Technical Data  Drive chain  Roller Chain Selection

10. Pin gear drive Pin gear drive

Pin gear drive

1. Speed considerations

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

(Examples: 50 m/min or less )

  • When a linear application is being considered: ：
    Switch to a winding method such as a roll drive
  • When a wrapping application is being considered: ：
    Change the chain mounting diameter to a smaller value
Pin gear speed factor Kv
Relative chain speed
m/min		 Pin gear speed factor
Less than 15 1.0
15 to 30 1.2
30 to 50 1.4

2. Sprocket considerations

Use a chain-type pin gear sprocket with 13 or more teeth;
18 teeth are recommended.

3. Examples of chain-type pin gear drive

Examples of chain-type pin gear drive

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).

Pin gear drive selection example

Pin gear drive selection example
SI Unit

Steps 1 Check machine and motor characteristics

  • Machine Cutting machine
  • Motor 1.5kW 4P 1750r/min
  • Motor moment of inertia ＩIm = 0.00425kg・m2
  • Starting torque Ts290%
  • Maximum (stalling) torque Tmax305%
  • Braking torque Tb180%
  • Reducer reduction ratio i181.9
  • Forward and reverse operation frequency Max. 900 times/hour
  • Sprocket pitch circle diameter (PCD)～Φ220mm～
  • Moment of inertia for the motor shaft converted load ＩIℓ = 0.00072kg・m2
  • There is no slack in the chain.

Steps 2 Calculate from load

Revolution speed of the pin gear drive sprocket n = 1750 × 1 181.9
= 9.6 (r/min)

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

Some impact assumed from cutting machine
......Service factor Ks = 1.3

Load is calculated from the torque on the drive side as the mass of the load is unknown.

Rated torque of the motor
Tn = 9.55 × kW n1 = 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 working tension F = 2T d 1000 = 2 × 1.49 220 1000 = 13.6 (kN)

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

Steps 3 Calculate from acceleration/deceleration time

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

As the load is unknown, the rated torque of the motor is Tn ＝ Tℓ and the load torque Tℓ ＝ 0.00819(kN·m)

Acceleration time ts = (Im + Iℓ) × n1 9550 × (Tm - Tℓ) = (0.00425 + 0.00072) × 1750 9550 × (0.0244 - 0.00819)
= 0.056 (s)

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

Deceleration time tb = (Im + Iℓ) × n1 9550 × (Tb + Tℓ)
= (0.00425 + 0.00072) × 1750 9550 × (0.0147 + 0.00819) = 0.040 (s)

Moment of inertia for the motor shaft converted load IIℓ = 0.00072 (kg・m2)

Fw = F= 13.6 (kN) [value from Step 2 ]

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

Motor shaft angular velocity ω = 2π × n1 = 2π × 1750 = 11000 (rad)

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

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

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

Steps 4 Calculate from inertia ratio R

Inertia ratio R = Iℓ Im = 0.00072 0.00425 = 0.17

According to Table 4, impact factor K ＝ 0.23 (There is no play in the drive transmission equipment as R < 0.2, R ＝ 0.2.)

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

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

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

Steps 5 Comparison of (1), (2), and (3)

Comparing (1), (2), and (3), an attachment chain for pin gears that meets 19.1kN of the maximum working load (2) is selected.

The maximum allowable load for RS120 attachment chain with pin gear use is 20.6 kN, which is acceptable.

A sprocket with 18 teeth (PCD = 222.49 mm) is selected as tentative from the pitch circle diameter of the pin gear sprocket approximate. Φ220.

Steps 2, 3, and 4 are calculated again here.

[Steps 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)

[Steps 3]

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

Design chain tension during deceleration

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

[Steps 4]

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

Design chain tension

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

Either design chain tension is within the maximum allowable load, so a chain with pin gear attachments and a pin gear sprocket can be used.

[Steps 6] Calculating number of links L

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

Corresponding standard length for 242 links (38.1 × 242 = 9220.2mm)
D + 2S = 2935 mm

{Gravity unit }

Steps 1 Check machine and motor characteristics

  • Machine Cutting machine
  • Motor 1.5kW 4P 1750r/min
  • GD2 of the motor GD2 = 0.017kgf・m2
  • Starting torque Ts290%
  • Maximum (stalling) torque Tmax305%
  • Braking torque Tb180%
  • Reducer reduction ratio i181.9
  • Forward and reverse operation frequency Max. 900 times/hour
  • Sprocket pitch circle diameter ～Φ220mm～
  • GD2 of the motor shaft converted load GD2ℓ = 0.00288kgf・m2
  • There is no slack in the chain.

Steps 2 Calculate from load

Revolution speed of the pin gear drive sprocket n = 1750 × 1 181.9
= 9.6 (r/min)

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

Some impact assumed from cutting machine
......Service factor Ks = 1.3

Load is calculated from the torque on the drive side as the mass of the load is unknown.

Rated torque of the motor
Tn = 974 × kW n1 = 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 working tension F = 2T d 1000 = 2 × 152 220 1000 = 1380 (kgf)

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

Steps 3 Calculate from acceleration/deceleration time

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

As the load is unknown, the rated torque of the motor is Tn = Tℓ and the load torque Tℓ = 0.835kgf·m

Acceleration time ts = (GD2m + GD2ℓ) × n1 375 × (Tm - tℓ) = (0.017 + 0.00288) × 1750 375 × (2.48 - 0.835)
= 0.056 (s)

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

Deceleration time tb = (GD2m + GD2ℓ) × n1 375 × (Tb + Tℓ)
= (0.017 + 0.00288) × 1750 375 × (1.5 + 0.835) = 0.040 (s)

Motor shaft converted load GD2 GD2ℓ = 0.00288 (kgf・m2)

Fw = F = 1380 (kgf) [value from Step 2 ]

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

Motor shaft angular velocity ω = 2π × n1 = 2π × 1750 = 11000 (rad)

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

Chain tension during deceleration Fb = GD2ℓ / 4 × ωb × i d (2 × 1000) × G + Fw
= 0.00288 / 4 × 4580 × 181.9 220 (2 × 1000) × 9.80665 + 1380
= 1940 (kgf)

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

Steps 4 Calculate from inertia ratio R

Inertia ratio R = GD2ℓ GD2m = 0.00288 0.017 = 0.17

According to Table 4, impact factor K ＝ 0.23 (There is no play in the drive transmission equipment as R < 0.2, R ＝ 0.2.)

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

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

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

Steps 5 Comparison of (1), (2), and (3)

Comparing (1), (2), and (3), an attachment chain for pin gears that meets 1940 kgf of the maximum working load (2) is selected.

The maximum allowable load for RS120 attachment chain with pin gear use is 2100 kgf, which is acceptable.

A sprocket with 18 teeth (PCD = 222.49 mm) is selected as tentative from the pitch circle diameter of the pin gear sprocket approximate. Φ220.

Steps 2, 3, and 4 are calculated again here.

[Steps 2]

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

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

[Steps 3]

Fb = GD2ℓ/4 × ωb × i d (2 × 1000) × G + Fw
= 0.00288/4 × 4580 × 181.9 222.49 (2 × 1000) × 9.80665 + 1380
= 1930 (kgf)

Design chain tension during deceleration

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

[Steps 4]

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

Design chain tension

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

Either design chain tension is within the maximum allowable load, so a chain with pin gear attachments and a pin gear sprocket can be used.

[Steps 6] Calculating number of links L

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

Corresponding standard length for 242 links (38.1 × 242 = 9220.2mm)
D + 2S = 2935 mm

Selection results

Chain Model No. RS120-2LK1+242L-JR
Sprocket model No. RS120-1□18TQ-G (Enter hub model in the blank. )

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).

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