A Treatise of Practical Mathematics, Part 2W. & R. Chambers, 1842 |
From inside the book
Results 1-5 of 25
Page 121
... describes a parabola . Thus , if a body is projected in the direction PT , it will describe a curvilineal path , as PVH , which will be a para- bola . 192. The velocity with which the body is projected is called the velocity of ...
... describes a parabola . Thus , if a body is projected in the direction PT , it will describe a curvilineal path , as PVH , which will be a para- bola . 192. The velocity with which the body is projected is called the velocity of ...
Page 123
... describe a semicircle MBP , and from its in- tersection with the tangent PT in B , draw BC parallel to the axis , and BA perpendicular to the im- petus MP . Then ( Geom . vol . II . Conic Sections , Par . XIII . ) AB = PC PH = r , and ...
... describe a semicircle MBP , and from its in- tersection with the tangent PT in B , draw BC parallel to the axis , and BA perpendicular to the im- petus MP . Then ( Geom . vol . II . Conic Sections , Par . XIII . ) AB = PC PH = r , and ...
Page 130
... describe the curve as before . Draw PT , bisecting the angle MPF , and measure angle TPH , which is e ; VD = h ' ; and t is found as before . In this case there will generally be two curves ; for in describing from P an arc with the ...
... describe the curve as before . Draw PT , bisecting the angle MPF , and measure angle TPH , which is e ; VD = h ' ; and t is found as before . In this case there will generally be two curves ; for in describing from P an arc with the ...
Page 131
... describe a circular segment DIP touching the plane PO , which may be done by drawing PC perpendi- cular to PO , and drawing DC , so that angle CDP = CPD . Then angle API = PDI ( Pl . Geom . III . 32 ) , and DPI = AIP ( Pl . Geom . I. 29 ) ...
... describe a circular segment DIP touching the plane PO , which may be done by drawing PC perpendi- cular to PO , and drawing DC , so that angle CDP = CPD . Then angle API = PDI ( Pl . Geom . III . 32 ) , and DPI = AIP ( Pl . Geom . I. 29 ) ...
Page 132
... describing the two curves PVO and PrO . These two di- rections make equal angles with the direction of the greatest range on the same plane . For the greatest range , the ver- tical line AI must evidently be a tangent to the segment DIP ...
... describing the two curves PVO and PrO . These two di- rections make equal angles with the direction of the greatest range on the same plane . For the greatest range , the ver- tical line AI must evidently be a tangent to the segment DIP ...
Contents
207 | |
209 | |
218 | |
225 | |
234 | |
241 | |
247 | |
259 | |
95 | |
103 | |
111 | |
112 | |
119 | |
127 | |
129 | |
137 | |
144 | |
151 | |
165 | |
172 | |
178 | |
187 | |
192 | |
200 | |
206 | |
267 | |
273 | |
283 | |
291 | |
299 | |
309 | |
316 | |
329 | |
336 | |
338 | |
344 | |
352 | |
363 | |
370 | |
376 | |
382 | |
389 | |
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...