Find: LBE [ft] A LAD =150 [ft] B LDE =160 [ft] R = 1,000 [ft] C D E

Παρουσίαση με θέμα: "Find: LBE [ft] A LAD =150 [ft] B LDE =160 [ft] R = 1,000 [ft] C D E"— Μεταγράφημα παρουσίασης:

1 Find: LBE [ft] A LAD =150 [ft] B LDE =160 [ft] R = 1,000 [ft] C D E
119 125 131 137 Find the length from Point B to Point E, in feet. [pause] In this problem, ---- R O

2 ? Find: LBE [ft] A LAD =150 [ft] B LDE =160 [ft] R = 1,000 [ft] C D E
119 125 131 137 were looking for the length from point B, on the curve, to a point E, just inside the curve. The problem provides the distance between ---- R O

3 ? Find: LBE [ft] A LAD =150 [ft] B LDE =160 [ft] R = 1,000 [ft] C D E
119 125 131 137 Point A and Point D, the distance between Point D and Point E, R O

4 ? Find: LBE [ft] A LAD =150 [ft] B LDE =160 [ft] R = 1,000 [ft] C D E
119 125 131 137 And, the radius of the curve, R. R O

5 ? Find: LBE [ft] A LAD =150 [ft] B LDE =160 [ft] R = 1,000 [ft] C D E
119 125 131 137 To solve this problem, we’ll first draw a line from Point O to Point B. R O

6 ? Find: LBE [ft] A LAD =150 [ft] B LDE =160 [ft] R = 1,000 [ft] C D E
119 125 131 137 We’ll create a point F where this line crosses the line segment D E. R O

7 ? Find: LBE [ft] A LAD =150 [ft] B LDE =160 [ft] F R = 1,000 [ft] C D
119 125 131 137 To solve for the length of line segment B E, we’ll solve for Triangle F E B, --- R O

8 ? Find: LBE [ft] A LAD =150 [ft] B LDE =160 [ft] F R = 1,000 [ft] C D
119 125 131 137 Where the length of line segment B E equals the length of line segment --- R O

9 Find: LBE [ft] A LAD =150 [ft] B LDE =160 [ft] F R = 1,000 [ft] C D E
LFE LBE = tan (AFBE) F E, divided by the tangent of angle F B E. Next we’ll create point G, --- R O

10 Find: LBE [ft] A LAD =150 [ft] B LDE =160 [ft] G R = 1,000 [ft] C E F
LFE LBE = tan (AFBE) which is a projection of Point B onto line segment A O. Since lines G O and B E --- R O

11 Find: LBE [ft] A LAD =150 [ft] B LDE =160 [ft] G R = 1,000 [ft] C E F
LFE LBE = parallel tan (AFBE) are parallel, then Angle G O B and Angle F B E are --- R O

12 Find: LBE [ft] A LAD =150 [ft] B LDE =160 [ft] R = 1,000 [ft] C E F D
equal LFE LBE = parallel tan (AFBE) equal. Angle G O B equals the arc sin of the --- AFBE = AGOB R O

13 Find: LBE [ft] A LAD =150 [ft] B LDE =160 [ft] G R = 1,000 [ft] C E F
LFE LBE = tan (AFBE) quotient, the length of line segment G B divided by the length of line segment B O. Line segment B O is the same length --- AFBE = AGOB R LGB AGOB=sin-1 LBO O

14 Find: LBE [ft] A LAD =150 [ft] B LDE =160 [ft] G R = 1,000 [ft] C E F
LFE LBE = tan (AFBE) as the radius, or 1,000 feet. The length of line segment G B ---- AFBE = AGOB R LGB AGOB=sin-1 LBO O

15 Find: LBE [ft] A LAD =150 [ft] B LDE =160 [ft] G R = 1,000 [ft] C E F
LFE LBE = tan (AFBE) is equal to the length of line segment D E, ---- AFBE = AGOB R LGB AGOB=sin-1 LBO LDE=LGB O

16 Find: LBE [ft] A LAD =150 [ft] B LDE =160 [ft] G R = 1,000 [ft] C E F
LFE LBE = tan (AFBE) which is 160 feet. Using these values, --- AFBE = AGOB R LGB AGOB=sin-1 LBO LDE=LGB O

17 Find: LBE [ft] A LAD =150 [ft] B LDE =160 [ft] G R = 1,000 [ft] C E F
LFE LBE = tan (AFBE) Angle GOB and Angle FBE equal --- AFBE = AGOB R LGB AGOB=sin-1 LBO 160 [ft] LDE=LGB O

18 Find: LBE [ft] A LAD =150 [ft] B LDE =160 [ft] G R = 1,000 [ft] C E F
LFE LBE = tan (AFBE) o 9.21 9.21 degrees. [pause] The length of line segment F E is equal to ---- AFBE = AGOB R o 9.21 LGB AGOB=sin-1 LBO 160 [ft] LDE=LGB O

19 Find: LBE [ft] AGOB=9.21 A LAD =150 [ft] B LDE =160 [ft] G
R = 1,000 [ft] C E F D LFE LBE = tan (AFBE) o the length of line segment D E minus the length of line segment --- AFBE=9.21 R LFE=LDE-LDF O

20 Find: LBE [ft] AGOB=9.21 A LAD =150 [ft] B LDE =160 [ft] G
R = 1,000 [ft] C E F D LFE LBE = tan (AFBE) o D F. The problem states the length of line segment D E equals --- AFBE=9.21 R LFE=LDE-LDF O

21 Find: LBE [ft] AGOB=9.21 A LAD =150 [ft] B LDE =160 [ft] G
R = 1,000 [ft] C E F D LFE LBE = tan (AFBE) o 160 feet. The length of line segment D F is equal to the length of line segment --- AFBE=9.21 R LFE=LDE-LDF O

22 Find: LBE [ft] AGOB=9.21 A LAD =150 [ft] B LDE =160 [ft] G
R = 1,000 [ft] C E F D LFE LBE = tan (AFBE) o D O times the tangent of Angle G O B. From before, we solved Angle G O B to be --- AFBE=9.21 R LFE=LDE-LDF LDF = LDO * tan (AGOB) O

23 Find: LBE [ft] AGOB=9.21 A LAD =150 [ft] B LDE =160 [ft] G
R = 1,000 [ft] C E F D LFE LBE = tan (AFBE) o 9.21 degrees. The length of line segment D O is equal to the length of line segment --- AFBE=9.21 R LFE=LDE-LDF LDF = LDO * tan (AGOB) O

24 Find: LBE [ft] AGOB=9.21 A LAD =150 [ft] B LDE =160 [ft] G
R = 1,000 [ft] C E F D LFE LBE = tan (AFBE) o A O minus the length of line segment A D. The length of line segment A O is equal to --- AFBE=9.21 R LFE=LDE-LDF LDF = LDO * tan (AGOB) LDO=LAO-LAD O

25 Find: LBE [ft] AGOB=9.21 A LAD =150 [ft] B LDE =160 [ft] G
R = 1,000 [ft] C E F D LFE LBE = tan (AFBE) o the radius of the curve, which is, AFBE=9.21 R LFE=LDE-LDF LDF = LDO * tan (AGOB) LDO=LAO-LAD O

26 Find: LBE [ft] AGOB=9.21 A LAD =150 [ft] B LDE =160 [ft] G
R = 1,000 [ft] C E F D LFE LBE = tan (AFBE) o 1,000 and the length of line segment A D is given as --- AFBE=9.21 R LFE=LDE-LDF LDF = LDO * tan (AGOB) LDO=LAO-LAD O

27 Find: LBE [ft] AGOB=9.21 A LAD =150 [ft] B LDE =160 [ft] G
R = 1,000 [ft] C E F D LFE LBE = tan (AFBE) o 150 feet, which makes the length of line segment D O equal to --- AFBE=9.21 R LFE=LDE-LDF LDF = LDO * tan (AGOB) LDO=LAO-LAD O

28 Find: LBE [ft] AGOB=9.21 A LAD =150 [ft] B LDE =160 [ft] G
R = 1,000 [ft] C E F D LFE LBE = tan (AFBE) o 850 feet. The length of line segment D F equals --- AFBE=9.21 R LFE=LDE-LDF LDF = LDO * tan (AGOB) LDO=LAO-LAD = 850 [ft] O

29 Find: LBE [ft] AGOB=9.21 A LAD =150 [ft] B LDE =160 [ft] G
R = 1,000 [ft] C E F D LFE LBE = tan (AFBE) o 850 feet times the tangent of 9.21 degrees, which is --- AFBE=9.21 R LFE=LDE-LDF LDF = LDO * tan (AGOB) LDO=850 [ft] O

30 Find: LBE [ft] AGOB=9.21 A LAD =150 [ft] B LDE =160 [ft] G
R = 1,000 [ft] C E F D LFE LBE = tan (AFBE) o 137.8 feet. The length of line segment F E equals ---- AFBE=9.21 R LFE=LDE-LDF 137.8 [ft] LDF = LDO * tan (AGOB) LDO=850 [ft] O

31 Find: LBE [ft] AGOB=9.21 A LAD =150 [ft] B LDE =160 [ft] G
R = 1,000 [ft] C E F D LFE LBE = tan (AFBE) o 160 feet minus feet, or --- AFBE=9.21 R LFE=LDE-LDF 137.8 [ft] LDF = LDO * tan (AGOB) LDO=850 [ft] O

32 Find: LBE [ft] AGOB=9.21 A LAD =150 [ft] B LDE =160 [ft] G
R = 1,000 [ft] C E F D LFE LBE = tan (AFBE) 22.2 [ft] o 22.2 feet. And finally, the length of line segment B E equals, --- AFBE=9.21 R LFE=LDE-LDF 137.8 [ft] LDF = LDO * tan (AGOB) LDO=850 [ft] O

33 Find: LBE [ft] AGOB=9.21 A LAD =150 [ft] B LDE =160 [ft] G
R = 1,000 [ft] C F D E LFE LBE = tan (AFBE) 22.2 [ft] o 22.2 feet, divided by the tangent of 9.21 degrees, which equals, --- AFBE=9.21 R LFE=LDE-LDF 137.8 [ft] LDF = LDO * tan (AGOB) LDO=850 [ft] O

34 Find: LBE [ft] AGOB=9.21 A LAD =150 [ft] B LDE =160 [ft] G
R = 1,000 [ft] C F D E LBE =136.9 [ft] LFE LBE = tan (AFBE) 22.2 [ft] o 136.9 feet. [pause] AFBE=9.21 R LFE=LDE-LDF 137.8 [ft] LDF = LDO * tan (AGOB) LDO=850 [ft] O

35 Find: LBE [ft] AGOB=9.21 A LAD =150 [ft] B LDE =160 [ft] G
R = 1,000 [ft] C F D E LBE =136.9 [ft] LFE LBE = tan (AFBE) 22.2 [ft] 119 125 131 137 When reviewing the possible solutions, --- o 9.21 R O

36 Find: LBE [ft] AGOB=9.21 A LAD =150 [ft] B LDE =160 [ft] G
R = 1,000 [ft] C F D E LBE =136.9 [ft] LFE LBE = tan (AFBE) 22.2 [ft] 119 125 131 137 The answer is D. o 9.21 R AnswerD O

37 ? Index σ’v = Σ γ d γT=100 [lb/ft3] +γclay dclay 1
Find: σ’v at d = 30 feet (1+wc)*γw wc+(1/SG) σ’v = Σ γ d d Sand 10 ft γT=100 [lb/ft3] 100 [lb/ft3] 10 [ft] 20 ft Clay = γsand dsand +γclay dclay A W S V [ft3] W [lb] 40 ft text wc = 37% ? Δh 20 [ft] (5 [cm])2 * π/4