ex dr. du = .cos x и .du 1/di du sin dz = ㅎ Hence = cos r. cos & u. cost = ㅎ and do 8 being the flight path, it's independent upon. hance flight path is eyclaid.. flight-path cos(4) = 8 Sin du cost utan r For 4 r(t-sint). ' Ⓡ αξ 4 case = ucos dn dł do = d и созд = dn cas u sin r Ӣ using = u² ban'} cost u. The general solution. for for dr dn. 19 = u² tand » dn = u² tax & do. integrating both sides). n = u² sec² + C₂ where C. & C₂ are integral constants.. For ६ = У = (1-cast) y=h0-710=π+28 (ha-n) [(π+28) -sin (π +28)] hon = r. [1-cos (π +28)] hon = r. c.1+. cos 28) n. = ho n r. (It cas 28) = ho - E (I+ cos 28) I.207 Sơn 20 It Cos 28 A non-dimensional model of a glider flying to minimise flight-time in the absence of lift and drag is COSY = u' = - sin " и ε = u cosy, n = usiny § where is the flight-path angle, u is the non-dimensional velocity, & is the non-dimensional range, and 7 is the non-dimensional altitude and ' (prime) denotes differentiation with respect to 7, the non-dimensional time. The glider travels from § = 0, ŋ = h0 to $ = L,n = h₁, where ho > h₁ > 0, and, at the start of the motion, the glider is pointing vertically downwards so that 7=-π/2.

Trigonometry (MindTap Course List)
8th Edition
ISBN:9781305652224
Author:Charles P. McKeague, Mark D. Turner
Publisher:Charles P. McKeague, Mark D. Turner
Chapter3: Radian Measure
Section3.5: Velocities
Problem 44PS: Velocity at the Equator Assuming the radius of the earth is 4,000 miles, use the information from...
Question
100%

How is the constant c determined?

ex
dr.
du
=
.cos x
и
.du
1/di
du
sin
dz
=
ㅎ
Hence = cos r.
cos &
u.
cost = ㅎ
and do
8 being the flight path, it's independent upon.
hance flight path is eyclaid..
flight-path
cos(4)
= 8
Sin du
cost
utan r
For
4
r(t-sint).
'
Ⓡ
αξ
4 case
= ucos
dn
dł
do
=
d
и созд
=
dn
cas
u sin r
Ӣ
using = u² ban'}
cost
u.
The general solution.
for
for
dr
dn.
19
=
u² tand » dn = u² tax & do.
integrating both sides).
n = u² sec² + C₂
where C. & C₂ are
integral constants..
For
६ =
У
= (1-cast)
y=h0-710=π+28
(ha-n)
[(π+28) -sin (π +28)]
hon = r.
[1-cos (π +28)]
hon = r. c.1+. cos 28)
n. = ho
n
r. (It cas 28)
=
ho -
E
(I+ cos 28)
I.207 Sơn 20
It Cos 28
Transcribed Image Text:ex dr. du = .cos x и .du 1/di du sin dz = ㅎ Hence = cos r. cos & u. cost = ㅎ and do 8 being the flight path, it's independent upon. hance flight path is eyclaid.. flight-path cos(4) = 8 Sin du cost utan r For 4 r(t-sint). ' Ⓡ αξ 4 case = ucos dn dł do = d и созд = dn cas u sin r Ӣ using = u² ban'} cost u. The general solution. for for dr dn. 19 = u² tand » dn = u² tax & do. integrating both sides). n = u² sec² + C₂ where C. & C₂ are integral constants.. For ६ = У = (1-cast) y=h0-710=π+28 (ha-n) [(π+28) -sin (π +28)] hon = r. [1-cos (π +28)] hon = r. c.1+. cos 28) n. = ho n r. (It cas 28) = ho - E (I+ cos 28) I.207 Sơn 20 It Cos 28
A non-dimensional model of a glider flying to minimise flight-time in the
absence of lift and drag is
COSY
=
u'
= - sin
"
и
ε = u cosy, n = usiny
§
where is the flight-path angle, u is the non-dimensional velocity, & is the
non-dimensional range, and 7 is the non-dimensional altitude and ' (prime)
denotes differentiation with respect to 7, the non-dimensional time.
The glider travels from § = 0, ŋ = h0 to $ = L,n = h₁, where ho > h₁ > 0,
and, at the start of the motion, the glider is pointing vertically downwards so
that
7=-π/2.
Transcribed Image Text:A non-dimensional model of a glider flying to minimise flight-time in the absence of lift and drag is COSY = u' = - sin " и ε = u cosy, n = usiny § where is the flight-path angle, u is the non-dimensional velocity, & is the non-dimensional range, and 7 is the non-dimensional altitude and ' (prime) denotes differentiation with respect to 7, the non-dimensional time. The glider travels from § = 0, ŋ = h0 to $ = L,n = h₁, where ho > h₁ > 0, and, at the start of the motion, the glider is pointing vertically downwards so that 7=-π/2.
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