390 
MR. LUBBOCK ON THE TIDES IN THE PORT OF LONDON. 
Hence by column B. Table V. 
22.80 = D+ 1.3788 E 
19.43 = D+ .6212 E 
Whence D = 16.68 and log. E = .64819, E — 4‘448 
When the moon passes the meridian at 
h. h. 
2 
the height of high water 
= 16.68 + 4-448{l + .3788} =22.80 
3 or 1 
do. 
do. 
= 16.68 + 4.448{cos8° 7' + .3788 cos 22°} =22.64 
4 or 0 
do. 
do. 
= 16.68 + 4.448{cos 15° 25' + .3788 cos 44° 30'}= 22. 1 7 
5 or 1 1 
do. 
do. 
= 16.68 + 4.448 {cos 20° 45' + .3788 cos 69° 30'} = 21.42 
6 or 10 
do. 
do. 
= 16.68 + 4.448 {cos 22° 2' - .3788 sin 8°} = 20.56 
7 or 9 
do. 
do. 
= 16.68 + 4.448 {cos 15° 45' -.3788 sin 44°} = 19.79 
8 
do. 
do. 
= 16.68 + 4.448 { 1 - .3788} = 19.43 
The following Table gives a comparison between theory and observation. 
Time of 
Moon’s 
Transit. 
Height of High Water. 
Error 
of 
Calculation. 
Observed. 
Calculated. 
h. 
Ft. 
Ft. 
Ft. 
0 
22.46 
22.17 
— 
.29 
1 
22.72 
22.64 
— 
.08 
2 
22.80 
22.80 
0 
3 
22.59 
22.64 
+ 
.05 
4 
22.10 
22.17 
+ 
•07 
5 
21.28 
21.42 
+ 
.14 
6 
20.37 
20.56 
+ 
• 19 
7 
19.56 
19-79 
+ 
.23 
8 
19 43 
19.43 
0 
9 
20.10 
19-79 
— 
.31 
10 
20.92 
20.56 
— 
.36 
11 
21.85 
21.42 
— 
.43 
I shall now endeavour to show that the expression for tan (2^ — 2 X), p. 38/, 
line 13, does not satisfy the observations. 
When the moon passes the meridian at 2 o’clock 6 — — X + \ = 0. 
tan (2 0 — 2 A ; ) 
— C' sin 2 l , 
■ ,,, -1 7 n nearly. 
1 + C cos 2 L J 
When the moon passes the meridian at 8 o’clock, & — 6 t — X + \ = 90° 
— C' sin 2 l 
tan (2 0, — 2 A ( ) = 
. m n 3 , o . 
1 + C cos 2 l 
m,n? 
nearly. 
