SHEET 2

FLATNESS

1. A table of upright drilling machines 550x 550 mm was tested for flatness error. A

mesh consists of 4 similar squares of 250 mm side length is drawn on the table.

A sensitive level of constant 0.04 mm/m is used to measure the level of each

point relative to the preceding one as listed on the following table. Using SEMI-

ANALYTICAL method, determine the maximum out of flatness.

Position ba cb ed fe

Reading (div) -3 2 4 1

Position hg ih da gd Reading (div) l 2 2 4

𝑐 = 0.04 mm/m = 0.01 mm/250mm

1- Reading

2- Accumulative

3-make the lower left corner =0

4-make the upper right corner =0

5-rotate about the diagonal

Maximum out of flatness = (3.5+1.5)× 0.01=0.05 mm = 50 microns

2. A Rotchdal arm tester having 200 mm leg distance was calibrated on three

points x, y, z on a line. Their reading at point Z was 0.1 mm. The same line is

divided into 4 equal distances to be checked using a sensitive level. The

readings were 1 and 3 μm respectively on the line xy and were -1 and -3 μm

respectively on the line yz

The calibrated Rotchdal arm is used to check the flatness of a surface shown in Fig.

The readings

Line abc Afe edc Bdh ehk bcl hjl cjk

Readings(μm) 15 -6 5 -10 -3 106 116 -83

Find the height readings relative to the plane abf, then find the out of flatness.

position x x1 y y1 z

reading relative to the preceding one 0 1 3 -1 -3

Accumulative (mm) 0 1 4 3 0

𝑅 = 2𝑞 − 𝑝 + 𝑦

Where:

R: level of the point under the dial indicator plunger

q: level of the point under the middle leg

p: level of the point under the end leg

y: dial indicator reading

get the dial reading

0 = 2 × 4 − 0 + 𝑦

𝑦 = −8 𝜇𝑚

Correction

𝑐 = 𝑅 − 𝑅𝑟𝑒𝑓 = 100 − (−8) = 108𝜇𝑚

Line abc Afe edc Bdh ehk bcl hjl cjk

Readings(μm) 15 -6 5 -10 -3 106 116 -83

Rcorrected=R-C -93 -114 -103 -118 -111 -2 8 -191

abe c = 2b -a + Rabe c -93

afe e = 2f - a + Rafe e -114

edc c = 2d - e + Redc d -52

bdh h = 2d - b + Rbdh h -222

ehk k = 2h - e + Rehk k -441

bcl l = 2c - b + Rbcl l -188

hjl l = 2j - h + Rhjl j -209

cjk k = 2j - c + Rcjk k -514

�̅� =3400

10= 340

𝑦 =1558.8

10= 155.88

𝑧̅ =−1319

10= −131.9

𝐸𝑟𝑟𝑜𝑟 𝑖𝑛 𝑟𝑒𝑎𝑑𝑖𝑛𝑔𝑠 =−514 − −441

−514= 0.142

Error within ±0.142 from the reading

𝑎 =∑ 𝑦′2 ∑ 𝑥′𝑧′ − ∑ 𝑥′𝑦′ ∑ 𝑦′𝑧′

∑ 𝑥′2 ∑ 𝑦′2 − (∑ 𝑥′𝑦′)2

𝑎 = −0.4854

𝑏 =∑ 𝑥′2 ∑ 𝑦′𝑧′ − ∑ 𝑥′𝑦′ ∑ 𝑥′𝑧′

∑ 𝑥′2 ∑ 𝑦′2 − (∑ 𝑥′𝑦′)2

𝑏 = −0.34669

𝑧𝑖′′ = 𝑎𝑥𝑖

′ + 𝑏𝑦𝑖′ = −0.4854𝑥′ − 0.34669𝑦′

𝛿𝑖 = 𝑧𝑖′ − 𝑧𝑖

′′

Maximum out of flatness = |126.4887| + |−27.178| = 153.667 microns

Calculations:

x Y z x' =x-�̅�

y' z' x'2 y'2 x'y' y'z' x'z' z'' delta

a 0 0 0 -340 -155.88 131.9 115600 24298.57 52999.2 -20560.6 -44846 219.078 -87.178

b 200 0 0 -140 -155.88 131.9 19600 24298.57 21823.2 -20560.6 -18466 121.998 9.901963

f 100 173.2 0 -240 17.32 131.9 57600 299.9824 -4156.8 2284.508 -31656 110.4913 21.40867

c 400 0 -93 60 -155.88 38.9 3600 24298.57 -9352.8 -6063.73 2334 24.91804 13.98196

e 200 346.4 -114 -140 190.52 17.9 19600 36297.87 -26672.8 3410.308 -2506 1.904621 15.99538

d 300 173.2 -52 -40 17.32 79.9 1600 299.9824 -692.8 1383.868 -3196 13.41133 66.48867

h 400 346.4 -222 60 190.52 -90.1 3600 36297.87 11431.2 -17165.9 -5406 -95.1754 5.075379

k 700 346.4 -441 360 190.52 -309.1 129600 36297.87 68587.2 -58889.7 -111276 -240.795 -68.3046

l 600 0 -188 260 -155.88 -56.1 67600 24298.57 -40528.8 8744.868 -14586 -72.162 16.06196

j 500 173.2 -209 160 17.32 -77.1 25600 299.9824 2771.2 -1335.37 -12336 -83.6687 6.568671

sum 3400 1558.8 -1319 0 -5.7E-14 0 444000 206987.9 76208 -108752 -241940

3. The flatness of a surface shown in Fig. was tested using a Rotchdal arm tester having 200 mm leg distance. The readings were:

Line Abc Afe edc Bdh ehk bcl hjl cjk

Readings(μm) 15 -6 5 -10 -3 106 116 -83

The Rotchdal arm tester was calibrated on three points x, y, z on a line. Its readings

at point Z was 0.1 mm while the reading of a sensitive level was 8 μm on line xy and

was -10 μm on line yz. Find the height readings relative to the plane ace.

position x y z

reading relative to the preceding one 0 8 -10

Accumulative (mm) 0 8 -2

𝑅 = 2𝑞 − 𝑝 + 𝑦

get the dial reading

−2 = 2 × 8 − 0 + 𝑦

𝑦 = −10 𝜇𝑚

𝑐 = 𝑅 − 𝑅𝑟𝑒𝑓 = 100 − (−10) = 110𝜇𝑚

Line abe Afe edc Bdh ehk bcl hjl cjk

Readings 15 -6 5 -10 -3 106 116 -83

Rcorrected=R-C -95 -116 -105 -120 -113 -4 6 -193

abe c = 2b -a + R abe b 47.5

afe e = 2f - a + R afe f 58

edc c = 2d - e + Redc d 52.5

bdh h = 2d - b + R bdh h -73

ehk k = 2h - e + R ehk k -259

bcl l = 2c - b + R bcl l -51.5

hjl l = 2j - h + R hjl j -65.25

cjk k = 2j - c + R cjk k -323.5

𝐸𝑟𝑟𝑜𝑟 𝑖𝑛 𝑟𝑒𝑎𝑑𝑖𝑛𝑔𝑠 =−323.5 − −259

−323.5= 0.142

Error within ±0.2 from the reading

�̅� =3400

10= 340

𝑦 =1558.8

10= 155.88

𝑧̅ =−290.75

10= −29.075

𝑎 =∑ 𝑦′2 ∑ 𝑥′𝑧′ − ∑ 𝑥′𝑦′ ∑ 𝑦′𝑧′

∑ 𝑥′2 ∑ 𝑦′2 − (∑ 𝑥′𝑦′)2

𝑎 = −0.29175

𝑏 =∑ 𝑥′2 ∑ 𝑦′𝑧′ − ∑ 𝑥′𝑦′ ∑ 𝑥′𝑧′

∑ 𝑥′2 ∑ 𝑦′2 − (∑ 𝑥′𝑦′)2

𝑏 = −0.19137

𝑧𝑖′′ = 𝑎𝑥𝑖

′ + 𝑏𝑦𝑖′ = −0.29175𝑥𝑖

′ − 0.19137𝑦𝑖′

𝛿𝑖 = 𝑧𝑖′ − 𝑧𝑖

′′

Maximum out of flatness = |73.21953| + |−99.9508| = 173.1703microns

Calculations:

x Y z x' =x-�̅�

y' z' x'2 y'2 x'y' y'z' x'z' z'' delta

a 0 0 0 -340 -155.88 29.075 115600 24298.57 52999.2 -4532.21 -9885.5 129.0258 -99.9508

b 200 0 47.5 -140 -155.88 76.575 19600 24298.57 21823.2 -11936.5 -10720.5 70.67576 5.899244

f 100 173.2 58 -240 17.32 87.075 57600 299.9824 -4156.8 1508.139 -20898 66.70547 20.36953

c 400 0 0 60 -155.88 29.075 3600 24298.57 -9352.8 -4532.21 1744.5 12.32576 16.74924

e 200 346.4 0 -140 190.52 29.075 19600 36297.87 -26672.8 5539.369 -4070.5 4.385188 24.68981

d 300 173.2 52.5 -40 17.32 81.575 1600 299.9824 -692.8 1412.879 -3263 8.355472 73.21953

h 400 346.4 -73 60 190.52 -43.925 3600 36297.87 11431.2 -8368.59 -2635.5 -53.9648 10.03981

k 700 346.4 -259 360 190.52 -229.925 129600 36297.87 68587.2 -43805.3 -82773 -141.49 -88.4352

l 600 0 -51.5 260 -155.88 -22.425 67600 24298.57 -40528.8 3495.609 -5830.5 -46.0242 23.59924

j 500 173.2 -65.25 160 17.32 -36.175 25600 299.9824 2771.2 -626.551 -5788 -49.9945 13.81953

sum 3400 1558.8 -290.75 0 -5.7E-14 0 444000 206987.9 76208 -61845.4 -144120

4. A Rotchdal arm tester having 200 mm leg distance was calibrated on three

points x, y, z on a line. Its readings at point Z was - 0.1 mm. The same line is

divided into 4 equal distance to be checked using a sensitive level. The readings

were 1 and 3 u.m respectively on the line xy and were -2 and -4 u.m respectively

on the line yz The calibrated Rotchdal arm is used to check the flatness of a surface shown in

Find the height readings relative to the plane ace.

position x x1 y y1 z

reading relative to the preceding one 0 1 3 -2 -4

y accumulative 0 1 4 2 -2

𝑅 = 2𝑞 − 𝑝 + 𝑦

get the dial reading

−2 = 2 × 4 − 0 + 𝑦

𝑦 = −10 𝜇𝑚

Correction

𝑐 = 𝑅 − 𝑅𝑟𝑒𝑓 = −100 − (−8) = −92 𝜇𝑚

line abc afe edc bdh ehk bcl hjl cjk

Readings(μm) 15 -6 5 -10 -3 106 116 -83

Rcorrected=R-C 107 86 97 82 89 118 128 49

abe c = 2b -a + R abe b -53.5

afe e = 2f - a + R afe f -43

edc c = 2d - e + Redc d -48.5

bdh h = 2d - b + R bdh h 38.5

ehk k = 2h - e + R ehk k 166

bcl l = 2c - b + R bcl l 171.5

hjl l = 2j - h + R hjl j 41

cjk k = 2j - c + R cjk k 131

Line abc afe edc Bdh ehk bcl Hjl cjk

Readings( μm) 15 -6 5 -10 -3 26 36 -43

𝐸𝑟𝑟𝑜𝑟 𝑖𝑛 𝑟𝑒𝑎𝑑𝑖𝑛𝑔𝑠 =166 − 131

166= 0.2108

Error within ±0.2 from the reading

�̅� =3400

10= 340

𝑦 =1558.8

10= 155.88

𝑧̅ =272

10= 27.2

𝑎 =∑ 𝑦′2 ∑ 𝑥′𝑧′ − ∑ 𝑥′𝑦′ ∑ 𝑦′𝑧′

∑ 𝑥′2 ∑ 𝑦′2 − (∑ 𝑥′𝑦′)2

𝑎 = −0.4854

𝑏 =∑ 𝑥′2 ∑ 𝑦′𝑧′ − ∑ 𝑥′𝑦′ ∑ 𝑥′𝑧′

∑ 𝑥′2 ∑ 𝑦′2 − (∑ 𝑥′𝑦′)2

𝑏 = −0.34669

𝑧𝑖′′ = 𝑎𝑥𝑖

′ + 𝑏𝑦𝑖′ = −0.4854𝑥𝑖

′ − 0.34669𝑦𝑖′

𝛿𝑖 = 𝑧𝑖′ − 𝑧𝑖

′′

Maximum out of flatness = |379.5954| + |−246.278| = 625.8734microns

6. A table of upright drilling machines 500x500 mm was tested for flatness error by

using A Rotchdale arm having 200 mm leg distance. The mesh shown in the

accompanying figure was drawn on the m/c table. The arm was first placed on a

reference straight edge and its reading was observed as 2.15 mm. the readings

on table were as listed on the following table.

Position a be bfg afe bdh geh

Reading (mm) 2.39 2.05 1.95 2.44 2.15

R corrected 0.24 -0.1 -0.2 0.29 0

abc c = 2b -a + R abe b -0.12

afe e = 2f - a + R afe f 0.1

bfg g = 2f - b + R bfg g 0.22

geh h = 2e - g + R geh h -0.22

bdh h = 2d - b + R bdh d -0.315

𝑎 = −0.4854

𝑏 =∑ 𝑥′2 ∑ 𝑦′𝑧′ − ∑ 𝑥′𝑦′ ∑ 𝑥′𝑧′

∑ 𝑥′2 ∑ 𝑦′2 − (∑ 𝑥′𝑦′)2

𝑏 = −0.34669

𝑧𝑖′′ = 𝑎𝑥𝑖

′ + 𝑏𝑦𝑖′ = −0.00072𝑥𝑖

′ + 0.000115𝑦𝑖′

𝛿𝑖 = 𝑧𝑖′ − 𝑧𝑖

′′

Maximum out of flatness = |0.385962| + |−0.34512| = 0.73108𝑚𝑚

7. A Rotchdal arm tester having 200 mm leg distance was calibrated on three points x, y, z on a line. Its readings at point Z was 0.1 mm. The reading of a sensitive level on the same line needs to divide the line into parts as shown. The readings were 1 and 3 μm respectively on the line xy and were -1 and -3 μm respectively on the line yz

x____ xl _____ y ____ yl _____ z

The calibrated Rotchdal arm is used to check the flatness of a surface shown in Fig. The readings were:

Line a be afe edc Bdh ehk bcl hjl cjk

Readings(μm) 15 -6 5 -10 -3 106 116 -83

Find the height readings relative to the plane abf, then find the out of flatness.

8. A Rotchdal arm tester having 200 mm leg distance was calibrated on a reference

straight edge and its reading was 45 μm .The calibrated Rotchdal arm is used to

check the flatness of a surface shown in Fig. The readings were:

Line abc Afe edc Bdh ehk bcl hjl cjk

Readings(μm) 15 -6 5 -10 -3 106 116 -83

Find the height readings relative to the plane abf.

R corrected -30 -51 -40 -55 -48 61 71 -128

abe c = 2b -a + R abe c -30

afe e = 2f - a + R afe e -51

edc c = 2d - e + Redc d -20.5

bdh h = 2d - b + R bdh h -96

ehk k = 2h - e + R ehk k -189

bcl l = 2c - b + R bcl l 1

hjl l = 2j - h + R hjl j -83

cjk k = 2j - c + R cjk k -264

𝐸𝑟𝑟𝑜𝑟 𝑖𝑛 𝑟𝑒𝑎𝑑𝑖𝑛𝑔𝑠 =−264 − −189

−264= 0.284

Error within ±0.284 from the reading

𝑎 = −0.13967

𝑏 = 0.24466

𝑧𝑖′′ = 𝑎𝑥𝑖

′ + 𝑏𝑦𝑖′ = −0.13967𝑥𝑖

′ + 0.24466𝑦𝑖′

𝛿𝑖 = 𝑧𝑖′ − 𝑧𝑖

′′

Maximum out of flatness = |126.4887| + |−45.2562| = 171.7449μm

9. The following figure shows the sensitive level readings in divisions which

were taken on a square grid drawn on a planer machine. The grid overall

dimensions were 800x600 mm. The level constant was 0.04 mm/m.

It required to determine the maximum out of flatness of the plane table

surface analytically and semi analytic.

semi analytic

𝑐 = 0.04 𝑚𝑚/𝑚 = 0.008𝑚𝑚/200𝑚𝑚

6 3 -2 -6 7

-3 9 2 -4 -1

-6 -2 -7 -3 1

0 -1 -2 1 -2

Accumulative:

-3 0 -2 -8 -1

-9 0 2 -2 -3

-6 -8 -1 -4 -3

0 -1 -3 -2 -4

Make the left down corner = 0

-3 1 0 -5 3

-9 -9 1 1 1

-6 -7 -7 -1 1

0 0 1 1 0

Make the right upper corner = 0

0 4 3 -2 6

-7 3 6 3 3

-5 -6 2 0 2

0 0 -1 1 0

Rotate about the diagonal

0 3.25 1.5 -4.25 3

-6 3.2 5.5 1.75 1

-3 -4.75 2.5 -0.2 1

3 2.25 0.5 1.75 0

∑ 12

𝑐𝑜𝑟𝑟𝑒𝑐𝑡𝑖𝑜𝑛 =12

20= 0.6

-0.6 2.65 0.9 -4.85 2.4

-6.6 2.6 4.19 1.115 0.4

-3.6 -5.35 1.9 -0.8 0.2

2.4 1.65 -0.1 1.15 -0.6

Maximum out of flatness = |4.19| + |−5.4|=10.3

10.3 × 0.008 = 0.0824 𝑚𝑚