 | Charles Talbot Porter - 1868 - 154 pagina’s
...89-25 6256-1 99-5 7775-6 79-25 4932-7 89-5 6291-2 99-75 7814-7 79-5 4963-9 89-75 6326-4 1007854-0 79-75 4995-1 906361-7 805026-5 90-25 6397-1 RICHARDS STEAM-ENGINE...and multiply it by the logarithm + 1 found as above ; the product will be the mean pressure through the stroke. EXAMPLE. — Suppose the length of the... | |
 | Charles Talbot Porter - 1874 - 294 pagina’s
...logarithm of the nearest smaller number ; as, for example, the logarithm of 1 '28 is .223 + -024 = '247. Then find the terminal pressure, by dividing the initial...and multiply it by the logarithm + 1 found as above ; the product will be the mean pressure through the stroke. Example. — Suppose the length of the... | |
 | Charles Talbot Porter - 1880 - 220 pagina’s
...steam is admitted; find in Table No. IV. the logarithm of the number nearest to the quotient, to which add 1, the sum is the ratio of the gain; then find...multiply it by the logarithm -|- 1, found as above; the product will be the mean pressure through the stroke. EXAMPLE. — Suppose the length of the stroke... | |
 | Emory Edwards - 1888 - 540 pagina’s
...logarithm of the number nearest to the quotient, to which add 1, the sum is the ratio of the gfain; then find the terminal pressure by dividing the initial...steam is admitted, and multiply it by the logarithm; + i, found as above; the product will be the mean pressure through the stroke. No. Logarithm. No. Logarithm.... | |
 | Emory Edwards - 1888 - 540 pagina’s
...logarithm of the number nearest to the quotient, to which add 1, the sum is the ratio of the grain; then find the terminal pressure by dividing the initial...which the steam is admitted, and multiply it by the logarithms + i, found as above; the product will be the mean pressure through the stroke. No. Logarithm.... | |
 | Charles Talbot Porter - 1888 - 324 pagina’s
...logarithm of the nearest smaller number ; as, for example, the logarithm of 1-28 is -223 + -024 = -247. Then find the terminal pressure, by dividing the initial...the stroke during which the steam is admitted, and muptiply it by the logarithm + 1 found as above; the product will be the mean pressure through the... | |
 | Emory Edwards - 1890 - 570 pagina’s
...steam is admitted ; find in the table the logarithm of the number nearest to the quotient, to which add 1, the sum is the ratio of the gain ; then find...multiply it by the logarithm + 1, found as above; the product will be the mean pressure through the stroke. Example 1. — Suppose the length of the... | |
 | Emory Edwards - 1891 - 554 pagina’s
...steam is admitted; find in the table the logarithm of the number nearest to the quotient, to which add 1 — the sum is the ratio of the gain; then find...which the steam is admitted, and multiply it by the logarithms + 1, found as above; the product will be the mean pressure through the stroke. No. Logarithm.... | |
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Table the logarithm of the number nearest to the quotient, and to this add 1 ; the sum is the ratio of the gain. Then find the terminal pressure, by dividing the initial pressure by the proportion of the stroke during which the steam is admitted, and... 