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I have three conveyors. I am trying to find the minimum speed needed for the second (middle) conveyor in order to avoid collisions with the pieces of wood. The conveyors 1 and 2 are perpendicular to one another, as are conveyor 2 and 3.

Information I have:

*Vx = 145 feet per minute; Conveyor 1 speed/ *W = 5 inches; Width of wood/ *G = 4 inches; Gap between pieces of wood/ *Vy = 180 FPM; Conveyor 3 speed/

Information I am looking for: *Speed needed for Conveyor 2 in order to avoid collisions

If anyone knows a formula or a resource they can point me to, that'd be awesome! I feel like I just can't find what I'm looking for online when I don't know exactly what I'm looking for.

EDIT Apparently my question was closed. I will add this diagram and see if someone can help now that I've added a visual representation of my words. I want to know how fast conveyor 2 needs to run in order for the pieces not to hit each other. All pieces of wood are the same size.

Diagram of conveyor set up with variables

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  • $\begingroup$ You need far more information than that. Has conveyor 3 evaporated? How do you fit a 5 inch piece of wood in a 4 inch gap? $\endgroup$
    – Greg Locock
    Commented Dec 28, 2022 at 23:31
  • $\begingroup$ No, conveyor 3 is listed in my post as Vy at 180 FPM, if I'm understanding you correctly. I'm trying to find a speed for the 2nd conveyor that will speed it up enough to where it won't hit. If the 2nd conveyor can accelerate enough, it will push the piece of wood out of the way before it hits, right? $\endgroup$
    – villasenorbritt
    Commented Dec 29, 2022 at 14:34
  • $\begingroup$ Please rewrite your question so that we don't have to read your mind as to what you want. A diagram would be worth a thousand posts. $\endgroup$
    – Greg Locock
    Commented Dec 29, 2022 at 21:11
  • $\begingroup$ I asked what the minimum speed of the 2nd conveyor would be so that the pieces do not collide and mentioned that they are perpendicular. Noted on the diagram, but you are implying I've not described my needs, just as you implied I had forgotten conveyor 3, which is not the case. $\endgroup$
    – villasenorbritt
    Commented Dec 29, 2022 at 22:14
  • $\begingroup$ Is this a homework problem? $\endgroup$
    – Drew ♦
    Commented Jan 7, 2023 at 7:00

1 Answer 1

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I'm going to estimate some numbers here that you haven't provided, but you should be able to adjust the math as needed.

Right off the bat we can forget about conveyor 3, since the rate at which staves are moved on to it is set by conveyor 1, and the speed of that is fixed.

So we need to figure out how fast conveyor 2 needs to move to put 4 inches between staves coming from conveyor 1.

The actual spacing of the staves on conveyor 2 will be 9in (4in gap + 5in stave width).

Now we need to know the rate (in staves per minute) that staves will be delivered onto this belt, from conveyor 1. Unfortunately we can't determine that from the numbers you have provided, so I'll make up some additional numbers.

Lets say there is 1 stave every 5 feet on conveyor 1, that equates to 29 staves per minute. Or 2.07s per stave.

You want conveyor 2 to travel 9in during this time, so the resulting speed is 0.75ft/.0345minutes = 21.7ft/min

Now lets determine whether the staves will collide during the transition.

Starting from the point where a stave has been moved onto conveyor 2 and has just begun moving to the right. It will have to move 5 inches (the width of 1 stave) before another stave collides with it. If the staves are 24 inches long then the gap on conveyor 1 is 36in.The staves cross this gap in 3ft / 145ft/min = 1.24seconds. The stave on conveyor 2 moves 21.7*12/60 = 4.34 inches/sec. So in 1.24 seconds it moves 5.3816 inches. So they will barely not collide. In practice they probably will. You could run conveyor 2 faster and increase the gap between staves to solve the problem, or leave larger gaps between staves on conveyor 1.

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  • $\begingroup$ Thank you very much for this! Very great, in-depth explanation on how to solve and approach this problem. $\endgroup$
    – villasenorbritt
    Commented Jan 11, 2023 at 17:46
  • $\begingroup$ Accepted as answer, not enough reputation to upvote :) $\endgroup$
    – villasenorbritt
    Commented Jan 11, 2023 at 17:46

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