Efficiency Formula for Cross Belt Sorter System
Author Information: Kizza Wong from ICONVEY Overseas Division
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Content section
In the process of communication with peers, we often discuss the efficiency of the equipment of the parcel sorting system. As an engineer, many times we need to make a quick estimate of the efficiency of a system.
ICONVEY has accumulated a lot of industry experience in the field of parcel sorting systems, and I would like to share here how I personally quickly estimate the efficiency of a cross-belt sorting system.
First, let’s define what sorting efficiency is: sorting efficiency refers to the number of packages that can be effectively sorted per unit time.
In a cross-belt sorting system, we need to make the following assumptions:
The distance between vehicles of the cross-belt sorter is: P (meters)
Cross-belt sorter main line running speedis: V (m/s)
The import capacity of the parcel single area (D1 or D2) is: N (pieces/hour)
Actual effective rate is: d(%)
The efficiency of the cross-belt sorter is: E (pieces/hour)
Let’s analyze the first and fundamentally simplest case.
Case 1
The sorting system has only one package inlet, and one vehicle corresponds to one belt
The incoming parcels of D1 are all picked out by Zone 1 and Zone 2
Conclusion:
E=N
E=3600*d*V/P
Explanation:
E=N is not difficult to understand, because the packages input are just sorted by the sorter and it happens that the efficiency of input and sorting is equal.
Why E=3600*V*d/L?
3600*V can be understood as 3600*V meters of belt will pass through the sorting port per hour, and 3600*V/L can be understood as 3600*V/L belt will pass through the sorting port per hour, which means that at the same time there are 3600*V/L parcels are sorted. That is to say, in a perfect case, the efficiency of the conveyor is to sort 3600*V/L packages per hour.
Finally (3600*V/L)*d represents the actual efficiency.
And then
Case 2
The sorting system has two inlets, one car corresponds to one belt,
D1’s parcels are all sorted out from the 1st compartment, D2’s are all sorted out from the 2nd compartment
Conclusion：E=2*K=2*3600*V*d/P
Explanation:Pls Contact info@iconveytech.com
Case 3:
The sorting system has two inlets, one car corresponds to one belt
The packages of D1 and D2 are sorted out from the 1st or 2nd grid, and the probability of all the grids is the same
Conclusion:
N=3600*V/P
N=DA+DB/2
N=DB+DA/2
(DA is the actual import package efficiency in Zone 1, DB is the actual import package efficiency in Zone 2)
E=DA+DB=4*3600*V*d/3*P
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Case 4:
The sorting system has two inlets D1 and D2, one vehicle has two belts (as shown in the figure below), the two belts can jointly sort a large package, and all the packages in D1 are picked out from the grid port in Zone 1, D2’s parcels are all sorted out from the 2-zone grid.
Conclusion:
N=2*3600*V/P
N=DA*B*2+DA*(1-B)
(B refers to the proportion of large packages)
E=2*DA=2*2*3600*V*d/P*(1+B)
Explanation: Pls contact info@iconveytech.com
The above are the situations corresponding to the four most basic models. As long as you master the principles of these four basic cases, you can easily estimate the sorting efficiency of any complex parcel sorting system.
If the sorting system you are concerned with is not the cross-belt sorter system described in this article, a new computational model may be required.