Technical Data Top chain Selection
The latter half of this page also introduces the following selection examples:
Click on each item to scroll to the main text.
- *WT0700 series Selection example
- *WT1500 series, WT1510 series and BTN5 Selection example
- *WT2525VG-K Selection example
Plastic modular chain (Mold-to-Width type) Selection example
WT0700 series Selection example
Nose bar on driven side
●Calculation formula (SI units: kN)
・Tension at return-way
[Tension at section A ：FA]
FA = m1・L・μ1・fn × 9.80665 × 10-3
・Tension at carry-way
[Tension at section B ：FB]
FB = FA + {(m1 + m2) L・μ1 + m2・LS・μ2} × 9.80665 × 10-3
・Chain load
F = FB
Note: LS = 0 when there is no accumulation of conveyed products.
●Calculation example (SI units)
| Usage conditions | |
|---|---|
| Chain Type | WT0705-W300-LFG (m1=5.9×0.3=1.77kg/m) |
| Chain width | 300mm |
| Layout | L=2m |
| Chain speed | V=15m/min |
| Conveyed object | 500-ml aluminum can (filled) |
| Conveyed product mass (per 1-meter unit of length) | m2=139kg/m2(523 g/piece ) ×0.3m=41.7kg/m |
| Rail | UHMW-PE (Plastic rail) |
| Accumulation distance | Ls=2m |
| Lubrication | For dry conditions |
| Operating Ambient Temperature | 20℃ |
| Coefficient of dynamic friction between chain and wearstrip | μ1=0.2 |
| Dynamic coefficient of friction between chain and conveyed product | μ2=0.2 |
| Nose bar coefficient | fn=1.8 |
・Tension at return-way
[Tension at section A ：FA]
FA = 1.77 × 2 × 0.2 × 1.8 × 9.80665 × 10-3 = 0.0125kN
・Tension at carry-way
[Tension at section B ：FB]
FB = 0.0125 + {(1.77 + 41.7) × 2 × 0.2 + 41.7 × 2 × 0.2} × 9.80665 × 10-3 = 0.35kN
・Determine acceptability
Max. allowable load ≧FB'
Converted to per meter of chain width
FB' = 1000 × FB 300 = 1.17(kN/m)
From the allowable load graph, maximum allowable tension is 2.5 (kN/m).
2.5(kN/m)≧1.17(kN/m)
Selected chain is acceptable.
Nose bar on front end
●Calculation formula (SI units: kN)
・Tension at return-way
[Tension at section A ：FA]
FA = 1.1m1・L・μ1 × 9.80665 × 10-3
・Tension at carry-way
[Tension at section B ：FB]
FB= [FA + {(m1 + m2) L・μ1 + m2・LS・μ2} × 9.80665 × 10-3] × fn
・Chain load
F = FB
Note: LS = 0 when there is no accumulation of conveyed products.
●Calculation example (SI units)
| Usage conditions | |
|---|---|
| Chain Type | WT0705-W300-LFG (m1=5.9×0.3=1.77kg/m) |
| Chain width | 300mm |
| Layout | L=2m |
| Chain speed | V=15m/min |
| Conveyed object | 500-ml aluminum can (filled) |
| Conveyed product mass (per 1-meter unit of length) | m2=139kg/m2(523 g/piece ) ×0.3m=41.7kg/m |
| Rail | UHMW-PE (Plastic rail) |
| Accumulation distance | Ls=2m |
| Lubrication | For dry conditions |
| Operating Ambient Temperature | 20℃ |
| Coefficient of dynamic friction between chain and wearstrip | μ1=0.2 |
| Dynamic coefficient of friction between chain and conveyed product | μ2=0.2 |
| Nose bar coefficient | fn=1.8 |
・Tension at return-way
[Tension at section A ：FA]
FA = 1.1 × 1.77 × 2 × 0.2 × 9.80665 × 10-3 = 0.0077kN
・Tension at carry-way
[Tension at section B ：FB]
FB = [0.0077 + {(1.77 + 41.7) × 2 × 0.2 + 41.7 × 2 × 0.2} × 9.80665 × 10-3] × 1.8 = 0.62kN
・Determine acceptability
Max. allowable load ≧FB'
Converted to per meter of chain width
FB' = 1000 × FB 300 = 2.07(kN/m)
From the allowable load graph, maximum allowable tension is 2.5 (kN/m).
2.5(kN/m)≧2.07(kN/m)
Selected chain is acceptable.
Nose bar on both ends
●Calculation formula (SI units: kN)
・Tension at return-way
[Tension at section A ：FA]
FA = m1・L・μ1・fn × 9.80665 × 10-3
・Tension at carry-way
[Tension at section B ：FB]
FB= [FA + {(m1 + m2) L・μ1 + m2・LS・μ2} × 9.80665 × 10-3] × fn
・Chain load
F = FB
Note: LS = 0 when there is no accumulation of conveyed products.
●Calculation example (SI units)
| Usage conditions | |
|---|---|
| Chain Type | WT0705-W300-LFG (m1=5.9×0.3=1.77kg/m) |
| Chain width | 300mm |
| Layout | L=2m |
| Chain speed | V=15m/min |
| Conveyed object | 500-ml aluminum can (filled) |
| Conveyed product mass (per 1-meter unit of length) | m2=139kg/m2(523 g/piece ) ×0.3m=41.7kg/m |
| Rail | UHMW-PE (Plastic rail) |
| Accumulation distance | Ls=2m |
| Lubrication | For dry conditions |
| Operating Ambient Temperature | 20℃ |
| Coefficient of dynamic friction between chain and wearstrip | μ1=0.2 |
| Dynamic coefficient of friction between chain and conveyed product | μ2=0.2 |
| Nose bar coefficient | fn=1.8 |
・Tension at return-way
[Tension at section A ：FA]
FA = 1.77 × 2 × 0.2 × 1.8 × 9.80665 × 10-3 = 0.0125kN
・Tension at carry-way
[Tension at section B ：FB]
FB = [0.0125 + {(1.77 + 41.7) × 2 × 0.2 + 41.7 × 2 × 0.2} × 9.80665 × 10-3] × 1.8 = 0.63kN
・Determine acceptability
Max. allowable load ≧FB'
Converted to per meter of chain width
FB' = 1000 × FB 300 = 2.1(kN/m)
From the allowable load graph, maximum allowable tension is 2.5 (kN/m).
2.5(kN/m)≧2.1(kN/m)
Selected chain is acceptable.
fn(Nose bar coefficient )
| Lubrication | Nose bar coefficient fn |
|---|---|
| Sliding Series | |
| For dry conditions | 1.8 |
| Soapsuds | 1.35 |
WT1500 series, WT1510 series and BTN5 Selection example
Nose bar on driven side
●Calculation formula (SI units: kN)
・Tension at return-way
[Tension at section A ：FA]
FA = m1・L・μ1・fn × 9.80665 × 10-3
・Tension at carry-way
[Tension at section B ：FB]
FB = FA + {(m1 + m2) L・μ1 + m2・LS・μ2} × 9.80665 × 10-3
・Chain load
F = FB
Note: LS = 0 when there is no accumulation of conveyed products.
●Calculation example (SI units)
| Usage conditions | |
|---|---|
| Chain Type | WT1506-K30-ALF (m1=6.7×0.762=5.1kg/m) |
| Chain width | 762mm |
| Layout | L=4m |
| Chain speed | V=15m/min |
| Conveyed object | 500-ml aluminum can (filled) |
| Weight of conveyed goods (per 1-meter unit of length ) | m2=139kg/m2(523 g/piece ) ×0.762m=106kg/m |
| Rail | Ultra high molecular weight polyethylene (Plastic rail ) |
| Accumulation distance | Ls=4m |
| Lubrication | For dry conditions |
| Operating Ambient Temperature | 20℃ |
| Coefficient of dynamic friction between chain and wearstrip | μ1=0.15 |
| Dynamic coefficient of friction between chain and conveyed product | μ2=0.14 |
| Nose bar coefficient | fn=1.35(bearing/roller type ) |
・Tension at return-way
[Tension at section A ：FA]
FA = 5.1 × 4 × 0.15 × 1.35 × 9.80665 × 10-3 = 0.04kN
・Tension at carry-way
[Tension at section B ：FB]
FB = 0.04 + {(5.1 + 106) × 4 × 0.15 + 106 × 4 × 0.14} × 9.80665 × 10-3 = 1.28kN
・Determine acceptability
Max. allowable load ≧FB'
Converted to per meter of chain width
FB' = 1000 × FB 762 = 1.68(kN/m)
From the allowable load graph, maximum allowable tension is 2.5 (kN/m).
10.5(kN/m)≧1.68(kN/m)
Selected chain is acceptable.
Nose bar on front end
●Calculation formula (SI units: kN)
・Tension at return-way
[Tension at section A ：FA]
FA = 1.1m1・L・μ1 × 9.80665 × 10-3
・Tension at carry-way
[Tension at section B ：FB]
FB= [FA + {(m1 + m2) L・μ1 + m2・LS・μ2} × 9.80665 × 10-3] × fn
・Chain load
F = FB
Note: LS = 0 when there is no accumulation of conveyed products.
●Calculation example (SI units)
| Usage conditions | |
|---|---|
| Chain Type | WT1506-K30-ALF (m1=6.7×0.762=5.1kg/m) |
| Chain width | 762mm |
| Layout | L=4m |
| Chain speed | V=15m/min |
| Conveyed object | 500-ml aluminum can (filled) |
| Weight of conveyed goods (per 1-meter unit of length ) | m2=139kg/m2(523 g/piece ) ×0.762m=106kg/m |
| Rail | Ultra high molecular weight polyethylene (Plastic rail ) |
| Accumulation distance | Ls=4m |
| Lubrication | For dry conditions |
| Operating Ambient Temperature | 20℃ |
| Coefficient of dynamic friction between chain and wearstrip | μ1=0.15 |
| Dynamic coefficient of friction between chain and conveyed product | μ2=0.14 |
| Nose bar coefficient | fn=1.35(bearing/roller type ) |
・Tension at return-way
[Tension at section A ：FA]
FA = 1.1 × 5.1 × 4 × 0.15 × 9.80665 × 10-3 = 0.03kN
・Tension at carry-way
[Tension at section B ：FB]
FB = [0.03 + {(5.1 + 106) × 4 × 0.15 + 106 × 4 × 0.14} × 9.80665 × 10-3] × 1.35 = 1.71kN
・Determine acceptability
Max. allowable load ≧FB'
Converted to per meter of chain width
FB' = 1000 × FB 762 = 2.24(kN/m)
From the allowable load graph, maximum allowable tension is 10.5 (kN/m).
10.5(kN/m)≧2.24(kN/m)
Selected chain is acceptable.
Nose bar on both ends
●Calculation formula (SI units: kN)
・Tension at return-way
[Tension at section A ：FA]
FA = m1・L・μ1・fn × 9.80665 × 10-3
・Tension at carry-way
[Tension at section B ：FB]
FB= [FA + {(m1 + m2) L・μ1 + m2・LS・μ2} × 9.80665 × 10-3] × fn
・Chain load
F = FB
Note: LS = 0 when there is no accumulation of conveyed products.
●Calculation example (SI units)
| Usage conditions | |
|---|---|
| Chain Type | WT1506-K30-ALF (m1=6.7×0.762=5.1kg/m) |
| Chain width | 762mm |
| Layout | L=4m |
| Chain speed | V=15m/min |
| Conveyed object | 500-ml aluminum can (filled) |
| Weight of conveyed goods (per 1-meter unit of length ) | m2=139kg/m2(523 g/piece ) ×0.762m=106kg/m |
| Rail | Ultra high molecular weight polyethylene (Plastic rail ) |
| Accumulation distance | Ls=4m |
| Lubrication | For dry conditions |
| Operating Ambient Temperature | 20℃ |
| Coefficient of dynamic friction between chain and wearstrip | μ1=0.15 |
| Dynamic coefficient of friction between chain and conveyed product | μ2=0.14 |
| Nose bar coefficient | fn=1.35(nose bar/roller type ) |
・Tension at return-way
[Tension at section A ：FA]
FA = 5.1 × 4 × 0.15 × 1.35 × 9.80665 × 10-3 = 0.04kN
・Tension at carry-way
[Tension at section B ：FB]
FB = [0.04 + {(5.1 + 106) × 4 × 0.15 + 106 × 4 × 0.14} × 9.80665 × 10-3] × 1.35 = 1.72kN
・Determine acceptability
Max. allowable load ≧FB'
Converted to per meter of chain width
FB' = 1000 × FB 762 = 2.26(kN/m)
From the allowable load graph, maximum allowable tension is 10.5 (kN/m).
10.5(kN/m)≧2.26(kN/m)
Selected chain is acceptable.
fn(Nose bar coefficient )
| Lubrication | Nose bar coefficient fn | |
|---|---|---|
| Sliding Series | bearing/roller type | |
| For dry conditions | 1.8 | 1.35 |
| Soapsuds | 1.35 | |
WT2525VG-K Selection example
Forward/reverse bottom drive selection example
●Calculation formula (SI units: kN)
・Tension at return-way
[Tension at section A ：FA]
FA = 1.1m1・L1・μ1 × 9.80665 × 10-3
・Tension at carry-way
[Tension at section B ：FB]
FB = 1.1{FA + (m1 + m2) × L・μ1 × 9.80665 × 10-3}
・Tension at return-way (drive )
[Tension at section C ：FC]
FC = FB + m1・L2・μ1 × 9.80665 × 10-3
・Chain load
F = FC
●Calculation example (SI units)
| Usage conditions | |
|---|---|
| Chain Type | WT2525VG-K36-G (m1=9.5×0.9144=8.7kg/m) |
| Chain width | 914.4mm |
| Layout | L=10m, L1=6m, L2=4m |
| Chain speed | V=10m/min |
| Conveyed object | cardboard sheet 900mm×1800mm×5mm 106.4kg/m2=861.8g/sheet x 200 sheets (height 1m) |
| Weight of conveyed goods (per 1-meter unit of length ) | m2=106.4kg/m2×0.9144m =97.3kg/m |
| Rail | Ultra high molecular weight polyethylene (Plastic rail ) |
| Lubrication | For dry conditions |
| Operating Ambient Temperature | 20℃ |
| Coefficient of dynamic friction between chain and wearstrip | μ1=0.25 |
・Tension at return-way (Driven Side )
[Tension at section A ：FA]
FA = 1.1 × 8.7 × 6 × 0.25 × 9.80665 × 10-3 = 0.14kN
・Tension at carry-way
[Tension at section B ：FB]
FB = 1.1 {0.14 + (8.7 + 97.3) × 10 × 0.25 × 9.80665 × 10-3} = 3.0kN
・Tension at carry-way
[Tension at section C ：FC]
FC = 3.0 + 8.7 × 4 × 0.25 × 9.80665 × 10-3 = 3.09kN
・Determine acceptability
Max. allowable load ≧FB'
Converted to per meter of chain width
FB' = 1000 × FB 914.4 = 3.34(kN/m)
From the allowable load graph, maximum allowable tension is 12.8 (kN/m).
12.8(kN/m)≧3.34(kN/m)
Selected chain is acceptable.
