A word of caution: These formulas won't make much sense if you are not using Netscape 3.0 or MSIE, and some of the super-/sub-scripts will not show up in MSIE regardless.

Sould anyone find an error in any of these formulas, or a lack of clarity as to their intended use, please e-mail me.

1. Weight Drop, Compounded:

C_vd = T_vd x (1/F) where:
C_vd is compounded drop, in feet;
T_vd is total weight drop required, in feet;
F is the number of falls.

2. Mechanical Advantage:

1. The accepted definition of a mechanical advantage is the ratio of effort to resistance, or, MA = R/E ; however, to compute a mechanical advantage ratio (really a disadvantage ratio, in that more weight is required with multiple falls), which is necessary to derive the total weight required in a compounded weight drive system, a permutation of the formula is required. Counting the number of falls at the weight and multiplying the design weight by the number of falls yields the same result.

R = FE, where:
R is the total weight required, in pounds;
F is the number of falls;
E is the actual weight required to drive the clock, in pounds.

2. The mechanical advantage of a winding jack, wherein a smaller gear (pinion) drives a larger gear is;
MA = Dt/Pt , where;

Dt is the number of teeth on the driven gear, and;
Pt is the number of teeth on the driving pinion (gear).

3. D/d ratio; which is the ratio of the sheave (or winding barrel) pitch diameter to the diameter of the rope;
D = (X d), where;

X is a manufacturer/federal specification constant for differing sizes and constructions of wire ropes, and;
d is the diameter of the rope in inches.

4. Bending stress: s_b = Ed_w /D, where:

s_b is in pounds;
E is the modulus of elasticity of the wire rope, which varies between 1 x 10⁷, and 1.4 x 10⁷.
1.2 x 10⁷ (12 ,000 ,000) is a frequently used average value;
d_w is the diameter of the component wire (for 6 x 19 rope, d_w is .063d);
D is the pitch diameter of the pulley in inches.

s_b = Ed_w /D, for bending stress; and
p_b = s_bA; where A = d²Q.

5. Bending load; p_b = s_b A; where A = d²Q; where:

p_b is in pounds;
d is the rope diameter;
A is the metal cross-sectional area of the rope; and
Q is a constant derived for a specific type of rope. For 6 x 19 wire rope with a fiber core, Q is .405.

N.B. In both 4. and 5. above, constants such as d_w, A, and Q, are provided by the rope manufacturer.

6. Radial pressure on a pulley or a barrel: P = 2T / (D d) , where;

P is radial pressure in pounds per square inch;
T is rope tension in pounds;
D is tread diameter of barrel or pulley in inches;
d is rope diameter in inches.

7. Calculating weight of solid stone driving weights;
W = (h x d x w x 160) / 1728, where:

W is in pounds, and;
h, d, & w, are in inches;
1728 is cubic inches in a cubic foot;
160 (lbs) is an empirically derived constant (see text).

8. Fleeting pulley/sheave placement;

F_sp = barrel depth/2 x .035, or;
F_sp = barrel depth x 1.25 (ft.), where;
F_sp and barrel depth are in feet.

8. To compute the effort and resistance of a lever, as in a barrel/great wheel:

L_ea x R
L_ra E

L_ea is the length of the effort arm;
L_ra is the length of the resistance arm;
R is the resistance (usually in pounds), and;
E is the effort (usually in pounds).

9. To compute frictional losses due to pulleys and wire bending loads: P = W/r where

P is the force at the winding drum, W is the weight (in pounds) of the driving weight.

r (to compute frictional losses in rewinding the clock [lifting the weight] ) is;

r = (1 - µ) ^{m + 1} + (1 - µ) ^{m + 2} +. . . + (1 - µ) ^{m + n} : where

m is the number of 180° bends the rope makes at the weight;
n is the number of parts of line.
µ is the loss coefficient (friction due to pulley bushing[s]),
expressed as a percentage, e.g., with a 2% friction loss; µ = .02.

An example: Assume a 50 pound weight is needed to drive a clock, requiring a triple fall (3 part) compound system:

Simple solution, disregarding r (no friction component): 50 x 3 = 150 + 10 (pulley weight) = 160 pounds; therefore 160/3 is 53.333 lbs. The weight itself must weigh 47 .666 pounds.

The complex solution:

Winding the clock: Let µ = .02, m = 1, and n = 3

1 - µ = .98, therefore; r = .98² + .98³ + .98⁴ (this is: .98 squared; plus .98 cubed, and so on.) therefore:

P = 160 / 2.82 = 56.74 lbs. (effective resistance to rewind effort)

Driving the clock:

P = 160 / 3.18 = 50 .32 lbs. (effective driving weight required)

These figures are ideal calculations--no real world modifiers such as a lack of lubrication, loading of the external hands, etc., are involved in the calculation. Still rather empirical, isn't it? We are, however, getting close to being able to determine the theoretical required weight to drive the hypothetical clock. Note, however, that it does take more effort to rewind a compound weight system.

10. To calculate the maximum length of rope (L) that can be wound on a winding barrel:

L = (A + D) x A x B x K, where:
A = nominal rope diameter,
D = diameter of barrel, inches,
B = width (or depth) of barrel, inches
K = appropriate factor from table below:

Rope Dia. (in)	Factor
3/32 (.094)	23.4
1/8 (.125)	13.6
3/16 ).1875)	6.14
1/4 (.25)	3.29
5/16 (.3125)	2.21
3/8 (.375)	1.58
7/16 (.4375	1.19
1/2 (.5)	.925

11. To calculate the theoretical maximum length of rope which can be wound on a barrel:

L = (D x pi) x (B/A) ÷ 12, where:
D = barrel diameter, inches
pi = 3.141592654
B = width of barrel, inches,
A = nominal rope diameter