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Ex. 2. The altitude 368, and elevation 40° 15', required the amplitude.

Anf. 1738.

PROBLEM VI.

The elevation and amplitude being given, and any other direction, to

find the amplitude for that direction.

EXAMPLE

The direction MAK, 50° 55'; its amplitude AK is 7000; any other direction, 32° 30' being given, to find the amplitude for that direction, the piece being the fame.

As the fine of twice the ift elev. 50° 15'
Is to the ist amplitude 7000,
So is the line of twice the ad elev. 32° 30'

9.99267 3.84510 9.95728 13.80238 3.80971

To the amplitude required, 64529

By Scale and Compasses.

The extent from 79° 30' to 65° on the line of fines, will reach backward, on the line of rumbers, from 7000 to 6452, the amplitude required.

Ex. 2. The angle of elevation is 28° 12'; its amplitud is 5100, and any other direction 37° 28', required the amplitude for that direction.

Anf. 5912.

PROBLEM VII.

The greatest altitude of a ball, with the elevation, and any other

altitude, not greater than the impetus, being given, to find the en levation with which the ball was projected.

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EXAMPLE

A cannon being fired at an angle of 24° 5', the greateft altitude of the ball 180 yards; another was fired off, and the greatest altitude of the ball was 400 yards; at what angle of elevation was the cannon fired off the second time?

As the first altitude 180 =
Is to the versed sine of twice the first elev. 48° 20'
So is the second altitude 400

2.25527 4.52249 2.60 206

7.12455 4.86928

To the versed fine of twice the ang,

of the second elevation,
The elevation required,

, } 74° 58

37° 28'

By Scale and Compasses. · The extent from 180, on the line of numbers, to 400, will reach from 48° 10' to 74° 56' on the line of verled lines.

Ex. 2. A ball was projected at an angle of 40° 30', its greatest altitude being 500 yards ; afterwards another was projected, whose altitude was 400, required the elevation of the piece.

Anf. 35° 31': Ex.

3. The greateft elevation of a ball being 450 yards, the elevation 36° required the elevation of another projection, the greatest altitude being 240 yards.

Anf. 25° 25'

PROBLEM VIII.

The elevation and amplitude being given, to find the time of the

flight.

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How long will a ball, fired off at an angle of 58', remain in the air, the amplitude being 528c feet ?

As

90°

= 10.00000 5280

3.73263 = 10.20421

58°

As radius
Is to amplitude
So is tangent elevation
To the square of 4 times the seconds

The square root of which is
Whereof the one-fourth is

8.450 3.92684 92 nearly. 23 seconds of time.

This Problem is necessary in adjusting the fusee of bombs, which are generally fired oit at an angle of 45°

It is common among gunners to find the angle between the object and the zenith, and take the complcment of half that angle for their elevation. And because a less charge of powder will ferve with this elevation than with any other, they find, by trial, what charge will reach the object,

PROBLEM IX.

The amplitude of the prejutile, with a given charge of powder being

given, to find what charge of powder will be necefiry to hit an object at any other distance, ( 110t greater than the orit most range) the elivation being the fame.

EXAMPLE.

If 16 lb. of powder will tot a cannon ball to the distance of 60c0 yards, required the neceffary charge to shoot the fame ball 5000,

with the same elevation.

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Or say, numerically,

As 6000 : 16:: 5000 : 13. 1b.

PROBLEMS on Afcents and Descents.

A projectile thrown on an afcent, with the velocity it would acquire in falling from B to A, in the direction AE, will strike the line AN in K, so that AK will be equal to 4CD. Suppofing the angle KAG a right angle, the angles GAB=GBA, and that a semicircle on G, as centre with the radius GB, cuts the line of direction in D, and that DC is parallel to AN, meeting AB in the point C.

Because the angles KAD, ADC, are equal, being the alternate angles, and AK touches the circle, and AD cuts it, the angles KAD, DBA, are equal; therefore the angle DBA= CDA, consequently the triangles ACD, ADB, are similar, having the angle at A common; therefore AC : AD:: AD: AB.

Again : Because the triangles ACD, PAK, are similar, AP : PK :: PK : 4AB; therefore AD=PK, consequently CD=AK.

4

Cor 1. Through D draw a line parallel to AB, cutting the circle in Dd, and join AD, then will the projectile, thrown in the direction Ad, strike the line NA in the point k, for CD= cd=AB=AK.

Cor. 2. Parallel to AB draw HL, a tangent to the circle in H, join AH, then shall AH be the direction which shall carry the projectile farthest on the line AN; because, when D coin

cides with H, CD is the greatest possible, and consequently AK (4CD) is the greatest distance the projectile can be carried to, with the velocity acquired in falling from B to A.

Cor. 3. It is plain that the angle HAN=HBA=HAB; therefore the direction AH bisects the angle BAN.

Cor. 4. The lines AD, Ad, make equal angles with AR, consequently the angles DAN, DAN, are equal, and the distance AK is invariably the same.

Cor. 5. When AK is given, and the direction required, take AR=AK, and through RD parallel to AB, meeting the cira

cle in D, d, draw AD, Ad, and there will be the directions.

PROBLEM I.

The horizontal distance, and the perpendicular height of the objet

above the level of proje&tion, also the impetus being given, to find the elevations.

EXAMPLES.

Let the horizontal distance be 7000, impetus 4200, and the horizontal height 744, required the directions.

3.84510 10.06 000 2.87157

9.02:47

As the horizontal distance 7000 =
Is to radius 90
So is the height of the object 744
To tangent angle of obliquity 6° 4' =
Half of which, added to 45°, makes 48° 2'.

Ax: AK : : AC : AG, that is,
7000 : 7040 :: 2100 :: 2112.

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