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Question 1 For each of the four vertical curves show in the attachment (click here for attachment) calculate the following: 1ATTACHMENTS:

Curve 1 L=200 ft Pl: Station 12+00, Elevation: 520 ft Curve 2 L = 300 ft Pl: Station 15+00, Elevation: 2250.00 ft 0 +4% Curve

Question 1 For each of the four vertical curves show in the attachment (click here for attachment) calculate the following: 1. K and r 2. station of BVC and EVC 3. elevation of point at a distance, L/4, from BVC and EVC 4. station of turning point 5. elevation of turning point 6. elevation of mid-point of each curve
Curve 1 L=200 ft Pl: Station 12+00, Elevation: 520 ft Curve 2 L = 300 ft Pl: Station 15+00, Elevation: 2250.00 ft 0 +4% Curve 3 Pl: Station 22+00, Elevation: 150.00 ft L = 200 ft 0 0 Curve 4 L=400 ft EVC: Station 10+00, Elevation: 50.00 ft
engineering   Civil-Engineering   
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L-200 H 2 L 0.03-Kool) +00 2oo 2 3. Shho elevation at %fvom BC 또 520 + 0.01 x 100 = S2ldt elevation Ch sfafionrom BVC - S2D-Selevahion station EvC Pele 520 + o.o3x loo satiana. dvom .3tation hşhat point on Curve* oite hishext point on cusve occun atPi k-는 300 32-31 -o0L-00661x 10 300 r = 82-80.01-001 = aas0-0.01x150 F2248.Sf. on Cuove at a distonce x The elevation ary poolosation ซึ่ point % from eve OL=75ft 229.063 ft elevation < point Y4 fom Eve point L/4 Eve = 300-7S : 225 ff . 2249.064 ft.PT station 22400 ks. :66.64 3 マ2+00-200 2. elevation oBVC slation 2 3+00 . poinl on e elevafion 481= ISO-3125-tt. ponf cor) highunt poinf is at bot, tangents, add both 91th . elevation tt tumirg point = 150.48 ft. flid pointSol 2 400 400 10 +00 - 4o0 6400. 3.tahion 8+00 ニ elevation ซึ่ Eve varim S2 ft elevation cf BVele PVI S20 02 xaoo Soft<p a+b+ eegvc 2X400 b 0.ol So S I fom EVC elevation ooft. 49.YS ft point te hihet point at %--2.5x16sk@oStool xamtso

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