The text's website has available for download the following files containing the STELLA models in this module: Fall.stm, FallFriction.stm, and FallSkydive.stm. (Right click, save the file, and open in STELLA.)
What is it like to skydive? Imagine ascending in a small plane to say 10,000 feet, when the jumpmaster opens the door. The jumpmaster asks you if you are ready to jump. You head for the door and walk out onto a step under the wing, holding on to a strut. You experience lots of wind and noise. Your heart is pounding wildly. The jumpmaster yells, “Go!” You arch your body and release your grip on the strut. Your adrenalin levels have never been higher as you plunge toward earth at 120 mph. Nevertheless, you are in control. For the next fifty seconds, simple body movements can alter your speed, direction and position. At three thousand feet, the landscape is fast approaching, and you pull your cord. As it deploys, your descent slows, and the mad rush of wind ceases, replaced by the rustling sounds of your canopy. Soon you gently settle to the ground.
The use of parachutes or parachute-like devices to slow the descent of jumpers from positions of considerable height may have begun with the 12th Century Chinese. However, the first evidence of a parachute in the western world appeared in the late 15th Century drawings of Leonardo da Vinci. His pyramid-shaped design was to be constructed of linen and a wooden frame. There is no record of Leonardo experimenting with his invention, but late last century it was demonstrated successfully.
Not much development of parachutes took place until late in the 18th Century, when hot air balloons were being shown across Europe. Andres-Jacques Garnerin, a French balloonist of dubious reputation, was one of the first persons to demonstrate a parachute without a rigid frame. He successfully descended from his balloon (which exploded) at about 3000 feet using a gondola suspended by an umbrella-shaped parachute.
Jumps using parachutes from airplanes began in the early 20th Century but were primarily used for rescuing observation balloon pilots. Barnstormers performed parachute-jumping demonstrations at air shows during the time between the world wars. During World War II, both sides exploited the capabilities of parachutes for dispersing men and supplies.
Sport parachuting (skydiving) probably has its roots in the first freefall conducted in 1914, but the sport really gained popularity only in the 1950’s and 1960’s.
In this module, using STELLA we model the motion of someone skydiving. Such a jump has two phases, a free-fall stage followed by a parachute state with greater air friction. In preparation for development of this model, we reconsider the main example in the module on "Rate of Change" and a follow-up exercise (Exercise 1) in the module "Fundamental Concepts of Integral Calculus" involve the motion of a ball someone throws straight up from a bridge. We model this motion with STELLA, first ignoring air friction and then refining the model to consider this additional force.
Acceleration, Velocity, and Position
The above-mentioned modules on calculus discuss how the instantaneous rate of change, or derivative, of position (s) with respect to time (t) is velocity (v), and the instantaneous rate of change of velocity with respect to time is acceleration (a). In derivative notation, we have the following:
In Example 1, we use these derivatives in modeling the main illustration from the module on "Rate of Change."
Example 1 To model in STELLA the motion of a ball that someone throws straight up from a bridge, we have stocks for the quantities that accumulate, the height (position) and velocity (velocity) of the ball. During the simulation, we can observe their changing values in a graph and table. A flow representing the change goes into velocity (change_in_velocity). Change in velocity is acceleration, and in this case, the acceleration is due to gravity. Therefore, a converter (acceleration_due_to_gravity) contains the constant for acceleration due to gravity, which with up being the positive direction is approximately -9.81 m/sec2. The converter connects to change_in_velocity, which has an equation equal to this constant (-9.81). Also, the flow for the change in height (change_in_position) is identical to the current velocity, velocity. Thus, we have a connector from velocity to change_in_position, and define the value of this flow to be velocity. Because velocity can be positive, zero, or negative, we make the flow a biflow. For flexibility in models that we derive from this one, we also make change_in_velocity a biflow. Moreover, we uncheck the Non-negative boxes for velocity and position, allowing each to take on negative values. To match the example in the earlier modules, we initialize velocity to be 15 m/sec and position to be 11 m, which is the height of the bridge. Figure 1 presents a diagram for the STELLA model of motion of the ball.
Figure 1 STELLA diagram motion of ball thrown straight up
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Output consists of a graph and a table of velocity and height versus time. For greater accuracy in the simulation, we specify the Runge-Kutta 4 integration technique under Run Specs…. The chapter on "Simulation Techniques" discusses this and the other integration techniques that STELLA employs. With a ∆t = 0.25 sec, we obtain a table of values that matches Table 1 in the module on "Rate of Change." The graph of velocity in Figure 1 agrees with a similar graph (Figure 2) in that module. As Figure 2 shows, the graph of velocity versus time is the line v(t) = 15 - 9.8t.
Figure 2 Graph of velocity and position of ball versus time
For some of the models, it is more convenient to consider speed than velocity. Speed gives the magnitude of the change in position with respect to time, while velocity expresses the magnitude with the direction. Thus, speed is the absolute value of velocity. To incorporate speed, we have a connector from the velocity stock to a new converter, speed, that stores the equation for the absolute value of velocity, ABS(velocity). The graph in Figure 3 shows speed and velocity decreasing in a linear fashion to 0 m/sec at about time 1.5 sec. Afterwards, speed steadily increases.
Figure 3 Graph of velocity, position, and speed of ball versus time
Before developing additional examples of falling and skydiving, we need to consider some formulas from physics--Newton's Second Law and approximations of friction. Newton's Second Law concerns force applied to a mass imparting acceleration. So that we can refine models to account for air friction, we also consider several approximations of such a force.
Newton's Second Law has far-reaching significance. In this text, we employ the law in modeling situations from the motion of skydivers to the motion of the planets. The law states that a force (F) acting on a body of mass (m) gives the body acceleration (a). Moreover, as the following models indicate, the acceleration is directly proportional to the force and inversely proportional to the mass:
a = F / m
or
F = ma
Newton's Second Law: A force, F, acting on a body of mass, m, gives the body acceleration, a, according to the following formula:
F = ma
We can apply this formula to obtain the relationship between weight and mass. Weight is a force and is not the same as mass. The acceleration involved is acceleration due to gravity, which is about -9.81 m/sec2 or -32 ft/sec2 for up being the positive direction. For example, an object that has mass of 20 kilograms (kg) has weight of -196.2 newtons, as the following shows:
weight = F = (20 kg)(-9.81 m/sec2) = -196.2 kg m / sec2 = -196.2 newtons
The metric unit for force is a Newton (N) or kg m / sec2.
| Definition | A Newton (N) is a measure of force, and 1 N = 1 kg m / sec2. |
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Kinetic friction or drag, too, is a force. This force between objects is in the opposite direction of a moving object and tends to slow motion. Thus, kinetic friction dampens motion of an object. When an object moves through a fluid, such as air or water, the fluid friction is a function of the object's velocity. For example, the faster we attempt to run in a swimming pool, the harder it is for us to do so. As our velocity increases, so does the friction of the water on our bodies.
Several models that estimate friction exist. In the module on "Empirical Models," we study how to derive our own model, such as a model for drag, from data. In this module, we consider two models for drag on a body traveling through a fluid.
For a small object traveling slowly, such as a dust particle floating through the air, we usually employ Stokes' friction, which states that friction on the particle is approximately proportional to its velocity,
F = kv
where k is a constant of proportionality for the particular object and fluid, and v is the velocity.
For a larger object moving faster through a fluid, we usually employ Newtonian friction, which states that drag is approximately as follows:
where C is a constant of proportionality (the coefficient of drag or drag coefficient) related to the shape of the object, D is the density of the fluid, and A is the object's projected area in direction of movement. For a particular situation, C, D, and A are constants, so that the drag is approximately proportional to the velocity squared. At 0 °C, the density of air at sea-level is 1.29 kg/m3. For shapes that are hydro-dynamically good, C < 1; for spheres, C is about 1; and for shapes that are hydro-dynamically inefficient, C > 1. Many objects have a coefficient of drag of about 1. Thus, through air with C = 1, Newtonian friction is approximately the following:
The density of water at 3.98 °C, where the fluid achieves its maximum density, is 1.00000 g cm-3, yielding a formula with a different coefficient. Table 1 summarizes the three models for fluid friction considered here.
Table 1 Summary of several models for fluid friction
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Name |
Formula |
Meanings of Symbols |
When to Use |
|
Stokes' friction |
F = kv |
k - constant v - velocity |
Very small object moving slowly through fluid |
|
Newtonian friction |
F = 0.5CDAv2 |
C - coefficient of drag D - density of fluid A - object's projected area in direction of movement v - velocity |
Larger objects moving faster through fluid |
|
Newtonian friction through air |
A - object's projected area in direction of movement v - velocity |
Larger objects with C = 1 moving faster through sea-level air |
The drag force is in the opposite direction of motion, and the sign of velocity indicates the direction. On the upward portion of the trajectory, drag and gravity both act downward; while on the downward part, drag is upward, and gravity downward. Thus, for the general formula for Newtonian friction, we take the absolute value of only one of the velocity terms and multiply the entire formula by –1, yielding -0.5CDAv|v|. The translation of this formula into STELLA is as follows:
-1.29 * drag_coefficient * density * projected_area * velocity * ABS(velocity)
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Example 2 Example 1 to model the motion of a ball thrown straight up does not account for air friction. To do so, we consider two forces on the ball, gravity and drag friction. The force due to gravity is its weight, which by Newton's Second Law is F = ma. Thus, in the STELLA diagram (Figure 3), we include a converter for weight with connections from converters for mass and acceleration_due_to_gravity. Newtonian friction for the air friction including direction is F = -0.65Av|v|. In the STELLA diagram, connectors go from the velocity stock and from a new area converter to a new converter for air_friction. The area converter stores the cross sectional, or projected, area of the object in the direction of motion. Presuming spherical objects, another converter stores the radius, and the equation in area is Pi * radius^2. Both the converters for force, weight and air_friction, connect to a new converter for total_force, which is the sum of the individual forces. Employing Newton's Second Law again with a = F / m, acceleration is total_force / mass. This acceleration provides the change in velocity for the flow into velocity.
Figure 4 STELLA diagram motion of ball under influence of air friction
Figure 3 contains a feedback loop. The initial value of air friction employs the initial velocity, here 0 m/sec; and air_friction contributes to the total_force, which acceleration uses. Acceleration is the change_in_position, which contributes to velocity. Then the current value of velocity "feeds back" into the air_friction converter for a new computation of that force.
To detect the influence of drag, we consider the ball of mass 0.5 kg and radius 0.05 m dropped (initial velocity = 0 m/sec) from a height of 400 m. Figure 5 presents the STELLA equations for the model in the order of execution.
Figure 5 STELLA equations to accompany diagram in Figure 4
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{ INITIALIZATION EQUATIONS }
{ RUNTIME EQUATIONS }
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Running the simulation for 15 seconds, we see in Figure 6 that the ball reaches a constant, or terminal speed, of about 31 m/sec. From about time 6 seconds on, the position graph is almost linear, so that acceleration is approximately 0 m/sec2.
Figure 6 Graph of position and speed of object under influence of friction
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Example 3 To model a skydive, we build heavily on Example 2 of a falling object. For simplicity, we consider someone jumping out of a stationary helicopter at 2000 m (about 6562 ft), and we ignore changes in air density. A project considers parachuting out of a moving plane, which imparts a horizontal velocity to the jumper. The model for a skydive out of a helicopter has two phases, one where the person is in a free-fall and another after the parachute opens when the larger surface area results in more air resistance. For our model, the main difference in these two phases is the projected area in the direction of motion, down. The cross-sectional area of a jumper in the stable arch position with arms arched back and legs bent at the knees is approximately 0.4 m2 (about 4.3 ft2). Parachutes vary in their designs, but 28 m2 (about 301 ft2) is a reasonable value. We trigger the pull of the ripcord by the height (position) above the ground, say 1000 m (about 3281 ft). Thus, the diagram must contain a converter (position_open) for this quantity and connectors from position to position_open and from position_open to projected_area. Figure 7 presents a STELLA diagram for this example with changes from Figure 4 on a ball's fall in color. The equation in projected_area is no longer a constant but is the following conditional statement:
IF (position > position_open) THEN 0.4 ELSE 28
Figure 7 STELLA diagram motion of skydiver under influence of air friction
Figure 8 shows graphs of the position and speed of a 90 kg (comparable to about 198 lb) skydiver versus time. Until a height of 1000 m, which occurs at about 27 seconds into the fall from 2000 m, the skydiver is in a free fall approaching a terminal speed of about 41.4 m/sec (about 93.0 mph). At 1000 m, the person pulls the ripcord, and in a very short amount of time, the parachutist's speed slows to a new terminal speed of 4.94 m/sec (about 11.1 mph).
Figure 8 Position and speed of skydiver
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Assessment of the Skydive Model
The shapes of the graphs of position and velocity in Figure 8 match the opening description of a skydive. However, our model exhibits a terminal speed of about 93.0 mph (about 41.4 m/sec), while actual, measured speeds of 110-120 mph are common. The example employs the sea-level density of air, while the air density at 10,000 ft (about 3,048 m) is about 73.8% (0.952 kg/m3) that of sea-level density. Adjusting the initial position to be 3048 m and using an air density of 0.952 kg/m2 with the Newtonian friction of F = 0.5CDAv2, the model indicates a terminal velocity of 48.15 m/sec (about 108 mph) for the free fall for less than 50 seconds. However, air density changes as the skydiver descends. Various projects explore refinements of the model to account for this variation. Another project also considers the skydiver jumping out from a moving plane as opposed to from a stationary helicopter.
1. a. Using the equations and values of Example 1, write differential equations with initial conditions for acceleration and velocity.
b. Using calculus and/or Mathematica, solve the differential equations of Part a to obtain velocity and position as functions of time.
2. Adjust Fall.stm of Example 1 so that the object falls with an initial velocity of zero. Compare the results with those in FallFriction.stm of Example 2, which accounted for friction.
3. a. Using the equations and values of Example 2, write a differential equation involving the derivative of velocity for when an object reaches terminal velocity. At terminal velocity, the forces acting on the body are equal.
b. Solve the equation of Part a using calculus and/or Mathematica.
4. Give the adjustments to the STELLA diagram in Figure 7 along with equations so that graphs new converters adjusted_position and adjusted_speed become horizontal lines at 0 after the parachutist lands.
5. Repeat Exercise 3 using Stokes' friction instead of Newtonian friction.
6. Suppose a raindrop evaporates as falls but maintains its spherical shape. Assume that the rate at which the raindrop evaporates (that is, the rate at which it loses mass) is proportional to its surface area, where the constant of proportionality is –0.01. Recall that the density (mass per volume) of water at 3.98 °C is 1 g/cm3. The surface area of a sphere is 4 π r2, and its volume is 4 π r3/3, where r is the radius. Project 8 models the motion of this raindrop under the influence of air resistance.
a. Assume that the initial radius is 0.3 cm. Determine the raindrop's initial mass.
b. Write a differential equation for the rate of change of mass as a function of r.
c. Write an equation for r as a function of mass.
7. Adjust Example 3 so that the parachute opening depends on time, not height above the ground?
8. Write a system of differential equations to represent Example 3.
1. Develop a model to estimate the total change in position of the car with velocities from Table 2 of "Fundamental Concepts of Integral Calculus." Employ an input graph instead of an equation to record Table 2's values for the change in position. Give the absolute and relative errors of your estimate in comparison to the exact value of 203 m.
2. Using Stokes' friction, develop a model for the motion of a dust particle floating down from a height of 50 m. Using comparative plots, determine its terminal speeds for various values of Stokes' constant of proportionality?
3. A bathysphere is a pressurized metal vessel in the shape of a sphere that allows people to explore the ocean to much greater depths than are possible by skin diving. A ship lowers and raises the sphere using a steel cable and communicates with its two occupants by telephone. In the 1930s, explorers William Beebe and Otis Barton developed the first bathysphere, which weighed 4,500 pounds and had a diameter of 4'9". In a subsequent one, they descended to about 3,000 ft in the ocean. Ignoring currents but not drag, model the sinking motion of a bathysphere. Assume that the boat reels out the steel cable fast enough so as not to affect the bathysphere's motion. [5] [8]
4. Table 2 below contains air densities at various altitudes. Using these values on an input graph, refine the model for Example 3 [1].
Table 2 Approximate air densities at various altitudes
|
Altitude (m) |
Density (kg/m3) |
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|
0 |
|
1.290 |
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610 |
|
1.216 |
|
|
1219 |
|
1.146 |
|
|
1829 |
|
1.078 |
|
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2438 |
|
1.014 |
|
|
3048 |
|
0.952 |
|
|
3658 |
|
0.894 |
|
|
4267 |
|
0.839 |
|
|
4877 |
|
0.786 |
|
5. Suppose an airplane is traveling in a straight line horizontally at 130 m/sec at a height of 600 m when a parachutist jumps out of the plane at an angle of 30° with the horizon. Model the motion of the skydive.
6. Model the motion of a meteor falling to the earth. Assume an initial height of 100,000 m, initial velocity of –10,000 m/sec, coefficient of drag of 2, mass of 500 kg, density of 8000 kg/m3 for iron or 3500 kg/m3 for stone [7]. Give graphs for position, velocity, and acceleration versus time. Give comparison graphs for velocity versus height for meteors of various masses. Similarly, give comparison graphs for acceleration versus height. NASA's Glenn Research Center gives the following model for air density using variables D - density (slugs/ft3), P – pressure (lbs/ft2), T – temperature (°F), and h – altitude (ft): D = , where
for h > 82,345 ft, T = -205.05 + 0.00164 h and P = 51.97;
for 36,152 < h < 82,345 ft, T = -70 and P = 473.1; and
for h < 36,152 ft, T = 59 – 0.00356h and P = 2116.
Note, if you wish to use Metric instead of English units, you can use the following: 1 slug = 14.5939 kg and 1 ft = 0.3048 m. [2]
7. Using NASA's Glenn Research Center gives the following model for air density at heights less than 36,152 ft (see Project 6), refine the model in Example 3.
8. a. Model the change in mass of the raindrop that Exercise 5 describes.
b. Model the motion of this raindrop taking into account air resistance.
9. Develop a model to compare the terminal velocities of objects of different masses, such as a mouse, cat, human, horse, elephant, etc. With the density of living protoplasm being almost constant across a wide variety of species, assume mass is proportional to the cube of a linear dimension, such as length or circumference; but surface area is proportional to the square of a linear dimension. How do the terminal velocities of more massive objects compare to those of less massive objects? Can a cat survive a fall from a tall building? [3]
1. Aber, James S. and Susan W., Emporia State University, “High-altitude Kite Aerial Photography,” Kite Aerial Photography, April, 2003. Available from: http://www.geospectra.net/kite/weather/h_altit.htm
2. Benson, Tom, Glenn Research Center, “NASA Glenn Learning Technologies Project,” Earth Atmosphere Model. Available from: http://www.grc.nasa.gov/WWW/K-12/
3. Diamond, J., "How cats survive falls from New York skyscrapers," Natural History, August, 1989
4. dos Santos, Carl P. E., Adventure Living, “Interesting Data (Adventure Living).” Available from: http://www.adventureliving.com/home/skydiving/data/index.html
5. EnchantedLearning.com, “Undersea-Related Inventors and Inventions.” Available from: http://www.enchantedlearning.com/inventors/undersea.shtml
6. Glenn Elert, ed., “Speed of a Skydiver (Terminal Velocity),” The Physics Factbook, 2002. Available from: http://hypertextbook.com/facts/JianHuang.shtml