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Unformatted text preview: Problem 3.2? Determine the sugereletratien reguired at the design sgeed. r'JrlseI cemgute the degree at currreI length at currreI and statiening at
the PC annd PT. ‘u’ := 100 R := 1000 J. := 33 [given] stap:=112513 r5 ;= 0.20 g:=32.2 Since the racetrack is singlelane. := R = 1000 Salve fcr required superelet‘aticn . 2
tut1.46? .
e+f3=;}1—fSe: rqu.3.34]
e = [1.413
salve fer degree at curre
13mm
D := —R D = 5.?3 degrees Ing. 3.35] use this and Equaticn 3.33 tc salve fer length cf curre rauuu : a: — L= _e_~. [Eq 3.39]
rD ran L ;= 103'; L: 523.5 ft calculate tangent length It's—1". tilt.“ I
T := Rtan: 2 Edeg :: T = 251943 IIEI:. 3.3I3]
Shape I= StEPI — T Shape = StEPC = staFrI := stapc + L staFrI = 113755.55 staF.T = 112346365 Pmbmm331
Determine the design sgeed used. Since the ramp is a single 12feet lane. center efreadwa‘y is center eftraveled path 1.:= 31] L := 528 11:13:: 13.4 [given]
using L and ﬂ. sewe fer R
: L1E3EI
L: —R_". R := R: 333.3 [EIZ. 3.33]
131] :3.
R; := R = 333.3
since Me and RV are knewn. we can use Equatien 3.43 te ﬁnd SSE]
IR, :'i I: Fit  W133":
SSE] := SIDdeg laces H SSE] = 250.1 11 [El 343]
frern Table 3.1. SSE] fer 35 mifh is 25th  cuwe is designed fer 35 mifh
Alternative Selutien
since Ms and Re are knewn. we can sewe Equatien 3.42 te ﬁnd SSD
f fassse ‘9
MS: 1 — ces: F. deg :: [EC]. 342]
SSE] = 250.1 ft
frem Table 3.1. SSE] fer 35 mifh is 251] ft  curve is designed fer 35 mifh
Cernering Check
1'.i":=351.2133? 1'.i"=51.3
fer 35 nw'h. f3 := [1.155 [Table 3.5]
g := 32.2
2
‘u’
e:=——fS e=E.E5
93s Se this cemhinatien efspeed. radius. and superelevatien is OK Problem 3.34
Determine the design sgeed used te design the curve. e := [1.116 {given}
. . . 12 since the read Is feurlane Wlil'l 1211 lanes. 11.13:: 52 — 12 — E ME = 34 11 tr'_.,.r EU mi“h Ru := 1341] [Table 3.5} see ;= are [Table 31}
r” Eggssg TH 11:13:: 1 — ees: deg :: ME = 30.134 11 (Ea. 3.42}
i *1 Jr} this is less than the required distance. try again tr'_.,.r TE] mi“h Ru := 2051] [Table 3.5} see ;= ran [Table 31}
r” Eggssg TH 11:13:: 1 — ees: deg :: ME = 32.4113 11 (Ea. 3.42}
i *1 Jr} this is less than the required distance. try again tr'_.,.r EU mi“h R133: 305” [Table 3.5} fansee ‘1‘} 11:13:: 1 — ees_ _ deg __ ME = 33.?65 11 (Ea. 3.42}
i *1 Jr} this rounds in 34 11. therefore the design speed is 31] mifh Determine the statiening and elevatien ef the P‘U’Cs and P‘U'Ts. Problem 3'9 EU mifh design speed Kc : 151 [Table 3.2]
Ks 3: 135 [Table 3.3]
A ;= In — 2 A = 2
calculate lengths cf crest and sag cuwes L3 ;= KSA L3 = 212 Calculate staticn and elevaticn cf PVT fcr crest cur5e.
staFm: = 1+51 ALE
20E] elevpyﬂc := 1m] — eleupVIE = 95.98 ﬁ Parr: ;= n + 302 Parr: = 302 staWTc = s + [:2 Calculate staticn ancl elevaticn cf PVT and PVC fcr sag cuwe. staFmS = 41+E‘l elevFllulIS := 1m: — [DUE4000] eleupVIE = 20 ft L L PUTS ;= 3': + 4mm] + 33 PUTS = 42s? staWTs = 42 + s?
ALS
elevFllugcS := 20 + 2m] eleanulCS = 22.?2 ft F'ch ;= PWS — LS F'ch = 4015 mama = 4n + 15 ...
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 Fall '08
 KATZ
 design sgeed, elevaticn cf PVT, tc salve fer, Eggssg TH, te ﬁnd SSE

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