1.3235 1.20 3.74 2.1520

Example. Art. 126.

66266.42:32.6036: true area;

and, by expanding, 4356 4408.96::32.6036: 32.9999+. Again, 662 66.42::32.6036 true area; and, by the tables, 3.7390878 3.6443362::1.5132655; 1.5185139-33 the answer.

LAYING OUT AND DIVIDING LANDS.

Example 1. Art. 143. Log. 124.25 2.0942964, log. 160 2.2041200, and their sum=4.2984164, which divided by 2=2.1492082, whose natural number=141.

Example 2. Log. 58.75=1.7690079, log. 160=2.2041 200, and their sum=3.9731279, which divided by 2=1.98 65639 whose natural number=96.953590900.

Example. Art. 144. Log. 77-1.8864907, log. 160= 2.2041200, and their sum-4.0906107. Log. 280=2.447 1580, which subtracted from 4.0906107, leaves 1.6434527 =44.

Example. Art. 145. 7: 3:52.5 22.5. 22.5 × 160= 3600 rods, whose square root=60 rods=the breadth of the parallelogram. 52.5x160-8400 rods, which divided by 60=140 rods the length of the parallelogram.

Example. Art. 146. 4×160=640 rods.

1224 144

÷4=36. 640+36=676, whose square root 26, and 26— 6 20 the answer.

LEVELLING.

=

Example 1. Art. 162. By Art. 161, h=√D2 + d2 D. 320×16=5280 the number of feet in a mile, which multiplied by 7940 (the diameter of the earth in miles)=41 923200 the diameter of the globe in feet. Hence, D2: 419232002=1757554698240000, and D2=43938867456 0000. d one mile as above 5280 feet; and d2=52802 =27878400, D2+d2=439388702438400; and √ D2 +d2 -20961600.664, which diminished by 20961600, (†D)= .664 (Art. 161.) √¿D2 + d2 -¿D. .664=,=ZZZ of 123984-500-7.988 the answer.

=

=

Example 2. 5280×2.5×13200, whose square=1742400 00-d2. By the preceding example, D2=439388674560 000, which added to 174240000=439388848800000=‡D2 +d2, whose square root=20961604.156=√‡D2+d2 — § D=209616200; therefore, √D2+d2 −¿D=20961604.1 56-20961600=4.156, .156-15=15% of 1.872, which added to 4-4 feet 1,872 inch the answer.

1000

Example 3. D2=79402 = 63043600, which divided by 4=15760900. d2=232=529, which added to 15760900= 15761429 D2+d2, whose square root=3970.0666, and this diminished by 3970=.066666 of a mile=888 of 320 351.6480=the answer.

666

Example 1. Art. 164. To find the perpendicular hight (Art. 6.) cos BAC (Fig. 2.): AB::sin BAC; BC; that is, cos 7° 4.5::sin 7° : BC; and, by the tables, 9.99675 07: 4.3758464::9.0858945; 3.4649902=2917.36107. By Art. 161, √ D2 +d2 – †D=√†79402 +4.52 7940— ✔63043600+20.25-3970=3970.00255-3970 = .002 55 of 5280-13.46400, which added to 2917.361 07-2930.82507=the answer.

=

255 100000

Example 2. By Art. 6, cos BAC (Fig. 6.): AB::sin BAC: BC; and, by the tables, 9.9993009: 4.3246939:: 8.7535278: 3.0789200 8=1199.28066. By Art. 161, √‡D2+d2 —†D √15160900+16-3970-3970.00201 3970 .00201= 201 of $280-10.61200, which added to 1199.28066=1188.66786 the answer.

100000

MECHANICS.

Preliminary remarks.

Sulhacten for

Mechanics is a mixed mathematical science, that treats of forces, motion, and moving powers, with their effects in machines, &c. This science is distinguished by Sir Isaac Newton, into Practical and Rational: the former treats of the mechanical powers, and of their various combinations; the latter comprehends the whole theory and doctrine of forces, with the motions and effects produced by them. Me.

chanics, says the celebrated Montucla, is the next of physico-mathematical sciences after arithmetic and geometry, having their certainty resting on the simplest foundations. It is a science also, the principles of which, when combined with geometry, are the most fertile and of the most general use in the other parts of the mixed mathematics. Mathematicians, therefore, who have traced out the developement of mathematical knowledge, generally place Mechanics immediately after the pure mathematics, although this method is departed from in the course under consideration. For reference to the principal writers on this science, the student is recommended to the article "Mechanics" in Hutton's Mathematical Dictionary. However, since that article was penned, several new works have appeared, something of which may be learned from the occasional allusions of Professor Olmsted in his compilations. As a popular work, the treatise on Mechanics contained in the Library of Useful Knowledge, has great value.

Example 1. Art. 21. As space-time velocity (Art. 19.) 9x17=153=the space described by the one body, and 5X 10=50 the space described by the other.

Example 2. By Art. 20, 540-6=90-the time of the motion of one body, and 320÷5-64 the time of motion of the other.

17

。,

595

Example 3., 7, and when reduced to a common denominator=88, 10, and 253. By (Alg. 360.) 280, 595, and 252=the relative velocities with which the bodies move.

Example 2. Art. 24. Q in A: Qin B::15; and, by reducing the terms to a common denominator, multiplying and dividing the numbers by 5 (Alg. 360. Cor. 1.) 914.

Example 3. As the weight of the three bodies are 5, 3, 2; and their momenta 1, 2, 3; their veiocities are as 1, 3, , which are (by reducing to a common denominator) as 6, 20, 45.

QUESTIONS FOR PRACTICE. Art. 39.

Example 1. Velocity is the distance which the bird flies in one second, therefore she flies 15 feet in one second, 900

feet in one minute, and 1296000 in one hour, which divided by 5280 (the number of feet in a mile)=245

miles.

Example 2. 25000÷8=3125=8 years 205 days.

Example 3. 6000÷10600 the distance which the wind blows in one day, which dividid by 24-25=the velocity per hour.

Example 4. Let x=the required weight of the lead: Then by Art. 12, 2240:x::(4004)2 (4000)2. Multip. extremes and means, 16032016 x= =35840000000, and x=2235.6.

Example 5. From Art. 13, it is manifest, that the man, after he was below the surface of the earth, would remain at rest in any part of the void. If by the question, it is intended that the earth be hollow only by a perforation through its center, it may be answered by saying, that the force necessary to sustain the man would be diminished as he descended, and that at the center it would become equal to nothing. (Art. 14.)

Example 6. By Art. 14, the weight of the ball would be nothing at the center. Let x=its weight equi-distant from the center and circumference; then, 4000: 44::2000;x. By multiplying extremes and means, 4000x=88000, and x=22 the answer.

Example 7. The stone would move with an uniformly accelerated velocity until it reached the center of the earth, and then it would move with an uniformly retarded velocity until it arrived at the surface on the opposite side, from whence it would return to the place where it was dropped in the same manner, and thus continue to fall alternately from one surface to the other. (Arts. 14, 15.) For veloci. ty=timespace (Art. 20.) and the velocity of the stone at the center of the earth would be equal to the product of the time of falling through the radius of the earth multiplied into the radius, and this would be exactly counterbalanced by the time space in falling from the center to the surface. The ball would, therefore, pass regularly from surface to surface.