SME
Machine Design
Gears and fasteners
Use this for “A 20-tooth pinion (m = 3 mm) at 1500 rpm drives a 60-tooth gear and carries 5 kW. Find d₁, d₂, n₂, V and Wₜ.”
Diagram
The pitch circles roll together: n₁N₁ = n₂N₂, 1,500 rpm × 20 = 500 rpm × 60. d = mN: d₁ = 60 mm = 3 mm × 20, d₂ = 180 mm = 3 mm × 60.
At the pitch point V = 4.71 m/s.
Wₜ
=
P
V
:
the
tangential
force
at
the
mesh.
This is an example. Type your own values to start a new problem.
d₁ (mm)
=
m (mm)
×
N₁
n₂ (rpm)
=
n₁ (rpm)
×
N₂
Wₜ (N)
=
P (kW)
V (m/s)
d₂
Gear pitch diameter
calculated · mm
Formulas
d₁
=
m
×
N₁
60 = 3 × 20
d₂
=
m
×
N₂
180 = 3 × 60
n₁
×
N₁
=
n₂
×
N₂
1,500 × 20 = 500 × 60
V
=
π
×
d₁
×
n₁
60,000
4.7124
=
π
×
60
×
1,500
60,000
Wₜ
=
1,000
×
P
V
1,061.033
=
1,000
×
5
4.7124
Assumptions
•
Spur gears with standard teeth; both have the same module m, so d = mN.
•
The pitch circles roll without slipping, and the power passes without loss.
•
d in mm and n in rpm give V in m/s; P in kW gives Wₜ in N.
•
20° full-depth teeth: a pinion under about 13–17 teeth undercuts unless the gear is small.
Step-by-step
WE KNOW
N₁ = 20, N₂ = 60, m = 3 mm, n₁ = 1,500 rpm, P = 5 kW
FIND
pinion pitch diameter (d₁), gear pitch diameter (d₂), gear speed (n₂), pitch-line speed (V), tangential tooth load (Wₜ)
Step 1 · Find pinion pitch diameter (d₁)
Use d₁ = m × N₁
The pitch diameter is the module times the teeth.
d₁
=
3
×
20
d₁
=
60
mm
Step 2 · Find gear pitch diameter (d₂)
Use d₂ = m × N₂
The pitch diameter is the module times the teeth.
d₂
=
3
×
60
d₂
=
180
mm
Step 3 · Find gear speed (n₂)
Use n₁ × N₁ = n₂ × N₂
Teeth pass the mesh at one rate: the driver’s speed times its teeth, over the driven gear’s teeth.
n₂
=
n₁
×
N₁
N₂
n₂
=
1,500
×
20
60
n₂
=
30,000
60
n₂
=
500
rpm
Step 4 · Find pitch-line speed (V)
Use V = π × d₁ × n₁ ÷ 60,000
The
pitch
circle
moves
πd₁
each
turn
and
n₁
turns
each
minute;
÷
60000
turns
mm/min
into
m
/s.
V
=
π
×
60
×
1,500
60,000
V
=
90,000π
60,000
V
=
4.71239
m/s
Step 5 · Find tangential tooth load (Wₜ)
Use Wₜ = 1,000 × P ÷ V
Power
is
force
times
speed
at
the
pitch
line:
the
power
in
watts
over
V
.
Wₜ
=
1,000
×
5
4.71239
Wₜ
=
5,000
4.7124
Wₜ = 1,061.033 N
Check
Put the numbers back in. Both sides should match.
60 = 3 × 20 ✓
180 = 3 × 60 ✓
1,500 × 20 = 500 × 60 ✓
4.71239
=
π
×
60
×
1,500
60,000
✓
1,061.033
=
1,000
×
5
4.71239
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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