DUB TO THE WEIGHT OE CONTINENTS, 
227 
The averages are not however estimated from a mean spheroidal surface, but from 
one which is far distant from the mean. 
The questions now to be determined are as follows—What is the proper greatest 
height and depression, estimated from a mean spheroid, which will bring out the above 
averages estimated from present sea-level, and what is the position of the mean 
spheroid with reference to the sea-level. 
From the solution of the problem considered in the note below/' it appears that, if 
* Conceive a series of straight harmonic undulations corrugating a mean horizontal surface, and 
suppose them to he flooded with water. This'will represent fairly well the undulations on the dried earth, 
and the water-level will represent the sea-level. 
Suppose that the average heights and depths of the parts above and below water are known, and 
that it is required to find the position of the mean horizontal surface with reference to the water-level, 
and the height of the undulations measured from that mean surface. 
Take an origin of coordinates in the water-level, the axis of x in the water-level and perpendicular to 
the undulations, and the axis of y measured upwards. 
Let 
y=h( cos a?—cos a ) 
be the equation to the undulations. 
1 (*+« 7i 
The average height of the dry parts is clearly ydx or -(sin a—a cos a). Similarly the average 
AClJ—a O' 
ll , Jl 
depth below water is -y_ 7 [sin (jr— a) — (jr—d) cos (tt— a)] or^—- sin a+ ( 7 r—a) cos a 
1t— a 
If the latter average be p times as great as the former 
ph cos «(- tan a —1 )=7& cos a( 
\a / V— 
a 
tan a 
This is an equation for determining a. 
Now I find that a=34° 30' gives p> = 8'983, which corresponds very nearly with_p=9 of the text above. 
This value of a corresponds with an average equal to *11657z for the height above water, and 1'0469/i 
for the depth below water. Now if we put 
l'04697i=3150 meters 
which gives '11657^=350 meters very nearly, 
we have 7^=3009 meters. 
The depth below water-level of the mean level is h cos 34° 30' or 2480 meters. 
The greatest height of the dry part above the water-level is 3009—2480 or 429 meters, and the 
greatest depth of the submerged part below water-level is 3009 + 2480 or 5489 meters. 
[After the proof-sheets of this paper had been corrected, Professor Stokes pointed out to me that, 
according to Rigaud (Cam. Phil. Soc., vol. 6), the area of land is about four-fifteenths of the whole area 
of the earth’s surface. Now, in the ideal undulations we are here considering the area above water is 
about one-tenth of the whole area; hence in this respect the analogy is not satisfactory between these 
undulations and the terrestrial continents. If I have not considerably over-estimated the average depth 
of the sea (and I do not think that I have done so), the discrepancy must arise from the fact that actual 
continents and sea-beds do not present in section curves which conform to the harmonic type; there must 
also be a difference between corrugated spherical and plane surfaces. 
The geological denudation of the land must, to some extent, render our continents flat-topped .—Added 
May Mh , 1882.] 
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