Cutting Speed Feed x, (sfpm) 800 700 700 600 600 500 500 450 400 x2 (ipr) 01725

Question

Cutting Speed Feed x, (sfpm) 800 700 700 600 600 500 500 450 400 x2 (ipr) 01725 1.00,0.90,0.74, 0.66 01725 00,1.20, 1.50, 1.60 01570 1.75,1.85, 2.00,2.20 02200 .20,1.50, 1.60, 1.60 01725 2.35,2.65, 3.00, 3.60 01725 6.40,7.80,9.80, 16.50 01570 8.80,11.00, 11.75, 19.00 02200 4.00, 4.70, 5.30, 6.00 01725 2150,24.50, 26.00,33.00 Tool Life, y (min) (a) Taylor's expanded tool life equation is yxx2-C. This relationship suggests that In(y) may well be approximately linear in both In(x,) and In(x,). Use a multiple linear regression program to fit the relationship to these data. What fraction of the raw vari- ability in In(y) is accounted for in the fitting process? What estimates of the parameters a1, o2, and C follow from your fitted equation? (b) Compute and plot residuals (continuing to work on log scales) for the equation you ft in part (a). Make at least plots of residuals versus fitted In(y) and both ln(x,) and In(x2), and make a normal plot of these residuals Do these plots reveal any particular problems with the fitted equation? (c) Use your fitted equation to predict first a log tool life and then a tool life, if in this machin ing application a cutting speed of 550 and a feed of.01650 is used (d) Plot the ordered pairs appearing in the data set in the (x, x2)-plane. Outline a region in the plane where you would feel reasonably safe using the equation you fit in part (a) to predict tool life.

Explanation / Answer

a. Taylor's expanded tool life equation is yx1a1x2a2 = C. This relationship suggests that ln(y) may well be approximately linear in both ln(x1) and ln(x2). Use a multiple linear regressions program to fit the relationship ln(y)=B0+B1ln(x1)+B2ln(x2) to these data. What fraction of the raw variability in ln(y) is accounted for in the fitting process? What estimates of the parameters a1, a2, and C follow from your fitted equation?

The data is exported to R and then the following program is run to get the answers.

> tt <- read.csv("clipboard",sep=" ")
> head(tt)
Cutting_Speed    Feed    y
1           800 0.01725 1.00
2           800 0.01725 0.90
3           800 0.01725 0.74
4           800 0.01725 0.66
5           700 0.01725 1.00
6           700 0.01725 1.20
> tt$logCS <- log(tt$Cutting_Speed)
> tt$logFeed <- log(tt$Feed)
> tt$logy <- log(tt$y)
> logy_lm <- lm(logy~logFeed+logCS,tt)
> summary(logy_lm)

Call:
lm(formula = logy ~ logFeed + logCS, data = tt)

Residuals:
     Min       1Q   Median       3Q      Max
-0.37780 -0.17145 -0.04072 0.12217 0.70253

Coefficients:
            Estimate Std. Error t value Pr(>|t|)   
(Intercept) 18.7496     1.5918   11.78 2.31e-13 ***
logFeed      -3.7379     0.3394 -11.01 1.37e-12 ***
logCS        -5.1209     0.1860 -27.53 < 2e-16 ***
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 0.2349 on 33 degrees of freedom
Multiple R-squared: 0.9595,    Adjusted R-squared: 0.9571
F-statistic: 391.1 on 2 and 33 DF, p-value: < 2.2e-16

The fraction of the raw variability in the log(y) accounted by the fitting process is 95.71%. The estimates of the parameters are 18.7496, -3.7379, and -5.1209 .

b. Compute and plot residuals (continuing to work on log scales) for the equation you fit in part (a). Make at least plots of residuals verses fitted ln(y) and both ln(x1) and ln(x2), and make a normal plot of these residuals. Do these plots reveal any particular problems with the fitted equation?

> residuals(logy_lm)
           1            2            3            4            5            6            7            8            9           10           11           12
0.305999512 0.200638996 0.004894419 -0.109515932 -0.377799313 -0.195477756 0.027665795 0.092204316 -0.170108362 -0.114538511 -0.036576969 0.058733211
          13           14           15           16           17           18           19           20           21           22           23           24
-0.075704802 0.147438749 0.211977270 0.211977270 -0.312771796 -0.192627484 -0.068574835 0.113746722 -0.244536749 -0.046711006 0.181547646 0.702525642
          25           26           27           28           29           30           31           32           33           34           35           36
-0.278007855 -0.054864303 0.011093664 0.491679403 -0.344918634 -0.183650486 -0.063506174 0.060546474 -0.175474134 -0.044853952 0.014569469 0.252980492
> par(mfrow=c(1,2))
> plot(tt$logFeed,residuals(logy_lm))
> plot(tt$logCS,residuals(logy_lm))

The residual plots don't indicate any outlier or systematic pattern.

c. Use your fitted equation to predict first a log tool life and then a tool life, if in this machining application a cutting speed of 550 and a feed of .01650 is used.

> predict(logy_lm,newdata=data.frame(logCS=log(550),logFeed=log(.01650)))
       1
1.778917
> exp(predict(logy_lm,newdata=data.frame(logCS=log(550),logFeed=log(.01650))))
       1
5.923437

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