MODULE 1 Mechanical Measurements

Transcription

1 MODULE 1 Mechanical Measurements 1. Introduction to Mechanical Measurements Why Measure? Generate Data for Design Generate Data to Validate or Propose a Theory For Commerce Figure 1 Why make measurements? We recognize three reasons for making measurements as indicated in Figure 1. From the point of view of the course measurements for commerce is outside its scope. Engineers design physical systems in the form of machines to serve some specified functions. The behavior of the parts of the machine during the operation of the machine needs to be examined or analyzed or designed such that it functions reliably. Such an activity needs data regarding the machine parts in terms of material properties. These are obtained by performing measurements in the laboratory.

2 The scientific method consists in the study of nature to understand the way it works. Science proposes hypotheses or theories based on observations and need to be validated with carefully performed experiments that use many measurements. When once a theory has been established it may be used to make predictions which may themselves be confirmed by further experiments. Measurement categories 1. Primary quantity. Derived quantity 3. Intrusive Probe method 4. Non-intrusive Measurement categories are described in some detail now. 1. Primary quantity: It is possible that a single quantity that is directly measurable is of interest. An example is the measurement of the diameter of a cylindrical specimen. It is directly measured using an instrument such as vernier calipers. We shall refer to such a quantity as a primary quantity.. Derived quantity: There are occasions when a quantity of interest is not directly measurable by a single measurement process. The quantity of interest needs to be estimated by using an appropriate relation involving several measured primary quantities. The measured quantity is thus a derived quantity. An example of a derived quantity is the determination of acceleration due to gravity (g) by finding the period (T) of a simple pendulum of length (L). T and L are the measured primary quantities while g is the derived quantity.

3 3. Probe or intrusive method: Most of the time, the measurement of a physical quantity uses a probe that is placed inside the system. Since a probe invariably affects the measured quantity the measurement process is referred to as an intrusive type of measurement. 4. Non-intrusive method: When the measurement process does not involve insertion of a probe into the system the method is referred to as being non-intrusive. Methods that use some naturally occurring process like radiation emitted by a body to measure a desired quantity relating to the system the method may be considered as nonintrusive. The measurement process may be assumed to be non-intrusive when the probe has negligibly small interaction with the system. A typical example for such a process is the use of laser Doppler velocimeter (LDV) to measure the velocity of a flowing fluid. General measurement scheme: Measured quantity Controller Detector and Transducer Signal Conditioner Computer Calibration or reference signal External power Output Figure Schematic of a general measurement system

4 Figure shows the schematic of a general measurement scheme. Not all the elements shown in the Figure may be present in a particular case. The measurement process requires invariably a detector that responds to the measured quantity by producing a measurable change in some property of the detector. The change in the property of the detector is converted to a measurable output that may be either mechanical movement of a pointer over a scale or an electrical output that may be measured using an appropriate electrical circuit. This action of converting the measured quantity to a different form of output is done by a transducer. The output may be manipulated by a signal conditioner before it is recorded or stored in a computer. If the measurement process is part of a control application the computer can use a controller to control the measured quantity. The relationship that exists between the measured quantity and the output of the transducer may be obtained by calibration or by comparison with a reference value. The measurement system requires external power for its operation. Some issues: 1. Errors Systematic or Random. Repeatability 3. Calibration and Standards 4. Linearity or Linearization Any measurement, however carefully it is conducted, is subject to measurement errors. These errors make it difficult to ascertain the true value of the measured quantity. The nature of the error may be ascertained by repeating the measurement a number of times and looking at the spread of the values. If the spread in the data is small the measurement is repeatable and may be

5 termed as being good. If we compare the measured quantity obtained by the use of any instrument and compare it with that obtained by a standardized instrument the two may show different performance as far as the repeatability is concerned. If we add or subtract a certain correction to make the two instruments give data with similar spread the correction is said to constitute a systematic error. The spread of data in each of the instruments will constitute random error. The process of ascertaining the systematic error is calibration. The response of a detector to the variation in the measured quantity may be linear or non-linear. In the past the tendency was to look for a linear response as the desired response. Even when the response of the detector was non-linear the practice was to make the response linear by some manipulation. With the advent of automatic recording of data using computers this is not necessary since software can take care of this aspect.

6 . Errors in measurements Sub Module 1. Errors accompany any measurement, however well it is conducted. The error may be inherent in the measurement process or it may be induced due to variations in the way the experiment is conducted. The errors may be classified as: 1. Systematic errors (Bias): Systematic errors due to faulty or improperly calibrated instruments. These may be reduced or eliminated by careful choice and calibration of instruments. Sometimes bias may be linked to a specific cause and estimated by analysis. In such a case a correction may be applied to eliminate or reduce bias. Bias is an indication of the accuracy of the measurement. Smaller the bias more accurate the data. Random errors: Random errors are due to non-specific causes like natural disturbances that may occur during the measurement process. These cannot be eliminated. The magnitude of the spread in the data due to the presence of random errors is a measure of the precision of the data. Smaller the random error more precise is the data. Random errors are statistical in nature. These may be characterized by statistical analysis.

7 We shall explain these through the familiar example shown in Figure 3. Three different individuals with different skill levels are allowed to complete a round of target practice. The outcome of the event is shown in the figure. Good Precision Good Accuracy Good Precision Poor Accuracy Poor Precision Poor Accuracy Figure 3 Precision and accuracy explained through a familiar example It is evident that the target at the left belongs to a highly skilled shooter. This is characterized by all the shots in the inner most circle. The result indicates good accuracy as well as good precision. A measurement made well must be like this case! The individual in the middle is precise but not accurate. Maybe it is due to a faulty bore of the gun. The individual at the right is an unskilled person who is behind on both counts. Most beginners will fall into this category. The analogy is quite realistic since most students performing a measurement in the laboratory may be put into one of the three categories. A good experimentalist has to work hard to excel in it!

8 Another example: 5 Standard Reference Individual Thermocouple Data Poly. (Individual Thermocouple Data) Output, mv Bias Error Temperature, o C Figure 4 Example showing the presence of systematic and random errors in data. The results shown in Figure 4 compare the response of a particular thermocouple (that measures temperature) and a standard thermocouple. The measurements are reported between room temperature (close to 0 C) and 500 C. That there is a systematic variation between the two is clear from the figure that shows the trend of the measured temperatures indicated by the particular thermocouple. The systematic error appears to vary with the

9 temperature. The data points indicated by the full symbols appear also to hug the trend line. However the data points do not lie on it. This is due to random errors that are always present in any measurement. Actually the standard thermocouple would also have the random errors that are not indicated in the figure. We have deliberately shown only the trend line for the standard thermocouple. Sub Module Statistical analysis of experimental data Statistical analysis and best estimate from replicate data: Let a certain quantity X be measured repeatedly to get X i, i=1,n (1) Because of random errors these are all different. How do we find the best estimate X b for the true value of X? It is reasonable to assume that the best value be such that the measurements are as precise as they can be! In other words, the experimenter is confident that he has conducted the measurements with the best care and he is like the skilled shooter in the target practice example presented earlier! Thus, we minimize the variance with respect to the best estimate X b of X. Thus we minimize: n [ i b] () S= X X i= 1

10 This requires that: b b = n S = [ X i X b] (-1) =0 X or X i=1 n i= 1 n X i (3) The best estimate is thus nothing but the mean of all the individual measurements! Error distribution: When a quantity is measured repeatedly it is expected that it will be distributed around the best value according to some distribution. Many times the random errors may be distributed as a normal distribution. If µ and σ are, respectively, the mean and the standard deviation, then, the probability density is given by 1 f(x) = e σ π 1 x µ σ (4) The probability that the error around the mean is (x-µ) is the area under the probability density function between (x-µ)+dx and (x-µ) represented by the product of the probability density and dx. The probability that the error is anywhere between - and x is thus given by the following integral: 1 v µ σ 1 F(x) = e dv σ π x (5) This is referred to as the cumulative probability. It is noted that if x the integral tends to 1. Thus the probability that the error is of all possible magnitudes (between - and + ) is unity! The integral is symmetrical with

11 respect to x=µ as may be easily verified. The above integral is in fact the error integral that is a tabulated function. A plot of f(x) and F(x) is given in Figure 5. Cumulative Probability density (x-µ)/σ Figure 5 Normal distribution and its integral Many times we are interested in finding out the chances of error lying between two values in the form ±pσ. This is referred to as the confidence interval and the corresponding cumulative probability specifies the chances of the error occurring within the confidence interval. Table 1 gives the confidence intervals that are useful in practice:

12 Table 1 Confidence intervals according to normal distribution Cumulative Probability Interval p The table indicates that error of magnitude greater than ±3.9σ is very unlikely to occur. In most applications we specify +1.96σ as the error bounds based on 95% confidence.

13 Example 1 Resistance of a certain resistor is measured repeatedly to obtain the following data. No R, kω What is the best estimate for the resistance? What is the error with 95% confidence? Best estimate is the mean of the data R = 9 = kω Standard deviation of the error σ: 1 Variance = Ri -R 9 Hence : σ = = = kω -4 Error with 95% confidence : Error = 1.96 σ = % = kω

14 Example Thickness of a metal sheet (in mm) is measured repeatedly to obtain the following replicate data. What is the best estimate for the sheet thickness? What is the variance of the distribution of errors with respect to the best value? Specify an error estimate to the mean value based on 99% confidence. Experiment No t, mm Experiment No t, mm The best estimate for the metal sheet thickness is the mean of the 1 measured values. This is given by ti tb = t = = = 0. mm 1 1 The variance with respect to the mean or the best value is given by (on substituting t for t b ) as 1 1 ti t ti 1 1 b = t σ = = 0. mm 1-5 = mm The corresponding standard deviation is given by

15 5 σ b = = mm The corresponding error estimate based on 99% confidence is Error = ±.58 σ b = ± ± mm Principle of Least Squares Earlier we have dealt with the method of obtaining the best estimate from replicate data based on minimization of variance. No mathematical proof was given as a basis for this. We shall now look at the above afresh, in the light of the error distribution that has been presented above. Consider a set of replicate data x i. Let the best estimate for the measured quantity be x b. The probability for a certain value x i within the interval x,x + dxto occur in the measured data is given by the relation i i i ( x ) b xi 1 p(x i) = e σ dxi (6) σ π The probability that the particular values of measured data are obtained in replicate measurements must be given by the compound probability given by n ( x ) ( ) b x x i b xi n n 1 1 i 1 p = e σ dx n i e = σ = dx n σ π i= 1 σ π i= 1 ( ) i (7) ( ) The reason the set of data was obtained as replicate data is that it was the most probable! Since the intervals dx i are arbitrary, the above will have to be maximized by the proper choice of x b and σ such that the exponential factor is a maximum. Thus we have to choose x b and σ such that

16 n ( x ) b xi 1 i 1 p' e = σ = (8) σ n has the largest possible value. As usual we set the derivatives p' x b p' = = 0 σ to get the values of the two parameters x b and σ. We have: n ( xi xb) n p' 1 i 1 e = σ = n ( xi x b) ( 1) 0 x + = b σ i= 1 This part should go to zero (9) Or n n ( xi x b) =0 or xb = xi = x (10) i= 1 i= n It is clear thus that the best value is nothing but the mean of the values! We also have: n ( x-x ) i b n p' n 1 i=1 σ = - + ( xi x b) e 0 n+1 n+3 = σ σ σ i = 0 This part should go to Zero (11) Or n i= 1 ( x x ) i b σ = (1) n This last expression indicates that the parameter σ is nothing but the variance of the data with respect to the mean! Thus the best values of the measured quantity and its spread is based on the minimization of the squares of errors with respect to the mean. This embodies what is referred to as the Principle of Least Squares.

17 Propagation of errors: Replicate data collected by measuring a single quantity repeatedly enables us to calculate the best value and characterize the spread by the variance with respect to the best value, using the principle of least squares. Now we look at the case of a derived quantity that is estimated from the measurement of several primary quantities. The question that needs to be answered is the following: A derived quantity Q is estimated using a formula that involves the primary quantities. a 1,a,...a n Each one of these is available in terms of the respective best values a 1, a,...anand the respective standard deviations σ, σ... σ. What is the best estimate for Q and what is the corresponding 1 n standard deviation σ Q? We have, by definition Q =Q(a 1,a,...a n) (13) It is obvious that the best value of Q should correspond to that obtained by using the best values for the a s. Thus, the best estimate for Q given by Q as Q =Q(a 1,a,...a n) (14) Again, by definition, we should have: σ Q N 1 ( i ) N (15) i = 1 = Q Q The subscript i indicates the experiment number and the i th estimate of Q is given by ( ) Q = Q a,a,...a (16) i 1i i ni

18 If we assume that the spread in values are small compared to the mean or the best values (this is what one would expect from a well conducted experiment), the difference between the i th estimate and the best value may be written using a Taylor expansion around the best value as N 1 Q Q Q Q a1i a i... ani N i= a1 a a σ = (17) where the partial derivatives are all evaluated at the best values for the a s. If the a s are all independent of one another then the errors in these are unrelated to one another and hence the cross terms. N ami aki = 0 for m k Thus equation i= 1 (17) may be rewritten as N 1 Q Q Q σ Q = a1i + a i ani (18) N i= 1 a1 a an N Noting that ( ) ji i= 1 a =Nσ j we may recast the above equation in the form Q Q Q Q = n a1 a an σ σ σ σ (19) Equation (19) is the error propagation formula. It may also be recast in the form Q Q Q Q = n a1 a an σ σ σ σ (0)

19 Example 3 The volume of a sphere is estimated by measuring its diameter by vernier calipers. In a certain case the diameter has been measured as D = ± m. Determine the volume and specify a suitable uncertainty for the same. Nominal volume of sphere: 3 3 D V =π = = m 6 6 The error in the measured diameter is specified as: D =± m The influence coefficient is defined as 5 3 V D I D = = π = = m D Using the error propagation formula, we have D V=I D= = m Thus V = ± m Alternate solution to the problem By logarithmic differentiation we have dv V dd =3 D

20 This may be recast as D V =± 3V = ± = ± m D This is the same as the result obtained earlier Example 4 Two resistances R 1 and R are given as 1000 ± 5 Ω and 500 ± 10 Ω. Determine the equivalent resistance when these two are connected in a) series and b) parallel. Also determine the uncertainties in these two cases. Given Data: R1 = 1000, σ 1 = 5;R = 500 σ = 10 All Values are in Ω Case a) Resistances connected in series: Equivalent resistance is Influence coefficients are: R s =R1+ R = =1500Ω R R I = = 1 ;I = = 1 s s 1 R1 R Hence the uncertainty in the equivalent resistance is ( ) ( ) ( ) ( ) s 1 1 σ = ± I σ + I σ = ± =± 6.93 Ω Case b) Resistances connected in parallel: Equivalent resistance is given by

21 RR R p = = = Ω R + R R p = RR 1 ( R + R ) 1 Influence coefficients are: R R R R I = = = = R R R p ( 1+ ) ( R1+ R) R R R R I = = = = p R ( R1+ R) ( R1+ R) Hence the uncertainty in the equivalent resistance is ( ) ( ) ( ) ( ) s 1 1 σ = ± I σ + I σ = ± = ± 5.4 Ω Thus the equivalent resistance is 1500 ± 6.9 Ω in the series arrangement and ± 5.4 Ω in the parallel arrangement.

22 Error estimation some results without proof Standard deviation of the means The problem occurs as indicated below: Replicate data is collected with n measurements in a set Several such sets of data are collected Each one of them has a mean and a variance (precision) What is the mean and standard deviation of the means of all sets? Population mean Let N be the total number of data in the entire population. Mean of all the sets m will be nothing but the population mean (i.e. the mean of all the collected data taken as a whole). Population variance Let the population variance be σ N i= 1 = ( x m) i N (1) Variance of the means Let the variance of the means be σ m. Then we can show that: ( N n) ( ) m σ = σ n N 1 () If n<<n the above relation will be approximated as

23 ( N n) ( ) ( 1-n/N) ( ) m σ = σ n N 1 σ = σ n1-1/n n Estimate of variance Sample and its variance How is it related to the population variance? Let the sample variance from its own mean m s be σ. e (3) Then we can show that: ( ) ( ) N n 1 σe = σ σ 1 n N 1 n (4) Error estimator The last expression may be written down in the more explicit form: σ n 1 e = ( x-m ) i ( n-1) s (5) Physical interpretation Equation (5) may be interpreted using physical arguments. Since the mean (the best value) is obtained by one use of all the available data, the degrees of freedom available (units of information available) is one less than before. Hence the error estimator uses the factor (n-1) rather than n in the denominator!

24 Example 5 (Example 1 revisited) Resistance of a certain resistor is measured repeatedly to obtain the following data. # R, kω What is the best estimate for the resistance? What is the error with 95% confidence? Best estimate is the mean of the data R = 9 = kω Standard deviation of the error σ e : Hence e = Ri R = σ e -4 σ = = k Ω Error with 95% confidence : Error = 1.96 σ = % e = kω

25 Sub Module Regression analysis: Now we are ready to consider curve fit or regression analysis. Suitable plot of data will indicate the nature of the trend in data and hence will indicate the nature of the relation between the independent and the dependent variables. A few examples are shown in Figure 6(a-c) y x Figure 6 (a) Linear relation between y and x

26 y x Figure 6(b) Linear relation between log x and log y

27 y x Figure 6(c) Non-linear relation between y and x The linear graph shown in Figure 6(a) follows a relationship of the form y=ax+b. The linear relationship on the log-log plot shown in Figure 6(b) follows b the form y= ax. The non-linear relationship shown in Figure 6(c) follows a 3 polynomial relationship of the form y= ax + bx + cx+ d. The parameters a, b, c, d are known as the fit parameters and need to be determined as a part of the regression analysis.

28 Linear fit is possible in all the cases shown in Table. Table y=ax+b Linear fit Plots as a straight line on a linear graph sheet y y b = ax Power law fit Plots as a straight line on a log-log graph bx = ae Exponential fit Plots as a straight line on a semi-log graph Linear regression: Let ( x 1, y 1),( x, y ),( x 3, y 3),...( x n, y n) be a set of ordered pairs of data. It is expected that there is a linear relation between y and x. Thus, if we plot the data on a linear graph sheet as in Figure (a) the trend of the data should be well represented by a straight line. We notice that the straight line shown in the figure does not pass through any of the data points shown by full symbols. There is a deviation between the data and the line and this deviation is sometimes positive, sometimes negative, sometimes large and sometimes small. If we look at the value given by the straight line as a local mean then the deviations are distributed with respect to the local mean as a normal distribution. If all data are obtained with equal care one may expect the deviations at various data points to follow the same distribution and hence the least square principle may be applied as under: Minimise s n n [ y y ] y ( ax + b) i f i i 1 1 = = n n (6) where y f = ax+ b is the desired linear fit to data. We see that s is the variance of the data with respect to the fit and minimization will yield the proper choice of

29 the mean line represented by the proper parameters a and b. The minimization requires that n s 1 s 1 = + = = + = a n a n yi ( axi b) xi 0; yi ( axi b) 0 (7) 1 1 These equations may be rearranged as two simultaneous equations for a and b n as given below: ( ) ( ) x a+ x b= x y ( i ) i i i i x a+ nb= y i (8) These are known as normal equations. The summation is from i=1 to n and is not indicated explicitly. The solution to these two equations may be obtained easily by the use of Kramer s rule. i y x n y i i i i i i i i i xy x x xy a =,b= n x n x i i i i x x x x i (9) We now introduce the following definitions: (30) i i i i i i x y xy x x x y x y x =,y =, σ = x, σ = y andσ = xy n n n n n The last of the quantities defined in (30) is known as the covariance. All the other quantities are already familiar to us from statistical analysis. With these definitions the slope of the line fit a may be written as σ a = (31) σ xy x The latter of the expressions in (8) may be solved for the fit line intercept b as y a b = (3) x

30 In fact the last equation indicates that the regression line passes through the point ( x,y ). The fit line may be represented in the alternate form Y = y yand X= x x. f f Example 6 Yf = ax where The following data is expected to follow a relation of the form y=ax+b. Determine the fit parameters by linear regression. x y It is convenient to make a table as shown below. The data given are in columns and 3. The other quantities needed to calculate the fit parameters are in the other columns. Data No. x y x y x y Column Sum: Column Mean σ x Slope of the fit line is: a = σ y The intercept is: b = 0.955

31 Sums are calculated column-wise and are shown in row 8. Various means are then in row 9. The variances are in rows 10, 11 and column. The regression parameters are then calculated using the results of the analysis presented earlier. The regression line is thus given by y f = 0.671x The data and the fit are compared in the following table. x y y f That the fit is a good representation of the data is indicated by the proximity of the respective values in the second and third columns. The plot shown in Figure 7 is further proof of this. Data Fit y and yf x Figure 7 Comparison of data with the fit

32 Goodness of fit and the correlation coefficient: A measure of how good the regression line as a representation of the data is deduced now. In fact it is possible to fit two lines to data by (a) treating x as the independent variable and y as the dependent variable or by (b) treating y as the independent variable and x as the dependent variable. The former has been done above. The latter is described by a relation of the form x = a'y+ b'. The procedure followed earlier can be followed through to get the following (the reader is expected to show these results): σ a' =,b' = x a'y σ xy y (33) The second fit line may be recast in the form 1 b' y' = x a' a' (34) The slope of this line is 1 a' which is not the same, in general, as a, the slope of the first regression line. If the two slopes are the same the two regression lines coincide. Otherwise the two lines are distinct. The ratio of the slopes of the two lines is a measure of how good the form of the fit is to the data. In view of this we introduce the correlation coefficient ρ defined through the relation Or σ xy x y slope of first Re gression line ρ = = aa ' = (35) slope of sec ond Re gression line σ σ σxy ρ=± (36) σσ x y

33 The sign of the correlation coefficient is determined by the sign of the covariance. If the regression line has a negative slope the correlation coefficient is negative while it is positive if the regression line has a positive slope. The correlation is said to be perfect if ρ = ± 1. The correlation is poor ifρ 0. Absolute value of the correlation coefficient should be greater than 0.5 to indicate that y and x are related! In Example 6 the correlation coefficient is positive. The pertinent parameters are x y xy σ = , σ = and σ = With these the.6167 correlation coefficient isρ= = Since the correlation coefficient is close to unity the fit represents the data very closely (Figure 7 has already indicated this). Polynomial regression: Sometimes the data may show a non-linear behavior that may be modeled by a polynomial relation. Consider a quadratic fit as an example. Let the fit equation be given by y = ax + bx+ c. The variance of the data with respect to f the fit is again minimized with respect to the three fit parameters a, b, c to get three normal equations. These are solved for the fit parameters. Thus we have s ( ) yi ax + bx+ c = n (37) Least square principle requires

34 ( ) s = yi ax + bx+ c xi = 0 a n ( ) s = yi ax + bx+ c xi = 0 b n ( ) s = yi ax + bx+ c = 0 c n These may be rewritten as (38) 4 3 i + i + i = i i a x b x c x x y 3 i + i + i = i i a x b x c x x y i + i + = i a x b x nc y (39) Normal equations (39) are easily solved for the three fit parameters to complete the regression analysis. Goodness of fit and the index of correlation: In the case of a non-linear fit we define a quantity known as the index of correlation to determine the goodness of the fit. The fit is termed good if the variance of the deviates is much less than the variance of the y s. Thus we require the index of correlation defined below to be close to ±1 for the fit to be considered good. [ y y ] f y y s ρ=± 1 =± 1 σ y (40) It can be shown that the index of correlation is identical to the correlation coefficient for a linear fit. The index of correlation compares the scatter of the data with respect to its own mean as compared to the scatter of the data with respect to the regression curve.

35 Example 7 The friction factor Reynolds number product fre for laminar flow in a rectangular duct is a function of the aspect ratio h A = where h is the height and w w is the width of the rectangle. The following table gives the available data: A fre Make a suitable fit to data. A plot of the given data indicates that a quadratic fit may be appropriate. For the purpose of the following analysis we represent the aspect ratio as x and the fre product as y. We seek a fit to data of the form y = ax + bx+ c. The following tabulation helps in the regression analysis. No. x y x x 3 x 4 x y x y E sum f The three normal equations are then given by a b c = a b c = a b + 10c =

36 These three simultaneous equations are solved to get the three fit parameters as a=58.354, b=-94.43, c=94.06 The following table helps in comparing the data with the fit. x y y f s =(y-y f ) s y Sum Mean The table also shows how the index of correlation is calculated. The column sums and column means required are given the last two rows of the table. Note that calculation of σ y requires sums of the form y y where y is available as the last entry in column. The index of correlation uses the mean values of columns 4 and 5 given by σ y = ands = The index of correlation is thus equal to ρ= 1 = The negative sign indicates that y decreases when x increases. The index of correlation is close to -1 and hence the fit represents the data very well. A plot of the data along with the fit given in Figure 8 also indicates this. The standard error of the fit is given by s = =± 1.1.

37 Data Quadratic fit Friction factor Reynolds number product Aspect ratio Figure 8 Comparison of the data with the quadratic fit In the above we have considered cases that involved one independent variable and one dependent variable. Sometimes the dependent variable may be a function of more than one variable. For example, the relation of the form b c Nu = a Re Pr is a common type of relationship between the Nusselt number (Nu, dependent variable) and Reynolds (Re) and Prandtl (Pr) numbers both of which are independent variables. By taking logarithms, we see that log(nu) = log(a) + b log(re) + c log(pr). It is thus seen that the relationship is linear when logarithms of the dependent and independent variables are used to describe the fit. Also the relationship may be expressed in the form z=ax+by+c, where z is the dependent variable, x and y are independent variables and a, b, c are the fit parameters. The least square method may be used to determine the fit

38 parameters. Let the data be available for set of n x, y values. The quantity to be minimized is given by ( ) i (41) s = z ax+ bx+ c i The normal equations are obtained by the usual process of setting the first partial derivatives with respect to the fit parameters to zero. i + i i + i = i i a x b x y c x x z i i + i + i = i i a x y b y c y y z a x + b y + nc= z i i i (4) These equations are solved simultaneously to get the three fit parameters. Example 8 The following table gives the variation of z with x and y. Obtain a multiple linear fit to the data and comment on the goodness of the fit. No. x y z The calculation procedure follows that given previously. Several sums are required and these are tabulated below.

39 No. x y z x x y y x z y z Sum The last row contains the sums required and the normal equations are easily written down as under: a b c = a b c = a b + 10c = 9.975

40 .000 Fit Parity Line Fit Data Figure 9 Parity plot showing the goodness of the fit These are solved to get the fit parameters as a=0.85, b=0.97, c= The data and the fit may be compared by making a parity plot as shown in Figure 9. The parity plot is a plot of given data (z) along the abscissa and the fit (z f ) along the ordinate. The parity line is a line of equality between the two. The departure of the data from the parity line is an indication of the quality of the fit. The above figure indicates that the fit is indeed very good. When the data is a function of more than one independent variable it is not always possible to make plots between independent and dependent variables. In such a case the parity plot is a way out.

41 We may also calculate the index of correlation as an indicator of the quality of the fit. This calculation is left to the reader! General non-linear fit: The fit equation may sometimes have to be chosen as a non-linear relation that is not either a polynomial or in a form that may be reduced to the linear form. In such a case the parameter estimation is more involved and requires the use of a search method to determine the best parameter set that minimizes the sum of the squares of the residual. The method is described in some detail below. Let us represent the fit relation in the form y f ( x;a,a,...a ) = where the f 1 m dependent variable is x and a 1 a m are m fit parameters to be determined by the regression analysis. As before we assume that n sets of x, y values are available. Consider the sum of the squares of the residual given by s ( a..a ) = S( a..a ) = y f ( x;a,a,...a ) (43) 1 m 1 m i 1 m i In general it is not possible to set the partial derivatives with respect to the parameters to zero to obtain the normal equations and thus obtain the fit parameters. In view of this let us look at what is happening to the sum of m squares near a starting parameter set a,a,...a. The sum of squares is evaluated using this parameter set in equation (43) to get S ( 1 m) = S a,a,...a.perturb each of the a s individually to get ( m) ( m) ( m),.. Sa,a,..a ( j + a..a j 0 m) Sa a,a..a,sa a,a..a,sa a,a..a ( 1 j + a..a j m), Sa,a,..a ( m a 0 m) Sa,a,..a +. Using these we may estimate the

42 partial derivatives by the use of finite difference approximation S as a j ( a 1,a,..am) ( 1 j + j m) ( 1 j m) S a,a,..a a..a S a,a,a..a = a j. There are m such partial derivatives and they are all likely to be non-zero (if they are all zero we are already at the optimum point where the sum of squares is possibly a minimum). S S S S The gradient vector is then given by the components,,..... The a 1 a aj a m magnitude of this vector is obtained by summing the squares of all the partial derivatives and then taking the square root of this sum. grad S m S = (44) j1 = a j We divide each of the partial derivatives occurring in the gradient vector by the magnitude of the gradient vector thus calculated to get the components of a unit vector that is aligned with the gradient vector. Thus S S S S a a 1 a j am,,..,.. grad S grad S grad S grad S (45) We now take a specific fraction (α, small) of each of these components to define a step along a direction opposite the gradient vector to get S S 1 0 a1 1 0 a a1 = a 1 α,a = a α,.. grad S grad S S S a 1 0 j 1 0 am aj = a j α,..am = am α grad S grad S (46) The calculation above is redone with the new values of the parameter set.

43 The calculation is continued till the magnitude of the gradient reaches zero or acceptably small value at which the calculation stops and the parameter set is assumed to have satisfied the least square principle. An example will make this procedure clear. Example 9 The data given in the following table is expected to follow a relation of the bx form y = ae + cx. Determine the fit parameters by general non-linear f regression. x ,8 y bx The sum of squares of the residual is given by S = i yi ( ae + cxi) partial derivatives needed are obtained analytically i= 1. The bxi bx S i bxi bxi ( ) ( ) ( ) S = y ae + cx e, = y ae + cx ax e, a i i i i i i= 1 b i= 1 10 i= 1 bxi ( ) S = y ae + cx x a i i i (47) The above means that the partial derivatives may be computed once the starting set of parameters is known or assumed. We start the calculation with the initial parameter seta=1,b=0.,c=0.1. The value of S turns out to be for this set of parameter values. Using (47) the partial derivatives are obtained respectively as-4.03, , the magnitude of the gradient vector is

44 0.5 = -4.03, -4.03, The components of then given by ( ) ( ) ( ) the unit vector u a,u b,u c are then given by ,, or ,-0.678, We shall choose α value of 0.0 to get the next trial values for the parameters as a b c ( ) ( ) ( ) a = a α u = = b = b α u = = 0.14 c = c α u = = 0.11 The S value for this parameter set turns out to be The calculations may be repeated as above. The results are summarized below. α a b c grad S S It is clear from the table that a large number of trials are involved in the regression analysis. The value of α needs to be reduced as we approach the optimum set. The final set of parameters for the present case is given by a=1.093,b=0.5093,c=0.663

45 Data Fit y x Figure 10 Comparison of data with the fit That the regression analysis has indeed converged to the proper fit parameters is seen from the excellent agreement between the data and the fit shown in Figure 10. The reader is left to determine the index of correlation and the standard error of the fit.

46 Sub Module Use of EXCEL for regression analysis EXCEL is a Microsoft product that comes along with the Office suite of programs. It is essentially a spread sheet program that provides a computing environment with graphic capabilities. The student is encouraged to learn the basics of EXCEL programming so that data analysis, regression analysis and suitable plots may all be done within the EXCEL environment. EXCEL work sheet provides a grid with cells in it. The cells form columns and rows as in a matrix. The columns are identified by alphanumeric symbols and the rows by numerals. For example, A1 refers to the cell in the first column and first row. Cell C5 will represent the cell in the 3 rd column (column number C) and the 5 th row (row number 5). Column identifiers will go from A Z and then from AA AZ and so on.. The cell can hold a number, a statement or a formula. A number or a statement is simply written by putting the cursor in the appropriate cell and keying in the number or the statement, as the case may be. A formula, however, is written by preceding the formula by = sign. The formula can contain a reference to many built in functions in EXCEL as well as the usual arithmetic operations. The formula can refer to the content of any other cell or cells. The formulas can be calculated repeatedly over a set of rows by simply copying down the formula vertically.

47 Figure 11 An extract of an EXCEL work sheet shows some of the things one can do! A B C D E F G This is a statement 4 In cell B4 is the formula "=A*B" i.e. product of two numbers 04 The formula in B4 is acted upon and the result alone appears in the cell B4 as 5 seen above.. 6 x x^ Sum of G7to G16 is obtained by entering the formula "=SUM(G7:G16) in Cell G SUM() is a built in function in EXCEL Data may be keyed into the cells in the form of columns as shown in the work sheet given as Figure 1 below. The plotting is menu driven and the plot may be displayed as a separate plot or within the work sheet. The latter is shown in the case given here. The data range for the plot is specified by simply blocking the Data cells shown by the blue background!

48 Figure 1 Another extract of an EXCEL work sheet, showing data and the corresponding plot. A B C D E E G H 1 x y Properties of cells, chart (plot is referred to as chart in EXCEL) are changed to suit the requirements with menu driven controls. Student should familiarize oneself by learning these through HELP available in EXCEL.

49 Figure 13 Another extract of an EXCEL work sheet, showing data and the corresponding plot along with the automatically generated fit. The inset in the plot gives the linear relation between y and x. The square of the correlation coefficient is also shown in the inset (symbol R ). A B C D E F G H 1 The following data is expected to follow a linear law. Obtain such a law by using the "Trend Line" option in EXCEL. 3 x y d Figure 13 shows how a trend line can be added to the plot. The inset in the plot shows the relationship that exists between the y and x data values. Correlation coefficient is very high indicating the fit to represent the data extremely well. Several examples of regression using EXCEL are presented below. The examples are self explanatory and I expect the student to work them out using EXCEL himself/herself.

50 Example 10 Linear fit example using EXCEL The EXCEL worksheet into which the data is input is shown below. The data is keyed in the cells as shown. The sums required are automatically calculated by using the SUM function. Variances and the covariance are calculated using their definitions given earlier. The slopes and the intercepts are calculated using the formulae derived by the least square method. The two regression lines are then given by: y = 14.15x y = 14.74x

51 Price per thousand pieces of a certain product (x) determines the demand (y) for the product. The data is given below. Fit a straight line to data and discuss the quality of the fit. No. x y x y xy y 1 y Mean: Sum Mean Variance of x= Variance of y= Covariance = Slope of first regression line= Intercept of first regression 88.4 line= Slope of second regression line= Intercept of first regression 97.6 line= Correlation coefficient = The correlation coefficient is calculated using the statistical parameters that have been already calculated.

52 Linear Fit to Data 140 y y1 y y x Figure 14 Plot of the resulting data using EXCEL The data generated has been plotted using EXCEL in the form of a chart in Figure 14.. The chart option used is scatter plot. The given data is shown using the red circles. The two lines of regression are shown by the brown and blue lines. Both of them pass through the mean x and mean y (indicated by the point ). The fit is good because the two regression lines are very close to each other.

53 Linear Fit Using "Trend line" Data Linear (Data) y 140 The regression equation is y = x x Figure 15 Plot of the data using EXCEL with Trend Line option The fit may also be done by using the trend line option available in EXCEL. We choose the linear trend line and get the plot shown in Figure 15. It is observed that the Trend Line option yields the first regression line that considers y to be a function of x. The required arithmetic is automatically performed by EXCEL. There is an option to automatically display the regression line equation on the chart.

54 Example 11 Exponential fit example using EXCEL The given in the first two columns are the time (t, s) and the corresponding temperature excess (T, C) over the ambient of a certain system. The data is expected to be well represented by an exponential law in the form T Obtain the fit using EXCEL. = A exp( Bt).

55 EXCEL work sheet appears as below. t (s) Data, Fit,(T) ln(t) ln(t) t t ln(t) [ln(t)] (T) mean t, s Mean of [ln(t)] Var t Covaria -nce Var of ln(t) slope B, 1/s interce pt, ln A τ s (T o )=A, o C ρ t (s) Data,(T) Fit,(T) Error It is noted that the data represents a linear law on the semi-log plot (the student is recommended to test this out by making a plot).

56 Comparison of data with fit Data Expon. (Data) Temperature Excess T, o C T = 7.5 e t Time t, s Figure 16 Comparison of data with fit The plot shown as Figure 16 indicates that the fit represents the data very well. Error bars are indicated based on 95% confidence limits using the feature available in EXCEL.

57 The standard error is calculated by using the following tabulation using EXCEL. t (s) Data, T C Fit,T C Error, C Square of error ( C) Standard Error 1.87 Error bar indicated on T is based on the standard error of 1.87 indicated in the above table. The problem may easily be solved by using the Trend Line option by choosing an exponential law from the menu. The quality of the fit may also be gauged by comparing the data with the values obtained by the use of the exponential least square fit. This is done by making a parity plot as given below as Figure 17. The distribution of the data around the parity line is a measure of the goodness of the fit. The points should be close to the parity line and must be distributed evenly on the two sides if there is no bias in the measurement. It is observed that the exponential fit is good based on both these counts!

58 Data Parity plot Parity line 60 Fit Data Figure 17 Parity plot for the exponential fit example. Sometimes it is instructive to show the error between the data and the fit. In this example the error in the temperature excess between the data and the fit is plotted as a function of time as shown in Figure 18. The error is evenly distributed on both the positive and negative sides indicating absence of bias. The error between the data and the fit is no more than 1.5 C Error plot 1.5 Error=(Data-Fit), o C Time, s Figure 18 Error distribution plot for the exponential fit example.

59 Example 1 Polynomial fit example using EXCEL The x y data set shown below is expected to follow a quadratic relationship. Obtain the fit by the least squares method. Discuss the relevant statistical parameters that characterize the fit. Make a suitable plot. x y(data) y(fit) (y-mean y) (y-y(fit)) Mean Index of correlation= Standard error = 1.06 Using EXCEL Trend Line polynomial (quadratic) option the fit is easily obtained as y(fit) =.3x x The index of correlation is calculated using the relation given previously and is shown in the table. Index of correlation of indicates that the fit is very good.

60 Plot shown below as Figure 19 indicates graphically the goodness of the fit. Note that the fit shown uses the Trend Line option with Polynomial Fit of EXCEL. Quadratic Fit to Data y(data) Poly. (y(data)) y =.318x x R = Data, Fit Figure 19 Polynomial fit examples showing the data and the fit. x The regression equation and its index of correlation are also given in the inset. Error bars are also indicated based on 95% confidence intervals.

61 Sub Module Design of experiments Goal of experiments: Experiments help us in understanding the behavior of a (mechanical) system Data collected by systematic variation of influencing factors helps us to quantitatively describe the underlying phenomenon or phenomena The goal of any experimental activity is to get the maximum information about a system with the minimum number of well designed experiments. An experimental program recognizes the major factors that affect the outcome of the experiment. The factors may be identified by looking at all the quantities that may affect the outcome of the experiment. The most important among these may be identified using a few exploratory experiments or from past experience or based on some underlying theory or hypothesis. The next thing one has to do is to choose the number of levels for each of the factors. The data will be gathered for these values of the factors by performing the experiments by maintaining the levels at these values. Suppose we know that the phenomena being studied is affected by the pressure maintained within the apparatus during the experiment. We may identify the smallest and the largest possible values for the pressure based on experience, capability of the apparatus to withstand the pressure and so on. Even though the pressure may be varied continuously between these limits, it

62 is seldom necessary to do so. One may choose a few values within the identified range of the pressure. These will then be referred to as the levels. Experiments repeated with a particular set of levels for all the factors constitute replicate experiments. Statistical validation and repeatability concerns are answered by such replicate data. In summary an experimental program should address the following issues: Is it a single quantity that is being estimated or is it a trend involving more than one quantity that is being investigated? Is the trend linear or non-linear? How different are the influence coefficients? What does dimensional analysis indicate? Can we identify dimensionless groups that influence the quantity or quantities being measured How many experiments do we need to perform? Do the factors have independent effect on the outcome of the experiment? Do the factors interact to produce a net effect on the behavior of the system?

63 Full factorial design: A full factorial design of experiments consists of the following: Vary one factor at a time Perform experiments for all levels of all factors Hence perform a large number of experiments that are needed! All interactions are captured (as will be shown later) Consider a simple design for the following case: Let the number of factors = k Let the number of levels for the i th factor = n i The total number of experiments (n) that need to be performed is n =Π n. If k = 5 and number of levels is 3 for each of the factors the total number of experiments to be performed in a full factorial design is 3 5 = 43. k i= 1 i k factorial design: Consider a simple example of a k factorial design. Each of the k factors is assigned only two levels. The levels are usually High = 1 and Low = -1. Such a scheme is useful as a preliminary experimental program before a more ambitious study is undertaken. The outcome of the k factorial experiment will help identify the relative importance of factors and also will offer some knowledge about the interaction effects. Let us take a simple case where the number of factors is. Let these factors be x A and x B. The number of experiments that may be performed is 4 corresponding to the following combinations:

64 Experiment No. xa xb Let us represent the outcome of each experiment to be a quantity y. Thus y1 will represent the outcome of experiment number 1 with both factors having their High values, y will represent the outcome of the experiment number with the factor A having the Low value and the factor B having the High value and so on. The outcome of the experiments may be represented as the following matrix: xa xb y 1 y3-1 y y4 A simple regression model that may be used can have up to four parameters. Thus we may represent the regression equation as y= p0 + paxa + pbxb + pabxaxb (48) The p s are the parameters that are determined by using the outcome matrix by the simultaneous solution of the following four equations: p + p + p + p = y 0 A B AB 1 p p + p p = y 0 A B AB p p p p = y 0 A B AB 3 p p p + p = y 0 A B AB 4 (49)

65 Figure 14 Interpretation of factorial experiment It is easily seen that the parameter p 0 is simply the mean value of y that is obtained by putting x A = x B = 0 corresponding to the mean values for the factors. Equation (49) expresses the fact that the outcome may be interpreted as shown in Figure 14. It is thus seen that the values of y- p 0 at the corners of the square indicate the deviations from the mean value and hence the mean of the square of these deviations (we may divide the sum of the squares with the number of degrees of freedom = 3) is the variance of the sample data collected in the experiment. The influence of the factors may then be gauged by the contribution of each term to the variance. These ideas will be brought out by example 13.