SME
Kinematics of Mechanisms · Problem type
Cycloidal rise
Use this for “A cam lifts its follower 20 mm with cycloidal motion over 120° of turn at 300 rpm. Find s, v and a at 45°.”
Chart
s(45) ≈ 5.249
Level h = 20 meets the curve at θ = 120
This is an example. Type your own values to start a new problem.
s (mm)
=
h (mm)
×
°
°
−
1
2π
×
sin
360°
×
°
°
v (mm/s)
=
h (mm)
×
ω (rpm)
βᵣ (rad)
×
1
−
cos
360°
×
°
°
a
Acceleration
calculated · m/s²
Formulas
βᵣ
=
β
×
π
180
2.0944
=
120
×
π
180
s
=
h
×
(
θ
β
−
(
1
2 × π
)
×
sin(360°
×
(
θ
β
)))
5.2492
=
20
×
(
45
120
−
(
1
2 × π
)
×
sin(360°
×
(
45
120
)))
v
=
(
h
×
ω
βᵣ
)
×
(1
−
cos(360°
×
(
θ
β
)))
512.132
=
(20
×
31.4159
2.0944
)
×
(1
−
cos(360°
×
(
45
120
)))
a
=
(2
×
π
×
h
×
ω
2
βᵣ
2
)
×
sin
360° × (
θ
÷
β
)
1,000
+19.993
=
(2
×
π
×
20
×
31.4159
2
2.0944
2
)
×
sin
360° × (45 ÷ 120)
1,000
Formulas are worked in rad/s. The step-by-step shows the conversions.
Assumptions
•
The rise starts at θ = 0 and ends at θ = β; the cam turns at a steady speed ω.
•
The acceleration starts and ends at 0, so the follower takes no shock at the ends of the rise.
•
The top speed, 2hω ÷ βᵣ at mid-rise, is higher than simple harmonic motion’s 1.57hω ÷ βᵣ.
Step-by-step
WE KNOW
θ = 45°, h = 20 mm, β = 120°, ω = 300 rpm
FIND
rise in radians (βᵣ), displacement (s), velocity (v), acceleration (a)
Step 1 · Convert to the formula’s units
The formulas don’t work directly in the units you chose, so convert first (to rad/s).
ω = 300 rpm = 31.4159 rad/s (1 rad/s = 9.5493 rpm)
Step 2 · Find rise in radians (βᵣ)
Use βᵣ = β × π ÷ 180
The speed formulas need the rise angle in radians: 180° is π rad.
βᵣ
=
120
×
π
180
βᵣ = 2.094395 rad
Step 3 · Find displacement (s)
Use s = h × (θ ÷ β − (1 ÷ (2 × π)) × sin(360° × (θ ÷ β)))
The
share
θ
β
of
the
rise,
less
a
sine
wave
that
starts
and
ends
the
climb
gently.
s
=
20
×
(
45
120
−
(
1
2 × π
)
×
sin(360°
×
(
45
120
)))
s
=
20
×
(
45
120
−
1
2π
×
sin(360°
×
0.375))
s
=
20
×
(
45
120
−
1
2π
×
sin(135°))
s
=
20
×
(
45
120
−
1
2π
×
0.70711)
s
=
20
×
(0.375
−
0.15915
×
0.70711)
s
=
20
×
(0.375
−
0.11254)
s
=
20
×
0.26246
s
=
5.2492
mm
Step 4 · Find velocity (v)
Use v = (h × ω ÷ βᵣ) × (1 − cos(360° × (θ ÷ β)))
Differentiate
s
by
time:
the
angle
grows
at
ω
,
so
β
must
be
in
radians
here.
v
=
(20
×
31.4159
2.094395
)
×
(1
−
cos(360°
×
(
45
120
)))
v
=
20
×
31.4159
2.0944
×
(1
−
cos(360°
×
0.375))
v
=
20
×
31.4159
2.0944
×
(1
−
cos(135°))
v
=
20
×
31.4159
2.0944
×
(1
−
(−0.7071))
v
=
20
×
31.4159
2.0944
×
1.7071
v
=
628.318
2.0944
×
1.7071
v
=
300
×
1.7071
v
=
512.132
mm/
s
Step 5 · Find acceleration (a)
Use a = (2 × π × h × ω² ÷ βᵣ²) × sin(360° × (θ ÷ β)) ÷ 1,000
Differentiate
once
more;
÷
1000
turns
mm/
s
2
into
m/
s
2
.
It
is
0
at
both
ends
and
at
mid-rise.
a
=
(2
×
π
×
20
×
31.4159
2
2.094395
2
)
×
sin
360° × (45 ÷ 120)
1,000
a
=
2π
×
20
×
31.4159
2
2.0944
2
×
sin(360° × 0.375)
1,000
a
=
2π
×
20
×
31.4159
2
2.0944
2
×
sin(135°)
1,000
a
=
2π
×
20
×
986.96
4.3865
×
0.70711
1,000
a
=
125.66
×
986.96
4.3865
×
0.70711
1,000
a
=
124,021.3936
4.3865
×
0.70711
1,000
a
=
28,273
×
0.70711
1,000
a
=
19,992
1,000
a
=
+19.993
m/
s
2
Check
Put the numbers back in. Both sides should match.
2.094395
=
120
×
π
180
✓
5.2492
=
20
×
(
45
120
−
(
1
2 × π
)
×
sin(360°
×
(
45
120
)))
✓
512.132
=
(20
×
31.4159
2.094395
)
×
(1
−
cos(360°
×
(
45
120
)))
✓
+19.993
=
(2
×
π
×
20
×
31.4159
2
2.094395
2
)
×
sin
360° × (45 ÷ 120)
1,000
✓
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