 | Royal Society of Edinburgh - Science - 1880 - 864 pages
...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) - &... | |
 | William Thomson Baron Kelvin - Mathematics - 1910 - 581 pages
...(27r/<r), the period of the rotation (27r/&>), and the time required to travel at the velocity cr/m across the canal. The more approximately nodal character...canal is solved by the following equations : — h = H [e-ty cos (mas - at) - e^ (cos mx + at)} u = H^— {e~ly cos (mx — at) + ely cos (mx + at)} ...(17).... | |
 | William Thomson Baron Kelvin - Mathematics - 1910 - 588 pages
...(2ir/<r), the period of the rotation (2ir/ta), 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-lv cos (mx - <rt) — el» (cos mx + <rt)} u = H — {e-lv cos (mx — <rt) + e** cos (mx... | |
 | William Thomson Baron Kelvin - Mathematics - 1910 - 588 pages
...oscillation (2ir/a), the period of the rotation (27r/o>), and the time required to travel at the velocity a/m across the canal. The more approximately nodal character...canal is solved by the following equations : — h = H {e~l» cos (mx — at) — ely (cos mx + at)} u = H — {e"lt cos (mx — at) + ety cos (mx + at)}... | |
 | Royal Society of Edinburgh - Science - 1880 - 874 pages
...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... | |
 | Royal Society of Edinburgh - Science - 1880 - 850 pages
...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... | |
 | Physics - 1880 - 502 pages
...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... | |
 | Sir Joseph Larmor - Electric power - 1929 - 596 pages
...(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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... 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... 