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Angle for hauling effect.

  1. Mar 4, 2015 #1
    I am trying to design a material transfer system ( the system at duration 0:57 to 1:34 ).
    The general layout is http://i.imgur.com/ajQ05qy.png
    It was suggested that angle be 5 degree. The reason is for hauling effect so the tray can revolve back. can you explain this? why 5 degrees? supporting references will be of great help.
    Thank You.
     

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  3. Mar 4, 2015 #2

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    Whatever inclination of the plane of the arm and counterweight is adequate to overcome the friction of the pivot bearing and yield a cycle time between loading and unloading the tray that is suitable for the process.
     
  4. Mar 4, 2015 #3
    There can be a range for it but how to approach it mathematically ?
    Could you please elaborate it in detail? How would you approach the problem ? Do you need any data? m1=3kg & m2=10kg
    I want to clear this concept and your help is highly appreciated.
     
  5. Mar 4, 2015 #4
    m 1 = Mass of empty pan in kg
    Pw = Weight of parts in N
    m 2 = Mass of counter weight needed to be attached in kg
    θ = Angle of inclination of the conveyor track
    x = Length of tray arm from axis of rotation in mm
    y = Length of counter weight arm from axis of rotation in mm
    a = Diameter of bearing housing in mm
    Height of counterweight from datum = 900 mm
    Height of pan from datum = 913 mm
    x = Length of tray arm from axis of rotation in mm =1000mm
    y = Length of counter weight arm from axis of rotation in mm = 300mm
    m1=3kg(assumption)
    m1(h + a sin θ) g x = m2 y h g
    m2 = 3 x 900 x 913 /300 x 900 =9.13kg
    How to proceed for assumption of θ?
    Can you explain me the following equation?
    Pwg ≥( m1 * x^2 + m 2 * y^2 ) sin2 θ / [ x(h-xsin2 θ)]
     
  6. Mar 4, 2015 #5

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    This is supposed to be giving you the minimum load (part weight) to unbalance the arm so it will swing and carry the part around the conveyor arc from station 1 to station 2 and hold it there. Unless the tray/pan and counterweight arms aren't 180 degrees opposed, there's no reason for that load to depend on the magnitude of theta other than that it be off vertical, so the form of the equation is somewhat arcane.

    There isn't enough information to to calculate the speed of the movement, or the shock absorbing requirements at the ends of the travel range. Speed increases as theta increases, but without knowing particulars of bushings/bearings on the pivot, viscosities of lubricants, moments of inertia of the loaded and unloaded assembly there's no way of coming up with numbers.
     
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