A Treatise of Practical Mathematics, Part 2W. & R. Chambers, 1842 |
From inside the book
Results 1-5 of 53
Page 3
... Altitude , Problems regarding Sideral and Mean Time , Problems regarding Culminations , Problems regarding Altitudes , Declinations , Latitudes , & c . , of Celestial Bodies , Methods of Determining Time , The Equation of Equal Altitudes ...
... Altitude , Problems regarding Sideral and Mean Time , Problems regarding Culminations , Problems regarding Altitudes , Declinations , Latitudes , & c . , of Celestial Bodies , Methods of Determining Time , The Equation of Equal Altitudes ...
Page 14
... altitudes of the principal triangles , and to calculate their areas by the simple rule in article 258 of Mensuration of Surfaces . - • Thus , by drawing the perpendicular BV on AC , and mea- suring it , the area of triangle ABC is AC BV ...
... altitudes of the principal triangles , and to calculate their areas by the simple rule in article 258 of Mensuration of Surfaces . - • Thus , by drawing the perpendicular BV on AC , and mea- suring it , the area of triangle ABC is AC BV ...
Page 23
... altitude of the triangle , and the quotient will be half the length of its base . ' A P B If a = area of the required part APC , 1 = AP , and α h = altitude of the triangle , then / = or / = h 2 a h ' Or , if A area of the given ...
... altitude of the triangle , and the quotient will be half the length of its base . ' A P B If a = area of the required part APC , 1 = AP , and α h = altitude of the triangle , then / = or / = h 2 a h ' Or , if A area of the given ...
Page 124
... altitude of the projectile . v2 The formulas to be used are h = r = 2h sin 2 e , t = 2 sin e√√ 2h 9 = 2 v g 2g ' sin e , and h'h sin2 e . Or r , t , and h ' , may sometimes be more easily found by logarithms , thus : - - Lr L2h + L ...
... altitude of the projectile . v2 The formulas to be used are h = r = 2h sin 2 e , t = 2 sin e√√ 2h 9 = 2 v g 2g ' sin e , and h'h sin2 e . Or r , t , and h ' , may sometimes be more easily found by logarithms , thus : - - Lr L2h + L ...
Page 125
... altitude and time of flight are found as in last problem . EXAMPLE . A ball was projected at an elevation of 54 ° 20 ′ , and was found to range 2000 feet ; required the initial velocity . h = r 2000 2 sin 2e 2 x · 947397 = 1055.5 and ...
... altitude and time of flight are found as in last problem . EXAMPLE . A ball was projected at an elevation of 54 ° 20 ′ , and was found to range 2000 feet ; required the initial velocity . h = r 2000 2 sin 2e 2 x · 947397 = 1055.5 and ...
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Common terms and phrases
actual range bastion body breadth bung diameter called cask centre constructed content in imperial cosec cosine course declination deflexion depth dial difference of latitude difference of longitude distance ditch divide divisor draw earth elevation equal EXAMPLE EXERCISES fathoms Find the content flank frustum gauge point glacis hence horizontal hypotenuse imperial bushels imperial gallons impetus inches inclination length logarithms longitude mean diameter measure meridian middle moon's multiply number of balls oblique observed opposite parallax parallel parapet perpendicular pile place of arms plane plane sailing pole potential range preceding primitive PROBLEM projectile projection proportional quadrant quotient radius ravelin redoubt refraction right angle right ascension sailing semi-diameter Severndroog Castle side sidereal sinē sine Sliding Rule small circle specific gravity spherical angle spherical excess spherical triangle spherical trigonometry square sun's surface true altitude ullage velocity vessel weight
Popular passages
Page 196 - Fig. 9. Case 1. Let AB, AC be each less than a quadrant. Let AE, AG be quadrants ; G will be the pole of AB, and E the pole of AC, and EC a quadrant; but, by prop. 12. CE is greater than CB, since CB is farther off from CGD than CE. In the same manner, it is shown...
Page 95 - To the square of the bung diameter add the square of the head diameter ; multiply the sum by the length, and the product again by .0014 for ale gallons, or by .0017 for wine gallons.
Page 96 - RULE. — To the square of the bung diameter add the square of the head diameter ; multiply the sum by the length, and the product by .0014 for ale gallons, or by .0017 for wine gallons.
Page 42 - A magnitude which has length, breadth, and thickness. Solution. The process by which the answer to a question is obtained. Specific gravity of a substance. The ratio of the weight of a given volume of it to that of an equal volume of water.
Page 192 - A sphere is a solid, bounded by one continued convex surface, every point of which is equally distant from a point within, called the centre. The sphere may be conceived to be formed by the revolution of a semicircle about its diameter, which remains fixed.
Page 227 - ZODIAC.— The Zodiac is an imaginary belt, or broad circle, extending quite around the heavens. The ecliptic divides the zodiac into two equal parts, the zodiac extending 8 degrees on each side of the ecliptic, and therefore is 16 degrees wide.
Page 196 - BC will be greater than a quadrant : for let AE be a quadrant, then E is the pole of AC, and EC will be a quadrant. But CB is greater than CE by Prop. 12.
Page 195 - Oj the same affection with the angles opposite to them, that is, if the sides be greater or less than quadrants, the opposite angles will be greater or less than right angles, and conversely.
Page 195 - IN a right angled spherical triangle, the sides are of the same affection with the opposite angles ; that is, if the sides be greater or less than quadrants, the opposite angles will be greater or less than right angles. Let ABC be a spherical triangle right angled at A, any side AB, will be of the same affection with the opposite angle ACB. Case 1.
Page 195 - ... will be greater than a quadrant. Let ABC be a right angled spherical triangle ; according as the two sides AB, AC are of the same or of different affection, the hypotenuse BC will be less, or greater than a quadrant. The...