The testis is small until the onset of puberty. In normal males it will then increase to adult size during adolescence.
testicular volume in cubic cm =
= (0.71 * ((length in cm)^2) * (width in cm)
where:
• The formula for an adult is given in the next section.
• The reason for squaring the length (rather than using the breadth) could be due to (1) available data was 2 dimensional rather than 3 or (2) a child's testis is more round than oval.
According to Figure 10.9 (page 325) in Hall et al (1989).
|
age in years |
10%ile |
50%ile |
90%ile |
|
0 - 8.5 |
1.0 cc |
1.64 cc |
2.2 cc |
|
9.0 |
1.0 |
1.64 |
2.2 |
|
9.5 |
1.0 |
1.64 |
2.2 |
|
10.0 |
1.0 |
1.64 |
2.3 |
|
10.5 |
1.0 |
1.64 |
3.0 |
|
11.0 |
1.0 |
1.8 |
4.0 |
|
11.5 |
1.2 |
2.5 |
5.5 |
|
12.0 |
1.6 |
3.3 |
7.0 |
|
12.5 |
2.2 |
4.5 |
10.0 |
|
13.0 |
3.1 |
6.6 |
12.5 |
|
13.5 |
4.0 |
9.0 |
15.0 |
|
14.0 |
5.3 |
11.95 |
17.0 |
|
14.5 |
6.6 |
12.5 |
18.0 |
|
15.0 |
8.0 |
14.0 |
18.9 |
|
15.5 |
9.0 |
15.0 |
19.6 |
|
16.0 |
10.0 |
15.8 |
20.0 |
where:
• 1 cubic inch = 16.39 cubic centimeters
The following equations were derived from the above table using JMP:
10th percentile curve from 11 to 14.5 years =
= (-0.01616 * ((age) ^3)) + (1.008658 * ((age) ^2)) - (16.1873 * (age)) + 78.52836
10th percentile curve from 14.5 to 16 years =
= (-0.4 * ((age) ^2)) + (14.44 * (age)) - 118.66
50th percentile curve from 10.5 to 13.5 years =
= (0.1244 * ((age) ^3)) - (3.575 * ((age) ^2)) + (34.23365 * (age)) - 107.7229
50th percentile curve from 15.5 to 16 years =
= (-0.5679 * ((age) ^2)) + (19.4746 * (age)) - 150.4175
90th percentile curve from 9.5 to 13 years =
= (0.0222 * ((age) ^3)) + (0.1476 * ((age) ^2)) - (8.8556 * (age)) + 53.9357
90th percentile curve from 13 to 16 years =
= (0.2222 * ((age) ^3)) - (10.4857 * ((age) ^2)) + (165.93 * (age)) - 860.733
Specialty: Urology
