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CONE.

N

C = $ndi = url.
S = url + TT p2 arry 92 + ha + a pl.

h .7854 d2 h p2 h
Х
4 3

3

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V =

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SPHERE.
S = π d2 = 4π2 = 12.5664 r2.
V = f* d3 = f 73 = .5236 d3 = 4.1888 r3.

-D

CIRCULAR RING.
D = mean diameter;
R = mean radius.

S = 472 Rr = 9.8696 D d.
V = 272 Rr2 2.4674 D da.

WEDGE.
V = wh(a + b + c).

PRISMOID. A prismoid is a solid having two parallel plane ends, the edges of which are connected by plane triangular or quadrilateral surfaces.

area one end; a = area of other end; m = area of section midway between ends; 1

perpendicular distance between ends.

V = 1l(A + a + 4m). The area m is not in general a mean between the areas of the two ends, but its sides are means between the corresponding lengths of the ends.

A ta Approximately, V= 1.

REGULAR PYRAMID.
P= perimeter of base;
A = area of base.

C= } Pl.
S - iPl + A.

Ah
V

3 To obtain area of base, divide it into triangles, and find their sum.

The formula for V applies to any pyramid whose base is A and altitude h.

FRUSTUM OF REGULAR PYRAMID.

a = area of upper base;
A = area of lower base;
р perimeter of upper base;
Р

perimeter of lower base.
C= 1l(P+p).
S 11(P + p) + A +a.

V th (A +a+V Aa).
The formula for V applies to the frustum of any pyramid.

LENGTH OF SPIRAL.
D + d

n number of coil;
- пп
2

7 length of spiral;
(R2 — 2).

pitch.

1

PRISM OR PARALLELOPIPED.

C = Ph.
S = Ph+2 A.

Ah.
For prisms with regular polygon as
bases, P length of one side X number of sides.

To obtain area of base, if it is a polygon, divide it into triangles, and find sum of partial areas.

FRUSTUM OF PRISM. If a section perpendicular to the edges is a triangle, square, parallelogram, or regular polygon, sum of lengths of edges

area of number of edges

right section.

REGULAR POLYGONS. Divide the polygon into equal triangles and find the sum of the partial areas. Otherwise, square the length of one side and multiply by proper number from the following table:

Name. No. Sides. Multiplier.
Triangle

3

.433 Square

4

1.000 Pentagon

5

1.720 Hexagon

6

2.598 Heptagon

7

3.634 Octagon

8

4.828 Nonagon

9

6.182 Decagon

10

7.694

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IRREGULAR AREAS. Divide the area into trapezoids, triangles, parts of circles, etc., and find the sum of the partial areas.

If the figure is very irregular, the approximate ve area may be found as follows: Divide the figure into trapezoids by equidistant parallel lines b, c, d, etc. The lengths of these lines being measured, then, calling a the first and n the last length, and y the width of strips,

a + m. Area

+6+c+ etc. + m 2

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d
C

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MECHANICS.

Let g

FALLING BODIES.
= 32.16 = constant acceleration due to the attrac-

tion of the earth;
t = number of seconds that the body falls;

velocity in feet per second at the end of the

ime t;
h = distance that the body falls during the time t.

2 h
Then, v = gt = V 2gh = 8.02

t vt

v2 h

= .015547 v?.
2

2 g
2h
12 h

.24938 V ñ
9

g

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V

PROJECTILES. The formulas under this and the preceding heading are rigidly true only for bodies moving in a vacuum or in space (as the stars and planets); they are approximately true for bodies moving in air, provided they are dense and the velocity is not very great. Fairly good results may be obtained by applying the formulas for projectiles in calculating the range of a jet of water issuing from a small orifice in the side of a vessel. Let g = 32.16 = acceleration due to gravity;

initial velocity in feet per second;

range; y vertical height of starting point above ground; A = elevation in degrees angle that the direction

of the projectile at the start makes with the

horizontal. Then the range, or distance from the starting point to the point where the projectile crosses a horizontal line through the starting point, is

sin 2 A.

V

If the body is projected in a horizontal direction, the range is the distance from the starting point to the point where the projectile strikes the ground, and

2 y VV

2A938 vị g.

g The range of a projectile fired in a horizontal direction, 30 ft. above the ground, with a velocity of 300 ft. per second, equals r = .24938 X 300 X V 30 409.77 ft.

T = V1

v

CENTRIFUGAL FORCE. F

centrifugal force in pounds; W= weight of revolving body in pounds;

distance from the axis of motion to the center of

gravity of the body in feet;
N number of revolutions per minute;
velocity in feet per second.

Wv2
F

= .00034 Wr No.

gr In calculating the centrifugal force of flywheels, it is customary to neglect the arms and take r equal to the mean radius of the rim; in such cases W is taken as one-half the weight of the rim. The result thus obtained, divided by a, is approximately the force tending to burst the iywheel rim.

EXAMPLE.-What is the force tending to burst a flywheel rim weighing 7 tons, making 150 rev. per min., and having a mean radius of 5 ft.? SOLUTION..00034 X (HX7X 2,000) 5 X 1502

85,227 lb. 3.1416

CENTER OF GRAVITY. The center of gravity of a body, or of a system of bodies, is that point from which, if the body or system were suspended, it would be in equilibrium.

If a line or a surface has two axes, or a solid has three axes of symmetry, the center of gravity lies at their point of intersection, and corresponds with the geometrical center of the figure.

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