property of lear siegler

Property of Lear Siegler

NAVIGATIONAL COMPUTERSLIDE 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) 3-04.240

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., dependentupon decimal placement.

SLIDE RULE VALUES

Time scale values: 70 minutes on the

outer ring of the rotating scale can be convertedto 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 2140.

Read this as? 74, 7.4, 740

Read this as? 37.5, 375, 3750

21.4 or 2140.

Questions?


Action: Identify components of the slide rule computer

Condition: Given a navigation computer


DISTANCE CONVERSIONINDICES

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 into3,600 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

-2000 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 fromthe hour index.

Used for True airspeed anddensity altitude computations.

INNER SCALE WINDOWS


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


True airspeed computations.

Altitude scale - • Located in window.• Altitude range from

-2000 ft to 80,000 ft in increments of 1000 ft.

INNER SCALE WINDOWS

DENSITY ALTITUDE WINDOW

Located above airspeed window and used to determinedensity 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?


Action: Compute simple proportion problems

Condition: Given a navigation computer and data


PROPORTIONS

A simple proportion is afractional 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 16on the inner scale is a 3/4 or three to fourrelationship.

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:

1. 6.7 lbs.2. 1 gallon3. 120 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?


Action: Convert distance measurements

Condition: Given a navigation computer and data


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?


Action: Determine ground speed

Condition: Given a navigation computer and situational data


Ground speed equals distancedivided 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 distancedivided 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 (nauticalmiles per hour). You must convert distance to nautical miles in order to solve for knots!

Problem:• 9 min to fly 28

KM. GS = X.


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.


3. Set 15.2 (outer scale) over 9 (inner scale.

4. Read GS, 101 kts, over the 60 index.

Questions?


Action: Determine time required



Time equals distance dividedby 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 174.

2. Read 115 minutes, under 333.

NOTE!115 min can be read as

1hr, 55 min, on inner scale of rotating scale.

Questions?


Action: Convert time-distance problems



Convert Time-Distance Problems

Problem:• If 50 minutes are required to travel 120

nautical miles, how many minutes are requiredto 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 130.

2. Read the distance, 174 NM, over the 1hr, 20 min mark.

Questions?


Action: Solve rate-time-distance problems using the 36 index



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.


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?


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.


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?


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?


Action: Determine rate of fuel consumption



Fuel Consumption

Rate of fuel consumption equals gallons offuel consumed divided by time.

Fuel Consumption

Problem:• What is the rate of fuel

consumption if 30 gallons of fuelare 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 fuelare 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?


Action: Compute true airspeed



Compute True Airspeed

True airspeed is calibrated airspeedcorrected for pressure and temperature.


Problem:• What is the TAS if the FAT

is - 15° C, the pressure altitude is8,000 ft, and the CAS is 125 knots?


NOTE!Use window marked

“FOR AIRSPEED AND DENSITY ALTITUDE COMPUTATION”

Problem:• FAT = - 15° C,

PA = 8,000ft, CAS = 125 kts, TAS = X.


1. Set 8,000 ft under - 15° in the airspeed computation window.

2. Over 125, inner scale, read TAS 137 kts.

Questions?


Action: Compute density altitude



Solving for Density Altitude

Density altitude is pressure altitudecorrected for temperature.

Compute Density Altitude

NOTE!Use the same window as for true altitude computations -

“FOR AIRSPEED AND DENSITY ALTITUDE COMPUTATION”


Problem:• What is the density

altitude if the FAT is - 15°C andthe pressure altitude is 8,000 ft?


PA = 8,000ft, DA = X.


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?


Action: Compute true altitude



Solving for True Altitude

True altitude is the altitude above MSL.True Altitude is determined bycorrecting indicated altitude

for temperature and pressure.

Compute True Altitude

NOTE!Use the window labeled

“FOR ALTITUDE COMPUTATIONS”


Problem:• What is the true altitude

(TALT) if the FAT is - 15°C, thepressure altitude is 2,000 ft and theindicated altitude (IALT) is 2,100 ft?


1. Set 2,000 ft under - 15° in the altitude computation window.

2. Above 2,100 ft (IALT), read 1910 ft (TALT).


PA = 2,000 ft, IALT = 2,100 ft.


PA = 2,000 ft, IALT = 2,100 ft.

Questions?

Questions?

For reference while working problems:

FM 3-04.240 Chapter 5, Paragraph 5-1 through 5-17

Property of Lear Siegler

property of lear siegler

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