MAE 100 Mathcad III

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MAE 100 Mathcad III

• Dr. Don Wallace
• Mr. Troy Skinner

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Trajectory Problem

• When an object is launched into the air it has a
  certain initial velocity and angle with respect
  to the horizontal as shown.

Vo
q
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Velocity Components

• The original vertical velocity is then Vo sin q
• The original horizontal velocity is Vo cos q.

Vo
sin(q)
Vo
cos(q)
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No Drag

• Under ideal conditions (no drag)
• the horizontal velocity remains constant and
• the vertical velocity decreases to zero under the
  influence of gravity and then reverses direction
  as the object falls.
• The resulting trajectory is parabolic.

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Drag

• More realistically a drag is exerted on the
  object by the air or other medium that the object
  is moving through.
• The amount of drag depends on a number of
  parameters including
• the frontal area of the object,
• the shape of the object,
• the density of the air (or other medium) and
• the velocity of the object.

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Newtons Law

• Newtons second law states that the acceleration
  caused by a force is proportional to the force
  and in the same direction as the force (i.e. F
  m a).
• For the purpose of this problem it will be
  assumed that the acceleration component caused by
  drag can be expressed as
• a -CD V2

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Acceleration Components

• Therefore the vertical and horizontal components
  of acceleration are
• Notice that the drag always opposes the motion so
  the direction of the velocity must be tracked in
  the vertical direction to determine if the
  velocity is up or down so that the drag can
  oppose it.

ax -CD Vx2
ay -g sign(Vy) CD Vy2
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Acceleration Components

• Therefore the vertical and horizontal components
  of acceleration are
• Notice that the drag always opposes the motion so
  the direction of the velocity must be tracked in
  the vertical direction to determine if the
  velocity is up or down so that the drag can
  oppose it.

ax -CD Vx2
ay -g CD Vy Vy
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Numerical Solution

• Since this would be hard to solve analytically a
  numerical approximation can be used to estimate
  the velocity and accelerations as the object goes
  through its trajectory.
• The change in velocity and position can be
  estimated by solving at time increments i which
  go from 0 to some number n

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Initial Conditions

• At time zero

t0 0
x0 0
y0 0
(Vx)0 V0 cos (q)
(Vy)0 V0 sin (q)
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Recursion Equations
ti ti-1 Dt

• New Time
• New Velocities
• New Positions

(Vx)i (Vx)i-1 ax Dt (Vx)i (Vx)i-1
(-CD (Vx)i-12)Dt
(Vy)i (Vy)i-1 ay Dt (Vy)i (Vy)i-1
(-g CD (Vy)i-1 (Vy)i-1 )Dt
xi xi-1 Dt (Vx)i (Vx)i-1/2
yi yi-1 Dt (Vy)i (Vy)i-1 /2
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Comments

• The velocity used in the position equation is the
  average of the two velocities at the beginning
  and ending of the time step.
• The smaller the time step the more accurate the
  solution will be.
• Similar equations are used for the velocity in
  the vertical and horizontal directions.

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Named variables
Function usage (sin, cos, sign, if, MATCH, INDEX)
Formatting for easy readability
Plot of the results
Named pages
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Mathcad
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Given Information
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Solution (Initial Conditions)
(Vx)0 V0 cos (q)
x0 0
At t0 0
(Vy)0 V0 sin (q)
y0 0
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Solution (Equations)
ti ti-1 Dt
(Vx)i (Vx)i-1 (-CD (Vx)i-12)Dt
xi xi-1 Dt (Vx)i (Vx)i-1/2
(Vy)i (Vy)i-1 (-g CD (Vy)i-1 (Vy)i-1 )Dt
yi yi-1 Dt (Vy)i (Vy)i-1 /2
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Result Values
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Results
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Plot
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Mathcad 2
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ax -CD Vx2
(Vx)0 V0 cos (q)
x0 0
y0 0
ay -g CD Vy Vy
(Vy)0 V0 sin (q)
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Results
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Plot
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Animation
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Number_1
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(No Transcript)
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2D Plot and Linear Curve Fit
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Number_2
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Symbolic
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Solve Blocks
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Number_3
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Initial Value Problem
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Boundary Value Problem
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Number_4
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ax -CD Vx2
(Vx)0 V0 cos (q)
x0 0
y0 0
ay -g CD Vy Vy
(Vy)0 V0 sin (q)
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Results
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Plot
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Animation