Tsubaki product selection

technical data Drive chain Roller Chain Selection

10. Chain type pin gear drive selection method Note: What is a chain type pin gear?

1. Speed Considerations

This selection method is applicable when the relative chain speed is 50 m/min or less.

(Example of action when speed is below 50m/min)

If you are considering a linear application:
Change to a winding method such as roll drive
If you are considering a wrapping application:
Chain mounting diameter reduced

Pin gear speed coefficient Kv
Relative Chain Speed m/min	Pin gear Speed Factor
Under 15	1.0
15以上30未満	1.2
30以上50未満	1.4

2. Sprocket Considerations

Please use a chain-type pin gear sprocket with 13 or more teeth.
18 teeth recommended.

3. Example of chain-type pin gear drive

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

Chain type pin gear drive selection example

SI units
Step 1: Check the machine and motor characteristics mechanicalcutting machine Motor1.5kW 4P 1750 r/min Motor moment of inertia IIm = 0.00425 kg m ² Starting torque Ts290% Maximum (stall) torque Tmax305% Brake torque Tb180% Reducer reduction ratio i181.9 Forward and reverse rotation frequencymax 900 times/h Sprocket pitch circle diameter (PCD)~Φ220mm~ Moment of inertia of the motor shaft converted load II _ℓ = 0.00072 kg m ² There is no play in the chain. Step 2 Calculate from the load Rotational speed of pin gear driving sprocket n = 1750 × 1 181.9 = 9.6 (r/min) Relative chain speed V = 220 × π × 9.6 1000 = 6.6 (m/min) ......Velocity factor Kv = 1.0 Assume some impact due to cutting machine ...... Service factor Ks = 1.3 Since the mass of the load is unknown, the applied tension is calculated from the drive torque. Motor rated torque Tn = 9.55 × kW n ₁ = 9.55 × 1.5 1750 = 0.00819 (kN・m) Pin gear drive sprocket shaft torque T = Tn × i = 0.00819 × 181.9 = 1.49 (kN・m) Chain tension F = 2T d 1000 = 2 × 1.49 220 1000 = 13.6 (kN) Corrected chain tension F'w = F × Ks × Kv = 13.6 × 1.3 × 1.0 = 17.7 (kN)..........(1) Step 3: Calculate from acceleration/deceleration time Acting torque Tm = Ts + Tb 2 × 100 × Tn = 290 + 305 2 × 100 × 0.00819 = 0.0244 (kN m) Since the load is unknown, the motor rated torque Tn = T _ℓ is used, Load torque T _ℓ = 0.00819 (kN・m) Acceleration time ts = (Im + I _ℓ) × n ₁ 9550 × (Tm - T _ℓ) = (0.00425 + 0.00072) × 1750 9550 × (0.0244 - 0.00819) = 0.056 (s) Motor brake torque Tb = 0.00819 × 1.8 = 0.0147 (kN・m) Deceleration time tb = (Im + I _ℓ) × n ₁ 9550 × (Tb + T _ℓ) = (0.00425 + 0.00072) × 1750 9550 × (0.0147 + 0.00819) = 0.040 (s) Moment of inertia of the motor shaft converted load II _ℓ = 0.00072 (kg・m ²) Fw = F= 13.6 (kN) [value obtained in step 2] Since tb < ts, find the chain tension during deceleration. Angular velocity of the motor shaft ω = 2 π × n ₁ = 2 π × 1750 = 11000 (rad) Motor shaft angular deceleration ωb = ω 60 × tb = 11000 60 × 0.040 = 4580(rad/ ^s2) Chain tension during deceleration Fb = I _ℓ × ωb × i 1000 × d (2 × 1000) + Fw = 0.00072 × 4580 × 181.9 1000 × 220 (2 × 1000) + 13.6 = 19.1 (kN) Corrected chain tension during deceleration F'b = Fb × Kv = 19.1 × 1.0 = 19.1 (kN)..........(2) Step 4 Calculate from inertia ratio R Inertia ratio R = I _ℓ Im = 0.00072 0.00425 = 0.17 From Table 4, impact coefficient K = 0.23 (There is no play in the transmission device. R < 0.2, so let's set R = 0.2.) Chain tension at start Fms = Ts × i d 2 × 1000 × 100 × Tn = 290 × 181.9 220 2 × 1000 × 100 × 0.00819 = 39.3 (kN) Chain tension during braking Fmb = Tb × i d 2 × 1000 × 100 × Tn × 1.2 = 180 × 181.9 220 2 × 1000 × 100 × 0.00819 × 1.2 = 29.3 (kN) From Fms > Fmb Corrected chain tension F'ms = Fms × K × Kv = 39.3 × 0.23 × 1.0 = 9.04 (kN)..........(3) Step 5 Compare (1)(2)(3) Compare (1), (2), and (3) and select a chain with pin gear attachments that has Maximum allowable load that satisfies the maximum applied tension (2) of 19.1 kN. Maximum allowable load when using RS120 attachment chain with pin gear Can be used up to 20.6kN. The pitch diameter of the pin gear sprocket is Φ220, so the sprocket has 18 teeth. Let's assume that (PCD = 222.49mm) is selected. Recalculate steps 2, 3 and 4. [Step 2] F = 2T d 1000 = 2 × 1.49 222.49 1000 = 13.4 (kN) F'w = F × Ks × Kv = 13.4 × 1.3 × 1.0 = 17.4 (kN) [Step 3] Fb = I_ℓ × ωb × i 1000 × d (2 × 1000) + Fw = 0.00072 × 4580 × 181.9 1000 × 220 (2 × 1000) + 13.4 = 18.8 (kN) Corrective chain tension during deceleration F'b = Fb × Kv = 18.8 × 1.0 = 18.8 (kN) [Step 4] Fms = Ts × i d 2 × 1000 × 100 × Tn = 290 × 181.9 222.49 2 × 1000 × 100 × 0.00819 = 38.8 (kN) Correction chain tension F'ms = Fms × K × Kv = 38.8 × 0.23 × 1.0 = 8.92 (kN) Since both corrected chain tensions are within Maximum allowable load, chains with pin gear attachments and sprockets for pin gears can be used. [Step 6] Calculate the number of links L Calculating the number of links L L= 180° tan^-1 P D + 2S = 180° tan^-1 38.1 2920 = 240.8 → 242 links Equivalent to 242 link standard length (38.1 x 242 = 9220.2 mm) D + 2S = 2935mm

SI units

Step 1: Check the machine and motor characteristics

mechanicalcutting machine
Motor1.5kW 4P 1750 r/min
Motor moment of inertia IIm = 0.00425 kg m ²
Starting torque Ts290%
Maximum (stall) torque Tmax305%
Brake torque Tb180%
Reducer reduction ratio i181.9
Forward and reverse rotation frequencymax 900 times/h
Sprocket pitch circle diameter (PCD)~Φ220mm~
Moment of inertia of the motor shaft converted load II _ℓ = 0.00072 kg m ²
There is no play in the chain.

Step 2 Calculate from the load

Rotational speed of pin gear driving sprocket n = 1750 × 1 181.9
= 9.6 (r/min)

Relative chain speed V = 220 × π × 9.6 1000 = 6.6 (m/min)
......Velocity factor Kv = 1.0

Assume some impact due to cutting machine
...... Service factor Ks = 1.3

Since the mass of the load is unknown, the applied tension is calculated from the drive torque.

Motor rated torque
Tn = 9.55 × kW n ₁ = 9.55 × 1.5 1750 = 0.00819 (kN・m)

Pin gear drive sprocket shaft torque
T = Tn × i = 0.00819 × 181.9 = 1.49 (kN・m)

Chain tension F = 2T d 1000 = 2 × 1.49 220 1000 = 13.6 (kN)

Corrected chain tension F'w = F × Ks × Kv = 13.6 × 1.3 × 1.0
= 17.7 (kN)..........(1)

Step 3: Calculate from acceleration/deceleration time

Acting torque Tm = Ts + Tb 2 × 100 × Tn = 290 + 305 2 × 100 × 0.00819
= 0.0244 (kN m)

Since the load is unknown, the motor rated torque Tn = T _ℓ is used,
Load torque T _ℓ = 0.00819 (kN・m)

Acceleration time ts = (Im + I _ℓ) × n ₁ 9550 × (Tm - T _ℓ) = (0.00425 + 0.00072) × 1750 9550 × (0.0244 - 0.00819)
= 0.056 (s)

Motor brake torque Tb = 0.00819 × 1.8 = 0.0147 (kN・m)

Deceleration time tb = (Im + I _ℓ) × n ₁ 9550 × (Tb + T _ℓ)
= (0.00425 + 0.00072) × 1750 9550 × (0.0147 + 0.00819) = 0.040 (s)

Moment of inertia of the motor shaft converted load II _ℓ = 0.00072 (kg・m ²)

Fw = F= 13.6 (kN) [value obtained in step 2]

Since tb < ts, find the chain tension during deceleration.

Angular velocity of the motor shaft ω = 2 π × n ₁ = 2 π × 1750 = 11000 (rad)

Motor shaft angular deceleration ωb = ω 60 × tb = 11000 60 × 0.040
= 4580(rad/ ^s2)

Chain tension during deceleration Fb = I _ℓ × ωb × i 1000 × d (2 × 1000) + Fw
= 0.00072 × 4580 × 181.9 1000 × 220 (2 × 1000) + 13.6
= 19.1 (kN)

Corrected chain tension during deceleration F'b = Fb × Kv = 19.1 × 1.0
= 19.1 (kN)..........(2)

Step 4 Calculate from inertia ratio R

Inertia ratio R = I _ℓ Im = 0.00072 0.00425 = 0.17

From Table 4, impact coefficient K = 0.23
(There is no play in the transmission device. R < 0.2, so let's set R = 0.2.)

Chain tension at start Fms = Ts × i d 2 × 1000 × 100 × Tn
= 290 × 181.9 220 2 × 1000 × 100 × 0.00819 = 39.3 (kN)

Chain tension during braking Fmb = Tb × i d 2 × 1000 × 100 × Tn × 1.2
= 180 × 181.9 220 2 × 1000 × 100 × 0.00819 × 1.2 = 29.3 (kN)

From Fms > Fmb
Corrected chain tension F'ms = Fms × K × Kv = 39.3 × 0.23 × 1.0
= 9.04 (kN)..........(3)

Step 5 Compare (1)(2)(3)

Compare (1), (2), and (3) and select a chain with pin gear attachments that has Maximum allowable load that satisfies the maximum applied tension (2) of 19.1 kN.

Maximum allowable load when using RS120 attachment chain with pin gear
Can be used up to 20.6kN.

The pitch diameter of the pin gear sprocket is Φ220, so the sprocket has 18 teeth.
Let's assume that (PCD = 222.49mm) is selected.

Recalculate steps 2, 3 and 4.

[Step 2]

F = 2T d 1000 = 2 × 1.49 222.49 1000 = 13.4 (kN)

F'w = F × Ks × Kv = 13.4 × 1.3 × 1.0 = 17.4 (kN)

[Step 3]

Fb = I_ℓ × ωb × i 1000 × d (2 × 1000) + Fw
= 0.00072 × 4580 × 181.9 1000 × 220 (2 × 1000) + 13.4
= 18.8 (kN)

Corrective chain tension during deceleration

F'b = Fb × Kv = 18.8 × 1.0 = 18.8 (kN)

[Step 4]

Fms = Ts × i d 2 × 1000 × 100 × Tn
= 290 × 181.9 222.49 2 × 1000 × 100 × 0.00819
= 38.8 (kN)

Correction chain tension

F'ms = Fms × K × Kv = 38.8 × 0.23 × 1.0 = 8.92 (kN)

Since both corrected chain tensions are within Maximum allowable load, chains with pin gear attachments and sprockets for pin gears can be used.

[Step 6] Calculate the number of links L

Calculating the number of links L
L= 180° tan^-1 P D + 2S = 180° tan^-1 38.1 2920
= 240.8 → 242 links

Equivalent to 242 link standard length (38.1 x 242 = 9220.2 mm)
D + 2S = 2935mm

{gravity unit}
Step 1: Check the machine and motor characteristics mechanicalcutting machine Motor1.5kW 4P 1750 r/min Motor GD ²GD ² = 0.017 kgf m ² Starting torque Ts290% Maximum (stall) torque Tmax305% Brake torque Tb180% Reducer reduction ratio i181.9 Forward and reverse rotation frequencymax 900 times/h Sprocket pitch circle diameter (PCD)~Φ220mm~ Motor shaft converted load GD ²GD ²_ℓ = 0.00288 kgf・m ² There is no play in the chain. Step 2 Calculate from the load Rotational speed of pin gear driving sprocket n = 1750 × 1 181.9 = 9.6 (r/min) Relative chain speed V = 220 × π × 9.6 1000 = 6.6 (m/min) ......Velocity factor Kv = 1.0 Assume some impact due to cutting machine ...... Service factor Ks = 1.3 Since the mass of the load is unknown, the applied tension is calculated from the drive torque. Motor rated torque Tn = 974 × kW n ₁ = 974 × 1.5 1750 = 0.835 (kgf・m) Pin gear drive sprocket shaft torque T = Tn × i = 0.835 × 181.9 = 152 (kgf・m) Chain tension F = 2T d 1000 = 2 × 152 220 1000 = 1380 (kgf) Corrected chain tension F'w = F × Ks × Kv = 1380 × 1.3 × 1.0 = 1790 (kgf)..........(1) Step 3: Calculate from acceleration/deceleration time Acting torque Tm = Ts + Tb 2 × 100 × Tn = 290 + 305 2 × 100 × 0.835 = 2.48 (kgf m) Since the load is unknown, the motor rated torque Tn = T _ℓ is used, Load torque T _ℓ = 0.835kgf・m Acceleration time ts = (GD ² m + GD ²_ℓ) × n ₁ 375 × (Tm - t _ℓ) = (0.017 + 0.00288) × 1750 375 × (2.48 - 0.835) = 0.056 (s) Motor brake torque Tb = 0.835 × 1.8 = 1.50 (kgf・m) Deceleration time tb = (GD ² m + GD ²_ℓ) × n ₁ 375 × (Tb + T _ℓ) = (0.017 + 0.00288) × 1750 375 × (1.5 + 0.835) = 0.040 (s) Motor shaft converted load GD ² GD ²_ℓ = 0.00288 (kgf・m ²) Fw = F = 1380 (kgf) [value obtained in step 2] Since tb < ts, find the chain tension during deceleration. Angular velocity of the motor shaft ω = 2 π × n ₁ = 2 π × 1750 = 11000 (rad) Motor shaft angular deceleration ωb = ω 60 × tb = 11000 60 × 0.040 = 4580(rad/ ^s2) Chain tension during deceleration Fb = GD ²_ℓ / 4 × ωb × i d (2 × 1000) × G + Fw = 0.00288 / 4 × 4580 × 181.9 220 (2 × 1000) × 9.80665 + 1380 = 1940 (kgf) Corrected chain tension during deceleration F'b = Fb × Kv = 1940 × 1.0 = 1940 (kgf)..........(2) Step 4 Calculate from inertia ratio R Inertia ratio R = GD ²_ℓ GD ² m = 0.00288 0.017 = 0.17 From Table 4, impact coefficient K = 0.23 (There is no play in the transmission device. R < 0.2, so let's set R = 0.2.) Chain tension at start Fms = Ts × i d 2 × 1000 × 100 × Tn = 290 × 181.9 220 2 × 1000 × 100 × 0.835 = 400 (kgf) Chain tension during braking Fmb = Tb × i d 2 × 1000 × 100 × Tn × 1.2 = 180 × 181.9 220 2 × 1000 × 100 × 0.835 × 1.2 = 2980 (kgf) From Fms > Fmb Corrected chain tension F'ms = Fms × K × Kv = 4000 × 0.23 × 1.0 = 920 (kgf)..........(3) Step 5 Compare (1)(2)(3) Compare (1), (2), and (3) and select a chain with pin gear attachments that has Maximum allowable load that satisfies the maximum applied tension (2) of 1940 kgf. Maximum allowable load when using RS120 attachment chain with pin gear Can be used up to 2100kgf. The pitch diameter of the pin gear sprocket is Φ220, so the sprocket has 18 teeth. Let's assume that (PCD = 222.49mm) is selected. Recalculate steps 2, 3 and 4. [Step 2] F = 2T d 1000 = 2 × 152 222.49 1000 = 1370 (kgf) F'w = F × Ks × Kv = 1370 × 1.3 × 1.0 = 1780 (kgf) [Step 3] Fb = GD²_ℓ/4 × ωb × i d (2 × 1000) × G + Fw = 0.00288/4 × 4580 × 181.9 222.49 (2 × 1000) × 9.80665 + 1380 = 1930 (kgf) Corrective chain tension during deceleration F'b = Fb × Kv = 1930 × 1.0 = 1930 (kgf) [Step 4] Fms = Ts × i d 2 × 1000 × 100 × Tn = 290 × 181.9 222.49 2 × 1000 × 100 × 0.835 = 3960 (kgf) Correction chain tension F'ms = Fms × K × Kv = 3960 × 0.23 × 1.0 = 911 (kgf) Since both corrected chain tensions are within Maximum allowable load, chains with pin gear attachments and sprockets for pin gears can be used. [Step 6] Calculate the number of links L Calculating the number of links L L= 180° tan^-1 P D + 2S = 180° tan^-1 38.1 2920 = 240.8 → 242 links Equivalent to 242 link standard length (38.1 x 242 = 9220.2 mm) D + 2S = 2935mm

{gravity unit}

Step 1: Check the machine and motor characteristics

mechanicalcutting machine
Motor1.5kW 4P 1750 r/min
Motor GD ²GD ² = 0.017 kgf m ²
Starting torque Ts290%
Maximum (stall) torque Tmax305%
Brake torque Tb180%
Reducer reduction ratio i181.9
Forward and reverse rotation frequencymax 900 times/h
Sprocket pitch circle diameter (PCD)~Φ220mm~
Motor shaft converted load GD ²GD ²_ℓ = 0.00288 kgf・m ²
There is no play in the chain.

Step 2 Calculate from the load

Rotational speed of pin gear driving sprocket n = 1750 × 1 181.9
= 9.6 (r/min)

Relative chain speed V = 220 × π × 9.6 1000 = 6.6 (m/min)
......Velocity factor Kv = 1.0

Assume some impact due to cutting machine
...... Service factor Ks = 1.3

Since the mass of the load is unknown, the applied tension is calculated from the drive torque.

Motor rated torque
Tn = 974 × kW n ₁ = 974 × 1.5 1750 = 0.835 (kgf・m)

Pin gear drive sprocket shaft torque
T = Tn × i = 0.835 × 181.9 = 152 (kgf・m)

Chain tension F = 2T d 1000 = 2 × 152 220 1000 = 1380 (kgf)

Corrected chain tension F'w = F × Ks × Kv = 1380 × 1.3 × 1.0
= 1790 (kgf)..........(1)

Step 3: Calculate from acceleration/deceleration time

Acting torque Tm = Ts + Tb 2 × 100 × Tn = 290 + 305 2 × 100 × 0.835
= 2.48 (kgf m)

Since the load is unknown, the motor rated torque Tn = T _ℓ is used,
Load torque T _ℓ = 0.835kgf・m

Acceleration time ts = (GD ² m + GD ²_ℓ) × n ₁ 375 × (Tm - t _ℓ) = (0.017 + 0.00288) × 1750 375 × (2.48 - 0.835)
= 0.056 (s)

Motor brake torque Tb = 0.835 × 1.8 = 1.50 (kgf・m)

Deceleration time tb = (GD ² m + GD ²_ℓ) × n ₁ 375 × (Tb + T _ℓ)
= (0.017 + 0.00288) × 1750 375 × (1.5 + 0.835) = 0.040 (s)

Motor shaft converted load GD ² GD ²_ℓ = 0.00288 (kgf・m ²)

Fw = F = 1380 (kgf) [value obtained in step 2]

Since tb < ts, find the chain tension during deceleration.

Angular velocity of the motor shaft ω = 2 π × n ₁ = 2 π × 1750 = 11000 (rad)

Motor shaft angular deceleration ωb = ω 60 × tb = 11000 60 × 0.040
= 4580(rad/ ^s2)

Chain tension during deceleration Fb = GD ²_ℓ / 4 × ωb × i d (2 × 1000) × G + Fw
= 0.00288 / 4 × 4580 × 181.9 220 (2 × 1000) × 9.80665 + 1380
= 1940 (kgf)

Corrected chain tension during deceleration F'b = Fb × Kv = 1940 × 1.0
= 1940 (kgf)..........(2)

Step 4 Calculate from inertia ratio R

Inertia ratio R = GD ²_ℓ GD ² m = 0.00288 0.017 = 0.17

From Table 4, impact coefficient K = 0.23
(There is no play in the transmission device. R < 0.2, so let's set R = 0.2.)

Chain tension at start Fms = Ts × i d 2 × 1000 × 100 × Tn
= 290 × 181.9 220 2 × 1000 × 100 × 0.835 = 400 (kgf)

Chain tension during braking Fmb = Tb × i d 2 × 1000 × 100 × Tn × 1.2
= 180 × 181.9 220 2 × 1000 × 100 × 0.835 × 1.2 = 2980 (kgf)

From Fms > Fmb
Corrected chain tension F'ms = Fms × K × Kv = 4000 × 0.23 × 1.0
= 920 (kgf)..........(3)

Step 5 Compare (1)(2)(3)

Compare (1), (2), and (3) and select a chain with pin gear attachments that has Maximum allowable load that satisfies the maximum applied tension (2) of 1940 kgf.

Maximum allowable load when using RS120 attachment chain with pin gear
Can be used up to 2100kgf.

The pitch diameter of the pin gear sprocket is Φ220, so the sprocket has 18 teeth.
Let's assume that (PCD = 222.49mm) is selected.

Recalculate steps 2, 3 and 4.

[Step 2]

F = 2T d 1000 = 2 × 152 222.49 1000 = 1370 (kgf)

F'w = F × Ks × Kv = 1370 × 1.3 × 1.0 = 1780 (kgf)

[Step 3]

Fb = GD²_ℓ/4 × ωb × i d (2 × 1000) × G + Fw
= 0.00288/4 × 4580 × 181.9 222.49 (2 × 1000) × 9.80665 + 1380
= 1930 (kgf)

Corrective chain tension during deceleration

F'b = Fb × Kv = 1930 × 1.0 = 1930 (kgf)

[Step 4]

Fms = Ts × i d 2 × 1000 × 100 × Tn
= 290 × 181.9 222.49 2 × 1000 × 100 × 0.835
= 3960 (kgf)

Correction chain tension

F'ms = Fms × K × Kv = 3960 × 0.23 × 1.0 = 911 (kgf)

Since both corrected chain tensions are within Maximum allowable load, chains with pin gear attachments and sprockets for pin gears can be used.

[Step 6] Calculate the number of links L

Calculating the number of links L
L= 180° tan^-1 P D + 2S = 180° tan^-1 38.1 2920
= 240.8 → 242 links

Equivalent to 242 link standard length (38.1 x 242 = 9220.2 mm)
D + 2S = 2935mm

Selection results

Chain model number	RS120-2LK1+242L-JR
Sprocket model number	RS120-1□18TQ-G (□ indicates the hub model)

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

10. Chain type pin gear drive selection method Note: What is a chain type pin gear?

1. Speed Considerations

2. Sprocket Considerations

3. Example of chain-type pin gear drive

Chain type pin gear drive selection example

Step 1: Check the machine and motor characteristics

Step 2 Calculate from the load

Step 3: Calculate from acceleration/deceleration time

Step 4 Calculate from inertia ratio R

Step 5 Compare (1)(2)(3)

Step 1: Check the machine and motor characteristics

Step 2 Calculate from the load

Step 3: Calculate from acceleration/deceleration time

Step 4 Calculate from inertia ratio R

Step 5 Compare (1)(2)(3)

Selection results