alexandramutch.files.wordpress.com€¦ · web viewthere is no way of calculating the exact...
Post on 23-Aug-2020
7 Views
Preview:
TRANSCRIPT
Calculus II
Year 12 Mathematics C | Mr Kirkman
Alexandra Mutch
Due: 20 August 2013
Alexandra Mutch
Table of Contents
Introduction...........................................................................................................................................................................2
Section 1: Use of Simpson’s Rule..................................................................................................................................3
Task 1................................................................................................................................................................................... 3
Section 2: Precision of Simpson’s Rule......................................................................................................................4
Task 2................................................................................................................................................................................... 4
Task 3................................................................................................................................................................................... 7
Task 4................................................................................................................................................................................... 9
Task 5................................................................................................................................................................................ 11
Task 6................................................................................................................................................................................ 13
Task 7................................................................................................................................................................................ 14
Section 3: Application of Simpson’s Rule...............................................................................................................17
Task 8................................................................................................................................................................................ 17
Task 9................................................................................................................................................................................ 25
1
Alexandra Mutch
Introduction
Simpson’s Rule is an approximation for area. It can be applied to area under curves to find the definite integral. It is commonly used by naval architects to calculate volume of centre of mass of ships, however, can be used to approximate any definite integral.
Johannes Kepler had used similar formulas more than 100 earlier, but Simpson’s Rule is credited to Thomas Simpson.
Simpson’s Rule:
A≑ w3 [E+4MO+2ME ]
Where A is approximated area, w is the width of strips, E is the sum of lengths of end strips, MO
is the sum of lengths of odd middle strips and ME is the sum of lengths of even middle strips.
A restriction of Simpson’s Rule is that the number of strips must be an even number.
2
Alexandra Mutch
Section 1: Use of Simpson’s Rule
Task 1
Approximate ∫sin (x2 ) . dxbetween 0 and π4 using 6 strips with Simpson’s Rule.
n = 6 strips
Strip width, w=b−an
w=
π4−0
6
w= π24 units
Sum of end strips, E = E1 + E2
E = 0 + 0.57847
E = 0.57847
Sum of odd strips, MO = M1 + M3 + M5
MO = 0.01713 + 0.15360 + 0.41539
MO = 0.58612
Sum of even strips, ME = M2 + M4
ME = 0.06849 + 0.27073
ME = 0.33922
Simpson’s Rule:
A≑ w3 [E+4MO+2M E ]
A≑
π243
[0.57847+4×0.58612+2×0.33922 ]
3
Strip x y
E1 0 0
M1π24 0.01713
M2π12 0.06849
M3π8 0.15360
M4π6 0.27073
M55π24 0.41539
E2π4 0.57847
Alexandra Mutch
A≑ π72
×3.60139
A≑0.15714 sq .units
Section 2: Precision of Simpson’s Rule
Task 2Determine the area under the line 4 x− y−2=0 where 2≤x ≤11.
(a) Basic area formula4 x− y−2=0
y=4 x−2
Fig. 1 – Diagram of Area Under 4x-y-2=0
*Not to scale
The diagram above shows that the total area under the line y=4 x−2 can be calculated by separating the area into a triangle (A1) and a rectangle (A2).
Area of A1
Area of a triangle, A1=12×b×h
A1=12×(11−2)×(42−6)
A1=12×9×36
A1=162 sq .unit s
4
42
6A2
A1
y
x112
y=4 x−2
Alexandra Mutch
Area of A2
Area of a rectangle, A2=L×W
A2=(11−2)×(6−0)
A2=9×6
A2=54 sq .units
Total Area
A1+A2=162+54
Atot=216 sq .units
(b) Definite integral
∫2
11
(4 x−2 ) . dx
¿ [2 x2−2x+c ]211
¿ [2(11)2−2(11)+c ]−[2(2)2−2(2)+c ]
¿ [220 ]−[4 ]
¿216 sq .units
(c) Simpson’s Rule
Approximate ∫(4 x−2) . dxbetween 2 and 11 using 10 strips
n = 10 strips
Strip width, w=b−an
w=11−210
w=0.9units
Sum of end strips, E = E1 + E2
E = 6 + 42
5
Strip x y
E1 2.0 6.0
M1 2.9 9.6
M2 3.8 13.2
M3 4.7 16.8
M4 5.6 20.4
M5 6.5 24.0
M6 7.4 27.6
M7 8.3 31.2
M8 9.2 34.8
M9 10.1 38.4
E2 11.0 42
Alexandra Mutch
E = 48
Sum of odd strips, MO = M1 + M3 + … + M9
MO = 9.6 + 16.8 + … + 38.4
MO = 120
Sum of even strips, ME = M2 + M4 + M6 + M8
ME = 13.2 + 20.4 + 27.6 + 34.8
ME = 96
Simpson’s Rule:
A≑ w3 [E+4MO+2M E ]
A≑ 0.93
[48+4×120+2×96 ]
A≑0.3×720
A≑216 sq .units
6
Alexandra Mutch
Task 3 y=x2+5 −1≤x ≤1
(a) Definite integral
∫−1
1
(x2+5 ) . dx
¿ [ x33 +5 x+c ]−1
1
¿ [(1)33 +5 (1)+c ]−[ (−1)33 +5 (−1)+c]¿ [5 13 ]−[−5 13 ]¿10 2
3sq .units
(c) Simpson’s Rule
Approximate ∫(x2+5) . dxbetween -1 and 1 using 10 strips
n = 10 strips
Strip width, w=b−an
w=1−(−1)10
w=0.2units
Sum of end strips, E = E1 + E2
E = 6 + 6
E = 12
Sum of odd strips, MO = M1 + M3 + … + M9
7
Strip x y
E1 -1.0 6.00
M1 -0.8 5.64
M2 -0.6 5.36
M3 -0.4 5.16
M4 -0.2 5.04
M5 0.0 5.00
M6 0.2 5.04
M7 0.4 5.16
M8 0.6 5.36
M9 0.8 5.64
E2 1.0 6.00
Alexandra Mutch
MO = 5.64 + 5.16 + … + 5.64
MO = 26.6
Sum of even strips, ME = M2 + M4 + M6 + M8
ME = 5.36 + 5.04 + 5.04 + 5.36
ME = 20.8
Simpson’s Rule:
A≑ w3 [E+4MO+2M E ]
A≑ 0.23
[12+4×26.6+2×20.8 ]
A≑ 115
×160
A≑10 23sq .units
8
Alexandra Mutch
Task 4y=(x+1)(x−1)(x−2) −1≤x ≤1
y=( x2−1 ) ( x−2 )
y=x3−2 x2−x+2
(a) Definite integral
∫−1
1
(x3−2 x2−x+2 ) .dx
¿ [ x44 −2x3
3 −x2
2 +2x+c ]−1
1
¿ [(1)44 −2 (1 )3
3− (1 )2
2+2(1)+c ]−[ (−1)44 −2 (−1 )3
3− (−1 )2
2+2(−1)+c]
¿ [1 112 ]−[−1 712 ]¿2 23sq .units
(c) Simpson’s Rule
Approximate ∫(x3−2 x2−x+2) . dxbetween -1 and 1 using 10 strips
n = 10 strips
Strip width, w=b−an
w=1−(−1)10
w=0.2units
Sum of end strips, E = E1 + E2
E = 0 + 0
E = 0
9
Strip x y
E1 -1.0 0.000
M1 -0.8 1.008
M2 -0.6 1.664
M3 -0.4 2.016
M4 -0.2 2.112
M5 0.0 2.000
M6 0.2 1.728
M7 0.4 1.344
M8 0.6 0.896
M9 0.8 0.432
E2 1.0 0.000
Alexandra Mutch
Sum of odd strips, MO = M1 + M3 + … + M9
MO = 1.008 + 2.016 + … + 0.432
MO = 6.8
Sum of even strips, ME = M2 + M4 + M6 + M8
ME = 1.664 + 2.112 + 1.728 + 0.896
ME = 6.4
Simpson’s Rule:
A≑ w3 [E+4MO+2M E ]
A≑ 0.23
[0+4×6.8+2×6.4 ]
A≑ 115
×40
A≑2 23sq .units
10
Alexandra Mutch
Task 5y= (x−2 ) ( x+1 )2(x−4) −1≤x ≤2
y= (x−2 )(x2+2 x+1)(x−4)
y=(x3−3 x−2)(x−4 )
y=x4−4 x3−3 x2+10 x+8
(a) Definite integral
∫−1
2
(x4−4 x3−3 x2+10 x+8 ) .dx
¿ [ x55 −x4−x3+5x2+8 x+c ]−1
2
¿ [(2)55 −(2 )4−(2 )3+5 (2 )2+8 (2)+c]−[(−1)55 −(−1 )4−(−1 )3+5 (−1 )2+8(−1)+c ]¿ [18.4 ]− [−3.2 ]
¿21.6 sq .units
(c) Simpson’s Rule
Approximate ∫ (x4−4 x3−3 x2+10 x+8 ) . dxbetween -1 and 2
using 10 strips
n = 10 strips
Strip width, w=b−an
w=2−(−1)10
w=0.3units
Sum of end strips, E = E1 + E2
E = 0 + 0
E = 0
11
Strip x y
E1 -1.0 0.0000
M1 -0.7 1.1421
M2 -0.4 3.8016
M3 -0.1 6.9741
M4 0.2 9.8496
M5 0.5 11.8125
M6 0.8 12.4416
M7 1.1 11.5100
M8 1.4 8.9856
M9 1.7 5.0301
E2 2.0 0.0000
Alexandra Mutch
Sum of odd strips, MO = M1 + M3 + … + M9
MO = 1.1421 + 6.9741 + … + 5.0301
MO = 36.4688
Sum of even strips, ME = M2 + M4 + M6 + M8
ME = 3.8016 + 9.8496 + 12.4416 + 8.9856
ME = 35.0784
Simpson’s Rule:
A≑ w3 [E+4MO+2M E ]
A≑ 0.33
[0+4×36.4688+2×35.0784 ]
A≑ 110
×216.032
A≑21.6032 sq .units
12
Alexandra Mutch
Task 6
The area under the functions in tasks 2 – 5 were approximated using Simpson’s Rule with significant increases in the number of strips from 10 to 100 and 1,000. The table below shows the calculated area and other values.
Task Degree a b n w E Mo Me A
2 x1 2 1110 0.9 48 120 96 216
100 0.09 48 1,200 1,176 2161,000 0.009 48 12,000 11,976 216
3 x2 -1 110 0.2 12 26.6 20.8 10.666667
100 0.02 12 266.66 260.68 10.6666671,000 0.002 12 2,666.67 2,660.67 10.666667
4 x3 -1 110 0.2 0 6.8 6.4 2.6666667
100 0.02 0 66.68 66.64 2.66666671,000 0.002 0 666.668 666.664 2.6666667
5 x4 -1 210 0.3 0 36.4693 35.0788 21.60348
100 0.03 0 360.045 359.91 21.61,000 0.003 0 3,600.00 3,599.99 21.6
The sum(seq(…)) function on graphics calculator was used to calculate the MO odd and ME
values.
For polynomials of degree 3 of less, the approximation was equal to the definite integral, no matter what the number of strips was. This shows that, for the given numbers of strips, the approximation was always 100% accurate for polynomials of degree 3 or less.
For the polynomial of degree 4, the approximated area using 10 strips had an error of approximately 0.016%. This is a relatively small error percentage, but was the only approximation that was not equal to the definite integral. When the number of strips was increased to 100, the approximation was equal to the definite integral. This indicates that the accuracy of the approximation increases as the number of strips increases with polynomials of degree 4 or higher.
This means that Simpson’s Rule is accurate for approximating area under curves of polynomial degree 3 or less. For approximating area under curves of polynomial degree 4 or more, a greater number of strips should be used for improved accuracy.
13
Alexandra Mutch
Task 7The error involved in using Simpson’s Rule varies as the fourth power of the width of strips. For
example, when the strip width is halved, the error is reduced by a factor of 116 . The validity of
this statement will be tested in this task.
Definite Integral
I=∫2
6
(7 x4−3 x3 ) . dx
I=[7 x55 −3 x4
4 +c ]2
6
I=[7(6)55 −3(6)4
4+c ]−[ 7(2)55 −3 (2)
4
4+c ]
I¿ [10,886.4−972 ]−[44.8−12 ]
I=9,881.6 sq .units
Simpson’s Rule – 2 Stripsn = 2 strips
Strip width, w=b−an
w=6−22
w=2 units
Sum of end strips, E = E1 + E2
E = 88 + 8,424
E = 8,512
Sum of odd strips, MO = M1
MO = 1,600
14
Strip x yE1 2 88M1 4 1,600E2 6 8,424
Alexandra Mutch
Sum of even strips, No even strips, therefore ME = 0
Simpson’s Rule:
I 2≑w3 [E+4MO+2M E ]
I 2≑23
[8,512+4×1,600+2×0 ]
I 2≑23×14,912
I 2≑9,94113sq .units
Simpson’s Rule – 4 Stripsn = 4 strips
Strip width, w=b−an
w=6−24
w=1 unit
Sum of end strips, E = E1 + E2
E = 88 + 8,424
E = 8,512
Sum of odd strips, MO = M1 + M3
MO = 486 + 4,000
MO = 4,486
Sum of even strips, ME = M2
ME = 1,600
15
Strip x yE1 2 88M1 3 486M2 4 1,600M3 5 4,000E2 6 8,424
Alexandra Mutch
Simpson’s Rule:
I 4≑w3 [E+4MO+2ME ]
I 4≑13
[8,512+4×4,486+2×1,600 ]
I 4≑13×29,656
I 4≑9,88513sq .units
Error
%Error=( Approximated areaActual area−1)×100
Area:
I = 9,881.6, I2 = 9,941 13 , I4 = 9,885 13
2 Strips
%Error I 2=( 9,941 139,881.6−1)×100
%Error I 2≈350579
%
4 Strips
%Error I 4=( 9,885 139,881.6−1)×100
%Error I 4=1754,632
%
When the strip width was halved (I2 to I4), the I4 proportion error of I2:
¿%Error I 4%Error I 2
¿( 1754,632 )( 350579 )
16
Alexandra Mutch
¿ 116
DiscussionTherefore the original statement that the error of the approximation varies as the fourth power of the width of strips is correct for this case.
Section 3: Application of Simpson’s Rule
Task 8The selected number of strips for this task is n = 4. This will be the simplest for calculations while including odd and even middle strips, along with end strips. 4 strips also satisfies the restriction that the number of strips must be even.
y = x
Definite Integral
∫c
d
x .dx
¿ [ x22 +k ]c
d
¿ [d22 +k ]−[ c22 +k ]¿ d
2−c2
2
Simpson’s Rulen = 4 strips
As the function is y = x, the y value of each strip will be equal to the x value.
The middle strips were calculated by adding one strip width each time. I.e. M1 = c + w, M2 = c + 2w, etc.
17
Strip x yE1 c c
M1 c+ d−c4
c+ d−c4
M2 c+ d−c2
c+ d−c2
M3 c+ 3 (d−c )4
c+ 3 (d−c )4
E2 d d
Alexandra Mutch
Strip width, w=b−an
w=d−c4
Sum of end strips, E = E1 + E2
E = c+d
Sum of odd strips, MO = M1 + M3
MO = (c+ d−c4 )+(c+ 3 (d−c )
4 )MO = 2c+(d−c)
MO = c+d
Sum of even strips, ME = M2
ME = c+d−c2
Simpson’s Rule:
A≑ w3 [E+4MO+2M E ]
A≑( d−c4 )3 [(c+d)+4×(c+d)+2×(c+ d−c
2 )]A≑ d−c
12[c+d+4c+4 d+2c+d−c ]
A≑ d−c12
[6 c+6d ]
A≑ c (d−c )2
+d (d−c )2
A≑ cd−c2
2+ d
2−cd2
18
Alexandra Mutch
A≑ d2−c2
2
19
Alexandra Mutch
y = x2
Definite Integral
∫c
d
x2 . dx
¿ [ x33 +k ]c
d
¿ [d33 +k ]−[ c33 +k ]¿ d
3−c3
3
Simpson’s Rule
n = 4 strips
Strip width, w=b−an
w=d−c4
y-values of strips
M1, M 1=(c+ d−c4 )
2
M 1=c2+2( cd−c2
4 )+ (d−c )2
42
M 1=c2+ cd−c2
2+ d
2−2cd+c2
16
M 1=16c2
16+ 8cd−8c
2
16+ d
2−2cd+c2
16
20
Strip x yE1 c c2
M1 c+ d−c4
9c2+d2+6cd16
M2 c+ d−c2
c2+d2+2cd4
M3 c+ 3 (d−c )4
c2+9d2+6cd16
E2 d d2
Alexandra Mutch
M 1=9c2+d2+6cd
16
M2, M 2=(c+ d−c2 )
2
M 2=c2+2( cd−c2
2 )+ (d−c )2
22
M 2=c2+cd−c2+ d2−2cd+c2
4
M 2=4 cd4
+ d2−2cd+c2
4
M 2=c2+d2+2cd
4
M3, M 3=(c+3 (d−c )4 )
2
M 3=c2+2(3 cd−3c24 )+ (3d−3c )2
42
M 3=c2+ 3cd−3 c2
2+
(3d−3c )2
42
M 3=16c2
16+ 24cd−24c
2
16+ 9d
2−18cd+9c2
16
M 3=c2+9d2+6cd
16
Sum of end strips, E = E1 + E2
E = c2+d2
Sum of odd strips, MO = M1 + M3
21
Alexandra Mutch
MO = ( 9c2+d2+6cd16 )+( c2+9d2+6cd16 )MO = 10c
2+10d2+12cd16
MO = 5c2+5d2+6cd
8
Sum of even strips, ME = M2
ME = c2+d2+2cd
4
Simpson’s Rule:
A≑ w3 [E+4MO+2M E ]
A≑( d−c4 )3 [(c2+d2)+4×( 5c
2+5d2+6 cd8 )+2×( c
2+d2+2 cd4 )]
A≑ d−c12 [ 2c2+2d22
+5c2+5d2+6cd
2+c2+d2+2cd
2 ]A≑ d−c
12 [ 8c2+8d2+8cd2 ]A≑ d−c
12[ 4c2+4 d2+4cd ]
A≑ c2d−c3
3+ d
3−c d2
3+ c d
2+c2d3
A≑ d3−c3
3
22
Alexandra Mutch
y = x3
Definite Integral
∫c
d
x3 . dx
¿ [ x44 +k ]c
d
¿ [ d44 +k ]−[ c44 +k ]¿ d
4−c4
4
Simpson’s Rulen = 4 strips
Strip width, w=b−an
w=d−c4
y-values of strips
M1, M 1=(c+ d−c4 )
3
M 1=( 9 c2+d2+6cd16 )×(c+ d−c4 )
M 1=9c3+cd2+6c2d
16+ 9c
2d−9 c3+d3−cd2+6c d2−6c2d64
M 1=36c3+4cd2+24 c2d
64+−9c3+d3−cd2+3c2d+5c d2
64
M 1=27c3+d3+27c2d+9c d2
64
23
Strip x yE1 c c3
M1 c+ d−c4
27c3+d3+27c2d+9c d2
64
M2 c+ d−c2
c3+d3+3 c2d+3c d2
8
M3 c+ 3 (d−c )4
c3+27d3+9c2d+27c d2
64E2 d d3
Alexandra Mutch
M2, M 2=(c+ d−c2 )
3
M 2=( c2+d2+2cd4 )×(c+ d−c2 )
M 2=c3+dc2+2c2d
4+ c
2d−c3+d3−c d2+2cd2−2c2d8
M 2=2c3+2dc2+4 c2d
8+−c3+d3−c2d+c d2
8
M 2=c3+d3+3c2d+3 c d2
8
M3, M 3=(c+3 (d−c )4 )
3
M 3=( c2+9 d2+6cd16 )×(c+ 3 (d−c )4 )
M 3=c3+9c d2+6 c2d
16+ 3c
2d−3 c3+27d3−27cd2+18c d2−18c2d64
M 3=4 c3+24 cd2+36c2d
64+−3c3+27d3−15c2d2−9c d2
64
M 3=c3+27d3+9c2d+27 cd2
64
Sum of end strips, E = E1 + E2
E = c3+d3
Sum of odd strips, MO = M1 + M3
24
Alexandra Mutch
MO = ( 27c3+d3+27c2d+9cd2
64 )+( c3+27 d3+9c2d+27c d2
64 )MO = 28 c
3+28d3+36c2d+36cd2
64
MO = 7c3+7 d3+9c2d+9c d2
16
Sum of even strips, ME = M2
ME = c3+d3+3 c2d+3c d2
8
Simpson’s Rule:
A≑ w3 [E+4MO+2M E ]
A≑( d−c4 )3 [(c3+d3)+4×( 7c
3+7d3+9c2d+9 c d2
16 )+2×( c3+d3+3c2d+3c d2
8 )]A≑ d−c
12 [c3+d3+ 7c3+7d3+9c2d+9 cd24+c3+d3+3c2d+3cd2
4 ]A≑ d−c
12 [ 4 c3+4 d34+7 c3+7 d3+9c2d+9c d2
4+c3+d3+3c2d+3c d2
4 ]A≑ d−c
12 [ 12c3+12d3+12d+12c d24 ]A≑ d−c
12[3c3+3d3+3d+3cd2 ]
A≑ c3d−c4+d4−c d3+c2d2−c3d+c d3−c2d2
4
A≑ d4−c4
4
25
Alexandra Mutch
DiscussionThis shows that Simpson’s Rule is 100% accurate for polynomials of degree 3 or less because the general form for the approximated area was equal to the general form for the definite integral in each of the three cases.
26
Alexandra Mutch
Task 9
It is proposed that a level road be constructed roughly North-South for the entire length of the map shown below. Justification of the selection of the road path will be outlined below along with mathematical reasoning for the optimum height of the road to minimise filling and removal costs.
Fig. 2 - Map of Area for Proposed Road
There is no way of calculating the exact height of the mountain between the topographical lines. Therefore it is assumed that the gradient of each of the areas between topographical lines is constant. The actual slope of the mountain would not be constant between the topographical lines, but rough and uneven, so any further calculations should be used as estimation only.
27
Alexandra Mutch
Fig. 3 - Selected Road Path
The road path (indicated in Fig. 3 in blue) was selected because it does not intersect the Dark Sea and does not go over the peaks of the mountain. It also follows relatively gradual mountain slopes which means dirt filling/removal would be easier (and probably cost less) and there would be less chance of landslide etc. during the dirt moving process.
The blue path would be a better choice than the red path as the red one goes right over the peak of the mountain. This means more dirt would need to be moved in order to create a level road which would be less cost effective. In addition, the red path crosses the river at less of an angle which means it will cover a greater area of the river than the blue path. This means the error (due to the requirement of a bridge) will be greater and any calculations would be less accurate.
The optimum height for the selected road will be determined by:
Calculating the cross sectional area underneath the selected road path. Multiplying the area by one ‘road width’ to find the total volume of dirt on the road path Calculating the point of equilibrium between the height of peaks and valleys of the
mountain
It is assumed that any rivers crossed by the road do not affect the height of the mountain at their locations. In reality a bridge would be built over the water or the river would be excavated further downward and out of the way of the road. This would mean less dirt would be required to fill the valley at the location of the river. Therefore calculations of the volume of dirt to be moved will be slightly more than the actual volume required.
28
Alexandra Mutch
Cross Sectional AreaFirst the height of the mountain at particular distances along the road (from the Northern border) will be measured and graphed. The distance from the Northern border of the topographical lined was measured to find the height of the mountain at each point.
Height of Mountain
Measured
Distance (mm)
Converted
Distance (km)
Height (km)
0 0 0.27.5 0.75 0.2
14.0 1.40 0.330.0 3.00 0.334.0 3.40 0.262.5 6.25 0.176.0 7.60 0.182.0 8.20 0.2
107.5 10.75 0.3131.0 13.10 0.3
It is assumed that the height of the mountain at the Northern and Southern edges of the map is equal to the nearest measurement. For example, the nearest measurement to the Northern border is height = 0.3km at 0.75km from the border. Therefore it is assumed that the height of the mountain at the Northern border is 0.3km and the height at the Southern border is 0.3km.
0 2 4 6 8 10 12 140
0.05
0.1
0.15
0.2
0.25
0.3
0.35
0.4
Graph 1 - Height vs Distance - Cross Section of Path
Distance from Northern Border (km)
Hei
ght
abov
e Se
a Le
vel (
km
)
29
Alexandra Mutch
In order to calculate the total cross sectional area underneath the road path, it has been sectioned into separate rectangles and triangles (shown in Fig. 4) which will be calculated separately and then summated.
0 2 4 6 8 10 12 140
0.05
0.1
0.15
0.2
0.25
0.3
0.35
0.4
Fig. 4 - Areas to be Calculated
A1
For areas 1 – 5, rectangle area calculations were used and for areas 6 – 10 triangle area calculations were used.
Calculation of A1
A=L×W
A=13.1×0.1
A=1.31 km2
The same process was used to find areas 2 – 5.
Calculation of A6
A=0.5b×h
A=0.5×(6.25−3.4)×0.1
A=0.1425k m2
The same process was used to find areas 7 – 10.
30
A2 A3
A4 A5
A6
A9
A8
A7
A
10
Alexandra Mutch
The areas were calculated to be:
SectionArea
(km2)A1 1.3100A2 0.3400A3 0.4900A4 0.1600A5 0.2350A6 0.1425A7 0.0300A8 0.0325A9 0.0200A10 0.1275Total 2.8875
The total cross sectional area under the road path was calculated to be 2.8875km2.
Volume of DirtIn order to calculate the volume of dirt under the selected road path, the cross sectional area underneath the path will be multiplied by one ‘road width’, or 11m. It is assumed that the road will consist of two lanes (one for each direction) each of width 3.5m and a 2m shoulder on either side of the lanes, totalling 11m.
It is assumed that the volume of dirt under the selected road path is a prism. Due to the small scale of the map, the height of the mountain one road width (0.11mm on the scaled diagram) from the original measurements would be the same height. The actual mountain would be uneven so calculations will be incorrect. However, due to the large size of the mountain relative to the road, this error is negligible.
Volume of dirt underneath road path:
V=Area×Width
V=2.8875×0.011
V=0.0317625 k m3
Height Optimisation The optimum height for the road is a height where all dirt taken off peaks can fill the valleys to create a flat, level road. This minimises dirt filling and removal costs.
31
Alexandra Mutch
It is assumed that there is no height restriction for the edges of the map. I.e. the height of the road does not have to match up with another road beyond the limits of the map.
Optimum Height Calculation
The volume of dirt must be moved to create a flat surface on the top. This will result in one rectangular prism of dirt or a rectangle of length 13.1km when the cross section is viewed. Therefore the optimum height will be the area divided by the length of the road.
Height= AreaLength
Height=2.887513.1
Height ≈0.2204 km
Therefore the road should be constructed at a height of approximately 220.4m above sea level to minimise filling and removal costs.
DiscussionDue to the assumptions made, the calculated cross sectional area, volume of dirt and optimum height are not completely accurate.
The assumption that the gradient of each of the areas between topographical lines is constant creates a large margin for error. The area between topographical lines has an error of ±50m
which is 16 of the largest height measurement used in calculations. If all of the assumed heights
of areas between topographical lines were out by +50m then this would add approximately 0.655km2 to the calculated cross sectional area or about 22.69%. This is a relatively large percentage of error and the calculations should be taken as a rough guide only.
32
top related