four_bar_kinematic_analysis_of_a_mechanical_bird_device.pdf

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    Four Bar Kinematic Analysis of a Mechanical Bird Device

    Evan A. Kontras

    University of Missouri

    Department of Mechanical and Aerospace Engineering

    November 2008

    Many important and practical machines utilize a four bar mechanism. Though simple,

    having only three links and a fixed reference or ground, the four bar mechanism has a wide

    variety of uses in engineering. Having only one degree of freedom, a four bar mechanism can be

    operated with a single input, yet a wide range of operations can be accomplished with different

    categories of the four bar. A common utilization of this mechanism is in transforming rotational

    motion into translational or oscillatory motion. In designing a flap wing style flying vehicle, the

    four bar mechanism is ideal.

    With many electric motors providing rotation, and the need to have a quickly oscillatinglink, the crank rocker category four bar has been selected for the vehicle design. A motor with

    6000 rpm capability was chosen. It will be assumed that the optimal wing speed is 30 flaps per

    second, therefore gears will have to be used to reduce the motor speed. This can be done using

    the simple relationship,

    (1)

    where r is the radius of each respective gear, and the angular velocity. A 3 mm gear on the

    motor hub in mesh with a 10 mm gear will result in 30 oscillations per second of L4, as shown in

    Fig. 1 below, which is a simplified schematic of the vehicle mechanism and structure.

    Figure 1. Crank Rocker Mechanism.

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    For simplicity the 10 mm gear will be represented in the four bar mechanism as L2, a

    strait line. A small, light design is desirable, therefore the other links were decided accordingly,

    taking into consideration the need for a moderate range of oscillation for link L4. This link is

    particularly important because it is where the wing of the vehicle connects to input motion. With

    link dimensions known, the position, velocity, and acceleration of a point on the four bar

    mechanism can be determined using the complex number method, utilizing

    d

    dt + (2)

    dt + + (3)

    where z, , and are the position, velocity, and acceleration respectively, the angular position, the angular velocity, and the angular acceleration. It should be noted that the terms in order

    from left to right in Eq. (3) represent the radial, coriolis, tangential, and normal components of

    acceleration. Using the complex number method is very helpful in understanding the kinematics

    of the four bar mechanism, but it can become quite tedious to repeat the calculations for many

    values of. However, using a spread sheet or similar computational program, multiple values of

    can be quickly and easily calculated. Also, the use of computer programs can provide for a

    quick means of plotting data as well. Using Microsoft Excel, the position, velocity, and

    acceleration of a point at the midspan of the wing was determined for crank angles between 0

    and 360. As the crank 360 the wing completes one flap. The plots showing how each

    parameter changes as the crank rotates are shown in the figures below.

    Figure 2. Displacement of Wing Center from Original Position.

    -300

    -250

    -200

    -150

    -100

    -50

    0

    50

    0 50 100 150 200 250 300 350 400

    Displacement(mm)

    Crank Angle (Degrees)

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    As expected with oscillating motion, the point at the center of the wing ends at the same

    point that it started. From crank angles between approximately 100 and 275, the wing is

    moving downward for more than 20 cm generating lift. The midpoint of the wing, which is

    initially at rest, reaches a peak velocity of 164 mm/s. Again, the velocity curve ends at the same

    value it begins, as shown in Fig. 3. The angular velocity of the wing is the same as that of link L4

    from Fig. 1.

    Figure 3. Angular Velocity of Wing.

    Finding the acceleration is more involved, though other important vehicle criteria can bedetermined once the components of acceleration are found. The low points on the angular

    acceleration curve correspond to points when the wing is changing direction. The maximum

    velocity occurs at the minimum acceleration as seen from Fig. 4.

    -200

    -150

    -100

    -50

    0

    50

    100

    150

    200

    0 50 100 150 200 250 300 350 400

    AngularVe

    locity(rad/s)

    Crank Angle (Degrees)

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    Figure 4. Angular Acceleration of Wing.

    It is at this point, approximately 175 of crank rotation, where the wing has experienced the

    longest time under acceleration and is traveling at 164 mm/s. After the maximum velocity, the

    acceleration changes direction, eventually stopping the wing and changing its direction of motion

    as well.

    Considerable forces can be developed along the wing due to the abrupt changes in both

    velocity and acceleration. A lift force is produced by each wing, proportional to the square of the

    angular velocity. For each wing, the lift force has been determined to be 60% of the total vehicle

    weight. Unlike the other links of the four bar mechanism analyzed for this vehicle, the wing itself

    is assumed to have mass. Each wing is one quarter of the total mass of the vehicle, with theentire mass being concentrated at the wing midpoint for simplicity. Inertial forces are produced

    as the mass of the wing rotates. Knowing the angular acceleration of the wing and the lift force

    generated,Newtons second law can be used to solve for the inertial force created by the

    accelerated mass of the wing. Balancing these forces is the torque on the other end of the link. As

    the angular acceleration of the wing changes, so too does the torque, as seen in Fig. 5.

    0

    2000

    4000

    6000

    8000

    10000

    12000

    14000

    16000

    18000

    0 50 100 150 200 250 300 350 400

    AngularAcceleration(rad/s^2)

    Crank Angle (Degrees)

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    Figure 5. Driving Torque for One Wing.

    Torque must originate at the motor, actually driving both wings. With a combined mass of halfthe vehicle, it is important that the motor have enough power to accelerate the wings as needed,

    without any extra gearing for mechanical advantage. Knowing the combined torque on each

    wing, the necessary toque the motor must output can be determined. Fig. 6 shows the energy

    required to move the wings during operation, which can provide information on how much

    power the motor must output.

    Figure 6. Energy for Complete Cycle.

    The motor must be able to reliably output the maximum energy, approximately

    400,000 kgf-mm. This is around 0.0146 hp, a power requirement many electric motors are

    capable of satisfying. Weight is one of the most important parameters for any flying vehicle, so

    the lightest motor available to produce 0.0146 hp is desirable. The weight of each structural

    -100000

    -50000

    0

    50000

    100000

    0 50 100 150 200 250 300 350 400

    Torque

    (N-mm)

    Crank Angle (Degrees)

    -300000

    -200000

    -100000

    0

    100000

    200000

    300000

    400000

    500000

    0 50 100 150 200 250 300 350 400Energy(kgf-mm)

    Crank Angle (Degrees)

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    component should also be considered. Due to the torque and inertial forces on the wing, a

    relatively high strength material will be used. The maximum torque is 75,000 -, which is

    equal to the force on the wing multiplied by the distance from the frame, which is half the

    wingspan. At this distance, a force of approximately 400 N is developed. Once a cross sectional

    area for the wing is determined, the stress can be found, and a proper material selected. The

    flapping wing flying vehicle design will use 3 mm diameter wing supports, with each connecting

    link, L2-L4 being 3x5 mm rectangles. The highest stresses are developed where the wing support

    meets the larger area connectors, with a maximum value of approximately 56 MPa. Plastics such

    as PVC and Nylon will yield under such stresses, so a more high strength material will be

    necessary. Most steels have a yield strength between 200 and 600 MPa, but unfortunately they

    are too heavy to be suitable for the flying vehicle. Aluminum, with a density nearly one third of

    steel, has an equally high strength, around 400 MPa. Due to its high strength to weight ratio and

    abundance, aluminum was selected as construction material for the vehicle. A simplified model

    of the four bar mechanism and aluminum frame for the flapping wing vehicle is shown below in

    Fig. 7.

    Figure 7. Basic Frame and Mechanism.

    To save weight, only wing supports will be used for the vehicle, with the wing itselfbeing made of a thin light weight fabric or plastic. The rear inside edge of each wing skin will

    connect to a rocker at the rear of the frame to control direction during flight. Also, a large

    adjustable tail section at the rear of the frame will control the altitude. Both direction control

    devices are actuated with linear servo motors. The entire vehicle will be powered with

    lightweight, rechargeable lithium batteries. A completed prototype is shown in Fig. 8, 9,

    10,and11.

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    Figure 8. Front View.

    Figure 9. Arial View.

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    Figure 10. Altitude Control.

    Figure 10. Directional Control.

    Animations, SolidWorks assembly files, and other perspective pictures are included on

    disk.

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    OMEGA 2= 628.3 rad/s

    AB

    C

    D

    NOTES:

    -Units are in millimeters

    10.0000

    20.0000

    40.0000

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