designing chemical rockets

8/4/2019 Designing Chemical Rockets

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To get an idea of the a ltitude you m igh t ge t your rocket to

reac h, y ou can u se som e trig on om e try . Ifyo u k no w th e d istan ce to

th e lau nch er fro m th e firin g site , an d m ak e a sm all dev ice to d eter-

m ine the m ax im um ang le to the peak of the rocket tra jec to ry , you

c an u se trig to so lv e fo r th e a ltitu de (sin e v alu e). R e fe r to F ig . 9 -1 3.

M e asu re off th e d istan ce to th e ro cke t from th e an gle m easu r-

m g pair o f sticks, m oun ted so one can p ivo t on a pro tracto r sca le .

F ig . 9 -1 0 . A comp lete mod el rock et s ys tem b uilt a rou nd th e E s tes C h alleng er-1 .



F ig . 9 -1 1 . T he th rill of lau nch ing a m od el rock et (cou r tes y of E s tes Ind us tr ies ).

Keep the bottom stick parallel to earth. When the rocket goes up,

just move the top stick to keep it aligned with the rocket and let it

remain at that position where the rocket reaches its peak. Then

read the angle.

Assuming that the rocket goes straight up, whichis not always

true, use these relationships:

Therefore, H = cos (angle)

Therefore the Altitude =H sin. angle(where H is in feet)

the sine and cosine may be obtained fromany trigonometry book or

set of trig tables. Youcan also get the average rocket speed. See

Fig. 9-13.

H cos. (angle) = 40 feet (in the example)

40



PEAK

/1/1

/ f

/ I/

H/

A NGLE ME A SURING /STIC KS ON A LA RGE /PROTRACTOR

/

,ANGLE

~,

I ALTITUDE

I

I

_ A LTITU DE (F T ) _ROC KE T VE LOC ITY - TIME (SE C ) - SPE ED (ft/s ec)

F ig . 9 -1 3. D eterm ining a ltitud e. U se a s top w atch to g et tim e from la unch topea k .

s tability . T his is s im ila r to th e sp inn ing o f a bu lle t cau sed by land s

an d g ro ov es in th e gu n ba rre l. It h as been ca lcu la ted th a t-a ll o th er

e ffe c ts be in g equa l-m ax im um g rou nd range from such a lau nch is

a ch iev ed by lau nch in g a t an an g le o f 45 d eg ree s . You can con firm

this with y ou r m o d el ro ck ets .

A E R O D Y N A M IC S O F R O C K E T S

T he con tro l p roblem o f g uid ing a ro ck e t is la rg ely co nn ec ted

w ith th e ae ro dy nam ic p ro blem , m u ch a s it is in th e case o f th e m o de l

a i rp lane . In th e ro ck e t, h ow eve r , w e a re co nce rn ed with mu c h

h igh er sp eed s , an d s in ce th e m otion is n ea r th e sp eed o f so und

(abo ut 700 m ile s p er h ou r) , w e m u st a lso co ns id er su ch th in gs a s th e

sho ck w ave s w h ich can a ffe c t th e con tro l p ro blem . T he sho ck

w ave s a re ju s t on e e ffe c t; th e re is ano th e r. Becau se th e ro ck e t

m oves so fas t, s lig h t v a r ia tio n s (no n sy rru n e try ) in th e bod y co n -



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F ig . 9 -1 5. U ng u id ed rock et ta rg et (cou rtes y of US A ir D efens e C e nter , F t. B lis s ,TX).

s tru ctio n can cau se lift an d d rag fo rc e s th a t a re ig no red a t low er

speeds .

T he concep t o f a sh o ck w ave is no t to o d iff icu lt. Ju s t im ag in e

th a t th e a tm osph e re is f illed w ith little m olecu les o f a ir . W hentrav e lin g a t subson ic sp eed s a p re ssu re w av e o r d is tu rban ce is se t

up in th e a ir , lik e a so un d w ave . T h is w av e trav e ls ah ead o f th e

m o vin g bod y, w arn in g th e m o lecu le s o f a ir th at th e bo dy is com ing .

T h e p res su re d is tu rban ce thu s p reced e s th e ro ck e t bo dy .

A s sp eed in c re a se s , th e bo dy ca tch es up w ith th e se p re ssu red is tu rban ce s . In fac t, w h en th e bo dy is m o vin g a t th e sp eed o f s ou nd

it is m ov ing w ith th e sam e speed a s th a t o f th e p re ssu re d is tu r-

ban ce . T h ere can be n o w arn ing th a t th e bo dy is com ing and th e

p re ssu re w av es ten d to bu ild up an d rem a in , lik e a barr ie r , a t th e tipo f th e ro ck e t.

T h e w av e th a t is fo rm ed is c a lled a sho ck o r m ach w ave. (The

m ach nu m be r is th e ra tio : sp eed o f body /sp eed o f sou nd .) Sin ce th e

sp eed o f sou nd is abo ut 700 m iles p er ho ur ( it v ar ie s with t empe ra -

tu re and ba rom etr ic p re ssu re ) , a ro ck e t w ou ld be go ing a t m ach 1

w hen it is g o in g 700 m iles p e r ho u r , m ach 2 a t 1 ,400 m ph .

Tw o k in d s o f p re ssu re w av e s a re se t a rou nd any p ro je c tio n

w hen it is trav e lin g a t, o r fa s te r th an , th e sp eed o f s o un d. T he se a re



th e n orm a l sho ck w av e th e th e o bliqu e sh ock w av e as illu stra ted inF ig . 9 -1 6. T he re will be an obliqu e shock w ave w hen the body is

m ov ing fa ste r th an th e sp eed o f sound , and th e m ach ang le o f F ig .

9 -16 is in d ica tiv e o f how m uch fas te r th an th a t o f sound th e ro ck e t

sp eed is . T he sm alle r th e ang le , th e g rea te r th e sp eed , e tc .

T he effec t th is h a s on th e con tro l p roblem is tha t su ch shock

wave s will fo rm at th e tip s o f the fin s and body . Sin ce th ey aregen era lly V shaped , th ey can encom pass p a rt o f th e f in in th e shock

w ave reg ion w hile leav ing th e re s t ou ts id e , a s show n in F ig . 9 -17.

T h e c on tro l fo rc es will be d iffe re nt fo r th e tw o d iffe re nt re gio ns , a nd

thu s w e cou ld have in s tability , and a ro ck e t w e cou ld no t con tro l.

W e des ire a fin th a t is en tire ly ou t o f the shock -w ave reg ion

w hen th e ro ck e t is m ov ing a t th e d es ign sp eed (F ig . 9 -17). T h is

,-aco un ts fo r th e m an y d iffe ren t sh ape s o f fin s, w h ich a re d es ign ed

from ca lcu la tio ns and experim en ts to in su re th a t th e fin s rem ain

ou ts id e th e sh ock -w av e reg io n.y

T he cro ss-sec tio n a l a re a o f a ro ck e t f in d iffe rs from th a t o f an

a irp lane . It m ay be sym m etrica l and rounded on each s id e , o r it m ay

be d iam o nd -sh ap ed , a s sh ow n in F ig . 9 -1 8. T here is a lso a d iffe ren t

PRE SSURE W AVE BA RRIE RBE NDS W HE N ROC KE T TRA VE LS

F A STE R THAN MAC H I

/

PRE SSUR E W A VE S PR EC E DE

ROC KE T WHE N ROC KE T TRA VE LS

A T SUBSONIC SPE EDS

/

CpA.PRE SSURE WA VE S SUI LD UPA T SPE ED OF SOUND (M AC H I)

NORMAL -+

SHOCK

OBLIQUESHOCK

F ig . 9 -1 6. C oncep t of a s hock w ave.



PA RT OF W ING IN

SHOC K A RE A

-0

RAKED W I N G NOT

IN SHOC K A RE A

F ig . 9 -1 7. Th e effect of s hock w aves on control,

concept of liftwith the rocket fin. After the airflowpasses throughthe shock wave, its pressure increases, while its velocity de-

creases. When the a ir passes through an expansion wave, its

velocity increases and its pressure decreases. Thus, as shown inFig. 9-18, there are two primary lift regions on this type of airfoil,

sides 1 and 2.

Finally, at the speeds at which the rocket travels, the body

itself contributes considerable lift which makes it possible to havejust tail surfaces for control and stability. In some rockets jets are

used to control direction (even the rear fins are omitted) and keep

the rocket following a stable trajectory.

A IR F LO W

E XPA NS IO N W A VE

1///I / ~

~'\ LOW PRE SSURE

\ ~l~"--O_----""\\ H IGH PRE SSURE

\ C ROSS SE C TION OF F IN

SHOCK

F ig . 9 -1 8. Sh ock and exp ans ion w aves on a fin. Th e lift is d ue to h ig h p res sure a t1 and low p res sure a t 2 .



MASS

F ig . 9 - 22 . A c ce le romete r p r in cip le .

If th e m o to r o f a p ar ticu la r m iss ile p ro vid es th ru st w h ich

acce le ra te s th e m iss ile con s tan tly a t 100 fee t p e r se co nd , w ha t

d is ta n ce ( d is p la c eme n t ) wil l be cove red w ith th is a cce le ra tio n fo r a

p erio d o f 1 0 seco nd s? A ssu m in g th at th e m iss ile s ta r ted fro m re s t,

a t th e end o f 10 seco nd s , it w ou ld h ave a f inal ve loc i ty , V r , a s

fo l lows :

V f = 100ft/sec /sec x 10 seco nd s = = 1,000 ft/sec,B ec au se th e m iss ile s ta rte d fro m res t, th e av erag e v elo city (Yav) is :

Va v 0 + 1.000 = 500 f t/s e c.2

T he d is tan ce trav eled du rin g th e 10 seco nd s (th e d isp la c em en t) is

fo und by m u ltip ly ing th e av e rag e v e lo c ity by th e tim e:

S = 500 ft/sec x 10 secon d s = 5 ,0 00 f t.

T hu s , d is tan ce trav e led , o r m iss ile d isp la cem en t, m ay be

fo un d by k no w in g th e acce le ra tio n an d th e len gth o f tim e it o ccu rs .

Ac tua ll y , a c ce le r a ti on will n ot be co ns tan t. Bu t th e d is tan ce m ay still

be found by th e m ath em atic a l op e ra tio n k now as in teg ra tio n . De -

v ice s to p e rfo rm in teg ra tio n a re in co rpo ra ted in th e o n-board m is-

s ile com p ute r . A cce le ra tio n s a re m easu re by acce le ro m ete rs , th ep rin c ip le o f w hich is bes t illu s tra ted by a pendu lum . F igu re 9 -22

sh ow s a p en du lu m su pp orted by a p iv ot th ro ug h a c ro ssp iec e . If th e

c ro ssp ie c e is a cce le ra ted to th e rig ht, a s sh ow n , th e p en du lum will

sw in g to th e le ft by in er tia l rea c tio n .

T h is sam e e ffec t o f in e rtia is exp e r ien ced in a ca r w hen th e

d riv e r s tep s on th e g as . A s th e sp eed o f th e ca r in c rea se s yo u a re

p ressed back aga in s t th e sea t.

Decrea s ing sp eed p rod uce s an op po site e ffe c t.



T E L E M E T R Y

T elem etry m ean s send in g in fo rm atio n from afa r. In th e te le -m etry sy stem , how ev er, w e p lace th e tran sm itte r in th e rem ote o r

m ov ing loca tion . In s tead o f send ing com m ands back , it send s s ig -

na ls to th e rece iv e r from m easu ring in strum en ts , w h ich w e can use

im m ed ia te ly o r reco rd fo r fu tu re ana ly sis .

O ne o f th e sim p lest te lem e try sy stem s is p robably th e D opp ler

sy stem (so n am e d be cau se o f th e p rin cip le in vo lv ed ) w h ic h will g ive

in fo rm atio n on speed and positio n in sp ace .

A sm a ll c ry sta l-c on tro lle d tra nsm itte r will sen d ou t a p u re

carrie r . W e can p ick th is up on a rece iv e r and a t th e sam e tim e w e

c an fee d in a sig na l f ro m a sig na l g en era to r, w h ic h is tu ne d to e xa ctly

th e sam e frequency as th a t o f th e tran sm itte r u n til th e tw o sign a ls

a re exac tly a t zero beat .

W hen w e pu t th is tran sm itte r in a rocke t and launch it, th ere isan e ffec t (th e D opp le r e ffec t) w h ich is th e sam e as th at no ticed w h ile

lis ten ing to th e ho rn o f an app roach ing ca r o r tra in . As it sp eed s

tow ard you , th e p itch o f th e m ov ing sound gene ra to r is h ighe r th an

w hen it is no t m ov ing . As the sound gene ra to r passes th e p itch

low ers . T he sam e is tru e o f th e tran sm itte r in th e rocke t o r m iss ile .

As the rocke t speed s aw ay , in c reas ing its d is tance from the launch -

ing site , th e frequen cy o f th e sig na l sen t ou t by its tran sm itte r will

arriv e a t th e rece iv e r low ered by an am oun t p ropo rtion a l to th e

sp eed o f th e rocke t. T h is will p ro duce a bea t no te w hich will be

p la in ly a ud ible to th e h um a n ea r if th e s ign al g en era to r is k ep t tu ned



ex ac tly to th e frequ en cy o f th e tran sm itte r be fo re th e rocke t w as

l aunched .

A s th e ro ck e t is la un ch ed th e sp eed o f th e m ov in g tran sm itte r

will cau se a frequency d iffe ren ce a t th e rece iv e r w hich w ill be

in d ica ted by an in c rea sin g to n e p itch . T he sound w e hea r s ta r ts a t a

v e ry low no te , in c rea se s to som e h ig h e r to n e as th e acce le ra tio n

c on tin ue s, d ec re ase s to ze ro a ga in , a nd th en be gin s in cre as in g u p to

som e h ig h p itch w he re it w ill abru p tly s to p w hen it c ra sh es in to th e

g ro und . A g raph o f th is e ffec t is sh ow n in F ig . 9 -28, w he re th e

chang in g p itch is sh ow n in re la tio n to th e tra jec to ry . As th e rocke t

sp eed in c rea se s and reach es a m ax im um , the p itch in c re a se s . T he

m oto r bu rn s ou t a t B , w hich is th e h ig h est p itc h . T he ro ck e t th en

co as ts o n up to its p eak a ltitu d e , d ec rea sin g in sp eed as it d o es so .

A s its sp eed d ec rea se s , th e p itch a lso d ec rea se s . A t th e pe ak , th e

ro ck e t is n e ith e r m ov in g aw ay no r re tu rn in g to u s , so th e p itch is

ze ro . W h en th e ro ck et, u nd er th e p ull o f g ra vity , be gin s its d esc en t,its sp eed ag a in in c rea se s , an d aga in w e h ea r th e to n e in c rea sin g in

p itch . T he m ax im um ve lo c ity o f th e ro ck e t is reached ju st a s it h its

BA LLISTIC TRA JE CTORY

FREQ

"ps

B 02500

2000

1500ABRUPT

1000 STOP

500

° A PEAK TIME

OF

TRAJECTORY

e

F ig . 9 - 28 . T he ra d ia l velocity of th e rock et ca n b e ca lcu la ted b y d ivid ing th e b ea tfreq uency b y th e freq uency of th e tra nsm itter (A ). B allis tic tra jectory is s h ow n in(8).



Tab le 9 .1 . Aud io Oscil la to r and Potentiometer Ca lib rat ion.

Potentiom eterwiper pos i tion Tone (cps )

Centered 60 01 0 d eg rees r ig h t 6502 0 d eg rees r ig h t 70 0

30 d eg rees r ig h t 7501 0 d eg rees left 55020 d eg rees left 50 030 d eg rees left 450

the earth . T herefo re , w e hear a t tha t m om en t the h ighest p itch and

then silence . By listen ing to the tone o f the receiver ou tpu t, w e can

te ll w hen the rocket has reached its peak , and ifw e clock the tim e it

tak es to reach the peak w e can d eterm in e rou ghly its a ltitu de , since

the d istance it w i l l f al l i s: ¥ 2S =-:2'

gt

(Where g is th e g rav ita tio nal co nstan t, 3 2.2 feet p er seco nd , an d tis

the tim e th e ro ck et is fa llin g.) A very close ap pro xim atio n is tha t it

w i l l fa ll f o r as long as it took to reach peak altitude . N ote that th is

situation wil l be true only fo r the h igh ang le overhead sho t. In

ball is ti c t ra je ct or ie s t he p itc h w i l l n ot d ecrease co m plete ly to zero at

peak altitud e sin ce th e rocket is still m o ving aw ay from the g rou nd

receiver .It is possible to determ ine the rocket speed from th is tone

in form atio n, k now in g th e frequen cy of th e rad io carrier. T he ton e

w e receive , in cycles per second , is exactly equal to the num ber of

w avelength s per second that the rocket is m oving aw ay from th e

rece iver on a rad ia l path . L et u s use , fo r exam ple , 144 MHz, which

h as a w av eleng th o f a bou t 6.86 ft. At ten seco nds after laun ching , if

th e tone is 1000 cycles per second , the rocket speed is 6.860 ft. per

second . The highest p itch tone at tha t in stan t (averaged over a

on e-secon d in terval) w ould giv e th e maximum ro ck et sp eed .

The nex t step w ould be to try to c olle ct o th er d ata lik e tempe r-

a tures and pressu res. T his can be done ifwe mod ula te th e tra nsm it-

ter with a num ber o f tones and m ake each tone vary with th e

quantity w e w an t to m easure . N ote that th is m eans w e m ust have

sma ll a udio o s cil la to rs c ap able o f havi ng f re quenc ie s c ontr oll ed with

p oten tiom eters , by v ariable cap acito rs, in du cto rs , o r by m e an s o f a

voltage .



PROPULS ION

The rocket engine furnishes the power which drives the rocketor missile into space. There are many types of jet propulsiondevices. Let us examine some basic principles of rockets, both thesolid and liquid propellant types.

There are two general types of solid fuel rockets, the re-stricted burning and the unrestricted burning types. In the re-

COMBUSTION SUPERSONIC- CHAMBER --NOZZLE-

IGNITERENOVIEW

C D UNRESTRICTED BURNING

stricted burning method (Fig. 9-31), the propellant is ignited on one

end only and the fuel bums like a cigarette. This type gives thrust

for a longer time than the unrestricted burning method.

In the unrestricted burning type rocket (Fig. 9-31), the fuel,

which is cast in solid form, is molded or slotted so that it bums

throughout its length. This type of rocket produces a much higherthrust (or push) for a much shorter period of time.Every rocket engine has a nozzle. This is the tapered and

flared exhaust section shown in Fig. 9-32. The nozzle has converg-ing and diverging sections; these tapers are very important ingetting the -nost power from the rocket. The converging sectioncauses the gas velocity to increase until sonic speed (the speed ofsound) is reached at the throat. The diverging section furtherincreases the velocity by allowing the gases to expand rapidly out

into the atmosphere. The expanding gas at the exit can reach avelocity higher than 2000 feet per second.

There is a popular belief that the rocket produces thrust whenthe exhaust gases push against the air. This is not true, because the

SOLID PROPELLANT

Fig. 9-32. Typical rocket engine.



Fig. 9-33. Action equals reaction.

rocket can operate in a vacuum and in space. What actually happens

is that the mass of gases pushed out of the rocket combustion

chamber at high speeds causes a reaction on the rocket. This

reaction is what produces the thrust.

A simple example of the force produced by such a reaction is

the kick ofa shotgun against one's shoulder when the lead pellets fly

out the barrel (Fig. 9..;33). Another way to show the effect of

reaction is to put on a pair of roller skates and throw some heavy

rocks or bricks forward. Your effort in throwing the bricks forwardwill produce a reaction that will move you backward. The important

point in the rocket is that, even though the gas molecules are very

small and have very little mass, they move at such trememdous

velocities that large reactions are produced.

The thrust (F), then, is obtained by increasing the velocityMof the gases from a zero value inside the combustion chamber to avery h igh value leaving the exhaust nozzle. This is called increasing

the momentum of the gases. Momentum is the quantity of motion

and is the product of the mass and the velocity (V) of the gases. Themass of the an object is its weight divided by the gravitational

constant 32.2.

The rocket-thrust equation can be written:

F = (Momentum at exit) - (Momentum in chamber)~

time to move from chamber to exit

or

F= (Mass x exit velocity) - (mass x chamber velocity)

timeor mathematically

W-xVe

F = g = W X Vet g x t

where Wis the weight in pounds, g is 32.2, t is time in seconds and

v; is the gas exhaust velocity in feet per second. The part of the



equat ion J Y : is th e am oun t o f p rop e llan t bu rn ed pe r second .t

N o w a n e xa m ple to c la rify th e th ru st e qu atio n: A ssu m e th at w e

h av e a p ro pe lla nt w e ig hin g 1 0 p o un ds a nd th at th is p artic ula r p ro pe l-lan t, w hen bu rned in a com bu stion cham ber w hich h as a p rop e rn ozzle , p ro du ces an exh au s t-g as ve lo c ity o f 1000 fee t p e r seco nd .

A ssu m e a lso th at th is bu rn s 5 p ou nds o f fu el p er secon d. W hat is th eth ru st p ro du ce d?

F = 5 (pound s) x _!_ x 1 ,0 00 ( ft/s ec )1 (second ) 32.2 1

F " : :: 5 000 = 155.28 pound s o f th ru st fo r 2 seconds32.2

Becau se w e h ave 10 p ound s o f f ue l an d bum on ly 5 . p ou nd s p er

second , th e eng ine will bu rn fo r 2 second s. If w e inc rea se th e

bu rn in g ra te (po un ds pe r secon d) th e th ru st will in cre ase , bu t las t

fo r a sh orte r tim e . C on ve rse ly , ifw e decrea se th e bu rn ing ra te th e

th rus t will d ec rea se , bu t la s t fo r a lo ng e r tim e .

C H E M I C A L R E A C T I O N S

Sin ce th e en erg y w h ich p rop els th e ro cke t com e s fro m chem i-

c als , le t u s a na ly ze c he m ic al re ac tio ns , p artic ula rly th e re ac tio n o f

zin c a nd su lfu r, w h ic h w e will u se a s th e p ro pe lla nt in th e ro cke t w ed esig n in th e n ex t ch ap te r.

T he en e rgy w hich p rop e ls th e ro cke t com es from a ch em ica l

reac tio n, o r fro m th e bu rn in g o f ch em ica ls . T h is p ro cess re lea se sene rgy in th e fo rm o f h ea t. T he h ea t in turn p ro du ce s k in etic

en e rg y , w hich is th e ene rgy o f m otion .

O ne s im p le chem ica l reac tion w h ich p ro du ces h ea t is th e co m -

b us tio n o f h y dr og en (H s ) and oxyg en (02) to fo rm w ate r p lu s h ea t.

T h e re is a w ay to show th is reac tion by m eans o f sym bo ls in an

equat ion : 2H 2 + 02 ~ 2lliO + hea t

T he arrow ind ica te s th a t th e hyd rog en m olecu le p lu s th e oxygen

m o lecu le ch ang e to w ate r (2H 2 0) p lu s h ea t.

Pe rh ap s y ou re ca ll m i xin g a n e le ctro ly te , u sin g d is tille d w a te r

and su lfu ric ac id . W hen the ac id w as pou red in to th e w ate r th e

m ix tu re a lso d eve lo ped h ea t. T his is an o the r ch em ica l reac tion

w h ic h p ro du ce s h ea t.



Ano th e r h ea t-p ro du c ing p roce ss , no t n o rm a lly tho ugh t o f a s

co m bu stio n, is th e reac tio n be tw een zin c (20) a nd su lfu r (S) to fo rm

z in c s u lf id e ( Z n S ) . T his can be sta te d as a ch em ica l equa tion :

Z n + S~ZnS + hea tT h is is th e bas is o f th e p ro p e llan t w e w ill u se in o u r m ode l

ro ck et d es ig n p ro ble m s.

P R O P O R T I O N O F P O W D E R S

W e wil l use tw o pow ders -zin c and su lfu r. T o know how m uch

o f on e to m ix w ith th e o th e r, w e m ust ap p ly th e law o f ba lanc ed

c h em ic a l e qu a tio n s.

W e firs t d e te rm ine th e fo rm u la w eig h ts o f th e ch em ica ls to be

u sed . T he fo rm ula w eig h t is equa l to th e sum o f th e w eigh ts o f th e

a tom s ind ica ted by th e fo rm u la . T he a tom ic w eigh t o f zin c is 65.38and th a t o f su lfu r 32.07.

N ow w e a lso know som eth in g e lse from th e ch em ica l law . In

th e case o f th e com bin a tio n o f h yd ro gen and oxyg en to fo rm w ate r

p lu s hea t, 2 a tom s o f hyd rog en com bin e w ith 1 a tom of oxygen , a s

d eno ted by th e 2 in fron t o f th e hyd ro gen sym bo l. In th e p re sen t

ca se , 1 a tom of zin c will co m bine w ith 1 a to m of su lfu r to p rod uce th e

zin c su lf id e an d h ea t. W e k no w th is becau se th ere a re n o n um bers in

fro n t o f th e p a rts o f th e equ a tio n . T h e fo rm ula w eig h t th en is:

T he num ber o f a tom s o f zin c tim es th e a tom ic w eig h t o f zinc,p lu s th e n um ber o f a tom s o f s ulfu r tim e s th e a to m ic w e ig ht o f su lfu r.

T he fo rm u la w eig h t can now be sta ted sym bo lica lly a s:

F orm u la w e ig ht = (1 x 65.38) + (1 x 32.07) = 97.45

T h is m ean s th a t w e w an t th e sam e ra tio fo r any am oun t o f

p ro p e llan t a s th e ra tio w e w ou ld g e t by m ix in g 65.38 p ou nd s o f zin c

wi th 32.07 p ou nd s o f su lfu r fo r a to ta l o f 97.45p ou nd s o f zin c s ulfid e

prope l l an t .

W e will no t u su a lly w an t a quan tity a s la rg e a s 97.45 pounds .

W e w an t to k now th e p ro po rtio n s o f zin c and su lfu r needed to mix ,

say , 1 p ound o f p rop e llan t. T he fo rm ula is :

65.38 + 32.07 = 1 p ou nd o f p ro pe llan t97.45 97.45

o r fo r a m ore p rac tica l am oun t:

0.67 po und s zinc p lu s 0.33 p ou nd s su lfu r = 1 p ou nd p ro pe lla nt

If w e w an t m ore p ro pe lla n t, w e can now m ultip ly th ese tw o

num bers (0.67 and 0.33 ) by th e num ber o f pound s o f p rop e llan t w e

wan t . Ifw e w an t 10 pound s o f p ro pe llan t w e w ou ld h av e to m ix 6.7

pounds zinc w ith 3 .3 p ound s o f su lfu r.



Ano the r u sefu l th ing to know is th e ra tio by w e igh t o f o ne

chemica l in th e fo rm ula to th e o the r. W e find th is by d iv id ing the

la rge r num ber by the sm alle r:

R· 65.38 204an o =32.07 .

T his m eans w e a lw ays u se 2.04 p arts by w eigh t o f zin c to eachp art o f su lfu r.

L IQ U ID P R O P E L L A N T R O C K E T S

T o ov ercom e w eigh t an d co oling p roblem s fo r lo ng -du ra tio nu nits , liq uid p ro p ella nts a re u se d. In m o s t liqu id p ro p ella nt ro ck ets ,

one o f th e p rop e llan ts en te rs th e rear o f th e eng in e , f low ing be t-

w een the w alls , coo lin g the inne r su rface , an d m ak in g th in -w alled

c omb u stio n c hambe rs p os sible .In th e p ro cess o f flow ing th rough an d coo ling the se su rfaces ,

th e p rope llan t itse lf is h ea ted ; it th en en te rs th e fo rw ard end o f the

c om bu stio n c ha m be r. T h is is k no w n a s reg en era tiv e co olin g. H ea t-

in g th e p ro pe lla nt re su lts in m o re e ffic ie nt c om b ustio n a nd be ca use

th e e ng in e is c oo le d, lo ng -d ura tio n bu rn in g is p oss ible . T h e o pe ra t-

in g p re ssu re in th e com bu stion cham ber is on ly 250 to 500 pound s

P R ES S UR E R ED UC ER

P RE S S U RE

FEED

-AND REGULATOR

AIR

IFUEL I ORI NERT

GAS , . . .~

OXID IZER -..;:;..--

COMBUST ION

CHAMBEh

t J J l

GAS

'-'TT"""--....I GENERATORPUMP

FEED

F ig . 9 - 34 . L iq uid - pr op e lla nt r ock ets .



per square inch (psi), and since the p ropellan ts are sto red ou ts ide

the com busion cham ber, the construc tion m ay be ligh ter in w eight

th an c omp a ra ble so lid p ro pe lla nt ro ck ets (c on sid erin g th ose be yo nd

a certa in im pu lse o r pow er).

P r o p e ll a n t S y s te m s

T wo m ethods are used fo r supp ly ing the rockets w ith fuel andan ox id izer from tanks. T hese are the p ressure feed system and the

pum p feed system (F ig . 9 -34).

T he p ressu re feed system is the less com plica ted of the tw o.

Pressu re is supplied to bo th the fuel and ox id ize r tanks. T he p res-

surized air (o r a n in ert ga s su ch as h eliu m ) is fed th ro ug h a red uctio n

va lve to the tw o tanks, a t a p ressu re o f abou t 500 psi, w hen the

ch am ber p ressu re is ap pro xim ate ly 3 00 p si. (A s a ru le o f th um b, th e

p ressu re on the p ropellan t tanks m ust be abou t 200 psi g rea ter than

th e o pera tin g p ressu re in the com bustion cham ber.) T o supp lyp ressu re fo r the du ra tion o f bu rn ing , w hich m ay be as long as 50 to

60 seconds, the p ressu re tank is o rig ina lly charged to at least 150

atm o sp heres (a bo ut 2200 p si). W ith a p ressure o f 5 00 p si o r g rea ter

on the p ropellan t tanks and 2200 psi in the a ir bottle , the se ta nks

m ust be o f heavy construc tion . As the size o f the rocket increases ,

th e im p u lse -w e ig ht ra tio d ec re as es, u ntil fin ally t he ' em p ty w e ig ht o f

a p ressure-feed rocket becom es a serious d isadvan tage com pared

to the em pty w eight o f a pum p-feed liqu id rocket w ith the sam eim pulse . T h is cond ition is reached at abou t 5 tons. H ow ever, the

pressure -feed system is the m ore econom ical o f the tw o fo r ligh t-

w e ig h t r oc k ets .

In the pum p-feed system , hyd rogen perox ide m ay be com -

bined w ith sod ium perm anganate as a cata ly st in a gas-gene ra ting

sy stem . C om bin in g in th e re ac tion ch am ber, th ey g en era te steam ,

w h ich is u sed to d rive the tu rbine , w h ich in turn drives the tw o

pum ps. Pressu re is fe lt on ly on the dow nstream lines. Con-

sequen tly , the fuel and ox id izer tanks can be o f m uch ligh te r con -

struc tion . T he pum p-feed system is far m ore com plica ted than the

p ressu re-feed system , bu t the re is a g rea t w eigh t sav ing , s ince

the re are no large a ir bottle s a n d s in c e th ic k -wa lle d fu e l a n d o x id ize r

tanks are no t necessary . Ano ther m ethod is to p lace a sm all tu rbine

in the exhaust je t, w hich d rives the pum ps, w hich in tum g iv e th e

necessary p ressure on the fuel and ox id izer lines . F igu re 9-35

show s a N ike H e rcu1es-a liqu id fue led rocket.



D E S I G N O F A M O D E L R O C K E T

We will now show how to design a large model rocket to meet

the following specifications:

• To have a thrust of 800 pounds for 0.5 second.

• To use zinc and sulfur as the propellant.

The procedure is as follows:

• Determine the amount of propellant needed.

• Determine the diameter of the burning surface.

• Determine the length of the propellant.

• Determine the thickness of the combustion chamber

walls.

• Determine the dimensions of the exhaust nozzle.

Certain other bits of infonnation are needed for the design,

especially those concerning the propellant. These can be calculated

but the mathematics are beyond the scope of this text. Instead we

wil l just state these as facts and use them in our design. Remember,

the factors in Table 10-1 apply to the zinc-sulfur propellant only.

As a point of information, some temperature conversion fac-

tors are:

Rankine = F + 4600Kelvin = C + 2730

Centigrade = 5/9 (F - 32°)

• Solu tion:

The weight of the propellant (Wp) required can be found byusing the thrust equation:



WF = - X v,or F = m x Ve

gt

where m is the mass of propellant burned per second.

For our problem we rewrite the equation:

F = W

p

x Vetbxg

where Wp is defined above, th is the burning time and g is the

gravitational constant 32.2 ft/sec'. Then:

Wp= F x tb x g

V e

,and inserting the values for these letters,

111

800 x 0.5 x 32.2 864lh 11 . dnp = 1490 = . s of prope ant require

Remember that we are using a ratio of 2.04 parts of zinc and 1

part ofsulfur (3.04 part total) so the proper weights ofeach will be:

Weight of zinc = ~:~: x 8.64 = 5.80 lbs of zinc powder.

Weight of sulfur can be either 8.64 - 5.80, or 2.84Ibs, or can

be calculated by using the ratio:

Weight of sulfur = 1 x 8.64 = lbs of sulfur powder.3.04

W e i g h t o f P r o p e l l a n t B u r n e d P e r S e c o n d

To determine the proper shape of the propellant, we must

determine the burning rate. We want all propellant burned in 0.5

Tab le 1 0 - 1 . C h a r acte ris tic s o f Z inc S ulfu r P ro pe lla nt.

Effective exhaust velocity (Ve)

Chamber pressure (Pc)

Chamber temperature (Te)

Molecular weight (Mw)

Spec ific h e at r atio (y)Density (0)

= 1490 Ibslft/sec.

= = 1,000 Ibs/sq. in.

=: 3060° Rankine

= 97.45 Ibs/mole

= 1.25

= = 161 Ibs/cu. ft.= 0.0932 Ibs/cu. in.



second and we knowhowmanypounds ofpropellant are necessary,

so the burning rate is simply:

total propellant weight _ 8.64 lbs of propellant

burning time - 0.5 second

= 17.28 lbs per second

T h e B u rn in g S u r f a c e

This can be determined easily since the volume of propellantburned per second equals the burning surface area (8) times the

burning rate (r). The volume burned per second also equals the

weight per second burned divided by the density (D). From the

above we can write two equations and using these, can find the

burning surface and from that the diameter ofpropellant required.

V=Sxr

We equate them:

W p .i,SXr= ~ x D

Insert the numerical values known, or found previously, and

solve this equation for (8):

S = 90 (in. per sec.) x 0.0932Ibs/in317.28 Ibs sec.

17.28S= 90 x 0.0932

= 2.06 square inches

So our propellant must have an end area of 2.06 inches.

Generally, the rocket chamber is in the form of a cylinder or

pipe, so this square-inch end area becomes an equivalent circular



area from w hich w e can sta te the tube d iam eter. Since the area o f a

c irc le is equal to 1 T r 2 , o r (rr) (d iameter)2

4

a ll w e have to do is to se t th is equa tion equal to 2.06 a nd s olv e fo r

th e d iam e te r.

s = ( 1 T ) (d iam eter)2 th en4 '

I4SV (;) = d iam e ter, an d

( 4) ( 2.0 6)3 .14 = 1.625 i nches

So our rocke t com bustion cham ber m ust have an in sid e d iam ete r o f

1.625 i nches .T he leng th requ ired is equal to th e burn ing ra te tim es th e to ta l

bu rn in g tim e :

90 in ches per second x 0.5 second = 45 i nches

At th is s tage , th en , w e know inwh at p ro po rtio n th e p ro pe lla nt

m ust be (zin c and su lfu r), and the size o f the com bustion cham ber

(tube or p ip e) th a t it will be pu t in. N ex t the th ickness o f th e p ipe

w alls m ust be de term ined . We don 't w an t ou r ro cke t to blow upw hen w e try to launch it

T he ten sile s treng th of th e m ateria ls to be used is ou r nex t

concern . If w e use a low -carbon stee l fo r th e p rope llan t tube th a t

has a ten sile s treng th of 62,400 psi (SAE 1020) w e d iv id e t ha t

ra ting by four and assum e tha t it can stand on ly 15,600 psi .

62,400 = 156004 '

T his g iv es a safe ty fac to r o f fou r. T his safe ty fac to r is neces-

sary beca use th e ex tre m ely h igh tem p era tu res tha t a re ge ne ra te d

by th e bu rn in g p ro pellan t red uce the stren gth o f th e ste el co nside r-

ably .

T he w all th ickness is d e te rm ined from th e equat ion:

cham ber p ressu re x rad iu s

a llowab le s tr es s



T he p ressu re in s ide th e ch am ber w as g iven as 1000 pound s pe r

squ are in ch . T he in s ide rad iu s o f the p ip e is 0.81 in ch , an d the stre ss

is th e fac to r w e com pu ted above , 15,600 pound s p er square in ch .

T he w all th ickness th en is :

1000 x 0.81 = 0 0519 ' h

15,600 . me

A s te el tu be w ith a w all t hick ness o f 0 .0625 in ch , a bo ut equ al to

16 gage , will do n ice ly . T he tube m ust h av e an ou ts id e d iam ete r o f

1.75 i nches and an in s ide d iam ete r o f 1.625 i nches will fulfill th e

requi rement .

T he final part o f th e ro ck e t to be ca lcu la ted is the nozzle . W e

m ust ca lcu la te th e th ro a t- and ex it-a rea d iam ete rs (F ig . 10-3 ).

T hese d im en sio ns are very im po rtan t s in ce they de te rm in e th e

m ax im um effic iency from th e exhaus t g ases and thu s max imumthrus t .

Addi t iona l informat ion is n eed ed to m ake th ese ca lcu la tio ns .

T h is is g iven in T able 1 0-2.T o find th e va lu e o f Q n eed ed , w e en te r th e table ho rizo nta lly

oppo site 800 pound s (the th ru s t v a lue in itia lly assum ed) to th e

co lum n headed by 1 .25, the v a lue o f gam m a (y) g iv en in T able 1 0-2.

At the in te rsec tio n w e obta in th e v a lue 1 . 54 to u se in th e fo n nu la :

Fo rc e = Q x At x P c

F irs t w e m ust so lv e th is formula fo r A t and w e ge t:

A t . (a re a o f th ro at) = ~.;;;..F........c x P c

T able 1 0-2. C oe ffic ie nt o f Thrust (Ct) .

Pressurey

(Pe) pSia 1.15 1.12 1.25 1.3 1.4

600 1.15 1.52 1.50 1.48 1.46700 1.58 1.54 1.52 1.50 1.48800--- - 1.60--- -1.56-- 0) - 1.54 1.52 1.50900 1.62 1.58 1.56 1.54 1.511,00 1.63 1.59 1.S7 1.55 1.521,100 1.64 1.60 1.58 1.56 1.531,200 1.66 1.62 1.59 1.56 1.541,300 1.67 1.63 1.60 1.57 1.541,400 1.68 1.64 1.61 1.58 1.55



CONVERGINGSECTION

Fig. 104. Engine dimensions.

T he fo rm u la fo r d e te rm in in g th e ex it a re a , Ae, is :

Ae = 8.2 A t

o r 8.2 tim es th e th ro a t a re a . W e fo un d th e th ro a t a rea to be 0.51

inch so , u s ing th e fo rm ula abo ve , th e ex it a re a is :

0.51 x 8.2 = 4.18 sq. in.

T o fin d t h e d iam e te r o f th e ex it end o f th e nozzle , w e aga in u seth e fo rm u la :

De = j 4 x:.18 = 2.31 in.

W e can no w d raw th e nozzle sh ow ing th e d im en sion s in F ig . 10-4.T o d ete rm in e th e len gth s o f th e co nv erg in g an d d iv erg in g sec tio ns

o f th e nozzle w e have u sed a co nve rg ing ang le o f 30° a n d a d iv e rg in g

an gle o f 15 deg ree s . W e use th e equ a tio n s :

. 0 .5 (Da=Dt)Lc on v ergm g le ng th ) = ta n 300

0.5 (De-Dt)La ( d iv e rg in g le n g th ) = tan 150

T h e n um e ric al v alu es Dean d D.w e re fo un d e ar lie r. F ro m a se t

o f trig o nom e tric ta ble s, tan 30° equ a ls 0 .577 and tan 1 5° e qu als

0.268.

t,= 0.5 (1.625-0.806) = o 711"n0.577 ..

L e t - 0.5 (2.31-0.806) = 2.80 in.- 0.268



END PIPE

A~ EASIEST TO MACHINEW~ HEAVIEST WEIGHT

E NO P IP E FITS RIGHT IN PIPE

AI~ MEDIUM MACHINING.., MEDIUM WEIGHT

SIX1"BOLTS PART IN PIPE, PART OUT4

E L L Y SPACED

~ DIFFICULT TO MACHINEV LIGHTEST WEIGHT

PART IN PIPE, PART OUTSOMEMETAL CUT OUT-

'-'~FORW!,RO CH~MBER P~~

\iii J. STEEL - - - - - w -B R A ZE A LL A R O U N D

F ig . 1 0 - 5. T y p ic al n oz z le a nd p lu g co ns tru ctio ns .

These are th e d im ension s show n inF ig . 1 0-4.N ow tha t w e have .com ple ted ou r ca lcu la tion s, w e find th a t w e

w ou ld u se tubing 1 .69 (o r 1 -11 /16) inches O .D ., .0625 inch th ick

w ith a conve rg ing ang le o f 30 deg rees and a d iverg ing ang le o f 15

deg rees . T he nozzle will have to be m ach ined in one o f th e th ree

pa tte rn s show n in F ig . 10-5. N ote th a t it is th en fastened in p lace in

th e tube , w ith m ach ine screw s, a t th e righ t leng th from the p lug a t

th e o the r end . T he p lug , show n in th e bo ttom of F ig . 10-5, shou ld be

a so lid p iece of O . 5-in. th ick stee l brazed in to th e tube end as show n.

A rounded o r bu lle t-sh aped nose sec tion o f hardw ood o r m e ta l cou ldbe fitted to th is p lugged end o f th e rocke t m o to r to fo rm an

aerodynam ic shape , o r th e rocke t m o to r m igh t be fitted in to a

rocke t body as show n in F ig . 10-6.

Fins

T he ta il f in s m ay be m ade from l/l6-in ch th ick stee l o r lh -in ch

th ick a lum inum and se t equ id is tan t a round the body as show n in F ig .

10-7. T he fin s can be fa stened by any o f seve ra l m ethod s (som e a re

show n in F ig . 10-7B) bu t th is m ust be done befo re any p rope llan t is

added .

HEAVY FORWARD

BU~KHEAO ",~r-*,_-__.( - : ? ( ~NSERT IN BODY

<J[--- -- ~n - J ' =lNOSE BODY MOTOR

FIN

F ig . 1 0 -6 . M otor a nd b od y a s s ep ara te u nits .



U se spec ia l c a re to g et th e fin s ex ac tly inlin e w ith th e lo n gitu d -

in a l ax is o f th e ro ck e t and m oun ted secu re ly so tha t flu tte r c anno t

c au se an erra tic flig ht. Sev ere flu tte r can cau se th e ro ck e t to bre ak

up . T he w id th o f th e f in s from tip to tip sh ou ld n ot be o ver 200 t imes

th e fin th ick ness . T his m ean s abou t 20 in ch es fo r th is ro cke t. T he

leng th m ay be from 18 to 24 inch es . In g enera l, the s id e -v iew area o f

th e ro ck et wi th f in s m ust be la rg e r th an th e fron t a rea , o n each s id eo f th e cen te r o f ba lan ce , fo r s tability in flig h t.

Des ig n of a Rock et M otor W hen th e Tub e Is Sp ecified F irs t

Befo re go ing ahead w ith th e laun ch ing and firin g , th e re is on e

ad ditio na l p ro blem in d es ig n. In th is p roblem assum e th a t you

a lread y h ave a len gth o f s tee l tu bin g o f th e fo llo w in g d im en sion s:

• Wa ll th ickn ess is 1 /16 in.• O u ts id e d iam e te r is 2 in.

• In s id e d iam ete r is 1% in . (1.875',)

• L eng th is 4 ft.

and you w an t to d e te rm in e th e th ru s t, th e bu rn ing tim e o f th e

prope l l an t in th e tube and th e nozzle d im ension s .

To determine th e burning tim e , w e m ust know th e bu rn in g

area an d th e len gth o f th e p rop ellan t in th e tu be . H w e allo t 3i nches

of tu be len gth fo r th e n ozzle an d n ose sec tio ns, it will le av e 4 5 in ch es

o f th e 48 fo r th e p rop e llan t. T he are a o f th e p rope llan t can be

c a lc u la te d f rom :

? T d 2

Area=S=T

= (3.14) x(1.875)2 = 2.76 in .

4

T he w eigh t o f th e p rop e llan t to be bu rn ed per se cond wil l be:

Area x b urn in g ra te x dens i ty

2.76 x 90 x 0.0932 = 23.15Ibs/sec .

T he bu rn ing tim e o f p rop e llan t in th e tube will be:

B urn in g tim e (ts) = l eng thbu rn in g ra te

45-90

= 0.5 sec .



~RWARO FASTENER

(jEAR FASTENE__,...R__,...

6BRAZE ALL ALONG

"_ - -

F ig . 1 0-7 . F in-as sem b ly d eta il (A ) and s trap m eth od (8) to avoid b ra z ing to

ch am b er or d rilling h oles excep t th os e to m ou nt noz zle. In (A). b olt in to th en oz z le o r mo to r e xte ns io n b u t n ot in to th e c ombu s io n c h ambe r. In (B). th e r ing isover th e for wa rd e nd of th e ta il. B ra ze to th e ta il a t p oint A. A t p oint B , s crew th e

e nd of th e ta il to th e n oz z le . A n d a t p oint C , b ra z e th e s tr ap s b etw een th e tail finsa t a n a p p rox im a te m id p oint, if d es ir ed .

F in a lly , to d e te rm in e how m uch p rope llan t is n eeded by

w e ig ht, m u ltip ly th e w e ig ht bu rn ed p er se co nd by th e bu rn in g tim e :

23 .15 x 0.5 = 1 1.58 lbs o f p ro pe lla nt

F rom th is w e determine th e w e ig ht ra tio s o f th e zin c a nd sul fur

p ro pe lla nt by :

Z in c p ow d er (lbs ) = ~ : ~ x 11 .58 = 7.771bs

1Sulfur ( lbs) = 3.04 x 11.58 = 3 .811bs

N ow w e can com pu te th e th ru s t. W e wil l o bta in , u sin g thefo rmula :

T hru s t (F ) =W eigh t o f p rope llan t bu rned pe r sec . x Exh au st v e lo c ity

G r av ita tio n a l c o n s ta n t



4:;"

1------- ~O.l4 "~----~

F ig . 1 0 -8 . D im ens ions of th e ch am b er a nd noz zle.

F = 23.1~2~21490 :::;1071 Ibs o f thrust

T he a rea o f th e nozzle th roa t is d e te rm in ed in exac tly th e sam e

m anne r a s fo r ou r firs t ex am p le . N ote th a t w e m ust u se a d iffe ren t

row in T able 10-2 to ge t th e th ru s t co effic ien t, 1.57.

A t = 1.5l~\,ooo = 0.682 sq. in .

T he ex it a rea o f th e nozzle, a ga in u sin g T able 10-2, is :

A e = 8.2 x A t = 5.59 sq. in .

T he th ro a t and ex it d iam e te rs a re th en ca lcu la ted as be fo re :

Dt= 4 x 0.628 - 0894 .3.14 . ID.

D / 4 " " X 5 . 5 9 2 66 .e=V~=' in .

F in ally , to d ete rm in e th e le ng th s o f th e co nv erg in g a nd d iv erg -

in g n ozzle s ec tio n s:

0.5 (1.875-0.894) 0 5·L:= 0.577 = .8 m ,

T • = 0.5 (2.66-0.894) = 3 29 O n.LA 0.268 . 1.

W e use exac tly th e sam e p rodedu re as in th e p rev io u s exam p le

w ith th e sam e co nve rg ing and d ive rg ing an gle s . It is sug gested th a t

on a ll no zzle s th e co m pu ted leng th s be in crea sed s ligh tly so th at th e

th roa t can be round ed . Do no t m ake th e leng th s sh o rte r . R ound ing

o f th e th ro at will a dd p erh ap s o ne-h alf in ch to th e le ng th . F ig ure 10-8



show s th e d im en sion s o f t h e ro ck et en gin e ba sed on th e se com pu ta -

tio ns . A ga in w e h av e sh ow n a O .5-in ch p lu g a t th e fo rw ard en d w h ich

m ust be brazed in p la ce .

I G N I T I O N S Y S T E M S A N D L A U N C H E R S

T he re is on ly one sa fe , app rov ed m e thod o f laun ch ing a ro ck et

and th a t is by e lec tric ity . T his a llow s th e p erson fir in g th e ro ck e t torem ain a t a sa fe d is tan ce from th e ro ck e t laun che r w hen th e ro ck e t

is ig nite d. T he p rin cip le o f e lec tric al ig nitio n is n ot c om p lex . A sm a ll

p ie ce o f n ich ro m e w ire is h ea te d by p ass in g a la rg e e lec tric al cu rre nt

th ro ug h it. Itwill h ea t to in cande scen ce . T his w ire is m oun ted on a

p la s tic o r w ooden p lu g (F ig . 10-9 ) ju s t la rg e enou gh to f it in to th e

nozzle end o f th e ro ck e t and p re ss ag a in s t th e p ro p e llan t un til th e

, exh au s t g a ses blow it ou t.

A re lay sy s tem is n ece ssa ry to p a ss a la rg e cu rren t th ro ugh th e

ign ite r w ire (it m u st h av e sho rt lead s to it) and a lso to m ake rem o teo pe ra tio n o f th e ig nite r p oss ible . T h is re la y sy ste m w ith its ba tte ry ,

w arn ing lig h t and sa fe ty sw itch is th e ro ck e t ig n itio n sy s tem . T he

re lay can ope ra te w ith a sm all cu rren t so w e can ru n long w ire s to

th e ig nitio n sy stem from a d is tan t bun ker and fire ou r ro ck e t m o to r

from a sa fe p os itio n .

A lw ays keep a short circu it across the ig niter un til th e las t

m om en t befo re you a re re ady to fire th e ro ck e t. N eve r connec t th e

ign ite r to th e ig n itio n sy s tem un til th e la s t m om en t befo re fir in g .

In th e ig n itio n sy s tem c ircu it show n in F ig . 10-10, s ta rtin g a tth e rig h t o f th e d iag ram , no te th a t th e lead s o f th e ig n ite r a re

con nec ted th rou gh heav y w ires to a sw itch (S) w h ich is sho rtin g th e

lin e to th e ign ite r . T his is a sa fe ty sho rt c ircu it. W hen w e a re ready

to f ire th e ro ck e t, th e sw itch is m oved dow n to po s itio n 2 so th a t th e

low er s id e o f th e lin e conn ec ts d ire c tly to th e n ega tiv e te rm in a l o f

th e 12-vo lt ba tte ry . T he ba tte ry m u st be la rg e eno ugh to supp ly th e

h eavy cu rren t requ ired fo r good ig nitio n .

- -+ - -SOLDE R IGNITE RTO LE AD -IN

S E A L W ITH GLUEORCEMENT

~DIAPHRAGM 1132"OR 1116"

B R I T T L E PLASTIC

F ig . 1 0 -9 . Typical igniter.



The left wire from the igniter connects to the normally open

relay contact. The relay is shown in the de-energized position. The

armature of the relay is connected to the positive terminal of the

12-volt battery. The normally closed relay contact is connected to

the positive terminal of the battery when the armature is de-

energized and the warning light is connected to this contact. This

will always indicate that the battery has been connected to ignitionsystem. Be careful when this light bums!

The relay has a 0.5 ,...,F capacitor across its windings. This

causes a delay in the releasing action of the relay to insure ignition

even though the firing button is depressed only momentarily.

B L A S T -O F F T E C H N IQ U E S

When the rocket is ready to fire the last person to leave the

launching area will manually change the position of the safe-arm

switch (S) to the a rm position. This same person should be the onlyone who is authorized to depress the firing switch back at the

bunker.

After closing the firing switch on the firing panel, the rocket

should blast off. It it does not, try three more times at five-minute

intervals (if the battery is weak this will allow it to recover some-

what to give another current jolt to the igniter). If the rocket still

does not fire, wait at least 30 minutes before anyone approaches it;

then one person goes to the launcher and, first, moves the safe-ann

to safe; next, disconnects the igniter wires; last, removes theigniter.

Check that:

• Firing switch is off

• Safety switch is at safe.

• Rocket igniter wires are shorted (wrapped together)

A typical rocket launching operations procedure is as follows:

• Place rocket on launcher. Send all extra persons from

rocket launch area.• Place relay box (ignition system) 10 feet from launcher.

• Connect battery to ignition system.

• Check that safety l ight is on.• Connect the firing panel to the ignition system.

• Connect an extra igniter to the ignition system leads. Do

not put this igniter in the rocket.

• Place safe-arm switch to arm.

• Operate the fire switch.



R ELA Y BOX

6FT HE AVYD U T Y W IR E

Fig. 10-10. One type of rocket ignition system.

• C h eck to see if th e te s t ig nite r h as bu rn ed th ro ug h. T h is isa te s t o f th e firin g c ircu it . If th e ig n ite r h a s bu rn ed

th ro ugh , go ah ead w ith th e nex t s te p . If no t, ch eck th efiring sys tem till it o p er ate s s atis fa cto rily .

• T u rn fire sw itch o ff .• Pla ce th e sa fe -an n sw itc h to sa fe .• Send a ll p e rso nne l bu t o n e from th e la un ch e r a rea . O th e r

pe rso nn e l sh ou ld be a t le a s t 200 fee t aw ay and und e r

cove r .

• Place th e rocke t ig n ite r in s id e th e ro ck e t no zzle u p ag a in s t

th e p ro p e llan t an d unsho rt th e le ad s .

• A tta ch the re lay box lead s to th e ro ck e t ig n ite r .• Place th e sa fe -an n sw itch to a rm .• T he la s t m an now goe s to th e sa fe a rea w hich is 200 fe et

away . H e g e ts u nd e r p ro tec tiv e cov e r.

• T h is la s t m an op e ra te s th e fire sw itch .

BL AST O FF !

designing chemical rockets

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