The millwright & engineer's pocket companion |
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Seite 5
... WATER WHEELS , PUMPS AND PUMPING ENGINES , STEAM ENGINES , TABLES OF SPECIFIC GRAVITY , & c . , & c . , & c . TO WHICH IS ADDED , AN APPENDIX ; CONTAINING THE CIRCUMFERENCES , SQUARES , CUBES , AND AREAS OF CIRCLES , SUPERFICIES AND ...
... WATER WHEELS , PUMPS AND PUMPING ENGINES , STEAM ENGINES , TABLES OF SPECIFIC GRAVITY , & c . , & c . , & c . TO WHICH IS ADDED , AN APPENDIX ; CONTAINING THE CIRCUMFERENCES , SQUARES , CUBES , AND AREAS OF CIRCLES , SUPERFICIES AND ...
Seite 12
... Water Wheels . 1. Of undershot wheels 121 2. Of breast and overshot wheels 121 To find the circle of gyration in a water wheel A table of angles for wind - mill sails The velocity of thrashing ma- chines , millstones , boring iron , & c ...
... Water Wheels . 1. Of undershot wheels 121 2. Of breast and overshot wheels 121 To find the circle of gyration in a water wheel A table of angles for wind - mill sails The velocity of thrashing ma- chines , millstones , boring iron , & c ...
Seite 120
... quotient is the power at that velocity . Suppose the wheel as above , at a velocity of 320 feet per minute . 7.35 x 320 x .0044 = 10.3488 horses ' power . ON THE MAXIMUM VELOCITY AND POWER OF WATER WHEELS . 120 VELOCITY OF WHEELS .
... quotient is the power at that velocity . Suppose the wheel as above , at a velocity of 320 feet per minute . 7.35 x 320 x .0044 = 10.3488 horses ' power . ON THE MAXIMUM VELOCITY AND POWER OF WATER WHEELS . 120 VELOCITY OF WHEELS .
Seite 121
... velocity is that of the water ; hence , to find the head of water proper for a wheel at any velocity , say , As the square of 16.083 , or 258.67 , is to 4 , so is the P C L square of the velocity of the wheel in feet per WATER WHEELS . 121.
... velocity is that of the water ; hence , to find the head of water proper for a wheel at any velocity , say , As the square of 16.083 , or 258.67 , is to 4 , so is the P C L square of the velocity of the wheel in feet per WATER WHEELS . 121.
Seite 122
... water required . EXAMPLE . - Required the head of water necessary for a wheel of 24 feet diameter , moving with a velocity of 5 feet per second . 5 x 3 2 = 7.5 feet velocity of the water . And ... water per second will 122 WATER WHEELS .
... water required . EXAMPLE . - Required the head of water necessary for a wheel of 24 feet diameter , moving with a velocity of 5 feet per second . 5 x 3 2 = 7.5 feet velocity of the water . And ... water per second will 122 WATER WHEELS .
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The Millwright and Engineer's Pocket Companion (Classic Reprint) William Templeton Keine Leseprobe verfügbar - 2017 |
Häufige Begriffe und Wortgruppen
1.-Multiply 1.-Required 12 inches 20 inches A B C D air pump avoirdupois beam bodies boiler breadth cast iron centre chord circle circular inch Circum circumference cistern column contain convex surface crane Cube Roots cubic feet cubic inches cylinder decimal depth Diam diameter in inches distance divide the product divisor engine equal EXAMPLE 2.-What EXAMPLE.-Required feet diameter feet per minute find the diameter foot force frustum half hence imperial gallons inches broad inches deep inches diameter inches long length of stroke lever nearly number of horses number of revolutions perpendicular height pinion piston pitch polygon pound pressure PROBLEM pulley pump's diameter quotient radius required the diameter revolutions per minute ring RULE RULE.-Multiply shaft side rods solid content square inch square root steam STRENGTH OF MATERIALS superficial Suppose TABLE teeth thickness tons velocity versed sine vulgar fractions water wheels wheel wrought iron وو
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Seite 89 - ... one of the most important, and at the same time, one of the least expensive and troublesome, which we possess.
Seite 72 - The areas of circles are to each other as the squares of their diameters.
Seite 34 - Multiply the divisor, thus augmented, by the last figure of the root, and subtract the product from the dividend, and to the remainder bring down the next period for a new dividend.
Seite 81 - To find the solidity of a spheroid. RULE. Multiply the square of the revolving axis by the fixed axis, and by *5236, and the product will be the solidity.
Seite 72 - Area of a circle is equal to the area of a triangle whose base equals the circumference and perpendicular equals the radius.
Seite 102 - These simple machines are the lever, the wheel and axle, the pulley, the inclined plane, the wedge, and the screw.
Seite 67 - NOTE. — 1. As 7 is to 22, so is the diameter to the circumference; or, as 22 is to 7, so is the circumference to the diameter.
Seite 102 - ... that there may be a balance between the power and the weight, the intensity of the power must exceed the intensity of the weight just as much as the distance of the weight from the prop exceeds the distance of the power.
Seite 34 - ... and to the remainder bring down the next period for a dividend. 3. Place the double of the root already found, on the left hand of the dividend for a divisor. 4. Seek how often the divisor is contained...
Seite 82 - ... by •5236 ; then say, as the square of the fixed axis is to the square of the revolving axis, so is the former product to the solidity.