SHEET 1

STRAIGHTNESS

1. With the aid of neat sketches Describe how the straightness of a machine guide

way may be experimentally determined:

a) Using spirit sensitive level.

b) Rocdale arm (arm comparator).

2. With the aid of neat sketches Describe how the straightness of a shaft may be

experimentally determined using a dial indicator

3. A Shaft of length 500 mm was tested for straightness error by using a dial

indicator. The shaft was divided into 10 equal spaces of 50 mm apart, and the

dial readings were observed at each position (11 readings). The readings were as

shown in the following table. Using the least square method determine the

maximum out of straightness in microns, and Plot the contour of the shaft.

x y x2 xy

y' =0.004099x +1.387838

𝛿 = 𝑦 − 𝑦′

0 0.01 0 0 -0.04682 0.05682

50 0.1 2500 5 0.03638 0.06362

100 0.15 10000 15 0.11958 0.03042

150 0.15 22500 22.5 0.20278 -0.05278

200 0.25 40000 50 0.28598 -0.03598

250 0.25 62500 62.5 0.36918 -0.11918

300 0.45 90000 135 0.45238 -0.00238

350 0.6 122500 210 0.53558 0.06442

400 0.4 160000 160 0.61878 -0.21878

450 0.75 202500 337.5 0.70198 0.04802

500 0.95 250000 475 0.78518 0.16482

∑ 2750 4.06 962500 1472.5

0

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0.7

0.8

0.9

1

0 100 200 300 400 500 600

y (m

m)

X (mm)

profile

profile

true straight line

𝑛 = 11

c = ∑ y ∑ x2 − ∑ 𝑥𝑦 ∑ x

n ∑ x2 − (∑ x)2 =

4.06 × 962500 − 1472.5 × 2750

11 × 962500 − 27502 = −0.048681

𝑚 =𝑛 ∑ 𝑥𝑦 − ∑ 𝑦 ∑ 𝑥

n ∑ x2 − (∑ x)2 =

11 × 1472.5 − 4.06 × 2750

11 × 962500 − 27502 = 0.0016636

𝒚 = 𝒎𝒙 + 𝒄 = 𝟎. 𝟎𝟎𝟏𝟔𝟔𝟑𝟔𝒙 − 𝟎. 𝟎𝟒𝟖𝟔𝟖𝟏 𝒎𝒎

Maximum out of straightness = |0.16482| + |−0.21878| = 0.3836𝑚𝑚 = 383.6microns

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Series1

4. A Shaft of length 300 mm was tested for straightness error by using a dial

indicator. The shaft was divided into 6 equal spaces of 50 mm apart, and the dial

readings were observed at each position (7 readings). The readings were as

follows:

2.7

2.9

3.1

3.3

3.5

3.7

3.9

4.1

4.3

0 50 100 150 200 250 300 350

y (m

m)

X (mm)

profile

profile

true straight line

x y x2 xy

y' =0.004216x+2.8965

𝛿 = 𝑦 − 𝑦′

0 2.9 0 0 2.8965 0.0035

50 3.15 2500 157.5 3.1073 0.0427

100 3.3 10000 330 3.3181 -0.0181

150 3.45 22500 517.5 3.5289 -0.0789

200 3.752 40000 750.4 3.7397 0.0123

250 4 62500 1000 3.9505 0.0495

300 4.15 90000 1245 4.1613 -0.0789

∑ 1050 24.702 227500 4000.4

𝑛 = 7

c = ∑ y ∑ x2 − ∑ 𝑥𝑦 ∑ x

n ∑ x2 − (∑ x)2 =

24.702 × 227500 − 1472.5 × 1050

7 × 227500 − 10502= 2.8965

𝑚 =𝑛 ∑ 𝑥𝑦 − ∑ 𝑦 ∑ 𝑥

n ∑ x2 − (∑ x)2 =

7 × 1472.5 − 24.702 × 1050

7 × 227500 − 10502 = 0.004216

𝒚 = 𝒎𝒙 + 𝒄 = 𝟎. 𝟎𝟎𝟒𝟐𝟏𝟔𝒙 + 𝟐. 𝟖𝟗𝟔𝟓 𝒎𝒎

Maximum out of straightness = |0.0495| + |−0.0789| = 0.1284𝑚𝑚 = 128.4microns

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5. The straightness of a slide was tested using a spirit level of 1.5 scale division and

a radius of curvature of 3 m. The readings taken in the forward direction (in

divisions) were

-3 +4 -1 +2 -3 +4 -3 -2 -7

The readings taken in the backward direction (in divisions) were

+3 -4 +1 -2 +3 -4 +3 +2 +7

Find out the out of straightness semi analytically. The distance between any two

points is 100 mm.

𝑐 =𝐷

𝑅=

1.5

3= 0.5𝑚𝑚/𝑚

𝑐 = 0.05𝑚𝑚/100𝑚𝑚

x

y y

average y

accumulative y

mm x2 xy

y' =-0.00022x+ 0.054545

𝛿= 𝑦 − 𝑦′

forward backward

0 0 0 0 0 0 0 0 0.054545 -0.05455

100 -3 3 -3 -3 -0.15 10000 -15 0.032545 -0.18255

200 4 -4 4 1 0.05 40000 10 0.010545 0.039455

300 -1 1 -1 0 0 90000 0 -0.01146 0.011455

400 2 -2 2 2 0.1 160000 40 -0.03346 0.133455

500 -3 3 -3 -1 -0.05 250000 -25 -0.05546 0.005455

600 4 -4 4 3 0.15 360000 90 -0.07746 0.227455

700 -3 3 -3 0 0 490000 0 -0.09946 0.099455

800 -2 2 -2 -2 -0.1 640000 -80 -0.12146 0.021455

900 -7 7 -7 -9 -0.45 810000 -405 -0.14346 -0.30655

sum 4500

-0.45 2850000 -385

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0

0.1

0.2

0 200 400 600 800 1000

y (m

m)

X (mm)

profile

true straight line

𝑛 = 10

c = ∑ y ∑ x2 − ∑ 𝑥𝑦 ∑ x

n ∑ x2 − (∑ x)2 =

−0.45 × 2850000 − (−385 × 4500)

10 × 2850000 − 45002 = 0.0545454

𝑚 =𝑛 ∑ 𝑥𝑦 − ∑ 𝑦 ∑ 𝑥

n ∑ x2 − (∑ x)2 =

10 × −385 − (−0.45 × 4500)

10 × 2850000 − 45002

= −0.00022

𝒚 = 𝒎𝒙 + 𝒄 = −𝟎. 𝟎𝟎𝟎𝟐𝟐𝒙 + 𝟎. 𝟎𝟓𝟒𝟓𝟒𝟓 𝒎𝒎

Maximum out of straightness = |0.227455| + |−0.30655| = 0.534005𝑚𝑚 = 534microns

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0 200 400 600 800 1000

de

lta

(mm

)

x (mm)

semi analytically

x

y y

average

y accumulati

ve

y mm

Y' delta yi-y'

forward backward

0 0 0 0 0 0 0 0

100 -3 3 -3 -3 -0.15 -0.05 -0.1

200 4 -4 4 1 0.05 -0.1 0.15

300 -1 1 -1 0 0 -0.15 0.15

400 2 -2 2 2 0.1 -0.2 0.3

500 -3 3 -3 -1 -0.05 -0.25 0.2

600 4 -4 4 3 0.15 -0.3 0.45

700 -3 3 -3 0 0 -0.35 0.35

800 -2 2 -2 -2 -0.1 -0.4 0.3

900 -7 7 -7 -9 -0.45 -0.45 0

Maximum out of straightness = |0.45| + |−0.1| = 0.55𝑚𝑚 = 550microns

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X (mm)

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Series2

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6. The squareness of the guide way of a milling machine was checked using sensitive

level. Its scale value is 0.02 mm/100mm. The readings taken on the guide way of the

bed are given in row "A" while the readings for the guide way of the column are

given in row "B". Find the out the out of squareness

A +1 +2 0 -3 +2 -1 +3

B -2 +1 -1 +2 0 -1 1

x

y

y accumulative

y mm

x2 xy y'

=(0.0000190476)x +(-0.000065895)

𝛿 = 𝑦 − 𝑦′

0 0 0 0 0 0 -0.000065895 0.000065895

100 1 1 0.02 10000 2 0.001838865 0.018161135

200 2 3 0.04 40000 8 0.003743625 0.036256375

300 0 3 0 90000 0 0.005648385 -0.00564839

400 -3 0 -0.06 160000 -24 0.007553145 -0.06755315

500 2 2 0.04 250000 20 0.009457905 0.030542095

600 -1 1 -0.02 360000 -12 0.011362665 -0.03136267

700 3 4 0.06 490000 42 0.013267425 0.046732575

sum 2800

0.08 1400000 36

𝑛 = 8

c = ∑ y ∑ x2 − ∑ 𝑥𝑦 ∑ x

n ∑ x2 − (∑ x)2 =

0.08 × 1400000 − (36 × 2800)

8 × 1400000 − 28002 = −0.000065895

𝑚 =𝑛 ∑ 𝑥𝑦 − ∑ 𝑦 ∑ 𝑥

n ∑ x2 − (∑ x)2 =

8 × 36 − (0.08 × 2800)

8 × 1400000 − 28002 = 0.0000190476

𝒚 = 𝒎𝒙 + 𝒄 = 𝟎. 𝟎𝟎𝟎𝟎𝟏𝟗𝟎𝟒𝟕𝟔𝒙 − 𝟎. 𝟎𝟎𝟎𝟎𝟔𝟓𝟖𝟗𝟓 𝒎𝒎

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0 200 400 600 800y (m

m)

X (mm)

A

Series1

Series2

x

y

y accumulative

y mm

x2 xy y'

=(0.0000285714)x 𝛿 = 𝑦 − 𝑦′

0 0 0 0 0 0 0 0

100 -2 -2 -0.04 10000 -4 0.00285714 -0.04285714

200 1 -1 0.02 40000 4 0.00571428 0.01428572

300 -1 -2 -0.02 90000 -6 0.00857142 -0.02857142

400 2 0 0.04 160000 16 0.01142856 0.02857144

500 0 0 0 250000 0 0.0142857 -0.0142857

600 -1 -1 -0.02 360000 -12 0.01714284 -0.03714284

700 1 0 0.02 490000 14 0.01999998 2E-08

sum 2800

0 1400000 12

𝑛 = 8

c = ∑ y ∑ x2 − ∑ 𝑥𝑦 ∑ x

n ∑ x2 − (∑ x)2 =

0 × 1400000 − (12 × 2800)

8 × 1400000 − 28002= 0

𝑚 =𝑛 ∑ 𝑥𝑦 − ∑ 𝑦 ∑ 𝑥

n ∑ x2 − (∑ x)2 =

8 × 12 − (0 × 2800)

8 × 1400000 − 28002 = 0.0000285714

𝒚 = 𝒎𝒙 + 𝒄 = 𝟎. 𝟎𝟎𝟎𝟎𝟐𝟖𝟓𝟕𝟏𝟒𝒙 𝒎𝒎

𝑚1

𝑚2≠ −1

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0 200 400 600 800y (m

m)

X (mm)

B

profile

true straight line

7. A straight edge is checked using a precision spirit level having scale value

0.5 mm/100 mm. The readings of the bubble in the forward and

backward directions for the five positions taken along the edges are as

shown in the following table, readings are taken as divisions. The interval

distance is 100 mm. Find out the out of straightness.

First Second Third Fourth Fifth

Forward 0 13 17 7 15

Backward 0 -11 -15 -5 -13

x

y y average

y accumulative

y mm

x2 xy y' =0.059x

+0.4 𝛿

= 𝑦 − 𝑦′

forward backward

0 0 0 0 0 0 0 0 0.4 -0.4

100 13 -11 12 12 6 10000 600 6.3 -0.3

200 17 -15 16 28 14 40000 2800 12.2 1.8

300 7 -5 6 34 17 90000 5100 18.1 -1.1

400 15 -13 14 48 24 160000 9600 24 0

sum 1000

61 300000 18100

0

5

10

15

20

25

30

0 100 200 300 400 500

y (m

m)

X (mm)

profile

true straight line

𝑛 = 5

c = ∑ y ∑ x2 − ∑ 𝑥𝑦 ∑ x

n ∑ x2 − (∑ x)2 =

61 × 300000 − (18100 × 1000)

5 × 300000 − 10002 = 0.4

𝑚 =𝑛 ∑ 𝑥𝑦 − ∑ 𝑦 ∑ 𝑥

n ∑ x2 − (∑ x)2 =

5 × 18100 − (61 × 1000)

5 × 300000 − 10002= 0.059

𝒚 = 𝒎𝒙 + 𝒄 = 𝟎. 𝟎𝟓𝟗𝒙 + 𝟎. 𝟒 𝒎𝒎

Maximum out of straightness = |1.8| + |−1.1| = 2.9𝑚𝑚 = 2900microns

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8. The straightness of a slide was tested using a spirit level of 1.5 mm scale value

and a radius of curvature of 3 m. The readings in the forward direction (in

divisions) are

-3 +4 -1 +2 -3 +4 -3 -2 -7

The readings in the backward direction (in divisions) are

+3 -4 +1 -2 +3 -4 +3 +2 +7

Find out the out of straightness semi analytically. The distance between any two

points is 100 mm.

𝑐 =𝐷

𝑅=

1.5

3= 0.5𝑚𝑚/𝑚

𝑐 = 0.05𝑚𝑚/100𝑚𝑚

semi analytically

x

y y

average

y accumulati

ve

y mm

Y' delta yi-y'

forward backward

0 0 0 0 0 0 0 0

100 -3 3 -3 -3 -0.15 -0.05 -0.1

200 4 -4 4 1 0.05 -0.1 0.15

300 -1 1 -1 0 0 -0.15 0.15

400 2 -2 2 2 0.1 -0.2 0.3

500 -3 3 -3 -1 -0.05 -0.25 0.2

600 4 -4 4 3 0.15 -0.3 0.45

700 -3 3 -3 0 0 -0.35 0.35

800 -2 2 -2 -2 -0.1 -0.4 0.3

900 -7 7 -7 -9 -0.45 -0.45 0

Maximum out of straightness = |0.45| + |−0.1| = 0.55𝑚𝑚 = 550microns

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X (mm)

profile

Series2

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9. A flat guide way of length 2000 mm was checked using beam comparator having

a dial indicator of scale value 0.01 mm. The distance between the beam legs is

200 mm. The guide way was divided into 10 divisions as shown and the recorded

readings are listed below. The beam was calibrated by placing it on a reference

straight edge and the indicator reading was observed as 3.15 mm. Find out the

maximum out of straightness.

a------b------c------d------e------f------g------h------i-------g------k

Beam Readings

position abc bed cde def efg fgh ghi hig igh

Reading mm 2.98 2.95 3.02 3.06 2.9 3.11 3.18 3.22 3.25

R corrected

= R-Rref -0.17 -0.2 -0.13 -0.09 -0.25 -0.04 0.03 0.07 0.1

𝑅 = 2𝑞 − 𝑝 + 𝑦

position a b c d e f g h i j k

x 0 200 400 600 800 1000 1200 1400 1600 1800 2000

y 0 0 -0.17 -0.54 -1.04 -1.63 -2.47 -3.35 -4.2 -4.98 -5.66

x y x2 xy

y' =0.066473x

-22.3036 𝛿 = 𝑦 − 𝑦′

0 0 0 0 0.876364 -0.87636

200 0 40000 0 0.264364 -0.26436

400 -0.17 160000 -68 -0.34764 0.177636

600 -0.54 360000 -324 -0.95964 0.419636

800 -1.04 640000 -832 -1.57164 0.531636

1000 -1.63 1000000 -1630 -2.18364 0.553636

1200 -2.47 1440000 -2964 -2.79564 0.325636

1400 -3.35 1960000 -4690 -3.40764 0.057636

1600 -4.2 2560000 -6720 -4.01964 -0.18036

1800 -4.98 3240000 -8964 -4.63164 -0.34836

2000 -5.66 4000000 -11320 -5.24364 -0.41636

∑ 11000 -24.04 15400000 -37512

𝑛 = 11

c = ∑ y ∑ x2 − ∑ 𝑥𝑦 ∑ x

n ∑ x2 − (∑ x)2 =

−24.04 × 15400000 − −37512 × 11000

11 × 15400000 − 110002= −22.3036

𝑚 =𝑛 ∑ 𝑥𝑦 − ∑ 𝑦 ∑ 𝑥

n ∑ x2 − (∑ x)2 =

11 × −37512 − −24.04 × 11000

11 × 15400000 − 110002 = 0.066473

𝒚 = 𝒎𝒙 + 𝒄 = −𝟎. 𝟎𝟎𝟑𝟎𝟔𝒙 + 𝟎. 𝟖𝟕𝟔𝟑𝟔𝟒𝒎𝒎

Maximum out of straightness = |0.553636| + |−0.87636| = 1.42999 𝑚𝑚

-6

-5

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0

1

2

0 500 1000 1500 2000 2500

y (m

m)

X (mm)

profile

true straight line

-1

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x (mm)

10. A flat guide way of length 2000 mm was checked using beam

comparator having a dial indicator of scale value 0.01 mm. The distance

between the beam legs is 200 mm. The guide way was divided into 10

divisions as shown and the recorded readings are listed below. Readings

were taken by a spirit level of constant 0.15 mm/m along the line abc

and the reading were listed below. Find out the maximum out of

straightness.

a---b---c---d---e---f---g---h---i---j---k

Bam Readings

position abc bcd cde def efg fgh ghi hij ijk

Reading mm

2.98 2.95 3.02 3.06 2.90 3.11 3.18 3.22 3.25

R corrected

2.92 2.89 2.96 3 2.84 3.05 3.12 3.16 3.19

Level readings

bc ab Position

5 R 3 R Forward Reading (Divs) 1 R 1 L Backward

+3 +1 Average

0.09 0.03 (n*c) mm

C=0.15mm/m=0.03mm/200mm

Correction = 0.06 mm

𝑅 = 2𝑞 − 𝑝 + 𝑦

position a b c d e f g h i j k

x 0 200 400 600 800 1000 1200 1400 1600 1800 2000

y 0 0 2.92 8.73 17.5 29.27 43.88 61.54 82.32 106.3 133.4

x y x2 xy

y' =0.066473x-

22.3036 𝛿 = 𝑦 − 𝑦′

0 0 0 0 -22.3036 22.3036

200 0 40000 0 -9.009 9.009

400 2.92 160000 1168 4.2856 -1.3656

600 8.73 360000 5238 17.5802 -8.8502

800 17.5 640000 14000 30.8748 -13.3748

1000 29.27 1000000 29270 44.1694 -14.8994

1200 43.88 1440000 52656 57.464 -13.584

1400 61.54 1960000 86156 70.7586 -9.2186

1600 82.32 2560000 131712 84.0532 -1.7332

1800 106.3 3240000 191340 97.3478 8.9522

2000 133.4 4000000 266800 110.6424 22.7576

∑ 11000 485.86 15400000 778340

-40

-20

0

20

40

60

80

100

120

140

160

0 500 1000 1500 2000 2500

y (m

m)

X (mm)

profile

true straight line

𝑛 = 11

c = ∑ y ∑ x2 − ∑ 𝑥𝑦 ∑ x

n ∑ x2 − (∑ x)2 =

485.86 × 15400000 − 778340 × 11000

11 × 15400000 − 110002= −22.3036

𝑚 =𝑛 ∑ 𝑥𝑦 − ∑ 𝑦 ∑ 𝑥

n ∑ x2 − (∑ x)2 =

11 × 778340 − 485.86 × 11000

11 × 15400000 − 110002 = 0.066473

𝒚 = 𝒎𝒙 + 𝒄 = 𝟎. 𝟎𝟔𝟔𝟒𝟕𝟑𝒙 − 𝟐𝟐. 𝟑𝟎𝟑𝟔 𝒎𝒎

Maximum out of straightness = |22.7576| + |−14.8994| = 37.657𝑚𝑚

-20

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x (mm)