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NAVIGATIONAL COMPUTER SLIDE RULE

Terminal Learning Objective At the end of this lesson the student will: Action: Identify the components of the navigation computer slide rule problems Condition: Given a computer and situational data Standard: In Accordance With (IAW) Field Manual (FM)

Enabling Learning Objective (ELO) #1 Action: Identify the scales, values, and spacing of the navigation computer slide rule Condition: Given a navigation computer Standard: IAW FM 1-240

SLIDE RULE SCALES Outer Scale: Stationary - Miles, Distance, or Quantity

SLIDE RULE SCALES Inner Scale: Rotating - Time or Rate

SLIDE RULE VALUES Values: Represent multiples of 10 on either scale, e.g. 70 can represent.07,.7, 7.0, 70, 700, 7000, etc., dependent upon decimal placement.

SLIDE RULE VALUES Time scale values: 70 minutes on the outer ring of the rotating scale can be converted to 1:10 (one hour, ten minutes) on the hours scale on the inner ring of the rotating scale 70 minutes 1 hr, 10 min

SLIDE RULE SPACING Spacing: spaces between numbers are not constant on either scale. From 60 to 15 (clockwise), unit values are 1. From 15 to 30 (clockwise), unit values are 2. From 30 to 60 (clockwise), unit values are 1/2.

DECIMAL PLACEMENT The units between 21 and 22 are represented by 5 spaces. The second unit past 21 can be read as 21.4 or Read this as? 74, 7.4, 740 Read this as? 37.5, 375, or 2140.

Questions?

Enabling Learning Objective (ELO) #2 Action: Identify components of the slide rule computer Condition: Given a navigation computer Standard: IAW FM 1-240

DISTANCE CONVERSION INDICES NAUT index - located at scale value 66, used for converting to nautical miles. STAT index - located at scale value 76, used for converting to statute miles. KM index - located at scale value 122, used for converting to kilometers.

INNER SCALE INDICES SPEED index - located at 60 on the inner scale and represented by a large black arrow. It is the hour index. 36 index - located at 36 on the inner scale and represented by a small arrow. It is the second index (divides one hour into 3,600 seconds).

INNER SCALE WINDOWS ALTITUDE COMPUTATIONS WINDOW Located immediately to the right of the hour index. True altitude computations. Temperature scale - Located inside of window. Temperature range from - 80° C to + 50°C.

INNER SCALE WINDOWS ALTITUDE COMPUTATIONS WINDOW True altitude computations. Altitude scale - Located under temperature window. Altitude range from ft to 34,000 ft in increments of 2000 ft, 35,000 ft to 80,000 ft use the same mark.

INNER SCALE WINDOWS AIRSPEED COMPUTATIONS WINDOW Located across from the hour index. Used for True airspeed and density altitude computations.

INNER SCALE WINDOWS AIRSPEED COMPUTATIONS WINDOW True airspeed computations. Temperature scale - Located above window. Temperature range from + 50°C to - 80°C. Negative temps are right, positive temps are left.

INNER SCALE WINDOWS AIRSPEED COMPUTATIONS WINDOW True airspeed computations. Altitude scale - Located in window. Altitude range from ft to 80,000 ft in increments of 1000 ft.

INNER SCALE WINDOWS DENSITY ALTITUDE WINDOW Located above airspeed window and used to determine density altitude. Uses airspeed window to set up temperature and altitude. Altitude range from - 10,000 ft to 80,000 ft, increments of 1,000 ft.

Questions?

Enabling Learning Objective (ELO) #3 Action: Compute simple proportion problems Condition: Given a navigation computer and data Standard: IAW FM 1-240

PROPORTIONS A simple proportion is a fractional relationship between numbers and is expressed as a ratio. Example: 20:40 represents a fraction of one-half (reduced to it’s lowest common denominator), or expressed as a ratio, 1:2 (one to two). 12:16, 12 on the outer scale over 16 on the inner scale is a 3/4 or three to four relationship.

Proportions In Problem Solving Problem: How many pounds are in 120 gallons of JP- 8? Solution: At least three factors must be known to solve for the unknown (X). There are 6.7 pounds in each gallon of JP-8. The three known are: lbs gallon gallons The ratio is: 6.7 : X 1 120

Proportions In Problem Solving Problem: How many lbs are in 120 gallons of JP- 8? 1. Set 6.7 on the outer scale over 1.0 on the inner scale. 2. Find 120 on the inner scale and read 805 lbs above 120 gallons.

Questions?

Enabling Learning Objective (ELO) #4 Action: Convert distance measurements Condition: Given a navigation computer and data Standard: IAW FM 1-240

Distance Measuring Units Problem: How many statute miles equals 90 nautical miles? How many kilometers equals 90 nautical miles?

Problem: X SM = 90 NM? X KM = 90 NM? 1. Set 90 NM under the NAUT index. 2. Read 104 SM under the STAT index. Distance Measuring Units 2. Read 166 KM under the KM index.

Questions?

Enabling Learning Objective (ELO) #5 Action: Determine ground speed Condition: Given a navigation computer and situational data Standard: IAW FM 1-240

Ground speed equals distance divided by time. Problem: What is the ground speed if it takes 35 minutes to fly 80 nautical miles?

Problem: 35 Min to fly 80 NM. GS = X. Ground Speed Problems 1. Set 35 (inner scale) under 80 (outer scale). 2. Read 137 kts (knots) over the speed index, (60 minutes).

Ground speed equals distance divided by time. Problem: What is the ground speed if it takes if it takes 9 minutes to fly 28 kilometers? NOTE! Ground speed is measured in knots (nautical miles per hour). You must convert distance to nautical miles in order to solve for knots!

Problem: 9 min to fly 28 KM. GS = X. Ground Speed Problems 1. Set 28 under the KM index. 2. Read 15.2 NM under the NAUT index. First, convert KM to NM.

Problem: 9 min to fly 28 KM. (15.2 NM). GS = X. Ground Speed Problems 3. Set 15.2 (outer scale) over 9 (inner scale. 4. Read GS, 101 kts, over the 60 index.

Questions?

Enabling Learning Objective (ELO) #6 Action: Determine time required Condition: Given a navigation computer and situational data Standard: IAW FM 1-240

Time equals distance divided by ground speed. Problem: How much time is required to fly 333 nautical miles at a ground speed of 174 knots?

Problem: Dis. = 333 NM, GS = 174. Time = X. Time Required Problems 1. Set 60 index under Read 115 minutes, under 333. NOTE! 115 min can be read as 1hr, 55 min, on inner scale of rotating scale.

Questions?

Enabling Learning Objective (ELO) #7 Action: Convert time-distance problems Condition: Given a navigation computer and situational data Standard: IAW FM 1-240

Convert Time-Distance Problems Problem: If 50 minutes are required to travel 120 nautical miles, how many minutes are required to travel 86 nautical miles at the same rate?

Problem: 50 min. = 120NM X min = 86NM. Time-Distance Problems 1. Set 50 min under 120 NM. 2. Read 36 min under 86 NM.

Convert Time-Distance Problems Problem: Your ground speed is 130 knots, you have flown for 1 hour and 20 minutes. How many nautical miles have you traveled?

Problem: GS = 130 knots, 1 hr, 20 min, distance = X Time-Distance Problems 1. Set speed index under Read the distance, 174 NM, over the 1hr, 20 min mark.

Questions?

Enabling Learning Objective (ELO) #8 Action: Solve rate-time-distance problems using the 36 index Condition: Given a navigation computer and situational data Standard: IAW FM 1-240

Solving problems using the 36 index. NOTE! Use the 36 index whenever the distance is less than 1/10th of the ground speed. Used whenever time must be calculated in seconds and minutes instead of minutes and hours.

Solving problems using the 36 index. Problem involving less than 60 seconds: What is the time required if the ground speed is 100 knots and the distance is 0.5 NM?

Problem: GS = 100 knots, distance = 0.5NM time required = X. Solving problems using the 36 Index 1. Set 36 index (3,600 sec) under 100 NM. 2. Read the time, 18 seconds, under the 0.5 NM mark.

Solving problems using the 36 index. Problem involving more than 60 seconds: What is the time required if the ground speed is 95 knots and the distance is 5 nautical miles?

Problem: GS = 95 knots, distance = 5NM time required = X. Solving problems using the 36 Index 1. Set 36 index (3,600 sec) under 95 NM. 2. Read the time, 190 seconds,(3 min, 10 sec), under the 5 NM mark.

Questions?

Enabling Learning Objective (ELO) #9 Action: Determine rate of fuel consumption Condition: Given a navigation computer and situational data Standard: IAW FM 1-240

Fuel Consumption Rate of fuel consumption equals gallons of fuel consumed divided by time.

Fuel Consumption Problem: What is the rate of fuel consumption if 30 gallons of fuel are consumed in 111 minutes (1 hour, 51 minutes)?

Problem: 30 gallons, 111 minutes (1+51), GPH = X. Fuel Consumption 1. Set 111 minutes under 30 gallons. 2. Read the rate, 16.2 gallons over the speed (60) index.

Fuel Consumption Problem: What is the rate of fuel consumption if 300 lbs of fuel are consumed in 111 minutes (1 hour, 51 minutes)?

Problem: 300 lbs, 111 minutes (1+51), PPH = X. Fuel Consumption 1. Set 111 minutes under 300 pounds. 2. Read the rate, 162 pounds over the speed (60) index.

Questions?

Enabling Learning Objective (ELO) #10 Action: Compute true airspeed Condition: Given a navigation computer and situational data Standard: IAW FM 1-240

Compute True Airspeed True airspeed is calibrated airspeed corrected for pressure and temperature.

Compute True Airspeed Problem: What is the TAS if the FAT is - 15° C, the pressure altitude is 8,000 ft, and the CAS is 125 knots?

Compute True Airspeed NOTE! Use window marked “FOR AIRSPEED AND DENSITY ALTITUDE COMPUTATION”

Problem: FAT = - 15° C, PA = 8,000ft, CAS = 125 kts, TAS = X. Compute True Airspeed 1. Set 8,000 ft under - 15° in the airspeed computation window. 2. Over 125, inner scale, read TAS 137 kts.

Questions?

Enabling Learning Objective (ELO) #11 Action: Compute density altitude Condition: Given a navigation computer and situational data Standard: IAW FM 1-240

Solving for Density Altitude Density altitude is pressure altitude corrected for temperature.

Compute Density Altitude NOTE! Use the same window as for true altitude computations - “FOR AIRSPEED AND DENSITY ALTITUDE COMPUTATION”

Compute Density Altitude Problem: What is the density altitude if the FAT is - 15°C and the pressure altitude is 8,000 ft?

Problem: FAT = - 15° C, PA = 8,000ft, DA = X. Compute Density Altitude 1. Set 8,000 ft under - 15° in the airspeed computation window. 2. Over DA index, read DA 6,200 ft. NOTE! Accurate results can only be obtained by using pressure altitude. Pressure altitude can be obtained by setting the altimeter to 29.92” of mercury and reading the pressure altitude directly from the altimeter.

Questions?

Enabling Learning Objective (ELO) #12 Action: Compute true altitude Condition: Given a navigation computer and situational data Standard: IAW FM 1-240

Solving for True Altitude True altitude is the altitude above MSL. True Altitude is determined by correcting indicated altitude for temperature and pressure.

Compute True Altitude NOTE! Use the window labeled “FOR ALTITUDE COMPUTATIONS”

Compute True Altitude Problem: What is the true altitude (TALT) if the FAT is - 15°C, the pressure altitude is 2,000 ft and the indicated altitude (IALT) is 2,100 ft?

Compute True Altitude 1. Set 2,000 ft under - 15° in the altitude computation window. 2. Above 2,100 ft (IALT), read 1910 ft (TALT). Problem: FAT = - 15° C, PA = 2,000 ft, IALT = 2,100 ft. Problem: FAT = - 15° C, PA = 2,000 ft, IALT = 2,100 ft.

Questions?

Enabling Learning Objective (ELO) #13 Action: Compute off-course corrections Condition: Given a navigation computer and situational data Standard: IAW FM 1-240

Solving Off-course Corrections The rule of 60 states that an aircraft headed 1° off course will be approximately 1 mile off course for each 60 miles flown.

TR TC 1° 1NM 60NM Solving Off-course Corrections At 60 NM, 1° off course = 1 NM off course Rule of 60

Solving Off-course Corrections Paralleling the course: For each mile off course, using the rule of 60, 1° of correction (for each mile off course) must be applied. Converging on the course: In order to converge on course at the destination, an extra correction, based on the rule of 60 must be applied.

Off Course Corrections Problem: An aircraft is 10 nautical miles to the left when 150 nautical miles from departure point A. How many degrees correction are required to parallel the course? If 80 nautical miles remain to destination B, how many additional degrees are required to converge? In what direction is the correction applied?

Rule of 60 - Parallel 10NM 150NM 80NM 1. Set 150 (inner scale) under 10 (outer scale). 2. Read 4° over 60 index, (correction to parallel).

Rule of Set 150 (inner scale) under 10 (outer scale). 2. Read 4° over 60 index, (correction to parallel).

Rule of 60 - Converge 10NM 150NM 80NM 1. Set 80 (inner scale) under 10 (outer scale). 2. Read 7.5° over 60 index, (to converge).

Rule of Set 80 (inner scale) under 10 (outer scale). 2. Read 7.5° over 60 index, (to converge).

Rule of 60 10NM 150NM 80NM To Parallel 4° +7.5° Converge =11.5° Aircraft is left of course, therefore correction must be to the right. Correction to right = add Correction to left = subtract

Solving Off-course Corrections The rule of 60 states that if the distance flown (or the distance to fly) is less than 60 NM, the correction to parallel (or converge to) the TC will be greater in degrees than the number of miles off course.

TR TC (TC = 60 NM, 1°off course = 1 NM off course), then (TC than miles of course). Rule of 60 1NM 60NM <60NM 1° >1°

Solving Off-course Corrections The rule of 60 states that if the distance flown (or the distance to fly) is more than 60 NM, the correction to parallel (or converge to) the TC will be less in degrees than the number of miles off course.

TR TC <1° 1NM 60NM (TC = 60 NM, 1°off course = 1 NM off course), then (TC >60 NM, ° off course < than miles of course). Rule of 60 1NM >60NM 1°

Distance Off Course To Parallel Distance Flown Distance Off Course To Converge Distance Left To Fly

Questions? For reference while working problems: FM Chapter 5, Paragraph 5-1 through 5-17

Property of Lear Siegler