A non-dimensional model of a glider flying to minimise flight-time in the absence of lift and drag is COS Y = u' ' U == - sin y {' = u cosy, n' = usin y where is the flight-path angle, u is the non-dimensional velocity, $ is the non-dimensional range, and ʼn is the non-dimensional altitude and' (prime) denotes differentiation with respect to 7, the non-dimensional time. The glider travels from § = 0, n = ho to § = L,nh₁, where ho > h₁ > 0, and, at the start of the motion, the glider is pointing vertically downwards so that y = −π/2. that u = ccosy for some constant c 0 and hence show that dy/dt = 1/c. 1 differential equations for dε/dy and dŋ/dy. the general solutions of the differential equations in (b). that the flight-path is given by a cycloid. is the constant c determined? (Hint for (d): let x = §, y = ho — ŋ and 0 = π+2y.) - @ dr. du cos Y и .du ** "/di dr/de ·sin du. cost = ㅎ → 42 cas * u. cost = £ и Hence u= c cost. αξ dz = ucosɣ and 畿 dn dł == 4 = cast = dz u cost = 4² cas σ u sin r utan r Ӣ using = u² tan'} cost .u. The general solution. for ds = u² is E-du²+c for dr. u² tan.. » dn. = u². tan & dr. bagating both sides) n = u² sec² + Cz where. C. & Cz are negal I being the flight-path, its independent upon 8 flight-path hence flight path is agclaid.. cos(4) For X.= For r(t-sint) = *-! ' (l-cast) y=h0-716-π+28 4-(4-4) E=7(π+28)-sin (π +28)] hon = r. [1-cos (π +28)] = r. c.1+. cos 28) .n. = ho- r.CH+ cas 28). n = ho- E π1 + 28 + sin 28 (I+ cps 28) how to think to Create this 33. It Cos 28
A non-dimensional model of a glider flying to minimise flight-time in the absence of lift and drag is COS Y = u' ' U == - sin y {' = u cosy, n' = usin y where is the flight-path angle, u is the non-dimensional velocity, $ is the non-dimensional range, and ʼn is the non-dimensional altitude and' (prime) denotes differentiation with respect to 7, the non-dimensional time. The glider travels from § = 0, n = ho to § = L,nh₁, where ho > h₁ > 0, and, at the start of the motion, the glider is pointing vertically downwards so that y = −π/2. that u = ccosy for some constant c 0 and hence show that dy/dt = 1/c. 1 differential equations for dε/dy and dŋ/dy. the general solutions of the differential equations in (b). that the flight-path is given by a cycloid. is the constant c determined? (Hint for (d): let x = §, y = ho — ŋ and 0 = π+2y.) - @ dr. du cos Y и .du ** "/di dr/de ·sin du. cost = ㅎ → 42 cas * u. cost = £ и Hence u= c cost. αξ dz = ucosɣ and 畿 dn dł == 4 = cast = dz u cost = 4² cas σ u sin r utan r Ӣ using = u² tan'} cost .u. The general solution. for ds = u² is E-du²+c for dr. u² tan.. » dn. = u². tan & dr. bagating both sides) n = u² sec² + Cz where. C. & Cz are negal I being the flight-path, its independent upon 8 flight-path hence flight path is agclaid.. cos(4) For X.= For r(t-sint) = *-! ' (l-cast) y=h0-716-π+28 4-(4-4) E=7(π+28)-sin (π +28)] hon = r. [1-cos (π +28)] = r. c.1+. cos 28) .n. = ho- r.CH+ cas 28). n = ho- E π1 + 28 + sin 28 (I+ cps 28) how to think to Create this 33. It Cos 28
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Transcribed Image Text:A non-dimensional model of a glider flying to minimise flight-time in the
absence of lift and drag is
COS Y
=
u'
'
U
==
- sin y
{' = u cosy, n' = usin y
where is the flight-path angle, u is the non-dimensional velocity, $ is the
non-dimensional range, and ʼn is the non-dimensional altitude and' (prime)
denotes differentiation with respect to 7, the non-dimensional time.
The glider travels from § = 0, n = ho to §
=
L,nh₁, where ho > h₁ > 0,
and, at the start of the motion, the glider is pointing vertically downwards so
that y = −π/2.
that u = ccosy for some constant c 0 and hence show that
dy/dt = 1/c.
1 differential equations for dε/dy and dŋ/dy.
the general solutions of the differential equations in (b).
that the flight-path is given by a cycloid.
is the constant c determined?
(Hint for (d): let x = §, y = ho — ŋ and 0 = π+2y.)
-
![@
dr.
du
cos Y
и
.du
**
"/di
dr/de
·sin du.
cost
=
ㅎ
→ 42
cas
*
u.
cost = £
и
Hence u= c cost.
αξ
dz
=
ucosɣ
and
畿
dn
dł
==
4
= cast
=
dz
u cost = 4²
cas σ
u sin r
utan r
Ӣ
using = u² tan'}
cost
.u.
The general solution.
for ds = u² is E-du²+c
for dr.
u² tan.. » dn. = u². tan & dr.
bagating both sides)
n = u² sec² + Cz
where. C. & Cz are
negal
I being the flight-path, its independent upon
8 flight-path
hence flight path is agclaid..
cos(4)
For X.=
For
r(t-sint)
=
*-!
'
(l-cast)
y=h0-716-π+28
4-(4-4)
E=7(π+28)-sin (π +28)]
hon = r.
[1-cos (π +28)]
= r. c.1+. cos 28)
.n. = ho- r.CH+ cas 28).
n
=
ho-
E
π1 + 28 + sin 28
(I+ cps 28)
how to
think to
Create
this 33.
It Cos 28](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa6c8ed7d-75cc-4e27-869e-3ad6a1efc0b4%2Fd85014d1-3d85-4b8e-afc7-f13455693da7%2Fbe9ak6g_processed.png&w=3840&q=75)
Transcribed Image Text:@
dr.
du
cos Y
и
.du
**
"/di
dr/de
·sin du.
cost
=
ㅎ
→ 42
cas
*
u.
cost = £
и
Hence u= c cost.
αξ
dz
=
ucosɣ
and
畿
dn
dł
==
4
= cast
=
dz
u cost = 4²
cas σ
u sin r
utan r
Ӣ
using = u² tan'}
cost
.u.
The general solution.
for ds = u² is E-du²+c
for dr.
u² tan.. » dn. = u². tan & dr.
bagating both sides)
n = u² sec² + Cz
where. C. & Cz are
negal
I being the flight-path, its independent upon
8 flight-path
hence flight path is agclaid..
cos(4)
For X.=
For
r(t-sint)
=
*-!
'
(l-cast)
y=h0-716-π+28
4-(4-4)
E=7(π+28)-sin (π +28)]
hon = r.
[1-cos (π +28)]
= r. c.1+. cos 28)
.n. = ho- r.CH+ cas 28).
n
=
ho-
E
π1 + 28 + sin 28
(I+ cps 28)
how to
think to
Create
this 33.
It Cos 28
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