Tsubaki product selection

technical data Drive chain Roller Chain Selection

6. allowable tension selection method

This is a selection method using Maximum allowable load.

1. Speed Considerations

This selection method is for when roller chains are used within the speed range shown in Table 1. If the chain is to be used above the upper speed limit in the table, please use the general selection method.

Table 1 Upper limit speed selection
Pitch mm	Upper speed limit m/min
Less than 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 maximum speed for Poly-steel chain is 70 m/min.

2. Considering impact

In harsh conditions, such as transmissions with large impacts, especially transmissions with large loads or transmissions where lateral loads may be present, please use F-type coupling links or 2-pitch offset links.

3. Strength of connecting links and offset links

When using M-type connecting links or offset links with the roller chains shown in Tables 2 and 3, multiply Maximum allowable load by the percentage shown in the table.

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 specifications	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 Series / APP Series	65%	-	-
Low-noise chain	65%	-	-

4. Sprocket considerations

When using Heavy duty drive chain, the chain tension increases. Therefore, commercially available cast iron sprockets may not have sufficient rib and hub strength. Use a material equivalent to S35C or higher. RS sprockets are strong enough to handle Heavy duty drive chain. For Heavy duty drive chain use sprockets with hardened tooth tips.

Please also refer to the formulas (here), coefficients (here) used for chain selection, and how to calculate the moment of inertia (here).

Example of selection using allowable tension selection method

conditions

Machine used	Conveyor Drive
Amount of material transported M	6000kg
Transport speed V _ℓ	30m/min
Conveyor roll outer diameter	380mm
Belt Thickness	10mm
Conveyor roll rotation torque	3.3kN･m{337kgf・m}
Motor specifications	11kW n1 = 1800r/min Starting torque Ts 200% Maximum (stall) torque Tmax 210% Brake torque Tb 200% Moment of inertia Im 0.088kg・m ² {Flywheel effect GD ² m 0.352kgf・m ²}

Reducer reduction ratio	1/50 (i = 50)
Drive shaft	Shaft diameter Φ66mm
driven axis	Shaft diameter Φ94mm
Center distance	500mm
Driven sprocket outer diameter	≦400mm
Start frequency	10 times/day
Shock type	With some shock.
Soft start/stop	none

SI units
Step 1: Check the motor characteristics ・Rated 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 (stall) torque Tmax = Tn × 2.1 = 0.058 × 2.1 = 0.122(kN・m) Brake torque Tb = Tn × 2.0 = 0.058 × 2.0 = 0.116(kN・m) ・Motor moment of inertia Im = 0.088(kgm ²) Step 2 Calculate from the load Driven shaft rotation speed n ₂ = V _ℓ × 1000 (Conveyor roll outer diameter + 2 × Belt thickness) × π = 30 × 1000 (380 + 20) × π = 23.9(r/min) Drive shaft rotation speed n = n ₁ /i = 1800 50 = 36(r/min) Chain reduction ratio = 23.9 36 = 1 1.51 PCD of driven sprocket d ₂ = 400mm Chain tension Fw = Conveyor roll rotation torque x 1000 x 2 d ₂ = 3.3 × 1000 × 2 400 = 16.5(kN) Select the chain provisionally. With some impact.... Service factor Ks = 1.3 Preliminary compensation chain tension = Fw × Ks = 16.5 × 1.3 = 21.5(kN) We tentatively select RS120-1, which Maximum allowable load of 30.4kN. Driven sprocket outer diameter < 400mm 31T Outer diameter 398mm PCD d ₂ = 376.60(mm) Number of teeth on 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 < 50m/min, so the allowable tension can be selected. Small sprocket rotation speed: 36 r/min, rotation coefficient Kn = 1.03 Number of teeth on small sprocket 21T....Tooth number factor Kz = 1.10 Chain tension Fw = Conveyor roll rotation torque x 1000 x 2 d ₂ = 3.3 × 1000 × 2 376.6 = 17.5 (kN) Correction chain tension F'w = Fw × Ks × Kn × Kz = 17.5 × 1.3 × 1.03 × 1.10 = 25.8(kN)...(1) RS120-1 can be used Maximum allowable load of 30.4kN. Check the transport speed (selection condition: 30m/min) V _ℓ = n ₂ × (Conveyor roll outer diameter + 2 × Belt thickness) × π 1000 = n ₁ × 21 31 × (Conveyor roll outer diameter + 2 × Belt thickness) × π 1000 = 36 × 21 31 × (380 + 2 × 10) × π 1000 = 30.6(m/min) Step 3: Calculate from acceleration/deceleration time Since the calculation in step 2 determined that the small sprocket (reducer output shaft sprocket) was RS120 21T, the following calculations will also select the same pitch and number of teeth. If the acceleration and deceleration times are known, then use those values in the calculations. Here, we will make the calculations under the assumption that the times are unknown. Acting 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 converted 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 (kgm ²) Motor moment of inertia Im = 0.088 (kg・m ²) Motor acceleration time ts = (Im + I _ℓ) × n ₁ 9550 × (Tm - T _ℓ) = (0.088 + 0.044) × 1800 9550 × (0.119 - 0.045) = 0.34(s) Motor deceleration time tb = (Im + I _ℓ) × n ₁ 9550 × (Tb + T _ℓ) = (0.088 + 0.044) × 1800 9550 × (0.116 + 0.045) = 0.15(s) Since tb < ts, the chain tension Fb during deceleration is greater than the chain tension Fs during acceleration, so this will be adopted below. deceleration αb = V _ℓ tb x 60 = 30.6 0.15 × 60 = 3.40(m/s ²) Chain tension during deceleration Fb = M × αb 1000 × (Conveyor roll outer diameter + 2 × Belt thickness) d ₂ + Fw = 6000 × 3.40 1000 × (380 + 2 × 10) 376.6 + 17.5 = 39.2(kN) Correction chain tension F'b = Fb × Kn × Kz = 39.2 × 1.03 × 1.10 = 44.4(kN)...(2) F'b = 44.4 (kN), so RS120-2 (Maximum allowable load 51.7 kN) Alternatively, RS120-SUP-2 (Maximum allowable load 66.7kN) can be used. If we consider RS140 18T (outer diameter 279mm d ₁ = 255.98) and 27T (outer diameter 407mm d ₂ = 382.88), which have the same PCD, they cannot be used as they violate the condition of driven sprocket outer diameter ≦ 400mm. The chain reduction ratio went from the required 3623.9 to 2618. Conveying speed = 30 × 3623.9 × 2618 = 31.3 m/min. If we consider 26T (outer diameter 393mm _d2 = 368.77), (2)F'b = 44.3(kN). RS140-1 cannot be used as it has Maximum allowable load of 40.2kN. RS140-SUP-1 can be used because it has Maximum allowable load of 53.9kN. The sprocket shaft hole diameter is maximum 89mm for 18T and maximum 103mm for 26T. It is possible to use a drive shaft with a diameter of Φ66 mm and a driven shaft with a diameter of Φ94 mm. Since the center distance is 500 mm, the number of sprocket teeth is 18T (d ₁ = 255.98). 26T (_d2 = 368.77) can be used. The number of links is 46. Step 4 Calculate from inertia ratio R Inertia ratio R = I _ℓ Im = 0.044 0.088 = 0.5 Since there is play in the transmission....Shock factor K = 1.0 Starting torque Ts = 0.116(kN・m) Chain tension due to starting torque Fms = Ts × i × 1000 × 2 d = 0.116 × 50 × 1000 × 2 255.63 = 45.4(kN) Brake torque Tb = 0.116(kN・m) Chain tension due to brake torque Fmb = Tb × i × 1.2 × 1000 × 2 d = 0.116 × 50 × 1.2 × 1000 × 2 255.63 = 54.5(kN) If Fmb > Fms, the larger Fmb is used. Compensating 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 corrective chain tension. F'mb = 61.7 (kN), so RS120-3 (Maximum allowable load 76.0 kN), Alternatively, RS120-SUP-2 (Maximum allowable load 66.7kN) can be used. Since the center distance is 500 mm, the number of sprocket teeth is 21T (d ₁ = 255.63). 31T (d ₂ = 376.60) can be used. The number of links is 54. RS160 15T (outer diameter 269mm d ₁ = 244.33) with the same PCD Considering 23T (outer diameter 400mm d ₂ = 373.07) (3)The maximum value is F'mb = 64.6(kN). RS160-1 cannot be used as it has Maximum allowable load of 53.0kN. RS160-SUP-1 can be used because it has Maximum allowable load of 70.6kN. The sprocket shaft hole diameter is maximum 95mm for 15T and maximum 118mm for 23T. It is possible to use a drive shaft with a diameter of Φ66 mm and a driven shaft with a diameter of Φ94 mm. Since the center distance is 500 mm, the number of sprocket teeth is 15T (d ₁ = 244.33). 23T (d₂ = 373.07) can be used. The number of links is 40.

SI units

Step 1: Check the motor characteristics

・Rated 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 (stall) torque
Tmax = Tn × 2.1 = 0.058 × 2.1 = 0.122(kN・m)

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

・Motor moment of inertia
Im = 0.088(kgm ²)

Step 2 Calculate from the load

Driven shaft rotation speed
n ₂ = V _ℓ × 1000 (Conveyor roll outer diameter + 2 × Belt thickness) × π
= 30 × 1000 (380 + 20) × π = 23.9(r/min)

Drive shaft rotation speed
n = n ₁ /i = 1800 50 = 36(r/min)

Chain reduction ratio = 23.9 36 = 1 1.51

PCD of driven sprocket d ₂ = 400mm
Chain tension Fw = Conveyor roll rotation torque x 1000 x 2 d ₂
= 3.3 × 1000 × 2 400 = 16.5(kN)

Select the chain provisionally.

With some impact.... Service factor Ks = 1.3

Preliminary compensation chain tension = Fw × Ks = 16.5 × 1.3 = 21.5(kN)

We tentatively select RS120-1, which Maximum allowable load of 30.4kN.

Driven sprocket outer diameter < 400mm 31T
Outer diameter 398mm PCD d ₂ = 376.60(mm)

Number of teeth on 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 < 50m/min, so the allowable tension can be selected.

Small sprocket rotation speed: 36 r/min, rotation coefficient Kn = 1.03

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

Chain tension Fw = Conveyor roll rotation torque x 1000 x 2 d ₂
= 3.3 × 1000 × 2 376.6 = 17.5 (kN)

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

RS120-1 can be used Maximum allowable load of 30.4kN.

Check the transport speed (selection condition: 30m/min)

V _ℓ = n ₂ × (Conveyor roll outer diameter + 2 × Belt thickness) × π 1000
= n ₁ × 21 31 × (Conveyor roll outer diameter + 2 × Belt thickness) × π 1000
= 36 × 21 31 × (380 + 2 × 10) × π 1000
= 30.6(m/min)

Step 3: Calculate from acceleration/deceleration time

Since the calculation in step 2 determined that the small sprocket (reducer output shaft sprocket) was RS120 21T, the following calculations will also select the same pitch and number of teeth.

If the acceleration and deceleration times are known, then use those values in the calculations. Here, we will make the calculations under the assumption that the times are unknown.

Acting 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 converted 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 (kgm ²)

Motor moment of inertia Im = 0.088 (kg・m ²)

Motor acceleration time
ts = (Im + I _ℓ) × n ₁ 9550 × (Tm - T _ℓ)
= (0.088 + 0.044) × 1800 9550 × (0.119 - 0.045)
= 0.34(s)

Motor deceleration time
tb = (Im + I _ℓ) × n ₁ 9550 × (Tb + T _ℓ) = (0.088 + 0.044) × 1800 9550 × (0.116 + 0.045) = 0.15(s)

Since tb < ts, the chain tension Fb during deceleration is greater than the chain tension Fs during acceleration, so this will be adopted below.

deceleration
αb = V _ℓ tb x 60 = 30.6 0.15 × 60
= 3.40(m/s ²)

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

Correction chain tension
F'b = Fb × Kn × Kz = 39.2 × 1.03 × 1.10 = 44.4(kN)...(2)
F'b = 44.4 (kN), so RS120-2 (Maximum allowable load 51.7 kN)
Alternatively, RS120-SUP-2 (Maximum allowable load 66.7kN) can be used.

If we consider RS140 18T (outer diameter 279mm d ₁ = 255.98) and 27T (outer diameter 407mm d ₂ = 382.88), which have the same PCD, they cannot be used as they violate the condition of driven sprocket outer diameter ≦ 400mm.

The chain reduction ratio went from the required 3623.9 to 2618.
Conveying speed = 30 × 3623.9 × 2618 = 31.3 m/min.
If we consider 26T (outer diameter 393mm _d2 = 368.77), (2)F'b = 44.3(kN).

RS140-1 cannot be used as it has Maximum allowable load of 40.2kN.

RS140-SUP-1 can be used because it has Maximum allowable load of 53.9kN.

The sprocket shaft hole diameter is maximum 89mm for 18T and maximum 103mm for 26T.
It is possible to use a drive shaft with a diameter of Φ66 mm and a driven shaft with a diameter of Φ94 mm.

Since the center distance is 500 mm, the number of sprocket teeth is 18T (d ₁ = 255.98).
26T (_d2 = 368.77) can be used. The number of links is 46.

Step 4 Calculate from inertia ratio R

Inertia ratio R = I _ℓ Im = 0.044 0.088 = 0.5

Since there is play in the transmission....Shock factor K = 1.0

Starting torque Ts = 0.116(kN・m)

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

Brake torque Tb = 0.116(kN・m)

Chain tension due to brake torque
Fmb = Tb × i × 1.2 × 1000 × 2 d
= 0.116 × 50 × 1.2 × 1000 × 2 255.63 = 54.5(kN)

If Fmb > Fms, the larger Fmb is used.

Compensating 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 corrective chain tension.

F'mb = 61.7 (kN), so RS120-3 (Maximum allowable load 76.0 kN),
Alternatively, RS120-SUP-2 (Maximum allowable load 66.7kN) can be used.

Since the center distance is 500 mm, the number of sprocket teeth is 21T (d ₁ = 255.63).
31T (d ₂ = 376.60) can be used. The number of links is 54.

RS160 15T (outer diameter 269mm d ₁ = 244.33) with the same PCD
Considering 23T (outer diameter 400mm d ₂ = 373.07)
(3)The maximum value is F'mb = 64.6(kN).

RS160-1 cannot be used as it has Maximum allowable load of 53.0kN.

RS160-SUP-1 can be used because it has Maximum allowable load of 70.6kN.

The sprocket shaft hole diameter is maximum 95mm for 15T and maximum 118mm for 23T.
It is possible to use a drive shaft with a diameter of Φ66 mm and a driven shaft with a diameter of Φ94 mm.

Since the center distance is 500 mm, the number of sprocket teeth is 15T (d ₁ = 244.33).
23T (d₂ = 373.07) can be used. The number of links is 40.

{gravity unit}
Step 1: Check the motor characteristics ・Rated 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 (stall) torque Tmax = Tn × 2.1 = 5.95 × 2.1 = 12.5(kgf・m) Brake torque Tb = Tn × 2.0 = 5.95 × 2.0 = 11.9(kgf・m) ・Motor GD ² ^GD2m = 0.352(kgf・m ²) Step 2 Calculate from the load Driven shaft rotation speed n ₂ = V _ℓ × 1000 (Conveyor roll outer diameter + 2 × Belt thickness) × π = 30 × 1000 (380 + 20) × π = 23.9(r/min) Drive shaft rotation speed n = n ₁ /i = 1800 50 = 36(r/min) Chain reduction ratio = 23.9 36 = 1 1.51 PCD of driven sprocket d ₂ = 400mm Chain tension Fw = Conveyor roll rotation torque x 1000 x 2 d ₂ = 337 × 1000 × 2 400 = 1690(kgf) Select the chain provisionally. With some impact.... Service factor Ks = 1.3 Provisional chain tension = Fw × Ks = 1690 × 1.3 = 2200 (kgf) We tentatively select RS120-1, Maximum allowable load of 3100 kgf. Driven sprocket outer diameter < 400mm 31T Outer diameter 398mm PCD d ₂ = 376.60(mm) Number of teeth on 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 < 50m/min, so the allowable tension can be selected. Small sprocket rotation speed: 36 r/min, rotation coefficient Kn = 1.03 Number of teeth on small sprocket 21T....Tooth number factor Kz = 1.10 Chain tension Fw = Conveyor roll rotation torque x 1000 x 2 d ₂ = 337 × 1000 × 2 376.6 = 1790 (kgf) Correction chain tension F'w = Fw × Ks × Kn × Kz = 1790 × 1.3 × 1.03 × 1.10 = 2640(kgf)...(1) RS120-1 can be used Maximum allowable load of 3100 kgf. Check the transport speed (selection condition: 30m/min) V _ℓ = n ₂ × (Conveyor roll outer diameter + 2 × Belt thickness) × π 1000 = n ₁ × 21 31 × (Conveyor roll outer diameter + 2 × Belt thickness) × π 1000 = 36 × 21 31 × (380 + 2 × 10) × π 1000 = 30.6(m/min) Step 3: Calculate from acceleration/deceleration time Since the calculation in step 2 determined that the small sprocket (reducer output shaft sprocket) was RS120 21T, the following calculations will also select the same pitch and number of teeth. If the acceleration and deceleration times are known, then use those values in the calculations. Here, we will make the calculations under the assumption that the times are unknown. Acting 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 Load side GD ² GD ²_ℓ = M × V _ℓ π × n ₁ ² = 6000 × 30.6 π× 1800 ² = 0.176 (kgf·m ²) Motor GD² GD²m = 0.352 (kgf・m ²) Motor acceleration time ts = (GD ² m + GD ²_ℓ) × n ₁ 375×(Tm - T _ℓ) = (0.352 + 0.176) × 1800 375 × (12.2 - 4.58) = 0.34(s) Motor deceleration time tb = (GD ² m + GD ²_ℓ) × n ₁ 375 × (Tb + T _ℓ) = (0.352 + 0.176) × 1800 375 × (11.9 + 4.58) = 0.15(s) Since tb < ts, the chain tension Fb during deceleration is greater than the chain tension Fs during acceleration, so this will be adopted below. deceleration αb = V _ℓ tb x 60 = 30.6 0.15 × 60 = 3.40(m/s ²) Chain tension during deceleration Fb = M × αb G × (Conveyor roll outer diameter + 2 × Belt thickness) d ₂ + Fw = 6000 × 3.40 G × (380 + 2 × 10) 376.6 + 1790 = 4000(kgf) Correction chain tension F'b = Fb × Kn × Kz = 4000 × 1.03 × 1.10 = 4530(kgf)...(2) F'b = 4530 (kgf), so RS120-2 (Maximum allowable load 5270 kgf) Alternatively, RS120-SUP-2 (Maximum allowable load 6800kgf) can be used. If we consider RS140 18T (outer diameter 279mm d ₁ = 255.98) and 27T (outer diameter 407mm d ₂ = 382.88), which have the same PCD, they cannot be used as they violate the condition of driven sprocket outer diameter ≦ 400mm. The chain reduction ratio went from the required 3623.9 to 2618. Conveying speed = 30 × 3623.9 × 2618 = 31.3 m/min. If we consider 26T (outer diameter 393 mm _d2 = 368.77), (2)F'b = 4520 (kgf). RS140-1 cannot be used as it has Maximum allowable load of 4100 kgf. RS140-SUP-1 can be used because it has Maximum allowable load of 5500 kgf. The sprocket shaft hole diameter is maximum 89mm for 18T and maximum 103mm for 26T. It is possible to use a drive shaft with a diameter of Φ66 mm and a driven shaft with a diameter of Φ94 mm. Since the center distance is 500 mm, the number of sprocket teeth is 18T (d ₁ = 255.98). 26T (_d2 = 368.77) can be used. The number of links is 46. Step 4 Calculate from inertia ratio R Inertia ratio R = GD ²_ℓ GD ² m = 0.176 0.352 = 0.5 Since there is play in the transmission....Shock factor K = 1.0 Starting torque Ts = 11.9 (kgf・m) Chain tension due to starting torque Fms = Ts × i × 1000 × 2 d = 11.9 × 50 × 1000 × 2 255.63 = 4660(kgf) Brake torque Tb = 11.9 (kgf・m) Chain tension due to brake torque Fmb = Tb × i × 1.2 × 1000 × 2 d = 11.9 × 50 × 1.2 × 1000 × 2 255.63 = 5590(kgf) If Fmb > Fms, the larger Fmb is used. Compensating 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 corrective chain tension. F'mb = 6330 (kgf), so RS120-3 (Maximum allowable load 7550 kgf), Alternatively, RS120-SUP-2 (Maximum allowable load 6800kgf) can be used. Since the center distance is 500 mm, the number of sprocket teeth is 21T (d ₁ = 255.63). 31T (d ₂ = 376.60) can be used. The number of links is 54. RS160 15T (outer diameter 269mm d ₁ = 244.33) with the same PCD Considering 23T (outer diameter 400mm d ₂ = 373.07) (3)The maximum is F'mb = 6620 (kgf). RS160-1 cannot be used as it has Maximum allowable load of 5400 kgf. RS160-SUP-1 can be used because it Maximum allowable load of 7200 kgf. The sprocket shaft hole diameter is maximum 95mm for 15T and maximum 118mm for 23T. It is possible to use a drive shaft with a diameter of Φ66 mm and a driven shaft with a diameter of Φ94 mm. Since the center distance is 500 mm, the number of sprocket teeth is 15T (d ₁ = 244.33). 23T (d₂ = 373.07) can be used. The number of links is 40.

{gravity unit}

Step 1: Check the motor characteristics

・Rated 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 (stall) torque
Tmax = Tn × 2.1 = 5.95 × 2.1 = 12.5(kgf・m)

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

・Motor GD ²
^GD2m = 0.352(kgf・m ²)

Step 2 Calculate from the load

Driven shaft rotation speed
n ₂ = V _ℓ × 1000 (Conveyor roll outer diameter + 2 × Belt thickness) × π
= 30 × 1000 (380 + 20) × π = 23.9(r/min)

Drive shaft rotation speed
n = n ₁ /i = 1800 50 = 36(r/min)

Chain reduction ratio = 23.9 36 = 1 1.51

PCD of driven sprocket d ₂ = 400mm
Chain tension Fw = Conveyor roll rotation torque x 1000 x 2 d ₂
= 337 × 1000 × 2 400 = 1690(kgf)

Select the chain provisionally.

With some impact.... Service factor Ks = 1.3

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

We tentatively select RS120-1, Maximum allowable load of 3100 kgf.

Driven sprocket outer diameter < 400mm 31T
Outer diameter 398mm PCD d ₂ = 376.60(mm)

Number of teeth on 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 < 50m/min, so the allowable tension can be selected.

Small sprocket rotation speed: 36 r/min, rotation coefficient Kn = 1.03

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

Chain tension Fw = Conveyor roll rotation torque x 1000 x 2 d ₂
= 337 × 1000 × 2 376.6 = 1790 (kgf)

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

RS120-1 can be used Maximum allowable load of 3100 kgf.

Check the transport speed (selection condition: 30m/min)

Step 3: Calculate from acceleration/deceleration time

Since the calculation in step 2 determined that the small sprocket (reducer output shaft sprocket) was RS120 21T, the following calculations will also select the same pitch and number of teeth.

If the acceleration and deceleration times are known, then use those values in the calculations. Here, we will make the calculations under the assumption that the times are unknown.

Acting 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 Load side GD ²
GD ²_ℓ = M × V _ℓ π × n ₁ ²
= 6000 × 30.6 π× 1800 ²
= 0.176 (kgf·m ²)

Motor GD² GD²m = 0.352 (kgf・m ²)

Motor acceleration time
ts = (GD ² m + GD ²_ℓ) × n ₁ 375×(Tm - T _ℓ)
= (0.352 + 0.176) × 1800 375 × (12.2 - 4.58)
= 0.34(s)

Motor deceleration time
tb = (GD ² m + GD ²_ℓ) × n ₁ 375 × (Tb + T _ℓ) = (0.352 + 0.176) × 1800 375 × (11.9 + 4.58) = 0.15(s)

Since tb < ts, the chain tension Fb during deceleration is greater than the chain tension Fs during acceleration, so this will be adopted below.

deceleration
αb = V _ℓ tb x 60 = 30.6 0.15 × 60
= 3.40(m/s ²)

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

Correction chain tension
F'b = Fb × Kn × Kz = 4000 × 1.03 × 1.10 = 4530(kgf)...(2)
F'b = 4530 (kgf), so RS120-2 (Maximum allowable load 5270 kgf)
Alternatively, RS120-SUP-2 (Maximum allowable load 6800kgf) can be used.

The chain reduction ratio went from the required 3623.9 to 2618.
Conveying speed = 30 × 3623.9 × 2618 = 31.3 m/min.
If we consider 26T (outer diameter 393 mm _d2 = 368.77), (2)F'b = 4520 (kgf).

RS140-1 cannot be used as it has Maximum allowable load of 4100 kgf.

RS140-SUP-1 can be used because it has Maximum allowable load of 5500 kgf.

The sprocket shaft hole diameter is maximum 89mm for 18T and maximum 103mm for 26T.
It is possible to use a drive shaft with a diameter of Φ66 mm and a driven shaft with a diameter of Φ94 mm.

Since the center distance is 500 mm, the number of sprocket teeth is 18T (d ₁ = 255.98).
26T (_d2 = 368.77) can be used. The number of links is 46.

Step 4 Calculate from inertia ratio R

Inertia ratio R = GD ²_ℓ GD ² m = 0.176 0.352 = 0.5

Since there is play in the transmission....Shock factor K = 1.0

Starting torque Ts = 11.9 (kgf・m)

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

Brake torque Tb = 11.9 (kgf・m)

Chain tension due to brake torque
Fmb = Tb × i × 1.2 × 1000 × 2 d
= 11.9 × 50 × 1.2 × 1000 × 2 255.63 = 5590(kgf)

If Fmb > Fms, the larger Fmb is used.

Compensating 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 corrective chain tension.

F'mb = 6330 (kgf), so RS120-3 (Maximum allowable load 7550 kgf),
Alternatively, RS120-SUP-2 (Maximum allowable load 6800kgf) can be used.

Since the center distance is 500 mm, the number of sprocket teeth is 21T (d ₁ = 255.63).
31T (d ₂ = 376.60) can be used. The number of links is 54.

RS160 15T (outer diameter 269mm d ₁ = 244.33) with the same PCD
Considering 23T (outer diameter 400mm d ₂ = 373.07)
(3)The maximum is F'mb = 6620 (kgf).

RS160-1 cannot be used as it has Maximum allowable load of 5400 kgf.

RS160-SUP-1 can be used because it Maximum allowable load of 7200 kgf.

The sprocket shaft hole diameter is maximum 95mm for 15T and maximum 118mm for 23T.
It is possible to use a drive shaft with a diameter of Φ66 mm and a driven shaft with a diameter of Φ94 mm.

Since the center distance is 500 mm, the number of sprocket teeth is 15T (d ₁ = 244.33).
23T (d₂ = 373.07) can be used. The number of links is 40.

Selection results

conditions	procedure	Model number	Sprocket	Number of links	Lubrication type
Start frequency less than 6 times	Step 2	RS120-1	21T×31T	54 Links	AII
Start frequency: 6 or more times Cushion start available.	Step 3	RS120-2	21T×31T	54 Links	AII
Start frequency: 6 or more times Cushion start available.	Step 3	RS140-SUP-1	18T×26T	46 Links	B
Start frequency: 6 or more times No cushion start.	Step 3 Step 4	RS120-3	21T×31T	54 Links	AII
		RS120-SUP-2	21T×31T	54 Links	B
		RS160-SUP-1	15T×23T	40 Links	B

Notes: 1. Lubrication type: Please check kW ratings table for each chain size and specifications.
2. Adjustment of the center distance is required for all shafts.

Please also refer to the formulas (here), coefficients (here) used for chain selection, and how to calculate the moment of inertia (here).

TSUBAKI Power Transmission Products Information Site

technical data Drive chain Roller Chain Selection

6. allowable tension selection method

1. Speed Considerations

2. Considering impact

3. Strength of connecting links and offset links

4. Sprocket considerations

Example of selection using allowable tension selection method

Step 1: Check the motor characteristics

Step 2 Calculate from the load

Step 3: Calculate from acceleration/deceleration time

Step 4 Calculate from inertia ratio R

Step 1: Check the motor characteristics

Step 2 Calculate from the load

Step 3: Calculate from acceleration/deceleration time

Step 4 Calculate from inertia ratio R

Selection results