Finding Gap Between Two Objects on Moving Conveyor

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SUMMARY

This discussion focuses on calculating the gap between two pieces of wood on a conveyor belt, specifically using a belt speed of 29 inches per second and a product length of 37 inches. The gap is determined using the formula g = (b * t) - x, where b is the belt speed, t is the time interval between items, and x is the length of the item. Participants clarify the importance of timing and provide examples to illustrate how to derive the gap based on different speeds and intervals.

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  • Ability to perform unit conversions, particularly between inches and feet.
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This discussion is beneficial for engineers, logistics professionals, and anyone involved in material handling or conveyor system design, particularly those looking to optimize item spacing and speed on conveyor belts.

Bri2012
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I am needing to find the gap between two pieces of wood on a conveyor belt. I am new t this and am having trouble knowing what to Google to find the right formulas. I found one answer on here that helped a little, but it was dealing with two conveyors, and I am not sure if I adjusted the formula correctly. I will include what I did as a reference and hopefully someone can let me know if I did this right.

L (product length) = 37 inches
Vx (Upstream Speed) = 145 FPM
G (gap) = ?

Time = 37/145 == .25517
Distance = 145 * .25517 == 45.93101
Gap = 45.93101 - 37 == 8.9 inches

Does this look correct?
 
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It seems that you are mixing feet and inches.

Welcome!
 
Shoot! I did not catch that, thank you! Besides that, am I finding the gap correctly?
 
The speed of your belt is 29 inches per second.
Don’t you need the time between two pieces of wood to be placed on the belt?
 
Timing is important.
For example, if one piece enters the conveyor per each second, pieces would overlap, as 37 inches of length > 29 inches displacement in one second.
For having zero gap, the faster one piece can follow the previous one is 37/29 seconds.
 
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Bri2012 said:
I am needing to find the gap between two pieces of wood on a conveyor belt.
The belt travels at a speed of; b inches per second.
An item is dropped onto the belt every; t seconds.
The pitch of the items on the belt is therefore; p = b * t
Each item has a length; x
The gap between items is; g = p - x
So the gap or space between two items is; g = ( b * t ) - x
Rearrange that to get what you need.
 
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How accurate do you need to be? I've worked the belts at UPS. When packages are placed on the belt, they have some finite acceleration (the belt accelerates them to belt speed over a finite distance).

I tried coming up with a linear drag model but I'm not sure.

let:

##w## be the belt speed

##\beta ( \mu, N )## is a generic coefficient for the time being

##v## is velocity of the package

##m## is the mass of the package

I propose something to the effect of:

$$ \beta ( w - v ) = m \dot v$$

Solving that leads to:

$$ v = w \left( 1 - e^{- \beta t }\right)$$

This model seems lacking because the packages technically never reach belt speed. However, if the things entering the belt are different in a predictable way (like parcels are entering the conveyor) this is going to be a factor in the spacing.
 
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  • #10
Belt Speed: 29 inches per second
Time in Seconds Item dropped: (InchesPerSecond/length of piece in inches) ---- (29/37) = 0.7838
Pitch: 29 * .7838 = 22.7302
Length: 37 inches
Gap: p - x = -14.2698
Get gap between two: (29 * .7838) - 37 = -14.2698***The last part I am hoping I did correctly is highlighted in red to calculate how often items are being dropped. Does the way I did that give me the time like I believe it does? Erobz, I will look into that further and let you know!

Truly appreciate all the help!
 
  • #11
Bri2012 said:
Belt Speed: 29 inches per second
Time in Seconds Item dropped: (InchesPerSecond/length of piece in inches) ---- (29/37) = 0.7838
Pitch: 29 * .7838 = 22.7302
Length: 37 inches
Gap: p - x = -14.2698
Get gap between two: (29 * .7838) - 37 = -14.2698***The last part I am hoping I did correctly is highlighted in red to calculate how often items are being dropped. Does the way I did that give me the time like I believe it does?Erobz, I will look into that further and let you know!

Truly appreciate all the help!
Something isn’t correct. How often is a new piece entering the belt? The timing is the independent variable here. Or you can calculate the time interval by specifying a gap length? Which do you want?
 
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  • #12
How often is a new piece entering the belt? The timing is the independent variable here. Or you can calculate the time interval by specifying a gap length? Which do you want?
Is there a mathematical way to determine the time? I was told that I could take FPM, divide by the length of the wood, and convert to seconds to get the time. But, I agree it is not seeming like the correct approach.

I want to be able to say if the speeds are this, you will need a gap this big, or if the gaps are this big, you will need speeds of at least this. I would like time to be a dependent variable if possible.
 
  • #13
Bri2012 said:
Is there a mathematical way to determine the time? I was told that I could take FPM, divide by the length of the wood, and convert to seconds to get the time. But, I agree it is not seeming like the correct approach.

I want to be able to say if the speeds are this, you will need a gap this big, or if the gaps are this big, you will need speeds of at least this. I would like time to be a dependent variable if possible.
For example:

If a piece comes in every ##t = 2 \rm{s}##, the gap between them ##g## would be:

$$g = 29 \rm{\frac{in}{s}} \cdot 2 \rm{s} - 37 \rm{in} = 21 \rm{in}$$

If you wanted the gap to be ## 36 \rm{in} ## then you have to solve for the time interval:

$$ 36 \rm{in} = 29 \rm{ \frac{in}{s}} \cdot t - 37 \rm{in} \implies t = \frac{36 \rm{in} + 37 \rm{in} }{ 29 \rm{\frac{in}{s}} } \approx 2.5 \rm{s} $$
 
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  • #14
Okay, I see. Thanks for taking the time to explain that to me. This is not stuff I normally do but am branching out and appreciate the kindness.
 
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