# Root Sum Squared Method

The root sum squared (RSS) method is a statistical tolerance analysis method.

In many cases, the actual individual part dimensions occur near the center of the tolerance range with very few parts with actual dimensions near the tolerance limits. This, of course, assumes the parts are mostly centered and within the tolerance range.

RSS assumes the normal distribution describes the variation of dimensions. The bell shaped curve is symmetrical and full described with two parameters, the mean, μ, and the standard deviation, σ.

The variances, not the standard deviations, are additive and provide an estimate of the combined part variation. The result of adding the means and taking the root sum square of the standard deviations provides an estimate of the normal distribution of the tolerance stack. The formula to combine standard deviations of the stack is

$$ \large\displaystyle {{\sigma }_{sys}}=\sqrt{\sum\nolimits_{i=1}^{n}{\sigma _{i}^{2}}}$$

Where σ_{i} is the standard deviation of the i’th part,

And, n is the number of parts in the stack,

And, σ_{sys} is the standard deviation of the stack.

[NOTE: Checkout the ebook Statistical Tolerance Analysis]

The normal distribution has the property that approximately 68.2% of the values fall within one standard deviation of the mean. Likewise, 95.4% within 2 standard deviation and 99.7% within 3 standard deviation.

## Simple example

Using the same example as with the worst case method, we have five plates which each will have different dimensions. For any given set of five, we do not know the five individual dimensions, yet we can estimate the what those dimensions will be using statistics.

On average the plates are 25mm thick. And assuming each part will be slightly different than the average value and the normal distribution describes the variation, we then need to estimate the standard deviation of the part thickness.

For this example let’s measure 30 plates and calculate the standard deviation. If we find the standard deviation is 0.33mm we know that most parts will have dimensions within the tolerance of 0.99mm if the parts follow a normal distribution (more on how to check this assumption later). This is our estimate of how the part thickness actually varies.

Stacking five blocks, the average thickness is 5 times the mean thickness or 125mm.

We expect approximately 99.7% of the stacks of five blocks to have the combined thickness to be within the range of plus or minus 3 standard deviations of the combined plates. In order to combine them we use the formula to add the variances and convert back to standard deviation with a square root.

In this case we add the five variances, 0.33^{2}, and take the square root of that sum.

$$ \large\displaystyle {{\sigma }_{sys}}=\sqrt{\sum\nolimits_{i=1}^{5}{0.33_{i}^{2}}}=0.7379$$

And, since approximately 99.7% of the values are within +/- 3σ, the range of combined thickness values for the stack of five plates should be within 125mm +/- (3 x 0.7379mm or 2.2137mm) or most fall between 122.79mm and 127.21mm.

To estimate the number of assemblies outside the desired tolerance we can use the system normal distribution values, in this case, the mean, μ, is 125, and standard deviation, σ, is 0.7379. Within Excell use the NORMDIST function. In general, construct the cell as follows:

=1-(NORMDIST(Mean+Tolerance, Mean, σ_{sys})-0.5)*2

Where the mean is of the combined means of the parts involved in the stack. In this example the system mean is 125mm.

The tolerance is the desired value, in this examples let’s assume we would like the total stack to be within 2mm of the mean, or a tolerance of 2.

The σ_{sys} is the standard deviation of the combined parts found using the root sum squared standard deviations of the parts involved.

We subtract 0.5 to find the one-sided probability of the result being below the maximum value (mean plus tolerance), and multiple the resulting probability by 2 to find the chance the final assembly is either above or below the desired tolerance.

In this example, for a tolerance of 2mm, we would expect 99.33% of assemblies to have a thickness within the 125mm+/-2mm. This implies that we should expect one assembly out of about 300 to result in a thickness either thinner than 123mm or thicker than 127mm. By varying the tolerance in the calculation we can estimate the scrap or defect rate and compare the cost of scrap/failure to the cost of tighter individual part tolerances.

The accompanying spreadsheet provides this example worked out using the above approach. See the RSS sheet. tolerance analysis examples

## Best practices and Assumptions

The normal distribution assumption relies on the process variation has many small perturbations that generally add to create the final dimension. It is best to actually measure approximately 30 samples to estimate the mean and standard deviation.

When gathering measurements is not feasible, then assuming the parts will have dimensions centered in the tolerance range and have plus or minus three standard deviations across the tolerance range is a conservative starting assumption. Of course, this implies the part creation process is capable of creating 99.7% of the parts within the tolerance specifications.

If measuring less than 30 parts to estimate the standard deviation, be sure to use the sample standard deviation formula.

$$ \large\displaystyle \sigma =\sqrt{\frac{\sum\nolimits_{i=1}^{N}{{{\left( {{x}_{i}}-\bar{x} \right)}^{2}}}}{N-1}}$$

Where N is the number of samples,

x_{i} is the i^{th} measurement,

And x̄ is the sample mean of the samples.

Related:

Worst Case Tolerance Analysis (article)

Variance (article)

Process Capability (article)

This quick introduction to three statistical analysis methods enables you to quickly determine or assess part tolerances. Plus, you will learn why tolerances are critical to achieving a reliability product or system.

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Marek Szymański says

Hello, thanks for this article – it put some light on tolerances area I was not aware of.

Thanks once more,

Marek

Fred Schenkelberg says

Hi Marek, thanks for the comment. Now that you are a site member, check out the ebook on tolerance analysis. cheers, Fred

Matthew says

Brilliant article.

Only thing I am unsure about is the -0.5 for one sided probability. Perhaps you could shed some light ?

Other than that – very helpful

Fred Schenkelberg says

Thanks Matthew – and the -0.5 in the NORMDIST function is to strip out the other half of the probability, leaving only a one-sided result. I suppose it could be done other ways without the subtraction and multiply by 2, and would have to experiments and make sure the function rpovides the results we think it does. cheers, Fred

Kurt says

Doesn’t your formula answer the question how many assemblies are “within” the tolerance band? That is how I see it. I think the correct formula for your question should be 1 – (NORMDIST(Mean+Tolerance, Mean, σsys)-0.5)*2. Did I miss something?

Fred Schenkelberg says

hum, I think you are right… other thoughts?

Kurt says

Yeah, my other thought is I still learned a lot from your post. Thanks.