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Conveyor Speed / Distance calculations

  1. Mar 1, 2012 #1
    I would like the formulae for calculating the distance created bewteen products transferring from one conveyor to another with a line of products moving at x M/min transferring onto a conveyor running y M/min. I need to be able to calculate what the gap will be between the products as this is critical to the design of the system. Urgent Help required PLEASE!!!!
  2. jcsd
  3. Mar 1, 2012 #2
    A very simplified approximation of product gap can be calculated if you assume infinite product acceleration from one conveyor to the other. If you are interested, I will help you find the solution, but I can't just give it to you.

    Vx = upstream conveyor speed (m/min)
    Vy = downstream conveyor speed (m/min)
    L = product length (m)
    G = gap between products (m)

    Note that the units of the conveyor speed must be the same, and the gap units will be the same as those for the product length. Please use lower case "m" for metre.

    If we assume that the product will accelerate from speed Vx to speed Vy at the instant that the leading edge of the product contacts conveyor y, and both products start fully on conveyor x, how much time will pass before the leading edge of the trailing product will contact conveyor y?
    Last edited: Mar 1, 2012
  4. Mar 1, 2012 #3
    Assume both products are touching on conveyor x and based upon a product width of 1m moving at 5m/min then 12 seconds would be the time elapsed for the leading edge of the trailing product to contact conveyor y
  5. Mar 1, 2012 #4
    FYI, I just changed the conveyor speed variables names to Vx and Vy.
    So the time is obviously the product length divided by upstream speed Vx, or:
    t = L/Vx
    Jumping ahead, in the same time "t", the leading edge of the leading product will travel
    a distance:
    D = Vy * t
    (Yes, I did just introduce variables t and D, but they are only temporary.)
    Making sense?
    I'm trying to move this along quickly...
    G = ?
  6. Mar 1, 2012 #5
    So D=2m so if i am understanding this G=D-L = 1m and that G is proportional to the original gap
  7. Mar 1, 2012 #6
    Yes, G = D - L
    Sorry, what was Vy? It must be 15 m/min, although I don't think you mentioned it.
    You can either use L = actual product length (assuming no gap) or Leff = effective product length which is equal to product length + initial gap.
    Yes, the final gap is proportional to product length (or effective product length).
    So what is G in terms for L, Vx and Vy? I thought that was what you wanted.
  8. Mar 2, 2012 #7
    So if G=D-L and we have the following form the calculations performed above

    t= L/Vx = 0.2min

    in the same time period "t" the leading edge of the product will move Vy*t to give product D = 10*0.2 = 2m. Therefore with G=D-L then G=2-1 so the gap between the products is 1m.

    Calculated in terms for L, Vx and Vy, I can see the following G= Vy/VxL-L am I correct?
  9. Mar 2, 2012 #8
    Not quite, it should be:
    G = L * (Vy / Vx) - L

    or more simply:

    G = L * [(Vy / Vx) - 1]

    To check:
    If Vy = Vx, G should be 0.

    Looks like I got D and G mixed up previously when I indicated that Vy = 15 m/min (I was wrong).
    Last edited: Mar 2, 2012
  10. Mar 5, 2012 #9
    Many thanks for your asistance. Much appreciated.
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