AGE OF THE EARTH—HOLMES 
367 
igltse 5 2 14\:/1 18 
° \ 
16 \ 
7 a 
1 
it? * 
a 13 
15 
19 
6 
2000 
3000 
237 
13 16 
17 
17 15 
t in Millions of Year 
4000 
Ficure 5.—Each of the curves numbered on the left corresponds (like AB in figure 4) to a 
pair of lead samples, of which No. 25 is one. The curves similarly based on No. 19 are 
numbered along the top. The numbers are those of Nier’s lead samples as listed in Holmes 
(1946). Each intersection, marked by a dot, provides a solution for x, and ¢,. 
years. For each assigned value of 
t, we have: 
ar—xr=b—y for a sample of lead of age tn 
and 
a’r' —xr'=b’ —y for a sample of lead of age 
whence 
b—6b’-+-a’r’—ar 
a 
in —F 
and 
y=b+rx—ar. 
Taking samples 25 and 18 (fig. 4) the 
following results are obtained: 
No. 25 No. 18 
Galena, Great Bear Galena, North Caro- 
Lake Pre-Cambrian lina Late Carbonif- 
erous 
a =15.93 18.43=a 
b =15.30 15.61=5b 
tm= 1,330 million years 220 million years=t, 
aes ae when t=2500 .16999=r’ 
x=14.11; y=14.87 
r=.30156 when ¢=3000 .22794=r’ 
{ x=12.40; y=14.24 
Nae he when t=3500 .30985=71’ 
x=10.59; y=13.18 
Plotting x against ¢, a curve AB is 
drawn; a point representing a solution 
for x and ?¢, lies somewhere on this 
curve or its continuation. Dealing in 
the same way with another pair of lead 
samples, such as No. 1 (Galena, Peru, 
Tertiary, 25 million years) and No. 19 
(Galena, Ivigtut, Greenland, Late 
Pre-Cambrian, 600 million years), a 
curve CD is similarly constructed. 
Where the two curves intersect at 
P, x=11.22 and t,=3,330 million years. 
In the same way we can plot y» 
against ¢ and construct two curves 
that intersect at y=13.71 and ¢,=3,330 
million years. The two pairs of curves 
based on the data for two pairs of lead 
samples thus yield a solution of the 
problem. 
Obviously a great many solutions 
can be obtained by this method. Fig- 
ure 5 illustrates the xt curves for pairs 
made up of No. 25 with each in turn 
of 14 younger lead samples, and for 
pairs made up of No. 19 with 11 
