Eng2Cg20survey1

July 28, 2018 | Author: Joy Mary | Category: Angle, Tangent, Trigonometric Functions, Curvature, Sine
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LESSON 1–SI MPLECURV RVE        1       e       g       a        P

1.Ar c Basi s a)Me t r i cSyst em 20/D =2πR/360⁰ D =1145. 916⁰/R  b)Engl i shSys t em 100/D =2 π  R/360 D =5( 1145. 916) 6⁰) /R

2.Chor dBasi s a)Met r i cSyst em Si nD/2=10/R R=10/Si nD/ D/2

 b)Engl i shSys t em Si nD/2=50/R R=50/Si nD/ D/2

El eme mentsofSi mpl eCurve

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 Thef ol l owi ngar et hee l ement sf oundi nasi mpl ec ur v e: P. C.=Poi ntofCur v at ur e

R =Radi usofCur v e

P. T.=Poi ntofTang enc y

T=Tang entDi s t anc e

P. I .=Poi ntofI nt er sect i on

D =Degr eeofCurv e

 TangentI /2=T/R 1.T=Rt angentI /2 E =RSecI /2–R E =ext ernaldi st ance M =mi ddl eor di nat e LC =Lengt hofCur v e

4.

C =2RSi nI /2

2.

E =R( SecI /2–I ) M =R –R c osI /2

3.

M =R( 1–CosI /2) Si nI /2=C/2R

5.

Lc/I=20/D Lc=20I /D

6.

Lc/I=100/D Lc=100I /D

El eme mentsofSi mpl eCurve

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 Thef ol l owi ngar et hee l ement sf oundi nasi mpl ec ur v e: P. C.=Poi ntofCur v at ur e

R =Radi usofCur v e

P. T.=Poi ntofTang enc y

T=Tang entDi s t anc e

P. I .=Poi ntofI nt er sect i on

D =Degr eeofCurv e

 TangentI /2=T/R 1.T=Rt angentI /2 E =RSecI /2–R E =ext ernaldi st ance M =mi ddl eor di nat e LC =Lengt hofCur v e

4.

C =2RSi nI /2

2.

E =R( SecI /2–I ) M =R –R c osI /2

3.

M =R( 1–CosI /2) Si nI /2=C/2R

5.

Lc/I=20/D Lc=20I /D

6.

Lc/I=100/D Lc=100I /D

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Fi gur e1.ASi mpl eCurv e

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 WORKSHEET1–SI MPLECURVE

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      g Name _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _       a        P Cour se/Year ___ ___ ___ ___ ___ ___ ___ ___ __Dat eSubmi t t ed___ ___ ___ _

 Thedeflect i onangl esoft woi nt er me di at epoi nt sAandB ofahi ghwaycur v ear e ⁰ 5 ⁰ 5 41 ’and 91 ’r es pect i v el y .The c hor d di s t ancebe t wee n poi nt s A and B i s

20. 00m.whi l et hel ongchor di s120. 00m.St at i oni ngofP. I .i s80+060.Fi nd t hes t at i oni ngofP . C.andP . T .

 WORKSHEET2– SI MPLECURVE         6       e       g       a        P

Name _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ Cour se/Year ___ ___ ___ ___ ___ ___ ___ ___ ___ _Dat eSubmi t t ed___ ___ ___ _  The bear i ng oft he bac kt angentofa si mpl e cur v ei s N70° 00’ E whi l et he f or war dt ange nt has a bear i ng of S82° 30’ E. t he degr ee of c ur v ei s 4. 5° . St at i o ni ngofPC i sa t10+345. 43.I ti spr o po se dt ode c r e a s et hec e nt r alang l eby changi ngt hedi r ect i on oft hef orwardt angentbyan angl eof7° 00’ ,i n such a  wayt hatt hepos i t i on oft hePToft hef or war dt angentandt hedi r ect i on oft he  bac kt ange ntshal lr e mai nt heunc hanged.De t er mi ne: a)Thenew r adi usoft hec ur ve  b)St at i oni ngofne w PC.

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 WORKSHEET3–SI MPLECURVE Name _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ Cour se/Year ___ ___ ___ ___ ___ ___ ___ ___ __Dat eSubmi t t ed___ ___ ____  Two t angent si nt er sect i ng atV wi t h bear i ngs N751 2’ E and S783 6’ E ar e ⁰ ⁰ connec t ed wi t h a 4⁰ si mpl ecur ve .Wi t houtchangi ngt hedi r ec t i on oft het wo t angent sandwi t ht hesameangl eofi nt er sect i on,i ti srequi r edt oshort en t he curv et o100. 00m.st art i ngf r om t heP. C. a)Byhow muc hshal lt heP. T.bemov edandi nwhatdi r ec t i on?  b)Whati st hedi s t ancebe t wee nt het wopar al l e lt angent s?

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 WORKSHEET4–SI MPLECURVE Name _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ Cour se/Year ___ ___ ___ ___ ___ ___ ___ ___ ___ Dat eSubmi t t ed___ ___ ___ _ I ti srequi r edt ol ayoutasi mpl ecurvebydeflect i on angl es.Thecurvei s ⁰ a t oconnectt wot angent swi t h an i nt er sec t i on angl e of32 ndaradi usof800

f t .Comput et he deflect i on angl es t o each f ul lst at i ons on t he curv e,i ft he t r ans i ti ss e tupatt heP . C.whi c hi sats t at i o n 25+57. 2.Whati st hes t at i o no f P . T . ?

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LESSON 2–COMPOUND CURVE Compound Cur v e co nsi s t s oft wo or mor e co nse cut i v es i mpl e curv es havi ngdi ffer entr adi us,butwhosecent er sl i eon t hesamesi deoft hecurve.I n acompoundcur ve ,t hepoi ntoft hec ommont ange ntwher et het wocur ve sj oi n i scal l ed t hepoi ntofcompound cur vat ur e( PCC) .Shown i n Fi gur e2 ar et he el e me nt so faCompoundCur v e .

El ement sofCompoundCur v e

Fi gur e2.CompoundCur v es

Exampl e :

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 The l ong c hor d ofa compound cur v ei s 425. 00 me t er sl ong and t he angl est hati tmakeswi t ht angent soft hecur vewer e20⁰  and24⁰  r e spe ct i v el y . Fi nd t her adi usR1  and R2 oft hecompound curv ei ft hecommon t ange nti s par a l l e lt ot hel ongc hor d.

C1  /Si n12⁰ =425/Si n158⁰ C1  =235. 88m.

C2  /Si n10⁰ =425/Si n158⁰ C2=197. 01m.

Si n10⁰ =235. 88/2R1 R1  =679. 15m.

Si n12⁰  =C2/2R2 R2  =473. 78m.

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 WORKSHEET5–COMPOUND CURVE Name _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ Cour se/Year ___ ___ ___ ___ ___ ___ ___ ___ ___ Dat eSubmi t t ed___ ___ ___ _  The hi ghway engi neerofa ce r t ai nr oad cons t r uct i on deci ded t o use a r adi usof100. 00m i nl ayi ngoutasi mpl ecurvehavi nganangl eofi nt er sect i on ⁰ 0 of362 ’ .Thest at i oni ngoft hevert exi s30+375. 20af t erver i f yi ngt heact ual

condi t i onsoft hepr oposed r out e,i twas f ound outt hatt hePT shoul d be mov ed outi n a par al l elt ange nthavi ng a per pendi cul ar di st ance of10. 00 met er swi t h an angl eofi nt er sect i on r emai ni ngt hesamewhi l et hecurv eshal l havet hesamePC.Det er mi ne: 1.Ther adi usoft henew cur v e.

2 . hest  T at i oni ngoft hene w PT.

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 WORKSHEET6–COMPOUND CURVE Name _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ Cour se/Year ___ ___ ___ ___ ___ ___ ___ ___ ___ Dat eSubmi t t ed___ ___ ___ _  Two t angent si nt er sec tatSt at i on 25 + 50.A compound c ur v el ai d on t he i rt ang e nt shast hef o l l o wi ngdat a. I 1⁰00’ 1 =3

I 6⁰00’ 2 =3

D1 =3⁰  00’

D2 =5⁰ 00’

a)Comput et heSt at i oni ngoft hePC,PCC andPToft hec ur v e.  b)I ft hePT i smov e50. 00 f t .out ,comput et hes t at i on oft hePT wi t ht he PCC ont hes amepoi nt .

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 WORKSHEET7–COMPOUND CURVE Name _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ Cour se/Year ___ ___ ___ ___ ___ ___ ___ ___ ___ Dat eSubmi t t ed___ ___ ___ _

 A common t ange nt ofa compound c ur v e makes an angl e wi t ht he t ange nt s of t he c ompound c ur v e of 25⁰30’and 30⁰00’r e s pe c t i v e l y .The st at i oni ngofA of10 + 362. 42.Thedegr eeofcurv eoft hefir stcurv ei s4⁰30’  whi l et hatoft hese condcur v ei s5⁰00’ .I ti srequi r edt ochanget hi scompound cur vewi t h asi mpl ecurv et hatshal lendatt hesamePTwhi l et hedi r ect i on of t het ange nt sr emai nst hesame.Fi nd t heradi usoft hi ssi mpl ecurv eand t he st at i oni ngoft henew PC.

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 WORKSHEET8–COMPOUND CURVE Name _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ Cour se/Year ___ ___ ___ ___ ___ ___ ___ ___ ___ Dat eSubmi t t ed___ ___ ___ _ Gi v e n ac ompound cur v e wi t h al ong c hor d equalt o 135. 00 me t er s f or mi ng an angl e of12⁰00’and 18⁰00’r espect i vel y wi t ht he t angent s.The common t ange nti s par al l elt ot he l ong chor d.Det er mi ne t he r adi ioft he compoundc ur v e.

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LESSON 3 –REVERSED CURVE  Thi ski nd ofcur v ei sf or med byt woci r cul arsi mpl ec ur v ewi t hc ommon t angent s butl i es on opposi t e si de.Rev er sed curv ei s usef uli nl ayi ng out pi pel i nes,flumes,l evees,and l ow speed r oadsand r ai l r oads.I n canal s,i ti s use d wi t ht r emendouscaut i onssi ncei tmakest hecanaldi fficul tt onavi gat e andc o nt r i but et oe r o s i o n. El ement sofARev er se dCur ve :

Fi gur e3.Rev er sedCurv e R1  andR2  =r a di io fc ur v at ur e D1 anD2  =degeeofcurv e  V1  andV2  =poi nt sofi nt e r s e c t i onoft ang e nt s θ= angl ebet weenconve r gi ngt angent s I 2 –I 1 =θ P . C.=po i nto fc ur v a t ur e

P . T .=poi ntoft ang e nc y P. R. C.=Poi ntofRever sedCurv e

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Lc=Lc1+Lc2=Lengt hofr ev er sedcurv e P=di st ancebet weenparal l elt angent s Fourt ypesofr eve r sedcurv epr obl emsare: 1.Reve r sedcurv ewi t hequalr adi iandparal l elt angent s. 2.Reve r sedcurv ewi t hunequalr adi iandparal l elt angent s 3.Rev er sedcurv ewi t hequalr adi iandconve r gi ngt angent s. 4.Rev er sedcurv ewi t hunequalr adi iandconve r gi ngt angent s. Me t hodofLayi ngout  Theme t hodofl at yi ngoutsi mpl ec ur v ei sappl i ed.Att hepoi ntwher et he c ur v er e v e r s e di ni t sdi r e c t i o ni sc al l e dt hePo i nto fRe v e r s e dCur v a t ur e( PRC) .  Af t er t hi s poi nt has been l ai d out f r om t he P . C.t he i ns t r ume nt i st hen t r nas f e r r e dt ot hi spo i nt .Wi t ht r ans i tatP . R. C.andar e adi nge qualt ot het o t al de fle c t i o n ang l ef r o mt heP . C.t ot heP . R. C.,t heP . C.i sbac ks i g ht e d.I ft hel i ne ofsi ghti sr ot at ed aboutt hever t i calaxi sunt lt hehori zont alr eadi ngbecome z er o,t hi sl i ne ofsi ghtf al l s on t he common t angent .The nexts i mpl e curv e coul d be l ai d outon t he opposi t e si de oft he t ange ntby deflect i on angl e me t hod.

Exampl e: Fr om t hefigureshown,t hetwo di ver gi ngtangent swer econnect ed  by a r ever sed cur ve wi t h bot h ar cs havi ng a 5˚cur ve.Det er mi ne t he St ati on ofP. I .i fIangl ei s41˚ ,Determi neal so theSt ati on ofP. T.i fTsi s measur edas550f t .TheSt at i on ofP. C.i s20 +40. 30.

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So l ut i o n:

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P. I .=20+40. 30+2, 758. 25=47+98. 55 P. T.=47+98. 55–550=42+48. 55

 WO  WORKSHEET9–REVERSED CURVE Name _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ Cour se/Year ___ ___ ___ ___ ___ ___ ___ ___ ___ Dat eSubmi t t ed___ ___ ___ _  The pe r pe ndi cul ardi s t ance be t we en t wo par al l e lt angent si s equalt o 8. 00 met er s,cent r alangl et o8˚ 00’ 00”and t her adi usofcurv at ur eoft hefir st cur v e equalt o 175. 00 me t er s.Fi nd t he r adi us oft he se cond cur v e oft he r ever sedcurve.

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 WO  WORKSHEET10–REVERSED CURVE Name _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ Cour se/Year ___ ___ ___ ___ ___ ___ ___ ___ ___ Dat eSubmi t t ed___ ___ ___ _  Twopar al l e lr ai l way200. 00me t er sapar twer et obeconnect edbyequal t urnout s.I ft hei nt erme medi at et angenti s400. 00met er sandt her adi usofcurve i s1100. 00met er s,det ermi mi net hecent r alangl e.

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 WO  WORKSHEET11–REVERSED CURVE Name _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ Cour se/Year ___ ___ ___ ___ ___ ___ ___ ___ ___ Dat eSubmi t t ed___ ___ ___ _  Theper pendi cul ardi s t ancebe t we en t wopar al l elt angent sofa r e v er sed cur v ei s 35. 00 me t er s.The az i mut h oft he bac kt ange nt oft he cur v ei s 270˚ 00’ 00”andt heazi mut hoft hecomm mmont angenti s300˚ 00’ 00” .I ft her adi us oft hebackcurv ei s150. 00met er sandt hest at i oni ngoft heP. R. C.i s10+140, findt hes t at i o no ft heP . T .

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 WORKSHEET12–REVERSED CURVE Name _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ Cour se/Year ___ ___ ___ ___ ___ ___ ___ ___ ___ Dat eSubmi t t ed___ ___ ___ _  A r e v er sed cur v e connect st wo conv er gi ng t angent si nt er sect i ng atan ang l eof30˚ 00’ 00” .Thedi s t anc eo ft hi si nt e r s e c t i onf r o mt heP. I .o ft hecur v ei s 150. 00 me t er s.The deflec t i on angl e oft he co mmon t ange ntf r om t he back t ange nti s20˚ 00’ 00” R,andt heaz i mut h oft hecommon t ange nti s320˚ 00’ 00” .  Thedegr eeofcur v eoft hesecondsi ml ec ur v ei s6˚ 00’ 00”andt hes t at i oni ngof

t hepoi ntofi nt er sect i on oft hefir stcur vei s4 +450.Det ermi net hest at i oni ng         2 o ft heP. C. ,t heP. R. C,andt heP. T .

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 WORKSHEET13–REVERSED CURVE Name _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ Cour se/Year ___ ___ ___ ___ ___ ___ ___ ___ ___ Dat eSubmi t t ed___ ___ ___ _  Two conv er gi ng t angent s hav e az i mut h of 330˚ 00’ 00” and 90˚ 00’ 00” r esect i vel ywhi l et hatoft hecommon t angenti s350˚ 00’ 00” .Thedi st ancef r om

t hepo i nti nt e r s e c t i o no ft het ang e nt st ot heP . I .o ft hes ec o ndc ur v ei s160. 00         3

me t e r swhi l et hes t at i o ni ngoft heP . I .o ft hefir s tc ur v ei sat10+432. 24.I ft he         2       e

      g r adi usoft hefir stcurv ei s285. 40met er s, det ermi net hest at i oni ngofP. R. C and       a        P

P . T .

LESSON 4 –SYMMETRI CALPARABOLI C CURVES Symme t r i cal par abol i c cur v e

i s

a

v er t i cal

parabol i c curv e wher ei n t he hori zont all engt h of t hecurv ef r om t hePC t o t hev e r t e xi se qualt ot he hori zont al l engt h f r om

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ByRat i oandPr oport i on:

2)Usi ngt hesquaredpr oper t yofparabol a:

3)Locat i onofHi ghestorLowestPoi ntoft heCurve : a)Fr om t heP. C.

Exampl e :

b)Fr om t heP. T.

On a r ai l r oad a 0. 80% gr ade me e t sa +0. 40% gr ade ats t at i on 90+00        5

 whose el ev at i on i s 100. 00 f t .The ma xi mum al l owabl ec hange i n gr ade per         2       e

      g s t at i o n ha v i nga l e ng t ho f100 f t .i so . 20.I ti sde s i r e dt o pl ac ea c ul v e r tt o       a        P

dr ai nt hefloodwat er sdur i nghea vydownpour .Wher emus tbet hel ocat i on of t hec ul v e r t ?Atwhate l e v at i o n mus tt hei nv e r to ft hec ul v e r tbes e ti ft hepi pe hasadi ame t e ro f3. 00f t .andt hebac kfil li s1. 00f t .hi g h.Ne g l e c tt het hi c kne s s o ft hepi pe . Fi gur e:

So l ut i o n: Lengt hperst at i on=100f t . r=r at eo fc hang epe rs t at i o n El e v a t i o no fP . T .=100+0. 04( 300)=101. 2f t .

 AC =101. 8–100=101. 8f t .  AB =BC =0. 90f t . n=6st at i ons

H =0. 90f t .

L=6( 100)=600f t

S1  =400f t . St at i oni ngofP. C.=90+00–300=87+00 St at i oni ngofP. T.=90+00+300=93+00

El evat i onofP. C.=100+0. 008( 300)=1 02. 4         6         2

      e  WORKSHEET14–SYMMETR I CALPARABOLI C CURVE       g       a        P

Name _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ Cour se/Year ___ ___ ___ ___ ___ ___ ___ ___ ___ Dat eSubmi t t ed___ ___ ___ _  ASymme t r i c alpar abol i ccur v ei sde si gnedt opasst hr oughpoi ntAats t at i on4 +50who see l e v at i o ni s76. 20f t .TheP . C.o ft hi spar abo l i cc ur v ei sats t at i o n2 +75who s ee l e v at i o ni s74f t .Thel e ng t ho ft hepar abo l i cc ur v ei s400f t .l o ng havi ngabackwardt angentgr adeof+2. 5%.I ti sr equi r edt odet er mi net he amountofcutandfil latst at i ons2+75,3+00,4+00,5+00,6+00,and6+ 75i ft hegr o unde l e v at i o nsar easf o l l o ws : St at i on 2+75 3+00 4+00

Gr oundEl e v at i on 74. 00 75. 881 76. 815

St at i on 5+00 6+00 6+75

Gr oundEl ev at i on 76. 925 75. 825 70. 940

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 WORKSHEET15–SYMMETRI CALPARABOLI C CURVE Name _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ Cour se/Year ___ ___ ___ ___ ___ ___ ___ ___ ___ Dat eSubmi t t ed___ ___ ___ _  Asymme t r i calpar abol i cc ur v ec onnect st wogr adesof+6% and4%.I ti s t opasst hr oughapoi ntpt hest at i oni ngofwhi chi s35+280andt heel evat i on i s198. 13me t e r s .I ft hee l e v at i o no ft hegr adei nt e r s e c t i o ni s200me t e r swi t h st at i oni ng35+300det er mi ne: a)Thel engt hoft hecur ve  b)St at i oni ngande l e v at i onsofP. C.andP. T. c)Thel ocat i onoft hehi ghestpoi ntoft hecurve . d)El ev at i onofst at i on35+260ont hecurv e

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 WORKSHEET16–SYMMETRI CALPARABOLI C CURVE Name _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ Cour se/Year ___ ___ ___ ___ ___ ___ ___ ___ ___ Dat eSubmi t t ed___ ___ ___ _  Ahor i z ont all ai dci r cul arpi pecul v er thavi ngane l ev at i onofi t st opt obe 85. 26f t .cr ossest heri ghtangl esunderapr oposed400f tpar abol i ccurve.The poi nto fi nt e r s e c t i o noft hegr a del i ne si sats t at i on12+80andi t sel e v at i o ni s 88. 50f twhi l et hecul ver ti sl ocat ed atst at i on13+20.Thebackwar dt angent hasagr adeof+3% andt hegr adeoft hef orwar dt angenti s–1. 6%.Undert hi s c ondi t i o ns ,whatwi l lbet hede pt ho fc o v e ro v e rt hepi pe ?

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 WORKSHEET17–SYMMETRI CALPARABOLI C CURVE Name _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ Cour se/Year ___ ___ ___ ___ ___ ___ ___ ___ ___ Dat eSubmi t t ed___ ___ ___ _  An unde r passr oadcr ossi ngar ei nf or cedconcr e t ebr i dgeal ongt heShaw Bl vd.hasadownwar dgr adeof–4% me e t i ngan upwar dgr adeof+8% att he  v er t exV ( el ev at i on 70. 00m)ats t at i on 7+700,e xact l yunder neat ht hece nt e r l i ne oft he br i dgehavi ng a wi dt h of10. 00 met er s.I ft he r equi r ed mi ni mum cl ear anceundert hebr i dgei s5. 00 met er sand t heel ev at i on oft hebot t om of t hebr i dg ei s78. 10me t e r s ,de t e r mi net hef o l l o wi ng : a)Lengt h oft he ve r t i calpar abol i c curv et hat shal lconnectt he t wo t angent s.  b)St at i oni ngande l e v at i onwher eacat c hbasi nwi l lbepl aced.

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 WORKSHEET18–SYMMETRI CALPARABOLI C CURVE Name _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ Cour se/Year ___ ___ ___ ___ ___ ___ ___ ___ ___ Dat eSubmi t t ed___ ___ ___ _  Av e r t i c alhi ghwaycur v ei st opasst hr oughar ai l r oadatgr ade.The cr ossi ngmustbeatSt at i on64+50andatanel evat i on724. 00f t .Thei ni t i al gr adeoft hehi ghwayi s+2% andmee t sa–3% gr adeats t at i on62+00atan el evat i onof732. 40f t .Ther at eofchangemustnotexceed1% perst at i on. a)Whatl engt hofcurv ewi l lmeett hecondi t i on?  b)Whati st hes t at i oni ngande l ev at i onoft hehi ghes tpoi ntoft hecur v e ?

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LESSON 5–UNSYMMETRI CALPARABOLI C CURVES nsi stofasymmet ri calparabol i ccurv ef r om PC t oPT.A, B • Co anot hersymmet ri calparabol i ccurv et angentt ot hatpoi ntAandPT •Usedi npr ovi deasmoot handcont i nuesc urv et r ansi t i onf r om PC t oPT

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Exampl e : Gi ven: g1 =7% L1=160m

g2  =4% L2  =120m

Requi r ed: a.El e v at i onoft he Cur veoft heunder pass  b.L2  i fe l e v at i o no fc ur v ei s22. 683m. c .St at i o ni ngoft heHPo ft hec ur v ef o r quest i onb. So l ut i o n:

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 WORKSHEET19–UNSYMMETRI CALPARABOLI C CURVE Name _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ Cour se/Year ___ ___ ___ ___ ___ ___ ___ ___ ___ Dat eSubmi t t ed___ ___ ___ _  Anunsymme t r i calpar abol i cc ur v ehasaf or war dt angentof8% andabacwar d t angentof+5%.Thel engt h oft hecurv eon t hel ef tsi deoft hecurv ei s40. 00 me t e r sl o ngwhi l et hato ft her i g hts i dei s60. 00me t e r sl o ng .I fP . C.i sSt at i o n6

+780andt heel evat i on i s110. 00 met e r sa)det er mi net hehei ghtoffil latt he        4 out cr op,b)Det ermi net hehei ghtofcur veatSt a.6+820.         3       e       g       a        P

 WORKSHEET20–UNSYMMETRI CALPARABOLI C CURVE Name _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ Cour se/Year ___ ___ ___ ___ ___ ___ ___ ___ ___ Dat eSubmi t t ed___ ___ ___ _

 A f or war dt ange nthavi ng a s l op e of 4% i nt er sect st he bac kt ange nt        5

havi ng asl ope+7% atpoi ntV atSt at i o n6+300havi nganel evat i onof230. 00         3       e

      g met er s.I ti sr equi r ed t o connec tt he t wo t ange nt s wi t h an unse mmet r i cal       a        P

par abol i c cur v e t hat s hal lpass t hr ough poi ntA on t he cur v e hav i ng an el ev at i on of227. 57 met er s ats t at i on 6 + 270.The l engt h ofcurv ei s60. 00 me t e r so nt hes i deo ft hebac kt ang e nt .A)I ti sr e qui r e dt ode t e r mi net hel e ng t h oft hecurv eon t hesi deoft hef orward t angent .B)Det er mi net hest at i oni ng ande l e v at i o no ft hehi g he s tpo i nto ft hec ur v e .

 WORKSHEET21–UNSYMMETRI CALPARABOLI C CURVE

        6         3 Name _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _       e       g       a        P

Cour se/Year ___ ___ ___ ___ ___ ___ ___ ___ ___ Dat eSubmi t t ed___ ___ ___ _ I n a ce r t ai nr oad const r uct i on under t aken by DPWH i twasdeci ded t o connecta f or war dt angentof3% and a back t angentof5% by200 me t er symmet ri calparabol i ccurve.I twas di scover ed t hatt hegr adei nt er sect i on at 10 + 100 whoseel ev at i on i s 100. 00 met er sf al lon a r ocky se ct i on wi t ht he exposed boul deratel evat i on 102. 67 met er s.To avoi dr ockky excavat i on,t he pr o j e c te ng i ne e rd e ci de dt oadj us tt hev e r t i c alar abo l i cc ur v ei ns uc hawa yt hat t hecur v ewi l lj us tc l e art her o c k wi t ho utal t e r i ngt hepo si t i o no fP . C.and t he gradeoft het angent s.Det ermi net hest at i oni ngandel evat i onofnew P. T.

       7         3

      e  WORKSHEET22–UNSYMMET RI CALPARABOLI C CURVE       g       a        P

Name _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ Cour se/Year ___ ___ ___ ___ ___ ___ ___ ___ ___ Dat eSubmi t t ed___ ___ ___ _  Af or war dt angentof+6% wasde si gnedt oi nt er sec tabac kt angentof 3% ata pr oposed under pas s al ong EDSA so as t o mai nt ai n a mi ni mum cl e ar ance al l owed under a br i dge whi c h cr os se s per pendi cul ar t o t he under pass.A 200. 00met er scurv el i eson t hesi deoft hebackt angentwhi l ea 100. 00met er scurvel i eson t hesi deoft hef r owar dt angent .Thest at i oni ngand el ev at i on oft hegradei nt er sec t i on i s12 + 530. 20 met er sand 100. 00 met er s r espect i vel y .The cent er l i ne oft he bri dge f al latst at i on 12 + 575. 20.The el ev at i on oft he under si de oft he br i dge i s 117. 48 met er s.Det er mi ne t he mi ni mum cl earanceoft hebr i dgei fi thasawi dt hof10. 00met er s.

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LESSON 6–SI GHTDI STANCE

NonPasssi ngSi ghtDi st ance No nPas s i ngs i g htdi s t anc ei st hes af es t o ppi ngdi s t anc eo fa  v ehi cl er unni ngatdes i gns pe ed.

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Passi ngSi ghtDi st ance

Pas s i ngs i g htdi s t anc ei st hedi s t anc er e q ui r e dt oo v e r t akes af e l yano t he r movi ngve hi cl ei nt hesamet r afficl ane.

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 WORKSHEET23–SI GHTDI STANCE Name _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ Cour se/Year ___ ___ ___ ___ ___ ___ ___ ___ ___ Dat eSubmi t t ed___ ___ ___ _  A v er t i c alsummi tc ur v e has a t ange ntgr ade s of2. 8% and 1. 6%.A mot ori stwhom ey esi ghti s4. 80 f t .abovet heroad way si ght ed t het op ofa  vi s i bl eobj ec t4. 20hi ghatt her i ghtsi deoft hesummi t .Cal cul at et hel engt hof t hec ur v ef o ras i g htdi s t anc eof432. 00f t .

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 WORKSHEET24–SI GHTDI STANCE Name _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ Cour se/Year ___ ___ ___ ___ ___ ___ ___ ___ ___ Dat eSubmi t t ed___ ___ ___ _  A gr adeascendi ngatt her at eof5% me e t sanot hergr adedesc endi ngat t her at eof4% att hever t exofel evat i on 20. 00m.andst at i oni ng5+000.Sol ve f ott hest at i oni ngand el evat i on oft hesummi toft hever t i calparabol i ccurve  whi c h wi l lconne ctt hegr adel i nesf orasaf edi s t anceof150. 00m. ,t hehe i ght oft heeyesoft hedri ver sabovet hepavementate ach endoft hesi ghtdi st ance  bei ng1. 50me t er s

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LESSON 7–PASSI NG SI GHTDI STANCEFOR VERTI CALSAG CURVE ATUNDERPASS 1.Whenpassi ngsi ghtdi st ancei sgr eat ert hant hel engt hoft hecurv e.

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Us i ngt hepr e v i o usr e l at i o noft hepar abo l i cc ur v e s . S>L  Wher e: S=l engt hofpassi ngsi ghtdi st ance L=l engt hofcurve h1  =he i ghtofdr i v e r ’ se ye h2  =he i g hto fo bj e c t C =ver t i calcl ear ancef r om t hel owestpoi ntofunderpasst ocurv e

2.Whent hepassi ngsi ghtdi st ancei sl esst hant hel engt hofcurv e.

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 WORKSHEET25–PASSI NG SI GHTDI STANCESFORVERTI CAL SAG CURVEATUNDERPASS Name _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ Cour se/Year ___ ___ ___ ___ ___ ___ ___ ___ ___ Dat eSubmi t t ed___ ___ ___ _ I n a ce r t ai n under pass ,t he ve r t i cal c l ear ance of t he 240. 00 me t er ar abol i c curv e wi t h P. C.ats t at i on 13 + 000 and el ev at i on 30. 00 met er si s 14. 50 f t .Thehe i g hto ft heo bj e c ta tt hei ns t anto fpe r c e pt i o ni s3. 50 f t .whi l e t hato ft hedr i v e r ’ se y ei s4. 50f t .I ft heappr o ac hg r a dei s4% andt hepas s i ng si ghtdi st ancei s320. 00 met er s,whati st hegradeoft hef orwardt angent ?At

 whatpoi nt( s t at i on and el ev at i on)of       5 t he cur v es houl dt he cat c h basi n be i ns t al l e d?

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 WORKSHEET26–PASSI NG SI GHTDI STANCESFORVERTI CAL SAG CURVEATUNDERPASS Name _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ Cour se/Year ___ ___ ___ ___ ___ ___ ___ ___ ___ Dat eSubmi t t ed___ ___ ___ _  Thev er t i calcl ear anceoft hepar abol i csagcur v eoft hene wl ycons t r uct ed Bal i nt awak Under pass i st o be det er mi ned i ft he maxi mum hei ght oft he dr i ve r ’ seyet hatc oul dut i l i zesuchunder passmeasur edf r om t hepavementi s

4. 50 f t . ,whi l et ha toft heo bj e c ta tt h ei ns t anto fpe r c e p t i o ni s 3. 50 f t .The         6

l engt h oft he parabol i c curv ei s 1, 152 . 00 f t .and t hatoft he passi ng si ght        4       e

      g di st ancei s1, 100. 00 f t .Thedesi gned gr adeatt heback t angenti s5% whi l e       a        P

t hef orwar dt angenthasadesi gnedgr adeof+3%.

 WORKSHEET27–PASSI NG SI GHTDI STANCESFORVERTI CAL SAG CURVE Name _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ Cour se/Year ___ ___ ___ ___ ___ ___ ___ ___ ___ Dat eSubmi t t ed___ ___ ___ _

 Ahi ghwaydesi gnedf orspeedsof 60mph,i sbei ngcons t r uct edov erahi l l        7

 wi t h a 3% asc endi nhg gr ade and        4 a 2% de scendi ng gr ade. The oi nt of       e

i nt e r s e c t i o no ft het wog r a de si sate l e v a      g t i o n 100. 00f t .andatt hats t at i o nt he       a        P

el evat i on oft hegr ound i s95. 00 f t .Whatwi l lbet hedept h ofcutatt hepoi nt  wher et het wogr ade si nt er secti ft hev er t i calcur v eus edi sdesi gnedf orasaf e passi ng si ght di st ance of 2, 100. 00 f t . Hei ght of obser ve r ’ s ey ef r om t he pavementi s4. 50f t .andt hatoft heobj ecti sal so4. 50f t .abovet hepavement .

 WORKSHEET28–PASSI NG SI G        8 HTDI STANCESFORVERTI CAL SAG C       4 U RVE       e       g       a Name _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _        P Cour se/Year ___ ___ ___ ___ ___ ___ ___ ___ ___ Dat eSubmi t t ed___ ___ ___ _  A v e r t i c alpar abol i csummi tcur v ewasdesi gnedi n or de rt ohav eac l ear si ghtdi st anceof120. 00met er s.Thegradel i nesi nt er sectast at i on 9+100at el ev at i on 160. 50met er s.Thecurv ewassodesi gnedsucht hatwhent hehei ght oft hedri over ’ seyei s4. 50 f t .abovet hepavementi tc oul dj ustseean obj ect  whosehei ghti s4. 20i nc hesabov et hepav eme nt .De t er mi net hent hemaxi mum speedt hatacarcoul dt r avelal ongt hi scurv e.Thegradel i neshasan upward gr adeof+5% andadownwar dgr adeof3%.

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LESSON 8–SPI RALCURVE  Aspi r alcur v ei sat r ansi t i oncur v ei nser t edbe t weenaci r cul arcur v eand t het ang e ntt ot hec ur v e .I ti sac ur v eo fv ar y i ngr a di usus e dt ogr adual l y i ncr easet hecurvat ur eoft her oadorr ai l r oadusedpri mari l yt or educe s ki ddi ngands t e e r i ngdi ffic ul t i e sbygr adualt r ans i t i o nbe t we e ns t r ai g ht l i ne andt urni ngmot i on.I tpr ovi desamet hodf oradequat el yf orsupere l evat i ng c ur v e s .

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El ement sofSpi r al TS = the point of change from tangent to spiral SC = the point of change from spiral to circular curve CS = the point of change from circular curve to spiral ST = the point of change from spiral to tangent SS = the point of change from one spiral to another

The symbols PC and PT, TS and ST, and SC and CS become transposed when the direction of stationing        1 is changed.

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a = the angle between the tangent at the TS and the chord from the TS to any point on the spiral A = the angle between the tangent at the TS and the chord from the TS to the SC  b = the angle at any point on the spiral between the tangent at that point and the chord from the TS  = the angle at the SC between the chord from the TS and the tangent at the SC c = the chord from any point on the spiral to the TS C = the chord from the TS to the SC d = the degree of curve at any point on the spiral ! = the degree of curve of the circular arc f = the angle between any chord of the spiral "calculated when necessary# and the tangent through the TS $ = the angle of the deflection between initial and final tangents% the total central angle of the circular curve and spirals & = the increase in degree of curve per station on the spiral ' = the length of the spiral in feet from the TS to any given point on the spiral 's = the length of the spiral in feet from the TS to the SC, measured in () e*ual chords o = the ordinate of the offsetted PC% the distance between the tangent and a parallel tangent to the offsetted curve r = the radius of the osculating circle at any given point of the spiral + = the radius of the central circular curve s = the length of the spiral in stations from the TS to any given point S = the length of the spiral in stations from the TS to the SC u = the distance on the tangent from the TS to the intersection with a tangent through any given point on the spiral

 = the distance on the tangent from the TS to the intersection with a tangent through the SC% the longer         2 spiral tangent

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v = the distance on the tangent through any given point from that point to the intersection with the tangent through the TS - = the distance on the tangent through the SC from the SC to the intersection with the tangent through the TS% the shorter spiral tangent  = the tangent distance from the TS to any point on the spiral / = the tangent distance from the TS to the SC y = the tangent offset of any point on the spiral 0 = the tangent offset of the SC 1 = the tangent distance from the TS to the offsetted PC "1 = /23, approimately#

= the central angle of the spiral from the TS to any given point = the central angle of the whole spiral Ts = the tangent distance of the spiraled curve% distance from TS to P$, the point of intersection of tangents 4s = the eternal distance of the offsetted curve

Spiral Formulas

The following formulas are for the eact determination of the functions of the ()5chord spiral when the central angle, , does not eceed 67 degrees. These are suitable for the compilation of tables and for accurate fieldwor&. "(#

"3#

"6#

";#  =

"8#

"7# A = " 28# 5 ).39:

5A

":# C = 's "Cos ).8

< ).))6 4sec 826

8

#

 seconds

"4sec "># / = C Cos A

= ( Tan  " #

"9# 0 = C Sin A

"((#

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"()#

"(3#

"(8# 1 = / 5 "+ Sin

#

"(6# o = 0 5 "+ -ers "-ers

"(7# Ts = "+ < o# Tan " $# < 1

#

= ( 5 Cos

#

"(;# 4s = "+ < o# 4sec " $# < o "4sec " $# = Tan " $#"Tan"? $##

Empirical Formulas

@or use in the field, the following formulas are sufficiently accurate for practical purposes when does not eceed (7 degrees. a=

28 "degrees#

A=

28 "degrees#

a = () &s3 "minutes#

S = () &S3 "minutes#

Spiral Lengths

!ifferent factors must be ta&en into account when calculating spiral lengths for highwa y and railroad layout.

 Highways. Spirals applied to highway layout must be long enough to permit the effects of centrifugal force to be ade*uately compensated for by proper superelevation. The minimum transition spiral length for any degree of curvature and design speed is obtained from the the relationship 's = (.;-82+, in which 's is the minimum spiral length in feet, - is the design speed in miles per hour, and + is the radius of curvature of the simple curve. This e*uation is not mathematically eact but an approimation based on years of observation and road tests.

Table ( is compiled from the above e*uation for multiples of 7) feet. hen spirals are inserted between the arcs of a compound curve, use 's = (.;-82+ a. + a represents the radius of a curve of a degree e*ual to the difference in degrees of curvature of the circular arcs.

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 Railroads Spirals applied to railroad layout must be long enough to permit an increase in superelevation not eceeding ( ? inches per second for the maimum speed of train operation. The minimum length is determined from the e*uation 's = (.(: 4-. 4 is the full theoretical superelevation of the curve in inches, - is the speed in miles per hour, and 's is the spiral length in feet.

This length of spiral provides the best riding conditions by maintaining the desired relationship  between the amount of superelevation and the degree of curvature. The degree of curvature increases uniformly throughout the length of the spiral. The same e*u ation is used to compute the length of a spiral between the arcs of a compound curve. $n such a case, 4 is the difference  between the superelevations of the two circular arcs. Spiral elements are readily computed from the formulas given above. To use these formulas, certain data must be &nown. These data are normally obtained from location plans or b y field measurements.

Exampl e :

 Thet ange nt sofaspi r alcur v ehas az i mut hof226˚and221˚r espec t i v e l y .        5  Themi ni mum l e ngt hoft hespi r ali s40. 00me t er s.wi t hami ni mum super        5       e el ev at i onof0. 10m/m wi dt hofr oadway .T hemaxi mum ve l oci t yt opassove r       g       a t hecur vei s70. 00km/hr .Ass umewi dt h       P oft her oadwayt obe9. 00me t er s. a)Det er mi net hedegr eeofsi mpl ec ur ve .  b)De t er mi net hel engt hoft hespi r alateac hoft hesi mpl ecur v e. c)Thesuperel ev at i onoft hefir st10. 00met er sf r om t heS. C.ont he 2 spi r al .Usee=0. 004K/R So l ut i o n: a)e=0. 004K2/R 2 0. 10=0. 004( 70) /R R =196. 00me t er s D =1145. 916/R D =1145. 916/196. 00 D =5. 85˚

b) Lc=0. 0036K3/R Lc=0. 0036( 70) 3/196. 00 Lc=63. 00met er s say60. 00met er( usemul t i pl eof10m. )

c#

e( = "(2;# ").()# = ).)(: m. "B () m. from TS on the Spiral# e3 = 7").)(:# = ).)>7 m. "B () m. from SC on the Spiral# e = ).)>7"9# = ).:;7 m. "super elevation at () m. from SC on the spiral#

 WORKSHEET29–SPI RALCURVE Name _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ Cour se/Year ___ ___ ___ ___ ___ ___ ___ ___ ___ Dat eSubmi t t ed___ ___ ___ _  Asi mpl ecur v ehavi ngar adi usof280. 00me t e r sc onnec t st wot angent s i nt e r s e c t i ngatanang l eof50˚ 00’ .I ti st ober e pl ac e dbyano t he rc ur v eha v i ng 80. 00met er sspi r al satendssucht hatt hepoi ntoft angencyshal lbet hesame. a)Det er mi net her adi usofnew ci r cul arc urv e  b)De t er mi net hedi s t anc et hatt hecur v ewi l lmov enear ert hev er t e x. c)Det ermi net hecent r alangl eoft heci r cul arcurve. d)Det er mi net hedeflect i onangl eatt heendoft hespi r al . e)Det er mi net heoffsetf r om t angentatt heendpoi ntoft hespi r al .

f ) Det er mi net hedi st anceal ong        6 t het angentatt hemi dpoi ntoft he s pi r al .        5       e       g       a        P

 WORKSHEET30–SPI RALCURVE Name _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ Cour se/Year ___ ___ ___ ___ ___ ___ ___ ___ ___ Dat eSubmi t t ed___ ___ ___ _  Thet wot angent sofasi mpl ec ur v ehasaz i mut hsof270˚ 00’and 10˚ 00’ r espect i vel y .I thas a r adi us of320. 00 met er s.I ti sr equi r ed t o change t hi s cur vet oaspi r alcurv et hatwi l lhaveval ueofp =2. 50 met er sand b =30. 00 met er sass howni nt hefigur e.Det er mi net hedi st anceonwhi cht henew cur ve mustbemove df r om t hever t exand i t sdi st ancef r om T. S.t ot heP. C.oft he s i mpl ec ur v e ,i fDE i spar a l l e lt oh.

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LESSON9–EARTHWORKS EARTHWORKS – t he c onst r uct i on of l ar ge open cut t i ngs or e xc av at i ons i nv o l v i ngbo t hc ut t i ngandfil l i ngofmat e r i alo t he rt hanr o c k.

EXCAVATI ON –i st hepr oce ssofl oose ni ng and r emov i ng ear t h orr oc kf r om         8 i t s or i g i nalpo si t i o ni nac utandt r ans po r t i ngi tt oafil lo rt oawas t ede po si t .        5       e       g       a        P

EMBANKMENT – t he t er m embankme nt de s cr i bes t he fil ladde d abov et he l ow poi nt sal ongt her oadwayt or ai set hel ev elt ot hebot t om oft hepavement s t r uct ur e mat er i alf or e mbankment c ommonl y comes f r om r oadway c ut s or desi gnat edborr ow ar eas

a.By Average End Areas  V=L/2( A1+A2)  Whe r e:  V=Vol umeofSect i onofEar t hwor ksbe t weenSt a.1and2,m³  A1,A2  =Cr osssect i onalareaofendst at i ons,m² L=Per pendi cul arDi st ancebet weent heendst at i on,m NOTES: 1.Theabov ev ol umef or mul ai se xac tonl ywhenA1=A2   buti sappr oxi mat eA1A2. 2.Consi deri ng t he f act s t hat cr osssect i ons ar e usual l y a consi derabl edi st ance apartandt hatmi nori nequal i t i esi nt hesur f ace oft he ear t h be t wee n se ct i ons ar e notc onsi der ed,t he me t hod ofend ar easi ssuffici ent l ypr eci sef orordi naryeart hwork. 3.By wher e heavy cut s or fil l s occur on sharp curves. The comput ed vol ume ofe ar t hwor k maybecor r ec t edf orcurv at ur eoutof o r di nar i l yt hec or r e c t i o ni sno tl ar g ee no ug ht obec ons i de r e d 4.A. By Pri smoi dalFormul a  V=L/6( A1 +AM +A2)

 Whe r e:         9  V=Vol umeofsect i onofear t hwor kbe t weenSt a1and2ofv ol ume        5       e ofpr i smoi d, m³       g       a        P  A1,A2  =c r o s s–s e c t i o nalar e a o fe nds e c t i o ns ,m²  AM  =Ar eaofmi dsect i onparal l elt ot heendsect i onsandwhi chwi l l  becomput edast heav er agesofr espe ct i v eenddi mensi ons,m³ NOTES: 1.A Pri smoi d i s a sol i d havi ng f or i t s t wo ends any di ssi mi l ar paral l el pl ane figur es oft he same number of si des,and al lt he si des oft he sol i d pl ane figur es.Al so,any pr i smoi d may be r es ol v ei nt o pr i sms ,pyr ami ds and we dges , havi ngacommon al t i t udest heper pendi cul ardi st ancebet ween t het wopar al l e le ndpl anec r o s s–s e c t i o n. 2.Asf arasvol umeofeart hwor ksareconcerned,t heuseof Pri smoi dalf ormul ai sj ust i fied onl yi fcr osssect i on ar et aken atshorti nt erval s,i sa smal lsurf acedevi at i onsareobserve d, and i ft he are as of successi ve cr osssect i on cl i ff or wi del y usual l yi t yi el ds smal l er val ues t han t hose comput ed f r om aver ageendar eas PRI SMOI DALCORRECTI ON FORMULA

CD =L/12 ( b1 –b2) ( h  – 1 h 2)  Whe r e: CD  =Pr i s moi dalCo r r e c t i on,I ti ss ubt r a c t e dal g e br a i c al l yf r o mt he  v ol umeasde t er mi ne dbyt heav er ageandt hear easme t hod t ogi vet hemor enear l ycorr ectvol umeasdet er mi nedbyt he Pr i smoi dalf ormul a,m³

        0 L=Per pendi cul ardi st ance        6 bet ween2paral l elandsect i ons,m       e  b1  =Di st ancebet weensl opes t akesatendsec t i onABC wher et he       g       a        P al t i t udei sh1,m  b2  =Di st ancebet weensl opes t akeatendsec t i onDEFwher et he al t i t udei sh2,m h1=Al t i t udeofendsect i onABC atSt a1,m h2  =Al t i t udeofendsect i onDEFatSt a2,m PROBLEM: 1.Gi ven t hef ol l owi ngcr osssect i on not esofaroadwaywi t h abase of6m and SS of1. 25: 1. 00,be t wee nt hevol umeoft hepr i smoi dbet wee n t het woendsec t i onsbyt hef ol l owi ngmet hods:a)endar eamet hod; b) Prism oidal form ul;c)end area m ethod and pri sm oidal correction form ua. St at i on 10+00 10+20

Cr ossSec t i onNot e s +6. 55 +2. 84 +2. 84 +7. 55 +3. 64 +1. 85

+6. 55 +3. 65

+2. 84 +0. 52

SOLUTI ON: Comput ef ort heareaateachst at i oncr osssect i onandatmi dsect i on Fi gur e

Chec kf orCutdi st ances DR1=DL1  =B/2+SHR=½ ( 6m)+1. 25( 2. 84) =6. 55m  Ar e abyme t hodoft r i angl eandr hombus  A1=BC +SC²=27. 12m² Fi gur e

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Checkf ort hedi st ances DR2  =B/2+SHR2=½ ( 6)+1. 25( 0. 52)=3. 65m DL2  =B/2+SHL2 =½ ( 6)+1. 25( 3. 64)=7. 55m  Ar e abyme t hodoft r i angl e  A2 =A  A +A L +A c+A d =½ ( 3) ( 3. 64)+½ ( 1. 85) ( 7. 55)+½ ( 1. 85) ( 3. 65)+½ ( 0. 52) ( 3)  A2  =16. 60m² Comput ef ort hedi mensi onsoft hemi dsec t i ons Fi gur e

DRm  =½ ( DR1+DR2) =½ ( 6. 55+3. 65) DRm =5. 10m.

HRm=½ ( HR1+HR2) =½ ( 2. 84+0. 52) HRm  =1. 68m.

DLm  =½ ( DL1 +DL2) =½ ( 6. 55+7. 55) DLm =7. 05

HLm  =½ ( HL1 +HL2) =½ ( 2. 84+3. 64) HLm =3. 24

HCm=1/2( HC1 +HC2)

=1/2( 2. 84+1. 85) HCm  =2. 345m Chec kf orCutdi st ances DRm  =B/2SHRm =1/2( 6)+1. 25( 1. 68) DRm=5. 10m

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DLm  =B/2SHLm =1/2( 6)+1. 25( 3. 22) DLm  =7. 05m  Ar e abyme t hodoft r i angl e  Am =Ae +Af+Ag +Ah =½ ( 3) ( 3. 24)+½ ( 7. 05) ( 2. 345)+½ ( 5. 10) ( 2. 345)+½ ( 3) ( 1. 68)  Am  =21. 68m COMPUTE FOR THE VOLUME OFEARTHWORK VOLUME OFCUTI N BETWEENTHETWO STATI ONS Fi gur e

1.ByEndAr e aMe t hod  Ve  =L/2( A1 +A2)  Whe r e: L=( 10+020)–( 10+000) =20m  A1=27. 12m²  A2  =16. 60m²  The n,  Ve =20/2 ( 27. 12+16. 60) =437. 20m² 2.ByPr i smoi dalFormul a  Vp  =L/6( A1 +4Am +A2)  Whe r e: L=20m;A1  =27. 12m² ;A2  =16. 60m² ;Am  =21. 67m²  The n

 Vp  =20/6[ 27. 12+4( 21. 67) +16. 60]=434. 13m³         3 1. Pri smoi dalFormul af or Co rrect i on         6       e Cp  =L/2( A1 +A2) ( b1 –b2)       g       a        P No t e : Re s ol v et heg i v e npr i s mo i di nt oas e r i e soft r i ang ul arp r i s mo i di nt oa s e r i e soft r i ang ul arpr i s moi d. Cp =Cpa +Cpb +Cpc +Ppd  Wher e: Cpa  =20/12( 2. 84–3. 64) ( 33)=0 Cpb  =20/12( 2. 84–1. 85) ( 6. 55–7. 55)=1. 65m³ Cpc  =20/12( 2. 84–1. 85) ( 6. 55–3. 65)=4. 785m³ Ppd  =20/12( 2. 84–0. 52) ( 33)=0  Then Cp  =1. 65+4. 785=3. 135m³ 2.Corrected Volum e  Vc =Ve -Cp  =437. 20–3. 135  Vc  =434. 065m

 WORKSHEET31–EARTHWORKS Name _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ Cour se/Year ___ ___ ___ ___ ___ ___ ___ ___ ___ Dat eSubmi t t ed___ ___ ___ _ Der i vet hepr i smoi dalcorr ect i on f ormul af ora t r i angul arend are asusi ngt he f or mul aV =L/6( A1 +AM +A2)

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 WORKSHEET32–EARTHWORKS Name _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ Cour se/Year ___ ___ ___ ___ ___ ___ ___ ___ ___ Dat eSubmi t t ed___ ___ ___ _ Deri vet hepri smoi dalf ormul af ordet ermi ni ngvol umesofr egul arsol i d.  V=L/6( A1 +AM +A2)

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 WORKSHEET33–EARTHWORKS Name _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ Cour se/Year ___ ___ ___ ___ ___ ___ ___ ___ ___ Dat eSubmi t t ed___ ___ ___ Usi ngt hepri smoi dalcorr ect i onf ormul a,findt hecorr ect edvol umeofcut  be t ween s t at i ons80. 00 me t e r sapar ti ft hear eaofi r r e gul arsec t i onsi nc utat st at i ons ar e 26. 00 sq. m.and 68. 00 sq. m.r es pec t i v el y .Base wi dt h = 8. 00 me t e r s ,s i des l o pe1: 1.

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 WORKSHEET34–EARTHWORKS Name _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ Cour se/Year ___ ___ ___ ___ ___ ___ ___ ___ ___ Dat eSubmi t t ed___ ___ ___  Thesi desl opeofar ai l r oadcutshowni s1: 1.Thewi dt hoft her oadwayi s 10. 00 me t er s.De t er mi ne t he cor r ec t ed v ol ume by appl yi ng t he pr i smoi dal corr ect i on.Di st ancebet weensect i onsi s100. 00met er s.

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 WORKSHEET35–EARTHWORKS Name _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ Cour se/Year ___ ___ ___ ___ ___ ___ ___ ___ ___ Dat eSubmi t t ed___ ___ ___ Gi vent hecr osssect i onnot esbel ow oft hegr oundwhi chwi l lbeexcavat edf ora r oadway ,comput et hevol umeofexc avat i onbe t wee nst at i ons47+00and48+ 00by:a)endare amet hod;andb)pr i smoi dalf ormul a.Ther oadi s30. 00f t .  wi dewi t hs l opesof1. 5: 1

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 WORKSHEET36–EARTHWORKS Name _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ Cour se/Year ___ ___ ___ ___ ___ ___ ___ ___ ___ Dat eSubmi t t ed___ ___ ___ Fr om t hef ol l owi ngcr osssect i onnot es,comput et hecorr ect edvol umesof c utandfil l ,t her o adbe dbe i ng20. 00f t .wi dei nc utand16. 00f t .i nfil l .The s i des l o pef o rbo t hc utandfil li s1: 1.Gi v et her e s ul t si nc ubi cy ar d.

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 WORKSHEET37–EARTHWORKS Name _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ Cour se/Year ___ ___ ___ ___ ___ ___ ___ ___ ___ Dat eSubmi t t ed___ ___ ___  Thef ol l owi ngcr osss ect i on not esar ef orar oad pass i ngahi l l ycount r y .  Ther oadsbedi s11. 00me t e r swi def ort hor oughcut ,10. 00me t er sf orsi dehi l l and 9. 00met er sf orfil l .Thesl opeoft hecuti s1: 1and1. 5: 1f orfil l .Fi nd t he  v ol umeofcutandfil lbyt heendar e ame t hod.

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