when adding tolerances, by divide the tolerance by sigma

Question

On a study question I came across for Engineering, we are to add the tolerances' to get the over all using he additive law of variance. On all  study question, it was self explanatory. On one question though example +/-(A), A was divided by 4 sigma before the formula was applied. Exp: 1 +/-.0040, the .0040 was divided by for giving .0010 the squared?  Other question this didn't apply.

Answer 1

Great question! The reason a tolerance is sometimes divided by 4 before using the additive law of variance is that tolerance and standard deviation are not necessarily the same thing.

In engineering tolerance analysis, we often use the Root Sum of Squares (RSS) method to combine independent dimensional variations.

Step 1: Understand the Difference Between Tolerance and Sigma

Suppose a dimension is specified as:

1.000 ± 0.0040

This means the acceptable dimension ranges from:

0.9960 to 1.0040

However, the tolerance of ±0.0040 does not automatically tell us the standard deviation (σ) of the manufacturing process.

We need additional information about how the process variation relates to its tolerance limits.

Step 2: Why Divide by 4?

Some engineering problems assume that the stated tolerance represents four standard deviations on either side of the nominal dimension.

In other words:

Tolerance = ±4σ

Under this assumption:

4σ = 0.0040

Divide by 4:

σ = 0.0040 / 4

σ = 0.0010

This is why your example divides 0.0040 by 4 before squaring it.

The calculation converts the specified tolerance into an assumed standard deviation.

Step 3: Apply the Additive Law of Variance

For independent random variables, the variance of their sum is the sum of their variances.

For example, suppose we have three independent dimensions with standard deviations:

σ1 = 0.0010

σ2 = 0.0020

σ3 = 0.0015

The combined variance is:

σtotal² = σ1² + σ2² + σ3²

Substitute:

σtotal² = (0.0010)² + (0.0020)² + (0.0015)²

σtotal² = 0.000001 + 0.000004 + 0.00000225

σtotal² = 0.00000725

Take the square root:

σtotal = √0.00000725

σtotal ≈ 0.00269

If the problem specifies a ±4σ tolerance for the combined dimension, multiply by 4:

Combined Tolerance = ±4(0.00269)

Combined Tolerance ≈ ±0.01077

Step 4: Why Don't All Problems Divide by 4?

Because different problems make different assumptions.

For example:

1. If the tolerance represents ±3σ, divide the tolerance by 3.
2. If the tolerance represents ±4σ, divide the tolerance by 4.
3. If the tolerance represents ±6σ, divide the tolerance by 6.
4. If the standard deviations are already provided, no conversion is needed.

Also, some engineering problems assume dimensions are uniformly distributed across their tolerance limits. In that case, the standard deviation is calculated differently.

For a uniform distribution between -T and +T:

σ = T/√3

Therefore, dividing by 4 is not a universal rule.

Final Answer

In your example:

1.000 ± 0.0040

The calculation:

0.0040 / 4 = 0.0010

indicates that the problem is treating the ±0.0040 tolerance as a ±4σ range.

The value 0.0010 is the assumed standard deviation, which is then squared when applying the additive law of variance.

Key Point: Always check whether the problem specifies a 3σ, 4σ, or 6σ convention, or a particular probability distribution. A tolerance by itself does not determine the standard deviation.