MASTER OF LANDSCAPE ARCHITECTURE DEPARTMENT OF LANDSCAPE ARCHITECTURE FACULTY OF DESIGN AND ARCHITECTURE UNIVERSITI PUTRA MALAYSIA LAr. ABD AZIZ OTHMAN LAN3501 SURVEY AND SITE ENGINEERING TUTORIAL #2 CONTOUR & SLOPE 1. Convert the following ratios to percentage slope: 50:1, 10:1, 8:1 & 3:1. 50:1 =(1/50) X100= 0.02X100= 2% 10:1 =(1/10) X100= 0.1X100= 10% 8:1 = (1/8) X100= 0.125X100= 12.5% 1:3 = (1/3) X100= 0.333X100= 33.3% 2. Give the correct ratio for the following slopes: = 200:20 = 10:1 = 8:4 = 1:2 = 10:2 = 5:1 3. Convert the following percentage slopes to degrees and minutes: 1.5%, 3.5%, 18% & 35%. 1.5% tan Δ = (1.5/100) tan Δ = tan -1 0.015 = 0.86 0 (0.86x60’) = 0 0 52’ 3.5% tan Δ = tan -1 (3.5/100) = tan -1 0.035 = 2.00 0 (2 + 0.0x60’) = 2 0 00’ 18% tan Δ = tan -1 (18/100) = tan -1 0.18 = 10.20 0 (10 + 0.20x60’) = 10 0 12’ 35% tan Δ = tan -1 (35/100) = tan -1 0.35 = 19.29 0 (10 + 0.29x60’) = 10 0 17’ 4. Compute the gradient for a 6:1 slope. G = (DE/L) G = (1/6) = 0.17 5. 107.50M of horizontal distance was measured along a 3% slope. Compute the difference in elevation. Slope (%) = (DE/L) X 100 3% = [(DE)/107.50] X 100 0.03 = (DE/107.50) = 3.225M 6. If a slope falls at a gradient of 0.06 for a distance of 10.00M, what is the difference in elevation. G = (DE/L) DE = G X L = 0.06 X 10.00 = 0.6M 7. At a 10% slope, what would be the change in elevation for the distance of 28.5M? G = (DE/L) DE = G X L = 0.1 X 28.5 = 2.85M 8. At an 8:1 slope, what would be the distance needed to obtain a 2M change in elevation? G = (DE/L) L = (DE/G) = 2.0/ ) = (1/8 2/0.125 = 16.0M 20 200 8 10 4 2 x y Δ tan Δ = (x/y) Δ = tan -1 (x/y)
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MASTER OF LANDSCAPE ARCHITECTURE DEPARTMENT OF LANDSCAPE ARCHITECTURE
FACULTY OF DESIGN AND ARCHITECTURE UNIVERSITI PUTRA MALAYSIA
LAr. ABD AZIZ OTHMAN
LAN3501 SURVEY AND SITE ENGINEERING TUTORIAL #2 CONTOUR & SLOPE 1. Convert the following ratios to percentage slope: 50:1, 10:1, 8:1 & 3:1.
2. Give the correct ratio for the following slopes: = 200:20 = 10:1 = 8:4 = 1:2 = 10:2 = 5:1 3. Convert the following percentage slopes to degrees and minutes: 1.5%, 3.5%, 18% & 35%.
1.5% tan Δ = (1.5/100) tan Δ = tan
-1 0.015 = 0.86
0 (0.86x60’) = 0
0 52’
3.5% tan Δ = tan
-1 (3.5/100) = tan
-1 0.035 = 2.00
0 (2 + 0.0x60’) = 2
0 00’
18% tan Δ = tan -1
(18/100) = tan -1
0.18 = 10.200 (10 + 0.20x60’) = 10
0 12’
35% tan Δ = tan -1
(35/100) = tan -1
0.35 = 19.290 (10 + 0.29x60’) = 10
0 17’
4. Compute the gradient for a 6:1 slope.
G = (DE/L) G = (1/6) = 0.17
5. 107.50M of horizontal distance was measured along a 3% slope. Compute the difference in
elevation. Slope (%) = (DE/L) X 100 3% = [(DE)/107.50] X 100 0.03 = (DE/107.50) = 3.225M
6. If a slope falls at a gradient of 0.06 for a distance of 10.00M, what is the difference in elevation.
G = (DE/L) DE = G X L = 0.06 X 10.00 = 0.6M
7. At a 10% slope, what would be the change in elevation for the distance of 28.5M? G = (DE/L)
DE = G X L = 0.1 X 28.5 = 2.85M 8. At an 8:1 slope, what would be the distance needed to obtain a 2M change in elevation?
G = (DE/L) L = (DE/G) = 2.0/ ) = (1/8 2/0.125 = 16.0M
20
200 8 10
4 2
x
y Δ
tan Δ = (x/y) Δ = tan
-1 (x/y)
MASTER OF LANDSCAPE ARCHITECTURE DEPARTMENT OF LANDSCAPE ARCHITECTURE
FACULTY OF DESIGN AND ARCHITECTURE UNIVERSITI PUTRA MALAYSIA
LAr. ABD AZIZ OTHMAN
9. The slope distance between two points is 29.705M and the difference in elevation between the points is 3.658M. Compute the horizontal distance.
c2 = a
2 + b
2
29.7052 = 3.658
2 +b
2
b2 =
29.705
2 – 3.658
2
b2 =
882.39 – 13.38
b2 =
869.01
b = Ѵ869.01
= 29.479M
10. The distance between two elevations 98.2M and 107.6M are 40.0M apart, what is the slope
percentage for the land between them? Slope (%) = (DE/L) X 100 = [(107.6 – 98.2)/40.0] X 100 = (9.4/40) X 100 = 23.5%
11. A 2% slope is required to drain a grass playing field which is 100M long. The lower end of the
field has an elevation of 62.5M. What is the elevation at the other end of the field? G = (DE/L)
DE = G X L = 0.02 X 100 = 2M Elevation at other end = 62.5 + 2.0 = 64.5M
12. A concrete walk has an elevation of 126.3M at one end. At a 5% slope, how far away will
elevation 127.0 be? G = (DE/L) L = (DE/G) = (127.0 – 126.3)/0.05 = 0.7/0.05 = 14.0M
13. An area of land has a slope at 5:1. If the horizontal length is 48.5M, what is the change in elevation from one end to the other end?
G = (DE/L) DE = G X L = 5:1 @ 0.02 X 48.5 = 9.7M
Use the following contour plan for the questions 14-15.
42.0 43.5 43.0 42.5
SCALE 1:100
+ A
+ B
+ C
a
b
c
1.35 1.65
Slope A = (DE/L)X100 = (0.5/3.0)X100 = 16.7% DE for A = (1.35/3.0)X0.5 = 0.225M Elev. A = 42.5 – 0.225
= 42.275M
Slope B = (DE/L)X100 = (0.5/3.8)X100 = 13.16% DE for B = (2.85/3.8)X0.5 = 0.375M Elev. B = 43.0 – 0.375 = 42.625M
2.85 0.95
Slope C = (DE/L)X100 = (0.5/5.0)X100 =10.0% DE for C = (2.30/5.0)X0.5 = 0.23M Elev. C = 43.5 – 0.23 = 43.27M
2.70
2.30
MASTER OF LANDSCAPE ARCHITECTURE DEPARTMENT OF LANDSCAPE ARCHITECTURE
FACULTY OF DESIGN AND ARCHITECTURE UNIVERSITI PUTRA MALAYSIA
LAr. ABD AZIZ OTHMAN
14. Compute the slope percentage at point A, B & C.
15. Determine the spot elevation for A, B & C.
16. Calculate the spot elevation (i.e., missing information marked by ‘?’) from the information
provided in each of the diagram below (arrows indicate direction of descending slope). Each
