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- .MCD 20000 0
- .CMD PLOTFORMAT logs=0,0 subdivs=1,1 size=5,15 type=l
- .CMD FORMAT rd=d ct=10 im=i et=3 zt=15 pr=3 mass length time charge
- .CMD SET ORIGIN 0
- .CMD SET TOL 0.001000
- .CMD MARGIN 0
- .CMD LINELENGTH 78
- .CMD SET PRNCOLWIDTH 8
- .CMD SET PRNPRECISION 4
- .CMD FILENAME data1 data1
- .CMD FILENAME data2 data2
- .TXT 1 0 1 45
- a1,45,43,44
- MOTION OF FALLING BODY WITH WIND RESISTANCE
- .TXT 2 0 3 70
- a3,70,68,185
- A parachutist jumps out of an airplane and opens his chute 5 seconds
- into his fall. Examine the characteristics of the flight, including
- position, velocity, acceleration, and "jerk."
- .TXT 4 0 2 45
- a2,45,43,64
- To start, set up N short time intervals dt.
- Also define units:
- .EQN 3 1 1 9
- N:20
- .EQN 0 16 1 12
- i:0;N
- .EQN 0 23 1 11
- sec:1T
- .EQN 0 11 1 10
- lb:1M
- .EQN 1 -50 1 14
- dt:.5*sec
- .EQN 0 16 2 12
- t[i:i*dt
- .EQN 0 23 1 10
- ft:1L
- .EQN 0 11 2 19
- g:-32*ft*sec^-2
- .EQN 2 -11 1 13
- lbf:lb*g
- .TXT 1 -40 1 32
- a1,32,30,31
- Initial conditions as follows:
- .EQN 2 1 2 15
- y[0:2000*ft
- .TXT 0 18 1 46
- a1,46,44,45
- Parachutist jumps out of plane at 2000 feet.
- .EQN 2 -18 3 13
- v[0:0*ft/sec
- .TXT 1 18 1 40
- a1,40,38,39
- Parachutist starts with zero velocity.
- .EQN 3 -18 1 13
- m:200*lb
- .TXT 0 18 1 33
- a1,33,31,32
- Mass of parachutist plus chute.
- .TXT 2 -19 3 72
- a3,72,70,141
- Force on parachutist results from gravity and air resistance.
- Gravitational force is constant; air resistance is proportional to
- velocity.
- .EQN 4 1 1 11
- Fg:m*g
- .EQN 0 13 1 14
- Fg=?lbf
- .EQN 2 -13 5 23
- α_nochute:.5*lbf/((ft/sec))
- .TXT 0 29 2 35
- a2,35,33,54
- The two α constants are a measure
- of air resistance.
- .EQN 6 -29 5 20
- α_chute:5*lbf/((ft/sec))
- .TXT 2 29 3 47
- a3,47,45,132
- The Ff function represents the frictional
- force for air resistance over time, which
- changes after 5 seconds when the chute opens.
- .EQN 4 -29 1 64
- Ff(t):α_nochute*(Φ(5*sec-t))+α_chute*(1-Φ(5*sec-t))
- .TXT 2 -1 1 69
- a1,69,67,68
- Now compute velocity, acceleration, position, and "jerk" over time.
- .EQN 2 1 4 32
- v[(i+1):v[i+(Fg+Ff(t[i)*v[i)/m*dt
- .TXT 2 39 2 34
- a2,34,32,66
- Change in velocity computed from
- acceleration (force over mass).
- .EQN 4 -39 2 20
- y[(i+1):y[i+v[i*dt
- .TXT 0 39 1 34
- a1,34,32,33
- Position computed from velocity.
- .EQN 3 -39 4 19
- a[(i+1):(v[(i+1)-v[i)/dt
- .TXT 2 39 1 38
- a1,38,36,37
- Acceleration computed from velocity.
- .EQN 3 -39 2 9
- a[0:g
- .EQN 2 -1 4 17
- j[i:(a[(i+1)-a[i)/dt
- .TXT 1 40 3 37
- a3,37,35,81
- "Jerk" computed from acceleration.
- (Jerk is the third derivative of
- position.)
- .EQN 4 -40 3 12
- j[0:0*g/dt
- .TXT 4 5 1 20
- a1,20,18,19
- POSITION OVER TIME
- .TXT 0 40 1 20
- a1,20,18,19
- VELOCITY OVER TIME
- .EQN 1 -45 17 30
- y[0&y[N&y[i{1,1,15,25,l}@t[N&t[0&t[i
- .EQN 0 38 17 33
- 0*ft/sec&v[11&v[i{1,1,15,25,l}@t[N&t[0&t[i
- .TXT 19 -31 1 24
- a1,24,22,23
- ACCELERATION OVER TIME
- .TXT 0 44 1 18
- a1,18,16,17
- "JERK" OVER TIME
- .EQN 1 -51 17 34
- a[12&a[0&a[i,0*g{1,1,15,25,l}@t[N&t[0&t[i
- .EQN 0 39 17 36
- j[11*1.1&j[12*1.1&j[i,0*g/sec{1,1,15,25,l}@t[N&t[0&t[i
- .TXT 10 -6 1 3
- a1,3,1,2
- 0
- .TXT 0 41 1 3
- a1,3,1,2
- 0
- .TXT 8 -74 1 77
- a1,77,75,76
- ---------------------------------------------------------------------------
- .EQN 1 44 28 11
- a[i/(ft/sec^2)=
- .EQN 0 16 28 11
- j[i/(ft/sec^3)=
- .EQN 1 -33 27 12
- v[i/(ft/sec)=
- .EQN 2 -26 25 7
- t[i/sec=
- .EQN 0 10 25 12
- y[i/ft{19075}=
-