Engineering Surveying 3rd Stage · Engineering Surveying 3rd Stage Areas . Units of Area •1...
Transcript of Engineering Surveying 3rd Stage · Engineering Surveying 3rd Stage Areas . Units of Area •1...
![Page 2: Engineering Surveying 3rd Stage · Engineering Surveying 3rd Stage Areas . Units of Area •1 hectare = 10,000 m2 •1 donum=2500 m2 •1 olk= 100 m2 •1 hectare = 4 donum=100 olk](https://reader035.fdocuments.us/reader035/viewer/2022081602/6130a1061ecc5158694437fc/html5/thumbnails/2.jpg)
Units of Area
• 1 hectare = 10,000 m2
• 1 donum=2500 m2
• 1 olk= 100 m2
• 1 hectare = 4 donum=100 olk
• 1 donum=25 olk
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Units of Area
• Example :
• 25360.22m2
• 25360.22/10,000=2.536022 hectare 2hac
• 5360.22/2500=2.144088 donum 2 donum
• 360.22/100=3.6022 olk 3 olk
• 60.22 m2
• 2hac, 2 donum, 3 olk , 60.22 m2
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Area
Uniform Area
Non-Uniform Area
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Uniform Area\ Triangle
Triangle
Right Angle Triangle
Inclined Triangle
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Uniform Area\ Triangle
• Area of Right Angle Triangle
• A= 1
2( ab*ac)
• Ac=h
h :ٌمثل الارتفاع العمودي على الماعدة او امتدادها
h
c
b a
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Uniform Area\ Triangle
• Area of Inclined Triangle
• A= 1
2(ab*h)
• A= 1
2( ab*h1)
• A= 1
2( cb*h2)
h :ٌمثل الارتفاع العمودي على الماعدة او امتدادها
h
a b
c
a b
c
h2
h1
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Area of Inclined Triangle
• لانون نصف المحٌط
• S=𝑎𝑏 + 𝑏𝑐 +(𝑐𝑎)
2
• A= 𝑠(𝑠 − 𝑎𝑏 )(𝑠 − 𝑏𝑐 )(𝑠 − 𝑐𝑎 )
(𝑏𝑐)
sin 𝐴=
(𝑐𝑎)
sin 𝐵=
(𝑎𝑏)
sin 𝐶
(𝑎𝑏)2 = (𝑏𝑐)2+ (𝑐𝑎)2- 2 (bc)(ca)cos(C)
(𝑏𝑐)2 = (𝑎𝑏)2+ (𝑐𝑎)2- 2 (ab)(ca)cos(A)
(𝑐𝑎)2 = (𝑎𝑏)2+ (𝑏𝑐)2- 2 (ab)(bc )cos(B)
A B
C
a b
c
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Uniform Area\ Circle
𝐴 = 𝜋𝑟2 𝑟
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Uniform Area\ Ellipse
• 𝐴 = 𝜋(𝑎)(𝑏)
a b
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Non-Uniform Area
Divide Area into
Triangle
E
A
B C
D
F G
C D
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Non-Uniform Area
Setting out offsets on a line at regular interval
Setting out offsets on a line at irregular interval
Setting out offsets on a line
D C
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Non-Uniform Area
• Setting out offsets on a line at regular interval
1. Average offsets formula باستخدام هذه الصٌغة ٌتم حساب
المساحة بصورة تمرٌبٌة
• 𝐴 =ℎ1+ ℎ2+ℎ3+⋯+ℎ𝑛
𝑛∗ 𝑛 − 1 𝑑
C D d d d d d
h1
h2 h3 h4 h5
h6
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Non-Uniform Area
2. Trapezoidal Rule باستخدام هذه االماعدة ٌتم حساب
المساحة بصورة دلٌمة لكن من شروطها ان تكون حدود مستمٌمة خطوط المطعة
• A=d[ℎ1+ℎ𝑛
2+ ℎ2 + ℎ3 + ⋯ + ℎ(𝑛 − 1)]
C D d d d d d d d
h1
h2 h3
h4 h5 h7
h6
h8
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Non-Uniform Area
3. Simpson’s Rule باستخدام هذه االماعدة ٌتم حساب المساحة بصورة غٌر منتظمة لكن من شروطها ان تكون حدود المطعة ادق من سابماتها
او منحنٌات وتطبك لعدد فردي من الاعمدةوعلى شكل الواس
𝐴 =𝑑
3[ℎ1+ ℎ𝑛 + 4(∑ℎ 𝑒𝑣𝑒𝑛 ) + 2(∑ℎ 𝑜𝑑𝑑 )]
عند جمع الاعمدة الفردٌة لاٌتم جمع العمود الاول والاخٌر : ملاحظة مهمة وفً الشكل السابك المثلث الاخٌر ٌحسب على حده لان المانون ٌطبك على
عدد فردي من الاعمده
C D d d d d d
h1
h2 h3 h4 h5
h6
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Non-Uniform Area
• Setting out offsets on a line at irregular interval
1. Using Special law
• 2A=[d1(ℎ1 − ℎ3)] +[(d1+d2)(ℎ2 − ℎ4)]
+ *(d1+d2+d3)(ℎ3 − ℎ5)] + *(d1+d2+d3+d4)(ℎ4 − ℎ6)]+ *(d1+d2+d3+d4+d5)(ℎ5 + ℎ6)]
• A=l2A/2l
C D d4 d5 d2 d3 d1
h1
h2 h3 h4 h5
h6
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Non-Uniform Area
2. Using Coordinate
C D d4 d5 d2 d3 d1
h1
h2 h3 h4 h5
h6
(E1,N1)
(E5,N5) (E4,N4)
(E3,N3) (E2,N2)
(E6,N6) E
N
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Non-Uniform Area N E
N1 E1
N2 E2
N3 E3
N4 E4
N5 E5
N6 E6
N1 E1
2A= (N1E2+ N2E3+ N3E4+ N4E5+ N5E6+ N6E1)-(N2E1+ N3E2+ N4E3+ N5E4+ N6E5+ N1E6)
A=l2A/2l
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Non-Uniform Area
3. Using Double – Meridian – Distance Method (D.M.D Method)
توضح الطرٌمة مباشرةً بأستخدام جدول على المثال التالً
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Example
5 5
5
5
10
15
8 6 5 4 9 8
4
8 5 5
9 10
8 9 7
11
Compute the area of he following figure (all dimensions are in meters)
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A1 A2
A3
A4
A5
5 5
5
5
10
15
8 6 5 4 9 8
4
8 5 5
9 10
8 9 7
11
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Solution • 𝐴 = 𝜋𝑟2
𝐴1 =𝜋 15 2
4= 176.71 m2
𝑠 =15 + 25 +(30)
2 = 35 m
𝐴2 = 35(35 − (15))(35 − (25))(35 − (30)) = 187.08 m2
A= 1
2( ab*ac)
A3= 1
2( 10*5)=25 m2
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A4
5 5 5
d= 5
A=d[ℎ1+ℎ5
2+ ℎ2 + ℎ3 + ℎ4]
A4=5[0+5
2+ 4 + 8 + 5] = 97.5m2
![Page 24: Engineering Surveying 3rd Stage · Engineering Surveying 3rd Stage Areas . Units of Area •1 hectare = 10,000 m2 •1 donum=2500 m2 •1 olk= 100 m2 •1 hectare = 4 donum=100 olk](https://reader035.fdocuments.us/reader035/viewer/2022081602/6130a1061ecc5158694437fc/html5/thumbnails/24.jpg)
A5
d1= 8 d2= 6 d3=5 d4=4 d5=9 d6=8
h1= 9 h2= 10
h3= 8 h4= 9 h5= 7 h6= 11
h7= 0
2A=[d1(ℎ1 − ℎ3)] +[(d1+d2)(ℎ2 − ℎ4)] + *(d1+d2+d3)(ℎ3 − ℎ5)]+ *(d1+d2+d3+d4)(ℎ4 − ℎ6)] + *(d1+d2+d3+d4+d5)(ℎ5 − ℎ7)] *(d1+d2+d3+d4+d5+d6)(ℎ6 + ℎ7)] 2A5=[8(9 − 8)] +[(14)(10 − 9)] + *(19)(8 − 7)] + *(23)(9 − 11)]+ *(32)(7 − 0)] *(40)(11 + 0)]=659 m2 A5=329.5 m2
A Total= A1+A2+A3+A4+A5 A Total= 815.79 m2 8 olk 15.79 m2
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(D.M.D Method)
A5
d1= 8 d2= 6 d3=5 d4=4 d5=9 d6=8
h1= 9 h2= 10
h3= 8 h4= 9 h5= 7 h6= 11
A B
C D E F G H
B
C
Latitude
Departure
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(D.M.D Method)
side Dep. D.M.D Lat. 2A
AB +40 0
BC -8 -11
CD -9 +4
DE -4 -2
EF -5 +1
FG -6 -2
GH -8 +1
HA 0 +9
∑ 0.00 0.00
A5
d1= 8 d2= 6 d3=5 d4=4 d5=9 d6=8
h1= 9 h2= 10
h3= 8 h4= 9 h5= 7 h6= 11
A B
C D E F G H
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(D.M.D Method) side Dep
. D.M.D Lat. 2A
AB +40 +40 0 0
BC -8 +72 -11 -792
CD -9 +55 +4 +220
DE -4 +42 -2 -84
EF -5 +33 +1 +33
FG -6 +22 -2 -44
GH -8 +8 +1 +8
HA 0 0 +9 0
∑ 0.00 0.00 -659
+ +
= + +
=
* =
A=l2A/2l A5=329.5 m2
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5 5
5
5
10
15
8 6 5 4 9 8
4
8
5
5
9 10
8 9 7
11
A Total= A1+A2-A3-A4+A5 A Total= 570.79 m2 5 olk 70.79 m2
A4
A3 A1
A2
A5