to ascertain a Standard of Weight and Measure . 145 
I therefore say, this interval was 5 inches correctly, to within 
less than the twenty thousandth part of an inch, on this scale. 
Measurement of the Cube , viz. of the Side 1. (See the Figure .) 
Q Q 
h * 
* 4 *4 5 8 $ ? 
a to c == 5 — >0115 ~ 
c to d = 5 — ,0105 
b to d 
5 
Inches. Inches. 
From a to 6 = 5 — ,0114 therefore = 4,9886 
- = 4,9885 
= 4.9895 
,0113 - ~ =4,9887 
The Side 2. 
From a to b = 5 — ,0106 
a to c = 5 — ,0098 
ctod = 5 — ,0102 - 
b to d = 5 — ,0112 
Mean. 
>= 
Inches. 
= 4,98882- 
= 4 > 9 8 94 
= 4 > 99° 2 
= 4,9898 
== 4,9888 
' 
>= 
4 > 9 8 955 
i . • / # ;? 
Height of the Cube, from Side l to Side 2. 
From a to a- 
Inches. 
— 5 — ,0110 - - 
Inches. 
= 4,9890“ 
b to b 
— 5 — ,0105 
= 4.9895 > 
c to c 
= 5 — ,0107 - - 
= 4.989s 
d to d 
= 5 — ■ ,0108 
= 4 ,g 892 _ 
/ 0 ■ a 0 
4’989 2 5 
• It cannot escape notice, that all these measures were something less than 5 inches, 
the quantity proposed : it arose from this, Mr. Trough ton informs me, he was more 
u 
MDCCXCVIII. 
