Statute Miles vs. Nautical · PDF fileStatute Miles vs. Nautical Miles Exercise 1: ... thousands of calculations for cargo weight and cubic space requirements, planners had to carefully

Math In Flight Unit 1

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There are many formulas used in flight. These formulas aid in navigation and airplane performance. The following examples show common calculations performed by flight personnel.

We measure long distances in miles. On the ground, these are referred to as statute miles (sm). In navigation, distance is measured in nautical miles (nm), which allows for the curvature of the earth. A mile on the ground (sm) is 5,280 feet. A mile in the air (nm) is 6076.1 feet.

To convert between the units, we use the formula

sm = nm • 1.15

This would mean a statute mile is smaller than a nautical mile. This would make sense because the farther we are from the earth’s center, the farther it is between two locations.

Example:

Find the distance in statute miles given a distance of 1000 nautical miles, the approximate distance from Miami, FL to Washington, DC.

Solution:

sm = nm • 1.15 sm = 1,000 • 1.15 sm = 1,150

The flight distance would be 1,150 statute miles.

Statute Miles vs. Nautical Miles

Exercise 1:

Find the distance in statute miles given the distance 2,600 nautical miles, the approximate distance from Washington, DC to Los Angeles, CA.

Exercise 2:

Find the distance in statute miles given the distance 1,700 nautical miles, the approximate distance from Albuquerque, NM to Orlando, FL.

Exercise 3:

Find the distance in staute miles given the distance 500 nautical miles, the approximate distance from Houston, TX to Birmingham, AL.


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Another important distance formula is:

distance = rate x timeor

d = r t

This formula is used to find the distance a plane will fly on a trip. It can also be used to find the speed of the plane when the dis-tance and time are given, or to find the time when the distance and speed are given.

In flight applications, distance is usually measured in miles. Rate or speed is usually measured in knots (nautical miles per hour). Time is usually measured in hours.

Example:

A jet travels at 690 knots (nautical miles per hour) for 10 hours. How many nautical miles will the plane cover?

Solution:

d = r t d = 690 • 10 d = 6,900

Since rate and time have both been mea-sured in hours, multiplication is all that is necessary.

The plane will cover 6,900 nautical miles.

Distance, Rate and Time

Exercise 1:

A plane travels 280 knots for 94 hours. How many nautical miles will the plane cover?

Exercise 2:

A plane covers 3,000 nautical miles in a 4-hour trip. What is the speed of the plane?

Exercise 3:

A jet covers 5700 nautical miles in 4.75 hours. What is the speed of the plane?

“Lucky Lady II”

On March 2, 1949 Lucky Lady II landed at Carswell Air Force Base (AFB), Tex-as. Captain James G. Gal-lagher and a 13-man crew took off from Carswell AFB, Texas, in a B-50 Superfor-tress named Lucky Lady II. Four days later the B-50 completed the fi rst nonstop, around-the-world fl ight in history, covering 23,452 miles in 94 hours and 1 min-ute. This was achieved by refueling the plane in fl ight. The fi rst refueling began over the Azores the morn-ing after takeoff. It took two hours, during which time the bomber and the tanker remained linked and had to maintain a tight formation. It was tiring work. The fl ight started and ended at Carswell AFB with refuel-ing accomplished over West Africa, the Pacifi c Ocean near Guam, and between Hawaii and the West Coast. This fi rst nonstop circum-navigation proved that vast distances and geographical barriers were no longer an obstacle to military air pow-er, thanks to aerial refueling. Bad weather, crew fatigue, and equipment problems complicated refueling in fl ight. Precise navigation (time, distance, and rate), expert fuel consumption cal-culations, and highly trained aviators made this feat a suc-cess.



The Wright brothers used a canard to control the front of the plane. A canard is a fl ap that directs the wind up or down so that the plane does not dive or climb out of control. The canard affected the center of gravity, the point where the weight of the plane is concentrated. Modern engineering proved that similar fl aps on the wings could replace the canard, moving the plane’s center of gravity toward the middle of the plane. This improved fl ight performance. The center of gravity is still calculated by fl ight engineers today. The formula is:

where c.g. is the center of gravity. This is important so that the plane is balanced side-to-side and front to back. Balancing the plane puts equal force on each wing and equalizes the force between the nose and the tail of the plane.

A moment is the force an object creates based on the object’s weight and position in the plane. A heavy item placed farther from the center of gravity will create a higher moment.

Example:

A pilot has added the weights and moments for the following:

Plane weight when emptyWeight of pilot and passengersBaggageOilFuel

The total weight of the items is 1,317.5 pounds. The total of their moments is 108,200. Find the center of gravity.

Solution:

The pilot then refers to a graph called the Center of Gravity Moment Envelope to see that 82.13 falls within the Normal category for the calculated weight and moment.

cg = Total momentsTotal weight

cg = Total momentsTotal weight

cg = 108,2001,317.5

cg = 82.13

Center of Gravity

Exercise 1:

Calculate the center of gravity for a total weight of 2,200 and a total moment of 90,000.

Exercise 2:

Calculate the total moment given a center of gravity of 78.5 and a total weight of 8,000.

Center of Gravity Moment Envelope for a Cessna 172N.

“Operation Vittles” aids

isolated Berlin

The Berlin Blockade (June 24, 1948 to May 11, 1949) became one of the fi rst ma-jor crises of the new Cold War, during which the Soviet Union blocked railroad and street access to West Berlin. The crisis abated after the Soviet Union did not act to stop American, British and French humanitarian airlifts of food and other provisions to the Western-held sectors of Berlin; this was referred to as Operation Vittles by the Americans and Opera-tion Plainfare by the British. At its height, The Berlin Air-lift delivered a record 12,940 tons in a 24-hour period. The Berlin Airlift, gradually reduced since May 12, 1949, offi cially ended on Septem-ber 30, 1949. 149 allied aircraft carried 2,343,301.5 tons of supplies on 277,264 fl ights, and U.S. planes car-ried 1,783,826 tons. This massive airlift involved thousands of calculations for cargo weight and cubic space requirements, planners had to carefully consider the lift capacity of each aircraft, the fuel consumption, as well as navigation, time and distance, and weather, to successfully conclude this famous operation. Clearly math helped make this airlift a success.



The early fl ights of the Wright brothers were typically 10 to 12 feet above the ground. It was 5 years later, in 1908, before their fl ying machine could reach an altitude of 100 feet. Today commercial jets fl y 30,000 to 40,000 feet. At such high altitudes, the atmospheric pressure is much different. Flight personnel calculate the atmospheric pressure, p, using

where α is the altitude of the plane, given in feet.

Example:

Calculate the atmospheric pressure for a plane flying at 6,000 feet. Round answer to the nearest hundredth.

Solution:

The atmospheric pressure is 11.61 pounds per square inch (11.61 lb/in2).

p = α

1,000( )2 α

1,000( )+ 40

-9.05 α

1,000( )2 65α

1,000– [ ]

p = α

1,000( )2

α1,000( )+ 40

-9.05 α

1,000( )2

65α1,000

– [ ]

+ 40 6,0001,000( )

2 6,0001,000( )

p =

-9.05 6,0001,000( )

2 65(6,000)

1,000– [ ]

p = -9.05 [36 - 390]

36 + 240

p = -9.05 [- 390]

276

p = 3,203.7

276

p = 11.61

p = -9.05 [(6) - 65(6)]

(6) + 40(6)2

2

Exercise 1:

Calculate the atmospheric pressure for a plane flying at 30,000 feet.

Exercise 2:

Calculate the atmospheric pressure for a plane flying at 42,200 feet.

Round answers to the nearest hundredth.

Atmospheric Pressure



One formula used to determine the height of an air-borne object being observed by two tracking stations, when given the distance between tracking stations (s) and the angles of inclination (α and β) is:

Example:

Two tracking stations 12.95 miles apart observe a plane flying over them. How high, in feet, is the plane above the ground if the plane’s angle of inclination at the first station is 30º and the angle at the second tracking sta-tion is 70º?

Solution:

The plane’s height is 9.4664 miles. To convert this to feet,

The plane is 49,982 feet above the ground.

�

h

sα β

h = 12.95

cot 30 - cot 70

h = 12.95

1.3680

h = 9.4664

h = s

cot α - cot β

h = 12.95

1.7320 - 0.3640

Exercise 1:

Two tracking stations 2.06 miles apart observe a plane flying over them. How high, in feet, is the plane above the ground if the plane’s angle of inclination at the first station is 20º and the angle at the second tracking station is 60º?

Exercise 2:

A plane is flying at 5,000 feet. Two tracking stations are observing it. How many miles apart are the tracking stations if the plane’s angle of inclination at the first station is 10º and the angle at the second tracking station is 80º?

Height of an Airborne Object

h = s

cot α - cot β

�

h

12.9530° 70°

h = 9.4664 • 5,280

h = 49,982.46 feet

=5,280 feet

1 mileh

9.4664



The cruising altitude of a plane is determined in part by its heading.

Example:

A plane is traveling northeast, heading 45º. Could its altitude be 27,000 feet?

Solution:

27,000 ÷ 1,000 = 27The answer is Yes, because 27 is an odd number.

359˚

180˚ 179˚

0˚

EvenThousands

OddThousands

Cruising Altitude

Exercises:

State yes or no if these headings and associated cruising altitudes would be acceptable.

Heading Cruising Altitude Yes/No

1. East 90º 25,000 feet

2. West 270º 21,000 feet

3. South-southeast 179º 21,000 feet

4. South 180º 21,000 feet

5. North-northwest 300º 32,000 feet

The SR-71 “Blackbird”

The Lockheed SR-71, un-offi cially known as the Blackbird, was an advanced, long-range, Mach 3 strate-gic reconnaissance aircraft. The aircraft holds a 15-mile course speed record of 2,193.167 MPH, set on July 27, 1976. The fi rst fl ight of an SR-71 took place on De-cember 22, 1964, and the fi rst SR-71 to enter service was delivered in January 1966. The USAF had SR-71 Blackbirds in service from 1966–1998. The SR-71 was one of the fi rst aircraft whose shape reduced radar signa-ture. Constructed largely of titanium, it was coated with high-heat emissive black paint and had precious met-als (e.g. gold) components. This combination helped retard the 1,100 F degree temperature from sustained supersonic fl ight. Designed as a strategic reconnais-sance aircraft, its equipment included electronic intelli-gence collection and radar surveillance systems, plus photographic equipment ca-pable of surveying 100,000 square miles of the earth’s surface in an hour. On Sep-tember 1, 1975 the SR-71 set a world record by fl ying from New York to London in 1 hour, 54 minutes, and 56 seconds.



Atmospheric pressure always decreases approximately 1” Hg (1 inch of mercury) per 1,000 feet of altitude gained. Lowering the altimeter setting lowers the altitude reading. If a pilot makes the following changes to the altimeter setting, what is the approximate change in altitude?

Example:

Setting changed from 30.18” to 29.76”.

Solution:

Altitude and Atmospheric Pressure

=0.42”

X1”

1000’

1 • X = 1,000 • 0.42

X = 420 feet lower

30.18 – 29.76 = 0.42

Exercise 1:


Exercise 2:


Exercise 3:


The F-22A Raptor

The F-22A Raptor is the Air Force’s newest fi ghter aircraft. Its combination of stealth, supercruise, ma-neuverability, integrated avionics and improved sup-portability represent an ex-ponential leap in War fi ght-ing capabilities. The Raptor performs both air-to-air and air-to-ground missions al-lowing full realization of operational concepts vital to the 21st Century Air Force. Its unique characteristics in-clude being able to sustain supersonic fl ight for periods of time without needing to engage afterburners; it is su-per-agile, stealthy, and has advanced integrated avion-ics - the heart of the avion-ics suite is a super computer that can process 10.3 bil-lion bytes per second. The F-22 construction is 39% titanium, 24% composite, 16% aluminum, and 1% thermoplastic by weight. Titanium is used for its high strength-to-weight ratio in critical stress areas, includ-ing some of the bulkheads, and also for its heat-resistant qualities in the hot sections of the aircraft. Carbon fi ber composites have been used for the fuselage frame, the doors, intermediate spars on the wings, and for the hon-eycomb sandwich construc-tion of the skin panels.



The rate at which a plane descends is referred to as the slope of descent. It is defined the same as the slope is in graphing:

Because the plane is descending, rise refers to the amount of descent.

Since the plane is descending, rise refers to the amount of descent. The slope of descent often is given as a percent.

Example:

Find the approximate slope of descent, expressed as a percent, if a plane is flying at 31,500 feet, intending to land 100 miles away. The elevation of the landing site is 1,500 feet.

Solution:

Using a diagram, we can determine

The plane will descend

31,500 feet – 1,500 feet = 30,000 feet

over 100 miles.

Ratios compare like units. To create the ratio of the slope, 100 miles must be converted to feet.

The ratio for the slope is

This means for every 5 feet the plane descends, it cov-ers 88 feet, or for every 5 miles the plane descends, it covers 88 miles. Since a ratio is a comparison of like units, we can use feet or miles to describe the ratio.

Since the slope of descent is often given as a percent, we can change our fraction to a decimal by dividing the numerator by the denominator. To convert 5/88 to a percent, convert to a decimal by dividing, then move the decimal point 2 places to the right.

The slope of descent would be 5.68%.

Slope of Descent

Slope = = Change in y (vert.) directionChange in x (horiz.) direction

riserun

Exercise 1:

Find the approximate slope of descent, expressed as a percent, if a plane is flying at 15,000 feet, intending to land 80 miles away. The elevation of the landing site is 2,000 feet.

Exercise 2:

Find the approximate slope of ascent, expressed as a percent, of a launched space shuttle. When the shuttle reached an altitude of 320 miles, it had covered 600 miles over the ocean.

100 miles

(not to scale)

1,500’

31,500’

rise (or descent)run

100 miles • = 528,000 feet 5,2801 mile

30,000 feet ÷ 6,000528,000 feet ÷ 6,000

= 588



As a plane climbs, the air around it becomes less dense. Resistance from drag (or push) and lift decreases. The actual airspeed (TAS – True Air Speed) is higher than the instruments indicate (IAS – Instrument Air Speed).

To calculate the True Air Speed, use the formula:

or

This states that IAS increases by 2% for every 1,000 feet of altitude.

Example:

At 8,000 feet, a plane has an Instrument Air Speed of 100 knots (nautical miles per hour). What is the True Air Speed?

Solution:

The True Air Speed is 116 knots.

Air Speed

TAS = IAS + 2%( ) • IASaltitude1,000

TAS = IAS + 0.02( ) • IASaltitude1,000

Exercise 1:

What is the True Air Speed of a plane at 20,000 feet whose instruments indicate an air speed of 300 knots?

Exercise 2:

What is the True Air Speed of a plane at 35,000 feet whose instruments indicate an air speed of 550 knots?

TAS = 100 + 0.02( ) • 1008,0001,000

TAS = 100 + 0.02(8) • 100

TAS = 100 + 0.16 • 100

TAS = 100 + 16

TAS = 116

TAS = IAS + 2%( ) • IASaltitude1,000

The X-15 sets records on

November 9, 1961

On a clear November morn-ing in 1961 USAF Major Robert M. White attained a top speed of 4,093 mph in an X-15 hypersonic rocket plane while fl ying at full throttle at an altitude of 101,600 feet. The X-15 set numerous speed and altitude records in the early 1960s, reaching the edge of space and bringing back valuable data that was used in the design of later aircraft and spacecraft. Following this achievement of the X-15’s, the emphasis of the pro-gram shifted from envelope expansion to fl ight research at high Mach numbers and altitudes. North American X-15 rocket plane was an experimental aircraft proj-ect. It was the fi rst aircraft to attain velocities of Mach 4, 5, and 6. The X-15 was a rocket-powered research aircraft, launched from a B-52 aircraft, and completed 199 missions between 1959 and 1968. It had no landing gear, but landed on skis and had reaction controls for at-titude control in space. At-mospheric pressure, altitude, slope of descent, and center of gravity all had to fi gure into the planning of each of these fl ights.



Calculate the following airplane speeds in still air.

Example:

With a tailwind, a plane travels 180 miles in 45 minutes. With a headwind, it travels 180 miles in one hour. Find the speed of the plane in still air (p).

Solution:

Since knots are nautical miles per hour, we should change the minutes to hours.

In the chart above, p represents the speed of the plane in still air; w represents the speed of the wind.

(Note: Because the distance in both situations (tailwind and headwind) is 180 miles, the two equations for dis-tance can be set equal.)

The plane traveled 180 nautical miles with the tailwind and 180 nautical miles with the head wind. Use either distance from the chart equal to 180, substituting 7w for p to fi nd w, the speed of the wind:

The wind speed is 30 knots.

Headwinds and Tailwinds

Chart 1Tailwind

Headwind

Rate

p + w

p – w

x

x

x

Time in hours

¾ hour

1

=

=

=

Distance

¾(p + w)

p – w

45 minutes x = = ¾ hour 1 hour

60 minutes45 ÷ 1560 ÷ 15

¾(p + w) = p – w 4[¾ (p + w)] = 4(p – w) to clear the fraction

3(p + w) = 4(p – w)3p + 3w = 4p – 4w3w = 4p – 4w – 3p3w = p – 4w3w + 4w = p7w = p

p – w = 1807w – w = 1806w = 180(1/6)6w = 180(1/6)w = 30



To verify this, use the tailwind distance.

To find the speed of the plane, use either distance from the chart.

Substitute 30 for w:

The speed of the plane is 210 knots.

This can be verified by using the other distance similarly:

Substitute 30 for w:

¾(p + w) = 180¾(7w + w) = 180¾(8w) = 1806w = 180(1/6)6w = 180(1/6)w = 30

p – 30 = 180p = 180 + 30p = 210

¾(p + 30) = 1804[¾(p + 30)] = (180)43(p + 30) = 7201/3[3(p + 30)] = (720)1/3

p + 30 = 240p = 240 – 30p = 210

¾(p + w) = 180

p – w = 180

Exercise 1:

With a tailwind, a plane travels 127 nautical miles in 30 minutes. With a headwind, it trav-els 107 nautical miles in 30 minutes. Find the speed of the plane in still air.

Exercise 2:

With a headwind, a plane travels 300 nautical miles in 45 minutes. With a tailwind, the plane travels 230 nautical miles in 30 minutes. Find the speed of the plane in still air.



Wind often causes a plane to drift from its heading or direction. To stay on course, the Wright brothers constantly adjusted the heading of their flying machine by pulling cables attached to flaps and the flexible wings. Pilots must calculate the effect the wind will have on the plane so that they can set the automatic navigation system at the correct heading. The diagram illustrates the effect of a crosswind, a wind blowing against the side of the plane.

In still air, a plane would travel due east along AC.

With a crosswind blowing in the direction of AB, the plane actually travels in the direction of AD. AD is called the course of the plane.

CAD is called the drift angle.

Example:

Find the course, the speed in the wind, and the drift angle of a plane headed at 120º when flying at 240 knots in still air if there is a crosswind of 20 knots blowing from direction 40º.

Solution:

Draw a diagram of the situation.

Using alternate interior angles:

Crosswinds

N

X

D

A

C

B

40º

120º240

20

drift angle

originalheading

alteredcourse

AXC is the drift angle.AXB = 40° + 180° – 120° = 100°

AXD = 180° – AXB

CAX = 80°AXD = 80°AXD = 180° – 100°

CAX = AXD

A

B

C

DACAC

ABABADAD ADAD

CAD



Using ∆XAC and the Law of Cosines c² = a² + b² - 2ab cos θ,

or, using the Law of Sines:

The course of the plane is 120º + 4.76º = 124.76º.The plane is traveling at 237.35 knots with a drift angle of 4.76º and a course of 124.76º.

XC = 237.35

XC² = 400 + 57,600 – 9,600(0.1736)XC² = 400 + 57,600 – 1667XC² = 56,333

XC² = 20² + 240² – (2 • 20) • 240 cos 80°XC² = AC² + XA² – 2AC • XA cos A

AXC = 4.76°

sin AXC = 20 • 0.9848237.3

= 0.0830

sin AXC20

= sin 80°237.3

= 0.9848237.3

=sin AXCAC

sin CAXXC

Exercise 1:

A pilot is to fl y on course 72º in a wind blowing 30 knots from direction 134º. If the plane’s speed is 280 knots in what direction must he head the plane and what will be the speed of the plane in the wind?

Exercise 2:

A plane is headed in direction 120º with a speed of 150 knots in still air. The course is 105º with a speed in the wind of 130 knots. Find the speed and direction from which the wind is blowing.

Round answers to the nearest hundredth.

Math In Flight Quiz


Quiz1. True or False? Statute miles are smaller than

nautical miles.

2. Using the formula d = r t fi nd the distance a plane travels after 4 hours at 500 knots (nautical miles per hour).

3. The cruising altitude of a plane is determined in part by its heading. Would a plane traveling at 28,000 feet be heading 0˚ to 179˚ or 180˚ to 359˚?

4. Find the altitude in miles that a space shuttle reached after covering 600 miles if the angle of inclination was 27˚ after launch.

5. With a tailwind, a plane traveling at 400 miles per hour can fl y 220 nautical miles in 30 minutes. With a headwind, the plane can only fl y 180 nautical miles in 30 minutes. What is the speed of the wind?

6. A plane is headed in direction 120˚ with a speed of 180 knots in still air. The wind is blowing from 178˚ at 50 knots. What will be the course of the plane and what will be the speed of the plane in the wind?

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Statute Miles vs. Nautical · PDF fileStatute Miles vs. Nautical Miles Exercise 1: ... thousands of calculations for cargo weight and cubic space requirements, planners had to carefully