SME
Kinematics of Mechanisms · Problem type
Pressure angle of a roller follower
Use this for “A roller follower rises 20 mm in simple harmonic motion over 120° of cam turn on a prime circle of radius 50 mm. Find the pressure angle at 60°.”
Chart
s′(60) = 15
This is an example. Type your own values to start a new problem.
s (mm)
=
h (mm)
2
×
(1
−
cos(180°
×
θ
°
÷
β
°))
s′ (mm/rad)
=
π
×
h (mm)
2
×
βᵣ (rad)
×
sin(180°
×
θ
°
÷
β
°)
tan(
φ
°)
=
s′ (mm/rad)
Rₚ (mm)
+
s (mm)
Formulas
βᵣ
=
β
×
π
180
2.0944
=
120
×
π
180
s
=
(
h
2
)
×
(1
−
cos(180°
×
(
θ
β
)))
10
=
(
20
2
)
×
(1
−
cos(180°
×
(
60
120
)))
s
′
=
(π
×
h
2 × βᵣ
)
×
sin(180°
×
(
θ
β
))
15
=
(π
×
20
2 × 2.0944
)
×
sin(180°
×
(
60
120
))
φ
=
tan⁻¹(
s
′
Rₚ +
s
)
14.0362
=
tan⁻¹(
15
50 + 10
)
Assumptions
•
An in-line (radial) roller follower: its line of motion passes through the cam’s centre.
•
Rₚ is the prime circle radius, measured to the roller’s centre at the bottom of the rise.
•
Simple harmonic rise, as on the main page; s′ is the rise per radian of cam turn.
•
Keep φ at 30° or less, or the follower pushes sideways and binds in its guide.
Step-by-step
WE KNOW
θ = 60°, h = 20 mm, β = 120°, Rₚ = 50 mm
FIND
rise in radians (βᵣ), displacement (s), cam slope (s′), pressure angle (φ)
Step 1 · 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 = 2.0944 rad
Step 2 · Find displacement (s)
Use s = (h ÷ 2) × (1 − cos(180° × (θ ÷ β)))
Over
the
rise
180°
θ
β
runs
from
0°
to
180°,
so
the
follower
climbs
half
a
cosine
wave
of
height
h
.
s
=
(
20
2
)
×
(1
−
cos(180°
×
(
60
120
)))
s
=
20
2
×
(1
−
cos(180°
×
0.5))
s
=
20
2
×
(1
−
cos(90°))
s
=
20
2
s
=
10
mm
Step 3 · Find cam slope (s′)
Use s′ = (π × h ÷ (2 × βᵣ)) × sin(180° × (θ ÷ β))
Differentiate
s
by
the
cam
angle
in
radians:
the
rise
per
radian
of
turn.
s
′
=
(π
×
20
2 × 2.0944
)
×
sin(180°
×
(
60
120
))
s
′
=
20π
2 × 2.0944
×
sin(180°
×
0.5)
s
′
=
20π
4.1888
×
sin(90°)
s
′
=
20π
4.1888
s
′
=
15
mm/rad
Step 4 · Find pressure angle (φ)
Use φ = tan⁻¹(s′ ÷ (Rₚ + s))
The
roller’
s
centre
is
Rₚ
+
s
from
the
cam’
s
centre;
it
climbs
s
′
for
each
radian
turned
there.
φ
=
tan⁻¹(
15
50 + 10
)
φ
=
14.0362°
Check
Put the numbers back in. Both sides should match.
2.0944
=
120
×
π
180
✓
10
=
(
20
2
)
×
(1
−
cos(180°
×
(
60
120
)))
✓
15
=
(π
×
20
2 × 2.0944
)
×
sin(180°
×
(
60
120
))
✓
14.0362
=
tan⁻¹(
15
50 + 10
)
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.