.

The flight computer can be used to solve problems of time, rate and distance. When two factors are known, the third can be found using the proportion:

Rate (speed) = Distance ÷ Time

Problem:

Find the time en route and fuel consumption based on the following information:

Wind..................................175° at 20 knots

Distance............................135 NM

True course.......................075°

True airspeed....................80 knots

Fuel consumption..............105.0 pounds/hour

Solution using the E6-B:

1. Using the wind face side of the E6-B computer, set the true wind direction (175°) under the true index.

2. Place a wind dot 20 units (wind speed) directly above the grommet.

3. Rotate the plotting disk to set the true course (075°) under the true index.

4. Adjust the sliding grid so that the TAS arc (80 knots) is under the wind dot. Note that the wind dot is 15° right of centerline, so the Wind Correction Angle (WCA) = 15°R.

5. Read the ground speed under the grommet: 81.0 knots.

6. a. Determine the time en route, using the formula:

Distance ÷ Ground Speed = Time

135 ÷ 81 = 1.67 hour

b. The E6-B may also be used to find the time en route:

i. Set the 60 (speed) index under 81 knots (outer scale).

ii. Under 135 NM (outer scale) read the time of 100 minutes or 1 hour 40 minutes (inner scale).

7. Find the amount of fuel consumed, using the formula:

Time x Fuel Consumption Rate = Total Consumed

1.67 hours x 105.0 pounds/hour = 175.35 pounds

The E6-B may also be used to find the amount of fuel consumed:

a. Set the 60 (speed) index under 105.0 pounds/hour.

b. Over 100 minutes or 1 hour 40 minutes (inner scale) read 175 pounds required (outer scale).

Solution using the CX-2:

1. Enter the Hdg/GS function to find the ground speed.

Wind direction......................175°

Wind speed.........................20 knots

Course.................................075°

True airspeed......................80 knots

Ground speed.....................81.0 knots

2. Enter the Leg Time function to find the time en route.

Distance.................................135 NM

Ground speed........................81.0 knots

Time......................................1:39:59

3. Enter the Fuel Burn function to find the amount of fuel consumed.

Time.......................................1:39:59

Fuel burn rate.........................105 pounds/hour

Fuel consumed.......................175.0 pounds

Problem:

Based on the following information, determine the approximate time, fuel consumed, compass headings, and distance traveled during the descent to the airport.

Cruising Altitude..........................10,500 feet

Airport elevation...........................1,700 feet

Descent to....................................1,000 feet AGL

Rate of descent............................600 ft/min

Average true airspee...................135 knots

True course.................................263°

Average wind velocity.................330° at 30 knots

Variation......................................7° east

Deviation.....................................+3°

Average fuel consumption..........11.5 gal/hr

Solution using the E6-B:

1. Find the time en route. The time to descend is based on the vertical distance from the cruising altitude of 10,500 feet down to the lower altitude of 2,700 feet MSL (1,000 feet AGL + 1,700 feet MSL).

10,500 feet cruising altitude

- 2,700 feet lower altitude

7,800 feet altitude change

Distance ÷ Rate = Time

7,800 feet ÷ 600 feet/min = 13 minutes

2. Find the amount of fuel consumed using the formula:

Time x Fuel Consumption Rate = Fuel Consumed

[13 minutes x 11.5 gal/hr] ÷ 60 min/hr = 2.5 gallons

3. Use the wind face of the E6-B to find the ground speed (120 knots) and wind correction angle (+ 12°R). Due to wind being present, the TAS is not the ground speed. The speed used to calculate the distance covered must be the ground speed.

4. Calculate the distance.

Time x Speed = Distance

[13 min x 120 knots] ÷ 60 min/hr = 26 NM

5. Determine the compass heading, by applying the wind correction angle that was found on the wind face of E6-B to the true course, which gives the true heading. The true heading is corrected by variation to give the magnetic heading. Finally, the deviation is used to correct magnetic heading to give the compass heading as follows:

263° TC

+ 12° R ±WCA

275° TH

- 7° E ±VAR

268° MH

+ 3° ±DEV

271° CH

Solution using the CX-2:

1. The time to descend is based on the vertical distance from the cruising altitude of 10,500 feet down to the lower altitude of 2,700 feet MSL (1,000 feet AGL + 1,700 feet MSL).

10,500 feet cruising altitude

- 2,700 feet lower altitude

7,800 feet altitude change

2. Enter the Leg Time function to find the leg time.

Distance.................................7,800 feet

Ground speed........................600 feet/minute

Leg time.................................13 minutes

3. Enter the Fuel Burn function to find the amount of fuel consumed.

Time.......................................00:13:00

Fuel burn rate.........................11.5 gal/hour

Fuel burned............................2.5 gallons

4. Enter the Hdg/GS function to find ground speed.

Wind direction.............................330°

Wind speed.................................30 knots

Course.........................................263°

True airspeed.............................135 knots

Ground speed............................120.4 knots

Heading......................................275°

5. Enter the Dist Fln function to find the distance covered during the descent.

Time............................................00:13:00

Ground speed.............................120.4 knots

Distance.......................................26.1 NM

6. The true heading was found in Step #4. The true heading is corrected by variation to give the magnetic heading, and the deviation is used to correct magnetic heading to give the compass heading:

275° TH

- 7° E ±VAR

268° MH

+ 3° ±DEV

271° CH

Finding Density Altitude

To find density altitude, refer to the right-hand window on the computer side of the E6-B.

Problem:

Find the density altitude from the following conditions:

Pressure Altitude.................................5,000 feet

True air temperature...........................+ 40°C

Solution using the E6-B:

1. Refer to the right-hand "Density Altitude" window. Note that the scale above the window is labeled air temperature (°C). The scale inside the window itself is labeled pressure altitude (in thousands of feet). Rotate the disc and place the pressure altitude of 5,000 feet opposite an air temperature of +40°C.

2. The density altitude shown in the window is 8,800 feet.

Solution using the CX-2:

1. Enter the Plan TAS function to find density altitude.

Pressure altitude.........................5,000 feet

Temperature...............................40°C

Enter a zero when prompted for calibrated airspeed.

2. Density altitude is 8,846 feet.

Finding Wind Direction and Velocity

The E6-B can be used to solve for an unknown wind. To determine wind direction and wind speed, the true course, wind correction angle, true airspeed and ground speed are necessary. The wind correction angle (WCA) may not be given directly in a problem, but can be determined from the true course (TC) and true heading (TH).

Problem:

Determine the wind direction and wind speed under the following conditions:

True course.....................................095°

True heading...................................075°

True airspeed..................................90 knots

Ground speed.................................77 knots

Solution using the E6-B:

1. Set the true course (095°) under the true index located at the top of the computer.

2. Move the sliding grid to place the ground speed (77 knots) under the center grommet.

3. Determine the wind correction angle. Above the true heading (075°) read the wind correction angle, 20°L. Draw the wind dot over the true airspeed arc (90 knots) and 20° (wind correction angle) to the left.

4. Finally, rotate the window until the wind dot is lined up directly above the grommet. The wind direction is read under the true index (020°). For convenience, the sliding grid may be moved so that 100 knots is placed under the grommet. The difference between the grommet and the wind dot indicates the wind speed (31 knots).

Solution using the CX-2:

1. Enter the Unknown Wind function.

Heading.................................075°

Ground speed........................77 knots

True airspeed........................90 knots

Course...................................095°

Wind direction........................019°

Wind speed............................32 knots

Off-Course Correction

Either the off-course correction equation or the E6-B can be used to find the total correction angle that is required to fly to the original destination.

The correction relation can be calculated by using the formula:

Distance Off x 60 + Distance Off x 60 = Degrees Total Correction

Distance Flown Distance Remaining

Problem:

Find the total correction angle to converge on the destination under the following circumstances:

Distance flown.................................48 miles

Distance remaining.........................120 miles

Miles off course................................4 miles

Solution using the off-course equation:

(4 miles) (60) + (4 miles) (60) = Total Correction

48 miles 120 miles

= 240 + 240

48 120

= 5° + 2° = 7° total correction angle to converge on destination.

Solution using the E6-B:

1. Set the following relationships on the computer face of the E6-B:

Miles off Course = Degrees to Parallel

Miles Flown

4 = 5° to Parallel

48

2. Miles off Course = Additional DEG to Converge

Miles Flown

4 = 2° Additional to Converge

120

5° to Parallel + 2° Additional to Converge = 7° Total Correction

VHF Omni-Directional Range (VOR)

All VOR stations transmit an identifier. It is a three-letter Morse code signal interrupted only by a voice identifier on some stations, or to allow the controlling flight service station to speak on the frequency. Absence of a VOR identifier indicates maintenance is being performed on the station and the signal may not be reliable.

All VOR receivers have at least the essential components shown in Figure 9-1. The pilot may select the desired course or radial by turning the Omni-Bearing Selector (OBS). The Course Deviation Indicator (CDI) centers when the aircraft is on the selected radial or its reciprocal. A full-scale deflection of the CDI from the center represents a deviation of approximately 10° to 12°.

The TO/FROM Indicator (ambiguity indicator) shows whether the selected course will take the aircraft TO or FROM the station. A TO indication shows that the OBS selection is on the other side of the VOR station. A FROM indication shows that the OBS selection and the aircraft are on the same side of the VOR station. When an aircraft flies over a VOR, the TO/FROM indicator will reverse, indicating station passage.

The position of the aircraft can always be determined by rotating the OBS until the CDI centers with a FROM indication. The course displayed indicates the radial FROM the station. The VOR indicator displays information as though the aircraft were going in the direction of the course selected. However, actual heading does not influence the display. See Figure 9-2.

VOR radials, all of which originate at the VOR antenna, diverge as they radiate outward. For example, while the 011° radial and the 012° radial both start at the same point, 1 NM from the antenna, they are 100 feet apart. When they are 2 NM from the antenna, they are 200 feet apart. So at 60 NM, the radials would be 1 NM (6,000 feet) apart. See Figure 9-3.

The VOR indicator uses a series of dots to indicate any deviation from the selected course, with each dot equal to approximately 2° of deviation. Thus, a one-dot deviation at a distance of 30 NM from the station would indicate that the aircraft was 1 NM from the selected radial (200 feet x 30 = 6,000 feet).

To orient where the aircraft is in relation to the VOR, first determine which radial is selected (look at the OBS setting). Next, determine whether the aircraft is flying to or away from the station (look at the TO/FROM indicator), to find which hemisphere the aircraft is in. Last, determine how far off course the aircraft is from the selected course (look at the CDI needle deflection) to find which quadrant the aircraft is in. Remember that aircraft heading does not affect orientation to the VOR.

VOR accuracy may be checked by means of a VOR Test Facility (VOT), ground or airborne checkpoints, or by checking dual VORs against each other. A VOT location and frequency can be found in the Airport/Facility Directory (A/FD) and on the Air-to-Ground Communications Panel of the Low Altitude Enroute Chart.

To use the VOT, tune to the appropriate frequency and center the CDI. The omni-bearing selector should read 0° with a FROM indication, or 180° with a TO indication. The allowable error is ±4°. VOR receiver checkpoints are listed in the A/FD. With the appropriate frequency tuned and the OBS set to the published certified radial, the CDI should center with a FROM indication when the aircraft is over the designated checkpoint. Allowable accuracy is ±4° for a ground check, and ±6° for an airborne check. If the aircraft is equipped with dual VORs, they may be checked against each other. The maximum permissible variation when tuned to the same VOR is 4°.

The pilot must log the results of the VOR accuracy test in the aircraft logbook or other record. The log must include the date, place, bearing error, if any, and a signature.


Figure 9-1. VOR indicators


Figure 9-2. VOR display


Figure 9-3. Radial divergence

Estimating Time and Distance To Station Using VOR

If the aircraft is flown across radials, the approximate time and distance to a station can be determined by using the elapsed time, change of radials (angle) during that time, and the true airspeed (TAS).

Time To Station = [60 x Elapsed Time ÷ Angular Change]

and

Distance To Station = TAS x Time To Station

Problem:

Determine the approximate time and distance to the station, given the following:

Heading 270°

Crossing 360° radial at 1237Z

Crossing 350° radial at 1243Z

TAS 130 knots

Solution:

1. Calculate elapsed time and angular change:

1243 Z 360°

- 1237 Z - 350°

6 minutes 10°

2. Calculate time to station using the relation:

Time To Station = [60 x Elapsed Time ÷ Angular Change]

(60)(6 min) = 36 minutes = 0.6 hour

10°

3. Calculate the distance to the station using the relation:

Distance = TAS x Time

130 knots x 0.6 hours = 78 NM

When the angle to the station is small, that is, the aircraft flies nearly along radials, another method of approximating time to a station can be used. The Isosceles Triangle Principle states that if two angles of a triangle are equal, the two sides opposite those angles are also equal. For example, see the problem following, and Figure 9-4.


Figure 9-4. Isosceles triangle solution

Problem:

While inbound on the 090° radial, a pilot rotates the OBS 10° to the left (080° radial) and turns the aircraft 10° to the right noting the time. Maintaining a constant heading, the pilot determines that the elapsed time for the CDI to center is 8 minutes.

Solution:

Based on this information the ETE (estimated time en route) can be found with the isosceles triangle method since the angle the OBS was changed (90° - 80° = 10°) is equal to the angle the aircraft was turned (010°). The time elapsed for 10° was 8 minutes, and due to the equal angles, the time en route is also determined to be 8 minutes. See Figure 9-4.

Horizontal Situation Indicator (HSI)

The Horizontal Situation Indicator (HSI) is a combination of two instruments: the heading indicator and the VOR. See Figure 9-5.

The aircraft heading displayed on the rotating azimuth card under the upper lubber line in Figure 9-5 is 330°. The course indicating arrowhead that is shown is set to 300°. The tail of the course indicating arrow indicates the reciprocal, or 120°.

The course deviation bar operates with a VOR/LOC navigation receiver to indicate either left or right deviations from the course that is selected with the course indicating arrow. It moves left or right to indicate deviation from the centerline in the same manner that the angular movement of a conventional VOR/LOC needle indicates deviation from course.

The desired course is selected by rotating the course indicating arrow in relation to the azimuth card by means of the course set knob. This gives the pilot a pictorial presentation. The fixed aircraft symbol and the course deviation bar display the aircraft relative to the selected course as though the pilot was above the aircraft looking down.

The TO/FROM indicator is a triangular-shaped pointer. When this indicator points to the head of the course arrow, it indicates that the course selected, if properly intercepted and flown, will take the aircraft TO the selected facility, and vice versa.


Figure 9-5. Horizontal Situation Indicator (HSI)

The glide slope deviation pointer indicates the relationship of the aircraft to the glide slope. When the pointer is below the center position, the aircraft is above the glide slope, and an increased rate of descent is required.

To orient where the aircraft is in relation to the facility, first determine which radial is selected (look at the arrowhead). Next, determine whether the aircraft is flying to or away from the station (look at the TO/FROM indicator) to find which hemisphere the aircraft is in. Next, determine how far from the selected course the aircraft is (look at the deviation bar) to find which quadrant the aircraft is in. Finally, consider the aircraft heading (under the lubber line) to determine the aircraft's position within the quadrant.

Automatic Direction Finder (ADF)

ADF equipment consists of a receiver that receives in the low-and-medium frequency bands and an instrument needle that points to the station. The ADF may be used to either home or track to a station. Homing is flying the aircraft on any heading required to keep the azimuth needle on 0° until the station is reached, which results in a curved path that leads to the station (if there is any crosswind at all). Tracking is following a straight geographic path by establishing a heading that will maintain the desired track, regardless of wind effect. When an aircraft is on the desired track outbound from a radio station with the proper drift correction established, the ADF needle will deflect to the windward side.

The azimuth needle, pointing to the selected station, indicates the angular difference between the aircraft heading and the direction to the station, measured clockwise from the nose of the aircraft. This angular difference is the relative bearing to the station, and may be read directly from the fixed-scale indicator.

Determining Bearings with ADF

To determine the magnetic bearing from an aircraft to a selected station, add the magnetic heading of the aircraft to the relative bearing to the station:

Magnetic Heading + Relative Bearing = Magnetic Bearing TO the Station

See Figure 9-6.


Figure 9-6. Magnetic bearing to the station

Problem:

An aircraft has a magnetic heading of 360° with a relative bearing to a radio beacon of 245°. Calculate the magnetic bearing TO that radio beacon.

Solution:

MH + RB = MB (TO)

360° + 245° = 605° - 360° = 245° (TO)

To find the magnetic bearing being crossed (intercepted) by an aircraft, or the magnetic bearing from a radio beacon (RBN), the reciprocal is taken of the magnetic bearing TO a station.

An aircraft is maintaining a magnetic heading of 275° and the ADF shows a relative bearing of 070°. The magnetic radial FROM that radio beacon can be calculated using the relation:

MH + RB = MB (TO) ± 180° = MB (FROM)

275 + 070° = 345° - 180° = 165°

Problem:

If the aircraft continues its present heading as shown in Figure 9-7, what will the ADF indicate when the aircraft reaches the magnetic bearing of 030° from the NDB?


Figure 9-7. Change in relative bearing

Solution:

1. The aircraft's present heading is 330° with a relative bearing of 270°.

MH + RB = MB

330° + 270° = 600° - 360° = 240° (TO)

2. The present bearing TO is the reciprocal of the bearing FROM.

240° - 180° = 060° (FROM)

3. The change from the present MB (060°) to the desired MB (030°) is a decrease of 030° (060° - 030°). As the magnetic bearing decreases by 030°, the relative bearing as indicated on the ADF also decreases. The present relative bearing (270°) is decreased by 030° giving a new relative bearing of 240° (270° - 030°).

Intercepting Bearings with ADF

Problem:

If the airplane is currently on the heading 035, what heading should the airplane turn to, in order to intercept a magnetic bearing of 240° FROM at a 030° angle (while outbound FROM)?

Solution:

1. The magnetic bearing of 240° FROM may be intercepted with a 030° angle at a heading of either

210° (240° - 030°) or 270° (240° + 030°).

2. A heading of 210° would take the aircraft further away from the desired bearing of 240°. Thus, a heading of 270° is the only option.

3. To turn from the current heading of 035° to the desired heading of 270° would require a left turn of 125°.

Estimated Time and Distance To Station Using ADF

Relative bearing changes can be used in two ways to estimate flying time, distance and fuel required to fly to a station (or NAVAID): wing-tip bearing change method, and the isosceles triangle technique.

Wing-Tip Bearing Change Method

The first technique is to use the time required for a bearing change when the station is approximately located off the wing tip. The approximate time, distance and fuel to a station can be determined by using the time and amount of wing-tip bearing change, the true airspeed (TAS) and the rate of fuel consumption.

Problem:

If a 15° wing-tip bearing change occurs in 8 minutes at a TAS of 90 knots with 8.6 gallons/hour rate of fuel consumption, what is the time, distance, fuel burn to the station?

Solution:

1. The time-to-station can be calculated using the relation:

Time = 60 x minutes = 60 x 8 = 32 minutes

degrees 15

2. The distance to the station can be calculated using the relation:

Distance = Time x TAS

32 minutes x 90 knots ÷ 60 minutes/hour = 48 NM

3. The total fuel burn can be calculated using the relation:

Fuel = Time x gallons/hour

32 minutes x 8.6 gph ÷ 60 minutes/hour = 4.58 gallons

Isosceles Triangle Technique

The second method of using ADF to determine time, distance and fuel to station, can be employed when the relative bearing is small. The method is based on the property of an isosceles triangle that if two angles of a triangle are equal, the two sides opposite those angles are themselves equal.

For example, using the isosceles triangle method while cruising at 100 knots and on a constant heading, a relative bearing of 10° doubles in 5 minutes and therefore, the time to the station is equal to the time required for the relative bearing to double. The wing-tip bearing change formula cannot be used since the bearing change was not off the wing tip.

Problem:

Time to Station = 5 minutes

Solution:

The distance to the station can be calculated using the relation:

Distance = Time x Speed

[5 minutes ÷ 60 minutes/hour] x 100 knots = 8.33 NM