Newton’s First Law



FORCES ACTING ON AN AIRPLANE IN STEADY STATE FLIGHT

I. Wings Level Constant Altitude Flight

Newton’s First Law. A body at rest remains at rest, and a body in motion remains in motion at the same speed and in the same direction, unless acted upon by an outside force.

For an aircraft in steady state straight and level flight, this means that thrust (T) = drag (D), and lift (L) = weight (W), assuming that the thrust acts along the flight path (direction of flight).

Formally, (Fx = (Fy = 0, where the x-axis (thrust-drag axis) is by convention along the direction of flight. An airplane in this situation will maintain a constant airspeed and altitude.

Scalar vs. Vector. A scalar quantity has only magnitude, while a vector quantity has both magnitude and direction.

Thrust, drag, lift and weight are all vectors, and are measured in pounds. Airspeed is a scalar, since it has magnitude but not direction. (Note: velocity—airspeed plus direction—is a vector; we will discuss this point later in more detail) Other examples of scalars are temperature, pressure, density, heartbeat, GPA, salary, and height.

Pilots usually measure airspeed in knots (nautical miles per hour). Since there are 6076 feet/nm, and 3600 seconds/hour, we can convert knots to feet/sec as follows:

(x nm/hour) (6076 ft/nm) / (3600 sec/hour) = x’ ft/sec.

That is, to convert knots to ft/sec (fps), multiply by 6076/3600. To convert to feet/minute (fpm), multiply by 6076/60.

Example. 300 kts = 300 ( 6076 / 3600 = 506.3 ft/sec = 300 ( 6076 / 60 = 30,380 ft/min.

Exercise. Convert 500 kts to fps and fpm.

Relationships in a Right Triangle: A right triangle has one right (90o) angle and two acute angles. The hypotenuse c of the triangle is the side opposite the right angle. Let ( be one of the acute angles. The other two sides are a and b, with a adjacent to (, as shown below. That is, b is the side opposite (, and a is the side adjacent to (.

By the Theorem of Pythagoras, c2 = a2 + b2. Also,

sin ( = opposite/hypotenuse = b/c;

cos ( = adjacent/hypotenuse = a/c;

tan ( = opposite/adjacent = b/a.

Note the following limits on the values of these functions, which derive from the fact that 0 < ( < 90o.

As ( increases, 0 < sin ( < 1; 1 < cos ( < 0; 0 < tan ( < (. This will help you avoid errors when using your calculators. Recall also that

tan a = opposite/adjacent = (opposite/hypotenuse) / (adjacent/hypotenuse) = sin a / cos a.

Example.

Note that 102 + 17.322 = 100 + 299.9 = 399.9 ( 202 = 400. Sin ( = 10/20 = 0.5, so ( =sin-10.5 = 60o. Cos ( = 10/20 = 0.5, so ( =cos-10.5 = 30o. The values of the acute angles can also be calculated using the tangent function. This triangle is the well known “30-60-90” or “one-two-square-root-of-three” triangle.

Key Point: If c is the hypotenuse of a right triangle with acute angle (, then the side opposite ( is c sin (, and the side adjacent to ( is c cos (.

Suppose the thrust vector of an airplane is not parallel the flight path. This is usually the case, since ordinarily an aircraft in straight and level flight has a small nose up pitch angle p. The wing develops a force called the aerodynamic force (AF), which we will study in detail later. The component of the AF perpendicular to the flight path is called lift; the component parallel the flight path is called drag. (Of course, drag created by the AF is not the only kind of drag.)

Key Point. Lift is always perpendicular the flight path; drag is parallel the flight path. We assume thrust developed by an airplane is parallel the longitudinal axis. Weight always acts parallel to the earth’s gravitational field, i.e., vertically downward.

Example, Suppose W = 3000#, T = 1000#, and p = 7o. Since (Fx = 0, F+x = F-x, and D = T cos p. Since (Fy = 0, F+y = F-y, and L + T sin p = W. Then,

F+x = F-x F+y = F-y

D = T cos p L = W – T sin p

= 1000 cos 7o = 3000 – 1000 sin 7o

= 992.5# = 3000 – 121.87 = 2878#

That is, 122 pounds of weight are being supported by the vertical component of thrust. This shows that lift is not necessarily equal to thrust in straight and level flight. However, the relative values of T, W, and p are not necessarily those which apply to light civil aircraft.

Exercise: Suppose W = 5000, L = 4800, p = 5o. Find D and T.

II. Wings Level Climbing Flight

Now consider the case of an aircraft in a wings-level steady state climb. The angle parameters are c, the climb angle (flight path); p, the pitch angle; and r, the pitch angle relative to the flight path. Again, L is perpendicular to the flight path, and drag parallel to the flight path.

Recall that the x-axis by convention lies along the flight path. Then since F+x = F-x, T cos r = D + W sin c; and since F+y = F-y, L + T sin r = W cos c.

Example. Suppose p = 32.5o, c = 25o, T = 2000#, and W = 3000#. Then r = p – c = 32.5 – 25 = 7.5o, and

F+x = F-x F+y = F-y

T cos r = D + W sin c L + T sin r = W cos c

D = T cos r – W sin c L = W cos c – T sin r

= 2000 cos 7.5 – 3000 sin 25 = 3000 cos 25 – 2000 sin 7.5

= 1982.89 – 1267.85 = 715.0#. = 2718.92 – 261.05 = 2458#

Note that thrust is supporting 3000 – 2458 = 542# of the weight of the aircraft.

To sum up, in a steady state climb:

• Part of the weight of the aircraft is supported by thrust

• Lift is less than weight

• Part of the weight of the aircraft acts in the same direction as drag

• If Tx and Ty are respectively the components of thrust parallel to and perpendicular to the flight path, and Wx and Wy respectively the components of weight parallel to and perpendicular to the flight path. Then

1. Lift = Wy - Ty; and

2. Drag = Tx - Wx

Exercise. Suppose weight is 115,000#, climb angle 20o, pitch angle 30o, and thrust 52,500#. Find lift and drag.

Exercise. Suppose weight is 100,000#, climb angle 10o, pitch angle15o, and thrust 40,000#. Find lift and drag.

III Wings Level Power Off Descending Flight

The situation for steady state descending flight can be analyzed using an approach similar to that for climbing flight. We consider the special case of power off gliding descents, i.e., no thrust descents.

Since F+x = F-x, D = W sin a; and since F+y = F-y, L = W cos a.

Example. Suppose the glide angle is 10o and weight is 3000#. Find lift and drag.

F+x = F-x F+y = F-y

W sin a = D L = W cos a

= 3000 sin 10o = 520.945# = 3000 cos 10o = 2,954.42#

Example. Suppose lift is 8000# and drag is 1200#. What is the angle of descent? The weight?

D / L = W sin a / W cos a = sin a / cos a = tan a D = W sin a

1200 /8000 = tan a W = D / sin a

a = tan-1 (1200/8000) = 1200 / sin 8.530765610

= tan-1 0.15 = 8.530765610o = 8089.499# = 8089#

How can we determine the glide angle for maximum glide distance?

Tan a = D / L. Since the tangent function of an acute angle increases as the angle increases, minimizing D / L will minimize a. That is, the minimum glide angle (and hence maximum glide distance) is achieved when the lift-drag ratio L / D is maximum, i.e., at (L / D)max.

(L / D)max corresponds to a unique angle of attack (AOA) for a given airplane. We will discuss AOA in greater detail later in the course. As weight increases, the airspeed for maximum range power off glide increases, but AOA remains constant. Note that maximum power off glide distance is independent of weight; e.g., an airplane full of fuel glides just as far as the same airplane empty of fuel.

If the glide angle and altitude are known (or can be calculated), the glide distance can be determined as follows.

Tan a = altitude / glide distance, so glide distance = altitude / tan a.

Example. Suppose altitude is 15,000 feet, and a = 10o. Then glide distance (in feet and nm) can be calculated as follows:

glide distance = altitude / tan a

= 15,000 / tan 10 = 85,069.2273 ft

= 85,069 ft / 6076 ft/nm = 14.0008 nm.

Exercise. Suppose weight is 20,000#, and drag is 2000#. Find power off glide angle, lift, glide distance in nm from 5000 feet altitude AGL, and glide distance in nm per thousand feet of altitude.

III Constant Altitude Turning Flight

Here we discuss flight where speed, altitude, bank angle, turn radius, and G force are constant.

The circumference of a c circle of radius r and diameter d = 2 r is c = 2( r = ( d.

Example. What is the rpm of a 7’ radius propeller with a tip speed of 975 ft/sec?

Diameter of prop d is 14’, so in one revolution a blade tip moves 14 ( = 43.98229617 ft/revolution.

Then (975 ft/sec) (60 sec/min) / (14 ( ft/revolution) = 1330.08 revolutions/min (rpm).

In general, RPM (rev/min) = tip speed (fps) (60 sec/min)/ (2 ( r ft/rev)

Exercise. Suppose RPM = 2500 and propeller radius is 4’. What is the tip speed?

Newton’s Second Law. A body of mass m acted upon by a force F experiences an acceleration a in the direction of the force directly proportional to F and inversely proportional to m.

That is, F (#) = m (slugs) ( a (ft/sec2). Acceleration a is the rate of change of velocity.

|Time(sec) |Velocity (ft/sec) |

|0 |0 |

|5 |10 |

|10 |20 |

|15 |30 |

|20 |40 |

If acceleration of object reflected in above table is constant, a = 40 ft/sec / 20 sec = 2 ft/ sec2. That is, each second the object increases velocity by 2 ft/sec.

Mass m is measured in m slugs = (F #) / (a ft/sec2) = (F/a) #-sec2 / ft. That is, the slug (#-sec2 / ft) is the unit of mass accelerated at the rate of 1 ft/sec2 when acted upon by a force of one pound.

Mass is independent of gravity. Applying Newton’s second law, F = ma or W = mg, where g is the acceleration due to earth’s gravity, i.e., 32.2 ft/sec2. Thus, m = W / g, and F = W a / g.

Important fact. g = 32.2 ft/sec2. Memorize this.

Example. Suppose an airplane weights 10,000# and initial thrust on takeoff is 5000#. What is the acceleration?

F = Wa/g, so a = Fg/W = (5000#) (32.2 ft/sec2) / 10,000# = 16.1 ft/sec2.

Exercise. Suppose 10,000# of thrust and initial acceleration of 20 ft/sec2. Find weight of aircraft.

Let ( (cap phi) be the bank angle of an airplane in steady state constant altitude turning flight. The forces acting on the airplane are as follows.

For steady state flight, W = L cos (. L sin ( is the force causing the radial acceleration according to Newton’s second law. This centripetal force is what causes the airplane to turn.

Note 1 : The diagram on p.188 of the text is misleading, since the forces are balanced and the airplane would not turn. Centrifugal force is just the force opposite and equal to centripetal force according to Newton’s third law. It causes the “G force” which you feel during turning flight.

Newton’s Third Law. For every action there is an opposite and equal reaction.

Note 2: L cos ( is the effective lift which must support the weight of the airplane. Since L cos 90o = 0, there is no effective lift in a 90o bank. (Discuss implications.)

Cos ( = effective lift / total lift = W / L. Since G force is defined to be G = L / W, cos ( = 1 / (L/W) = 1 / G. so G = 1 / cos (.

As L increases, L sin ( increases, and the G force on the a/c increases, while radius of turn decreases in accordance with Newton’s Second law.

|G |Cos ( = 1/G |( (in deg) |

|1 |1 |0 |

|2 |0.5 |60 |

|3 |1/3 |70.53 |

|4 |0.25 |75.52 |

|5 |0.2 |78.46 |

|6 |1/6 |80.41 |

Important fact. G force is independent of weight or speed in a steady state level turn. It depends solely on the bank angle. No 90o steady state turn is possible.

Example. Suppose bank angle is 72o; what is the G force? Suppose G force is 3.75; what is bank angle.

G = 1 / cos 72 = 3.24. Cos ( = 1 / 3.75, so ( = cos-1 (1 / 3.75) = cos-1 0.26666667 = 74.53o.

Radius of Turn as a Function of Bank Angle and Airspeed.

It can be shown that a body circling a point at a distance of radius r with a velocity V experiences a radial acceleration of V2/r. Applying F = m a, we have

CF = m V2 / r = [pic]

Since W = L cos (, and CF = L sin (, r = [pic].

Note: V must be expressed in ft/sec. (Recall that multiplying by 6076/3600 converts knots to fps.)

Example. Suppose airspeed is 300 kts at 45o angle of bank. Then the radius of turn is

r = V2 / g tan (

=[pic]

Exercise. Suppose 380 KTAS at 50o angle of bank. Find radius of turn in feet and in nautical miles.

Rate of Turn as a Function of Bank Angle and Airspeed.

Since 2 ( r is the circumference of a circle of radius r, an a/c circling a fixed point at a distance of radius r feet travels

(2 ( r feet/revolution) / (360 degrees/revolution) =

(( r / 180) feet/degree

Then, if the a/c has a speed of V ft/sec, rate of turn RT is

TR = [pic], [pic]

As with radius of turn problems, V must be converted to fps if it is given in knots.

Example. What is the turn rate in the previous problem?

TR = [pic]deg/sec.

Example. Suppose a/s is 380 knots. What angle of bank is required for a SRT (3o/sec)? What is the turn radius and G force developed?

TR = [pic]deg/sec.

[pic]

( = 46.20293974o.

G = 1 / cos ( = 1/ (cos 46.20293974) = 1.444865088.

[pic]ft (or 12,248.99/6076 = 2.01596 nm).

Exercise. Suppose 90 KTAS. What angle of bank is required for a SRT? What is the G force? What is the turn radius in feet and nautical miles?

PROPERTIES OF THE ATMOSPHERE

I. Pressure.

Static Pressure P is the weight per unit area of a column of air. At sea level on a standard day (temperature = 15o C or 59o F),

P0 = 2116 #/ft2 = (2116 #/fts) / (144 in2/ft2) = 14.69444 #/in2.

(Alternative measure of standard pressure is 29.92 inches of mercury; the millibar is a unit of pressure used by meteorologists.)

Static pressure decreases as altitude increases. About ½ of the atmosphere by weight is below 18,000 feet, so the standard pressure at that altitude is approximately ½ the standard pressure at sea level.

The pressure ratio ( (small delta) at a given altitude with ambient pressure P is defined to be

( = P / P0

where P0 is as defined above. For example, (18000’ = 0.4992, reflecting the fact that about half the atmosphere lies below 18,000 feet. Note that in general, 0 ( ( ( 1; ( > 1 only when ambient pressure exceeds standard pressure at sea level.

II. Temperature.

To convert centigrade to Fahrenheit, C = 5/9(F-32). To convert Fahrenheit to centigrade, F = 9/5C + 32.

Example. 40oC = 9/5(40)+32 = 104oF. 70oF = 5/9(70 – 32) = 21.11oC.

Exercise. Convert 30 and 50 degrees centigrade to Fahrenheit. Convert 80 degrees Fahrenheit to centigrade.

Must use absolute temperature (absolute zero = -273oC) in temperature calculations! -273o C = 9/5(-273) + 32 = -459.4 ( -460o F (absolute zero in Fahrenheit).

Add –273 (-460) to centigrade (Fahrenheit) to get Kelvin (Rankine). K is the symbol for Kelvin, R for Rankine.

Sea level standard temperature:

T0 = 59o F = 59+460 = 519o R.

= 15o C = 15 + 273 = 288o K.

Temperature decreases with altitude until the tropopause (36, 089’), then stays constant at –69.7o F until about 85,000’.

The Temperature ratio ( (cap theta) at a given altitude where the temperature is T is defined as

( = T / To,

where T0 is as above.

Example. At the tropopause, ( = (-69.7+460) / (59 + 460) = ( (5/9) (-69.7 – 32) + 273) / (15+273) = 0.752

III. Density.

Density ( (small rho) is mass per unit volume. (Most important of the three properties in aerodynamics.)

( = mass / unit volume = slugs/ft3 = (#-sec/ft) / ft3 = (#-sec)/ft4

The density at sea level on a standard day is (0 = 0.002377 slugs/ft3.

The density ratio ( (small sigma) at an altitude where the density is ( is defined to be

( = ( / ( 0

where ( 0 is as above.

Example. (22000’ = 0.001183 slugs/ft3 in a standard atmosphere (to be defined shortly), so

(22000’ = (22000’ / ( 0 = 0.001183 /0.002377 = 0.497686159 ( 0.5.

From this we may conclude that about ½ the atmosphere by density lies below 22,000’. (Compare this to 18,000’ for pressure.)

Density of a gas is proportional to P/T; i.e. ( = k P / T for some k > 0, where the value of k depends on the properties of the gas. Thus,

( = ( / ( 0 = (kP/T) / (kP0/T0) = (P/P0) / (T/T0) = ( / (.

In English, this equation says: “Density ratio is equal to pressure ratio divided by temperature ratio.”

These and related ideas are summed up in the Standard Atmosphere Table (p. 18 text). The ICAO (International Civil Aviation Organization) adopted values in the table (based on observations and averages) to promote uniformity in measuring and comparing aircraft performance worldwide.

IV. Pressure and Density Altitude

Pressure altitude is altitude corresponding to a given static pressure in a standard atmosphere; i.e., is observed altitude corrected for non-standard pressure, assuming no altimeter error. (Set 29.92 plus or minus altitude correction in altimeter pressure window.)

If pressure is 972 #/ft2, then ( = 972 /2116 = 0.4594, which corresponds to approximately 20,000’ pressure altitude in a standard atmosphere.

Density altitude is pressure altitude corrected for non-standard temperature. (Recall that ( = k P / T, so that density is directly proportional to pressure and inversely proportional to temperature.)

Density altitude is used to construct aircraft performance charts for various altitudes, since air density most directly relates to performance parameters such as thrust, lift and drag.

We will return to these subjects later in the course.

ICAO Standard Atmosphere Table

PITOT-STATIC AIRSPEED INDICATORS

I. Air Behavior in a Bernoulli Tube

Consider a circular tube with a uniform gradual construction in it. Suppose air flows through this tube without being compressed (called incompressible flow, or subsonic flow in aerodynamics). Then the air must “speed up” in the “narrower” part of the tube, and the speed is inversely proportional to the cross-sectional area of the tube.

More precisely, let A1 and A2 be the cross-sectional areas (in ft2) of the tube at point 1 and point 2 respectively, and let V1 and V2 be the corresponding airflow velocities (in ft/sec) at these two points. Bernoulli discovered that

A1 V1 = A2 V2.

If A1 < A2, then V2 > V1; i.e., the air speeds up at point 2. Suppose A1 = 1 ft/sq2 and A2 = 0.2 ft/sq2, with V1 = 100 ft/sec, as shown in the Bernoulli tube depicted below. Then

A1 V1 = A2 V2, or V2 = (A1 V1) / V2 = (1 ft2) (100 ft/sec) / (0.2 ft2) = 500 ft/sec.

Now consider the dynamic air pressure q at points 1 and 2. As opposed to static air pressure, dynamic air pressure is pressure due to the force exerted by moving air. It can be shown that dynamic air pressure is proportional to the square of the air’s velocity. Specifically, dynamic pressure is given by

q = ( V2 / 2 (recall that ( is air density)

The unit for q is [pic], which is the same as for static pressure P. Let H stand for total pressure. Bernoulli also discovered that as dynamic pressure increases in a Bernoulli tube, static pressure decreases, and vice versa. That is, total pressure remains constant:

H = P + q.

This principle may be stated as follows:

Bernoulli’s Principle: Static pressure is inversely proportional to air velocity as velocity changes, and total pressure—static pressure plus dynamic pressure—remains unchanged.

Assume the Bernoulli tube depicted above is situated at sea level in a standard atmosphere. Then ( = 0.002377 slugs/ft3, and H = P = 2116 #/ft2. (Total pressure H and static pressure P are the same when air velocity V = 0, since then q = 0)

Then the dynamic pressures q1 and q2 at points 1 and 2 respectively is as follows:

q1= ( V12 / 2 = 0.002377 (100)2 / 2 ( 12 #/ft2; and

q2= ( V22 / 2 = 0.002377 (500)2 / 2 ( 297 #/ft2.

Then, applying Bernoulli’s Principle,

P1 = H1 – q1 = 2117 – 12 = 2104 #/ft2 at point 1, and

P2 = H2 – q2 = 2117 – 297 = 1819 #/ft2 at point 2.

This decrease in static pressure due to increased air velocity explains why an airfoil develops lift (as discussed in detail later in the course), and is also the basis for the pitot-static airspeed system.

II. Pitot-Static Airspeed Systems

These systems measure airspeed in terms of the difference between total and static pressure acting on an aircraft. As airspeed increases, dynamic pressure increases and static pressure decreases, while total pressure remains constant according to Bernoulli’s principle. Thus low static pressure is related to high airspeed, and higher static pressure to low airspeed.

Total pressure is measured at the tip of a pitot tube, where a stagnation point exists and the air velocity is zero. At this stagnation point, the static pressure is equal to the total pressure H, since H = P + q.

III. Types of Airspeed

IAS Indicated airspeed is the airspeed shown on the airspeed indicator.

CAS Calibrated airspeed is indicated corrected for pitot/static system errors. These are of two types:

a) errors in the a/s indicator itself

b) position error due to interference with airflow at the static port. Figure 2.6 (text p. 23) shows that typical errors are plus or minus 5-7 knots, with the larger errors occuring at very high airspeeds. (Good design minimizes the problem.)

EAS Equivalent airspeed is CAS corrected for compressibility effects. Such effects occur when ram air effects on the pitot tube cause the air entering it to be compressed. This error always results in an a/s reading that is too high. Fig 2.7 (text p. 23) shows this problem causes a/s indicators to read as much as 28 kts fast. Problems are worst at altitudes 20-40 M and airspeeds in the range 300-500 kts.

TAS True airspeed is EAS corrected for non-standard density (i.e., for altitudes where density differs from standard density at sea level). Useful primarily for navigation; has little or nothing to do with the way the aircraft responds to control input from the pilot’s point of view.

The following derivation shows an interesting and important relationship between EAS and TAS:

q = ( V2 / 2 and ( = ( / (0, so ( = ( (0, and q = ( (0 V2 / 2 = ((0/2) (( V2).

That is, dynamic pressure is directly proportional to (( V2), since ((0/2) is a constant. It follows that

(TAS)2 ( = (EAS)2 (0 = (EAS)2, or (TAS)2 = (EAS)2 / (, and TAS = [pic].

This helps explain why the ICAO Standard Atmosphere Table has a [pic] column. Note: [pic] is a value that you will meet again in AS310 Performance. It is called SMOE, which is an acronym for standard means of evaluation.

-----------------------

a

D

T

W

L

b

c

(

(

20

10

17.32

(

D

T cos p

W

L

T sin p

T

p

flight path

(

c

c sin (

c cos (

flight path

(x-axis)

p

T

T sin r

D

T cos r

W

L

c

r

W sin c

W cos c

p = pitch angle

c = climb angle

r = p – c = relative pitch angle

c

flight path

L

W cos a

W sin a

W

a (glide angle)

a

a

glide path (x-axis)

D

L cos (

L sin (

L

W

(

glide distance

D

altitude AGL

a

glide path (x-axis)

L

L = W cos a

D = W sin a

W

a

a

L sin (=CF

Total L

W

(

Effective L

L L L30 L L45 L L60

L

V

r

CF

0o 30o L30 = 1.1547 L 45o L45 = 1.414 L 60o L60 = 2 L

W

(

L cos (

( r / 180 feet

1o

r not to scale

Point 1 Point 2

A1 = 1 ft2 A2 = 0.2 ft2

V1 = 100 ft/sec V2 = 500 ft/sec

q1 ( 12 #/ft2 q2 ( 297 #/ft2

P1 ( 2104 #/ft2 P2 ( 1819 #/ft2

Stagnation point of airflow at tip of pitot tube

(thus total pressure H is measured inside pitot tube) static port (dynamic

pressure here is P)

Pitot Tube

pressure here is P,

pressure here is H = q + P, static pressure, which

with q increasing as airspeed remains constant

increases as q increases

degree of diaphragm displacement

measures H – P = q, the dynamic pressure

due to airflow velocity (airspeed)

[pic]

[pic]

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