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This paper presents a computer-aided analysis to calculate sag in chain drives exactly, which is commonly used in mechanical power transmission. With sprocket rotating, sag in chain drives is different. Sag is a function of meshing position of chain and sprocket. Because sag in chain drives is a variable, the maximum and minimum sag can be obtained by numerical calculation. Corresponding tensions in slack chain are obtained. A computer programme to calculate sag and tensions in slack chain is programmed.

Chain drives are widely used in mechanical engineering. It is useful to calculate sag exactly in practice. In the past, calculation of sag in chain drives is approximate calculation under a series of assumable conditions. These assumable conditions are far different from practical condition of chain drives. In some methods sprocket’s polygon is replaced with pitch circle. But sag in chain drives is variable due to polygonal action. These methods to calculate sag in chain drives have bigger error. This paper presents a computer-aided analysis to calculate sag in chain drives exactly. Because sag in chain drives is a variable, the maximum and minimum sag can be obtained by numerical calculations. Corresponding tensions in slack chain are obtained. A computer programme is made.

Slack chain sink due to chain weight, slack chain can be considered a slick cure that called catenary, catenary’s equation [

Suppose end of slack chain drives are (x_{1},y_{1}) and (x_{2},y_{2}), shown on

Suppose

Form references [

In above equations:

C―center distance

R_{1}, R_{2}―pitch radius of driving or driven sprocket;

Z_{1}, Z_{2}―number of teeth of driving or driven sprocket;

p― chain pitch

s―length of slack chain

L_{c}― chain length

L_{t}―tight chain length

K_{1}, K_{2}―coefficient determined in reference [

From reference [

From Equations (1), (5) and

To multiply Equation (6) by Equation (7), and order

Equation (8) is exceeding equation. Equation (8) can be solved with Newton iterative method. Numerical result of a can be gained.

When slack chain is tight, end point of slack chain can be determined with θ_{t}, ψ_{t}_{ }

In Equation (9), θ_{t}, ψ_{t} are angle displacement of driving and driven sprocket.

Because polygonal action and fixed center distance, line is not integer number of pitch, whose end determined by θ and f points are (X_{1},Y_{1}) and (X_{2},Y_{2}). Only when slack chain has sag, length of slack chain can be integer number of pitch.

F_{1}―Force between chain link on sprocket and roller, N F_{2}―Force between teeth of sprocket and roller, N T―Force between roller chain and roller, N α―pressure angle

When the angle between sprocket and chain link is outside concave, force of roller shown in

When the angle between sprocket and chain link is outside protruding, force in roller is in balance force of roller shown in

To sum up, the angle between sprocket and chain link should be outside protruding and not outside concave.

When he angle between sprocket and chain link is outside protruding,

From equation of slack chain of chain drive, we have

From

When slack chain is tight, L, H and s can be gained by Equation (2)-(4). Points of slack chain (X_{1},Y_{1}) and (X_{2},_{ }Y_{2}) can be calculated by Newton iterative method with Equations (6), (2) and (1).

A program to calculate coordinate value of end point is programmed based on

Equation of line between two end points of slack chain is shown in

Equation of slack chain:

Sag in Chain Drives:

f―Sag in Chain Drives, mm

When x belong to area (X_{1},Y_{1}), X can has a serious of values. So we can have accurate sag in chain drive enough.

As shown in _{1}, T_{2} are tensions of end point slack chain. q is unit weight of roller chain (N). s_{1} is length of slack chain between (X_{1},Y_{1}) and (0,a). s_{2} is length of slack chain between (0,a) and (X_{2},Y_{2}). H_{0} is tension at point (0,a).

By Equations (1), (5) and Equation (16), we have

A Chain Drive is given. Chain pitch p = 50.8 mm. Number of Sprocket Z_{1} = 20, Z_{2} = 50. Number of Chain link L_{P} = 116, Unit Weight of Chain q = 0.1 N/mm.

Calculating results are given in

C (mm) | f_{max} (mm) | T_{1} (N) | T_{2} (N) | f_{min} (mm) | T_{1} (N) | T_{2} (N) |
---|---|---|---|---|---|---|

2042.96 2042.92 2042.80 2042.60 2042.40 2042.20 2042.00 2041.80 2041.60 2041.40 2041.20 2041.00 | 9.777 12.713 18.630 25.427 30.716 35.357 39.452 43.205 46.612 49.747 52.787 55.606 | 5300.73 4078.05 2785.06 2043.70 1694.03 1473.53 1322.37 1208.89 1121.67 1052.00 993.54 943.02 | 5276.61 4053.94 2760.92 2019.57 1669.92 1449.42 1298.26 1184.75 1097.55 1027.87 968.43 918.91 | 0.000 0.252 15.789 23.425 29.360 34.135 38.409 42.218 45.714 48.929 51.984 54.860 | 6344.76 4006.83 3280.10 2215.90 1771.22 1525.59 1357.67 1236.66 1143.33 1069.29 1007.47 955.57 | 6368.87 4530.91 3255.98 2191.82 1747.12 1501.50 1333.59 1212.57 1119.29 1045.21 983.40 931.49 |

1) When driving sprocket is running and tight chain is straight, sag in chain drives and tensions of end point of slack chain are changing.

2) A little change in centre distance should make bigger change of sag in chain drives. So when centre distance in chain drive is not adjustable, it must be very careful to design centre distance in chain drives.

Changfa Rong, (2016) Calculation on Variable Sag in Chain Drives. Open Access Library Journal,03,1-6. doi: 10.4236/oalib.1102632