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Manufacturing Processes · Problem type

Tolerance stack-up: worst case and RSS

Use this for “Four parts in a row are ±0.05, ±0.10, ±0.05 and ±0.02 mm. Find the gap’s worst-case and RSS tolerance.”

Diagram

±0.05±0.1±0.05±0.024 dimensions in a chain (mm)worst case ΣTᵢ = ±0.22 mmRSS √(ΣTᵢ²) = ±0.124 mm
Worst case: Twc = ΣTᵢ = 0.05 mm + 0.1 mm + 0.05 mm + 0.02 mm = ±0.22 mm.
RSS:
√
ΣTᵢ²
=
±0.124
mm,
likelier
when
the
parts
vary
at
random.
This is an example. Type your own values to start a new problem.
Twc (mm)
=
T₁ (mm)
+
T₂ (mm)
+
T₃ (mm)
+
T₄ (mm)
Tᵣₛₛ
RSS tolerance
worked out (not typed)

Formulas

Twc
=
T₁
+
T₂
+
T₃
+
T₄
0.22 = 0.05 + 0.1 + 0.05 + 0.02
Tᵣₛₛ
=
√
T₁
2
+
T₂
2
+
T₃
2
+
T₄
2
0.1241
=
√
0.05
2
+
0.1
2
+
0.05
2
+
0.02
2

Assumptions

•
Each dimension is its nominal size ± Tᵢ; the gap’s tolerance does not need the nominal sizes.
•
Worst case: every part sits at its limit at once.
•
RSS: the parts vary at random and independently of each other.
•
Fewer parts: leave the rest at 0.

Step-by-step

WE KNOW
T₁ = 0.05 mm, T₂ = 0.1 mm, T₃ = 0.05 mm, T₄ = 0.02 mm
FIND
worst-case tolerance (Twc), RSS tolerance (Tᵣₛₛ)
Step 1 · Find worst-case tolerance (Twc)
Use Twc = T₁ + T₂ + T₃ + T₄
Every tolerance adds at its worst, so the gap can be off by their sum.
Twc = 0.05 + 0.1 + 0.05 + 0.02
Twc = 0.22 mm
Step 2 · Find RSS tolerance (Tᵣₛₛ)
Use Tᵣₛₛ = √(T₁² + T₂² + T₃² + T₄²)
Random variations add as squares: take the root of the sum of squares.
Tᵣₛₛ
=
√
0.05
2
+
0.1
2
+
0.05
2
+
0.02
2
Tᵣₛₛ
=
√
0.0025 + 0.01 + 0.0025 + 0.0004
Tᵣₛₛ
=
√
0.0154
Tᵣₛₛ = 0.1241 mm
Check
Put the numbers back in. Both sides should match.
0.22 = 0.05 + 0.1 + 0.05 + 0.02 ✓
0.1241
=
√
0.05
2
+
0.1
2
+
0.05
2
+
0.02
2
✓
Topic check
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