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Chng 1

Nhng kin thc chung v trc a1.1. i tng v nhim v ca trc a

Trc a l mt khoa hc nghin cu hnh dng, kch thc tri t v biu din b mtri t di dng bnh hoc bn . Ngo i ra trc a cn gii quyt h ng lot cc vn trong o c cc cng trnh nh chuyn bn thit k ra thc a, quan st ln, bin dng ccng trnh: thy in, thy li, xy dng... v i tng nghin cu ca trc a rt rng ngta chia trc a ra nhiu chuyn ng nh khc nhau nh:

Trc a cao cp:Nghin cu hnh dng, kch thc tri t, nghin cu vic xydng li trc a quc gia, nghin cu hin tng a ng hc, gii cc b i ton trc a trnb mt tri t v trong v tr.

Trc a cng trnh:Nghin cu vic kho st, tham gia thit k, thi cng cc cngtrnh, quan st ln, bin dng ca cc cng trnh....

Trc a nh:nghin cu vic xy dng bn bng nh my bay, nh mt t, nhv tinh, dng nh thay th cho cc phng php truyn thng quan st ln v bin dngca cc cng trnh xy dng.

Ng y nay xu hng chung ngi ta gi cc ng nh: trc a, bn , vin thm lGeometics. Ngi ta coi Geometics gm kin thc ca cc mn trc a cao cp, trc a cntrnh, trc a nh vin thm, bin tp bn , h thng thng tin t, tin hc.

Trong chng trnh mn hc n y chng ta nghin cu nhng kin thc c bn ca mntrc a ph thng. Phm vi nghin cu l o c trn phm vi nh ca b mt tri t. Cc sliu o c (chiu d i, gc...) c tin h nh trn mt phng v biu din chng ln mt ph(khng tnh n nh hng cong ca b mt tri t). V vy nhim v c bn ca mn

n y l trang b cho sinh vin nhng kin thc c bn v trc a trn mt phng, bit cch xdng bnh , bn t gip sinh vin vn dng kin thc ca mnh v o lnh vchuyn mn trong ng nh qun l v quy hoch t ai.1.2 Cc n v thng dng trong trc a

Trong trc a thng phi o cc i lng hnh hc nh chiu d i, gc bng, gcng... v cc i lng vt l nh: nhit , p xut....1.2.1 n v o chiu d i

Nm 1791 t chc o lng quc t ly n v o chiu d i trong h thng SI l mvi quy nh: "Mt mt l chiu d i ng vi 4.10-7 chiu d i ca kinh tuyn i qua Paris" v ch to ra mt thc chun c d i 1m bng thp khng g, c gi n n rt nh t tVin o lng Paris.

T sau th k 19, chnh xc ca thc chun khng cn p ng c yu cu olng cc phn t v cng nh. V th nm 1960 quy nh n v o chiu d i l : "Mt mtchiu d i bng 1.650.763,73 chiu d i ca bc sng bc x trong chn khng ca nguyn t Kripton - 86, tng ng vi qu o chuyn ri ca in t gia 2 mc nng lng 2P 10 v


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5d5". 1 mt (m) = 10 decimt (dm) = 102 centimet (cm) = 103 milimet (m.m) = 106 micromet(m) = 109 nanomt (Nm).

n v o din tch thng dng l mt vung (m2), kilomt vung (km2) v hecta (ha).1 km2 = 106 m2 = 100 ha, 1 ha = 104m2

Ngo i ra mt s nc cn dng n v o chiu d i ca Anh l :1foot = 0,3048m, 1inch = 25,3 mm

1.2.2. n v o gcTrong trc a thng dng 3 n v o gc l : Radian, , Grad.1- Radian: K hiu l Rad l 1 gc phng c nh trng vi tm ca 1 vng trn v

chn 1 cung trn ng trn vi chiu d i cung trn ng bng bn knh ca ng trn . ln ca gc bt k s bng t s gia d i cung chn bi gc v bn knh vng

trn.Gc trn l gc trn ng trn chn cung trn c chiu d i bng chu vi hnh trn.

Chu vi hnh trn c chiu d i l : 2R nn gc trn c ln l : 2Rad. Radian l n v ogc c dng trong tnh ton, c bit l khi s dng cc phng php ni suy cc gi trh m lng gic.

2. : k hiu l (o) l gc tm ng trn chn 1 cung trn c chiu d i bng 1/360chu vi hnh trn. 1 chia th nh 60 pht, 1 pht chia th nh 60 giy, k hiu l : 0 ' "

V d: gc c vit A = 120025'42''Tuy nhin gc c th vit bng , pht v phn mi pht. Gc trn c th vit l :

A = 120o25'7 3. Grad : k hiu l Gr l gc tm chn cung trn c d i bng 1/400 chu vi ng

trn. 1 Grad chia th nh 100 pht Grad (miligrad), 1 pht Grad chia th nh 100 giy Grad(decimiligrad), k hiu tng ng l : c, cc

V d: Gc B = 172gr 12c 27cc 4. Quan h gia cc n v:T nh ngha ba loi n v o gc, ta c quan h:

1 gc trn = 2Rad = 360o = 400 grT suy ra cc quan h chuyn i cc n v o gc. Khi tnh ton

2 Rad = 360o suy ra:Rad = o180

o = Rad 180

t cc h s:

o = 0180 = 57o17'44''8


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' = 60x180 = 3438'

'' = 60x60x180 = 206265''

Tng ng vi cng thc trn ta c cng thc chuyn i gia v Radian:

o = o. Rad

' = '. Rad

'' = ''. Rad Trong cc b i ton k thut khi tnh ton gi tr cc h m s lng gic ca cc gc nh

c th dng quan h tng ng, ngha l ch ly s hng bc nht trong cng thc khai trih m lng gic th nh chui s.

sin = + .......!5!3

53

++

tg = + .......!5!3

53

++

Ngha l gi tr gc nh tnh bng giy c ly bng gi tr gc tnh bng Radian.

1'' sin 1'' ''206265

Rad''1 = 0,00000 4848 Rad

V d: C 1 gc nh = 15'' chn 1 cung trn c bn knh R = 1000m. Ta c th suyra chiu d i cung trn chn bi cung l :

C= R. '' = 1000000 mm ''206265''15 = 72,7mm

1.3. Khi nim v cc mt c trng cho hnh dng ca Tri t

Khi nghin cu hnh dng tri t ngi ta thy rng tri t c dng elp quay, dt 2cc, b mt t nhin ca tri t rt phc tp. Din tch b mt tri t l : 510575.103 km2 trong i dng chim 71,8%, lc a chim 28,2%. cao trung bnh ca lc a so vmc nc i dng khong gn 900m. Nh vy b mt hnh hc tri t khng th biu dibng mt phng trnh ton hc n o c. Tuy nhin trong mt s trng hp tnh ton gnng ngi ta coi tri t c dng hnh cu, bn knh l : 6371 km.

Trong o v bn cc s liu o c c tin h nh trn mt cong, khi biu dinchng li thc hin trn mt phng. x l cc s liu o c ngi ta a ra cc loi mtdng trong trc a nh sau:1.3.1. Mt Geoid v Kvazigeoid.

1. Mt Geoid

Mt Geoid l b mt tri t gii hn bi mt ng th i qua im tnh cao. V


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xc nh mt Geoid c xc nh gn vi mc nc bin trung bnh. Th trng trng tGeoid c vit l : Wo.

2. Mt Kvazigeoid

V nhng bin i phc tp ca gi tr trng trng, xc nh chnh xc Geoid ngocc tr o trc a trn b mt tri t cn cn c hiu bit y v cu to ca tri l im khng d l m c. Vin s Nga Molodenxki a ra l thuyt xc nh mt gnmt Geoid, ng bng ch chnh lch so vi Geoid t 2 n 4 cm, vng ni chnh khng qu2m v c gi l mt Kvazigeoid. Nhiu nc trn th gii trong c Vit Nam dng mtKvazigeoid l m mt c s xc nh cao quc gia gi l cao thng(1).1.3.2. Mt Ellipsoid tri t v mt Ellipsoid quy chiu

1. Mt Ellipsoid tri tNh ta bit mt Geoid hoc Kvazigeoid l khi i din cho tnh cht vt l

Tri t. N c lin quan cht ch n tr o trc a nhng khng th dng l m c s xton hc cc tr o trc a v khng th dng phng trnh ton hc n o biu th mc (mt khng c phng trnh ton hc).

T l thuyt v khi cht lng quay quanh trc, ngi ta ngh n vic biu din tohc ca Tri t phi l 1 khi Ellip quay, dt 2 cc gi l Ellipsoid tri t. Bi v mtEllipsoid l mt ton hc, nn thc hin cc tnh ton trn mt n y trong Trc a cao c xy dng cc cng thc quy chiu cc tr o (gc, chiu d i...) ln mt n y.

Khi Ellip c trng tm v xch o trng vi trng tm v xch o ca tri t, ckhi lng bng khi lng tri t quay quanh trc to ra b mt gn vi mt Geoid trnphm vi to n cu gi l Ellipsoid chung hay Ellipsoid tri t. Kch thc ca Ellipsoid trit c c trng bi bn trc ln a, bn trc nh b, dt f.

f= aba

Hnh 1.1

1 * Ngo i cao thng cn c cao chnh, cao ng lc

b

a


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C nhiu nh khoa hc trn th gii xc nh kch thc Ellipsoid (bng 1.1).Bng 1.1

Tn Ellip soid Nm xc nh Bn trc ln a (m) dt f Everest

Hayford

Karaxovski

Reference

W.G.S-84

1830

1909

1940

1967

1984

6.377.276

6.378.286

6.378.245

6.378.100

6.378.137

1:300,80

1: 297,00

1: 298,3

1: 298,25

1: 298,257

2. Mt Ellipsoid quy chiu

Trc khi c Ellipsoid chung do yu cu x l ton hc ca mi quc gia tnh raEllipsoid cho ph hp vi l nh th ca nc mnh, c th dng Ellip soid ca nc khcnhng c 2 trng hp c nh v cho ph hp nht vi Geoid ca l nh th nc mnh. Ellipsoid c s dng ring ca tng nc gi l Ellip soid quy chiu. Trong h toa HN-72Vit Nam ly Ellipsoid Kraxovski (1940) l m Ellip soid quy chiu. Hin nay Vit Nam c hta mi VN-2000 ly Ellip soid W.G.S-84 l m Ellip soid quy chiu. Gc ta t tikhun vin Vin nghin cu a chnh trn ng Ho ng Quc Vit - H Ni.1.4. cao tuyt i, tng i, chnh cao

nghin cu b mt g gh ca tri t phc v cho vic xy dng cc cng trnh:thy li, thy in, giao thng, xy dng.... v nghin cu b mt tri t ngi ta a ra ccnh ngha v cao.1.4.1. nh ngha

cao ca mt im l khong cch thng ng t im n mt thy chun.Mt thy chun: mt thy chun l mt c phng vung gc vi ng dy di ti

mi im.Nh vy, v nh ngha trn b mt tri t c v s mt thy chun. C qua 1 i

trn b mt Tri t c 1 mt thy chun i qua. phn bit cc mt thy chun ngiphn chng th nh 2 loi:

Mt thy chun i a:

Mt thy chun i a l mt nc bin, i dng trng thi trung bnh, yn tnh cng chnh l mt Kvazigeoid.

Mt thy chun gi nh:Mt thy chun gi nh l mt thy chun i qua mt im bt k. im n y c

nh cao gi l cao gi nh.


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1.4.2. cao tuyt i cao tuyt i ca 1 im l khong cch thng ng t im n mt th

chun i a. cao n y c gi l cao thng. im gc cao c cao l 0m. VitNam ly mc nc bin trung bnh ti Trm nghim triu Hn Du - Sn - Hi Phng l mim cao gc. cao c ghi trn bn l cao tuyt i.1.4.3. cao tng i

cao tng i ca mt im l khong cch thng ng t im n mt thchun gi nh gi l cao gi nh.

phc v vic nghin cu b mt tri t, phuc v vic xy dng cc cng trnh thyli, thy in, giao thng... trong phm vi l nh th ca mt nc ngi ta xy dng li cao gm nhiu cp, gi l li cao Nh nc.

Tuy nhin cc im cao Nh nc vn rt tha tht khng p ng c cho tt ccc cng trnh. V vy trn khu vc nh khi xy dng cc cng trnh ngi ta c th tnh ton

theo n v cao tng i ( cao gi nh).Khi mun chuyn cao tng i v cao tuyt i ngi ta phi o ni ca

(c trnh b y trong chng 2 ca gio trnh n y).1.4.4. Chnh cao

Chnh cao l hiu cao ca 2 im.Gi s im A c cao l HA, im B c cao l HB (hnh 1.2). Chnh cao ca 2im A v B l :

hAB = HB - HA

Nu cao im B ln hn cao ca im A th hAB > 0Nu cao ca im B nh hn cao ca im A th hAB < 0Nh vy chnh cao c du.Khi bit cao ca im A l HA, bit chnh cao hAB ta c th tnh c cao ca

im B l :HB = HA + hAB

Chnh cao hABc xc nh bng nhiu phng php khc nhau: phng php o caohnh hc, phng php o cao lng gic, phng php o cao bng thit b GPS (GlobalPositioning System)... (c trnh b y chng 2).1.5. Bnh , bn v mt ct

1.5.1. Bnh Bnh l hnh chiu thu nh ca 1 phn nh b mt tri t ln giy theo mt t l

nht nh (khng tnh n nh hng cong ca b mt tri t).

nh 1.2


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Nh ta bit phc v cc mc ch khc nhau. V d khi cn kho st, thit kmt khu vc nh ngi ta cn biu din cc yu t (nh a hnh, a vt) ln trn giy themt t l nht nh. Khi ngi ta coi b mt tri t trong khu vc o v l phng. Cc yt o c (chiu d i, gc...) c xc nh coi nh xc nh trn mt phng v khi biu di

chng cng c tin h nh trn mt phng, v vy khng c s bin dng. Cc yu t cbiu th theo mt t l nht nh gi l t l bnh . T l bnh thng ln: t l 1:51:1000, 1:2000. Bnh thng biu din 1 khu vc nh, thng khng c im ta cao Nh nc. Ta v cao trn bnh thng l gi nh. Thc t cho thy mt khvc c din tch khong 20km2 ngi ta c th biu din n di dng bnh , ngo i phm vi phi biu din di dng bn .1.5.2. Bn

Bn l hnh chiu thu nh ca mt phn hay to n b tri t ln giy theo mt t lnht nh (c tnh n nh hng cong ca b mt tri t).

V vy im khc nhau c bn gia bnh v bn l ch cc yu t o c trbnh c coi nh o trn mt phng v vic biu din n cng c tin h nh trn mphng, coi nh khng c s bin dng. Cn trong bn cc yu t o c c thc hin trnmt cong, khi biu din chng li tin h nh trn mt phng v vy khng th trnh c bin dng. Do ngi ta phi tnh ton s bin dng bng cch chiu cc yu t o (khixy dng li ta , cao Nh nc) ln mt Ellipsoid quy chiu, t mt Ellipsoid quychiu c chiu ln mt phng trung gian (mt nn hoc mt tr). T mt trung gian ra mt phng. Qua qu trnh thc hin php chiu ngi ta tnh ton s hiu chnh v gcchiu d i. H thng ta , cao trong o v bn c thng nht trong tng quc gia.Bn Vit Nam trc y c thng nht trong 1 h ta HN-72, Ellip soid quy chiu lEllipsoid Kraxovski. Hin nay Vit Nam ang s dng h ta VN-2000, Ellip soid quy

chiu l Ellip soid W.G.S-84.Bn c chia l m 2 loi l bn a l chung v bn chuyn . T l b

theo mc ch s dng, bn c t l 1/200, 1/500, 1/1.000, 1/1.000.000...1.5.3. Mt ct a hnh

Khi kho st cc tuyn ng mng mng ngo i bnh hoc bn cn phi lpmt ct dc v ngang tuyn. Mt ct phc v cho vic thit k, tnh ton khi lng o khc vi bnh , bn biu din mt t trn mt phng ngang, cn mt ct a hnh lchiu ca mt ct dc hoc ngang ca mt tuyn a hnh ln mt phng thng ng

Hnh 1.3a) a hnh

b) Mt ct dc tuyn


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biu din a hnh bng mt ct dc, ta ra thc a ng cc cc theo s thay ca a hnh (cc 1, 2, 3, 4, 5 - hnh 1.3a). Sau tin h nh o cao v khong cch gia ccc. Trn giy k li ly trc thng ng l m trc cao (H), trc nm l m trc khong cchngang theo t l ng v t l ngang biu th cc im 1, 2, 3, 4, 5 (hnh 1.3b). Ty thuc v

dc a hnh chn t l ng v ngang cho ph hp. Thng t l ng ln hn t l ngangV d t l ng l 1/500, t l ngang l 1/1000.1.6. T l bn , chnh xc ca t l bn

1.6.1. T l bn Khi th nh lp bn (hoc bnh ) kt qu o c c thu nh li 100, 1000 ln

biu th trn giy. Mc thu nh ph thuc v o din tch khu vc, yu cu mc chi tica i tng biu th, mc ch s dng bn . Mc thu nh gi l t l bn .

T l bn l t s gia on ab trn bn v on thng AB tng ng ngo i thca, k hiu t l bn l 1/M.

ABab

M1 =

T l bn c biu th bng phn s c t s bng 1.

V d: Bn t l1000

1M1,

5001

M1 == ....

Nh vy khi bit c chiu d i on ab trn bn , chiu d i tng ng AB ngothc a ta c th tnh c t l bn . V d o trn bn c on thng ab = 5cchiu d i AB tng ng ngo i thc a l AB = 100m. Vy t l bn l :

20001

10000cm5cm

100m5cm

ABab

M1

==== Trong thc t khi bit t l bn 1/M, bit chiu d i on thng ab trn bn ta c

th tnh c chiu d i on AB ngo i thc a v ngc li.V d 1: Bit t l bn 1/10000 chiu d i on ab l 2cm, tnh chiu d i AB ng

thc a.

Theo nh ngha:ABab

M1 = suy ra: AB = ab.M

= 2cm.10000 = 20000 cm = 200m.V d 2: Bit t l bn l 1/2000, on AB = 100m. H y biu din on AB ln tr

bn . T nh ngha ta c:

ABab

M1 = suy ra; ab = 5cm

200010000cm

2000100m

MAB ===

Nh vy biu din on AB ngo i thc a ln bn t l 1/2000 l on ab = 5cm.


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1.6.2. chnh xc ca t l bn Bng thc nghim ngi ta thy rng i vi mt ngi bnh thng phn bit

2 im khong cch nhn l 20cm, th khong cch nh nht gia 2 im l 0,1mm. T ngi ta a ra nh ngha v chnh xc ca t l bn .

chnh xc ca t l bn l khong cch ngo i thc a tng ng vi 0,1mmtheo t l bn .

V d: bn t l 1:10000 chnh xc ca n l : 0,1mm x 10000 = 1000mm = 1m.Vy chnh xc ca bn t l 1/10000 l 1m, tng t chnh xc ca bn t

l 1/500 l 0,1mm x 500 = 50mm = 5 cm.Nh vy bn c t l c ng ln chnh xc c ng cao v ngc li. Ngo i ra bn

c t l c ng ln mc chi tit c ng cao, biu th c vt c din tch vng nh. Bn ct l c ng nh tnh khi qut v a hnh v a vt c ng cao. Ty theo mc ch s dngngi ta s dng bn c t l thch hp.

1.7. Thc t l thun tin cho vic s dng bn di mi t bn ngi ta thng dng thc

t l. C 2 loi thc t l l thc t l thng v thc t l xin.1.7.1. Thc t l thng

Gi s dng thc t l thng cho bn t l 1:5000Trn on thng c bn AB = 2cm tng ng vi t l bn l 100m ngo i thc a

(hnh 1.4). Ta t cc on lin tip c d i 2cm, 4cm, 6cm tnh t im gc 0, tng nvi chng l : 100m, 200m ngo i thc a. Trn on c bn AB ta chia l m 10 phn nh bngnhau. Nh vy mi on nh c on d i 2mm tng ng vi 10m ngo i thc a (hnh 1.4).

Hnh1.4Cch s dng thc t l thng:

Dng compa o chiu d i on thng ab trn bn t l 1/5000. Gi nguyn khu compa m v o thc c c gi tr thc a ca on thng AB = 240m.1.7.2. Thc t l xin

nng cao chnh xc xc nh chiu d i trn bn , bn t l ln ngi ta

thng dng thc t l xin.1. Cch dng thc t l xin

Gi s cn dng thc t l xin cho bn t l 1/2000. Trn na on thng ta lyon c bn l AB = 2cm ng vi chiu d i 40m ngo i thc a. Bt u t 0 t cc on ltip c chiu d i l : 2cm, 4cm, 6cm, 8cm tng ng vi gi tr thc al 40m, 80m, 120m,160m (hnh 1.5).


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Hnh 1.5Dng cc vung tng ng c kch thc 2 x 2cm, vung th nht chia cc cnh

vung th nh cc phn bng nhau (n = 10, m = 10). Theo chiu ngang k cc ng songsong v u nhau.

Theo chiu ng k cc dng xin song song v u nhau. Vi cch dng nh trn tac IKB ng dng viNOB. Ta c t s.

BO

NO.BKIK

BO

BK

NO

IK == (1.1)

mm

ABNO,

n1

BOBK == (1.2)

Thay (1.2) v o (1.1) ta c:

IK =10.10

40n.m

AB = = 0,4m

v cc on tip theo c chiu d i tng ng ngo i thc a l : 0,8m, 1,2m, 1,6m, 2,0m...2. S dng thc t l xin

o trn bn on ab bng compa. Gi nguyn khu compa m v o thcc c chiu d i CD ngo i thc a l :

CD = 80m + 8m + 1,6m = 89,6m1.8. Phng php biu din a hnh bng ng ng mc

Trn bn a hnh thng th hin 2 yu t l a hnh v a vt. Vic biua hnh c th c thc hin bng cc phng php: phng php t m u, phng php kvn... Nhng thng dng v chnh xc nht l phng php biu din a hnh bng ng mc (ng bnh , ng ng cao).

Gi s c a hnh (mt qu i). Ta tng tng dng cc mt phng E1, E2, E3 (ccmt phng n y song song vi mt thy chun) ct qu i theo cc mt phng ngang . C

mt phng n y cch u nhau mt khong l h (hnh 1.6). Cc vt ct nhn c c chithng ng xung mt phng ngang E. Hnh chiu ca chng l cc ng cong khp kc gi l ng ng mc (ng bnh ) h c gi l khong cao u ca ng nmc. C 3 loi ng ng mc:

ng ng mc con, ng ng mc ci v ng ng mc ph.

C D


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Hnh 1.6ng ng mc con : L ng ng mc biu th bng nt nh v trn

khng ghi cao.ng ng mc ci : L ng ng mc biu th bng nt ln hn ng ng

mc con v trn n c ghi cao.ng ng mc ph : vng ng bng a hnh bng phng v vy khong cch

gia cc ng ng mc ln, ni suy cc im cao trn bn c d d ng gia 2ng ng mc con ngo i ra k thm mt ng ng mc ph, ng ng mc ph cth hin bng nt t trn khng ghi cao.1. Cc tnh cht ca ng ng mc:

- Cc im nm trn cng mt ng ng mc c cng cao ngo i thc a.- Cc ng ng mc l cc ng cong trn tru, lin tc khp kn.- Ni n o c ng ng mc c ng tha th a hnh c ng thoi v ngc li n

ng mc c ng mau a hnh c ng dc. Nu chng trng nhau th c vch ng.- Cc ng ng mc khng bao gi ct nhau (tr trng hp a hnh h m ch).Nhng yu t a hnh khng biu th c bng cc ng ng mc nh vch ni

b mng c dng k hiu khc biu th . phn bit gia ni v h ngi ta thngdng k hiu nt ch hng dc hoc ghi ch cao a hnh.

- Khong cch ngn nht gia 2 ng ng mc l on vung gc ti ch dc nht.


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2. Nguyn tc chn khong cao u cu ng ng mcChnh cao gia 2 ng ng mc k nhau gi l khong cao u, k hiu l h. Vic

chn h phi m bo tnh kinh t, k thut. Tr s h c ng nh th mc biu th a hnh bn c ng chnh xc. Tuy nhin i hi khi lng o c ngo i thc a c ng nhiu v gi

th nh c ng cao. Ngo i ra tr s h cn ph thuc v o dc a hnh khi o. T l bn n dc a hnh ln th khng th bin th cc ng ng mc vi khong cao u nh v khi cc ng ng mc s chng ln nhau. Cho nn vic tnh ton khong cao u cang ng mc phi da v o 2 yu t l : dc a hnh v t l bn . Trong thckhong cao u ca ng ng mc c quy nh c th trong quy phm o v bn hnh (bng 2.2).

Bng 2.2Khong cao n h (m)

a hnh1:500 1:2000 1: 10000

Vng bng phng ( dc v < 20

) 0,5 1,0 2,0Vng i (2o 150) 1,0 5,0 5,01.9. K hiu quy c ca bn

Trn bn a hnh (hoc a chnh) thng biu th 2 yu t l : a hnh vvt. a hnh c biu th bng cao im hoc bng ng ng mc.

Cc yu t a vt c biu th trn bn bng cc phng php khc nhau i via vt c kch thc ln nh sng, h, ng quc l, khu cng nghip... th phi biu thng v tr, kch thc ca n theo t l bn . Tc l chng c biu th theo ta ph

(x, y) hoc theo ta cc (B, S). Khi xc nh n trn bn theo t l ta c th tnh kch thc, din tch ca chng ngo i thc a v cch biu th nh vy c gi l biu da vt theo t l.

i vi vt c kch thc nh m khng th biu din n theo t l c ngi ta biuth phi t l. V d: ging nc, ct in, ng mn, a gii.... khi ngi ta xc chnh xc tm ca n v dng k hiu quy c biu th. Cc k hiu quy c c trnh br trong cun "K hiu bn a hnh" hoc cun "k hiu bn a chnh". Cc k hquy c n y c xy dng trn c s khoa hc. Tc l khi nhn v o k hiu ngi ta lintng n a vt c hnh dng tng t. iu gip cho ngi dng d nh, d s dng.

Ngo i ra th hin ni dung ca a vt cn phi ghi ch bng ch hoc bng s nha danh l ng, x , tn sng ni, su lng h, hng dng chy... Tuy nhin cc ghi chcng cn tun theo cc quy nh sau:

Ch ghi ch trn bnh , bn phi vit song song vi cnh khung trn hoc di.Tn gi ca sng, sui, mm ni cn vit dc theo hng ca chng, cao ca ng ngmc c ghi ch ngt qu ng v u ch c quay ln pha cao. Khi th nh lp bnh hobn cn tun theo cc quy nh ty theo loi bn (bn a chnh hoc bn a hnh).


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1.10. Cc h ta thng dng trong trc a

xc nh v tr mt im trn mt t trong trc a ngi ta thng dng nhiuta khc nhau. H ta a l, h ta trc a, h ta vung gc phng GausKruger, h ta UTM.

Trong gio trnh n y, chng ta xt h ta thng dng trong trc a thc h nh.1.10.1. H ta a l

H ta a l cn c gi l h ta thin vn.Ta a l ca mt im trn mt t c xc nh bi kinh v v a l v

nh ngha:V a l: v a l ca mt im

l gc hp bi ng dy di i qua im v mt phng xch o. V a l c khiu l . bin i t O90o v 2 pha Bcv Nam tnh t xch o (hnh 1.7).

Kinh a l:

Kinh a l ca mt im l gc nhdin hp bi mt phng kinh tuyn gc v mtphng kinh tuyn i qua im , kinh al c k hiu l. bin i t 0 180o v2 pha ng v Ty tnh t kinh tuyn gc(2*).

Nu im nm pha ng kinh tuyn gc th im c kinh ng. Nu i nm pha Ty kinh tuyn gc th im c kinh Ty. Ta a l ca 1 im c th

trn bn hoc xc nh trc tip ngo i thc a bng vic "o thin vn".Vit Nam ho n to n nm pha Bc bn cu v pha ng kinh tuyn Greenwich ch

nn to n b l nh th Vit Nam u c v Bc v kinh ng.V d: Ct c H Ni c ta a l l :

= 21002'B,= 105o50'Trn cc t bn a hnh thng biu th li kinh v tuyn v ta a l ca c

gc khung ca t bn . Chnh lch v v kinh ca gc khung l :

=N -M, = N - M T s chnh lch v ta ca cc gc khung bn ta c th xc nh c ta

a l ca bt k im n o trn bn .1.10.2. H ta vung gc Gauss - Kruger

T th k 19 nh ton hc K.F. Gauss a ra php chiu hnh bn v sau Kruger ho n thin. V vy gi l php chiu Gauss - Kruger. Php chiu c m t nh sau:

2 Kinh tuyn gc l kinh tuyn i qua u thin vn Greenwich - Lun n

Hnh 1.7


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Ngi ta chia qu t th nh 60 mi, hoc 120 mi, mi mi l 6o hoc 3o nh utheo th t t Ty sang ng tnh t kinh tuyn gc. Mi mi c chia th nh 2 phn unhau bi kinh tuyn gia mi gi l kinh tuyn trc. T tm O ca qu t chiu ln mt trsau ct mt tr theo ng sinh v tri ra mt phng. Mt phng n y gi l mt phng

hnh Gauss.H ta c xy dng trn mt phng ca mi chiu 6o trong mt phng chiu hnh

Gauss c gi l h ta Gauss- Kruger. Trong nhn hnh chiu ca kinh tuyn trc l mtrc X, ca xch o l m trc Y. Nh vy nu tnh t gc v pha Bc x lun lun mang ddng, v pha Nam mang du m. Cn tr s y v pha ng mang du dng, v pha mang du m. Bn cu Bc c x > 0, y c th m c th dng. khi tnh ton trnh y m quy c im gc O ca ta xo = 0, yo = 500 km, ngha l tnh tin kinh tuyn trc v phaTy 500 km (hnh 1.8)

tin vic s dng trn bn a hnh ngi tak sn li ta vung gc Gauss bng nhng ng

song song vi trc OX v OY to th nh li km. Chiud i cnh ca li vung c th tnh n nh hngca bin dng tng ng vi t l bn . V d vibn t l 1:10000 chn vung 10 cm x 10cm, bn t l 1:25000 chn vung 4 cm x 4cm, bn tl 1:50.000 chn vung 2cm x 2cm.

Pha ngo i khung bn c ghi tr s X v Y ca cc ng song song. phn bit ngayc ta ca im nm mi chiu th my v cch im gc O bao nhiu ngi ta quynh cch vit ho nh y v ghi km theo th t mi chiu.

V d: ta im Lng Trung (H Ni) l : 2325464,246; 48.505973,362 c ngha lim cch xch o pha Bc: 2325464,246 m v mi th 48 v pha ng kinh tuyn g105o.

(505973,362 - 500000,000 = 5973,362m) tnh tr s kinh ca kinh tuyn gia mi th n (*) n o ta dng cng thc

N = 6o.n - 3o Ho ng Sa nm mi th 49 kinh tuyn gia ca mi n y c kinh l : =

6o.19 - 3= 111oLi khng ch trc a mt phng XY Vit Nam trong h ta HN-72 c x

dng theo h ta vung gc Gauss - Kruger.Trong Ellip soid quy chiu l Ellip soid Kraxovski.

(*) i vi mi nm pha ng th n phi hiu l s mi tr i 30. V d: n= 49, th hiu l n= 49 30 = 19

nh 1.8


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1.10.3. H ta vung gc UTM (N.E)Php chiu UTM (Universal Transerse Mercator) khc vi php chiu Gauss l ch

Ellip soid quy chiu ct mt tr ch khng tip xc vi mt tr ti kinh tuyn gi (kinh tuytrc). iu l m hn ch s bin dng 2 kinh tuyn bin. Da trn c s ca php chiu

ngi ta xc nh h ta gi l h ta UTM.Trong php chiu UTM hnh chiu ca

kinh tuyn gia v xch o l 2 ngthng vung gc vi nhau c chn l mtrc ta (hnh 1.9). Trong M l imcn xc nh ta . O' l giao im hnhchiu kinh tuyn trc O'Z v xch o O'E.im F l hnh chiu ca M ln kinh tuyntrc. Cung LM l hnh chiu ca v tuynqua M. Cung ZM l hnh chiu ca kinh

tuyn qua M. l lch kinh tuyn (gn kinh tuyn). Ta UTM ca im Mc xc nh bi tung Hnh 1.9

NM (North) v ho nh (EM) (East) ging nh quy nh trong php chiu Gauss ngi tari 0' n 0, 00' = 500 km tc l gc ta l 0 v vy ta :

EM = E' + 500 km.Trc nm 1975 qun i M s dng h ta UTM vi Ellip soid quy chiu l Ellip

soid Everest th nh lp bn a hnh min Nam Vit Nam. Do khi s dng bn ncho thng nht ta phi chuyn h ta UTM (XUTM v YUTM) sang h ta Gauss - Kurger

(XG, YG).Trong h ta VN-2000 ta cng dng php chiu UTM nhng Ellip soid quy chiu l

Ellip soid W.G.S-84. Gc ta nm ti khun vin Vin Nghin cu a chnh - H Ni.1.10.4. H ta gi nh

Khi o v bnh khu vc nh, khng c h ta Nh nc ta c th ginh mt h ta vung gc. Trong gcta 0 l ty . Ta c th gi thit ta im gc 0 l xoyo (hnh .1.10). Trc x trngvi kinh tuyn t. Trc n y c th xc nhc bng cch t my kinh v ti 0 xcnh kinh tuyn t ti . Trc y vung gcvi trc x. x > 0, y > 0 ta chn gc ta nm v tr thch hp. Khi ta ccim trong khu o thuc h ta gi nhxoyo la chn. Hnh 1.10


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1.10.5. H ta ccTrong o v im chi tit bng phng php to n c thng dng h ta n y.Gi s trn mt phng chn im A l

im cc, v 1 h ng AB l m trc cc. V tr

im 1 c xc nh bi gc1 cnh cc S1 (hnh1.11) gc cc1 l gc tnh t hng trc cc ABtheo chiu kim ng h n hng cnh cc S1,cn cnh cc S1 l chiu d i ngang tnh t imgc A n im 1. Tng t nh vy xc nhim 2 cn bit gc cc2 v cnh cc S2...Trong o v im chi tit im A l im t my kinh v, cn im B l im nh

xc nh im chi tit i n o cn phi xc nh gc cci v chiu d i cnh ccSi tng ng.1.11.1. nh hng ng thng

Khi biu th mt on thng ln trn bnh hoc bn khng nhng bit chiu don thng m cn phi bit phng hng ca n. Vic xc nh hng ca 1 ng thng sovi hng gc gi l nh hng ng thng. Hng gc (hng chun) c th l hng cakinh tuyn thc hoc kinh tuyn t hoc kinh tuyn trc. Ty theo hng ca kinh tuyn gcm c cc loi gc khc nhau.1. Gc phng v

Gc phng v ca mt ng thng l gc ngang hp bi hng bc ca kinh tuyn vi hngca ng thng. Gc phng v c tnh theo chiu kim ng h, bin i t 00 360o.

Nu kinh tuyn l kinh tuyn thc gc phng v gi l gc phng v thc khiu l A thc (A thc bin i 03600)

Nu kinh tuyn l kinh tuyn t th gc phng v gi l gc phng v t, khiu l A t (A t bin i t 0o 360o).

Kinh tuyn thc: Kinh tuyn thc l ng giao ca mt phng qua trc tri t v bmt tri t.

Kinh tuyn t : Kinh tuyn t l ng giao ca mt phng qua trc ca kim namchm v b mt tri t. Ti mt im trn b mt tri t hng ca kinh tuyn thc v ktuyn t khng trng nhau. Gc hp bi kinh tuyn thc v kinh tuyn t gi l t thin cakim nam chm, k hiu l .

Nu kim nam chm lch v pha ngkinh tuyn thc th mang du +, nu kimnam chm lch v pha Ty kinh tuyn thcth mang du -.

Quan h gia gc phng v thc vphng v t l : Athc = At

nh 1.11

nh 1.12

-

+

1

1

2 2


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T N

BA l nghchTa c: BA = AB + 1800 -> AB = BA -

1800. Tuy nhin, c trng hp phi + 1800.V vy tng qut ha l1800.3) Gc 2 phng

V gc phng v, gc nh hng bini t 00 3600 qu trnh tnh ton khngthun li. V vy ngi ta dng gc 2 phng.

Hnh 1.15Gc 2 phng ca mt ng thng l gc ngang hp bi hng Bc - Nam gn nht

ca kinh tuyn vi hng ca ng thng. K hiu l R, gc 2 phng bin i t 00 900 c km theo tn ca gc phn t (hnh 1.16).

Tng ng vi gc phng v thc c

gc 2 phng thc k hiu l R'

thc. Tngng vi gc phng v t c gc 2 phng t,k hiu l R' t. Tng ng vi gc nhhng c gc 2 phng k hiu l R: Quan hgia gc nh hng v gc 2 phng cch ra bng 1.3.

Cn c v o bng 1.3 ta c th chuyn t gc nh hng sang gc 2 phng v ngcli. Thc t khi tnh ton bng cc thit b ths ta c th tnh chuyn t gc nh hng

sang gc 2 phng xt du caX ,Yhoc x,y theo cc gc phn t.

V d: Trong h ta gi nh (hnh1.7) du ca gia s ta c th hin trongbng 1.4.

Hnh 1.6

Bng 1.3.Gc phn t Gc nh hng v 2 phng

IIIIIIIV

RBD = 1RND = 1800 - 2

RNT = 3 - 1800 RBT = 3600 - 4

Bng 1.4.Gia s ta Gc phn

t Gc

2 phng X YIIIIIIIV

RBDRND

RNT RBT

+--+

++--

1

3

4

2

nh 1.7

y

x


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Thc t cho thy, khi bit gc phngv t ca mt ng thng, bit t thin v gn kinh tuyn ta c th tnh c gcnh hng ca ng thng .

V d: Tnh gc nh hng ca ngthng AB bit gc phng v t ca n l At =300 15', t thin = 5', gn kinh tuyn = - 3' . V hnh biu din n?

Da v o nh ngha ta v c cckinh tuyn v gc phng v t ca ngthng AB. (hnh 1.8).

T hnh v ta c:

Athc= At + = 30015' +5' = 30020'

= Athc + = 300

20' + 3'

= 300

23'

V < 900. ng thng nm gc phn t th nht = RB = 30023'

1.12. B i ton xc nh ta vung gc phng

Trong trc a thng phi gii 2 b i ton c bn trong h ta vung gc phng l :1- B i ton trc a thun:

Ni dung b i ton trc a thun l :Trong h ta vung gc phng

(hnh 1.19). Bit ta im A l xA, yA, chiud i on AB l SAB. Gc nh hng cnh AB

l AB (hoc gc 2 phng cnh AB l RB).Tnh ta im B.Nhn v o hnh v 1.9 ta c :

xAB = SAB . CosAB = SAB.CosRByAB = SAB . SinAB = SAB.SinRB

Ta im B l :

xB =xA + xAB.yB= yA + yAB.

2- B i ton trc a ngc

Bit ta im A l xAyA, ta im B l xByB, tnh chiu d i on thng AB; gcnh hng cnh AB.Trn hnh 1.9 ta c:

xAB=xB - xA .yAB = yB yA .SAB =

22 AB AB y x +

nh 1.8

nh 1.9

x

y

KT trc

// KT trc


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xy

arctgRxy

tgR =

=

Du cax, y quyt nh tn gi ca gc phn t.

Trong trng hp n yx > 0, y > 0 do ng thng nm gc phn t th nht Rmang tn gi l RB . Nh vy gc nh hng bng gc 2 phng (tc l RB = )3. Tnh chuyn gc nh hng

Gi s rng c ng chuyn (hnh 1.10)

Hnh 1.10Bit gc nh hng cnh 1,2 l12 , bit cc gc o1, 2 Vn l phi tnh

chuyn gc nh cnh 1, 2 n cc cnh khc ca ng chuyn.Nhn v o hnh v ta c:23 = 12 + 1800 - 134 = 23 + 1800 -2 = 12 + 1800 - 1 + 1800 - 2.34 = 12 + 2.1800 (1+2). (1.3)

Tng qut ha (1.3)c = + n.1800 -

n

i1

(1.4)

Nu trc X trng vi kinh tuyn trc th l gc nh hng, nu trc X trng vikinh tuyn t th gc l gc phng v t. Trong trng hp chung gi l gc nh hng.

Cng thc (1.4) l cng thc tnh chuyn gc nh hng khi gc o bn phi ng o.Trong :c l gc nh hng cnh cui

l gc nh hng cnh ul tng gc o bn phi ng o.

n l s gc okhi gc o bn tri ng o cng thc (1.4) bin i l :

c = - n.1800 + (1.5).Trong l tng gc o bn tri ng o.Nh vy, khi tnh chuyn c gc nh hng n cc cnh, bit chiu d i cc cnh t

dng b i ton trc a thun c th tnh c gia s ta v ta cc im k tip trong chuyn.

n

1i

n

1i

n

1i

12

12 23

34 23

1

2


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Chng 2

o cao2.1 Mc ch, ngha v cc phng php o cao.

o cao l vic xc nh yu t hnh hc c bn th nh lp bn a hnh pv cho vic nghin cu hnh dng qu t v s vn ng ca n theo phng thng ng

Trong chng 1, chng ta l m quen vi cc khi nim, cao tuyt i, caotng i, hiu s cao, mt thu chun gc, mt thu chun gi nh, trong chng n ych tm hiu cc phng php thng dng xc nh cao trong mng li cao nh nc t hng IV tr xung ( cao tnh theo phng dy di t im cn xc nh n mt thu chgc. Vit Nam, mt thu chun gc l mt i qua im mc Hn Du cnh Khch sn VnHoa Sn Hi Phng) hoc cao tng i ( cao so vi mt thu chun quy c) phcv ch yu cho cng tc th nh lp bn a hnh.

Da v o nguyn l hnh hc hoc vt l, cng nh thit b o v chnh xc m nhng phg php xc nh chnh cao nh sau:1. Phng php o cao hnh hc theo nguyn l tia ngm nm ngang, ngha l trong phm vio v hp ngi ta coi tia ngm song song vi mt thu chun v vung gc vi phng dydi. Dng c o l my v mia thu chun.2. Phng php o cao lng gic theo nguyn l ca tia ngm nghing. Dng c o l mykinh v, my to n c.3. Phng php o cao thu tnh theo nguyn tc bnh thng nhau ca cht lng. Dng c ol my thu tnh v thng c s dng trong trc a cng trnh.4. Phng php o cao p k da v o s thay i p sut khng kh theo cao. Dng c och yu l p k.5. Phng php o cao radio theo nguyn l phn x ca sng in t. Dng c o l cc myo cao radio c t trn my bay.6. Phng php o cao bng h nh v to n cu GPS. cao cc im trn mt t c xnh thng qua cc s liu thu t v tinh.7. Phng php o cao c hc theo nguyn l hot ng ca con lc t trc tip trn t xc nh cao theo mt tuyn xc nh.

Trong phm vi gii hn ca chng trnh ch trnh b y hai phng php truyn thngl o cao hnh hc v o cao lng gic phc v ch yu cho cng tc o v th nh lp b a hnh t l ln.2.2 Nguyn l v cc phng php o cao hnh hc. 2.2.1 Nguyn l o cao hnh hc.

Gi s cn xc nh chnh cao gia hai im A v B trn mt t (hnh 2.1). Ti im n y t hai ng thu tinh thng ng trn c khc cc vch chia v ghi s. Ni 2 nthu tinh bng ng cao su to th nh mt h thng bnh thng nhau.

Ta nc t t v o ng B v theo

ng dn nc s chy sang ng A. Khikhng nc na th mc nc hai ngthu tinh s bng nhau. ng ni t mtnc ca hai ng A v B s l ng nmngang. c s vch trn ng A c s c a v c s vch trn ng B c s c b th chnh cao gia hai im A v Bl :

a

A

B

b

hAB

nh 2.1

2O


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Ob = R + i + qOJ = R + iJb = S

Do ta vit c:[(R+i) +q]2 = (R+i)2 + S2

Hay:S2 = 2(R + i)q + q2

V i v q rt nh so vi bn knh tri t Rnn i lng i v q2 c th b qua, nn ta c cngthc:

S2 = 2R.q R

Sq

2

2

= (2.8)

Nhn xt: S hiu chnh do nh hng cong qu t lun lun t l thun vi d i giahai im.

Nu ly bn knh l R = 6371 km, ln lt cho khong cch tng s c kt qu th hin bng 2.1Bng 2.1

S (m) 50 100 113 500 1000 5000q (mm) 0,2 0,8 1,0 19,6 78 1692

2.3.2. S hiu chnh do nh hng chit quang.Lp kh quyn bao quanh qu t c t trng khng u nhau v thay i theo chi

cao, c ng gn mt t th t trng c ng ln.Gi s khng c lp kh quyn th t A n B tia ngm i thng theo hng AB. Thc

t do tia ngm n im B khi qua cc lp kh quyn c t trng khc nhau s b nh hnkhc x m to th nh ng cong (hnh 2.6). Cung ca ng cong hng mt lm v phamt t, ng cong n y gi l ng cong chit quang.

Tht vy, mt ta nhn thy im B ti B theo hng tip tuyn AB ca ng conchit quang ti A. Hin tng tia sng b lch gi l hin tng khc x hay chit quang.Gc r gia hng thc AB v hng AB gi l gc chit quang. Tt c cc tia khc x unm trong mt mt phng thng ng.

Do nh hng ca chit quang m chng ta cm thy tt c nhng im ngm cnng cao ln so vi v tr thc ca n. Sai s do nh hng chit quang c ng ln khi khongcch t my ti mia tng dn.

Hnh dng thc ca ng cong chitquang khng th xc nh c th, nhng nukhong cch nhng im o khng xa lm nh trong trng hp o cao hnh hc, th ta c th xemng cong chit quang c dng cung trn c bnknh OA = OB = R1 (hnh 2.7).

Khong chnh BB c coi l sai s chitquang. Nu coi d i cung AB bng tia ngm AB:

AB = AB = S.Xt tam gic OAB, ta c:

OB2 = OA2 + AB2

O

RR

J S b'

biA

B

q

nh 2.5

nh 2.6

r

O

A

B'

B


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(R1 + )2 = R21 + S2 2R1. + 2 = S2

V2 rt nh so vi bn knh R1 nn coi1

2

R = 0, khi :

1

2

2 RS= (2.9)

Kt qu nghin cu chng minh rng i lng R1 rt kh xc nh chnh xc, nph thuc v o khong cch gia hai im A v B, v o tnh cht a hnh gia chng, v o nh v p sut khng kh, v o chiu cao tia ngm so vi mt t. T s gia bn knh tri vi bn knh R1 ca ng cong chit quang gi l h s chit quang K:

KR

RR

RK 1

1==

Thay v o cng thc (3.9) ta c th vit licng thc di dng:

R2SK

2= (2.10)

H s chit quang K khng c nh vthng thay i tu theo thi gian trong ng y. Trs trung bnh ca K l 0,14, do cng thc(2.10) c dng:

R2S

.14,02

= (2.11)

So snh i lng n y vi s hiu chnh do cong tri t tnh theo cng thc (2.8)chng ta thy s hiu chnh do nh hng chit quang trong iu kin khong cch gingnhau s gn bng 1/7 s hiu chnh q ca cong qu t, ngha l :

q.71= (2.12)

Mt iu khc nhau c bn gia hai s hiu chnh trn l : S hiu chnh do congqu t c xc nh r v tr s v du ca n, cn s hiu chnh v chit quang thxc nh trc v iu kin thc t.

* Ch : gim bt nh hng ca sai s chit quang n cc kt qu o chnh caocn lu mt s im sau y:

- Do nh hng ca mt tri nn ng cong chit quang ban ng y v ban m chng li ngc chiu nhau. V vy nn chn thi gian o thch hp sau lc mt tri mc vtrc lc mt tri ln khong 1,5h.

- C ng gn mt t sai s chit quang c ng ln, nhng t cao 1,52m n tng in nh. V vy cn t my sao cho tia ngm cch mt t t 1,5m tr ln.

- Cn phi o hai chiu vi khong thi gian khc nhau trong ng y.- Nu a hnh dc th sai s chit quang i vi mia sau ln hn mia trc, v vy c

gng chn trm my gia cch u hai mia. 2.3.3 S hiu chnh chung cho nh hng ca cong qu t v chit quang.

O

A S B'

R1

B

nh 2.7


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Gi s hng ca trc ngm my thu chun trng vi ng cong song song vi mtng thu, th trn cc mia dng thng ng ti A v B (hnh 2.8) ta c c cc tr s a v b.Chnh cao cn tm gi thit s bng:

h = a - bNhng thc t trc ngm trng vi tip tuyn ca ng cong n y, do trong tr s

o s c sai s do nh hng ca cong qu t qa v qb, v nh th, thay v o tr s a v b tacn tnh tr s a+qa; b+qb. Nhng tr s n y s tip tc b nh hng bi chit quang gy ral m cho tia ngm b cong i v cc tr s ni trn gim ia v b. Do tr s trn mia thct s l :

a' = a + qa - a b' = b + qb - b

Gi qa - a = f a v qb - b = f b l s hiu chnh chung do nh hng ca cong qu tv chit quang i vi cc tr s trn mia t cch my v hai pha trc v sau vi khongcch Sa v Sb.

T cng thc (2.8) v (2.11) ta c:

RS

.43,0R2S

.14,0R2S

f

222

== (2.13)S hiu chnh do nh hng chung ca

cong qu t v chit quang khng kh lun lunt l thun vi bnh phng khong cch t myti mia.

i vi o cao pha trc s hiu chnhn y c th ln n 1mm khi khong cch t myti mia S = 120m.` Cc tr s trn mia c th tnh d d ng saukhi hiu chnh li nh hng ca cong qut v chit quang, ngha l :

a = a - f a b = b - f b V chnh cao gia hai im A v B s l :

hAB = a - b = a - f a - b + f b = (a - b) +(f b - f a)Nu k hiu hiu s cc s hiu chnh lf = f b - f a, th:

hAB = a - b +f (2.14)Khi ta tin h nh o cao hnh hc t gia, my s t khong gia hai mia. Nu Sa = Sb

th c th tnh f a = f b, do f = 0. Nh vy o cao t gia s kh c c nh hng do cong qu t v chit quang. l u im ca o cao hnh hc t gia so vi phng phpo cao pha trc.2.4 Cu to my v mia thu chun.

Da theo cu to, my thy chun thng chia th nh 2 loi: My thu chun c ngknh c nh v my thu chun c ng knh ri. Ring loi my c ng knh ri cng chial m nhiu loi ph thuc v o v tr ca ng thu, tuy nhin loi n y t c s dng trong snxut v chnh xc thp.

My thu chun c ng knh c nh, tt c cc b phn ca n c gn cht v onhau v s lin kt gia chng rt chc chn. My thu chun hin i c b phn i quangtrong, do c th nhn thy r v tr ng thu ngay khi ngm mia. i vi cc my thuchun t ng tia ngm ngang th s lin h gia ng thu v ng knh ht sc cht ch.

B

nh 2.8

a'a

A

qA

h

b'b

qB

B


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Trong nhng loi my n y hnh ca bt thu khng nhng hin ngay trn trngngm m cn c tc dng a trc ngm v v tr nm ngang c tr s trn mia.

Da v o chnh xc, my thu chun c chia l m 3 loi: Loi c chnh xc caodng xc nh cao hng I v hng II vi sai s trung phng mh = 0,5mm/1km thuchun, loi my c chnh xc trung bnh dng o cao hng III v hng IV vi mh =

3mm/1km v loi my thu chun k thut c mh = 10mm/1km dng tng d y caocho cc li khng ch cp thp. 2.4.1 Cu to my thu chun.

Hnh 2.9 m t chung nhng b phnch yu ca my thu chun gm: 1 - knh vt,2 - knh iu quang, 3 - m ng ch thp, 4 -knhmt, 5 - ng thy d i, 6 - my, 7 - c cnbng. Cc trc hnh hc ca my bao gm:Trc ng thy d i LL, trc ngm ng knh CC,trc quay my VV.a. My thu chun cn bng nh vt nghing v ng thu d i.

Tiu biu cho nhm my n y l my thu chun HB - 1 do xng my trc aMaxcva ch to. Xt v c im cu to ca my gm:

- M ng ch ch thp khng c c iu chnh.- Hnh nh hai u bt nc ca ng thu d i trng nhn qua h thng lng knh

bit trn ng thu d i (hnh 2.10). iu chnh bt nc v o gia (hai nhnh parabol trn trng nhn chp nhau) nh

c vt nghing ca my. Gi tr khong chia trn ng thu l 17-23/2mm. ng thu d i c ch to c bit khi thay i nhit th chiu d i bt n

khng thay i.

My thy chun HB - 1 dng o thu chun hng III, IV v thu chun k thut. phng i ng knh l 31x. Trng nhn ca ng knh l 1020, khong cch ngn

nht c th nhn r vt l 3m.b. My thu chun t cn bng trc ngm.

Nguyn l chung ca loi my l da v o tnh t cn bng ca con lc hoc ca b mcht lng di tc dng ca trng lc. trng thi yn tnh dy treo con lc trng vi phngdy di, cn b mt ca cht lng vung gc vi phng .

nh 2.10

nh 2.9


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Mia thu chun dng o thu chun hng III, IV thng l mia hai mt. Mt mt ckhong chia u sn , trng gi l mt . Mt khc c khong chia u sn en, trngl mt en ca mia. i vi mt en, tr s khng ca mia trng vi y mia, cn mt thtr s tng ln hn 4000mm bt u t mt s n o (thng l 4500). Vi cch nh s trnmia nh vy m tr s ly hai mt mia s khc nhau, nhng hiu s tr s ly theo mi mt s

bng nhau, do c th kim tra kt qu o trong qu trnh o c. S chnh s c gia hamt en v ca mt mia c gi l h ng s mia, k hiu l K.K = a- aen (2.17)

Khi o thu chun t gia thng phi dng mt cpmia nht nh, do mi mia c hng s ring nn mi cp miacng tnh c hng s cp mia:

K = K1 - K2 (2.18)V c sai s khc vch trn mia nn thng thng tr

s K khng ng bng 100 m s b sai lch i mt v imm.

Khi tia ngm nm ngang, theo dy ch gia cam ng ch thp chng ta c s trn hai mt ca hai mia sc mt cp s c khc nhau, nhng tr s chnh cao phinh nhau:

h = aen1- ben 2= a 1- b 2 K (2.19)Vi cch l m n y ta lun kim tra c s c trn tng trm my khi o thu chun.

2.5 Kim tra, kim nghim my thu chun. 2.5.1 Kim tra my thu chun.

My thu chun trc khi em ra s dng cn phi kim tra, xem xt mt s iu cnthit sau y:

1. Knh vt, knh mt ca my c b mc khng.2. Cc c iu chnh c l m vic tt khng.

3. Hnh nh bt nc d i c i xng th nh ng parabol ho n chnh khng.4. Cc c c nh v vi ng quay c nh nh ng, chnh xc khng.5. Cc b phn ca gi ba chn c y khng.6. B phn ngm ca my quay quanh trc c nh nh ng khng.7. Bt nc trn ca my khi c cn bng ph hp vi bt nc d i cha.8. M ng ch ch thp v v tr chun cha.

2.5.2 Kim nghim v hiu chnh my thu chun.My thu chun cn tho m n cc iu kin hnh hc c bn sau y:

1. Trc ng thu d i cn vung gc vi trc quay my.t ng thu d i song song vi hai c cn bng bt k, dng hai c cn n y a bt

ng thu v v tr gia ng thu. Sau quay my i 1800 nu bt thu d i vn v tr gia

ca ng thu th iu kin c tho m n. Nu bt thu b lch i th dng hai c cn bmy a bt ng thy v 1/2 khong lch, mt na khong lch cn li dng vt hiu chnhng thu a v nt.

2. Dy ch ngang ca li ch ch thp phi vung gc vi trc quay my. kim nghim iu kin n y ngi ta t my ti mt im, cn bng my th

chnh xc, quay ng knh ngm v o mt im c nh cch my t 5060m dng vt vi ngngang quay ng knh qua tri v qua phi ca im c nh. Nu im lun nm trn dch ngang th iu kin trn t yu cu.

Hnh 2.13


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Trong trng hp ngc li ta tin h nh iu chnh bng cch vn lng cc c h m cali ch thp sau xoay m ng ch thp n khi iu kin t yu cu th vn cht c h m

3. Trc ngm ng knh cn song song vi trc ng thu d i.y l iu kin c bn ca my thu chun nhm m bo xc nh chnh cao ca

mt trm my c chnh xc.Hin nay c nhiu phng php kim nghim v hiu chnh, di y ta xt h

phng php:a. Phng php 1:

Chn hai im A v B trn mt t cch nhau khong 50m ng hai cc. Trc tin tmy ti A dng mia thng ng ti im B (hnh 2.14a).

Cn bng my chnh xc, o chiu cao my ti mm l i1. Nu trc ngm ng knhkhng song song vi trc ng thu d i th s c trn mia theo dy ch gia khng phi l b0 m l b1 cha sai s x, t hnh 2.14a ta c th vit:

hAB = i1 - b0 = i1 - (b1 - x) = i1 - b1 + x (2.20)i ch my cho mia (hnh 2.14b), tin h nh cn bng my chnh xc, o chiu cao

ca my l i2 v c s theo dy ch gia trn mia l b2.V chnh cao gia hai im khng i, nn s c b2 cha tr s sai s x, ngha l :

hAB= b0 - i2 = (b2 - x) - i2 = b2 - x - i2 (2.21)Gii phng trnh (2.20) v (2.21) vi n s x ta c:

222121 iibb x

++= (2.22)

Sai s x khng c vt qu4mm, nu vt qu cn phi hiu chnh. Cch hiuchnh nh sau:

Gi s my vn t ti B, mia dng ti A, ta dng c vi ng nghing a dy chgia v s c ng trn mia l :

b0 = b2 - x (2.23)Lc n y bt ng thu d i s

khng cn v tr gia, ta dng inhhiu chnh nng ln hoc h xung chiu chnh ng thu bt thy d iv o v tr gia. Sau nhng bc l mtrn, ngi ta phi o kim tra li, nux 4mm th ta hiu chnh xong,nu khng li tin h nh iu chnh liti khi t yu cu.

i1

A

BhAB

x

B

b'0

hAB

nh 2.14

i 2b0b1

b2

x

A(a) (b)

nh 2.15

hAB

xxb

a

A

B


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Nu ta t my chnh gia hai im A v B (hnh 2.15) th sai s gc i do trc ngkhng song song vi trc ng thu d i khng nh hng n kt qu xc nh chnh cao giahai im A v B v s c trn mia A v trn mia B cng cha sai s x, ngha l xa = xb. T hnh 2.15 ta c:

hAB = (a - xa) = (b - xb) = a - b (2.24)Nh vy khi o thu chun, t my chnh gia hai im mia, mc d iu kin

bn ca my khng t yu cu, song khng nh hng ti kt qu xc nh chnh cao.b. Phng php 2.

Trn mt t bng phng ng hai cc A v B cch nhau 45m (hnh 2.16) gia haiim A v B ngi ta t my thu chun, k hiu l I1 v trn ng AB ko d i t trmmy I2, cch im B mt khong bng 1/10 AB. Khong cch I1A v I1B khng c vt qu2dm.

v tr t my I1 cn bng my chnh xc ri c s trn hai mia theo dy ch gia la1 v b1, chnh cao gia A v B c tnh theo cng thc:

hAB 1 = a1 - b1 Sau chuyn my ti I2, tin h nh cn bng my v c s trn hai mia theo dy ch

gia c s c a2 v b2. Chnh cao gia hai im A, B k hiu l hAB 2 s tnh theo cng thc:hAB 2 = a2 - b2Nu khng c sai s gc i do trc ngm khng song song vi trc ng thy d i th:

hAB1 = hAB2 Nu c sai s gc i th hAB1 hAB2, hay:

h = hAB1 - hAB2 0Khi gc i c tnh theo cng thc:

h D

i = ."" (2.25)

Ta tin h nh hiu chnh nh sau: Tnh s c ng cho mia t ti A theo cng thc:a2 = a2 +1,1.h (2.26)

Khi tnh c s c ng ngi ta dng c vi ng ng a dy ch ngang ca lch thp v tr s ng a2 trn mia ti A. Khi bt ng thu s nghing v ngi ta dnginh hiu chnh nng ln hoc h xung c hiu chnh ng thu a bt thu d i v o vgia.

i vi cc my thu chun t ng ngi ta dng c hiu chnh m ng ch thp dchchuyn dy ch ngang sao cho s c a2 = a2. Sau khi iu chnh phi kim tra li.

hAB

nh 2.16

B

a1 b1

a2 b2

i"

i"

D = 45mA4,5m

I 1

I2


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Ch : i vi o thu chun hng III, IV ca nh nc, khi tnh c gc i" nh hn20" th khng cn hiu chnh. Trng hp ngc li phi hiu chnh iu kin n y.

Cch ghi s v tnh ton mt ln kim tra n gin th hin bng 2.2Bng 2.2

Trm my Mia Cng thc tnhS c thc

mt en K hiu Tnh tonA a1 1460 h +4B b1 1325 S 45000I1

hAB = a1 - b1 +135 i" 18"A a2 1616B b2 1485I2

hAB = a2 -b2 +131a2 1620

4. Kim nghim s n nh ca trc ngm khi thay i tiu c.

Khi t mia cch my vi nhng khong cch khc nhau s c r nt ta phi thayi tiu c. Nhng do thu knh iu quang b r v chuyn ng khng chnh xc trn trngm, gy nn sai s m ta phi kim nghim.

Cch kim nghim nh sau:Trn mt b i t bng phng ng mt lot cc 1, 2, 3, , 8 nm trn mt cung trn

bn knh bn knh 40 - 50m.u tin ta t my ti J1 v ln lt t mia ti cc im 1, 2, , 8 (hnh 2.17). Tin

h nh cn bng my, dng c vi ng nghing a bt thu d i v o gia. Ngm my n miaiu chnh tiu c cho nh r nt v c s c ln lt trn cc mia dng cc im l a1, a2, , a8. Theo cc s c tnh chnh cao gia cc cc theo cng thc;

h12 = a1 - a2 h23 = a2 - a3 (2.27) h78 = a7 - a8

Sau ngi ta chuyn myn im J2 l im khng cch ucc im trn cung trn. Tin h nhcn bng my chnh xc v ln ltc s trn cc mia c s c l b1,b2, , b8.

R r ng khi c s trn mia ti cc im phi iu chnh li tiu c ca my. Cngtheo s c n y ta tnh c chnh cao gia cc im:

h'12 = b1 - b2

h'23 = b2 - b3 (2.28) h'78 = b7 - b8

Chnh cao tnh c theo cng thc (2.27) v (2.28) khng c vt qu4mm, s sai lch ln v gi tr chng t chnh xc ca thu knh thp khi iu chnh tiu c. Nhmy n y cn phi a v o xng sa cha.

J1 1 J28

7

6

5 4

3

2

nh 2.17


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2.6 o cao hng IV.Li cao hng IV thuc li cao nh nc, c pht trin tng d y t hng III.Li cao hng IV c dng l m c s cao o v a hnh. Trn ng o cao

hng IV c 57 km cn chn mt mc trn mt t. Hai u ca ng o cao hng IV cni n cc im cao ca li cao cp cao hn hoc l cc im nt ca h thng o cao hng IV.

My thu chun dng o cao hng IV phi tho m n cc yu cu sau:- H s phng i ca ng knh phi ln hn 24x tr ln. Gi tr khong chia trn ng

thu d i c th ti 25"/2mm.- Li ch ch thp c 3 ch nm ngang.C th dng my thu chun NI 003, HB, KONI - 007 v cc loi my c chnh xc

tng ng o.Mia dng o cao hng IV l mia hai mt (Mt en v mt ). Trn mi mia c gi

tr khong chia l 1cm. Gia hai mt en v mt ca mt mia c hng s l 4687 ho4787. Mi cp mia dng o cao c hng s mia l 100. Sai s ngu nhin ca tng dm trnmia khng c vt qu1mm.

t mia cn dng cc st hoc gi mia hnh tam gic.ng o cao hng IV ph hp v khp kn ch o theo mt chiu v p dng phng

php o cao t gia. Chiu d i trung bnh t my ti mia l 100m.Nu h s phng i t 30x tr ln th chiu d i c th tng ln 150m. S chnh

khong cch t my n hai mia khng vt qu 5m. 2.6.1 Th t o v tnh ton trn mt trm my.

t my gia hai mia, sau khi cn bng my chnh xc, quay my ngm v mia sau,c ba s c trn mia mt en theo th t: Ch di (1), ch trn (2), v ch gia (3) v ghv o s o (bng 3.3). Quay my n mia trc v c s trn mt en tng t nh mia sauCh di (4), ch trn (5), ch gia (6). nguyn my v quay mt mia trc, c s theody ch gia (7). Quay my v mia sau v c s c mt theo dy ch gia (8).

Nh vy o xong mt trm my, trc khi chuyn my sang trm o tip theo cnphi kim tra tnh ton ngay s o:- Kim tra s c trn mia theo hng s K ca mia sau v mia trc:

(9) = K1 +(3) - (8)(10) = K2 +(6) - (7)

Yu cu s c trn mt en + K so vi s c trn mt ca tng mia khng vtqu 3mm.

- Tnh chnh cao theo s c mt en v mt :(11) = (3) - (6)(12) = (8) - (7)

Yu cu (13) = (11) - (12)100- Tnh chnh cao trung bnh on o gia mt en v mt :

(14) = {(11) + (12)}/2- Tnh khong cch t my ti mia sau v mia trc:

(15) = (1) - (2)(16) = (4) - (5)

- Tnh s chnh khong cch t my n cc mia:(17) = (15) - (16)


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- Tnh s chnh tch lu trn mt on o gm nhiu trm o theo nguyn tc:(18) = (17) + (18) ca trm o trc.

Sau khi tnh ton v kim tra xong trm 1, nu cc s liu o u nh hn tr s chophp th c chuyn my sang trm 2 v cng vic o s lp li nh trn.

Cc kt qu o v tnh ton v tnh ton c ghi v o s o cao hng IV bng (2.3).Bng 2.3 Mu s o cao hng IVo t III - XT1 (L) n III -XT2(T)

Bt u: 7h30Kt thc: 9h30Thi tit: Rm, mtHnh nh: R, n nh

Ng y 20 thng 10 nm 1996Ngi o: Phm Bch TunNgi ghi: Nguyn Th HinNgi kim tra: Nguyn Khc Thi

Ch di

Ch diMia

sau Ch trn

Miasau Ch

trn

S c trn mia

K.cch sau K.cch trc

TTtrmo

S S

Khiumia

Mt en Mt

K+en -

Chnhcao TB(mm)

Ghi ch

2001 (1) 1115 (4) S 1651 (3) 6124 (8) +1(9) K1=44741300 (2) 0414 (5) T 0764 (6) 5339 (7) -1(10) K2=4574701 (15) 701 (16) S-T 887 (11) 785 (12) +2(13) +886

1

0 (17) 0 (18) (14)2657 1933 S 2317 6891 01979 1258 T 1595 6069 0678 675 S-T +722 +822 0 +722

2

+3 +30935 2771 S 0707 5179 +20479 2620 T 2546 7120 0456 451 S-T -1839 -1941 +2 -1840

3

+5 +81255 1900 S 0921 5497 -20590 1231 T 1566 6039 +1665 669 S-T -645 -54 -3 -643,5

4

-4 +4(1)

6848(21)7719(24) 5596(28) 23691(31) +1 -875.5

(2)4348(22)

5223(25) 6471(29) 24567(32) (34) (35)

2500 (23) 2496(26) -875(30) -876(33) -875,5+4 (27) (36)


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f h f h cho php Khi o cao hng IV sai s khp cho php xc nh theo cng thc:

f h cho php= )km(L20 (mm) (2.33)Trong L l chiu d i ng o tnh theo n v km.

Khi f h f h cho phptnh s hiu chnh cho chnh cao gia cc mc. Gi Vhi l s hiuchnh cho on chnh cao th i, th s hiu chnh n y c tnh:

ih

hi l.Lf

V = (2.34)

Trong : li - l chiu d i on o th i gia hai mc caoL - l chiu d i c ng o

- Tnh chnh cao sau bnh sai theo cng thc:

hiii Vhh += (2.35)- T cao im A vi cc chnh cao sau bnh sai ta ln lt tnh cao cho cc mc

trn ng o theo cng thc:

Hi+1 = Hi + hi, i+1 (2.36)Vic tnh ton ng o cao hng IV c th hin bng 2.4Bng 2.4

N0 imo on o li (km)

Chnhcao

(mm)Vhi(mm) ih (mm)

caomc (m) Tnh ton

A 251,7681 2,8 9473 +9 +9482

P 1 261,2502 2,7 7524 +8 +7532

P 2 268,7823 1,6 -2876 +5 -2871

P 3 265,9114 4,7 3771 +14 +3785

B 269,696 11.8 +17892 +36 +17928

ho=+17892

hLT =+17928

f h=-36mm

f hcp=69mm

2.7 o cao k thuto cao k thut dng xc nh cao cc im trn mt t ca mt khu vc phc

v cho cc cng trnh xy dng trong cng nghip, nng nghip, giao thng vn ti. v.v..Phng php o cao k thut cng ging nh o cao hng IV, nhng khi o k thut

khng cn c s theo dy ch bin m ch c s theo dy ch gia ca li ch ngm hiu s cao, cn khong cch t my ti mia c c lng gn bng nhau. Sai s khphiu s cao trng hp im u v im cui ng o l cc im cao ca li khch cp cao hn s c xc nh theo cng thc (2.31). Sai s khp hiu s cao cho phpc xc nh theo cng thc:


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f h cho php= )(30 km L (mm) (2.37)Trong L l chiu d i ng o cao tnh theo km.Trng hp ng o cao khp kn, sai s khp hiu s cao c tnh theo cn

thc (2.32). Sai s khp hiu s cao cho php c xc nh theo cng thc (2.37).

Trng hp ng o cao m im u v im cui khng phi l im cao li khng ch cp cao hn, khi o cao hai ln trn ng o th sai s khp hiu s cao l chnh lch gia kt qu ca hai ln o. Sai s khp hiu s cao cho php c xc theo cng thc (2.37).

Cc s liu v cao cc im cp cao hn, kt qu o v tnh c ghi trong s cao bng 2.5. S o cao k thut c dn ra l m v d bng 2.5 l trng hp o cao dctuyn. Cc im C0, C1, C2, C23 l cc im cc trn ng trc cng trnh cch nhau 100mmt.

Bng 2.5S c trn mia (mm)N0

trmo

N0 imo Sau Trc Trung

gian

Hiu s cao(mm)

Hiu s caotrung

bnh(mm)

caotrmmy

(m)

cao ccim(m)

1 2 3 4 5 6 7 8 91 M1

C0 0951(1)5736(4)4785(5)

1401(2)6183(3)4782(6)

-450(7)-447(8)

-2-448(9)

100.000(15)99.550(16)

2 C0 C1

C1+70

093357194786

176065444784 1248

-827-825

-1-826 100,483

99.55098.72399.235

3C1 C2

02565039

4783

29587744

4786

-2702-2705

-1-2704

98.72396.018

4 C2 C3

C3.PH+20C3TR+20

187866604782

246172474786 0148

2978

-583-587

-2-585

97,896 96.01895.43197.74894.918

27172(10)-36298

9126(12)

36298(11) -9123(13)

-4563(14)

+6-4569(17)

27 C23 x

23727156

4784

08415624

4783

+1531+1532

-1+1532 107.090

28 xM2

222770094782

034851324784

+1879+1877

-2+1878

107.090108.066

23792-184045388

18404 +5388 +2694 +6+2688


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42

Chiu d i hng o:h = +9004 m mf h cho php = 30 4.2 =45mm;L = 2.4 km

h

d C

f

H H =mm

mm

38

8966

+

+

im C1+70 l im cng, ti im n y c s thay i v a hnh, im n y cch C1 khong cch l 70m. im C3PH+20 v C3TR+20 l cc im nm trn mt ct ngang vunggc vi ng trc cng trnh, cch im C3 v pha phi v pha tri 20m. im x l imchuyn c tc dng c tc dng chuyn tip cao t im C23 n im mc M2. S c trnmia ti cc im cng v cc im mt ct ngang c ghi v o ct s c trung gian, cn ccs c cn li c ghi v o ct s c sau v s c trc. Ht trm my s 4 l ht trang s,cui trang s cng c tnh kim tra nh i vi s o cao hng IV.

Ht trm my s 28 kt thc ng o cao, tnh sai s khp hiu s cao theo cngthc (2.37), s hiu chnh hiu s cao theo cng thc (2.34), tnh hiu chnh hiu s c cao ca cc im c s c ghi ct s c trc c tnh theo hiu s cao hiuchnh. cao ca cc im c s c ghi ct s c trung gian c tnh theo cao trmmy. Khi tnh cao trm my, dng s c sau trn mia mt en.2.8 o cao vung.

Phng php o cao vung c p dng vng t t b che khut.Trong khu vc o, dng my kinh v v thc thp xy dng h thng cc ng

thng gc to th nh cc vung c cnh t 10m n 100m. Ti cc nh vung ng cc g mt t. K hiu cc ng thng nm ngang bng cc ch s rp, cc ng thng nbng cc ch ci. Cch k hiu n y thun li cho gi tn cc nh vung. Th d cc nh vung ngo i cng c tn gi l 1a, 6a, 6g, 1g.

Tu thuc v o cnh cc vung ln hay nh b tr trm my cho thch hp. Thngthng c hai cch b tr trm my sau y:

Trng hp cc vung c cnh bng hoc ln hn 100m, ngi ta b tr trm my

gia cc vung. (Hnh 2.19)My thu chun t trm my I, dng cc c cn my a bt thu d i v o v tr giang. S dng hai mia o cao. Mia dng thng ng ti 1a v 1b, c s trn mia c ccs c m v n. Tip theo mia c chuyn n t ti 2a v 2b, c s c trn mia m1, n1.

Chuyn my sang trm my II. c s c trn mia t tai 2a v 2b c cc s cm2, n2, Cc cc c s hiu 2a, 2b, 3a, 3b, ..., 5a, 5b, 6b, ..., 6e, 5g, ... 2g, 1e,.... 1b, 2b c gil cc im lin h.

Cc cc c s hiu 1a, 6g, 6a, 1g gi l cc im trung gian.Kim tra kt qu o cao bng ng thc:m1 - n1 = m2 - n2

Hay: m1 + n2 = m2 + n1 (2.39)Ngha l tng cc s c cho ca cc im lin h phi bng nhau.Do nhng sai s khng th trnh khi trong qu trnh o nn chnh lch cho php

ca cc tng s n y l5mm.Ti cc trm my u phi kim tra trc khi chuyn sang trm my khc. Nu trm

my n o kim tra thy sai lch ln hn5mm th cn o li ngay trm my .Cc s c trn mia c ghi v o s o cao hoc tt nht l ghi trc tip ln s ti

cc cc.


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Kt thc qu trnh o titrm XIX cn bnh sai kt qu ocao.

Bnh sai theo hai ng ocao khp kn.

ng th nht l 1a 6a 6g1g 1a v ng th hai 2b 5b 5e 2e2b. Tnh sai s khp hiu s caotheo cng thc (2.31), sai s khphiu s cao theo cng thc(2.37), s hiu chnh theo cng thc(2.34).

Sau khi bnh sai hiu s cao, tnh cao nh vung theocng thc (2.36).

Cc vung pha trong XVII, XVIII, XIX c bnh sai nh i vi ng o cao cim u v im cui l im khng ch cp cao hn.

Trong khu vc nu c mc cao gn th ngi ta o dn chuynv mt nh trong li vung. Nutrong trng hp khng c im cao bit trc, ngi ta cho mt imcc 1a mt cao quy c ( caogi nh) tnh cao cc cc khctrong li.

Trng hp cc vung c cnh nh hn 100m, gi s cc vung c cnh bng 40m,cn b tr trm o sao cho ti mt trm my o c nhiu nh vung.(Hnh 2.20)

Trc tin dng my kinh v v thc thp b tr cc vung trn mt t, ti cc nh vung ng cc nh du.

Trong khu vc o t bn trm my I, II, III, IV s c ht cc im cc. Cc im4b - 4c, 5 - 6, 4e - 4g, 2 - 3 l cc cnh lin h. S c trn mia c ghi trc tip trn s . S c trn cc im lin h c ly l m trn n mm, cn cc s c trn cc imtrung gian c ly l m trn n cm. Ti cc im lin h, s c trn mia c kim tra theocng thc (2.39).

X l kt qu o cao c thc hin nh sau: Theo s thc a, lp ng o cai vi cc im lin h 4b, 6, 4g, 2 (hnh 2.21).

Tnh hiu s cao i vi ng o n y:h1 = 1275 - 1154 = +121mm = 0,121mh2 = 1506 - 2489 = - 983mm = - 0,983mh3 = 1048 - 1477 = - 429mm = - 0,429mh4 = 2067 - 0782 = +1285mm = 1,285m.

Cc s liu tnh bnh sai c ghi v o bng 2.6.

nh 2.19

n

2 n21 n1


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Bng 2.6Hiu s cao

N0 imTnh (m) hiu chnh (m)

cao (m)

4b +1 72,000

+0,121 +0,1226 +2 72,122-0,983 -0,981

4g +1 71,141-0,429 -0,428

2 +2 70,713+1,285 +1,287+1,406 +1,409-1,412 -1,409

f = -0,006m 0,000m

Chiu d i cnh vung bng 40m, nn to n b ng o cao ni lin cc im lin h4b, 6, 4g, 2 c chiu d i l 0,46km. Tnh sai s khp hiu s cao cho php:

f h cho php= 46,050 mm = 34mmTnh s hiu chnh cho cc hiu s cao v tnh hiu s cao c hiu ch

Ly cao quy c ca im cc 4b l 72,000m, ln lt tnh cao ca cc im lin h 64g, 2 theo cng thc (2.36).

nh 2.20


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Sau khi c hiu s cao cc im lin h, tin h nh tnh cao trm my (vit tl CTM trn hnh 2.21).

H11 = 72,000m + 0,782 m = 72,782m.H21 = 70,713m + 2,067 m = 72,780m.

m781,722m780,72m782,72

Htb1 =

+=

cao cc trm my khc tnh tng t. Dng cao cc trm my tnh caocho tt c cc cc. cao cc cc ghi trc tip ln s .

2.9 Th nh lp bnh theo kt qu o cao vung.Trn giy v c cc nh vung cao cc nh n y c ghi l m trn n ph

trm mt.Theo cao cc nh vung v ng ng mc. Dng phng php ni suy

xc nh v tr cc ng ng mc.C hai phng php ni suy l phng php gii tch v phng php th:

2.9.1. Phng php gii tch.Gi s trn mt vung c cao cc nh nh trn hnh 2.22 cn v cc ng ng

mc c khong cao u h = 0,5m; chiu d i cc cnh vung S = 40m.

S

A1N 1

M 1

B A

H A-H Bh 2 h1 S 1 S 2 S 3

N M

nh 2.23

71,5

72,0

72,5

71,18

71,85

71,72

71,80 S' 3

S' 2

S' 1

S'' 2

S'' 1 S''' 2 '''

S 3 S 2 S 1

nh 2.22

CTM

CTM CTM

CTM

I

II III

IV

72,781

72,276 73,629

71,14124691048

70,71314772067

72,000 12750782

115472,1221506

72,190

nh 2.21


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Trn cnh AB s c cc ng ng mc c cao 72,0m v 72,5m. Cn xc nh vtr ca cc ng ng mc n y. Mun th chng ta dng li v tr tng i ca cc im AB thc a nh trn hnh 2.23. Trn hnh n y im M l v tr ng ng mc c 72,0m, im N l v tr ng ng mc c cao 72,5m. Cc hiu s cao tng ng:

h1 = HM - HB h2 = HN - HB

T cc tam gic ng dng AA1B, MM1B v NN1B c:

BA

11

HHh

SS

= ; BA122

HHhh

SS

= ; ( )

BA

2BA3

HHhHH

SS

=

Rt ra:

S.HH

hS

BA

11 = ; S.HH

hhS

BA

122

= ; S.HH

h)HH(S

BA

2BA3

=

C th c:

m6,740.05,120,0

S1 ==

m0,1940.05,150,0

S2 ==

m4,1340.05,135,0

S1 ==

Kim tra S = 40,0mSau khi tnh c chiu d i cc on thng, theo t l ca bnh , t B dc theo hng

n im A t on S1, tip theo t on S2, S3.i vi cc cnh vung khc

cng l m tng t. Sau khi xc nh cv tr cc ng ng mc, ni cc im c

cng cao li, s c ng ng mcbiu din a hnh khu vc o. 2.9.2. Phng php th.

Dng giy bng can k cc ngnm ngang v thng ng to th nh hthng cc vung. Ghi cao cho ccng nm ngang.

Ly li th d xc nh v tr ngng mc 72,0 v 72,5 trn cnh vungAB. t t giy bng can cho cc ngnm ngang song song vi AB (Hnh 2.24).

Dng cao cc im A, B ln t giy bng can c cc im A1, B1.ng ni A1B1 trn giy bng can gp ccng nm ngang c cao 72,0m v72,5m. Dng cc giao im n y xungng AB, s xc nh c v tr ca ccng ng mc c cao 72,0m v72,5m trn ng AB.

89

10

2 3 4

6 5

87

90

72

73

B A71,8072,072,572,85

B

A

nh 2.24

nh 2.2571,5

72,0

72,5

73,0 A

B


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Hoc trn t giy bng can k ccng thng song song, ghi cao cho ccng thng n y. t t giy bng can lnbnh (hnh 2.25), xoay t giy bng can im A v B nm ng cao.

t cnh thc thng ni lin haiim A, B. Dng kim chm cc giao imca ng AB vi cc ng thng songsong c cao 72,0m v 72,5m ln trnbnh , s c v tr ca cc ng ngmc c cao 72,5m v 72m.

Xin dn ra y th d v kt quo cao vung v biu din a hnh bngng ng mc trn bnh (hnh 2.26).

Trn hnh 2.26 bnh c t l 1:1000, ng ng mc c khong cao u h=0,5m.2.10. o cao lng gic. 2.10.1 Nguyn l o cao lng gic.

Khi tin h nh tng d y im khng ch cao phc v cng tc o v chi tit nhngvng i ni, c dc ln m p dng phng php o cao hnh hc xc nh cao, skhng kinh t, tc chm. V th ngi ta dng phng php o cao lng gic xc nh cao cc im s tin li hn.

o cao lng gic ngi ta dng cc my trc a c b n ng xc nh khongcch v gc ng tnh chnh cao gia cc im, hay ni cch khc l o chnh cao gia ccim theo nguyn tc tia ngm nghing.

Gi s cn xc nh chnh cao gia hai im A, B ngo i thc a. Ti A ngi tat my kinh v, ti B ngi ta dng mia thng ng (hnh 2.27).

o chiu cao my AJ = i, hng ng knh n im M trn mia, khong cch t B nM k hiu l l. Khong cch nm ngang t A n B l S. Gi s c mt tia ngm ngang hp

vi tia ngm nghing mt gc nghing v v ct mia dng im B l N. on MN s bngS.tgv.T hnh 2.27 ta c:

hAB + l = i + S.tgvHay:

hAB = S.tgv + i - l (2.40)Khi khong cch gia hai

im AB ln, th kt qu o chnh caophi hiu chnh do cong tri t vchit quang khng kh.

Tht vy do nh hng ca chit quang, tia ngm t ng knh ti mc tiu khng phil ng thng JN m theo ng cong chit quang JN'. ng cong chit quang quay b lmv pha tm tri t (hnh 2.28). Khong cch NN' = r gi l s hiu chnh do chit quang tngm.

Gi s c mt mt thu chun i qua J v ct mia thng ng ti P. ng nm ngangtip tuyn vi mt thu chun qua J v ct mia thng ng ti B, on MP=q gi l s hichnh do cong tri t.

T hnh 2.28 ta c:i + q + MN = hAB + l + r

A

B

N

hAB

A nh 2.27

i

M

l J v S

nh 2.26


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49

Mu s o ng chuyn o cao lng gicTrm my B

Loi my: Theo 010A S my:T im: A n im C

Ng y o: 26 11 1997Bt u: 10h30 Kt thc: 11h30

Thi tit: Nng nhNgi o: o n Thanh Hng

Ngi ghi: V Khc LunNgi kim tra: Trn KhiBng 2.7

Mia sau A Mia trc CS liu trm o

Ln o 1 Ln o 2 Ln o 1 Ln o 2Ch trn 2382 2581 2319 2520

Khong cchCh di 0821 1020 0883 1082

156,1 156,1 143,6 143,6Khong cch ngang S=K.l.cos2V

156,0 156,0 143,5 143,5S c ln 1 900429 900384 88032'4 88027'2S c ln 2 900430 900385 88032'5 88027'2

B ntri Trung bnh 900430 900384 88032'4 88027'2

S c ln 1 269018'0 269022'2 2710284 2710332S c ln 2 269018'0 269022'3 2710284 2710333

o gcng V B n

phi Trung bnh 269018'0 269022'2 2710284 2710332

Sai s M0 +0'5 +0'3 +0'2 +0'2Gc ng V -0042'5 -0043'1 +1028'2 +1033'0S.tgV (m) -1,929 -1,728 +3,638 +3,888 cao mc tiu l (m) 1,60 1,80 1,60 1,80Chiu cao my i (m) 1,50 1,50 1,50 1,50Chnh cao h = StgV+i-l (m) -2,03 -2,03 +3,58 +3,59Chnh cao trung bnh (m) -2,03 +3,58

2.10.3 chnh xc ca phng php o cao lng gic .T cng thc (2.41) ta thy chnh xc nh chnh cao hAB ph thuc v o chnh

xc cc i lng S, v, i, l v f. chnh cao o bng phng php o cao lng gic thngc xc nh ti cm.

Sai s o chiu cao my v chiu cao mc tiu khng c vt qu 1cm, nn c thb qua. S hiu chnh f do cong tri t v chit quang tia ngm khong cch nh hn

300m cng khng vt qu 1cm, do c th b qua.Nh vy chnh xc o chnh cao ch cn ph thuc v o chnh xc o gcnghing v v chiu d i nm ngang S.

S nh hng ca cc sai s nu trn n kt qu o chnh cao c th phn tch t cng thc:

hAB = S.tgvLy lg hai v, tm o h m v chuyn v dng sai s trung phng, ta c:


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50

2V

22S

22h m.V

hm.

Sh

mAB

+

=

Vcos

S.

"

mVtg.mm 4

22V22

Sh AB

+=

Vcos

S.

"m

Sm

Sm4

22V

2S2

h AB

+

= (2.45)

Gi s gc ng v khng vt qu 300; S = 100m, mv khng ln hn 1'0 v sai s o

khong cch khng vt qu400

1 , p dng cng thc (2.45) ta tnh c mh= 2cm.

Trn c s trong quy phm quy nh hn saih = 2mh=4cm/100m cho trng hpo hai chiu (o i - o v).

Sai s khp hiu s cao ca ng chuyn cao lng gic khng c vt quf h cho php:

f h cho php= hm).n.S.04,0( Trong : S - Chiu d i trung bnh ca cnh

n - S cnh trong ng chuyn


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Chng 3

o v bnh 3.1. Khi nim v o v bnh

Nh chng ta bit, bnh l hnh chiu thu nh ca mt phn nh b mt tri ln giy theo mt t l nht nh (khng tnh n nh hng cong ca b mt tri t). Nvy biu din mt khu vc nh b mt tri t ln giy trong qu trnh o c ngi ta b mt tri t (ti khu vc o v) l phng. Cc i lng o (Gc, chiu d i...) c tih nh trn mt phng v khi biu din chng cng tin h nh trn mt phng, v vy chkhng b bin dng. y l im khc nhau c bn gia bnh v bn . Cng vic tin th nh lp bnh mt khu vc l : Xy dng li khng ch o v

Xy dng li khng ch o v:

Li khng ch o v l tp hp cc im trn mt t c lin h vi nhau theo mtquy lut ton hc nht nh. Li khng ch o v bao gm li mt bng v li cao.Thng thng cc im khng ch mt bng u c cao. Li khng ch mt bng v cao trong o v bnh c ta gc l ta gi nh. Tuy nhin n vn tun th theo quynh ca li o v khi o v bn .

Li mt bng:

Ty theo a hnh a vt khu vc o v ngi ta b tr li mt bng theo dng tamgic nh. Tc l cc im ca li lin h vi nhau theo dng tam gic cnh ngn. Trong o cc gc trong tam gic v chiu d i cnh khi tnh hoc l o chiu d i tt c cc cnhTrong thc t hin nay dng n y t c s dng.

Thc t ngi ta thng dng dng ng chuyn. C 2 dng ng chuyn c bn,

l ng chuyn kinh v khp kn v ng chuyn ph hp. (hnh 3.1)

nh 3.1a) ng chuyn khp kn a) ng chuyn ph hp


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T 2 dng ng chuyn n y ngi taphi tin h nh nhiu dng khc nhau nh ng chuyn kinh v c mt im nt, 2 imnt...

Vic la chn dng n o ty thuc v okhu vc o. Trn khu vc o cng c th kthp c 2 dng ng chuyn kinh v khp knv ng chuyn ph hp. (hnh 3.2).

Trong li o v thng o cc gc vchiu d i cc cnh. Gc phng v cnh u(nu l ng chuyn kinh v khp kn ).Gc phng v cnh u v cui(nu l ng chuyn kinh v ph hp). Ta im (x1y1) l gi nh. Trong o v bnh li n y thng l li c lp. Li o v gm 2 cp l : ng chuyn kinh v cp 1 v ng chuyn kinh v cp 2, i khi c th m rngbng phng php giao hi hoc phng thm im ph. y l li trc tip o v im ctit (im a hnh, a vt).

Li cao: Khi khu vc o v khng c im cao Nh nc ta c th gi nh cao ban u ( cao gi nh). Vic la chn cao ban u sao cho tng i ph hp vi cao khu vc (c th tham kho trn bn a hnh hin c). Vic tnh chuyn cao n im ca li c dng thy chun tia ngm ngang (s dng my kinh v) hoc thy chunlng gic. Thng cc im ca li o v u c cao ( cao gi nh).

Trong o v bn li o v c pht trin t li a chnh cp 2 tr ln. Ta , cao cc im ca li c thng nht theo h ta v cao nh nc. Nhng trong ov bnh th ta v cao c tnh theo ta , cao gi nh. V vy, li o v b

thng l li c lp (ta v cao n y ch c ngha trong khu vc o v).3.2. My kinh v quang hc

tin h nh o gc, chiu d i trong li o v ta c th dng cc lai dng c okhc nhau: My kinh v quang hc, my to n c in t (Electric total station) ... v cc dngc o c khc. 3.2.1. Cu to my kinh v quang hc

My kinh v quang hc l mt dng c o c th o c tt c cc i lng o trongtrc a (gc ngang, gc nghing, khong cch v chnh cao) v vy ngi ta cn gi n lmy to n c.

C rt nhiu loi my kinh v khc nhau, v c bn chng c hnh dng chung tngi ging nhau. Tuy nhin, v chi tit chng c th khc nhau i cht. Khi s dng mt loimy n o ta cn nghin cu k v n (nghin cu v cu to, chnh xc ca my v cch cs).1- Cu to chung : Hnh 3.3 l cu to ca my 2T5K. Theo th t ghi trn hnh ta c: c cnmy (1); my (2); hp b n ngang (3); ng thy d i (4); ng thy trn (5); ng knhngm (6); ng knh c s (7); hp b n ng (8); thc ngm s b (9); quai sch (10);nm xa b n (11); c h m v vi ng dc (12); c h m v vi ng ngang (13); di tm

nh 3.2


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quang hc (14); c t v tr hng (15); gng phn chiu (16).Ngo i ra my cn c t trn gi 3 chn gi l chn my. Chn my c th thay

cao ty theo chiu cao ca ngi o.. Gia my v chn my c lin kt vi nhau bng 1c ni. (hnh 3.3).

Hnh 3.32- Cc b phn chnh ca my kinh v quang hc

ng ngm: ng ngm trong my trc a l loi knh vin vng cho php ngm xa v chnh xc. Trn hnh 3.4 l s cu to ca ng knh.

Hnh 3.4Gm 3 ng thp hnh tr ng tm lng v o nhau. Mt u ca ng tr ngo i (1) c

gn vi knh vt (2) cn u kia l vng iu quang (3) c r nh xon khp vi r nh xon cng tr trong (4) c gn knh iu quang (5) v m ng ch ch thp (6). Khi xoay vng iquang (3) th knh iu quang (5) di chuyn dc theo trc CC ca ng knh. ng tr th 3

gn vi knh mt (7). Trc CC ca ng knh phi i qua tm ca knh vt (2), tm ca kiu quang(5) v tm ca knh mt (7) cng vi giao im ca m ng ch thp (6). Trc CC trc quang hc ca my. Trong trng hp l tng trc n y l mt ng thng. Thc t khidi chuyn knh iu quang (5) trc quang hc b thay i iu l m nh hng trc tn s c, nht l khi o thy chun ngi ta phi kim nghim iu kin s thay i ngm khi thay i tiu c.

M ng ch ch thp: Cu to ca m ng ch ch thp l mt tm knh mng t trong


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174055'2 hay 1740 55' 12''.B n ng: Tng t b n ngang, b n ng cng l m bng thy tinh hoc pha

l trong sut ch khc l ngo i kiu khc vch lin tc t 0 3600 cn c cc kiu khc ixng 00- 00, 00 1800, 900 2700. Ty theo cch khc vch m ta c th c c tr s gc

ng. V d trong my Theo 20A ngi ta khc kiu vch i xng 900

2700

. Ngha l khi ng knh v tr nm ngang (b n thun) s c c trn b n ng l 90000'00''. Nuo knh ngm li im c gi tr c c l 270000'00'' (ng knh v tr nm ngang). Trnhnh 3.6 gi tr c c trn b n ng l 87004'00'' v c gc V = 90000'00'' - 87004'00''= 2056'00''.

ng thy : ng thy l b phn quan trng trong my kinh v quang hc. Da v o vtr bt nc trong ng thy ngi ta xc nh c b n nm ca my v tr nm ngancha. Da v o c tnh ca cht lng l bt kh ca n lun lun v tr cao nht ngi ta cto ng thy. cn bng my nhm a b n ngang v v tr nm ngang ta phi dng 3 ccn (cn my) khi mi v tr ca b n bt thy u gia chng t b n ngang tr ngang. C 2 loi ng thy l ng thy trn v ng thy d i.

ng thy d i: Hnh 3.7 m t cu to ca ng thy d i. N l mt ng thy tinh hnhtr cong bt kn (1) bn trong cha te hoc cn cha mt t khong khng kh gi l bt nc(2). ng thy tinh c gn c nh trong hp kim lai hnh tr (3). Mt cong ca ng thytinh d i l mt cung bn knh R (hnh 3.8) im gia 0 ca cung l im chun nhnkhng nh du m ngi ta khc i xng qua n.

Hnh 3.7 Hnh 3.8

gi n cch gia cc vch u nhau v thng c gi tr l = 2mm. Hai u ng thy tinh c thnng hoc h xung nh c iu chnh (4) khi cn iu chnh bt nc ta dng que hiu ch(5) vn c n y. Nu ng thy tinh nghing i mt gc n o th bt nc cng tri i sovi im chun 0 mt cung d tng ng (hnh 3.8). Nu cung d ng bng tr s gi n cch lgia hai vch chia (d=l) th gc tm tng ng l'' v l v '' rt b nn ta c:'' = (l/R)*''. Gc '' c gi l nhy ca ng thy, tc l n biu th kh nng di chuyn nhanh hay


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chm ca bt nc chim v tr cao nht. Nu ng thy c bn knh R c ng ln th gi tr''c ng nh v kh nng cn bng c ng chnh xc. nhy'' cn c gi l gi tr khongchia vch trn ng thy tinh theo n v gc v c biu th bng t s '' /l.V d: l = 2mm, R = 21m,

''

=206265''

ta c :''

= 20''/2mm. ng thy trn : V ca ng thy trn l hnh tr

ng bng kim loi (1) bn trong ng thy tinh (2) c mttrn l hnh bn cu vi bn knh khong 0,5m (hnh 3.9) trnmt chm cu khc hai vng trn ng tm (3) v khi btnc nm tm chm cu th my c cn bng tng i.

ng thy trn c c iu chnh (5). ng thy trn c nhy '' khong 3' n 5'. V vy n ch dng cn bngmy s b.

3.2.2. o gc bng v gc ng bng my kinh v quang hc1- Bn cht ca gc bng v gc ng trong trc a

Hnh 3.10Gi s c 3 im ABC ngo i thc a (ba nh ni). Qua A ta dng mt mt phn

ngang P. Qua AB ta dng mt mt phng thng ng V1 (V1 vung gc vi P). Qua AC tadng mt mt phng thng ng V

2(V

2vung gc vi P). Mt phng V

1ct P theo ng Ab,

mt phng V2 ct P theo ng Ac. Nh vy gc bng trong trc a l gc bAc thuc mphng P. N cng chnh l gc hnh chiu ca gc ngo i thc a (ABC) ln mt phngang, k hiu l. Vy bn cht ca gc bng trong trc a l gc hnh chiu ca gc ngo ithc a ln trn mt phng ngang (hnh 3.10).

Gc V1, V2 l gc ng trong trc a. N chnh l gc hp bi tia ngm AB hoc ACvi mt phng ngang P.

nh 3.9


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Vy gc ng l gc hp bi tia ngm vi mt phng ngang, k hiu l V.Nu tia ngm pha trn mt phng ngang th V mang du +, nu tia ngm pha d

mt phng ngang th V mang du - . Nu tia ngm trng vi mt phng ngang th V = 0.Da trn c s ngi ta thit k my kinh v c b n nm nm ngang o gc

bng, b n ng vung gc vi b n nm o gc ng V.2- o gc bng

C nhiu phng php o gc bng: phng php o gc n, phng php o lp,phng php o to n vng, phng php o t hp. Vic s dng phng php o n o tythuc v o s hng o, chnh xc ca gc o. Khi trm c 2 hng ngi ta thng dngphng php o gc n hoc phng php o lp. Khi trm o c t 3 hng tr ln dngphng php o to n vng. Khi trm o c qu nhiu hng dng phng php o t hp.3- o gc ng

Gc ng l gc hp bi tia ngm vi mt phng

ngang, k hiu l V.Nu tia ngm nm pha trn mt phng ngang V mangdu +, nu tia ngm nm pha di mt phng ngang Vmang du 4- B n v cch c s

- Cu to b n ngang:

B n ngang c cu to l mt a trn bng thy tinh hoc pha l c bit trong sut, trn c 3600 (hoc 400 grat) , gia mi khong c th chia th nh khong pht.Gi tr mt khong pht ph thuc v o s lng phn khong trong mt .

V d trn hnh 3.12. Mt c chia th nh 6 khong ln, 1 khong ln l 60'/6 = 10'(nh s t 0,1,....,6). Trong mt khong ln chia th nh 10 khong nh, mi khong nh l 1'v khi c s ta c lng n 1/10 khong nh tng ng vi 0,1' ( tc l 6'') v ta ni rng chnh xc ca my t'' = 6''.

- Cch c s:

Gi tr c c trn my l gi tr ca hng o. Datrn tr s trn b n ngang ta c c gi tr ca hngl 1740 55' 2. Cc my nh Theo 20A c cu to b n nh trn.

Mi loi my c khc vch khc nhau. Ty thuc v ocu to ca n m ta c cch c s cho ph hp. Mun vytrc khi c s ta cn nghin cu cch khc vch trn b n, chnh xc ca my c c cch c s ng. nh 3.12

nh 3.11

86 87


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5- B n ng v cch c s

- Cu to b n ng:

Tng t nh b n ngang, b n ng cng bng thy tinh hoc pha l trong sutch khc l cch khc vch theo kiu khc i xng 00- 00, 00 1800, 900 2700. (hnh3.12).Trn hnh ta hiu l 0 l 0'... 6 l 60' (tng ng vi 10). Nh vy t 0 n 1 ta hiu l 10' chial m 10 khong, mi khong l 1', ta c c 1/10 khong ng vi 0,1' tc l 6''.

- Cch c s:

Nhn v o b n ng ta c c: 870 04' 5 tc l 870 04' 30''. y l gi tr hng v1 im M n o . Gi tr hng nm ngang l 900 2700. Ta hiu l : v tr b n tri (b n thun) hng nm ngang l 900 = 89059' 60''. Khi gc nghing V l : V = 89059' 60'' - 870

04'30'' = 2055'30''.3.3. Cc phng php o gc ngang

Ty thuc v o s hng trn mt trm o ngi ta c th dng cc phng php ogc ngang:

- Phng php o gc n, phng php o lp: dng cho trm o c 2 hng.- Phng php o to n vng: phng php o t hp dng cho trm o c t 3 hng

tr ln. Khi s hng trn mt trm qu nhiu ngi ta dng phng php o t hp.1- Phng php o gc n:

Gi s ta cn o gc AOB. Ti O ta t my, cn my v nh tm my. Ti A,B tadng tiu. Thng thng vic nh tm v cn my c tin h nh ng thi. Dng 3 c cca my iu chnh cho nh ca im O v o tm vng trn nh ca ng di tm quhc. My c cn bng nghi l iu kin: Trc ng thy d i phi vung gc vi trc quay

ca my c thc hin. Cn vic nh tm my tc l l m cho tm my trng vi tmmc. Tuy nhin trong thc t 2 tm khng th trng nhau, nhng n phi nh hn hocbng sai s cho php. Sai s cho php k hiu l e, e c tnh theo cng thc e = (t''.S)/2''.Trong : S l khong cch t my n tiu ngm.

'' = 206265''.t'' l chnh xc ca my o.

V d: t'' = 30'', S =100m, ta c :

mm3,7"206265.2mm100000".30

e ==

iu c ngha l : Tm my v tm mc trongtrng hp n y khng c lch nhau qu 7,3mm.

Sau khi nh tm, cn my ta tin h nh o gc 2v tr b n :

V tr b n tri (b n thun) (hnh 3.13)V tr b n phi (b n ngc). nh 3.13


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khc phc sai s do vch khc khng u trn b n ta phi o gc nhiu ln mi ln o thay i tr s hng m u: 1800 / n (n l s ln o). V d n = 3 ln o. Ln uta t b n l 00, ln th 2 l 600, ln th 3 l 1200. Trnh t o gc n ca mt ln o ctin h nh nh sau:

- Na ln o thun knh:M c h m ngang v c h m dc a ng knh ngm tiu A (hnh 3.13). Kha c

h m dng c vi ng ngang v dc iu chnh cho nh ca tiu ngm trng vi ch ng cady ch ch thp (Thng hng m u t v tr 00) c s: 00 00' 06'' (ghi v o ct 3 bng3.1). Nu dng my kinh v in t ta n phm: OSET trn m n hnh tr s hng OA l 00 00'00''. M c h m ngang v dc quay my theo chiu kim ng h ngm B. Tng t, ngmchnh xc B v c c gi tr hng B l : 162048' 12'' (ghi v o ct 3 dng 2). Nh vy gcAOB o c mt na vng o thun.

- Na ln o o knh:

o ng knh, quay my ngc chiu kim ng h ngm li im B c s c: 3200 48' 00''. Tip tc quay my ngc chiu kim ng h ngm A c s c : 1800 00' 12'' (ghiv o ct 4). Nh vy kt thc mt vng o. Ln o th 2 c tin h nh nh ln o 1 nhnhng ban u t b n 900.

Trong khi o ngi ta phi tnh ngay tr s ngm chun 2C = L (R1800).Trong L l tr s c triR l tr s c phi

V d: Dng 1 ct 5 tr s 2C = - 6'' (xem bng 3.1) (Sai s n y s c trnh b y trong phnkim nghim my). Sau khi tnh c gi tr trung bnh ca hng ta ly gi tr hng sau tr i gi tr hng trc c gi tr ca gc o trong mt vng o.V d: Vng 1: Gc o l 1620 47' 57''.

Vng 2: Gc o l 1620 47' 45''.

Tnh chnh lch gia 2 na vng o : = 1620 47' 57'' - 1620 47' 45'' = 12''.Quy nh: cho php= 2t"

t" l chnh xc ca my

Mu s o gc bng phng php o gc n:Ng y o : 18 2 2003 My Theo 20 ABt u: 8h 30 pht Ngi o: Nguyn Vn TunKt thc: 9h 5 pht Ngi ghi: L Tun AnhThi tit : tt Ngi kim tra: Trn QuytS hiu trm o : 0


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Bng 3.1.

S c b n ngangLno

imngm L R

2C Tr s hngtrung bnh

Gc o Gc o trungbnh

1 2 3 4 5 6 7 81

AB

00 00' 06''

162048'12''180000'12''

342048'00'' -6

+1200 00' 09'' 162048' 06''

162047' 57''

2AB

90000'12''

252047'54''270000'12''72048'000

0-6

90000'12''

252047'57'' 162047' 45''

162047' 51''

y t = 6'', 2t = 12''.Nh vy, gc o t yu cu. Ly gi tr trung bnh c: = 162047' 51''.

2- Phng php o to n vng:

Khi o gc trong li tam gic ngi ta thng dng phng php o to n vng hockhi o ng chuyn m ti trm o s hng o ln hn hoc bng 3 hng.

Gi s ti trm o 0 c 4 hng o l ABCD. Trc khi o ta phi tnh ton v tr b n trong mi vng o theo cng thc =1800 / 'nTrong n l s vng o. V d n = 3 th v tr b n l 00, 600,1200. Mt vng o y gm 2 na vng o.

Na vng o th nht:

t my ti 0 (hnh 3.13*), cn my, nh tm my, t b n 00

ngm A, B, C, D,A theo chiu kim ng h, c s b n ngang (Bng 3.2).Na vng th 2:

o knh, ngm A, D, C, B, A quay my ngc chiu kim ng h, c s b n ngang, kt qu ghi v o bng 3.2. (khi o ch c iu quang 1 ln).Theo kt qu o ta tnh s ci chnh cho cc hng o nh sau:

T = T T = 0' 58'' - 01'05'' = -7'' P = P P = 1' 07'' - 1' 11'' = -4'' TB = (T + P )/2 = (- 7 4)/2 = - 5''5.

Trong : T l s c b n tri ti A vng ngm thun.T l s c b n tri ti A vng ngm ngc.P l s c b n phi ti A vng ngm thun.P l s c b n phi ti A vng ngm ngc.

S hiu chnh cho hng c tnh theo cng thc:

nh 3.13*


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k = )1k (m

TB

Trong : k l th t hng om l s hng o (m=4).

Mu s o gc bng phng php to n vng:

Trm o : 0 Loi My: Theo 20 ANg y o: 20 11 2003 S my: 40102Bt u: 7h 30 pht Ngi o: Lu Vn HiKt thc: 8h 40 pht Ngi ghi: Tun AnhThi tit : tt Ngi kim tra: Ho ng V

Bng 3.2.S c I , IITn

hngVngo

S ctrn b n a1 a2

a1+a2 2

T+P2

Gi tr hng0 ' ''

1 2 3 4 5 6 7 8T 00 01' 07'' 01' 03'' 01' 05'' 01' 08'' A P 1800 01' 12'' 01' 10'' 01' 11'' 0 0

0 00' 00''

T 820 24' 30'' 24' 28'' 24' 29'' 24' 38'' B P 2620 24' 44'' 24' 48'' 24' 46'' +1 820 23' 31''

T 1250 33' 02'' 33' 00'' 33' 01'' 33' 08'' C P 3050 33' 13'' 33' 15'' 33' 14'' +3 1250 32' 03''

T 2750 30' 31'' 30' 35'' 30' 33'' 30' 42'' D P 950 30' 50'' 30' 52'' 30' 51'' +4 2750 29' 38''

T 00

00' 59''

00' 57''

00' 58''

01' 02''

A P 1800 01' 09'' 01' 05'' 01' 07''

Sau khi tnh gi tr quy 0 ca tng hng o trong mt vng o ta tnh c gi trtrung bnh ca tng hng o trong mt vng o (Bng 3.3).

Gi s ti 0 ta o 3 vng, kt qu mi vng o c quy 0. T ta tnh c gi trtrung bnh hng t 3 vng o (Bng 3.3)

Bng 3.3B C DS

vng

o

V tr b n 620 12' V 1070 48' V 2370 37' V

| V |

1 2 3 4 5 6 7 8 9123

00 00' 600 00' 1200 00''

52''47''54''

1''-4''3''

11''16'' 18''

-4''+1''+3''

40'' 46'' 43''

-3''

+3''

0''

8'' 8''

6''51'' 0 15'' 0 43'' 0 22''


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nh gi chnh xc:

=1,25 364

2225,1

)13(3422

25,1)1n(nm

V =

=

Trong : V l sai s n l s vng o, n= 3m l s hng, m= 4

M= 233

n==

3- Phng php o lp

Phng php o lp c ng dng khi o gc ring bit vi yu cu chnh xc caov n hn ch c sai s c s trong qu trnh o. Thc cht l o mt gc nhiu ln nhnch c s tr s u v cui ca tr o. Hn nay, phng php n y t c s dng v cc mkinh v, c bit l my kinh v in t hn ch c sai s c s n mc ti a.3.4. o khong cch bng my kinh v quang hc

3.4.1. Xy dng cng thc xc nh khong cch bng dy ch ca my kinh v quang hc.Xc nh khong cch ngang bng li ch ca my kinh v quang hc c ch ra 2

trng hp:1. Trng hp ng knh nm ngang:

Gi s qua ng knh my kinh v ngm mia c c s c dy ch trn, ch dhiu ca 2 s c l l0 . F l tiu im ca knh vt my kinh v, khong cch t mia n tiuim knh vt l L, khong cch t tiu im knh vt n knh l tiu c ca knh vKhong cch t knh vt n trc quay ca my l, khong cch t ch trn n ch ditrong ng knh l p (hnh 3.14).

Khong cch S t my n mia c tnh theo cng thc S = L + f+.T 2 tam gic ng dng: MFN v mFn ta c: L/f = lo /p suy ra: L = lo.f/p.

t: f/p = k ta c: S = k. lo + f +

Hnh 3.14.


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cch ngang t my n mia khi tia ngm nghingvi mt phng ngang mt gc V.

Gi s my t ti I, mia t ti K, cn xc nh khong cch nm ngang S. Trongtrng hp n y ng knh nghing vi mt phng nm ngang mt gc l V. Ngm mia, c s

trn mia, ch trn ct mia ti M, ch di ct mia ti N, khong cch MN = l (l hiu s cch trn v ch di khi mia dng thng ng). chng minh cng thc, ta gi s quay miavung gc vi tia ngm ti A. Ch trn ct mia ti M1, ch di ct mia ti N1. Khong cchM1N1 = l0 ( l hiu s c ch trn v ch di khi mia dng vung gc vi tia ngm OA).

Theo chng minh trn: OA = 100l0 + (3.3)V khong cch t mia n my qu xa, ta coi 3 tia ngm song song vi nhau, khi OMvung gc vi mia xoay ti M1, ON vung gc vi mia xoay ti N1. Nh vy tam gic vungM1AM v tam gic vung NAN1 c gc MAM1 = NAN1 =V (gc c cnh tng ng vunggc).Ta c:

M1A = MA. cos VN1A = NA.cos V

M1A + N1A = (MA + NA). cos VM1A + N1A = l0; MA + NA = l

Suy ra:l0 = l.cosV (3.4)

Thay (3.4) v o (3.3) ta c:

OA = 100.l.cosV + (3.5)V gc nghing V thng l nh, cho nn coi:= .cosV (3.6)Thay (3.6) v o (3.5) Ta c:

OA = 100.l.cosV +. cosV = (100l +). cosVKhi khong cch ngang t my n mia OB = S c xc nh:

S = OA.cosV = (100.l +). cos2V (3.7)Xt tam gic vung B1AO Ta c:

OB1 = OA/ cosV = {(100.l +). cosV}/cosV = 100.l +

S = OB1 OB = 100.l + - (100.l +). cos2

VS = (100.l +)(1- cos2V) = (100.l +).sin2V (3.8)S = 100.l + - S (3.9)Da v o cng thc (3.8). Ta c th lp bng tnh sn gi trS. Thc t cho thy khi

gc V = 30 tng ng vi khong cch S = 100, 200, 300 ta cs l : 0,27; 0,50; 0,80; khi


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M c h m b n , quay b n cho ng thy d i ni chiu song song vi ng ni 2c cn 1, 2. Xoay c cn theo chiu ngc nhau cho bt thy tp trung v o gia (v tr a).Xoay b n i 1 gc 900 so vi v tr u, sao cho ng thy vung gc vi ng ni 2 c cn1, 2, bt thy b lch, dng c cn th 3 xoay c cn n y theo hng thun hoc ngc cho

bt thy v o gia (l m i l m li mt v i ln). Nu bt thy vn gia xoay b n i 1800

sovi v tr b. Nu bt thy vn gia (v tr c) th chng t trc ng thy d i vung gc vi trcquay ca my. Ngc li ta phi hiu chnh ng thy. (hnh 3.18).

Phng php hiu chnh nh sau:

Dng c cn th 3, a bt thy v na khong lch, dng c iu chnh a bt thyv na khong lch cn li, l m i l m li mt v i ln cho n khi mi v tr ca my bthy u gia, y chnh l ng tc cn my trc khi o. 3.5.2. Ch ng phi thng ng, ch ngang phi nm ngang:

y l iu kin quan trng m my kinh v quanghc cn tha m n. iu kin n y c nh hng n chnhxc o gc v o khong cch. kim tra iu kin n y tatin h nh nh sau:

Sau khi cn bng my chnh xc, a ng knhngm 1 dy ch mnh c treo mt qu di ni lng gi vquan st nu thy ch ng trng vi cnh ca dy di thiu kin trn tha m n. Ngc li iu kin trn khngtha m n, ta iu chnh bng cch ni lng c 8 (hnh3.19).

dch chuyn m ng ch thp sao cho ch ng trng vi dy di. Sau vn cht

c li. kim tra ch ngang ta a ng knh ng

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