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209. 'The square of the time is proportional to the tan
gent of the elevation'; also t2 =
For t2 sin e
4 sin2 e
2 sine cos e g
or t24 sin2 e
and r2h sin 2 e,
and hence t2 = 4 sin2 e⚫
1. In what time will a shell range 3250 feet at an elevation of 32° ? Ans. 11.25 seconds. 2. What is the time of flight for the greatest range for any impetus ? =√r nearly.
GEOMETRICAL CONSTRUCTION OF THE FOUR PRECEDING
1. For the second problem, after calculating h, draw a horizontal and vertical line, PH, PM; make PM=h, and angle HPT e; draw G MBF perpendicular to PT; make BF BM; M draw FE parallel to PM, and ME to PH; then ME is the directrix, and F the focus of the parabola, which can now be A constructed by article 372, (Conic Sections, P Part I.)
Measure PH, and it
g sin 2 e
will give the horizontal range in numbers. Also measure HT, and find the time due to it as a space fallen through, and it will give the time of flight. Measure VD, and it will be the greatest altitude.
2. For the third problem, make angle HPT=e, and PH = r; draw the vertical lines HT, PM, and make PB = PT; draw MBF perpendicular to PT, and EF to PH, and also bisecting it; then F is the focus, and ME parallel to PF is the directrix; then describe the curve as before.
Measure DV, and it will give the greatest altitude; and PM is the impetus, and the velocity due to it is the initial velocity; and the time of flight is found as in the preceding
3. In the fourth problem, make PH=r; calculate h, the space due to v, when v is given, and make the vertical PMh; bisect PH perpendicularly by EF, which is the axis; make PF = PM, and F is the focus; and draw ME parallel to PH for the directrix; then 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 radius PM, it will generally cut EF in two points F and F. If the other curve be described, of which F' is the focus, and ME the directrix, it will give the other values of e, h', and t.
4. In the fifth problem, make HPT=e, and calculate the height h" due to t (190); and make the vertical HT equal to it, which can be done by drawing the vertical PG=h", and GT parallel to PH to cut PT in T, and then draw TH parallel to PM. Bisect PH perpendicularly by EF; make PB1PT; draw MBF perpendicular to PT, and ME parallel to PH; then F is the focus, and ME the directrix; and the curve can be described as before.
Measure PH for the range, and PM for the impetus, and the velocity due to it (190) is the velocity of projection.
2. Projectiles on Inclined Planes.
Let PH be a horizontal plane passing through the point of projection P, PO an inclined plane, and O an object on it, PT the direction of projection, PD the impetus, PVO
the curve described, and DR its directrix; then PO is the range, and TO the height due to the time of flight.
On PD describe a circular segment DIP touching the plane PO, which may be done by drawing PC perpendicular 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); and triangles PDI and PIA are similar.
Let h, v, e, and t, denote the same quantities as in 200; also let i OPH, the inclination of plane to horizon, r=range PO, h' = TO, or height due to t.
Then angle API = PDI = e—i, DPI = PIA = 90° and DIP IAP = 180° . = - APD = 180° — (90° — i) : 90°+i. Also (Geom. vol. II., Conic Sections, I. 13) r = PO=4 PA, and h' TO=4IA.
210. When the plane PO is descending instead of as
cending, the same construction and remarks are applicable; but then
angle API=PDI=e+i, DPI PIA = 90° — e, and DIP=IAP = 180°- APD=180°-(90°+i)
211. There is a certain direction PT' for which the range is the greatest for a given plane PO, and it bisects the angle DPO.
It is also proved in dynamics that there are two directions for which the range on an inclined plane is the same, as happens also for projectiles on horizontal planes (201). Thus, if PO is the plane, there are two directions PT, Pt, for which the projectile will reach the same object O, after
describing the two curves PVO and PrO. These two directions make equal angles with the direction of the greatest range on the same plane. For the greatest range, the vertical line AI must evidently be a tangent to the segment DIP, for then PA4r is a maximum. Hence, also, ID=IP, and angle IPD = IDP = IPA.
If E, e, and e', denote the angles for the maximum range, and for the directions PT, Pt, that give the same range PO,
DPO (90° — i) — 45° —ți. Hence, ET′PH = 45° — i+i=45° +†i; tPOTPD = 90°. b,
and since therefore, tPO-e-i90°-e, or e' 90° +i—e; also, TPt 2 (E-e) = 90° + i-2e90°-(2e-i).
For a descending plane,
E=45°-i, tPO = 90°―e = TPD, tPOe'i 90°-e, or e' 90°—i—e, and TPt 2 (E-e)=90°-i-2e 90°-(2e+i).
212. PROBLEM I.-Given the impetus, the elevation, and the inclination of the plane, to find the range and the time of flight.
The formulas are, r =
4h cos e sin (e—i)
or, Lr Lh+L cos e +2 L sec i + L sin (ei)-40. When the plane is ascending, e-i is to be taken, and when descending, e+i; these angles being the elevation
2 sin (e-i)2h; or,
above the oblique plane. Also t =
Lat Lsin (e-i) + L sec i + L2h — (20+4L 32.2.)
EXAMPLE. The inclination of the plane is 11° 25′, the elevation 33° 40', and the impetus 2500 feet; required the range and time of flight.
and r = h
Lir Lh+L cose +2 L seci + L sin (e-i)-40 =L2500+Lcos 33° 40'+2 Lsec 11°25′+Lsin 22° 15'-40 3.397940+9·920268+20·017359+9.578236-40
L819-98, and r=3279-9.
Lat Lsin (e-i) + L sec i + L2 h − (20+ L 32.2) =Lsin 22°15′+L sec 11°25'+L5000-(20+1L32.2) 9.578236+10-008679 +1-84948520-753927
=682473 = L 4·81, and t = 9.62 seconds.
In the triangle DIP (p. 131), sin I: sin D DP: PI, or, since I 90° + i, cosi : sin (e-i) = h: PI And in triangle API, sin A: sin I= PI: PA, or, since A = 90° + i, and I = 90° —e, cosi : cos e and compounding these two proportions, cosi: cos e sin (e-i) = h:r,
cos e sin (e i)
=h cos e sin (e—i) · sec2 i.
Lh+Lcos e+L sin (e-i)+2L sec i-40.