SME
Machine Design · Problem type
Gear-train value of a compound train
Use this for “Gears of 20, 60, 18 and 54 teeth form a compound train, with 60 and 18 on one shaft. The input turns at 1800 rpm. Find e and the output speed.”
Diagram
Gears
2
and
3
share
a
shaft.
e
=
ΠN
driving
ΠN
driven
=
20
×
18
60 × 54
=
0.111.
n
out
=
e
×
n
in
=
200
rpm,
the
same
way
round
as
gear
1.
This is an example. Type your own values to start a new problem.
e
=
×
×
nₒᵤₜ (rpm)
=
e
×
nᵢₙ (rpm)
Formulas
e
=
N₁
×
N₃
N₂
×
N₄
0.1111
=
20 × 18
60 × 54
nₒᵤₜ
=
e
×
nᵢₙ
200 = 0.1111 × 1,800
Assumptions
•
Gears 2 and 3 are keyed to one shaft, so they turn together.
•
Gears 1 and 3 drive; gears 2 and 4 are driven.
•
Each mesh reverses the turning, so after two meshes the output turns the same way.
Step-by-step
WE KNOW
N₁ = 20, N₂ = 60, N₃ = 18, N₄ = 54, nᵢₙ = 1,800 rpm
FIND
train value (e), output speed (nₒᵤₜ)
Step 1 · Find train value (e)
Use e = (N₁ × N₃) ÷ (N₂ × N₄)
Multiply the driving gears’ teeth, and divide by the driven gears’ teeth.
e
=
20 × 18
60 × 54
e
=
20
×
18
3,240
e
=
360
3,240
e
=
0.1111
Step 2 · Find output speed (nₒᵤₜ)
Use nₒᵤₜ = e × nᵢₙ
The
output
turns
e
times
as
fast
as
the
input.
nₒᵤₜ = 0.1111 × 1,800
nₒᵤₜ = 200 rpm
Check
Put the numbers back in. Both sides should match.
0.1111
=
20 × 18
60 × 54
✓
200 = 0.1111 × 1,800 ✓
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Lewis bending stress in a gear tooth
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Bolt preload and tightening torque
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