The document provides formulae for various mechanical systems including friction, screw threads, clutches, disk brakes, belt drives, bearings, gears, flywheels, rotors, and reciprocating machinery. Key formulae include equations for frictional force, torque required to tighten a nut, axial force in a clutch or brake given a pressure distribution, transmitted power in a belt drive, bearing characteristic number, gear module and ratios, flywheel energy fluctuations given input and output torques, out-of-balance forces and moments in rotors, and primary and secondary out-of-balance forces and moments in reciprocating systems.

1. Dynamics of Machines
Formulae
FRICTION
For solid friction the frictional force, F, at the point of
slip is given by:
NF µ=
where µ is the coefficient of friction and N is the
normal contact force
SCREW THREADS
The torque, T, required to tighten a nut against an axial
force, W, is given by:
2sincos
cossin md
WT 





−
+
=
φµφ
φµφ
where φ is the helix angle, µ is the coefficient of
friction and dm is the mean thread diameter. The torque
to loosen the nut is given by:
2sincos
cossin md
WT 





+
+−
=
φµφ
φµφ
The efficiency of tightening, η, is given by:
T
Wt
π
η
2
=
where t is the pitch of the thread.
CLUTCHES AND DISK BRAKES
The axial force, W, associated with a radial pressure
distribution, P(r), between one pair of contacting
surfaces in a clutch or disk brake (assuming that the
brake pad is a sector of an annulus) is:
( )∫=
2
1
R
R
drrrPW θ
where θ is the angle subtended by the brake pad (θ
= 2π for a clutch), r is the radial position, R1 is the
inner radius of the brake pad or clutch and R2 is its
outer radius.
Similarly, the maximum torque, T, at the point of slip
due to one pair of contacting surfaces in a clutch or
disk brake is given by:
( )∫=
2
1
2
R
R
drrPrT θµ
where µ is the coefficient of friction.
In a new clutch or disk brake the pressure distribution
can be assumed to be equal to a constant pressure, P0,
so:
( ) 0PrP =
For a worn clutch or disk brake a common assumption
is that the pressure distribution is inversely
proportional to radial position:
( )
r
k
rP =
where k is a constant.
BELT DRIVES
For a belt or rope wrapped around a drum or pulley, the
maximum possible ratio of tensions between the tight,
T1, and slack, T2, sides at slip is given by:
L
e
T
T µφ
=
2
1
where µ is the coefficient of friction and φL is the angle
of lap. If the rope and pulley are rotating then dynamic
effects should be included and the previous equation
becomes:
L
e
mvT
mvT µφ
=
−
−
2
2
2
1
where m is the mass per unit length of the belt and v is
the belt velocity.
The power, P, transmitted by a belt drive is given by:
( )21 TTvP −=
and the initial belt tension, T0, is related to the running
belt tensions by:
2
21
0
TT
T
+
=
BEARINGS
The bearing characteristic number, n, for a journal
bearing may be defined (other definitions possible) as:
P
n
γω
=
where γ is the viscosity of the lubricant, ω is the
angular velocity of the shaft and P is the nominal
pressure (the radial load divided by projected area).
GEARS
The module, m, of a gear wheel is defined as:
π
P
N
D
m
p
==
where Dp is the pitch circle diameter, N is the number if
teeth and P is the circular pitch. In order for a pair of
gears to mesh correctly, they must have the same
module.
For a pair of meshing spur gears:

2. 1
2
1
2
2
1
N
N
D
D
p
p
==−
ω
ω
where ω1,2 are their angular velocities (measured in the
same sense on both gears), D1,2 are their pitch circle
diameters and N1,2 are their numbers of teeth.
If a gearbox with multiple input and output shafts
denoted by the subscript i, is assumed lossless then:
0=∑i
iiT ω
where the torque, Ti, and angular velocity, ωi, on a
particular shaft are measured in the same direction.
An epicyclic gearbox operating at constant speed must
be in moment equilibrium, hence:
0=++ csa TTT
where Ta, Ts and Tc are the torques on the annulus, sun
and carrier respectively, measured in the same sense.
FLYWHEELS
For a flywheel rotating from angle θ1 to θ2, subjected to
input and output torques, Tin and Tout respectively:
( ) ( ) ( )[ ] θθθωω
θ
θ
dTTI outin∫ −=−
2
1
2
1
2
22
1
where I is the moment of inertia of the flywheel and ω1
and ω2 are its angular velocities at angles θ1 and θ1
respectively.
The maximum fluctuation in kinetic energy (MFKE)
over one revolution is defined as:
( ) ωωωω ∆≈−= meanIIMFKE 2
min
2
max2
1
where ωmean, ωmin and ωmax are the mean, minimum and
maximum angular velocities respectively and ∆ω =
ωmax - ωmin. The maximum percentage fluctuation in
speed (MPFS) is defined as:
%100minmax
×
−
=
ω
ωω
MPFS
ROTORS
The amplitude of the out-of-balance rotating force, F,
generated by a rotor represented as a number of point
masses, mi, is given by:
( ) ( )222
sincos iiiiii rmrmF θθω ∑∑ +=
where ω is the angular velocity and ri and θi are the
radial and angular positions of the ith
mass.
The amplitude of the out-of-balance rotating moment,
M is given by:
( ) ( )222
sincos iiiiiiii armarmM θθω ∑∑ +=
where ai is the axial position of the ith
mass.
RECIPROCATING MACHINERY
The oscillating force, F, required to accelerate and
decelerate a slider of mass m in a slider-crank
mechanism is usually approximated by:






+−= t
q
tmrF ωωω 2cos
1
cos2
where r is the length of the crank, ω is the angular
velocity of the crank, t is time and q is the ratio of
connecting rod length, d, to crank length (i.e. q = d / r).
The two terms are usually considered separately as
primary and secondary out-of-balance forces. The
amplitude of the primary out-of-balance force, F1, is
therefore:
2
1 ωmrF =
and the amplitude of the secondary out-of-balance
force, F2, is given by:
q
mr
F
2
2
ω
=
In a machine with multiple parallel sliders driven or
driving a common crankshaft, the amplitude of the total
out-of-balance primary force, F1
Total
, is given by:
( ) ( )222
1 sincos ∑∑ += iiiiii
Total
rmrmF θθω
where mi, ri and θi are the slider masses, crank lengths
and relative angular positions of the cranks for each
slider. Similarly, the amplitude of the total out-of-
balance secondary force, F2
Total
, is given by:
22
2
2
2sin2cos






+





= ∑∑
i
iii
i
iiiTotal
q
rm
q
rm
F
θθ
ω
where qi are the connecting rod to crank length ratios
for each slider. The amplitudes of the total out-of-
balance primary moment, M1
Total
, and secondary
moment, M2
Total
, are respectively:
( ) ( )222
1 sincos ∑∑ += iiiiiiii
Total
armarmM θθω
and
22
2
2
2sin2cos






+





= ∑∑
i
iiii
i
iiiiTotal
q
arm
q
arm
M
θθ
ω
where ai is the axial position of the slider along the
crankshaft.