 | Andrew Mackay - Mathematical instruments - 1811 - 162 pages
...110 inches on A. The REMARK. The above rule, generally, gives the periphery less than the truth.: and the square root of half the sum of the squares of the diameters, multiplied by 3.14, in general, makes the periphery greater ; hence, the mean of these two... | |
 | William Galbraith - Astronomy - 1827 - 412 pages
...product of the transverse axis into the conjugate axis multiplied by 0-785398. Periphery. — Multiply the square root of half the sum of the squares of the two axes by 3.141593, the product will be the periphery nearly. Examples for Exercise. 1. Required... | |
 | Thomas Hornby (land surveyor.) - Surveying - 1827 - 318 pages
...the rule (AB + CD) x 3'1416 -(30 + 20) x ^ = 78.5400 chains, the circumference nearly. RULE. Multiply the square root of half the sum of the squares of the two axes by 3. 1416, and the product will be the circumference nearly. — (This is Hutton's 3rd Rule.)... | |
 | John Bonnycastle - Geometry - 1829 - 256 pages
...the circumference of an ellipse, the transverse and conjugate diameters being known. RULE.* Multiply the square root of half the sum of the squares of the two diameters by 3.1416, and the product will be the circumference nearly. * Demon. • Let <=transverse diameter, r=conjugate,... | |
 | Thomas Curtis - Aeronautics - 1829 - 810 pages
...+ 10-7 5 190-7500. Answer. PROB. XV. To find the circumference of an ellipse. Rule 1. — Multiply the square root of half the sum of the squares of the two diameters, and the product will be the circumference nearly. Demonstration. — If t =: the transverse, с the... | |
 | Francis Walkingame - 1833 - 204 pages
...an Ellipse, the transverse and conjugate diameters being given. RULE. Multiply the square Fig. 10. root of half the sum of the squares of the two diameters...and the product will be the circumference nearly.* (1) What is the circumference of an ellipse whose transverse diameter is 24, and conjugate 18 ? (2)... | |
 | Samuel YOUNG (of Manchester.) - 1833 - 272 pages
...area of the ring formed by these circles ? PROBLEM XIV. To find the circumference of an Ellipse. RULE. The square root of half the sum of the squares of the 2 diameters multiplied by 3- 141 6 will be the circumference, nearly. (1) The transverse diameter is... | |
 | Tobias Ostrander - Measurement - 1834 - 182 pages
...PROBLEM v. The transverse and conjugate diameters are given, to find the circumference. Rule — Multiply the square root of half the sum of the squares of...transverse diameter of an ellipse is 24, and the conjugate 20— Required its circumference. Ans. 69,398 — . 2. The axes of an ellipse are 60 and 40 — What... | |
 | William Galbraith - Astronomy - 1834 - 454 pages
...product of the transverse axis into the conjugate axis multiplied by 0.785398. Periphery. — Multiply the square root of half the sum of the squares of the two axes by 3.141593, the product will be the periphery nearly. I. IMPERIAL LAND MEASURE. Marked, Square... | |
 | Nathan Scholfield - 1845 - 894 pages
...18. (12 + 9) + 314159 = 21 X 3,14159 = 65,9735 equal the circumference nearly. OR RULE 2. Multiply the square root of half the sum of the squares of the two axes by *, and the product will be nearly = the circumference. Ex. Taking the same example as before,... | |
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Multiply the square root of half the sum of the squares of the two axes by *, and the product will be nearly = the circumference. Ex. Taking the same example as before, we hare /24' + IT \/ = — X 3,14159 = 66,6433= the circumference nearly. 
