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" ... the canal. The more approximately nodal character of the tides on the north coast of the English Channel than on the south or French coast, and of the tides on the west or Irish side of the Irish Channel than on the east or English side, is probably... "
Philosophical Magazine - Page 113
1880
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Proceedings, Volume 10, Issue 1878

Royal Society of Edinburgh - Science - 1880
...represented by this factor, taken into account along with frictional resistance, in virtue of which the tidw of the English Channel may be roughly represented...to south. The problem of standing oscillations in n endless rotating canal is solved by the following equations — h = H {c-'' cos (mx - crt) - &...
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Proceedings of the Royal Society of Edinburgh, Volume 10

Royal Society of Edinburgh - Science - 1880
...period of the rotaV0"/ tion I — \ and the time required to travel at the velocity - across \ <a J m the canal. The more approximately nodal character...rotating canal is solved by the following equations — h = H {(-* cos (mx - at) - (! (cos mx + at)} (17) If wo give ends to the canal we fall upon the unsolved...
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Proceedings of the Royal Society of Edinburgh, Volume 10

Royal Society of Edinburgh - Science - 1880
...period of the rotation ( — |, and the time required to travel at the velocity — across \ <a J m the canal. The more approximately nodal character...rotating canal is solved by the following equations — • (17) If we give ends to the canal we fall upon the unsolved problem referred to above of tesseral...
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Philosophical Magazine

Physics - 1880
...follow from the interpretation of this factor with different particular suppositions as to (O_v — ], the period of the rotation ( — ), and the time required...canal is solved by the following equations: — h = H ^ e~'y cos (mx — at) — e*,(cos mx + at) J ; -. =0. * If we give ends to the canal, we fall...
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Mathematical and Physical Papers, Volume 1

Joseph Larmor - Electricity - 1929
...(27T/0-), the period of the rotation (27r/a>), and the time required to travel at the velocity <r/m across the canal. The more approximately nodal character...canal is solved by the following equations : — h = H {e-'v cos (mx - crt) — el (cos mx + <rf)\ u = H*-— {e~ly cos (mx — at) + ely cos (mx +...
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