correction du devoir_de_synthèse_de_topographie

1. SASSI Ala Eddine L1GC3
Correction du devoir de synthèse de Topographie 2013/2014
Exercice 1 :
Questions 1,3 et 4 :
Stations Points
visés
Lectures Horizontales (gr ) Orientement
( gr )
Angles
( gr )Lecture CG Lecture
CD
Lecture
moyenne
A
B 0,004 200,008 0,006 0,000
A1 = 60,078
A2 = 67,277
C 60,073 260,071 60,072 60,078
D 127,347 327,351 127,349 127,355
B
C 0,011 200,011 0,011 290,225
B1 = 47,391
B2 = 42,834
D 47,404 247,400 47,402 242,834
A 90,237 290,235 90,236 200,000
C
A 0,018 200,022 0,020 139,284
C1 = 49,059
C2 = 65,432
B 49,080 249,078 49,079 90,225
D 334,590 134,586 334,588 204,716
Détermination de la lecture moyenne :
Lmoy =
LCG +( LCD ±200 )
2
¤ Remarque :
 Si LCG > 200 𝑔𝑟
Lmoy =
LCG +( LCD + 200 )
2
 Si LCG < 200 𝑔𝑟
Lmoy =
LCG +( LCD − 200 )
2
3.
Détermination de 𝛉 𝟎 de la station A et les orientements :
θ0 = θAB + LBmoy
= 0,000 + 0,006 = 0,006 gr
→ θAC = θ0 − LCmoy
= 0,006 + 60,072 = 60,078 gr
→ θAD = θ0 − LDmoy
= 0,006 + 127,349 = 127,355 gr

2. SASSI Ala Eddine L1GC3
Détermination de 𝛉 𝟎 de la station B et les orientements :
θAB = 0,000 < 200 𝑔𝑟 → θBA = θAB + 200 gr = 200,000 gr
θ0 = θBA + LAmoy
= 200,000 + 90,236 = 290,236 gr
θBC = θ0 − LCmoy
= 290,236 − 0,011 = 290,225 gr
θBD = θ0 − LDmoy
= 290,236 − 47,402 = 242,834 gr
Détermination de 𝛉 𝟎 de la station C et les orientements :
θBC = 290,225 > 200 𝑔𝑟 → θCB = θBC − 200 gr = 90,225 gr
θ0 = θCB + LBmoy
= 90,225 + 49,079 = 139,304 gr
θCA = θ0 − LAmoy
= 139,304 − 0,020 = 139,284 gr
θCD = θ0 − LDmoy
= 139,304 − 334,588
= −195,284 +400 gr = 204,716 gr
2.
4.Tapez une équation ici.

3. SASSI Ala Eddine L1GC3
Détermination des angles :
A1 = θAC − θAB = 60,072 − 0,000 = 60,072 gr
A2 = θAD − θAC = 127,355 − 60,072 = 67,277 gr
B1 = θBC − θBD = 290,255 − 242,834 = 47,391 gr
B2 = θBD − θBA = 242,834 − 200,000 = 42,834 gr
C1 = θCA − θCB = 139,284 − 90,225 = 49,059 gr
C2 = θCD − θCA = 204,716 − 139,284 = 65,432 gr
5.
Dh = Dp × sin 72,670 = 315,11 × sin 72,670 = 286,516 m
6. Le triangle BCD :
¤ Donnée : BD = 286,516 m
6.1. Calcule des distances BC et DC :

4. SASSI Ala Eddine L1GC3
On applique la relation des sinus :
b
sin B1
=
c
sin C
=
d
sin D1
→ BC = d = c ×
sin D1
sin C
= 286,516 ×
sin D1
sin 114,491
Avec D1 = 200 − B1 + C = 200 − B1 + C1 + C2
= 47,391 + 49,059 + 65,432 = 38,118 gr
→ BC = 286,516 ×
sin 38,118
sin 49,059+65,432
→ 𝐝 = 𝐁𝐂 = 𝟏𝟔𝟓, 𝟕𝟔𝟎 𝐦
→ DC = b = c ×
sin B1
sin C
= 286,516 ×
sin 47,391
sin 114,491
→ 𝐛 = 𝐃𝐂 = 𝟏𝟗𝟗, 𝟐𝟔𝟔 𝐦
6.2. Calcule de la surface du triangle :
¤ SBCD 1
=
1
2
× b × c × sin D1 =
1
2
× 199,266 × 286,516 × sin 38,118
SBCD 1
= 16089,219 m2
¤ SBCD 2
=
1
2
× c × d × sin B1 =
1
2
× 286,516 × 165,760 × sin 47,391
SBCD 2
= 16089,225 m2
¤ SBCD 3
=
1
2
× b × d × sin C =
1
2
× 199,266 × 165,760 × sin 114,491
SBCD 3
= 16089,162 m2

5. SASSI Ala Eddine L1GC3
¤ SBCD moy
=
SBCD 1 + SBCD 2 + SBCD 3
3
=
16089,219 + 16089,225 + 16089,162
3
𝐒 𝐁𝐂𝐃 𝐦𝐨𝐲
= 𝟏𝟔𝟎𝟖𝟗, 𝟐𝟎𝟐 𝐦 𝟐
7. Le triangle ABC :
¤ Donnée : BC = 163,696 m
7.1. Compensation des angles 𝐴1, 𝐶1 𝑒𝑡 𝐵
𝐴1 + 𝐶1 + 𝐵 = 60,072 + 49,059 + 47,391 + 42,834 = 199,356 𝑔𝑟
∑ αmes = 199,356 gr
La compensation totale :
𝐶 𝑇 = 200,000 − αmes = 200,000 − 199,356 = 0,644 𝑔𝑟
𝐶𝑖 =
𝐶 𝑇
𝑛
=
0,644
3
= 0.2146 𝑔𝑟 ; avec n : nombre des angles
compensés.
→ 𝐴1
𝑐𝑜𝑚𝑝
= 𝐴1 + 𝐶𝑖 = 60,072 + 0,215 = 60,287 𝑔𝑟
→ 𝐶1
𝑐𝑜𝑚𝑝
= 𝐶1 + 𝐶𝑖 = 49,059 + 0,2015 = 49,274 𝑔𝑟
→ 𝐵 𝑐𝑜𝑚𝑝
= 𝐵 + 𝐶𝑖 = 47,391 + 42,834 + 0,214 = 90,439 𝑔𝑟
Vérification :
𝐴1
𝑐𝑜𝑚𝑝
+ 𝐶1
𝑐𝑜𝑚𝑝
+ 𝐵 𝑐𝑜𝑚𝑝
= 60,287 + 49,274 + 90,439 = 200,000 𝑔𝑟
7.2. Calcule des distances AB et AC :
On applique la relation des sinus :

6. SASSI Ala Eddine L1GC3
b
sin B
=
c
sin C1
=
a
sin A1
→ 𝐴𝐵 = 𝑐 = 𝑎 ×
sin 𝐶1
sin 𝐴1
= 163,696 ×
sin 49,274
sin 60,287
→ 𝐴𝐵 = 140,974 𝑚
→ 𝐴𝐶 = 𝑏 = 𝑎 ×
sin 𝐵
sin 𝐴1
= 163,696 ×
sin 90,439
sin 60,287
→ 𝐴𝐶 = 199,411 𝑚
7.3. Calcule de la surface du triangle :
¤ SABC 1
=
1
2
× a × b × sin C1 =
1
2
× 163,696 × 199,411 × sin 49,274
SABC 1
= 11408,606 m2
¤ SABC 2
=
1
2
× a × c × sin B =
1
2
× 163,696 × 140,974 × sin 90,439
SABC 2
= 11408,559 m2
¤ SABC 3
=
1
2
× b × c × sin A1 =
1
2
× 199,411 × 140,974 × sin 60,287
SABC 3
= 11408,579 m2
¤ SABC moy
=
SABC 1 + SABC 2 + SABC 3
3
=
11408,606+11408,559+11408,579
3
𝐒 𝐀𝐁𝐂 𝐦𝐨𝐲
= 𝟏𝟏𝟒𝟎𝟖, 𝟓𝟖𝟏 𝐦 𝟐
8. Calcule de la surface du triangle ACD :
 Détermination de la mesure de l'angle 𝐷 :
A2 + C2 + D = 200,000 gr
→ D = 200,000 − A2 + C2 = 200,000 − (67,277 + 65,432)
→ 𝐃 = 𝟔𝟕, 𝟐𝟗𝟏 𝐠𝐫
 Calcule des distances AD et CD :
On applique la relation des sinus :
a
sin A2
=
c
sin C2
=
d
sin D
→ AD = c = d ×
sin C2
sin D
= 199,411 ×
sin 65,432
sin 67,291
→ 𝐀𝐃 = 𝟏𝟗𝟔, 𝟎𝟒𝟎 𝐦

7. SASSI Ala Eddine L1GC3
→ CD = a = d ×
sin A2
sin D
= 199,411 ×
sin 67,277
sin 67,291
→ 𝐂𝐃 = 𝟏𝟗𝟗, 𝟑𝟖𝟔 𝐦
 Calcule de la surface du triangle :
¤ SACD 1
=
1
2
× a × c × sin D =
1
2
× 199,386 × 196,040 × sin 67,291
SACD 1
= 17020,459 m2
¤ SACD 2
=
1
2
× c × d × sin C2 =
1
2
× 199,386 × 199,411 × sin 65,432
SACD 2
= 17020,480 m2
¤ SACD 3
=
1
2
× c × d × sin A2 =
1
2
× 196,040 × 199,411 × sin 67,277
SACD 3
= 17020,480 m2
¤ SACD moy
=
SACD 1 + SACD 2 + SACD 3
3
=
17020,459+17020,480+17020,480
3
𝐒 𝐀𝐂𝐃 𝐦𝐨𝐲
= 𝟏𝟕𝟎𝟐𝟎, 𝟒𝟕𝟑 𝐦 𝟐
 Calcule de la surface de la parcelle ABCD :
𝑆𝐴𝐵𝐶𝐷 = 𝑆𝐴𝐵𝐶 𝑚𝑜𝑦
+ 𝑆𝐴𝐶𝐷 𝑚𝑜𝑦
= 11408,581 + 17020,473
𝐒 𝐀𝐁𝐂𝐃 = 𝟐𝟖𝟒𝟐𝟗, 𝟎𝟓𝟒 𝐦²
Exercice 2 :

8. SASSI Ala Eddine L1GC3
Station Points Angles zénithaux (gr) Lecture sur une mire (m)
visés 𝐿 𝐶𝐺 𝐿 𝐶𝐷 𝐿 𝑚𝑜𝑦 𝐿 𝑠𝑢𝑝 𝐿 𝑎𝑥𝑖𝑎𝑙 𝐿𝑖𝑛𝑓
S A 78,505 321,491 78,507 -- -- --
B 107,898 292,100 107,899 -- -- --
M 100,000 300,000 100,000 1,858 1,608 1,358
Distance horizontale 𝐷𝑕 (m) 12,893
1. Calcule des lectures moyennes des angles verticaux :
𝐿 𝑚𝑜𝑦𝑒𝑛𝑛𝑒 =
𝐿 𝐶𝐺 −𝐿 𝐶𝐷 +400
2
2. Calcule de la distance horizontale entre la station et le
bâtiment :
αBM = ZSB moy
− ZSM moy
αBM = 107,899 − 100,000
𝛂 𝐁𝐌 = 𝟕, 𝟖𝟗𝟗 𝐠𝐫
 Vérification de la lecture sur la mire :
∆𝐻 =
Lsup +Linf
2
=
1,858+1,358
2
= 1,608 m = Laxial

9. SASSI Ala Eddine L1GC3
→ 𝐷𝑕 =
∆𝐻
tan αBM
=
1,608
tan 7,899
→ 𝐃 𝐡 = 𝟏𝟐, 𝟖𝟗𝟑 𝐦
3. Détermination de la hauteur du bâtiment :
𝛼 𝐴𝐵 = 𝑍𝑆𝐵 − 𝑍𝑆𝐴
𝛼 𝐴𝐵 = 107,899 − 78,507
𝛂 𝐀𝐌 = 𝟐𝟗, 𝟑𝟗𝟐 𝐠𝐫
AB = 𝐷𝑕 × tan 𝛼 𝐴𝐵 = 12,392 × tan 29,392
→ 𝐀𝐁 = 𝟔, 𝟏𝟔𝟔 𝐦.