SME
Kinematics of Mechanisms · Problem type
Slider acceleration (the engine formula)
Use this for “The same slider-crank (r = 50 mm, L = 200 mm, ω = 100 rad/s) at 40°. Find the piston’s acceleration.”
Chart
a(40) ≈ −404.7
This is an example. Type your own values to start a new problem.
a (m/s²)
=
−
r (mm)
×
(
ω (rad/s)
)
2
×
(cos(
θ
°)
+
×
cos(2
×
°)
L (mm)
)
÷
1000
Formulas
a
=
−
r
×
ω
2
×
(cos(
θ
)
+
r
÷
L
×
cos(2
×
θ
°))
÷
1,000
−404.7
=
−50
×
100
2
×
(cos(40°)
+
50
÷
200
×
cos(2
×
40°))
÷
1,000
Assumptions
•
The crank turns at a steady ω, so it has no angular acceleration.
•
L ÷ r ≥ 3: the terms the two-term series drops then stay under 1%.
•
An in-line slider; θ from the line of stroke, and a is + away from the crank.
Step-by-step
WE KNOW
θ = 40°, ω = 100 rad/s, r = 50 mm, L = 200 mm
FIND
slider acceleration (a)
Step 1 · Find slider acceleration (a)
Use a = −r × ω² × (cos(θ) + r ÷ L × cos(2 × θ°)) ÷ 1,000
Differentiate
the
slider’s
position
twice,
keeping
two
terms
of
its
series;
÷
1000
turns
mm/
s
2
into
m/
s
2
.
a
=
−50
×
100
2
×
(cos(40°)
+
50
÷
200
×
cos(2
×
40°))
÷
1,000
a
=
−50
×
100
2
×
(cos(40°)
+
50
÷
200
×
cos(80°))
÷
1,000
a
=
−50
×
100
2
×
0.766 + 50 ÷ 200 × 0.1736
1,000
a
=
−50
×
100
2
×
0.766 + 0.25 × 0.1736
1,000
a
=
−50
×
100
2
×
0.766 + 0.0434
1,000
a
=
−50
×
100
2
×
0.8094
1,000
a
=
−50
×
10,000
×
0.8094
1,000
a
=
−500,000
×
0.8094
1,000
a
=
−
404,700
1,000
a
=
−404.7
m/
s
2
Check
Put the numbers back in. Both sides should match.
−404.7
=
−50
×
100
2
×
(cos(40°)
+
50
÷
200
×
cos(2
×
40°))
÷
1,000
✓
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