Sailplane performance, stability and control

Delft University of Technology

Ship Hydromechanics Laboratory

Library.

Mekelweg 2, 2628 Cl) Delft

The Netherlands

Phone: +31 15 2786873 - Ear +31 15 2781836

Sailplane Performance, Stability and Control

By Hiroshi SATo Professor of Aeronautics (Received Dec. 14, 1959) Contents Page Introduction 36

Chapter 1 Gliding and Soaring 37 1-1 Gliding Speed Polar Curve 37

1-2 Circling Performance 40

1-3 Climb in Thermals 42

1-4 Nose Dive and Dive Brakes 43

Chapter 2 Cross-Country Soaring 46

2-1 Optimum Glide Speed and Effective Cruising Speed 46

2-2 Cross-Country Flight of Laminar-Wing Sailplane 48

2-3 Optimum Glide Speed Diagram 51

Chapter 3 Launching Peformance 52

3-1 Rubber-Rope Catapult 52

3-2 Winch Launching 56

3-3 Aero-Towing Take-Off 62

Chapter 4 Aero-Towing 65

4-1 Performance of Tow-Train 65 4-2 Aero-Tow Performance Estimation 68

4-3 Configuration of Towline 70

Chapter 5 Static Stability and Control of Sailplane 73 5-1 Static Longitudinal Stability 73 5-2 Elevator Stick-Force in Pull-Up or in Turn 78

5-3 Longitudinal Control of Aero-Towed Sailplane 82 5-4 Static Directional, Lateral Stability and Control 84

Chapter 6 Dynamic Stability of Sailplane 87

6-1 Dynamic Longitudinal Stability 87

6-2 Dynamic Lateral Stability 92

Chapter 7 Dynamic Stability of Sailplane in Aero-Tow 96

7-1 Longitudinal Stability 97

7-2 Directional Stability 99

Conclusion 102

References 102

Faculty WbMT

Dept. of Marine Technology Mekelweg 2, 2628 CD Delft

The Netherlands

36

List of Symbols

a angle of attack (deg, rad)

at angle of attack of tail plane (deg, rad)

a, stabilizer angle of attack (deg, rad)

a.c. aerodynamic center

A aspect ratio= b'IS

Aw wing aspect ratio

A, tail aspect ratio

wing span (m)

sideslip angle, or towline angle (deg,

rad)

dihedial angle (deg, rad) wing chord (m)

D drag coefficient

C DS parasite drag coefficient

CD induced drag coefficient of wing C,.. profile drag coefficient of wing

C wing drag coefficient

CL lift coefficient

Cy, tail plane lift coefficient

Cm pitching moment coefficient

Ch hinge moment coefficient

c.g. center of gravity

total drag of sailplane (kg)

8 rudder angle (deg, rad)

downwash angle (deg, rad) efficiency factor

E.A.S. equivalent airspeed = (I.A.S. corrected for instrument and position error)

km/h force (kg)

acceleration due to gray ity=9.8 (m/s')

height above sea-level (m, km)

propeller efficiency

nt tail efficiency=q,/q= (v,lv)2

setting of wing chord to fuselage

datum (deg, rad)

mass moment of inertia (kg. m)

I.A.S. indicated airspeed (km/h)

0 angle of pitch, or flight-path angle

(deg, rad)

tail arm, tail a.c. to c.g. (m)

lift (kg), or rolling moment (kg. m) horizontal tail lift (kg)

mass (kg.

pitching moment (kg. m)

m.a.c. mean aerodynamic chord (m)

coefficient of friction, or relative density factor

Hiroshi Sem

0

(Vol. XX

yawing moment (kg. m) normal acceleration load factor

angle of roll (bank), or sag angle of towline (deg, rad)

rolling velocity (deg/sec, rad/sec)

dynamic pressure= (p/2)v2(kg/m2)

yawing velocity (deg/sec, rad/sec) radius (m)

air density (kg. s2. in-4) wing area (m2)

St horizontal tail area (m2)

a air relative density=p/po

time (sec)

T.A.S. true airspeed (km/h)

v, V velocity (m/s, km/h)

VI rate of climb (m/sec, m/min)

vs sinking speed (m/s)

vy, velocity of,wind (m/s)

VI tail volume ratio=(S1/S)(4/c)

weight of sailplane (kg) angle of yaw (deg, rad)

Additional symbols are introduced in each

section.

Introduction

Sailplane in Research and Sport

Lilienthal (1890), Wright (1900) and other

aero-pioneers started studying aviation by

means of gliders'''. Later on, Lippisch began

the research on tailless gliders in 1920, and

his efforts resulted in the world's first rocket airplane Me-163 (1941), and then the delta

wing of today.

Moreover, sailplanes are now effectively used for boundary-layer studies,

as measurements are made possible by them

in the turbulence-free and noise-free air

flowo), (4), (6).

While soaring has made rapid progress by

the help of meteorology, the meteorological

science has been also benefited much by soar-ing. Soaring pilots together with scientists are

making efforts to study the air flow around

the mountains, thermal up-currents, and late-ly the air stream in the stratosphere' (8)' (9).

Since about 1920, soaring sport began to

develop in full scale, and in those early days,

efforts were made to stay in the air longer

using mainly the slope up-currents. Thus the

re-1960) Sailplane Performance, Stability and Control 37

aching 57 hours in France 1954. The years

of thermal up-currents have begun from about

1928, promoting the progress of cross-country

and altitude soaring. The distance record of

860 km was established in U.S.A. in 1951.

High-altitude soaring in the standing wave on the lee side of mountains has also made rapid

progress since 1933,

and the

world height record of 13,500 m was set up in California

in 1952.

Progress in Sailplane Design

Since 1948, with the introduction of

tri-angular-course speed contests, the low drag laminar-flow airfoils have been used more and more in order to increase glide ratio at high cruising speed. While the glide ratios of

normal-wing sailplanes are about 30 at most, the ratios of laminar-wing sailplanes amount to 35-40.

Besides the adoption of laminar airfoils, various aerodynamic improvements such as

variable-camber wings, and parasite-drag

re-duction have lead to the prore-duction of the

sailplanes of high performance and good

fly-ing qualities.

After the war, many dual-control training sailplanes have been

built, and now it

is

generally realized that the best instruction is

obtained in double-seater trainers.

As to structural design, wooden construction

is still general, but all-metal sailplanes are

produced on a small scale, and the use of

plastic is also progressing.

As a

result of many improvements in

design and operation, sailplane performances

are making rapid progress, as if there were no bounds.

Special Features of Sailplane

Performance

The methods of sailplane launching

most usually employed are rubber-rope cata-pult, auto-towing, winch launching, and

aero-towing, each of which offering its own par-ticular launching problems.

During circling in a thermal; the

sail-plane flies at a low speed, but between two thermals it glides at a relatively high speed.

Soaring is the constant repetition of such

low-speed circling and high-low-speed gliding.

Aero-towing of the sailplane is a kind of powered flight, and brings with it various

special performance and stability problems.

As regards sailplane performance analysis, not a few papers have been published, but it seems the entire field has not yet been

suf-ficiently covered. The purpose of the present work is to summalize the methods of calculat-ing sailplane performance, stability and con-trol in a systematized form.

Chapter 1

Gliding and Soaring

In still air, a motorless aircraft glides along

an inclined flightpath losing gradually its

height. But when in an up-current, it gains altitude and flies longer. This kind of flight is called soaring. A glider is intended

main-ly for gliding, and a sailplane for soaring. 1-1 Gliding Speed Polar Curve

Gliding performance , of a sailplane is

con-veniently expressed in

the form of

speed

polar curve, the sailplane's sinking speed

plotted against the gliding speed at sea level. Let CL=sailplane lift, CD=its drag, CD-=

wing profile drag, CDs=parasite drag, A=wing

aspect ratio=Y/S, e=wing efficiency factor4=,

0.8 (glider), 1.0 (sailplane). The drag polar

of a _sailplane may be approximated by

CLa

CD=- r,

C DI=CD..-1- CDS (1-1) 7rAe

and gliding speed v(m/s), V(km/h) and

sink-ing speed vs(m/s) are

From (1-2), polar curve v =

4)/

C L S

V=14.4)/1 W

C L S CD

Vs=V

Cy

-3) and -4) the equation

becomes =Ki(1±V+ Ks

'10'

10)

\ V)

CD, where K1=1.34 (W /S)S) K2=1.84 (WI S) _i. Ae (1-2) (1-3) (1-4) of speed (1-5)

(WIS)ICDf= parasite loading") (WIS)I(Ae)=WIY=span loading

From (1-5) the minimum sinking speed (v.).s.

(m/s) and the corresponding glide speed

Ii",(km/h) will be

(vs)mi. (3KIK23)11

K

V,= 10(3;)

_3K)

From (1-1), -2) and-3) drag to lift ratio

glide ratio will be

CD

(V

(10V

"

Hiroshi SATO (VOL XX

The maximum glide ratio (CL/CD).,,,, and the

corresponding glide speed 173(km/h) become

5L =in (kik)-1/2=0.89

D ma2, Df

V 2=10 (1. 114

For normal glider-airfoils, CD.40.01. Parasite

drag C Ds will be expressed as

1.1

C

DS=7

Ecf

where the factor 1.1 accounts for 10% drag

increase due to interference. S=wing area,

C=drag coefficient of any glider part, f=drag

area of that part.

CDR cannot be computed from (1-10) unless

the dimensions of each glider part are given. Therefore, for preliminary work the following

approximate formulas for C Ds may be

use-ful(.

Primary glider (braced wing, open seat)

CD8-0.40/S+ 0.012 (1-11)

Secondary glider (braced wing, nacelle-seat)

CDs'=0.20/S+ 0.012 (1-12)

Training sailplane (strutted wing, box

fuse-lage)

CD8=v0.14/S+ 0.006 (1-13)

Performance sailplane (cantilever wing,

stream-lined fuselage)

CD8=,0.071S+ 0.004 (1-14)

(1-7)

or

Speed Polars of Various Sailplanes

It has been so far assumed that the wing profile-drag C,. is constant, but actually it varies with CL. Therefore, the foregoing ap-proximate performance formulas may be

use-ful only for preliminary work. In order to

get more accurate results, the speed polars of

sailplanes with various C Ds and A are

obtain-ed, using the CL,.. test data in Fig. 1-1, where Reynolds number Re41x106 is nearly

the full-scale value.

Fig. 1-2 a, b and c give the speed polars

of sailplanes with Go-535 wings and W/S=16

kg/ma. If WIS increases to e.g. 18 kg/ma,

1.8 1.4 as a4 az -0.2 0 .01 .02 .03 .04 .05 .0 . Loco 23012 Go-549 GO-535 Go-532 " Fig. 1-1 -5653L .. _,--- sov.____ _---_,.... 23012. GO-549

I

7 I R. = 140' -_ - ' 23012. 1 \Cal ---53- i"---532 ----1: ....- _-_--C Df k,=0A8 k2=0.66 (WIS)

(1-8) multiply v, and V given here by V18/16 .

Fig. 1-3 a, b and c give the speed polars of

sailplanes with Go-549 wings and WIS= 20 kg/ma.

1960) Sailplane Performance, Stability and Control 39 V7 60 V knylh Fig. 1-2a 171 Fig. 1-2b 8 14 12 18 LID 15 V.

(RANGE) MAX. GLIDE RATIO

(FLOAT) MIN. SINK SPEC

Fig. 1-2e

Win Gö-549. A.1 5, 147S.20 k Im2

Fig. 1-3a 35 30 25 LID 20 IS 10 5 00.535. ..410 W1S .16 1.21m2

ipinc.

Ei

MENIIIME

JAZINKEN.z.

W A

41

M

mil,'

M.

w

.

1m

=

MEM

NINA

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III

D

-Add

pm

"ppm-rappAoppr 0100 .0125 0150 0,75 C°s 4 A

A

liarlaing.

2 0

-A -AMIIIM

sr-MI-6

AM

ran4 1 Go-535 Gii-535. A .a W1S.16 kglrn2

EIN

III

_All

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161.

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Ilv:PME

v

=111=1116111a 12

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WV

0111'..,.4 1111 60 .. . nn inn tar 1.2 icr nun as, as a4 - 66-535 WS =16kg1m2 --171i6 Cos 020 .010 .015 . 8 10 12 IL 16 ta x Go575, Ali WIS. 76 kg!,,'.? _{moo C.}

-Ell

CO 020

Mal

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Loinakenor

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r

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POI Moil

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gi

irmari

60 80 ... .. 100 120 2 18 2 16 2 LID 2 10 1 8 1 8 1 1 V km/., 100 Fig. 1-2c 8 10 12 g 16 18 20 A Fig. 1-2d Vs 1

40 ,Hiroshi SATO (VOL XX 4 4 3 3 2 2

Win Go-50, A.20, W15.20 1,712

Fig. 1-3b

Win Go-5.19. 445, W15.20 kg/n2

Fig. 1-3c

MAX. GLIDE RATIO

35 30 LID 25 20 15 10 10 15 20 A Fig. 1-3d 3(1

MIN. SINK SPEED

L, v2

tan 0=

gR

Fig. 1-4

(1-15)

L= WO+ tana0 =(1-16)

_{cos 0}

At sea level p=1/8 (kg. s2. m-4) from (1-2)

and-4)

1 W C D

Vo=

4/

-e;

v,0=- v0

CL (1-17)

Assuming a constant CL both in str. glide and in turn, from (1-16) and -17)

vo

v-

, -V cos 0 and from (1-15) V a sin 0 = (1-19) gR

Since a sailplane gains the power required for its steady glide solely from its own weight

(1-18) 3

11

cbs A

igka

_i

liggra_ii

1.K.7-.111

..01.-.E.

...vesialq,amm

-am

,

,

A

2

t

_II

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_Ii

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mr-N

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II

r

_,

rhigiIIMIPPAII

₁

_'Alligl

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1111...M

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80 - -00

1 Go-549 WIS.20 kg1m2 1,0 1.0 aft

L

N,Nz.z. 'ihmi...._... _as L72 .. . ,

...

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erill

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WA

Ir

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...,...

WOW .010

: LIDma .012 : A

Fa.

Frail Cos

.

-A i A

A

.. ,

..

10 1 A Fig. 1-3e 1-2 Circling Performance

Let R=circling radius (m), 0=angle of bank

(deg). v, vs = speed, sink in turn (m/s).

vo, vso=speed, sink in straight glide (m/s).

The equilibrium of the forces acting on a

sailplane

in a turn may be approximately

Vs 2 A VS 1 r:71 Fig. 1-5 SO 60 70 80 90 100 1 0 Fig. 1-6 120 13d' 4 2 LID 18 16

WVs=C D iV3S, WZso=CDfiVeS

From these and (1-18)

vso

Vs - (1-20)

cos 01/cos 0

For a given sailplane as in Fig. 1-5,

com-pute 2,0 and v,0 from (1-17), 0 from (1-19), v

and v, from (1-18) and -20). Then the speed

polars in circling are obtained as in Fig. 1-6.

From this, (v8),1 and the corresponding V

and 0 are plotted against R as shown in Fig. 1-7 (broken lines).

From (1-18), -19) and -20), assuming a

con-stant CL as before 90

v-[1- (v0/gR)1114 vso vs-[1- (ve/gR)9314 70 40 40 _{50 V km/h 6} Fig. 1-8

Mg

/AM

Sri"

1.4stKA (1-21) (1-22) Olympia 9 30 40

ismott

m/s' , ro

-MffillIME

,

Mfonard

IIIERVIE

MINIM

60 \ . _

\

\

CL=1.0 Pe: .... 20-iIll IIII °20 LO 60 n ... RO 100 121 1111

mi.

6 C040

Ealliiii.

mlh 1 1 ? Olympia-M E

5-\

_\

3 Vs mi& Cel. 0

-

--1960) Sailplane PerfOferlinde, Stability and Control 41

40 50 v kmth 6 70 Fig. 1-9

a

so so 100 20 40 _{6 R m} Fig. 1-7 100 120

42 Hiroshi SATO (Vol. XX

These are plotted in Fig. 1-8 and 1-9. Using these, the speed polars in circling can be

re-adily obtained from the straight glide polars.

In case of normal sailplanes, as (1)30).in in

a straight glide occurs commonly at CL1.0, the circling in thermals may be also usually

performed at a like CL. At CL=1.0, from (1-17)

vo=4/W'

_s

v.° =

4CDI/

_s

(1-23)

and from (1-19) and -20)

sin

cb= gR.S)

16 (W\ 1.63 On

R ksj

(1-24) DA 8 70 77 74 Fig. 1-11 al . BREGUET 901 Fig. 1-13 1-3 Climb in Thermals

Using the speed polars of sailplanes with

Go-535 wings (Fig. 1-2), (vs),is, are plotted against R as in Fig. 1-14. Now, assume an

arbitrary thermal, whose up-current velocity

profile

is something like a cosine curve as

shown in Fig. 1-15 (broken line).

The difference between the up-current ve-locity (Fig. 1-15) and the sinking veve-locity of

the sailplane (Fig. 1-14) at the same R gives the rate of climb of the sailplane relative to

the ground (Fig. 1-15, full lines). From Fig.

1-15, the effect of A and CDs of the sailplanes on the maximum climb vs in the thermal can be obtained as in Fig. 1-16.

Similarly, for sailplanes with Go-549 wings

(Fig. 1-3), the maximum climb vc in the

thermal are obtained as in Fig. 1-17. From

this, the effect of A and W/S of the sailplanes on vc is known as in Fig. 1-18. 80 90

NM

=WM

Turn CL.1.0

M MIN

13231001

'1111

MEM

MEM

*

MMIlliiiimaral

le

140 Ix 80 L4 4 Olympia.Meise

En

Turn ce,1.0 ,11 D MN_

_goo

,ELIN

HEIM

111:11111112111 ,

.

ini

_{sirageMliald-iG}

8025 74

.,

MN

NM 111,,2 VII TA 1 19 vs

=1

M

CMII 0

EWA%

glIONI Ili

mai

ea

104 A '90 Bregue1901 :14°

mom

Turn CL.1.0

a

'NM

MN

Man

1

°68 ,MGMMI

NEM

tiEntlNEEN

MIIIIMiiiiall:

70 77 7476 78 80 87 12 1.0 P2IMMITI

MIS

IMM ,mi

ii

COMI

upriwrga

/111%-d

=

Ali

on%

EU

MI=

vs=1.25vs0=5C9/17 (1-25)

Fig. 1-7 (full lines) and Fig. 1-10 give the circling performance of Olympia Meise and

Fig. 1-11, those of Breguet-901. Fig. 1-12,

the comparison of vs when the two planes

turn at CL=1.0 and with the same R.

72 ₇₆

1960) Sailplane Performance, Stability and Control 43 5 4 Vt 34 vc 2,E 6 20 40 _Pm Fig. 1-15 8 10 12 U A Fig. 1-16 Vc 20

MAX. RATE OF CLIMB IN THE THERMAL 4 10 15 A Fig. 1-17 G6-549

M.,....,,_.

Iv_

.oac .008 412 15kglm2

=11

w-1-! 404 408 PIor_ rA/14111111111111

WA

,o12 Cos

Mr

,W74,1 20 kg1m2 _AU

=4

111.mmww-I le.

478412

11111 ill

_r

_{25 kglm}

au

ill

,v,s .16 leglm: Go-535

il

11111,

MIS

Nam%

11

=111111111..P7m

mufti rea NMI NUM

'

IVIIII

_litinam=2

III

iiii

till

... -... .... N

_\

_{WIS .16 kg1m2} Go-535 N

\

_\

\

_{\ASSUMED DISTRIBUTION OF} \ Cos

\

\

\

THERMAL STRENGTH A.18 020 \ \

\

.0I0 ' \ IS \ \ .4.8

11

1, \\1\ \

\

... Gii -549 Cos=.008

...11

. EN

_mis . _

mmi

3,

Es

..

. . . . . G6-535 cc3 .cno p . odOr . P _.sio 45 420 - fnls .

ANN-,

/18 kg, m2 P 15 20 25 hIS Irglm2 Fig. 1-18 1-4 Nose Dive and Dive Brakes

In the diving speed calculation, the varia-tion

of air density with the height fallen

should be taken into account. The equation of vertical dive motion will be

W dv W--= D f±e s (1-26) g dt ° 2 30 R m Fig. 1-14 3 Vc 1

where CD0=sailplane drag in dive without

using dive brakes. Let

B = CDogr) =S-(1±) (1-27)

2 W q r

2

where qr =dynamic pressure in terminal dive

(WIS)

q=

4,D0 Then (1-26) becomes

dv_ dv

(it- -y cw-c-g-By2

For a given sailplane, B varies with the air density p (Fig. 1-19).

But within a small

height interval,

it may be assumed as

con-stant. Integrating (1-29) by assuming B con-stant, gives the distance fallen x (m) and time

t (sec) during which the sailplane speed

changes from v1 to v2 (m/s) as

x= -

1 [

log, (g- Be) 1v2

2B

At the beginning of the dive, the sailplane

drag is small compared with its weight W.

Thus, neglecting the drag term By' in (1-26)

and -29), x and t from v1=0 to v, will be

obtained as 1 2 VX 12) 2g(V2

t. -1" (v2-0

Consider a sailplane, W=290 kg, S=15.8 m2,

CD0=.021, qr=870 kg/m2, the terminal dive speed Yr = 118 m/s = 425 km/h. The height

fallen from v1=0 to v2=20 m/s is from (1-32)

x (20)' =20.5m

2g

Next, estimate x: v curve and the value of

B for the following interval (v1=20, v2=40).

Putting

this B into

(1-30) and finding x,

check whether the value of B is correct for this height interval. Repeating this process

gives a group of curves showing the speed

change during the dive as in Fig. 1-20 (group

without notation yBK).

If a dive

is started from a large height, the speed exceeds the

terminal dive speed vT = 118 m/s.

2 r1°-'6 7 03 4 40"46 7 Fig. 1-19 oo

66

8 0 20 Fig. 1-20 V mis 20 40 60 EV 100 120

Till

Dive Brakes 100 200 300 V kWh Fig. 1-21 6

wpm

triltmem

mosman

1118121111

,1111011y to

00 207 Fig 1-22 3 2 40 60 80 100 120 V rms

The variation of q during the dive is

obtained from Fig. 1-20, as shown in Fig. 1-22 (curve group without notation qm,-).

These q's also exceed the terminal qr, if the sailplane continues diving from a large height(12)'

5

ZO 403 600

q6000' Fig. 1-23

To keep the dive speed of the sailplane

within the safety limit, dive brakes are corn-go or or

x= -

12-[logia(1 -11 v g JJVi 0.5 [ . 1+vi/Blg 102

'

(1-30) (1-31)

t-

_gB 1.15[l 1- vl/B/g 1+ vi/BIgiv2

t-

_gB 10g10

1v i/Blg

44 _{Hiroshi SATO (Vol. XX}

400 500 Il 3 Avn 2 1003 800 (1-28) (1-29) (1-32) (1-33)

1960) Sailplane Performance, Stability and Control 45

monly used. Typical dive brakes consist of

small drag plates (or spoilers) which can be extended normal to the air flow from the top and bottom surfaces of the wing (Fig. 1-24).

Spoilers should be installed on the wing so that neither the tail nor ailerons lie in their

wake. Their chordwise location is usually (0.4-0.5)c, the aspect ratio bBlh=r4, and the gap h'= (0.2-0.3)h.

Now, let CB= drag coefficient of spoiler= L8 2.2, SB=total area of spoilers4bBh, CDO= sailpane drag in dive (brakes retracted),

(in= (VV/S)/C,0=terminal dive q (brakes

re-tracted), vo..=zpax. dive speed (brakes re-tracted). CDB=sailplane drag in dive (brakes

extended), qr,=(WIS)ICDB=terminal dive q

(brakes extended), VB.= max. dive speed

(brakes extended). wing surface Fig. 1-24 CDBCDO+L,BSB S _SBCDO(CDB _1) _(1-34) S CDO CDB =11TO _( VO-rna.. y _(1-35) CDO QTB Vitinaa

From these, the total brake area SB required

to limit VBynaz can be estimated. For the

example sailplane, CD0=.021, 42,0=870 kg/m2, CB=1.8. From (1-35) and -34), SB required

to limit q,B to about q,,,/4 will be

CDB q4

SB .021 (4_1)=.035

CDO QTB S 1.80

Since CDB=4CD0, multiplying B (1-27) by

4, obtain vBK- and 03K- (brakes extended) just in the same way as before (Figs. 1-20, -21, -22 and -23, vBK and qBK curves). As seen in

these Figs. vB.a.0-5voma. and (a) =-n

?nem BIC --ITO, -/4

=220 kg/m2.

Airworthiness Requirements for

Dive Speed

The sailplane should be pulled out before its dive speed reaches the safety limit.

Air-worthiness regulations specify the design dive

speed according to the sailplane category.

Japanese and British regulations specify the design diving speed in the form Vp=aVs, where Va=stalling speed, while German

re-gulations specify the design dynamic pressure

in the form qp=b (WIS). From these two

a2

QD=

_{\ S}

(1-36)

Using (1-36), Japanese and German regula-tions can be compared as follows :

Japanese regu. for normal sailplanes, a=4.5,

=1.3,---1.4, safety factor 1.5. From (1-36) Ultimate qp=1.5(14-16)VVIS=(21-24)WIS

Japanese regu. for aerobatic sailplanes,

a=5.5, CL.,=as above, safety factor 1.5.

Ultimate qp= 1.5(21-23) W/S= (31-35) W/S

German regu. for normal sailplanes, CD0=

.035--.025, safety factor 2.

Ultimate QD=2.7D= 2 x 0.5( W/S)/CDo

= (28-40) W/S

The ultimate qp required, for German

sailplanes is very large compared with that for Japanese sailplanes.

Use of Spoilers in Landing

Spoilers steepen the glide angle without

speed increase, and they are very useful as an aid to landing in small fields. Let

CL, CD=sailplane lift,

drag at

a

given a

(spoilers retracted), CLB, CDB=sailplane lift, drag at the same a (spoilers extended). The

lift decreases and the drag increases when the

spoilers are extended.

21)B c

CLB=CL (1 K ), CDBCD+4-'B--S

(1-37)

where c=wing chord at spoiler. For normal-type spoilers k may be taken as about 1.0"8),

and CB about 2.0.

Consider a sailplane like KU-11, W=217 kg,

S=14 m2, spoiler SB=0.45 m2, bB=0.7 m, h=0.16

m, c=1.2 m. Assuming k=1.0, CB= 2.0, from

(1-37)

CLB=0.88CL, CDB=CD+ 0.064 (Fig. 1-26) When the spoilers are opened in a steady

glide, the glide ratio changes from CL/CD to

CLB/CDB, usually without trim change. Let

46 Hiroshi SATO (VOL XX

Fig. 1-25

angle (spoilers extended).

Then the a

in-crease will be 4a=0D-0, and tan 0=CDICL. Draw 0 :CLICD curve (Fig. 1-25) and a:CLICD curves (Fig. 1-26) to the same scale, and

superpose the 0 axis on the a axis. The in-tersections of these curves show how CL/CD change into CLB/CDB when the spoilers are opened (Fig. 1-26, broken str. lines). Fig.

1-27 gives the change of the sailplane's speed

polar due to the use of spoilers.

Chapter 2

Cross-Country Soaring 2-1 Optimum Glide Speed and Effective

Cruising Speed

A cross-country soaring may be divided into

a number of units as shown in

Fig. 2-1.

Leaving a thermal A, the sailplane glides to meet another thermal, in which it circles up

to B.

Let v,=sailplane's rate of climb in

thermal (m/s), v =inter-thermal glide speed (m/s), vs=sinking speed corresponding to v

(m/s), tAD=time taken from A to B (sec),

D=distance between two thermals (m).

tAB=H(Vg+Vc)_ D(vs+v,)

vsvc

v \

v,

Therefore, the effective cruising speed v, (m/s) or V, (km/h) is 15 0 -4 -2 0 2 4 6 Fig. 1-26 SO 60 70 _{v 801h}90 100 110 kin kmm 0 100 110 Fig. 1-27

ve(

v0

Vol,

v° (2-1) vs+

\

vs+v,J)17

As before, the sailplane's drag polar will be

approximated by (1-1). Let Vi=glide speed

(km/h) for (LID),., vsi=sinking speed (m/s)

corresponding to V1, and any glide speed be

expressed as (n)V1(23)''''. From (1-5) 10 \ 14=-Ifi(n1V01)3+ K2 (n17,) from (1-9) 01/2(CDrAe)4 10.82 (W/S)1/2 V,=10(K:\1/4-

_ki

(2-2)

from the above two

---

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1960) _{Sailplane Performance, Stability and Control} 47 B Fig. 2-1 3.4 (CD1)1"( W/S)" vs,= 2 (K,K22)1/4 - (2-3) a1/2(Ae)314 from (2-2) and -3) v,=(ns +

1)

n 2 (2-4) from (2-1) and -4) nV, (n3+1In)v3112+v,

V,-

2(v,Iv81)V1 (2-5) (v,Ivsi)(21n)+(lIn)2+ n2

From the condition for max. cruising speed,

aVelan= 0

n4-(v1Iv31)n-1= 0 (2-6)

Using n obtained from (2-6), the optimum inter-thermal speed Vo, is known as nVi. It

is easily shown that the inter-thermal Vo, is

equal to the Vo, in a down current of

veloci-ty vc.

Given the true-airspeed polar of a sailplane

for altitude H, take vc axis in .the opposite direction to vs axis, using the same scale in

both axes.

From a point

v, on the

axis,

draw a tangent to the polar, and the contact point on the polar gives Vo,(TAS) when the

climb in thermal is vc.

The relation between the true airspeed

Vi(TAS) and the equivalent (or indicated) air

speed Vit(EAS) for max. LID at an altitude of air density p is

VI(TAS)= L

Vit j/8! (2-7) P VJTAS) 1701,, (TAS) 12 V, (TAS) V1t Vo, (TAS)_ n pa vT). -n (2-8)

Similarly, the relation between the true

cruis-ing speed Ve(TAS) and Vii(EAS) is

Ve(TAS)_ V Ty- V,(TAS)

m

Vi(TAS) Vic

V,,

- V a

Ve(TAS) m

-(2-9)

For the Pilot Vot(EAS) is most important.

V ov(TAS)-= Vopt(EAS)//a

From this and (2-7)

1/0(EAS) Vo,(TAS) _.n

V,, VI(TAS)

Obtain n from (2-6)

for various values of

(v,Ivs,). Using this n, from (2-8), N at

alti-tude H is computed. Similarly, from (2-5)

m, and from (2-9) M at altitude H is

obtain-ed (Fig. 2-2). 9 8 7 6 5 2 10 1 H '72_{3 knk} (2-10)

Principal Dimensions of Sailplane and

Cross-Country Performance

The wing loading WIS (kg/m2) of a single-seater sailplane may be given by an empirical

formulam

W 145 + 0.2A S

S S

Putting this WIS into (2-2) and -3), V,(km/h)

and vs, (m/s) for max. LID will be

V -

10.82 (145/S+ 0.2A V S )1/2 1/ (CDfAe)" 10.82 (145/V S+ 0.2b2)1/2 V-07 (CD1.eb2)1" (2-11) oo 1 14 N 2 2.5 3 Fig. 2-2

3.4 (Cpf)"(145/S+ 0.2 AV-S-)ilz

V81

6 (AO"

3.4 (CDf)114(145+0.2bY)112S1/4

(eb2)/4

Since CDf=CDS+CD.0, CD-:---,.009 for normal sailplane profiles, and CDs= 0.07/S+ .004 for

normal sailplanesm

C11= 0.07/S+ 0.013

Putting this Cpf into (2-11) and -12) 10.82 (145+ 0.2bY S )1.12

1/ a (0.07+ .013S)4(en"

3.4(0.07 + .013S)'4(145+

0.2b2)"2vs,-The thermal strength may be usually w,=

1-3 m/s. Assuming the radius of circling in thermals R - 760 m, and the sinking speed in

thermals as about 1.2v31

W34

(2-16) V gi VII

Obtain vs, from (2-15) at a cruising altitude, e.g. H=1.5 km, v3/v3, from (2-16) for wu=1,

6 2t 0

w2 mis

km.h 20 V. An30₂₅ 20 10 15 b20m 20 30 Fig. 2-3

2, 3 m/s, V11 from (2-14) putting a=1, and 171, from Fig. 2-2. Fig. 2-3 shows the effect of the sailplane data, span b, wing area

(2-12) _{S, and aspect ratio A, on the cruising speed} V,.

2-2 Cross-Country Flight of Laminar-Wing Sailplane

One of the post-war trends of sailplane

design is the use of low-drag laminar-flow

airfoils. In case of laminar airfoils, drag

in-crease is not so large, even if the thickness increases up to 18-20 %, while the optimum

thickness of conventional airfoil is only 12 %.

Thicker laminar airfoils are structurally and aerodynamically advantageous.

Sailplane's Reynolds number at sea level is

652114 23012 Fig. 2-4 1 Re-3.62

j/' x /06

_(2-17) V CL S

Let c=geom. mean chord of wing=S/b

1 j 1

W

109 (2-18) Re-3.62 V CL V A X

Consider a sailplane like Breguet-901, W=315

kg, A=20, c=.865 m. From (2-18), R, are

Fig. 2-5 ...___Az 30 I ---"--.-- 2520 Ve

..---

15

I _..._:;;;:;:--- _---.10 b m 7 10 15 20 25 30

'AVAINIIIIIN

skyAiAS

AIIIMINIMI

lhl11.111=

xf, '

7i

Ei: reguet / , r I,-I I

1

_-, i , ? t A .".iiic-.1-? Ai

.02;di" .06

.06. .0

48 Hiroshi SATO (Vol. XX

(2-13)

(2-14) (2-15)

1 1

1960) Sailplane Performance," Stability and Control 49

computed as in Fig. 2-5.

Taking into account the effect of Re upon

CD., the CD. of a laminar airfoil is

approxi-mated by

CD= (.005-.006) +iCoor2 (2-19)

where (.005-.006) is the min. profile drag of a laminar airfoil at Re .73 x 102 (22), and in the

following calculation, the smaller value .005

will be used.

By careful aerodynamic refinements, the

parasite drag CDs may be reduced tom

CD31-,-0.04/S+ .0022 (2-20)

Since the wing-lift distribution of modern

sailplane is very near the optimum or elliptic,

the efficiency

factor may be taken as

Then the wing induced drag CDt=C21(7rA).

The sailplane drag CD=CDt+CD-+CDs, or CD= (1+ 0.01A)CAL2 + cpr CDf=.0072+0.04 from (2-21) A 1/2

()ma

x = .886 (1+ 0.01A)C Di 1

V, (km/h) and yr, (m/s) for max. LID will be 10.82( 1 +0.01A r (Wy/2

AC (2-23)

V a \

Df S

IVV\ '21 1

vs,- 1/1-40. (CD.r)" Is] k A+ 0.003/4 (2-24)

The equation of speed polar becomes

v8=K,()3+ ic,(11--°T) CDf K1= 1.34 a - 1.34-f-WIS .1f2= 1.84(1 + 0.01A)WIS aA 1 =1.84(1 + 0.01A) a

where V (km/h) and vs (m/s). Using these K, and K2, from (2-23) and -24)

V1=io(&)im, = 2(Kilf-23)1/4 (2-26)

(2-25)

Relation Between (Ve)min, Ve, and

Assuming the weight of a pilot with

para-chute as 85 kg, the gross

weight W of a

single-seater sailplane will be)

W-1-:140+ kbY k=0.15-0.20 (2-27)

In the following calculation, let k=0.16 (Sky,

1-23D, Skylark 3, Breguet 901 ...), then WIS

(kg/m2) becomes W

_-14

0

0-+ .16A (2-28)

S

From (2-23), -24) and -28) V, and vs, for max.

LID will be

Eliminating A from (2-22) and -24), at sea

level (a=p1p0= 1) 2.83 (wis)1/Z

(LID). C

vsi= (2-31) From (2-22) 1 0.786 A C1(LID)Using CDf(2-21), W/S(2-28), and A(2-32),

the relation between (L/D),.a

and v,, for a

laminar-wing sailplane is obtained from (2-31) as in Fig. 2-6. In case of laminar-wing sailplanes, vs, does not differ much from

(vs)min

Assume an average wing are S=16 m2,

and compute A from (2-32) for various values

of (LID)ma, then b from b=1/AS, and WIS

from (2-28). As seen in Fig. 2-7, for a

sailplane of S=16 m2 to

have (LID),.=38

for instance, A=21.5, b=18.5 m, and W/S= 23

kg/m2 are needed.

Next, compute 1714 from (2-29) putting a=1, and yin at H=1.5 km or 1/1/ a =1.075 from

(2-30). Assuming the sailplane's sinking

air-speed when circling in a thermal is 1.5 x 7

1.5 x (vs)min, and letting tv= thermal strength (m/s) as before 0.01 (2-32) 10.82 { 1/A + 0.01 }1/4 X Va 0.04/S+ .0072 x ri-L4-/-3+ 0.16A-1/-01/2 (2-29)

\ S

= 3.4 f(04 + 0072.v )1/4x S x +0.16A"1/)1/2(-2!"-i +0.01)3/4 (2-30) (2-21) (2-22)

50 Hiroshi SAT° (Vol. XX .90 .85 ao 75 20 VS1 \12m2 S

7i\

1

(LID)max

I/

Jitti

25 30 35 Fig. 2-6 40 25 (LID)max Fig. 2-7 vc/v8i--= wavn 1.5

From this and Fig. 2-2, the relation between

(LID), and V, for a given w. -is known as in Fig. 2-8 (full-line W-curves).

Next, consider the case when the sailplane

weight W is increased to kW by the use of

ballast.

In this case (L/D),. remains

un-changed, but V,, and v31 vary directly as V k

6 5 4 Vc 3 2 vs 2 100 90 80 70 Ve 6 5 40 30 m 20 _{25 30 35 40 45} (LID )max Fig. 2-8 Fig. 2-9

v.f ...

... .. ,..6 7 1 ... 1071J1 14 .

I ...

... f V., ?P. 1 1.66 6P ,m7P. ... Olympia. H1500 m Fig. 2-10

Assuming k=1.32, V, at H=1.5 km are plotted

against (L/D)Tha as in Fig. 2-8 (broken-line .

1.32W-curves). As seen here, for a sailplane

of (L/D).=30 for example,

ballasting is

I i =1 , 6 k m

man

ENENNOM

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.-1960) Sailplane Performance, Stability and Control 51

useless, unless thermal strength wu is larger than 2.5 m/s.

2-3 Optimum Glide Speed Diagram

As an example, consider a sailplane like

Olympia-Meise, WIS= 18 kg/m', b= 15 m, S=

15 m2, CDf=.018. Assuming H=1.5 km as

be-fore, draw the speed polars at H=0 and 1.5 km as given in the lower part of Fig. 2-9.

Since vg1=0.79 (H=1.5 km), 0.73 m/s (H=0),

and VI, = 65 km/h, Vovt(TAS) and Ve(TAS)

curves are drawn from Fig. 2-2, as in Fig.

2-9.

Now, assume the climb in thermals vs=3.5

m/s for example. The intersections of the horizontal line through this vs and ve, Vox curves give Ve(TAS) = 73.5 km/h, Vos(TAS)= 120 km/h, and the corresponding vs=2.23 m/s.

But as

mentioned before, the pilot wants

Vs,(EAS), which is given by the intersection of the inclined broken line through the origin

0, and (H=0) curve, as V0,(EAS)=111

km/h.

In this way, the scales of Vs(EAS), v

and V, will be graduated as shown in the

upper left part of Fig. 2-9. Thus, the

opti-mum inter-thermal speed diagram or airspeed selector, Fig. 2-10 is made.

The usage of this diagram is as follows : let the estimated climb in thermals be vc=3

m/s for example. Then the Vs,(EAS) called

for will be given as 106 km/h, and V;(TAS)

69 km/h.

If the sink at this time is 2 m/s

as given by the v, scale, maintain the Vox

as it is. But if the variometer indicates

(2+ w)m/s, there may be a down current of

w m/s. Then correct vs as (3+ w) m/s and proceed as before.

Fig. 2-11 gives the speed polars of Breguet-901 (laminar wing), Sky (normal wing), and Meise (normal wing). Fig. 2-12, their

opti-mum glide speed diagrams.

Besides the above-mentioned linear-scale

speed diagram, the ring-scale diagram is also

used, and the latter can be easily made from the former.

As said before, the inter-thermal Vc, when

the thermal climb is v is equal to the Vopt

in a down-current of wa=vs.

First, using v

Vcs, and vs of the linear scale, make a table,

Voxversus (vc-F vs) as follows :

Vopt km/h 87.6 99.6 1 2 Fig. 2-11

9. 1

? . 4 , 190

tp

I ,1t5. ? . ,28, 4P, 6P , 88 Meise H-1500m 9 f ? , 4 , 8P 9P "P 417,9P 6P 6P Sky H-1500 . ? m ,9P .190 . , r1.40 VGP km/h Vs ms q o1 v 719 89 9P 14° Ve kWh Efreguet H-1500 m Fig. 2-12 a (vc+vs)m/sec 1 Ve Vop Vs Ve Ve km/h vo=3

(vc+vs) or (wrz+vs) is the variometer

read-ing in the down current. Therefore, when

ve=0, glide at V", (column a in

the table)

corresponding to the variometer reading (column b). When vc=1 m/s, use col. a and

col. c, which is shifted downward against col.

b by 1 m/s.

AIM

4

OR Sky

gm 111:11:1111

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52 . Hiroshi Sero (Vol. XX 5 4 3 ye Is 1 Fig. 2-13

Now, as shown in Fig. 2-13, a rotatable ring scale is put on the outside of the

vari-ometer. Using the dial for col. b, graduate the

V, scale from col. a on the inner side of the ring. On the outer side of the ring, graduate

vc with the scale equivalent to the dial, from a fixed mark A as vc=0, 1, 2, ... m/s.

When vc=2 m/s, turning the ring clockwise,

set 2 m/s of v scale on the mark A, and

glide at the Vop, pointed by the variometer. In this case, the Variometer automatically takes care of the down current. The larger

the climb v, the larger the 1709 and the

smaller the LID or dl h (Fig. 2-14), where

d = distance, h= height.

111111111

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dlh

Wd= 0 Fig. 2-14 Chapter 3

Launching Performance of Sailplane

3-1 Rubber-Rope Catapult

A rubber cord

of some 60 m length is

hooked at its middle on the glider anchored to the ground.

The ends of the cord are

pulled out by the launching crew

in a

V-form. When the cord is stretched to

about double its original length, the glider is released and it shoots into the air.

The process of launching may be divided

into 3 stages : (I) from start to take-off, (2) from take-off until rubber cord falls off, (3)

after that.

From Start to Take-Off

Let v = ground speed of glider (m/s),

w= wind velocity (m/s), p= coefficient of ground friction, 0=ground inclination (deg). /0=1/2

total

cord length (m), E=tension on hook

when both ends of cord are stretched by 1 m (kg/m), va=running speed of crew (m/s).

Assume that the glider is released when the

cord is stretched by I m. The cord tension

T (kg) acting on the hook will be

- x+ vat)

where t= time after start (sec), x=distance

(m) covered in time t. The glider's ground

friction R (kg) and the air drag D (kg) are

R= Wcos 0 -"XL S(v+

C S(v+ w)2

The equation of motion of the glider during ground run

W dv

_{=T+ Wsin 0-D-R}

g dt

dv _{g {(E1+ Wsin -1.1Wcos 0)- E(x)+}

dt W

+ v aE (t) + (AC L- CD) S(v + w)2} (3-1)

From start to take-off, approximately a

uni-formly accelerated motion may be assumed.

dx

kt, dv. .

From (3-1) and 2)

Putting t=0 in (3-3), k (m/s') becomes k= {(E1+Wsin to- pWcos 0)+

+ (pC L- C D) SO} (3-4)a

Putting this

k into the right-hand

side of (3-3)

dv

dt =(k)+ (k,)t + (k3)t2 (3-5) where ki (m/s3) and k, (mIs4) are

k1=17 {v aE + (pCL- C D)pSkw} (3-4)b

k2=÷ i(pC

L- CD)

El

(3-4)c

Ground speed v (m/s), and distance run x (m)

are

v=(k)t+( k2' )ta-F(Lnt3 (3-6)

x=(4)P+(k6i)t3+(ile;)t4

The glider's take-off ground speed in the wind

of w (rills) is

41/(WIS)/CLw

Obtain first k, k1, and k, from (3-4)a, 4)b,

and 4)c, then the take-off time t, from (3-6), and the take-off distance x0 from (3-7).

From Take-Off until Rubber Cord falls off

Measuring the time

t and the distance x

from the moment of take-off, the tension T

(kg) of the rubber cord after take-off will be T= El E(x,+ x)+ Eva(tta+

The equation of motion for this to. stage will

be from (3-1),

putting g=0, and using T

above given

dv _{- W{El + W sin 0 - E(x,+ x)+}g

dt

dv

dt =k- g {(E1+ W sin 0

EA-(02+ van)

+ (ficL

-2 Wcos 0) C D) S(kt +w)2} (3-3) (3-7) _(3-10) ±Eva(tto+t)CD iS(V+ w)2} (3-8)

As before, for a small interval of time after

take-off dv x4v,at+

(r)t2 (3-9)

dt 2 From (3-8) and -9) dv _w-g _{{El+ Wsin}

0-dt

-E(x,a+

v,at+ t2)+ Eva(t,a+

t)-- C S(v,a+ kt+ w)2} (3-10)

From this, k at the moment of take-off (t=0)

k= {El+ W sin 0 Ex,+

Evatta--C D S(v,a+ w)2} (3-11)a

Putting

{Ev,a Eva+ C DpSk(v,a+ w)} (3-11)b

= _ {-}E + CD Se} (3-11)c

Putting k (3-11)a into the right-hand side of

dv

= (k) + 1)t + (k2)t 2 (3-12)

V = (Vto)

(k)t+(jL)t2 +(k; )t3

(3-13)

x=(v,)t+G)P-FC0t3+(ik22)t4 (3-14)

Since these t and x are measured from the moment of take-off, .the time i and the

dis-tance i from the start are

t,+t,

i=x,0-Fx

and the tension of the rubber cord at this

time is

Therefore, when x-=1+ v at , T=0 and the cord falls off from the hook. The glider's speed

reaches its max. value just before the cord

drops.

Now, divide the time t from the take-off to

into some small intervals t1, t2, t3,

First, for the interval t1, compute v1 and x1

from (3-11), -13) and -14).

In the next

in-terval t2, usingt

- 1= -to+ t-1, 1= TO+X 11x and v,

for to, x0, and v. in (3-11), obtain k, lei and

k2. _{Then (dvIdt)2, v2 and x2 from (3-12), -13)}

and -14). Repeat this process until the

con-dition for (dvIdt=0) is satisfied.

Usual-ly, two or three intervals may be sufficient.

1

Effect of Rubber-Rope Size

Consider a sailplane like KU-11, W=224 kg, S=14 m2, WIS= 16 m2. The condition for max.

acceleration in take-off, (mCL-CD)ma,, of this sailplane is shown in Fig. 3-1. Assume the plane maintains the attitude CL=1.4, CD= .085

during take-off, and the t.o. airspeed v0= 13.5

m/s or 49 km/h.

Rubber cord No. 1 (small) diameter 19 mm, 10=30 m, E=8 kg/m.

Rubber cord No. 2 (big) diameter 23 mm,

/0=30 m, E=12 kg/m. Let va=3 m/s, w=0,

arid /2=0.2. The result of the calculation is

given as follows :

10

46

,02

Hiroshi SATO (VOL XX

Effect of Ground-Inclination and Wind

First, assuming w=0, p=0.2, 1=30 m, and

big cord, the effect of ground-inclination 0 is obtained as shown in Fig. 3-2. As the

ground-inclination 0 increases, the time t to reach becomes shorter, and larger.

Next, assuming 0=0, it= 0.2, 1=30 m, and big cord, the effect of wind w is obtained as

in Fig. 3-3. As the wind speed increases, the ground run shortens rapidly.

17

EWAN

MIRE

11E/111

MEM

Fig. 3-2

11111111=1.1

111111FAWM

Mal=

ard1111111

_{28 30 32} Fig. 3-3

Approximate Calculation of

Rubber-Cord Launching

Since the sailplane's speed is comparatively small at the beginning of the launching,

omit-ting the last term of (3-1), the equation of

ground run becomes

dv

W {(E1-1- Wsin 0 - Wcos

0)-dt

-E(x)+v.E(01 (3-15)

The equation of motion from t.o. to v,,2. is as before

dv

dt wg {(E1+ Wsin 0)- C S(11

--E(x)+vaE(t)} (3-15)a

The drag term in (3-15)a can be expressed as

CD-P-S(v w)5=kDW

2

and kD 10.35+ 0.65( Vt

vni )2} 5-12 (3-16)

where v,,, = max. speed, vt=t.o. speed. Usually

v,Iv,= 1.3-1.6, and CDICL=1110-1/30.

- S(V w)2=(0.20..005) W= k,,W 2

The ground friction in (3-15) is usually

Wcos 04(0.20-0.04)W-12W u 20/I

P

imonivio

ERNE

011ie

,ailltIlk

co f6

immonom

ussimmii

arall-maap.mil

Rubber cord From start to v. t (sec) x (m) (M/S) VmaxlVto small 3.0 34.6 19.0 1.4 big 2.3 32.5 22.9 1.7 08 1,0 1.2 Fig. 3-1 1,6

1960) Sailplane Performance,

Therefore, combining (3-15) and (3-15)a, the equation of motion, relative to the ground, up

to Vm will be expressed as

dv

= -fi- {(El-kDW)- E(x)+ vaE(01

dt W

Here, for simplicity, the level ground (0=0)

is assumed. Now, put A (m1s3), B (sec-1),

and C (miss) as follows :

A= g (El - k nW)= gk

B=

' C = g Va Ew =

W

The above equation of motion becomes

dv d'x

_=A-Bax+Ct

dt d- t3

Solving this and using the condition, when t=0, v and x=0, the distance x (m) in t (sec) after the start, and the ground speed v (m/s) at that time, are given as

1 C .

x -

(A+ Ct)-B3 sm Bt Ba (3-17) A cos Bt .139 (3-18)' v= v2(1 -cos Bt)+-a-sin Bt (3-19)

From the condition-for v m,dvIdt=0, the time 4m and the distance xm up to v,. are

tin tan, (AcE )1

, A

vi-1-4....t.,)=+ Vat,. B2

-(va+ --e-A2 )cos Btm

= v B 11)sinBtm (3-21)

1 C

Neglecting the small term vaCI A, and putting sin Bt,:--1.0, the max. ground speed vm will be approximately

vmva+ A (3-22)

From (3-22), the max. airspeed in the wind

of w (m/s) will be

v,,,=w+v+

-

EW (3-23)

Since the t.o. airspeed v,=4-1/(W/S)/CL (m/s)

. Stability and Control 55

-0.78 {0.32(w+ va) + EIW - k Di/ W 1E}

v, _{V (WI S)I CL}

(3-24)

First, estimating kn as (.20-.15) for gliders and (.10-.06) for sailplanes, obtain the 1st

approximation of vm/v, from (3-24). Then,

putting this v,/v, into (3-16), obtain k0. And

using this kJ., obtain the 2nd approximation

of v,/v,.

Usually the 2nd or the 3rd

ap-proximation will suffice.

Height Gained in Rubber-Cord

Launching

Let 0=0, vc=glider's rate of climb (m/s), v=glider's airspeed (m/s), Wi=sum of glider

weight and cord weight= W+ Wo (kg). The

equation of glider's vertical motion from t.o. to cord-drop will be W dv.₌ p 2

-

try JV VVI

g dt

2 Putting p S Ki=C L

g

2 W Kr= Wi (3-25)

the above equation of motion becomes

dt

Since the airspeed v is the sum of the ground

(3-20) speed (3-19) and the wind speed w

v=w+ va(1 - cos BO+ sin Bt (3-27)

Substituting (3-27) into (3-26) and integrating

v.= Icif(ajt + (a.) sin Bt+ (a.) cos Bt+

+ (a.) sin 2Bt+ (a,) cos 2Bt} + C, (3-28)

a,=(w+ v.)2+1 v2+ -1 (LI )2- IC2

2

2 B

2 a2= -- (w+Va)Va a.=

-1 (w+ v.)4(3-29)

a.=

-val A

a,- +

v. 2B2

Since the time t in (3-27) is measured from

start, letting tt=time from start

to t.o., the integration constant C, in (3-28) can be

de-termined from the condition, vc=0 at tt.

C,= -

(cot, + (a2) sin Bt,+ (a3),cos Btt+ (a4) sin 2Bt,+ (a6) cos 2Bt1l (3-30)

As mentioned before, the-glider gains just before the cord falls dowri. Therefore, the

height gained h, (m) during the time from t.o. to cord-drop may be obtained using (3-28)

as

h,=- vcdt=K,[=ltB-1AcosBt+

Bt-et 2 B

(14 a, C t

- cos 2Bt+ sin 2Bt + ,t]et

(3-31)

2B 2B K,

where is given in (3-20).

After the cord has fallen off, the equation

of glider's horizontal motion will , be,

dv

g dt

- - (CD cos 8+ CL sin (3)1)-Sva2 (3-32)

where v = glider's airspeed, j9= angle of 'climb.

Assume 13constant during climb, cos ft.:=71 and put

pS

AL= (C D+ CL sin 8)g -T7v (333)

Integrating (3-32) and using the condition

v=v,, t=0

v-ALvt+ 1

x=c vdt-Llloge()

v

Using K1(3-25) as in (3-26), the equation of

glider's vertical motion after the cord-drop will be dv, _{= K,v2 -g (3-36)} dt Putting v (3-34) info (3-36) dv, _K1Vm2 dt 1)2 g

Integrating the above and using the condition,

at t=0, vc=v=rate of climb at

I-"X±(v,-v)- gt (3-38) From (3-25) and -33)

K1_

CL CD±CL sin 13 (3-37) (3-39)

The final height gained h (m) will be

h=hi+c vcdt=h1-7(1)x+

0

(vc.± KA:v4 ( )

where x from (3-35) x= log10 (44124t +1) A

From (3-34), t (sec) required to gain a steady glide speed v,

1

(

From (3-40), -40)a, and -41), the height gained

until the glider begins a steady glide, will be

obtained.

Now, consider a glider of mass Wig gains

a speed v, larger than

its t.o. speed v,

dur-ing launchdur-ing. The excess kinetic energy

being converted into potential energy, the

height h (m) is gained. Then (vm -10= --Hg7:-,gh

h= -L

{( y_

W 2

2g v,

For rapid estimation, (3-42) will suffice.

3-2 Winch Launching

In this method of launching, steel cable of

some 1000 m length is hooked on a sailplane,

and winding the cable onto the winch drum, the plane is launched high into the air just after the fashion of a kite.

(3-42)

56 - 'Hiroshi Seto (Vol. XX

(3-34) (335) (3-40) (3-40)a proper (3-41)

1960) Sailplane Performance, Stability and Control 57

Climbing Angle 0 and Towline Angle 3

Let 0=angle bet. sailplane's airspeed and horizontal, 8=angle bet, cable tension FA and

horizontal, Cmg=wing pitching moment about sailplane c.g. From the equilibrium of forces and moments on the sailplane during

launch-Let FA =cable tension (kg), d=distance bet.

c.g.

and release hook, r =angle bet. d and

wing chord, di= perpendicular drawn from c.g.

on FA.

di= d sin 1(3 + 0)- (r - a)} (3-46)

Eliminating F, from (3-43) and -44)

(D + Wsin 0) tan (0+0+ Wcos 0 +Lt

(3-47)

Eliminating F, and di from (3-44), -45) and

-46)

In= {tan (8+ 0) cos (r a)

--sin (r - a)} (D + W sin 0)+Lti (3-48) where =Cmg SV2f From (3-47) and -48) tan (19+ 0)- 1 (L-L1- Wcos 0 + A

1+B/ D+ Wsin 0

(3-49) where d .

A= sin (r -a),

B=d

cos (r -a)

1,

Given a and v, 8 corresponding to 0 is obtain-ed from (3-49). The max. possible a

main-+L.,1}

Usually L, takes its max. value

sailplane gains the max. height.

(3-50) when the {BL- AD- BW+ Lei} CL=_{(1+ B)S,} {Cmg ± +BCL-ACD--T47}

_{q S}

Let i,=tail incidence against wing chord, trail-ing edge up (+), e=down wash angle of wtrail-ing,

a,=tail angle of attack

at=a -it-e

(3-53)

The max. a that can be taken during winch

towing should be determined so that, at such

angle of attack, the necessary CL, is obtainable by full up-elevator.

1 S

Fig. 3-5

Climbing Path and Cable Speed

As shown in Fig. 3-5, winding up the cable 0A0 with the winch placed at 0, the sailplane is launched along a path Ao Ai A2 ... Am A. Let Om =climbing angle at Am, Om=towline angle at Am, 0= sag angle of cable. The pro-jection of flight path Am An on the ground is

B 0 tan 0.-4.)- A,,,B

B,B-

(3-54)

tan (1/2)(0,+ 0)+ tan (fl)

The height difference bet. and Am is

AC= B,B tan

(0,+0.) (3-55)

The height at An is

An.B= AnCn+ AniBm (3-56)

1 (L,),...=1+B

sin (1/2)(0m+ On) _(3-57)

AC

When the position of Am(B,,,O, AmB,,,) is tained during the launch may be determined

as follows. From (3-48) and -49)

The ratio, cable speed v, to sailplane speed

V is

1

{BL-AD- W (A sin 0+ B cos 0)+ k= 271 V

AmBm ABn 1.

1+ B _{sin (13.-15)} sin

(P.-ing at constant a and v (Fig. 3-4)

L= F, sin (9+ 0) + Wcos 0+ (3-43) D=F,i cos (f3 + 0)- Wsin 0 (3-44)

L111= FAdi+ Cmg. Sec (3-45)

(3-51)

58 Hiroshi SATo (Vol. XX

known, the position of the neighboring A. will be determined from (3-54), -55) and -56). Since A0B0=0, from (3-54)

B00 tan (81-95)

BoB,- (3-54)a

tan (1/2)(00+ 0,) + tan (131- )

From (3-55) and-56)

AiBT= AiCi= BoB, tan-1 (00+ 01) (3-55)a

2

B00 in (3-54)a is the length of the cable, and

the sag angle (1) may be assumed as about

60 (43), (46). _{Thus Ai from (3-54)a and -55)a,}

then A2, A3 ... will be determined one after

another.

Next, consider the case when the sailplane

is towed into the wind of

(m/s)

at

air-speed v and climb angle 0 (relative to the

air).

Then, the climb angle 0, (relative to

the ground) will be

v sin 0

tan g

v cos 0-v. (3-58)

Using this 0 g in place of 0 the towing flight path in the wind will be obtained just in the

same way as before. And to begin with,

as-VW 43-04

A,

Z

Fig. 3-6

suming the sag angle 95A=5---6°, obtain an

ap-proximate flight path. The ratio k= vil v in the wind of v,,, will be from Fig. 3-6

k=1 {cos (0+

A)

1) cos OA

- vw cos (13 -95A) (3-59)

v4= air speed of i, Oti=angle bet. v., and cable

portion i, cb,= angle bet. AO and cable

por-tion i.

Vt. 7 ,

o 1

, {v/ cos 95i+ vw cos (13 - .4)}

cos 00

(3-61)

tan (01-q5j' . + vw sin (19-95A)- vt sin 95i

coS q5.1+ vw cos (8- OA)

(3-62)

Let d=cable diameter (m), w=cable weight

(kg/m), qt=(p12)v?= q on portion i (kg/m'), 4F0=force acting on portion i of length 41= 1/n

(kg).

4F4= 41 {(C D cos Oil -C L sin

Ozhicl--w sin (ft-04+ cbi)} (3-63)

where CL, CD= lift, drag coeff. of cable making

angle

eu with wind.

Divide the cable length 1 into n equal portions, and the cable tension Fi (kg) at i will be

n

Fl= FA+EzIFt=FA+ 1 dE(GD cos

Ou-t n

LSin

wisin

(1, A+ t) (3-64)

n

And the cable tension at the winch end F0

(kg)

The component sailplane speed v normal to

AO joining the plane and the winch v=v sin (0 + OA)

-- v, sin (19--04)+ vi sin OA (3-60)

Let /=cable length when sailplane is at A, /i=distance bet. point i and winch, (4//)v= component speed of point i normal to AO,

1960) Sailplane Performance, Stability and Control 59

n

Fo= FA+ E n cos O(i C L sin

0)qi-n a=1.

n

E sin 0- OA+ 10 (3-65) n t=i

The deflection angle 40, of cable portion i (Fig. 3-8)

41

sin WOO- {(C, sin Ott+ CL cos 01,)q,c1+

F,

+ w cos (19 -OA + (101 (3-66)

Fig. 3-8

From Figs. 3-7 and -8

flit E 40k (3-67)

k=t

Therefore, (1- OA+ in (3-64), -65) and -66)

becomes

C9-9sA+oi)=(j-40,c) (3-68)

Further, in the (n+2)-side polygon, 0 1 2 3... n A, (Fig. 3-8), the following two equations in interior angles 00 and OA will be obtained.

00 + 0A = E 40, (3-69)

1=1.

OA= 1 (2i - 1)40 , (3-70)

2n i=i

Assuming 40, nearly constant, from (3-69) and

-70) IS 00= OA, 40=2 A Thus, (3-66) will be sin (40 nFl[(CDsin 01,+ C L cos 0,,)q,d + + w cos113

-

(n-i +1)041 (3-71)

As said before, assuming 0=5-6° as a 1st

approximation, obtain 1,, (Fig. 3-5), then 40,

from (3-71), the 2nd approximation OA and 00 from (3-69) and -70). In most cases, the 2nd

approximation will suffice.

13800

Fig. 3-10

Winch Towing of Sailplane ASO-G Sailplane data. b=13.6 m, S=17 m2, A=10.9,

c= 1.4 m, 1, = 3.5 m, St= 2.4 m2, W= 330 kg, L/D= 19.4, v 3=0.96 m/s. Release hook position

(Fig. 3-10). Nose (N), middle (M), c.g. (G).

Sailplane's launching airspeed v=19, 22, 25 m/s. Wind speed v,,,= 0, 5, 10 m/s. Sailplane's angle of attack in tow a=4° (N), 6° (M), 7° (G).

Angle of Climb 8, OR

Fig. 3-11 gives the curves (0 vs. 13)

obtain-ed from (3-49), assuming (1) v=22 m/s, N-, M-, and G-tow, (2) M-tow, v=19, 22, and 25 m/s. Assuming v constant, M- and G-tow give larger 0 than N-tow.

Next, obtain Og from (3-58) assuming M-tow, v=22 m/s, and v=0, 5 10 m/s (Fig. 3-11).

Og increases with v,.

Cable Tension FA and Elevator Angle e The cable tension F4 at v=22 m/s is given

by (3-44)

for N-, M- and G-tow as in Fig.

3-12.

In case of M- or G-tow, the cable

TWO-SEATER

SAILPL4NE

7150.4-LJ

im 7

tension EA is about twice as large as that of N-tow.

Fig. 3-13 gives L, required to maintain the max. a during towing, computed from (3-45)

using the above F4. And the up-elevator

angle e required to obtain the above L, is

shown in Fig. 3-14. 711 3 10 7. 2 30. 40 50 60 70 80 79° Fig. 3-11 B. Fig. 3-12 10 20 30 40 50 60 Fig. 3-13 2E0 00 200 00 Winch 402 m &Wawa Fig. 3-16 800

Cable speed vi and Power Required The cable speed vt is computed from (3-59)

as in Fig. 3-17 and -18, where 8-04=angle

bet. horizontal and line joining winch and

sailplane. As the wind speed 14, increases, Allow V r25/>.../

,...---

_---22 _.---22 ---N 79 M 22

12111

mu

8-7:pz

._

modiu,

Inshm

iwkivkli

-\\mq

7sAmb_No

Nola.

W.330 kg

1111Elk.aj.

,ffing111111

II

wimp'

EMI

MIL 111E'LN

-Millrikm

10 20 30 40 50 60 70 8

IMO

V0

IN

rea41-iiii

I/ REM-Ellii

y,"111 22

Iniallihtil

VI

IMO Ilik.,

-

EBIAMIN

INIZIEMEN

MMIII

I SUUUUliall

VI

IMO

Il

V.2.5mls _...

11111111

11

-.

roll

OW

A 1- 1

60 Hiroshi SATO (Vol. XX

10 20 30 40 50

B.

Fig. 3-14

Height Gained

Assuming 4 mm tow-cable of 1000 m length,

no wind, v=22 m/s, the climbing path for

each N-, M- or G-tow is obtained as in Fig.

3-15.

In case of G- or M-tow, the height

about twice as high as N-tow can be gained.

Fig. 3-16 shows the effect of wind on

M-tow, where v=22 m/s and load factor n=1.69 is assumed. The height gained increases with

the wind, but

if v is kept constant, n also

remains constant, regardless of the wind speed.

winch Olean= Glider

Fig. 3-15 . n.1.69 402 m 622 BOO 24 2 1 4

1960) Sailplane Performance, Stability and Control 61 1 8 6 0 t VI .6 8

smaller cable speed v, will be needed to give the sailplane a definite airspeed, v=22 m/s for

instance.

Fig. 3-19 gives the cable tension F0 at the

winch end, computed from (3-65) assuming

the cable weight w=6/100 kg/m. The effect

of wind speed v on the cable tension F0 is

small.

Using the above v1 (m/s) and F0 (kg), the

required power is computed from (HP= F0v,/75) as in Fig. 3-20.

In case of G- or

M- tow, more height can be gained, but more

power will be required than in case of

N-tow. The stronger the wind, the less power

is required to

obtain a

definite launching

speed.

Figs. 3-21, -22 and -23 give the sag angles

of the cable, OA (sailplane end) and 00 (winch

end), obtained from (3-69), -70) and -71). In

V022'%. IC?o 20 20 40 I3° Fig. 3-21 M-tow. Vw. 0 Fig.B.3-22 14-tow. V 221% /3° Fig. 3-23 -

ENE VW °

=

IIIB.

-is

-'114till

1111ENIE

IIIi

Hofinnz

A* v;. 1111111111111

IMM

IIIMINAINI

MN=

111 MK

--B_;:mIg

41 WI

immilum

m

imam

I W 330 kg

16111111=1

WM

MEER

II

all'IM

NOM

--v-2---4______ _ , tU V45071, -- I I?

:6MMu

4 0 1

--

--- ---ii----::::::-

---:---.7--NEM

NB

m--I

Ettmmil

NArarn

MS

--.. .40 lito .. . -...-:.: ---7 Jo LO sn so 20 40° 60 80 Fig. 3-18 20 40 0-0A so Fig. 3-19 20 40° 60 so Fig. 3-17 MIOve. 40° so 8 134 Fig. 3-20 so ao so fro 2 1 6 4 1 6 HP 4 20 7. a 0 6 2

case of M- or G-tow, the sag angles are

smaller than those in case of N-tow. They

decrease as the launching speed of the

sail-plane v increases, but they increase with the

wind speed vw.

3-3 Aero-Towing Take-Off

For aero-towing, the cable of 40---80m length is commonly used. When the sailplane

takes off

first and gains 3-4 m height, the

sailplane pilot, pushing the stick gently and

slackening the cable, should help the airplane to take off. Towing take-off may be divided

into three stages, (1) from start to sailplane

t.o.

(2) from that to airplane t.o.

(3) after

that.

Let W, Wg=airplane, sailplane weight (kg),

We= cable weight (kg), W= Wa+ Wg + Wc, T = TO-a(p12)v2= airplane thrust (kg)("),

To=static thrust (kg), a=drag area

account-ing for thrust decrease (m2), v=ground speed

of tow-train (m/s), w=wind speed (m/s), va,

vg=airplane, sailplane to. airspeed (m/s), and g=coeff. of ground friction=0.03-0.10.

From Start to Sailplane T.O.

Since the airspeed is (v+w), the equation

of motion for this t.o. stage

dv = {To-(PaWa+ AzgWg)}

g dt

-P- {cc. pacL.)s. +₂

+ (Cpg-ggCLg)S,+ a} (v +

Put 2=11(m/s2) and B1(m) as follows : A,- g { To- (PaWa+ Pit W g)}

B,= W 2g { (CDa -paCLOS, + + (C DA. PgC Lg)S g a} dv dv =v = A,- Bi(v+w)2 dt d

Since the sailplane's to. ground speed is

(vg-w), integrating the above from v=0 to (vg-w) gives t.o. time ti(sec) and distance run x(m) as

1.15 (1/ AilB + v g) 1111B1

-t1-

loglo

VAIBI (VAI/B1 -vg) (1/A1/B1 + w) (3-76) _1.1510 (Ail B,-wa) Bi 6'1)(41 Bi-vg2) + 1.15w km (i/Ai/Bi +w) (1/ AilB1 - v l*1° Ai/B1- Ailg7 + vg) (3-77)

From Sailplane T.O. to Airplane T.O.

As the sailplane, after taking off, tends to

climb higher, the sailplane pilot should keep

its height 3-4 m above the airplane, by re-ducing CLg inversely as (v+w)2. However, as the variation of sailplane's LID during this

t.o. stage is not so large, assume CLgICDg approximately as a constant(O®. _{Then the} sailplane drag Dg'-7(CDgICLg)Wg, and the

equation of motion for this t.o. stage will be

g dt

Lg

dv

Wa rC Wg}

f± (C/Ja PaCLa) Sa a} (v + w)2 (3-78)

2

where v=ground speed, as before.

Now, putting A

2= w

g {To-

PaWa

W el Lg B - g --e-{(Cpa-ilaCLa)Sa+a

_2W2

2 (3-78) becomes dv dv (3-81) =V A,- B2(v + dt d

Integrating this from sailplane's t.o. ground

speed (vg-w) to airplane's (v.-w) gives 12(sec)

and x2(m) for this stage as

1.15 (1/ A2IB2 + v a)(1/ A2IB2 - v i)

t,-.11213210810(1/A2/B2 -va)(vA2/B2+vg) (3-82) (3-79) (3-80) - 1.15 r iogio (A2032-v + B2 L (4421 B a- v (3-74) A2IB2 + v a) (1/A2/B2 - vg) l + w B21Aziogio (1/A2/B2 va)(vAs/B2 + vg) (3-75) (3-83)

As seen in (3-82), t2 is independent of the wind.

After Airplane T.O.

It is customary to calculate the total

dis-tance required for the airplane to clear a

I5-m obstacle. In this stage, assume the

air-62 Hiroshi SATO (Vol. XX

(3-72)

1960) Sailplane Performance, Stability and Control 63

plane as making steady climb at

constant airspeed v8(m/s) and rate of climb u3(m/s)(49).

Then, the time to(sec) and the distance x3(m)

for this t.o. stage will be

ts = 15 (3-84)

Us

xo= to(vo w)= 15

w)

u9 Vs

Here, again to _is independent of the wind.

The total aero-tow t.o. time t(sec) and

dis-tance x(m) over a 15m obstacle

t=ti+t2+t3 (3-86)

x=x1-Ex2d-x3 (3-87)

Towing Take-Off in Still Air

Assuming w--=0 in the above equations, t.o.

time to and distance xo in still

air will be

obtained from (3-76) and 77) as 1.15 (1/A1/B1 + vg) 10g10 1/111B/ (1/ AIM/ Vg) 1.15 logio(l+vgi/Bi/Ai ) _(3-88) VAIB, vg i/B1/Ai ) xi.o= B1 1.15 log/0 (A,IB, vg2) 1.15 logi0(1 vg2---LB (3-89) B, A,

In most cases, since vgl/B1/A, <1, Ai< 1,

developing the logarithms

ti

1 SG

+

1(

vg Bi y+ vA,Bi g V A, .1 3 \ V A,

+Ike /TY+

1 .1(v 2 .B1)± 1(vg2 B1 y+ ' 2B1

(\ g Ai]

2 \

Ali

+-1(v2 B')3

+

J

Adopting the first three terms in (3-90) or 91) will give a sufficiently accurate result,

but for approximation, only the 1st term will

be adopted

Vg

A1'

xVg2

/0 2A,

(3-85)

(3-92)

These give a little smaller values than (3-90)

and 91).

As before, for the 2nd t.o. stage in still air

1.15 * t2 t2

-- X

A2B2 (1+ vai/B2/Aa ) (1vg-VH2/A2) x log10

(1 vai/B2/A2) (I+ vo/B2/A2)

X2.0 B

_{1.215 logio((AA2IB2 vg22)}

_{2IB2v )}

1.15 (1v 2B21212)

log10 g.

B2 (1 VajBai A2)

Since usually

vai/B2/A2 <1, v. aBol A2< 1

vgi/B2/442<1, 242B21 A2< 1

developing the logarithms

t2 1 (v.-14)(1/ 1:)+ A2B2( 31 (vas vffs)(1/BA22)3+ (Va5 _{Vg5)(/BAl22)5} 5 x2'°= 142{(va2 vg2)(t) (v.4 vg4) (1122)2 A (Vaa vga) (Ds +...

Again, adopting three terms will give a suf-ficiently accurate result, but here only one

term will be adopted for approximation

2 2

V g Va. V g

X3.0 0-97)

A2 2A2

which gives rather smaller values than (3-95)

and 96).

From (3-84) and 85), for the 3rd stage in still air

15 Vs

t3.&c4 s =

xs.otsvs=15

143 Us

The total towing t.o. time to(sec) and distance

x0(m) over a 15m obstacle, in still air will

be

t0=t1.0-1-t2.0+t3.0 (3-99)

x0=x1.0+x2.0+x3.0 (3-100)

Towing Take-Off in Wind

Since (vgw), (v. w)= sailplane, airplane

t.o. ground speed, and (v3w)---ground speed

of both planes in climb, putting (vgw) for

(3-98) (3-90) (3-91) (3-93) (3-94) (3-95) (3-96)