redifined approach for the modeling of sky diving

7/31/2019 Redifined Approach for the Modeling of Sky Diving

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VARUN VARMA P 10BEC1065

Abstract The present model is used to propose an extended model of the real timescenario. The modified model is shown to be more physically realistic and isno more complicated than the original problem. This discussion gives equalemphasis to both the modelling and the analysis of the problem.

The below data is concerned only for LINES FIRST deploymentmechanism. The present situation is an initial value problem concerned withthe constraints by THE NEWTONS SECOND LAW of motion.

This model has been developed to compute various characteristics of thesteady descent of a parachute system. The model demonstrates the variationof drag force with variable positions of the body in the free fall.It explains the variation of the size of the canopy with varying creeping (or)gust force .It takes an overall consideration of the time varying gust forcebefore, after and during the deployment of the canopy.

NOTE:In the "lines- first release, the parachute remains in a deployment bag untilthe RISES and the SUSPENSION lines are fully extended.


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Keywords

REYNOLDS NUMBER, CREEPING FLOW,CO-EFFICIENT OF DRAG FORCE(GUSTFORCE),TERMINAL VELOCITY,SNATCH FORCE IVP(INITIAL VALUEPROBLEM),IMPACT VELOCITY

Introduction

Skydiving , also known as parachuting, is the action of exiting an aircraft or jumping off a tall structure, and returning to earth with the aid of a parachute.It may or may not involve a certain amount of free-fall, a time during which theparachute has not been deployed and the body gradually accelerates toterminal velocity.

A skydiver begins a jump at a specif ic height, x , abovethe ground and falls towards Earth under the influence of gravity Having aclear vision on how the drag force is related with velocity, by having apredetermined knowledge on Reynolds number and have known the type of body or size of the object and the constant of proportionality when the chuteis closed (free fall) and open(final descent),For eg ()

Shape Reynolds number CdHemispherical shell Re > 1.33

Disc Re > 1.10Flat strip Re > 1.95Cylinder < Re < 2* 1.95

Re > 5* ~0.35Sphere < Re < 2* 0.45

Re > 3* ~0.20


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THE ENGINEERING ASPECTS INCLUDE

Determining the descent time. The latest time that the parachute can be opened while keeping the

impact velocity below a specified threshold. Compare the motions for jumps when the parachute is opened after a

fixed amount of time, at a specified altitude, and when a given velocity isattained.

Find the corresponding model with quadratic air resistance, with

coefficients selected so that the pre- and post deployment terminalvelocities are the same as for the linear model; how do the two motionscompare and lots more.

The purpose of this assignment is to present an analysis of the traditionalparachute problem that coordinates graphical solutions with the theory forinitial value problems for a system of first-order ODEs.

Mathematical model

MATHEMATICAL ASSUMPTIONS:

THE GRAVITATIONAL FORCE CONSTANT IS ALWAYS ASSUMED TO BE~9.8 FOR DIFFERENT ALTITUDES.

THE CO-EFFICIENT OF DRAG

K1 , 0 t < t0K =

K2 , t t0

when the drag force is assumed to be directly proportional to the VELOCITY;


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Re > .

Hence creeping flow is not valid.

Therefore the DRAG on a body having a cross sectional area A is given by:

Fd =1\2 (Cd A )

We consider only the lines-first deployment scheme which can be modelled inthree distinct stages, Starting at time t when the ripcord is pulled.First, the suspension lines are released and become fully extended.

At this time, t = t , the snatch force pulls the skydiver from the spread eagleposition into an upright position and the canopy begins to inflate.

At t = t the canopy is fully inflated, i.e., the first time when the cross sectionalarea of the canopy reaches its projected steady-state value.

At time t t the skydiver attains terminal velocity the skydiver attainsBetween times t and t the momentum of the surrounding air massoverinflates the canopy before returning to the steady-state area for finaldescent (t > t3).

Hence the total drag force is given by:Fd = Fd b + Fd e = (C A + C A)

where the superscripts b and e are used to distinguish the drag coefficients

and cross-sectional areas of the skydiver's body and equipment.

ASSUMPTIONS:

This model ignores the drag force produced by theSuspension lines and assumes that the body and equipmentare rigidly connected.

For simplicity, it will be assumed that the length of the suspension linesincreases linearly over the interval [t ,t].


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The cross sectional area of the canopy cannot be modelled by assuming alinear increase in diameter as experimental data suggests that

A is the increase in area during the inflation.And

A is the increase in cross sectional area during over inflation.

THE EXACT MATHEMATICAL MODELLING:

b0 tt 0 b0 t 0 < t t 1

Ab(t) = b 1 t 1 < t t 2 b1 t 2 < t t 3b1 t t 3

Cdb (t) = 1.95 t = t 0 1.95 t 0 < t t 1 0.35*h t 1 < t t 2 0.35*h t 3 < t t 30.35*h t t 3

Where h is the height of the parachutist.

Ae(t) = 0.0 t t 0 b1 t 0 < t t 1

A (t) t 1 < t t 2 A (t) t 2 < t t 3 1 t t 3


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Cde(t) = 0.0 t t 0 0.35*l*(t-t0/ t1- t0) t 0 < t t 1 1.33 t 1 < t t 2 1.33 t 2 < t t 3 1.33 t t 3

m dv/dt = -mg+K .

Where

K =1/2 *(Cd bA + CdeA)

= 1.95 b0

t t0

* 1.95 b 0 +0.35*l(t-t 0/ t 1- t 0) t 0


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h=Height of the skydiverm=Sum of mass of sky diver + parachute + harnessl=Length of suspension linesa=Cross section area of Canopy

A,(t)= exp ((t -t)/(t -t) A,(t)=(1+sin((t-t)/(t -t)) =0.15(Experimentally proven value) =1.95b+0.35 *b*(1-h)/1.33=log (a/)

Solution to the model:

Considering the standard military parachute and its values for determiningthe solution of the equation, we have its parameters as:

Diameter of canopy=7.47mCross sectional area of canopy=a=*7.47*7.47/4=43.8m Length of rip cord(suspension lines)=l=8.96mLength(height) of the sky diver=h=1.78m (510) Cross sectional area of the sky diver in upright position=b=0.1m Cross sectional area of the skydiver in stable spread eagle position=b=0.5m

Mass of the sky diver=86.36kgMass of the parachute and suspended lines=10.84kg

Total mass=86.36+10.84=97.2KgDeployed time=t=10s

t=10.5s t=11.5s

t=13.2s =1kg/m By substituting the above values in the above equations, we got theFollowing relation for k(drag coefficient) in terms of t

0.5*0.976 0


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A numerical solution of the equation is created and graph is plotted using .

The values of the parameters taken are H L m

43.8 0.5 0.1 1.78 8.96 97.2 10 10.5 11.5 13.2


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ANALYSIS OF THE MODEL:

To find the solution of this model we assume that A,(t) is continuous on ( t,t 2)and A,(t) is continuous on (t 2,t 3).

There is exactly one continuous solution to the equation for t >0.

The solution of the initial value problem for finding velocity and displacementin the time interval (0< t t ) acts as the initial conditions for the equation of motion during the time period (t


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On solving consecutive differential equations with different boundaryconditions ultimately lead to the steady state terminal velocity after time(tt 3).

The solutions obtained from the equation during different time intervals are

bound to be continuous in (0, ) but may fail to beDifferentiable in any of t 1,t 2,t 3. To investigate the smoothness of the solutionsat the end points of different stages of the jump, acceleration can be obtainedfrom the equation:

a= =-g+(1/m)k

since g and m are constants and v is continuous, the acceleration is continuous

when k is continuous.

For k to be continuous at , the required condition is1.95 + 0.35 l =0.35 h + 1.33 A,(t1)

For k to be continuous at ,A,(t2)= A,(t2)

If A, is continuous on (t1, t2) with boundary conditions

And A, is continuous on (t2, t3) with boundary conditions

Then k is continuous on (0, )

The time derivative of the acceleration is the jerk, j = da/dt

differentiating the main equation produces which immediately gives conditionsunder which the acceleration is differentiable.

Let :

Then the conditions of continuity are satisfied when


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Where =relative increase in cross sectional area above a projected area.And =a

The analysis of the solution concludes with an estimate of the time when the

skydiver returns to solid ground. Notice that once the motion approachesterminal velocity, the position is essentiallylinear. Simplifying even further, the motion appears to be piecewise linearwith slope given by the free-fall and final descent terminal velocities,respectively.

Now we considered the five values of time (i.e)5s,10.3s,11s,12.5s,14s and calculated the corresponding values of thedrag coefficient(k) and the velocity(v) and tabulated as follows:

S.NO TIME(s) K(DRAGCOEFFICENT)

VELOCITY(m/s)

1 5 0.488 38.72

2 10.3 0.582 40.083 11 4.255 14.97

4 12.5 33.36 5.34

5 14 29.15815 5.71


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Conclusion:

0

5

10

15

20

25

30

35

40

0 10 20 30 40 50

k ( t

)

v(t)

k(t) v/s v(t)

Series1

05

1015202530354045

0 5 10 15 20

v ( t

)

TIME(s)

v(t) v/s t

Series1

0

5

10

15

20

25

30

35

40

0 5 10 15 20

k ( t

)

TIME(s)

k(t) v/s t

Series1


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It can be concluded that as a body falls its velocity increases and hence thedrag force increases, however after reaching a certain velocity known as theterminal velocity, the drag force becomes constant and so does the velocity.

References:

redifined approach for the modeling of sky diving

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