Tsubaki Products

Technical Data Drive chain Roller Chain Selection

6. Allowable load selection method

Selection based on maximum allowable load

1. Speed considerations

This selection method is used when the roller chain will be used within the range of speeds shown in Table 1. When using at speeds higher than the upper limit shown in the table, use the General Selection Method to select the chain.

Table 1 Selection speed limit
Pitch mm	Speed limit m/min
Under 12.70	120
12.70	100
15.875	90
19.05	80
25.40	70
31.75	60
38.10	50
44.45	50
50.80	50
57.15	40
63.50	40
76.20	40
101.60	30
127.00	30

The speed limit for Poly Steel Chain is 70 m/min.

2. Impact considerations

For transmission with large impacts and other extreme conditions, in particular large-load transmission and transmission where a thrust load may operate, use F-type connecting links or two-pitch offset links.

3. Strength of connecting links and offset links

When using M-type connecting links or offset links for roller chains shown in Tables 2 and 3, multiply the maximum allowable load by the percentage given in the tables.

Table 2 Strength of M-type connecting links
RS roller chain	RS15, RS25, RS37, RS38, RS41, BF25-H	80%
RS roller chain BS/DIN standard	RF06B, RS56B, RS56B	80%
Cold Resistant Roller Chain KT Series	all sizes	80%

Table 3 Strength of offset links
	Offset link
	1-pitch	2-pitch	4-pitch
RS roller chain	65%	100%	-
RS roller chain BS/DIN standard	60%	60%	-
Super chain	-	-	85%
RS roller chain NP Series	65%	-	-
RS roller chain NEP/APP Series	65%	-	-
Low-noise chain	65%	-	-

4. Sprocket considerations

When heavy-duty drive chain is used, chain tension will increase. For this reason, commercially available cast iron sprockets may not have sufficient rim or hub strength. Use S35C or equivalent material. RS sprockets have the strength capable of handling heavy-duty drive chain. For heavy duty drive chain, use sprockets with hardened teeth.

Please also refer to the calculation formulas (click here) and factors (click here) used for chain selection, as well as calculating moment of inertia (click here).

Selection example using the allowable load selection method

Conditions

Machine to be used	Conveyor drive
Weight of conveyed goods M	6000kg
Load speed V_ℓ	30m/min
Conveyor roller diameter	380mm
Belt thickness	10mm
Conveyor roller rotation torque	3.3kN･m{337kgf・m}
Motor	11kW n1 = 1800r/min Starting torque Ts 200% Maximum (stalling) torque Tmax 210% Braking torque Tb 200% Moment of Inertia Im 0.088kg・m² {Flywheel effect GD²m 0.352kgf・m²}

Reducer reduction ratio	1/50 (i = 50)
Drive shaft diameter	Shaft diameter Φ66mm
Driven shaft diameter	Shaft diameter Φ94mm
Distance between shafts	500mm
Driven sprocket diameter	≦400mm
Starting frequency	10 times /day
Type of impact	Moderate shock
Soft start/stop	None

SI Unit
Steps 1 Check motor characteristics ・Braking torque Tn = 9.55 × kW n1 = 9.55 × 11 1800 = 0.058(kN・m) ・Starting torque Ts = Tn × 2 = 0.058 × 2 = 0.116(kN・m) ・Maximum (stalling) torque Tmax = Tn × 2.1 = 0.058 × 2.1 = 0.122(kN・m) ・Braking torque Tb = Tn × 2.0 = 0.058 × 2.0 = 0.116(kN・m) ・Motor moment of inertia Im = 0.088(kg・m²) Steps 2 Calculate from load Driven shaft revolution n₂ = V_ℓ × 1000 (Conveyor roller diameter + 2 × Belt thickness ) × π = 30 × 1000 (380 + 20) × π = 23.9(r/min) Drive shaft revolution n = n₁/i = 1800 50 = 36(r/min) Chain reducer ratio = 23.9 36 = 1 1.51 If the driven sprocket PCD d₂ = 400mm Chain load Fw = Conveyor roller rotation torque × 1000 × 2 d₂ = 3.3 × 1000 × 2 400 = 16.5(kN) Tentatively select the chain. With moderate shock .......Service factor Ks = 1.3 Tentative design chain tension = Fw × Ks = 16.5 × 1.3 = 21.5(kN) Tentatively select RS120-1 with a maximum allowable load of 30.4 kN. 31T from driven sprocket ＜ 400mm Outer Diameter 398mm PCD d₂ = 376.60(mm) Number of teeth of drive sprocket = 31 1.51 = 21T PCD d = 255.63(mm) Chain speed = P × Z'× n 1000 = 38.1 × 21 × 36 1000 = 28.8m/min ＜ 50 m/min, so it is possible to select by allowable load. Small sprocket RPM 36r/min・RPM Kn = 1.03 Number of teeth of small sprocket 21T....Number of teeth factor Kz = 1.10 Chain load Fw = Conveyor roller rotation torque × 1000 × 2 d₂ = 3.3 × 1000 × 2 376.6 = 17.5 (kN) Design chain tension F'w = Fw × Ks × Kn × Kz = 17.5 × 1.3 × 1.03 × 1.10 = 25.8(kN)...(1) RS120-1 (Max. allowable load: 30.4kN) can be used. Check the conveyance speed (Selection criteria 30m/min) V_ℓ = n₂ × (Conveyor roller diameter + 2 × Belt thickness ) × π 1000 = n₁ × 21 31 × (Conveyor roller diameter + 2 × Belt thickness ) × π 1000 = 36 × 21 31 × (380 + 2 × 10) × π 1000 = 30.6(m/min) Steps 3 Calculate from acceleration/deceleration time The small sprocket (reducer output shaft sprocket) was decided as RS120 21T from the calculations in Step 2.Thus, calculate using the same pitch and number of teeth. If the acceleration/deceleration time is known, use that value for the calculation. The following is calculated assuming it is unknown. Working torque Tm = Ts + Tmax 2 = 0.116 + 0.122 2 = 0.119(kN・m) Load torque T_ℓ = Fw × d 2 × 1000 × i = 17.5 × 255.63 2 × 1000 × 50 = 0.045(kN・m) Motor shaft conversion 　Moment of inertia on the load side I_ℓ I_ℓ = M × V_ℓ 2 × π × n1 ² = 6000 × 30.6 2 × π × 1800 ² = 0.044(kg・m²) Motor moment of inertia Im = 0.088(kg・m²) Acceleration time of the motor ts = (Im + I_ℓ) × n₁ 9550 × (Tm - T_ℓ) = (0.088 + 0.044) × 1800 9550 ×(0.119 - 0.045) = 0.34(s) Deceleration time of the motor tb = (Im + I_ℓ) × n₁ 9550 × (Tb + T_ℓ) = (0.088 + 0.044) × 1800 9550 × (0.116 + 0.045) = 0.15(s) As tb ＜ ts, chain tension during deceleration Fb is larger than chain tension during acceleration Fs. Thus, use the following. Deceleration αb = V_ℓ tb × 60 = 30.6 0.15 × 60 = 3.40(m/s²) Chain tension during deceleration Fb = M × αb 1000 × (Conveyor roller diameter + 2 × Belt thickness ) d₂ + Fw = 6000 × 3.40 1000 × (380 + 2 × 10) 376.6 + 17.5 = 39.2(kN) Design chain tension F'b = Fb × Kn × Kz = 39.2 × 1.03 × 1.10 = 44.4(kN)...(2) RS120-2 (maximum allowable load 51.7 kN) or RS120-SUP-2 (maximum allowable load 66.7 kN) can be used because F′b = 44.4(kN). Considering RS140 18T (outer diameter 279 mm d₁ = 255.98) and 27T (outer diameter 407 mm d₂ = 382.88) with similar PCD results conflict with the driven sprocket outer diameter≦ 400 mm, they cannot be used. Chain reduction ratio becomes 2618 from the required 3623.9, and conveyance speed = 30 × 3623.9 × 2618 = 31.3m/min, but upon examination 26T (outer diameter 393mm d₂ = 368.77) (2) is F'b = 44.3(kN) RS140-1 cannot be used because its maximum allowable load is 40.2 kN. RS140-SUP-1 can be used because its maximum allowable load is 53.9 kN. Since the sprocket bore diameter of 18T is up to 89 mm, and for 26T is up to 103 mm, it can be used with a drive shaft diameter of 66 mm and driven shaft diameter of 94 mm. With the distance between shafts at 500 mm, a sprocket with 18T (d₁=255.98) and 26T (d₂=368.77) can be used. Number of links will be 46 links. Steps 4 Calculate from inertia ratio R Inertia ratio R = I_ℓ Im = 0.044 0.088 = 0.5 There is clearance in the drive equipment····Shock factor K = 1.0 Starting torque Ts = 0.116(kN・m) Chain tension from starting torque Fms = Ts × i × 1000 × 2 d = 0.116 × 50 × 1000 × 2 255.63 = 45.4(kN) Braking torque Tb = 0.116(kN・m) Chain tension from braking torque Fmb = Tb × i × 1.2 × 1000 × 2 d = 0.116 × 50 × 1.2 × 1000 × 2 255.63 = 54.5(kN) Since Fmb＞Fms, use the larger Fmb. Design chain tension F'mb = Fmb × K × Kn × Kz = 54.5 × 1.0 × 1.03 × 1.10 = 61.7(kN)...........(3) Comparing (1),(2), and (3), (3) has the largest design chain tension. Since F′mb=61.7 (kN), RS120-3 (maximum allowable load 76.0 kN) or RS120-SUP-2 (maximum allowable load 66.7 kN) is usable. With the distance between shafts at 500mm, a sprocket with 21T (d₁=255.63) and 31T (d₂=376.60) can be used. Number of links will be 54 links. Considering RS160 15T (outer diameter 269mm d₁ = 244.33) and 23T (outer diameter 400mm d₂ = 373.07) with similar PCD, (3) F'mb = 64.6 (kN) will be largest. RS160-1 cannot be used because its maximum allowable load is 53.0 kN. RS160-SUP-1 can be used because its maximum allowable load is 70.6 kN. Since a sprocket bore diameter with 15T is up to 95mm, and 23T is up to 118mm, it can be used for a drive shaft diameter of 66mm, and driven shaft diameter of 94mm. With the distance between shafts at 500mm, a sprocket with 15T (d₁ = 244.33) and 23T (d₂ = 373.07) can be used. Number of links will be 40 links.

SI Unit

Steps 1 Check motor characteristics

・Braking torque
Tn = 9.55 × kW n1 = 9.55 × 11 1800 = 0.058(kN・m)

・Starting torque
Ts = Tn × 2 = 0.058 × 2 = 0.116(kN・m)

・Maximum (stalling) torque
Tmax = Tn × 2.1 = 0.058 × 2.1 = 0.122(kN・m)

・Braking torque
Tb = Tn × 2.0 = 0.058 × 2.0 = 0.116(kN・m)

・Motor moment of inertia
Im = 0.088(kg・m²)

Steps 2 Calculate from load

Driven shaft revolution
n₂ = V_ℓ × 1000 (Conveyor roller diameter + 2 × Belt thickness ) × π
= 30 × 1000 (380 + 20) × π = 23.9(r/min)

Drive shaft revolution
n = n₁/i = 1800 50 = 36(r/min)

Chain reducer ratio = 23.9 36 = 1 1.51

If the driven sprocket PCD d₂ = 400mm
Chain load Fw = Conveyor roller rotation torque × 1000 × 2 d₂
= 3.3 × 1000 × 2 400 = 16.5(kN)

Tentatively select the chain.

With moderate shock .......Service factor Ks = 1.3

Tentative design chain tension = Fw × Ks = 16.5 × 1.3 = 21.5(kN)

Tentatively select RS120-1 with a maximum allowable load of 30.4 kN.

31T from driven sprocket ＜ 400mm
Outer Diameter 398mm PCD d₂ = 376.60(mm)

Number of teeth of drive sprocket = 31 1.51 = 21T PCD d = 255.63(mm)

Chain speed = P × Z'× n 1000 = 38.1 × 21 × 36 1000
= 28.8m/min ＜ 50 m/min, so it is possible to select by allowable load.

Small sprocket RPM 36r/min・RPM Kn = 1.03

Number of teeth of small sprocket 21T....Number of teeth factor Kz = 1.10

Chain load Fw = Conveyor roller rotation torque × 1000 × 2 d₂
= 3.3 × 1000 × 2 376.6 = 17.5 (kN)

Design chain tension F'w = Fw × Ks × Kn × Kz
= 17.5 × 1.3 × 1.03 × 1.10 = 25.8(kN)...(1)

RS120-1 (Max. allowable load: 30.4kN) can be used.

Check the conveyance speed (Selection criteria 30m/min)

V_ℓ = n₂ × (Conveyor roller diameter + 2 × Belt thickness ) × π 1000
= n₁ × 21 31 × (Conveyor roller diameter + 2 × Belt thickness ) × π 1000
= 36 × 21 31 × (380 + 2 × 10) × π 1000
= 30.6(m/min)

Steps 3 Calculate from acceleration/deceleration time

The small sprocket (reducer output shaft sprocket) was decided as RS120 21T from the calculations in Step 2.Thus, calculate using the same pitch and number of teeth.

If the acceleration/deceleration time is known, use that value for the calculation. The following is calculated assuming it is unknown.

Working torque Tm = Ts + Tmax 2 = 0.116 + 0.122 2 = 0.119(kN・m)

Load torque T_ℓ = Fw × d 2 × 1000 × i = 17.5 × 255.63 2 × 1000 × 50
= 0.045(kN・m)

Motor shaft conversion 　Moment of inertia on the load side I_ℓ
I_ℓ = M × V_ℓ 2 × π × n1 ²
= 6000 × 30.6 2 × π × 1800 ²
= 0.044(kg・m²)

Motor moment of inertia Im = 0.088(kg・m²)

Acceleration time of the motor
ts = (Im + I_ℓ) × n₁ 9550 × (Tm - T_ℓ)
= (0.088 + 0.044) × 1800 9550 ×(0.119 - 0.045)
= 0.34(s)

Deceleration time of the motor
tb = (Im + I_ℓ) × n₁ 9550 × (Tb + T_ℓ) = (0.088 + 0.044) × 1800 9550 × (0.116 + 0.045) = 0.15(s)

As tb ＜ ts, chain tension during deceleration Fb is larger than chain tension during acceleration Fs. Thus, use the following.

Deceleration
αb = V_ℓ tb × 60 = 30.6 0.15 × 60
= 3.40(m/s²)

Chain tension during deceleration
Fb = M × αb 1000 × (Conveyor roller diameter + 2 × Belt thickness ) d₂
+ Fw = 6000 × 3.40 1000 × (380 + 2 × 10) 376.6 + 17.5 = 39.2(kN)

Design chain tension
F'b = Fb × Kn × Kz = 39.2 × 1.03 × 1.10 = 44.4(kN)...(2)
RS120-2 (maximum allowable load 51.7 kN) or RS120-SUP-2 (maximum allowable load 66.7 kN) can be used because F′b = 44.4(kN).

Considering RS140 18T (outer diameter 279 mm d₁ = 255.98) and 27T (outer diameter 407 mm d₂ = 382.88) with similar PCD results conflict with the driven sprocket outer diameter≦ 400 mm, they cannot be used.

Chain reduction ratio becomes 2618 from the required 3623.9, and conveyance speed = 30 × 3623.9 × 2618 = 31.3m/min, but upon examination 26T (outer diameter 393mm d₂ = 368.77)
(2) is F'b = 44.3(kN)

RS140-1 cannot be used because its maximum allowable load is 40.2 kN.

RS140-SUP-1 can be used because its maximum allowable load is 53.9 kN.

Since the sprocket bore diameter of 18T is up to 89 mm, and for 26T is up to 103 mm, it can be used with a drive shaft diameter of 66 mm and driven shaft diameter of 94 mm.

With the distance between shafts at 500 mm, a sprocket with 18T (d₁=255.98) and 26T (d₂=368.77) can be used. Number of links will be 46 links.

Steps 4 Calculate from inertia ratio R

Inertia ratio R = I_ℓ Im = 0.044 0.088 = 0.5

There is clearance in the drive equipment····Shock factor K = 1.0

Starting torque Ts = 0.116(kN・m)

Chain tension from starting torque
Fms = Ts × i × 1000 × 2 d
= 0.116 × 50 × 1000 × 2 255.63 = 45.4(kN)

Braking torque Tb = 0.116(kN・m)

Chain tension from braking torque
Fmb = Tb × i × 1.2 × 1000 × 2 d
= 0.116 × 50 × 1.2 × 1000 × 2 255.63 = 54.5(kN)

Since Fmb＞Fms, use the larger Fmb.

Design chain tension
F'mb = Fmb × K × Kn × Kz
= 54.5 × 1.0 × 1.03 × 1.10 = 61.7(kN)...........(3)

Comparing (1),(2), and (3), (3) has the largest design chain tension.

Since F′mb=61.7 (kN), RS120-3 (maximum allowable load 76.0 kN) or RS120-SUP-2 (maximum allowable load 66.7 kN) is usable.

With the distance between shafts at 500mm, a sprocket with 21T (d₁=255.63) and 31T (d₂=376.60) can be used. Number of links will be 54 links.

Considering RS160 15T (outer diameter 269mm d₁ = 244.33) and 23T (outer diameter 400mm d₂ = 373.07) with similar PCD, (3) F'mb = 64.6 (kN) will be largest.

RS160-1 cannot be used because its maximum allowable load is 53.0 kN.

RS160-SUP-1 can be used because its maximum allowable load is 70.6 kN.

Since a sprocket bore diameter with 15T is up to 95mm, and 23T is up to 118mm, it can be used for a drive shaft diameter of 66mm, and driven shaft diameter of 94mm.

With the distance between shafts at 500mm, a sprocket with 15T (d₁ = 244.33) and 23T (d₂ = 373.07) can be used. Number of links will be 40 links.

{Gravity unit }
Steps 1 Check motor characteristics ・Braking torque Tn = 974 × kW n1 = 974 × 11 1800 = 5.95(kgf・m) ・Starting torque Ts = Tn × 2 = 5.95 × 2 = 11.9(kgf・m) ・Maximum (stalling) torque Tmax = Tn × 2.1 = 5.95 × 2.1 = 12.5(kgf・m) ・Braking torque Tb = Tn × 2.0 = 5.95 × 2.0 = 11.9(kgf・m) ・GD² of the motor GD²m = 0.352(kgf・m²) Steps 2 Calculate from load Driven shaft revolution n₂ = V_ℓ × 1000 (Conveyor roller diameter + 2 × Belt thickness ) × π = 30 × 1000 (380 + 20) × π = 23.9(r/min) Drive shaft revolution n = n₁/i = 1800 50 = 36(r/min) Chain reducer ratio = 23.9 36 = 1 1.51 If the driven sprocket PCD d₂ = 400mm Chain load Fw = Conveyor roller rotation torque × 1000 × 2 d₂ = 337 × 1000 × 2 400 = 1690(kgf) Tentatively select the chain. With moderate shock .......Service factor Ks = 1.3 Tentative design chain tension = Fw × Ks = 1690 × 1.3 = 2200(kgf) Tentatively select RS120-1 with a maximum allowable load of 3100 kgf. 31T from driven sprocket ＜ 400mm Outer Diameter 398mm PCD d₂ = 376.60(mm) Number of teeth of drive sprocket = 31 1.51 = 21T PCD d = 255.63(mm) Chain speed = P × Z'× n 1000 = 38.1 × 21 × 36 1000 = 28.8m/min ＜ 50 m/min, so it is possible to select by allowable load. Small sprocket RPM 36r/min・RPM Kn = 1.03 Number of teeth of small sprocket 21T....Number of teeth factor Kz = 1.10 Chain load Fw = Conveyor roller rotation torque × 1000 × 2 d₂ = 337 × 1000 × 2 376.6 = 1790 (kgf) Design chain tension F'w = Fw × Ks × Kn × Kz = 1790 × 1.3 × 1.03 × 1.10 = 2640(kgf)...(1) RS120-1 (Max. allowable load: 3100kgf) can be used. Check the conveyance speed (Selection criteria 30m/min) V_ℓ = n₂ × (Conveyor roller diameter + 2 × Belt thickness ) × π 1000 = n₁ × 21 31 × (Conveyor roller diameter + 2 × Belt thickness ) × π 1000 = 36 × 21 31 × (380 + 2 × 10) × π 1000 = 30.6(m/min) Steps 3 Calculate from acceleration/deceleration time The small sprocket (reducer output shaft sprocket) was decided as RS120 21T from the calculations in Step 2.Thus, calculate using the same pitch and number of teeth. If the acceleration/deceleration time is known, use that value for the calculation. The following is calculated assuming it is unknown. Working torque Tm = Ts + Tmax 2 = 11.9 + 12.5 2 = 12.2(kgf・m) Load torque T_ℓ = Fw × d 2 × 1000 × i = 1790 × 255.63 2 × 1000 × 50 = 4.58(kgf・m) Motor shaft conversion 　GD² of the load side GD²_ℓ= M × V_ℓ π × n₁ ² = 6000 × 30.6 π × 1800 ² = 0.176(kgf・m²) GD² of the motor GD²m = 0.352(kgf・m²) Acceleration time of the motor ts = (GD²m + GD²_ℓ) × n₁ 375×(Tm - T_ℓ) = (0.352 + 0.176) × 1800 375 × (12.2 - 4.58) = 0.34(s) Deceleration time of the motor tb = (GD²m + GD²_ℓ) × n₁ 375 × (Tb + T_ℓ) = (0.352 + 0.176) × 1800 375 × (11.9 + 4.58) = 0.15(s) As tb ＜ ts, chain tension during deceleration Fb is larger than chain tension during acceleration Fs. Thus, use the following. Deceleration αb = V_ℓ tb × 60 = 30.6 0.15 × 60 = 3.40(m/s²) Chain tension during deceleration Fb = M × αb G × (Conveyor roller diameter + 2 × Belt thickness ) d₂ + Fw = 6000 × 3.40 G × (380 + 2 × 10) 376.6 + 1790 = 4000(kgf) Design chain tension F′b = Fb × Kn × Kz = 4000 × 1.03 × 1.10 = 4530(kgf) … (2) RS120-2 (maximum allowable load 5270 kgf) or RS120-SUP-2 (maximum allowable load 6800 kgf) can be used because F'b = 4530(kgf). Considering RS140 18T (outer diameter 279 mm d₁ = 255.98) and 27T (outer diameter 407 mm d₂ = 382.88) with similar PCD results conflict with the driven sprocket outer diameter≦ 400 mm, they cannot be used. Chain reduction ratio becomes 2618 from the required 3623.9, and conveyance speed = 30×3623.9×2618 = 31.3m/min but upon examination 26T (outer diameter 393mm d₂=368.77) (2) is F′b=4520(kgf) RS140-1 cannot be used because its maximum allowable load is 4100 kgf. RS140-SUP-1 can be used because its maximum allowable load is 5500 kgf. Since the sprocket bore diameter of 18T is up to 89 mm, and for 26T is up to 103 mm, it can be used with a drive shaft diameter of 66 mm and driven shaft diameter of 94 mm. With the distance between shafts at 500 mm, a sprocket with 18T (d₁=255.98) and 26T (d₂=368.77) can be used. Number of links will be 46 links. Steps 4 Calculate from inertia ratio R Inertia ratio R = GD²_ℓ GD²m = 0.176 0.352 = 0.5 There is clearance in the drive equipment····Shock factor K = 1.0 Starting torque Ts = 11.9(kgf・m) Chain tension from starting torque Fms = Ts × i × 1000 × 2 d = 11.9 × 50 × 1000 × 2 255.63 = 4660(kgf) Braking torque Tb = 11.9(kgf・m) Chain tension from braking torque Fmb = Tb × i × 1.2 × 1000 × 2 d = 11.9 × 50 × 1.2 × 1000 × 2 255.63 = 5590(kgf) Since Fmb＞Fms, use the larger Fmb. Design chain tension F'mb = Fmb × K × Kn × Kz = 5590 × 1.0 × 1.03 × 1.10 = 6330(kgf)...........(3) Comparing (1),(2), and (3), (3) has the largest design chain tension. Since F′mb = 6330 (kgf), RS120-3 (maximum allowable load 7550 kgf) or RS120-SUP-2 (maximum allowable load 6800 kgf) is usable. With the distance between shafts at 500mm, a sprocket with 21T (d₁=255.63) and 31T (d₂=376.60) can be used. Number of links will be 54 links. Considering RS160 15T (outer diameter 269mm d₁ = 244.33) and 23T (outer diameter 400mm d₂ = 373.07) with similar PCD, (3) F'mb = 6620 (kgf) will be largest. RS160-1 cannot be used because its maximum allowable load is 5400 kgf. RS160-SUP-1 can be used because its maximum allowable load is 7200 kgf. Since a sprocket bore diameter with 15T is up to 95mm, and 23T is up to 118mm, it can be used for a drive shaft diameter of 66mm, and driven shaft diameter of 94mm. With the distance between shafts at 500mm, a sprocket with 15T (d₁ = 244.33) and 23T (d₂ = 373.07) can be used. Number of links will be 40 links.

{Gravity unit }

Steps 1 Check motor characteristics

・Braking torque
Tn = 974 × kW n1 = 974 × 11 1800 = 5.95(kgf・m)

・Starting torque
Ts = Tn × 2 = 5.95 × 2 = 11.9(kgf・m)

・Maximum (stalling) torque
Tmax = Tn × 2.1 = 5.95 × 2.1 = 12.5(kgf・m)

・Braking torque
Tb = Tn × 2.0 = 5.95 × 2.0 = 11.9(kgf・m)

・GD² of the motor
GD²m = 0.352(kgf・m²)

Steps 2 Calculate from load

Driven shaft revolution
n₂ = V_ℓ × 1000 (Conveyor roller diameter + 2 × Belt thickness ) × π
= 30 × 1000 (380 + 20) × π = 23.9(r/min)

Drive shaft revolution
n = n₁/i = 1800 50 = 36(r/min)

Chain reducer ratio = 23.9 36 = 1 1.51

If the driven sprocket PCD d₂ = 400mm
Chain load Fw = Conveyor roller rotation torque × 1000 × 2 d₂
= 337 × 1000 × 2 400 = 1690(kgf)

Tentatively select the chain.

With moderate shock .......Service factor Ks = 1.3

Tentative design chain tension = Fw × Ks = 1690 × 1.3 = 2200(kgf)

Tentatively select RS120-1 with a maximum allowable load of 3100 kgf.

31T from driven sprocket ＜ 400mm
Outer Diameter 398mm PCD d₂ = 376.60(mm)

Number of teeth of drive sprocket = 31 1.51 = 21T PCD d = 255.63(mm)

Chain speed = P × Z'× n 1000 = 38.1 × 21 × 36 1000
= 28.8m/min ＜ 50 m/min, so it is possible to select by allowable load.

Small sprocket RPM 36r/min・RPM Kn = 1.03

Number of teeth of small sprocket 21T....Number of teeth factor Kz = 1.10

Chain load Fw = Conveyor roller rotation torque × 1000 × 2 d₂
= 337 × 1000 × 2 376.6 = 1790 (kgf)

Design chain tension F'w = Fw × Ks × Kn × Kz
= 1790 × 1.3 × 1.03 × 1.10 = 2640(kgf)...(1)

RS120-1 (Max. allowable load: 3100kgf) can be used.

Check the conveyance speed (Selection criteria 30m/min)

Steps 3 Calculate from acceleration/deceleration time

The small sprocket (reducer output shaft sprocket) was decided as RS120 21T from the calculations in Step 2.Thus, calculate using the same pitch and number of teeth.

If the acceleration/deceleration time is known, use that value for the calculation. The following is calculated assuming it is unknown.

Working torque Tm = Ts + Tmax 2 = 11.9 + 12.5 2 = 12.2(kgf・m)

Load torque T_ℓ = Fw × d 2 × 1000 × i = 1790 × 255.63 2 × 1000 × 50
= 4.58(kgf・m)

Motor shaft conversion 　GD² of the load side
GD²_ℓ= M × V_ℓ π × n₁ ²
= 6000 × 30.6 π × 1800 ²
= 0.176(kgf・m²)

GD² of the motor GD²m = 0.352(kgf・m²)

Acceleration time of the motor
ts = (GD²m + GD²_ℓ) × n₁ 375×(Tm - T_ℓ)
= (0.352 + 0.176) × 1800 375 × (12.2 - 4.58)
= 0.34(s)

Deceleration time of the motor
tb = (GD²m + GD²_ℓ) × n₁ 375 × (Tb + T_ℓ) = (0.352 + 0.176) × 1800 375 × (11.9 + 4.58) = 0.15(s)

As tb ＜ ts, chain tension during deceleration Fb is larger than chain tension during acceleration Fs. Thus, use the following.

Deceleration
αb = V_ℓ tb × 60 = 30.6 0.15 × 60
= 3.40(m/s²)

Chain tension during deceleration
Fb = M × αb G × (Conveyor roller diameter + 2 × Belt thickness ) d₂
+ Fw = 6000 × 3.40 G × (380 + 2 × 10) 376.6 + 1790 = 4000(kgf)

Design chain tension
F′b = Fb × Kn × Kz = 4000 × 1.03 × 1.10 = 4530(kgf) … (2)
RS120-2 (maximum allowable load 5270 kgf) or RS120-SUP-2 (maximum allowable load 6800 kgf) can be used because F'b = 4530(kgf).

Chain reduction ratio becomes 2618 from the required 3623.9, and conveyance speed = 30×3623.9×2618 = 31.3m/min but upon examination 26T (outer diameter 393mm d₂=368.77)
(2) is F′b=4520(kgf)

RS140-1 cannot be used because its maximum allowable load is 4100 kgf.

RS140-SUP-1 can be used because its maximum allowable load is 5500 kgf.

Since the sprocket bore diameter of 18T is up to 89 mm, and for 26T is up to 103 mm, it can be used with a drive shaft diameter of 66 mm and driven shaft diameter of 94 mm.

With the distance between shafts at 500 mm, a sprocket with 18T (d₁=255.98) and 26T (d₂=368.77) can be used. Number of links will be 46 links.

Steps 4 Calculate from inertia ratio R

Inertia ratio R = GD²_ℓ GD²m = 0.176 0.352 = 0.5

There is clearance in the drive equipment····Shock factor K = 1.0

Starting torque Ts = 11.9(kgf・m)

Chain tension from starting torque
Fms = Ts × i × 1000 × 2 d
= 11.9 × 50 × 1000 × 2 255.63 = 4660(kgf)

Braking torque Tb = 11.9(kgf・m)

Chain tension from braking torque
Fmb = Tb × i × 1.2 × 1000 × 2 d
= 11.9 × 50 × 1.2 × 1000 × 2 255.63 = 5590(kgf)

Since Fmb＞Fms, use the larger Fmb.

Design chain tension
F'mb = Fmb × K × Kn × Kz
= 5590 × 1.0 × 1.03 × 1.10 = 6330(kgf)...........(3)

Comparing (1),(2), and (3), (3) has the largest design chain tension.

Since F′mb = 6330 (kgf), RS120-3 (maximum allowable load 7550 kgf) or RS120-SUP-2 (maximum allowable load 6800 kgf) is usable.

With the distance between shafts at 500mm, a sprocket with 21T (d₁=255.63) and 31T (d₂=376.60) can be used. Number of links will be 54 links.

Considering RS160 15T (outer diameter 269mm d₁ = 244.33) and 23T (outer diameter 400mm d₂ = 373.07) with similar PCD, (3) F'mb = 6620 (kgf) will be largest.

RS160-1 cannot be used because its maximum allowable load is 5400 kgf.

RS160-SUP-1 can be used because its maximum allowable load is 7200 kgf.

Since a sprocket bore diameter with 15T is up to 95mm, and 23T is up to 118mm, it can be used for a drive shaft diameter of 66mm, and driven shaft diameter of 94mm.

With the distance between shafts at 500mm, a sprocket with 15T (d₁ = 244.33) and 23T (d₂ = 373.07) can be used. Number of links will be 40 links.

Selection results

Conditions	Steps	Model No.	Sprocket	No. of links	Lubrication method
Starting frequency less than 6 times	Steps 2	RS120-1	21T×31T	54 links	AII
Staring frequency 6 times or more with cushioned start	Steps 3	RS120-2	21T×31T	54 links	AII
Staring frequency 6 times or more with cushioned start	Steps 3	RS140-SUP-1	18T×26T	46 links	B
Staring frequency 6 times or more without cushioned start	Steps 3 Steps 4	RS120-3	21T×31T	54 links	AII
		RS120-SUP-2	21T×31T	54 links	B
		RS160-SUP-1	15T×23T	40 links	B

Note) 1.Refer to the kilowatt ratings tables for the lubrication method of each chain size and type.
2.All shaft distances need to be adjusted.

Please also refer to the calculation formulas (click here) and factors (click here) used for chain selection, as well as calculating moment of inertia (click here).

TSUBAKI Power Transmission Products Information Site

Technical Data Drive chain Roller Chain Selection

6. Allowable load selection method

1. Speed considerations

2. Impact considerations

3. Strength of connecting links and offset links

4. Sprocket considerations

Selection example using the allowable load selection method

Steps 1 Check motor characteristics

Steps 2 Calculate from load

Steps 3 Calculate from acceleration/deceleration time

Steps 4 Calculate from inertia ratio R

Steps 1 Check motor characteristics

Steps 2 Calculate from load

Steps 3 Calculate from acceleration/deceleration time

Steps 4 Calculate from inertia ratio R

Selection results