Corrections

(updated March 9, 1999), for the 1991 Edition of:

Applied Drilling Engineering, by Adam T. Bourgoyne Jr., Keith K.

Millheim, Martin E. Chenevert and F. S. Young Jr., SPE Textbook Series,
Vol.2
e 2.203a2D - 1

p.33, Eq.1.19
change to:

e 2.303a2D - 1

change to:

+
pH = log[H ]
+
pH = - log[H ]

p.44, Eq.2.4

p.48, right column, line 4 from bottom

change to:
p.49, left column, line 7:
change to:

... molecular weight ...

... atomic weight ...
... added to 100 g of
... added to 100 cm3 of

p.50, left column, lines 12 - 13 from bottom ... molecular weight ...
change to:
... atomic weight ...
p.51, right column, line 9
change to:

... CaCO3 ...

... CaSO4 ...

p.52, Table 2.5, right column, line 3 from bottom

279,500 mg/L

change to:

283,300 mg/L

p.52, Table 2.5, right column, line 2 from bottom

change to:

283,300 mg/L
297,200 mg/L

p.60, right column, line 24 from bottom

change to:

... 30 lbm/100 sq ft...

... 30 lbf/100 sq ft...

p.69, left column, line 22 from bottom:

given by qufm , qufw , and qufB ...

change to:
p.82, Exercise 2.3

given by qufum , qufuw , and qufuB ...

+
... Mg ...

++
... Mg ...

change to:
p.82, Exercise 2.6
change to:

... answer: 3.18; 3.00; 3.00; 157,024; 175,500; 15.7; 65.4.

... answer: 3.198; 3.00; 3.00; 157,440; 175,500; 15.7; 65.4.

p.83, Exercise 2.30

... answer:

change to:

... answer:

6.8 lbm/min of clay, 2.7 lbm/min of

deflocculant, 7.4 gal/min of water,
and 3.01 lbm/min of bariye.
10.02 lbm/min of clay, 4.01 lbm/min
of deflocculant, 9.78 gal/min of
water, and 20.81 lbm/min of barite.

p.88, left column, line 15 from bottom:

2.5 mL of free water. ...

change to:

3.5 mL of free water. ...

p.94, right column, lines 18-20 from bottom:

the thickening time is inceased,

not reduced!
(with increasing diatomaceous earth concentrations)

p.134, left column, line 6

change to:

s2)
1 lbf (454 g/lbf)(980 cm/
100 sq ft
100(30.48 cm/ft)2
s2)
1 lbf (454 g/lbf)(980 cm/
100 sq ft
(30.48 cm/ft)2

p.135, left column, line 4 from bottom

change to:

... (r1)1/(r2)2 ...

... (r1)1/(r2)1 ...

p.136, Fig.4.25. Incorrect velocity profile.

Curvature should be in the opposite direction.

change to:

... dr y ...
dy
... d y ...
dy

p.142, right column, line 6 from bottom

q = vdA = vWdy..

p.141, right column, line 8 from bottom

change to:

q=

vdA =
o

p.143, left column, line 10

change to:

vWdy ...
o

... lot ...

... slot ...

2
pf = dpf L = f v
25.8de
dL
2
dp
f L = f v
pf =
25.8de
dL
0.75

p.152, First equation in 2nd column

change to:

... p

p.161, left column, the numerator of line 15

change to:
p.165, right column, line 4 from bottom

...
0.25
... p ...

q = vdA = vWdy
h

q=
change to:

vdA =
o

vWdy
o

3 dp
f v PWh
q = - Wh
12 dL
2

p.165, Eq.4.90a

3 dp
f v PWh
q = Wh
12 dL
2

change to:
p.166, right column, denominator of Eq.4.94

6d 4 - 4(d2 - d1)2(d22 - d12 )

- 6d 4 - 4(d2 - d1)2(d22 - d12 )

change to:
p.166, denominator of line 2 from bottom

6(10)4 - 4(12 - 10.75)2(122 - 10.752)

- 6(10)4 - 4(12 - 10.75)2(122 - 10.752)

change to:
p.166, right column, the last line

= + 0.489 ft/s

change to:

p.167, left column, line 3

= - 0.4865 ft/s
=

2(0.489 + 1.0/2)
= 0.00101 psi/ft
1,000(12 - 10.75)2

change to:
p.170, right column, line 19
change to:
p.182, right column, the numerator of line 6

2(- 0.4865 + 1.0/2)

= 0.0000175psi/ft
1,000(12 - 10.75)2

(qa)1 = fa(qt)1 = (0.8445 fa ) ft/s.

(qa)1 = fa(qt)1 = (0.8445 fa ) ft3/s.
827(0.79)(0.25)

change to:

928(9)(0.25)

p.185, Exercise 4.24(e)

... answer: ... K = 1 eq cp.

change to:

... answer: ... K = 1 eq poise.

p.138, right column, line 5 from bottom:

... r - 0 yields
... r 0 yields

change to:

v = mean flow velocity, ft/s

p.140, right column, line 12

v = mean flow velocity, ft/s

change to:
p.140, right column, line 7 from bottom

... = 4,680 psi.

change to:

... = 4,680 psig.

p.327, Eq.7.17b

change to:

M =

2( z )max
+ F a y + F sc x
Ed n

M =

2( z )max EI
+ F a y + F sc x
Ed n

p. 354 Equation 8.6 for the maximum inclination angle, for the build-and-hold case, is
not limited to X3 < r1. It is also valid for X3 r1.
p. 357 Equation 8.22, in the first and third line, (r1 - r2) should be changed to (r1 + r2).
The correct version is as follows:
=180 - arc tan

D4 - D1
X4 - r 1 + r 2

- arc cos

r 1 +r 2
D4 - D1

p.361, right column, line 8 from bottom

change to:
p.361, right column, line 6 from bottom

sin

...

arc tan

D4 - D1
x4 - r 1 + r 2

cos 20 + 20
2
...
...

sin 20 + 20
2
...

D3 = ... = 99.996 ft,

change to:

D3 = ... = 99.966 ft,

p.361, right column, line 5 from bottom

change to:

D = 8,099.99 + 99.996 = 8,199.996 ft.

D = 8,099.996 + 99.966 = 8,199.962

ft.
M4 = ... cos 20 + 21 = 1.53 ft.
2

p.362, left column, the first line

M4 = ... sin 20 + 21 = 1.53 ft.

2

change to:

p.362, left column, line 4

ft.

D = 8,199.996 + 99.90 = 8,299.896

change to:

D = 8,199.962 + 99.90 = 8,299.862

ft.
M5 = ... cos 21 + 22 = 2.24 ft.
2

p.362, left column, line 9

M5 = ... sin 21 + 22 = 2.24 ft.

2

change to:
p.362, left column, line 12
ft.

D = 8,299.896 + 99.81 = 8,399.706

change to:

D = 8,299.8962 + 99.81 = 8,399.672 ft.

p.362, left column, line 9 from bottom

change to:

r1 =

180 100 ft
+
= 5,730 ft .

r1 =

180 100 ft
*
= 5,730 ft .

Di = D i /2 cos i -1 +cos i F i

p.365, Eq.8.37
change to:

di = D i /2 cos i -1 +cos i F i

p. 370 Equations 8.50, 8.51 and 8.52 are not Ouija Board Nomograph solutions. Eqs.
8.51 and 8.52 are usually highly inaccurate.
5

If Eq. 8.50 is further simplified by letting sin and cos 1 we then get the
following Ouija Board Solution:
= arc tan

sin
+ cos

Another form of the Ouija Board Nomograph is:

= arc cos N cos -

The Ouija Board solution is not very accurate if or is large.

p.370, right column, line 6 from bottom

change to:
p.371, right column, line 6
change to:
p. 372, Fig. 8.34, Step Three
change to:

2.4
i =
100 ft = 4.00 /100ft .
Lc
60 ft
=

2.4
i =
100 ft = 4.00/100ft .
Lc
60 ft

Eq. 8.51
Eq. 8.55
= 10 o
= 12

p.373, left column, line 4

... from N40E to N50E ...

change to:

... from N40E to N52E ...

p.373, left column, line 6

... is N30E , ...

change to:

... is N28E , ...

p.374, right column, line 25 from bottom

change to:

... (N14W or 346o) ...

... (N15W or 345o) ...

p.374, right column, line 23 from bottom

... be at 97o right ...

change to:
p. 375, Fig. 8.36

... be at 79o right ...

The new direction is 228 + 79 = 307 (not 325 ).

Without reverse torque, the tool face angle should be :

(228 + 97) = 325 (not 345).
To allow for the 20 reverse torque the total face angle should be set at 97 +
20 = 117 (as stated in the text). This would
correspond to a tool face
setting of 228 + 117 = 345 (not 346).
p.376, right column, line 20 from bottom:

S66E and 4.75o ...

change to:
p.376, right column, line 2 from bottom:
change to:
p.395, Fig.8.69
p.428, Fig.8.124 (ordinate)
change to:
p.429, right column, the first line:
change to:

S66E and 4.75 ft ...

o
is 66 ...
o
is 62 ...
#39 is printed as #37.
... (lb/sq in.)
... (lbf.in2)

... the collar in lbm/cu ft in air,

... the collar in lbm/linear ft in air,

p.443, The titles of Fig.8.149 and Fig.8.150 should be interchanged.

p.453, Exercises 8.2

... Answer. (1)...26o; (2)...4,380 ft, ... 11,255 ft;

(3)...3,402; 937.73; 1,206; and 2,474.94 ft,

4)...5,636.19; 6,248.19, and 9,141.22ft,...
change to:

... Answer. (1)...25.3o; (2)...4,137 ft, ... 11,235 ft;

(3)...366; 1,013; 1,273; and 2,502 ft,
(4)...5,649; 6,258, and 9,134 ft,...

Case 1 Time to Drill = 0.016( DM) - 91 Maximum Inclination Factor

Case 2 Time to Drill = 0.013( DM) - 95 Maximum Inclination Factor
Case 3 Time to Drill = 0.015(
DM) - 42 Maximum Inclination Factor

p.455, Table 8.27

change to:
Case 1 Time to Drill = 0.016( DM) - 91 Maximum Inclination Factor
Case 2 Time to Drill = 0.013( DM) - 95 Maximum Inclination Factor
Case 3 Time to Drill = 0.015( DM) - 42 Maximum Inclination Factor
p.475, left column, lines 4-5:
change to:
p.475, left column, lines 6-7:
change to:
p.475, right column, line 15
change to:
p.482, Eq. (B-15b)
change to:

... by subtracting the 600=rpm dial reading

from the 300-rpm dial reading; ...
... by subtracting the 300-rpm dial reading
from the 600-rpm dial reading; ...
... by subtracting the 300-rpm dial reading
from the plastic viscosity.
... by subtracting the plastic viscosity
from the 300-rpm dial reading.
...1.725 ...
...1.7245 ...

o = ...
= ...