1- Arc definition curve with Da = 3°35’25”, I = 12°56’51”, and PI station = 8...
1- Arc definition curve with Da = 3°35’25”, I = 12°56’51”, and PI station = 8 + 56.25
L =
T =
E =
M =
LC =
PC Station =
PT Station =
2- Arc definition curve with Da = 1°23’47”, I = 19°25’15”, and PI station = 22 + 25.10
L =
T =
E =
M =
LC =
PC Station =
PT Station =
3- Arc definition curve with Da = 5°46’45”, I = 21°14’45”, and PI station = 15 + 52.19
L =
T =
E =
M =
LC =
PC Station =
PT Station =
Homework Answers
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1- Arc definition curve with Da = 3°35’25”, I = 12°56’51”, and
PI station = 8...
ON2ohtal curves .The deflecti n angle used when sighting back on the curve during a move up was w...
Please complete the chart and explain how to solve number 16.
Thank you.
oN2ohtal curves .The deflecti n angle used when sighting back on the curve during a move up was wrong, causing the curve to get wasn't enough, and the construction equipment removed the stakes. Using the wrong backsight when laying out a curve by QUESTIONS AND PROBLEMS Calculate the missing curve parts in the following Table. Curve# 1 of Curve E M.O. LC 1 18°54' 357.25 2 2930...
HomewUI A circular-horizontal curve to the right has the following characteristics: Δ or l = 8"24...
HomewUI A circular-horizontal curve to the right has the following characteristics: Δ or l = 8"24°00" PI @ Sta. 64+27.46 Dare = 20000" Coordinates of the PI are 1234.56N & 2468.10E R 2864.79 ft The Azimuth of the back tangent is 76 26'30" T = 210.38 ft. (Azimuth from PC to PI) L =420 ft. Draw a sketch and calculate the following: a) The station of the POC b) The coordinates of the PC c) The station of the PT...
5. For a horizontal curve, A=82'30', the station of the PC is 6+14.00, and terrain conditions...
5. For a horizontal curve, A=82'30', the station of the PC is 6+14.00, and terrain conditions require the minimum radius permitted by the specifications of, 100 ft. (are definition). Calculate (15 points): a. The stationing of PI and PT, and the external (E) and middle ordinate distances (M) and chord length (LC) for this curve. b. Compute sub-deflection angles and sub-chords to stake out this curve. Use quarter-stations 1251 (Note: In the example we solved together in class, we used...
A horizontal curve (arc definition) having an intersection angle of 80° and LC = 600 ft....
A horizontal curve (arc definition) having an intersection angle of 80° and LC = 600 ft. Compute R, L, T, E and M, using the equations in the slides (show your calculations). Survey class
Example 10-5 Consider the simple horizontal curve given in Example 10-1, with a tangent distance T=...
Example 10-5 Consider the simple horizontal curve given in Example 10-1, with a tangent distance T= 229.27 ft and an arc length L 391.53 ft. If the station of the PI is established as 7 +47.64, find the stations of the PC and the PT
Name: Grade: /10 Dec 3, 2018 w ..CA-Horizontal Curve VI FORT WAYNE Tabulate R or D,...
Name: Grade: /10 Dec 3, 2018 w ..CA-Horizontal Curve VI FORT WAYNE Tabulate R or D, T, L, E, M, PC, PT, deflection angles, and incremental chords to lay out the circular curves at ful stations (100 ft) for the Highway curve with R 1200 ft, I-30°00, and PI station 45+ 50.00 ft. Station Chord Defl. Increment Defl. Angle
SRV 111-SURVEYING II-TEST Highwaur CURVEK THE PI IS AT 53+62, I IS 26 20', FOR A...
SRV 111-SURVEYING II-TEST Highwaur CURVEK THE PI IS AT 53+62, I IS 26 20', FOR A HORIZONTAL CIRCULAR AND A DOF 5' HAS BEEN SELECTED. COMPUTE THE FOLLOWING VALUES OF R, T, LC, E, M, AND L ALONG WITH THE PC AND PT STATIONS. COMPUTE THE DEFLECTION ANGLES AND INCREMENTAL CHORD DISTANCES TO STAKE THE CURVE FOR 100 FT STATIONS. Tncucie el horco KEEP ALL CALCULATI ONS NEAREST 0.01 FT 5729.58 50 P-5729.58 = 1145.92ft =R
Find L please! -3 % high elevation on curve is 206.00 Find L. 2. PC station...
Find L please!
-3 % high elevation on curve is 206.00 Find L. 2. PC station = 10+00.00 PC elev = 200.00 G1=+2% G2=-3% at station 15+00 elevation must be 205.00 Find L 3. PI elev 185.00 GI=+2% G2--3% high elevation on curve is 176.00 Find L 4. P1 station = 16+25.00 P1 elev = 185.00 G1 +2% G2--3% at station 15+00 elevation must be 175.00 Find L
24.7 Assignment 5 Due Date: April 24, 2018 Tabulate R or D, T, L, E, M,...
24.7
Assignment 5 Due Date: April 24, 2018 Tabulate R or D, T, L, E, M, PC, PT, deflection angles, and incremental chords to layout the circular curves at full stations (100ft or 30m) in problems 24.7 through 24.14 24.7 Highway curve with D 2°30', I 10°30', and PI station 36+44.50 ft 24.10 Highway curve with R 900m, I 1230', and PI station 4+200.600 m
Assume PI sta = 18+00, PI elev. = 100.00, L = 18 stations, incoming grade =...
Assume PI sta = 18+00, PI elev. = 100.00, L = 18 stations, incoming grade = +2%, outgoing grade = -4% 1. The elevation at station 25+00 is (a) 71.33 (b) 72.33 (c) 73.33 (d) 74.33 2. The elevation change from the PI location on the curve to the PI elevation is (a) -27.00 (b) -13.50 (c) +13.50 (d) +27.00 3. Extend the incoming grade to the PT. The elevation change from this point to the actual PT elevation is...
Please complete the chart and explain how to solve number 16.
Thank you.
oN2ohtal curves .The deflecti n angle used when sighting back on the curve during a move up was wrong, causing the curve to get wasn't enough, and the construction equipment removed the stakes. Using the wrong backsight when laying out a curve by QUESTIONS AND PROBLEMS Calculate the missing curve parts in the following Table. Curve# 1 of Curve E M.O. LC 1 18°54' 357.25 2 2930...
HomewUI A circular-horizontal curve to the right has the following characteristics: Δ or l = 8"24°00" PI @ Sta. 64+27.46 Dare = 20000" Coordinates of the PI are 1234.56N & 2468.10E R 2864.79 ft The Azimuth of the back tangent is 76 26'30" T = 210.38 ft. (Azimuth from PC to PI) L =420 ft. Draw a sketch and calculate the following: a) The station of the POC b) The coordinates of the PC c) The station of the PT...
5. For a horizontal curve, A=82'30', the station of the PC is 6+14.00, and terrain conditions require the minimum radius permitted by the specifications of, 100 ft. (are definition). Calculate (15 points): a. The stationing of PI and PT, and the external (E) and middle ordinate distances (M) and chord length (LC) for this curve. b. Compute sub-deflection angles and sub-chords to stake out this curve. Use quarter-stations 1251 (Note: In the example we solved together in class, we used...
Example 10-5 Consider the simple horizontal curve given in Example 10-1, with a tangent distance T= 229.27 ft and an arc length L 391.53 ft. If the station of the PI is established as 7 +47.64, find the stations of the PC and the PT
Name: Grade: /10 Dec 3, 2018 w ..CA-Horizontal Curve VI FORT WAYNE Tabulate R or D, T, L, E, M, PC, PT, deflection angles, and incremental chords to lay out the circular curves at ful stations (100 ft) for the Highway curve with R 1200 ft, I-30°00, and PI station 45+ 50.00 ft. Station Chord Defl. Increment Defl. Angle
SRV 111-SURVEYING II-TEST Highwaur CURVEK THE PI IS AT 53+62, I IS 26 20', FOR A HORIZONTAL CIRCULAR AND A DOF 5' HAS BEEN SELECTED. COMPUTE THE FOLLOWING VALUES OF R, T, LC, E, M, AND L ALONG WITH THE PC AND PT STATIONS. COMPUTE THE DEFLECTION ANGLES AND INCREMENTAL CHORD DISTANCES TO STAKE THE CURVE FOR 100 FT STATIONS. Tncucie el horco KEEP ALL CALCULATI ONS NEAREST 0.01 FT 5729.58 50 P-5729.58 = 1145.92ft =R
Find L please!
-3 % high elevation on curve is 206.00 Find L. 2. PC station = 10+00.00 PC elev = 200.00 G1=+2% G2=-3% at station 15+00 elevation must be 205.00 Find L 3. PI elev 185.00 GI=+2% G2--3% high elevation on curve is 176.00 Find L 4. P1 station = 16+25.00 P1 elev = 185.00 G1 +2% G2--3% at station 15+00 elevation must be 175.00 Find L
24.7
Assignment 5 Due Date: April 24, 2018 Tabulate R or D, T, L, E, M, PC, PT, deflection angles, and incremental chords to layout the circular curves at full stations (100ft or 30m) in problems 24.7 through 24.14 24.7 Highway curve with D 2°30', I 10°30', and PI station 36+44.50 ft 24.10 Highway curve with R 900m, I 1230', and PI station 4+200.600 m
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