Mathematical Modeling Project in Mathematica Cynthia Wenz MA 354

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1 clw.ma 1 Mathematical Modeling Project in Mathematica Cynthia Wenz MA 354 Introduction and Problem Statement When a parachute jump is made from an airplane, the first part of the descent is made in free fall at a relatively high altitude. After a certain time the parachutist then opens the parachute to enable a soft landing to be completed. The important and interesting question for the jumper is to decide when the parachute should be opened. Obviously it must not be done too late or the landing speed will be too great and the parachutist may face injury and possibly death. On the other hand, at a great altitude with a thin atmosphere a free fall can generate great speeds which can add exhileration. There may also be more practical reasons for not opening the parachute too early such as the proximity of the plane and the other jumpers. It is therefore of interest to model the motion of the parachute jump and to determine if there is an optimum position at which to open the chute to maximize enjoyment and maintain safety. If we take an airplane travelling horizontal at a given height and speed the most important simulation of a parachute jump will be to make accurate assessments of the air resistance and the obvious change in air drag when the parachute is opened. The idea of terminal speed becomes an important factor. At time t=0 when the parachutist initially leaves the plane his speed is zero and the retarding force of air resistance is also zero. The acceleration will simply be g, 9.8 m/s 2, in the downward direction. As the speed of the jumper increases, the drag force increases and the acceleration becomes less than g. Eventually, the speed is great enough for the drag force, given by bv n to equal the force of gravity mg, so the acceleration is zero. The object then continues moving at a constant speed v t, called its terminal speed. constants Let Fnet = mg - bv n = ma where bv n is the drag force with b and n as Then if in a state of free fall when a = 0 because the terminal velocity has been

2 clw.ma 2 achieved, bvtn = mg vt= (mg/b)^(1/n) Therefore the larger the constant b, the smaller the terminal speed. In regard to the parachute problem, it is reported that the terminal speed of human free fall is about 120 miles/hour. The parachutist hopes to make a soft landing, a landing that is essentially equal to falling off a 12ft wall. This means that the terminal velocity with parachute open must be the same as the velocity acquired on falling off the wall. More simply put it is useful to achieve a terminal speed during the parachute-open portion of the jump that is equivalent to the speed attained when one falls off a 12 ft wall. Simple energy equations on falling through a height h( and neglecting air resistance over a small distance, e.g. 12 ft) yield Clear[ h, g, m] g=9.8065; (*meters/sec^2*) h=(12 * (12/39.37)); (*meters*) Solve [(1/2) * m * v^2 == m * g * h, v] {{v -> }, {v -> }} So the velocity one hopes to land at is approximately 8.47 m/s. If we allow for air resistance proportional to the square of the velocity or v air resistance constant k is given by v=8.47; (*meters/sec*) Solve[ k== (g/v^2), k] 2 = g/k, then the {{k -> }} So the air resistance constant for the landing at terminal speed is approximately.1367 m-1. The air resistance constant for the free fall part can be calculated from the terminal velocity v = g/k which is the case when the air resistance is taken to be directly proportional to velocity. This is valid since the atmosphere is fairly thin at a great height and the falling human body is relatively compact. If we take the terminal velocity to be 120 miles/hour, which is equal to m/s then,

3 clw.ma 3 v = Solve [k == g/v, k] {{k -> }} (*meters/sec*) So the air resistance constant for the free fall will be approximately.1828 s-1. Now that the issue of air resistance is clarified and the constants are derived, the model of the parachute fall can be formulated.

4 clw.ma 4 Mathematical Model In order to give a better view of the situation, I offer the rather crude representation given below. The rectangle and the line extending from it represent the plane and its continuing path as it flies away. The green line is that of the parachutist and his descent. The rather odd circle is a representation of the angle theta which will become important in building equations for the motion of the parachutist. height ground It will be necessary to define a table of symbols and units. Description Type Symbol Units Horizontal distance travelled by parachutist Variable x(t) m Vertical distance dropped by parachutist Variable y(t) m Time Variable t s Velocity of parachutist Variable v m/s Mass of parachutist plus equipment Parameter m kg Air resistance force Variable R N Air resistance constant Parameter k s -1 Acceleration due to gravity Constant g m/s 2 Angle of inclination of path of parachutist Variable theta deg Initial height Parameter h m Initial horizonatal velocity Parameter u m/s

5 clw.ma 5 To formulate a model equations of motions for the parachutist must be defined. Application of Newton s law gives two equations: one for the horizontal motion and one for the vertical. For the horizontal motion, -R cos ϑ = m x [t] and for vertical motion, -R sin ϑ + mg = m y [t] where, R = mkvn. Ιn essence these equations are a representation of the standard F = ma. They are formulated with respect to the component forces of the vertical and horizontal directions. Τhen from calculus we know, v[t] = Sqrt [ (x [t])^2 + (y [t])^2] x [t] = v[t] cos ϑ y [t] = v[t] sin ϑ The exponent n will be taken as 1 for free fall and 2 for the effect once the parachute has opened. Using the above relations we can easily eliminate v and ϑ to get x [t] = -k x [t] (x [t]^2 + y [t]^2) ^ ((n-1)/2) y [t] = -g - k y [t] (x [t]^2 + y [t]^2) ^ ((n-1)/2) These two differential equations make up the general mathematical model for the parachute drop. It is important to note that the assumption is made that the motion takes place in one plane. In other words this model is only two dimensional. For the first part of the motion, the free fall, the assumption is made that n=1 which leads the two equations to reduce to x [t] = -k x [t]

6 clw.ma 6 y [t] = -g - k y [t] To find the equations of motions Mathematica is used to solve the previous differential equations. Clear[ solution1 ] solution1 = DSolve[ {x [t] == -k x [t], x[0] == 0, x [0] == u}, x, t] u u {{x -> Function[t, ]}} k k t E k Clear[ solution2 ] solution2 = DSolve[ {y [t] == -g - (k y [t]), y[0] == 0, y [0] == 0}, y, t] g g g t {{y -> Function[t, ]}} 2 k t 2 k k E k The two equations solved for above are x[t] = ( u/k ) * [1 - exp( -kt) ] y[t] = - (gt/k) + [ (g/k2) * [ 1- exp( -kt) ] So long as the assumptions made on air resistance hold and the parachute is not opened, then these two equations give the path of the falling person. Through some manipulation of the equations above it is possible to solve for y directly in terms of x and eliminate t. Clear[ x, k, t, y, g] Solve [ x == (u/k) - (u/(k * Exp[ k*t])), t] u Log[ ] u - k x {{t -> }} k t = Log[u/(u - k*x)]/k; Solve [y == -(g t / k) - (g / (Exp[k t] k^2)) - (g / k^2), y] u 2 g u - g k x + g u Log[ ] u - k x {{y -> -( )}} k u

7 clw.ma 7 The above shows that y can be explicitly expressed in terms of x by eliminating t and this is equivalent to y = - (gx/ku) - ((g/k 2)ln( 1 - (kx/u) )) While it is of interest to see the graph of the motion, it is better to do so later on since there is a differing k value that is dependent on time. It is of interest to find the velocity of the jumper as the drop takes place which can be done by taking the de of the equations for the motion which is given by x [t] and y [t]. Clear [ u, k, t] D[ (u/k) - ((u/k) * Exp[-k t]), t] u ---- k t E D[ -(g t / k) - (g / (Exp[k t] k^2)) - (g / k^2), t] g g -(-) k k t E k So the equations describing velocity for the first part of the fall are x [t] = u/e^(k*t) y [t] = - g/k - g/(e^(k*t)*k) Since v2 = x [t] 2 + y [t] 2 v2 is equivalent to 2 g g 2 u (-(-) ) k k t 2 k t E k E The graph of the equation shows the velocity of the jumper while he is in free fall.

8 clw.ma 8 g= 9.8; (*meters/sec2*) k=.1828; (*s-1*) u= 125; (*meters/sec*) Plot[ Sqrt[%13], {t, 0, 10.18}, PlotRange -> {{0,11},{45,125}} ] It can be seen from the graph and its behavior that as time elapses v has a minimum value. In other words, after the jumper leaves the plane his velocity decreases until it reaches a minimum. If one considers that initially he had zero vertical velocity and the horizontal velocity of the plane it makes sense that since he loses the planes propulsion with time that his velocity will reach a minimum before it starts to increase again. This can also be checked through differentiation. If the derivative of the velocity equation is set equal to zero, the time for the minumum can be found and through substitution so can the velocity. D[(g/k - g/(e^(k*t)*k))^2 + u^2/e^(2*k*t), t] g g 2 g ( ) k k t 2 E k 2 k u k t 2 k t E E By taking this derivative and setting it equal to zero we can find the time of this minimum velocity. Solve [ (2*g*(g/k - g/(e^(k*t)*k)))/e^(k*t) - (2*k*u^2)/E^(2*k*t) == 0, t] g + k u Log[ ] 2 g {{t -> }} k

9 clw.ma 9 The speed at this time, t, is obtained by substituting the above into the equation for v simplifying. 2 and t = Log[(g^2 + k^2*u^2)/g^2]/k; Solve [v^2 ==(g/k - g/(e^(k*t)*k))^2 + u^2/e^(2*k*t), v] k u k u {{v -> -(u Sqrt[ ])}, {v -> u Sqrt[ ]}} g + k u g + k u Simplify[ u*(1 - (k^2*u^2)/(g^2 + k^2*u^2))^(1/2) ] 2 g u Sqrt[ ] g + k u The minimun speed is given above. Now the initial value of v is u, the horizintal speed of the plane and the terminal speed is g/k for the free fall; so the miniumum speed of the parachutist is logically less than the that of the initial value of u and less than the terminal speed. and 2 u Sqrt [ g ] < u (initial speed) g2 + k2u2 2 u Sqrt [ g ] < g/k( terminal speed) g2 + k2u2 Taking k =.1828 s-1 and u = 125 m/s (the initial horizontal speed of the the parachutist, the same as that of the plane) we can find the time and the position of the jumper at the minimum velocity as well as that velocity itself. k =.1828; u = 125; g = ; Solve [ v == u*(g^2/(g^2 + k^2*u^2))^(1/2), v] {{v -> }} Solve [t == Log[(g^2 + k^2*u^2)/g^2]/k, t] Solve[True, ]

10 clw.ma 10 t = ; Solve [ x == ( u/k ) * (1 - Exp [-k t] ), x] {{x -> }} Solve[ y == (g t/k) - ( (g/k^2) * ( 1- Exp[ -k t] )), y ] {{y -> }} So the parachutist reaches a minimum speed of 49.31m/s after 10.18s and he has travelled 577.5m in the horizontal direction and fallen m down. The minimum speed obtained by the jumper during free fall is then around 110 miles per hour. Regardless of the value for n and k the parachutist will still experience decreasing velocity before speeding back up again. This time of minimum velocity is perhaps the most appropriate time to open the parachute. When the parachute is opened, there will be a sharp jerk due to the abrupt change in the resistance force. This force can be considerable (and probably unpleasant for the parachutist), which is why the decision can now be made to open the parachute when the speed is a minimum. We can calculate this change in force which causes the jerk from the difference between the values of R before and after the opening. Before the opening, R = mk1v = m * * ( where m is the mass of the jumper) and after the opening, R = mk2v2 = m *.1367 * (49.31) *49.31*m m.1367*49.31^2*m The difference is about 320 * mass (Newtons) which is large, but on the other hand the parachute will not be opened instantaneously; so this change may be spread over several seconds. To simulate the remainder of the this descent, the same equations need to be solved when k=.1367m-1 and n=2. These will represent the motion once the parachute is opened. Once again these are the equations.

11 clw.ma 11 x [t] = -k x [t] (x [t]^2 + y [t]^2) ^ ((n-1)/2) y [t] = -g - k y [t] (x [t]^2 + y [t]^2) ^ ((n-1)/2) So with n=2 and the new value for k NDSolve is used to interpolate the functions over values for time. Since seconds has elapsed in model, initial conditions will be given at seconds. Then the final part of the descent can be graphed. Clear [ g, k, t, v, x, y, solution] k =.1367; g = ; solution = NDSolve[ {-k x [t] ((x [t]^2 + y [t]^2)^.5) == x [t], -g -(k y [t] ((x [t]^2 + y [t]^2)^.5)) == y [t], x[10.18] == 577.5, x [10.18] == 19.44, y[10.18] == , y [10.18] == }, {x[t],y[t]}, {t,10.18,40}] {{x[t] -> InterpolatingFunction[{10.18, 40.}, <>][t], y[t] -> InterpolatingFunction[{10.18, 40.}, <>][t]}} ParametricPlot[ Evaluate[ {x[t],y[t]} /. solution ], {t, 10.18,40}, PlotRange -> {{0,600},{0,600}}, PlotStyle -> RGBColor[1,0,0], AxesLabel -> {ground, height}] {Rectangle[{0,500},{50,550}]} {Rectangle[{0, 500}, {50, 550}]} Show[ %27, Graphics[{%}]] height ground -Graphics- This view is not the best but it does show how far the parachutist went in the initial 10 seconds of free fall. A better picture can be seen by zooming in.

12 clw.ma 12 ParametricPlot[ Evaluate[ {x[t],y[t]} /. solution ], {t, 10.18,40}, PlotRange -> {{550,600},{100,300}}, PlotStyle -> RGBColor[1,0,0], AxesLabel -> {ground(meters), height(meters)}] height meters ground meters Graphics- This new view shows the actual jerk that the parachutist recieves when the chute opens and that it does in fact give quite a force. It now becomes possible to graph the motion during the first ten seconds and the combine the two graphs with the differing air resistance to achieve one picture of the total motion. Clear[ solution1, k, u ] k=.1828; u = 125; solution1 = NDSolve[ {x [t] == -k x [t], x[0] == 0, x [0] == u}, x, {t,0,10.18}] {{x -> InterpolatingFunction[{0., 10.18}, <>]}} Clear[ solution2, k, u, g ] g = ; k =.1828; solution2 = NDSolve[ {y [t] == g - (k y [t]), y[0] == 500, y [0] == 0}, y, {t,0,10.18}] {{y -> InterpolatingFunction[{0., 10.18}, <>]}}

13 clw.ma 13 ParametricPlot[ {x[t]/. solution1[[1]], y[t]/.solution2[[1]]}, {t,0,10.18}, PlotRange -> {{0,600},{0,600}}, PlotStyle -> RGBColor[0,1,0], AxesLabel-> {ground,height}] height ground ParametricPlot[ Evaluate[ {x[t],y[t]} /. solution ], {t, 10.18,40}, PlotRange -> {{0,600},{0,600}}, PlotStyle -> RGBColor[1,0,0], AxesLabel-> {ground, height}] height Graphics- -Graphics ground

14 clw.ma 14 This is not such a great view but it does give a perspective of the distance travelled during free fall and that travelled with the chute open. Show [%45, PlotRange -> {{550,610},{175,230}}] height ground Show[ %44, %43] height ground -Graphics- -Graphics- The above graph shows the actual change in position once the parachute is open. Another topic of interest is the velocity of the jumper through the fall. The graph below represents the change in velocity with time and requires some explanation. The green curve is the velocity of the jumper in free fall until he pulls the chute. We saw this graph earlier in the model. At seconds the parachutist opens the chute and the the red line represents the change in velocity with time as he continues his descent. He will hit the terminal speed with the chute open at approximately 14 seconds and he will land at 33 seconds. The smaller

15 Graphicsclw.ma 15 dashed line represents the terminal speed with the chute of 8.47m/s or the speed acquired when falling off a 12 ft wall. The larger dashed line represents the terminal velocity he would have achieved had he not pulled the chute at seconds and had continued in free fall. He would have built back up his speed and fallen with a velocity of m/s or 120 miles/hour. It is my opinion that he would not have survived the impact with the ground at that speed and it was wise to pull the chute. Show[%174,%167,%172] velocity time

16 clw.ma 16 Conclusion While most potential parachutists don t consider the actual mechanics behind a jump when they decide whether or not to step out of a plane that is 2000ft in the air, I am sure that the instructors, pilots, and manufacturers take this into account. Most parachutes have secondary back- up chutes that contain a gaseous cartridge triggered at a certain altitude. This automatic release prevents the jumper from reaching terminal speed again once he has reached his minimum velocity. Sometimes malfunctions cause the release of this chute and sometimes the crazy person, I mean parachutist, just wants to wait until the last second. Parachutists, according to the Encyclopedia Brittanica, are typically instructed to pull the cord at a time that corresponds to the minimum velocity they will achieve so the formulas in this model are of importance in knowing how to ensure safety and enjoyment. In review these equations are the ones describing the motion of the fall and may be solved using the methods demosnstrated in this model in accordance with the values of k and n. x [t] = -k x [t] (x [t]^2 + y [t]^2) ^ ((n-1)/2) y [t] = -g - k y [t] (x [t]^2 + y [t]^2) ^ ((n-1)/2) If the free fall part is considered the equations below come into play and can be used to calculate the most comfortable time to pull the chute and the velocity at that time. These equations are only dependent upon u, the plane s horizontal speed, and two known constants k and g; therefore, this model has some application in the real world. x[t] = ( u/k ) * [1 - exp( -kt) ] y[t] = - (gt/k) + [ (g/k2) * [ 1- exp( -kt) ] time of minimum velocity = Log[(g^2 + k^2*u^2)/g^2]/k v(minimum velocity) = u*(g^2/(g^2 + k^2*u^2))^(1/2) Obviously the height is an important factor but with these equations the height is easily determinable. The horizontal motion also plays a greater factor than one might think since the jumper initially travels only with a horizontal velocity. The time of the minimum speed is in essence when the horizontal speed is zero and only the vertical velocity is left so these equations are also very useful in determining precisely where a person will land. This is quite handy to know unless you enjoy crashing into trees however small the speed. In conclusion this model takes some of the simpler principles of physics and applies them to a real-life situation.

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