Kline Mcclintock Uncertainty

Last Rev.: 11 JUN 08

Kline-McClintock Uncertainty : MIME 3470

KLINE-MCCLINTOCK1 METHOD

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and the uncertainty is calculated as:

(EXPERIMENTAL UNCERTAINTY) ~~~~~~~~~~~~~~

The first thing the student must be aware of is that this procedure does not calculate an error; instead, it states what the possible error —the uncertainty—is in a final result based on experimental measurements or a tolerance in fabrication. So what causes the uncertainty? Basically, it is due to the coarseness of measuring tools. A simple six-inch scale graduated in increments of 0.1 inches can only be read accurately to the nearest 0.1 of an inch. The uncertainty is taken as one half of a graduation or 0.05 inches. An alternate dimensional uncertainty is a tolerance as used in drawing of machine parts. If a dimension is 8mm  0.1mm, then the 0.1mm is the uncertainty in the dimension. The Kline-McClintock Method determines the uncertainty of a calculation given certain measurements and the tolerances on those measurements. To exemplify how this method works, assume one is to determine the volume of a cube containing a cylindrical hole (see the figure below). D

  dV  V       dL   

      

L 

2

2

 3     2

2



2

L       

 0.2

 

2D    L 4  

2

      3 5  2  

2



2

2



 184.5870  5.5516

2

1/ 2

D 

1/ 2

 

2

 

  67.9314   0.2    23.5619  0.1 2

 0.1



1/ 2

 

2

 

2 1/ 2

 13.7891in . 3

This value is not a percent!

 D V  L3     2 

L2 L1

2

 D V  L3    L  2  If one measures D and L with measurements

(A.1)

L  5"0.2"



 0.2    100  4.000%  5 

D  3"0.1"



 0.1    100  3.333%  3 

what is the anticipated possible error (either positive or negative) when the volume is calculated using Equation A.1? In general, if n measurements xn are being made, each with a measurement tolerance of n, and a function F is calculated using the measured values, then the uncertainty or tolerance in the calculation can be determined as: 2    dF  dF    1  2     F     dx    dx1  2    For the proposed example,

2

L 2

5

 125  35.3429  89.6571in3 . Thus, one could have a percent error as large as

To make this a simple explanation of the procedure, assume that L = L1 = L2= L3. The volume, ideally, would then be:

2

 2 

1/ 2



2

.  

2

1  0.2 1  0.1

x2  D 1

2

2



2

 3   5 3     2

L

D 

1/ 2



2

The ideal volume calculation is

L3

 D F  V  L3     2 x1  L

2

 dV     dD 

 D 3L2     2

   3 5   



2

Kline, S. J., and F. A. McClintock. “Describing Uncertainties in Single-Sample Experiments.” Mechanical Engineering, Vol. 75, No. 1, January 1953: 3-8.

  13.7891     100  15.38%  89.6571  even though the errors in individual measurements are less than 4%.

Last Rev.: 11 JUN 08

Kline-McClintock Uncertainty : MIME 3470

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