498 Proceedings of the Royal Society 
dynamical estimate (from the measured strain) of what the cur- 
vature must have been between the ship and the water, I find 
that its inclination in the water, when the ship’s speed was nearly 
6J miles per hour, must have been about 6|°, that is to say, the 
incline was about 1 in 8|-. Thus the length of cable, from the 
ship to the bottom, when the water was 2 miles deep, must have 
been about 17 miles. 
The whole amount (14 cwt.) of fluid resistance to the motion of 
this length of cable through it, is therefore about -81 of a cwt. per 
mile. The longitudinal component velocity of the cable through 
the water, to which this resistance was due, may be taken, with but 
very small error, as simply the excess of the speed of paying out 
above the speed of the ship, or about 1 mile an hour. Hence, 
to haul up a piece of the cable vertically through the water, at 
the rate of 1 mile an hour, would require less than 1 cwt. for over- 
coming fluid friction, per mile length of the cable, over and above 
its weight in water. Thus fluid friction, which for the laying 
of a cable performs so valuable a part in easing the strain with 
which it is paid out, offers no serious obstruction, indeed, scarcely 
any sensible obstruction, to the reverse process of hauling back, if 
done at only 1 mile an hour, or any slower speed. 
As to the transverse component of the fluid friction, it is to 
be remarked that, although not directly assisting to reduce the 
egress strain, it indirectly contributes to this result ; for it is the 
transverse friction that causes the gentleness of the slope, giving 
the sufficient length of 17 miles of cable slipping down through the 
water, on which the longitudinal friction operates, to reduce the 
egress strain to the very safe limit found in the recent expe- 
dition. In estimating its amount, even if the slope were as 
much as 1 in 5, we should commit only an insignificant error, if 
we supposed it to be simply equal to the weight of the cable 
in water, or about 14 cwt. per mile for the 1865 Atlantic cable. 
The transverse component velocity to which this is due may be esti- 
mated with but insignificant error, by taking it as the velocity of 
a body moving directly to the bottom in the time occupied in 
laying a length of cable equal to the 17 miles of oblique line 
from the ship to the bottom. Therefore, it must have been about 
2 miles in 17-f- 6^^= 2‘61 hours, or ’8 of a mile per hour. It is not 
