ﻣﺴﺎﺋﻞ ﻓﻰ اﻟﻤﺴﺎﺣﺔ اﻟﻤﺴﺘﻮﻳﺔ
ﻣﺴﺎﺋﻞ ﻓﻰ اﻟﻤﺴﺎﺣﺔ اﻟﻤﺴﺘﻮﻳﺔ -1ﻗﻴﺲ ﺧﻂ ﺑﺸﺮﻳﻂ ﻣﻦ اﻟﺘﻴﻞ ﻓﻜﺎن ﻃﻮﻟﻪ ) ، (93.75 mﻓﺈذا ﻋﻠ ﻢ أن اﻟﻄ ﻮل اﻹﺳ ﻤﻰ ﻟﻬ ﺬا اﻟ ﺸﺮﻳﻂ ه ﻮ ) ، (20.0 mواﻟﻄ ﻮل اﻟﺤﻘﻴﻘﻰ ﻟﻪ ) (20.01 mاﺣﺴﺐ اﻟﻄﻮل اﻟﺤﻘﻴﻘﻰ ﻟﻠﺨﻂ. اﻟﺤﻞ ﻳﺴﺘﺨﺪم اﻟﺸﺮﻳﻂ ﻟﻠﻘﻴﺎس اﻟﻤﺒﺎﺷﺮ ،وﻳﻤﻜ ﻦ أن ﻳﻜ ﻮن ﻣ ﻦ اﻟﻜﺘ ﺎن أو ﻣ ﻦ اﻟ ﺼﻠﺐ ،وﻟﻜﻨ ﻪ ﻳ ﺼﻌﺐ اﺳ ﺘﺨﺪاﻣﻪ ﻓ ﻰ ﺗﻴ ﺎرات اﻟﻬﻮاء اﻟﺸﺪﻳﺪة ﻟﺼﻌﻮﺑﺔ ﺷﺪﻩ أﻓﻘﻴ ًﺎ. اﻟﻄﻮل اﻟﺤﻘﻴﻘﻰ ﻟﻠﺨﻂ = اﻟﻄﻮل اﻟﻤﻘﺎس ×
اﻟﻄﻮل اﻟﺤﻘﻴﻘﻰ ﻟﻠﺸﺮﻳﻂ )أو اﻟﺠﻨﺰﻳﺮ( اﻟﻄﻮل اﻹﺳﻤﻰ ﻟﻠﺸﺮﻳﻂ )أو اﻟﺠﻨﺰﻳﺮ 20.01 = 93.7968 m * True length = 93.75 20.0
∗∗∗∗∗∗ -2ﻓﻘﺪت ﻋﻘﻠﺔ ﻣﻦ اﻟﻤﺘﺮ اﻟﺜﺎﻣﻦ ﻣﻦ اﻟﺠﻨﺰﻳﺮ ،وآﺎن ﻃﻮل اﻟﺨﻂ اﻟﻤﻘﺎس ) . (88.50 mاﺣﺴﺐ اﻟﻄﻮل اﻟﺤﻘﻴﻘﻰ ﻟﻠﺨﻂ. اﻟﺤﻞ اﻟﺠﻨﺰﻳﺮ ﻋﺒﺎرة ﻋﻦ ﻗ ﻀﺒﺎن رﻓﻴﻌ ﺔ ﻣ ﻦ اﻟﺤﺪﻳ ﺪ أو اﻟ ﺼﻠﺐ ﺗﺘ ﺼﻞ ﺑﺒﻌ ﻀﻬﺎ ﻋ ﻦ ﻃﺮﻳ ﻖ ﺣﻠﻘ ﺎت ،وﻳﻨﺘﻬ ﻰ ﻃﺮﻓ ﺎ اﻟﺠﻨﺰﻳ ﺮ ﺑﻤﻘﺒﻀﻴﻦ ﻣﻦ اﻟﻨﺤﺎس ،واﻟﻄﻮل اﻟﻤﻌﺮوف ﻟﻠﺠﻨﺰﻳﺮ هﻮ ) ،(20.0 mوﻃﻮل آﻞ ﻋﻘﻠﺔ هﻮ ) (20.0 cmﺑﻤﺎ ﻳﺘﺒﻌﻬﺎ ﻣﻦ ﺣﻠﻘﺎت. اﻟﻄﻮل اﻹﺳﻤﻰ ﻟﻠﺠﻨﺰﻳﺮ = ، 20.0 mاﻟﻄﻮل اﻟﺤﻘﻴﻘﻰ ﻟﻠﺠﻨﺰﻳﺮ = 19.80 m = 20.0 – 0.20
19.80 = 87.615 m 20.0
* true length = 88.50
∗∗∗∗∗∗ -3ﻗﻴ ﺴﺖ ﻣ ﺴﺎﺣﺔ أرض ﺑﺎﺳ ﺘﺨﺪام ﺷ ﺮﻳﻂ ﻃﻮﻟ ﻪ ) (30.0 mﻓﻜﺎﻧ ﺖ اﻟﻤ ﺴﺎﺣﺔ ) ،(656.0 m2ﻓ ﺈذا ﻋﻠ ﻢ أن اﻟﻄ ﻮل اﻟﺤﻘﻴﻘ ﻰ ﻟﻠﺸﺮﻳﻂ هﻮ ) (29.99 mﻓﺄوﺟﺪ اﻟﻤﺴﺎﺣﺔ اﻟﺤﻘﻴﻘﻴﺔ ﻟﻘﻄﻌﺔ اﻷرض. اﻟﺤﻞ 2
اﻟﻤﺴﺎﺣﺔ اﻟﺤﻘﻴﻘﻴﺔ = اﻟﻤﺴﺎﺣﺔ اﻟﻤﻘﺎﺳﺔ ×
)اﻟﻄﻮل اﻟﺤﻘﻴﻘﻰ ﻟﻠﺸﺮﻳﻂ أو اﻟﺠﻨﺰﻳﺮ( )اﻟﻄﻮل اﻹﺳﻤﻰ ﻟﻠﺸﺮﻳﻂ أو اﻟﺠﻨﺰﻳﺮ(
2
2
⎞ ⎛ 29.99 2 ⎜ * True area = 656.0 ⎟ = 655.562 m . ⎠ ⎝ 30.0
∗∗∗∗∗∗ -4اﺷﺘﻖ اﻟﻌﻼﻗﺔ ﻟﺘﺼﺤﻴﺢ اﻟﺨﻄﺄ اﻟﻨﺎﺗﺞ ﻋﻦ اﻟﻘﻴﺎس ﻋﻠﻰ أرض ﻣﺎﺋﻠﺔ ﺑﺎﻧﺘﻈﺎم ﺑﺪﻻﻟﺔ: أ -زاوﻳﺔ ﻣﻴﻞ اﻷرض θ ب -ﻓﺮق اﻟﻤﻨﺴﻮب ﺑﻴﻦ اﻟﻨﻘﻄﺘﻴﻦ. 1
Prepared By:/ Amr A. El-Sayed, Civil Eng Dept., El-Minia Univ., Eg.
ﻣﺴﺎﺋﻞ ﻓﻰ اﻟﻤﺴﺎﺣﺔ اﻟﻤﺴﺘﻮﻳﺔ اﻟﺤﻞ أ -ﺑﺪﻻﻟﺔ زاوﻳﺔ اﻟﻤﻴﻞ θ درﺟﺔ اﻹﻧﺤﺪار )أو ﻣﻴﻞ اﻷرض( 1:nﺣﻴﺚ 1اﻟﻤﺴﺎﻓﺔ اﻟﺮأﺳﻴﺔ n ،اﻟﻤﺴﺎﻓﺔ اﻷﻓﻘﻴﺔ.
H1 θ
V1
V2
L2
L1
H2 L 2.n 2 Or the Horizontal distance (H) = L . cos θ
= )E (Correction
ب -ﺑﺪﻻﻟﺔ ﻓﺮق اﻟﻤﻨﺴﻮب )اﻟﺒﻌﺪ اﻟﺮأﺳﻰ(: Data is the inclined distance (L), and level difference (V). Assume that L2 = V2 + (L-E)2, where E is the correction value. ⎛ ⎞ V2 V4 ⎟ E = L − ⎜⎜ L − − ⎠⎟ 2.L 8.L3 ⎝ V2 V4 =E + 2.L 8.L3
-5ﻗﻴﺲ ﺧﻂ ﺑﺸﺮﻳﻂ ﻣﻦ اﻟﺼﻠﺐ ﻃﻮﻟﻪ ) (20.0 mﻓﻜﺎن ﻃﻮﻟﻪ ) (100.0 mوآﺎن ﻣﻘﺪار اﻟﺘﺮﺧﻴﻢ ) (5.0 cmﻣﻦ اﻟﻄﻮل اﻟﺤﻘﻴﻘ ﻰ ﻟﻠﺸﺮﻳﻂ ) (20.01mوآﺎﻧﺖ درﺟﺔ اﻟﺤﺮارة أﺛﻨﺎء اﻟﻘﻴﺎس ) (18.0 0Cوأﺛﻨﺎء اﻟﻤﻌﺎﻳﺮة ) (20.0 0Cوﻣﻌﺎﻣﻞ ﺗﻤﺪد اﻟﺸﺮﻳﻂ (9*10- )7ﻟﻜﻞ درﺟﺔ ﻣﺌﻮﻳﺔ ،وزاوﻳﺔ ﻣﻴﻞ اﻟﺸﺮﻳﻂ ) (60 0اﺣﺴﺐ اﻟﻄﻮل اﻟﺤﻘﻴﻘﻰ ﻟﻠﺨﻂ. اﻟﺤﻞ -1اﻟﻌﻼﻗﺔ اﻟﺘﻰ ﺗﺴﺘﺨﺪم ﻟﺘﺼﺤﻴﺢ اﻟﺘﺮﺧﻴﻢ ﻓﻰ اﻟﺸﺮﻳﻂ: 2
2
8.S 32.S − − .... 3.D 15.D 2
H = D−
Where: Horizontal distance length of the measuring tape sag of the tape
2
Prepared By:/ Amr A. El-Sayed, Civil Eng Dept., El-Minia Univ., Eg.
H D S
ﻣﺴﺎﺋﻞ ﻓﻰ اﻟﻤﺴﺎﺣﺔ اﻟﻤﺴﺘﻮﻳﺔ وﺑﺎﻟﺘﺎﻟﻰ ﻳﻤﻜﻦ ﺣﺴﺎب اﻟﺨﻄﺄ اﻟﻨﺎﺗﺞ ﻣﻦ اﻟﺘﺮﺧﻴﻢ ﻣﻦ اﻟﻤﻌﺎدﻟﺔ: 2
4
8.S 32.S + 3.D 15.D 3
= E = D−H
وﻳﻤﻜﻦ إهﻤﺎل اﻟﺤﺪ اﻟﺜﺎﻧﻰ ﻣﻦ اﻟﻤﻌﺎدﻟﺔ ﻷﻧﻪ ﺻﻐﻴﺮ ﺟﺪًا.
Correction of sag: 8 * (0.05) 2 = 0.033 cm. 3 * 20.0
=E
-2اﻟﺨﻄﺄ اﻟﻨﺎﺗﺞ ﻣﻦ اﺧﺘﻼف درﺟﺔ اﻟﺤﺮارة: ) E = α .H ( d 1 − d
Where: E α d1 d
Correction due to temperature change coefficient of thermal expansion degree of temperature at measuring degree of temperature at calibration.
*E = 9*10-7 -6ﻋﻨﺪ ﻋﻤﻞ رﻓﻊ ﻟﺒﺮآﺔ ﻣﻴﺎﻩ أﺣﻴﻄﺖ ﺑﻤﺜﻠﺚ ABCأﻃﻮال أﺿﻼﻋﻪ آﺎﻵﺗﻰ : AB = 400.0 m , AC = 250.0 m. أﺧﺬت اﻟﻨﻘﻄﺔ Dﻋﻠﻰ اﻟﺨﻂ BCﺑﺤﻴﺚ آﺎن ﻃﻮل CD = 150.0 m ،BD = 125.0 mاﺣﺴﺐ ﻃﻮل اﻟﺨﻂ ،ADأذا آﺎن : أ – ﻃﻮل اﻟﺠﻨﺰﻳﺮ اﻟﻤﺴﺘﻌﻤﻞ ﺻﺤﻴﺤ ًﺎ. ب – اﻟﺠﻨﺰﻳﺮ اﻟﻤﺴﺘﻌﻤﻞ ﺑﻪ ﻋﻘﻠﺔ ﻣﻔﻘﻮدة ﻓﻰ اﻟﻤﺘﺮ ).(15 اﻟﺤﻞ
C m 15 0.0
m
25 0.0
θ
D m 12 5.0
A 40 0. 0 m
3
B
Prepared By:/ Amr A. El-Sayed, Civil Eng Dept., El-Minia Univ., Eg.
ﻣﺴﺎﺋﻞ ﻓﻰ اﻟﻤﺴﺎﺣﺔ اﻟﻤﺴﺘﻮﻳﺔ AD =
( AB) 2 .DC + ( AC ) 2 .DB − BD.DC BC
AD =
(400) 2 .150 + (250) 2 .125 − 125.0 *150.0 = 311.339 m 275
or, ( AC ) 2 + ( BC ) 2 − ( AB) 2 cos(θ ) = 2 * AC * BC cos(θ ) =
(250) 2 + (275) 2 − (400) 2 = 2 * 250 * 275
(AD)2 = (AC)2 + (CD)2 – 2*AC*CD*cos (θ) وﺿﺢ آﻴﻔﻴﺔ اﻟﺘﻐﻠﺐ ﻋﻠﻰ اﻟﻌﻘﺒﺎت اﻟﺘﻰ ﻧﻮاﺟﻬﻬﺎ ﻓﻰ اﻟﻄﺒﻴﻌﺔ-7
: ﻟﻠﺘﻐﻠﺐ ﻋﻠﻰ اﻟﻌﻮاﺋﻖ ﻓﻰ اﻟﻘﻴﺎس ﻳﻤﻜﻦ اﺳﺘﺨﺪام اﻟﻄﺮق اﻵﺗﻴﺔ-ب -1
1
2 e
d
c
A
3
d
e B A
B
c
f c
4
e
90.0°
f
d
c A A
B
B d
Prepared By:/ Amr A. El-Sayed, Civil Eng Dept., El-Minia Univ., Eg.
4
ﻣﺴﺎﺋﻞ ﻓﻰ اﻟﻤﺴﺎﺣﺔ اﻟﻤﺴﺘﻮﻳﺔ -8وﺿﺢ آﻴﻔﻴﺔ ﻗﻴﺎس ﺑﺮج ﻳﻤﻜﻦ اﻟﻮﺻﻮل إﻟﻰ ﻗﺎﻋﺪﺗﻪ وﻻ ﻳﻤﻜﻦ اﻟﻮﺻﻮل إﻟﻰ ﻗﻤﺘﻪ ﺑﺎﺳﺘﺨﺪام أدوات اﻟﻘﻴﺎس اﻟﻄﻮﻟﻰ.
E
C H F
G
A D
B
-9وﺿﺢ آﻴﻔﻴﺔ ﻗﻴﺎس ﺑﺮج ﻻ ﻳﻤﻜﻦ اﻟﻮﺻﻮل إﻟﻰ ﻗﺎﻋﺪﺗﻪ ،وﻻ إﻟﻰ إرﺗﻔﺎﻋﻪ ﺑﺎﺳﺘﺨﺪام أدوات اﻟﻘﻴﺎس اﻟﻄﻮﻟﻰ ﻓﻘﻂ.
5
Prepared By:/ Amr A. El-Sayed, Civil Eng Dept., El-Minia Univ., Eg.
ﻣﺴﺎﺋﻞ ﻓﻰ اﻟﻤﺴﺎﺣﺔ اﻟﻤﺴﺘﻮﻳﺔ
اﻟﻤﺴﺎﺣﺔ ﺑﺎﻟﺒﻮﺻﻠﺔ :ﺗﻌﺮﻳﻔﺎت اﻟﺸﻤﺎل اﻟﻤﻐﻨﺎﻃﻴﺴﻰ-1 اﻟﺸﻤﺎل اﻟﺠﻐﺮاﻓﻰ-2 زاوﻳﺔ اﻹﺧﺘﻼف-3 إﻧﺤﺮاف اﻟﺨﻂ-4 اﻹﻧﺤﺮاف اﻟﺪاﺋﺮى-5 اﻹﻧﺤﺮاف اﻟﻤﺨﺘﺼﺮ-6
Summation of internal angles of a closed polygon = (n-2)*180
………. (1)
Where: n
number of polygon’s sides.
Summation of internal angles of a closed polygon (with the help of member’s bearings) = Σ of back bearings - Σ of front bearings + L*360 ………. (2) Where: L number of points whose front bearing is greater than the back bearing of the previous member. .ﻋﺪد اﻟﻨﻘﺎط اﻟﺘﻰ ﺑﻬﺎ اﻹﻧﺤﺮاف اﻷﻣﺎﻣﻰ ﻟﻠﺨﻂ اﻟﻼﺣﻖ أآﺒﺮ ﻣﻦ اﻹﻧﺤﺮاف اﻟﺨﻠﻔﻰ ﻟﻠﺨﻂ اﻟﺴﺎﺑﻖ Error of internal angles (Δ) = (n-2)*180 –{ Σ of back bearings - Σ of front bearings + L*360 } Allowable error = K . 2.n Where:
K
constant
The amount of correction for each direction (e) =
Δ 2.n
Examples:
1- A closed polygon (abcd) was created. The bearings of all its members was measured by a compass, and they were as follows: Member Front Bearing Back Bearing
Ab 1800 ...`
bc 2600 40`
Ca 610 50`
cd 3500 30`
Da 850 40`
…0 20`
810 …
2400 40`
1700 40`
2650 30`
Prepared By:/ Amr A. El-Sayed, Civil Eng Dept., El-Minia Univ., Eg.
6
ﻣﺴﺎﺋﻞ ﻓﻰ اﻟﻤﺴﺎﺣﺔ اﻟﻤﺴﺘﻮﻳﺔ
2- The following observations were taken by a compass for a closed traverse. It is required to adjust the bearings of that traverse by the exact method, and the sides of the traverse by “Bowditch” method (length of side’s method).
Side
Length
Front Bearing
Back Bearing
AB BC CD DE EA
96.23 121.15 79.16 102.77 94.75
64o 16` 128o 17` 201o 03` 288o 46` 324o 16`
244o 20` 307o 51` 22o 36` 107o 59` 144o 12`
Solution :ﺧﻄﻮات اﻟﺤﻞ : ﺗﺼﺤﻴﺢ اﻟﺨﻄﺄ اﻟﺰاوى-1 :ﻣﻦ اﻟﻤﻌﺮوف أن ﻣﺠﻤﻮع اﻟﺰواﻳﺎ اﻟﺪاﺧﻠﻴﺔ ﻷى ﻣﻀﻠﻊ ﻣﻐﻠﻖ ﺗﺴﺎوى ………. (1)
= 180 (n-2)
where n = total number of angles, or sides of the polygon وﻳﻤﻜﻦ ﻣﻌﺮﻓﺔ ﻗﻴﻤﺔ اﻟﺰواﻳﺎ اﻟﺪاﺧﻠﻴﺔ ﻟﻠﻤﻀﻠﻊ ﺑﻤﻌﻠﻮﻣﻴﺔ اﻹﻧﺤﺮاف اﻷﻣﺎﻣﻰ ﻟﺨﻂ واﻹﻧﺤﺮاف اﻟﺨﻠﻔ ﻰ ﻟﻠﺨ ﻂ اﻟ ﺴﺎﺑﻖ آﻤ ﺎ ه ﻮ ﻣﺒ ﻴﻦ :ﺑﺎﻟﺮﺳﻢ
B
128°17'0"
244°20'0"
A
144°12'00" 64°16'0"
C
201°03'0"
307°51'0"
E
107°59'0"
324°16'00"
22°36'0"
288°46'0"
D
Angle (abc) = 244o 20` - 128o 17` Angle (cde) = 360 – (288o 46` - 22o 36`)
Prepared By:/ Amr A. El-Sayed, Civil Eng Dept., El-Minia Univ., Eg.
7
ﻣﺴﺎﺋﻞ ﻓﻰ اﻟﻤﺴﺎﺣﺔ اﻟﻤﺴﺘﻮﻳﺔ وﺑﺎﻟﺘﺎﻟﻰ ﻳﺼﺒﺢ : ﻣﺠﻤﻮع اﻟﺰواﻳﺎ اﻟﺪاﺧﻠﻴﺔ ﻟﻠﺸﻜﻞ = ﻣﺠﻤﻮع اﻹﻧﺤﺮاﻓﺎت اﻟﺨﻠﻔﻴﺔ – ﻣﺠﻤﻮع اﻹﻧﺤﺮاﻓﺎت اﻷﻣﺎﻣﻴﺔ +ل × 360 ………… )(2 ﺣﻴﺚ أن )ل( = ﻋﺪد اﻟﻨﻘﻂ اﻟﺘﻰ ﺑﻬﺎ اﻹﻧﺤﺮاف اﻷﻣﺎﻣﻰ ﻟﻠﺨﻂ اﻟﻼﺣﻖ أآﺒﺮ ﻣﻦ اﻹﻧﺤﺮاف اﻟﺨﻠﻔﻰ ﻟﻠﺨﻂ اﻟﺴﺎﺑﻖ ﻼ ﻧﻘﻄﺔ ) (dاﻟﺨﻂ اﻟﻼﺣﻖ ﻟﻬﺎ هﻮ ) (deواﻟﺨ ﻂ اﻟ ﺴﺎﺑﻖ ﻟﻬ ﺎ ه ﻮ ) ،(cdﻓ ﺈذا آ ﺎن اﻹﻧﺤ ﺮاف اﻷﻣ ﺎﻣﻰ ﻟﻠﺨ ﻂ ) (deأآﺒ ﺮ ﻣ ﻦ ،ﻣﺜ ً اﻹﻧﺤﺮاف اﻟﺨﻠﻔﻰ ﻟﻠﺨﻂ ) (cdﻓﺈن ﻧﻘﻄﺔ ) (dﺗﻀﺎف إﻟﻰ اﻟﻨﻘﻂ اﻟﻤﺴﺘﺨﺪﻣﺔ ﻹﻳﺠﺎد ﻗﻴﻤﺔ )ل(. وﻟﺤ ﺴﺎب ﻗﻴﻤ ﺔ )ل( ﻓ ﻰ اﻟﻤﺜ ﺎل اﻟ ﺴﺎﺑﻖ ﻧﺒ ﺪأ ﺑ ﺎﻹﻧﺤﺮاف اﻷﻣ ﺎﻣﻰ ﻟﺜ ﺎﻧﻰ ﺧ ﻂ ) (bcوﻧﻘﺎرﻧ ﻪ ﺑ ﺎﻹﻧﺤﺮاف اﻟﺨﻠﻔ ﻰ ﻟﻠﺨ ﻂ اﻟﺴﺎﺑﻖ ) (abأى ﻧﻘﺎرن ` 1280 17ﻣﻊ ` ،2440 20وهﻜﺬا آﻤﺎ هﻮ ﻣﺒﻴﻦ ﻣﻦ اﻹﺗﺠﺎهﺎت ﻓﻰ اﻟﺠﺪول اﻟﺘﺎﻟﻰ:
Back Bearing
Front Bearing
Line
Point
`244o 20
`64o 16
AB
A
`307o 51
`128o 17
BC
B
`22o 36
`201o 03
CD
C
`107o 59
`288o 46
DE
D
`144o 12
`324o 16
EA
E
`826o 58
`1006o 38
Summation
وهﻜﺬا ﻧﺠﺪ أن ﻋﺪد اﻟﻨﻘﻂ اﻟﺘﻰ ﻳﺰﻳﺪ ﻓﻴﻬﺎ اﻹﻧﺤﺮاف اﻷﻣﺎﻣﻰ ﻟﻠﺨﻂ اﻟﻼﺣﻖ ﻋ ﻦ اﻹﻧﺤ ﺮاف اﻟﺨﻠﻔ ﻰ ﻟﻠﺨ ﻂ اﻟ ﺴﺎﺑﻖ ﻳ ﺴﺎوى ) (2هﻤ ﺎ ل=2 ﻧﻘﻄﺘﻰ ) ،(d, eﻣﻊ ﻣﻼﺣﻈﺔ أن اﻹﻧﺤﺮاف اﻷﻣﺎﻣﻰ ﻟﻠﻀﻠﻊ اﻷول ﻳﻘﺎرن ﺑﺎﻹﻧﺤﺮاف اﻟﺨﻠﻔﻰ ﻟﻠﻀﻠﻊ اﻷﺧﻴﺮ. اﻟﺨﻄﻮة اﻟﺘﺎﻟﻴﺔ هﻰ إﻳﺠﺎد ﻗﻴﻤﺔ اﻟﺨﻄﺄ اﻟﺰاوى ) (Δﻣﻦ اﻟﻤﻌﺎدﻟﺔ:ﻣﺠﻤﻮع اﻟﺰواﻳﺎ اﻟﻤﺤﺴﻮﺑﺔ ﻣﻦ اﻟﻤﻌﺎدﻟﺔ ) – (1ﻣﺠﻤﻮع اﻟﺰواﻳﺎ اﻟﻤﺤﺴﻮﺑﺔ ﻣﻦ اﻟﻤﻌﺎدﻟﺔ )(2
} = 180 (5-2) – { (826o 58` - 1006o 38`) + 2 * 360 `= 540.0 – 540o 20 `= -00o 20 زﻳﺎدة اﻹﻧﺤﺮاﻓﺎت اﻷﻣﺎﻣﻴﺔ و إذا آﺎن اﻟﺨﻄﺄ )ﺳﺎﻟﺐ( ﻓﺈﻧﻪ ﻳﺠﺐ اﻹﻧﺤﺮاﻓﺎت اﻷﻣﺎﻣﻴﺔ و إذا آﺎن اﻟﺨﻄﺄ )ﻣﻮﺟﺐ( ﻓﺈﻧﻪ ﻳﺠﺐ ﺗﻘﻠﻴﻞ
Δ
ﺗﻘﻠﻴﻞ اﻹﻧﺤﺮاﻓﺎت اﻟﺨﻠﻔﻴﺔ زﻳﺎدة اﻹﻧﺤﺮاﻓﺎت اﻟﺨﻠﻔﻴﺔ
واﻟﺨﻄﺄ اﻟﻤﺴﻤﻮح ﺑﻪ ﻳﻤﻜﻦ ﺣﺴﺎﺑﻪ ﻣﻦ اﻟﻤﻌﺎدﻟﺔ: Allowable error = k . 2.n , k is constant, n: number of internal angles. هﺬا اﻟﺨﻄﺄ ﻳﻮزع ﻋﻠﻰ ﻋﺪد اﻹﺗﺠﺎهﺎت اﻟﻤﺮﺻﻮدة وهﻰ ﺗﺴﺎوى )ﻋﺪد اﻷﺿﻼع × 10 = 2 × 5 = (2اﺗﺠﺎهﺎت ﻗﻴﻤﺔ اﻟﺘﺼﺤﻴﺢ = ﻗﻴﻤﺔ اﻟﺨﻄﺄ ÷ ﻋﺪد اﻹﺗﺠﺎهﺎت ،وﻳﻔﻀﻞ أن ﻳﻜ ﻮن اﻟﺘ ﺼﺤﻴﺢ ﻟﻜ ﻞ إﺗﺠ ﺎﻩ ﻣﻘ ﺪار ﺻ ﺤﻴﺢ ،ﻟ ﺬﻟﻚ إذا آﺎﻧ ﺖ ﻗﻴﻤ ﺔ اﻟﺘﺼﺤﻴﺢ ﻟﻺﺗﺠﺎة اﻟﻮاﺣﺪ آﺴﺮ ﻓﺈﻧﻪ ﻳﺘﻢ اﻵﺗﻰ: Δ = )error for each direction (e ), (fraction 2.n
ex: assume that n = 5, Δ = 46,
8
Prepared By:/ Amr A. El-Sayed, Civil Eng Dept., El-Minia Univ., Eg.
ﻣﺴﺎﺋﻞ ﻓﻰ اﻟﻤﺴﺎﺣﺔ اﻟﻤﺴﺘﻮﻳﺔ
46 = 4.6 2*5 -1ﻧﻌﻴﻦ ﻗﻴﻤﺘﻴﻦ ﺻﺤﻴﺤﺘﻴﻦ ﻟﻠﺨﻄﺄ )(eأﺻﻐﺮ ﻣﺒﺎﺷﺮة ﻣﻦ ﻗﻴﻤﺔ اﻟﻜﺴﺮ ﺛﻢ ﺗﻠﻚ اﻟﻘﻴﻤﺔ اﻟﺼﺤﻴﺤﺔ 1 + اﻟﻘﻴﻤﺘﻴﻦ اﻟﺼﺤﻴﺤﺘﻴﻦ ﻟﻠﺘﺼﺤﻴﺢ هﻤﺎ ).(4, 4+1 e1 = 4, and e2 = 5 -2ﻧﻌﻴﻦ ﻋﺪد اﻹﺗﺠﺎهﺎت اﻟﺘﻰ ﺗﺄﺧﺬ اﻟﺘﺼﺤﻴﺢ )(e2 d1 = Δ - 2.n.e1 )= 46 – 2*5*4 = 6 directions (takes the correction 5 -3ﻋﺪد اﻹﺗﺠﺎهﺎت اﻟﺘﻰ ﺗﺄﺧﺬ اﻟﺘﺼﺤﻴﺢ )(e1 d2 = 2.n – d1 )d2 = 10 – 6 = 4 directions (takes the correction 4 =e
ﺑﺎﻟﻌﻮدة إﻟﻰ اﻟﻤﺜﺎل اﻷﺻﻠﻰ: Δ `= -20.0 / 10 = -2.0 2.n
=e
Δ = -20.0`,
ﻳﺘﻢ زﻳﺎدة ﺟﻤﻴﻊ اﻹﻧﺤﺮاﻓﺎت اﻷﻣﺎﻣﻴﺔ ﺑﻤﻘﺪار )` (2.0وﺗﻘﻠﻴﻞ ﺟﻤﻴﻊ اﻹﻧﺤﺮاﻓﺎت اﻟﺨﻠﻔﻴﺔ ﺑﻤﻘﺪار )`.(2.0
Corrected Back Bearing
Corrected Front Bearing
Back Bearing
Front Bearing
Line
`244o 18
`64o 18
`244o 20
`64o 16
AB
`307o 49
`128o 19
`307o 51
`128o 17
BC
`22o 34
`201o 05
`22o 36
`201o 03
CD
`107o 57
`288o 48
`107o 59
`288o 46
DE
`144o 10
`324o 18
`144o 12
`324o 16
EA
`826o 48
`1006o 48
ﻣﺠﻤﻮع اﻟﺰواﻳﺎ اﻟﺪاﺧﻠﻴﺔ ﺑﻌﺪ اﻟﺘﺼﺤﻴﺢ =
o.k.
9
Summation
= 826o 48` - 1006o 48` + 2 * 360 = 540o
Prepared By:/ Amr A. El-Sayed, Civil Eng Dept., El-Minia Univ., Eg.
ﻣﺴﺎﺋﻞ ﻓﻰ اﻟﻤﺴﺎﺣﺔ اﻟﻤﺴﺘﻮﻳﺔ -2ﺗﺼﺤﻴﺢ اﻹﻧﺤﺮاﻓﺎت ﺑﺎﻟﻄﺮﻳﻘﺔ اﻟﺪﻗﻴﻘﺔ ﻟﻠﺠﺎذﺑﻴﺔ اﻟﻤﺤﻠﻴﺔ:
δa
F1
A
B δb F2 ﻓﻰ اﻟﺸﻜﻞ اﻟﻤﺒﻴﻦ ) (δa , δbاﻟﻔﺮق ﺑﻴﻦ اﻟﺸﻤﺎل اﻟﺠﻐﺮاﻓﻰ واﻟﺸﻤﺎل اﻟﻤﻐﻨﺎﻃﻴﺴﻰ ﻋﻨﺪ اﻟﻨﻘﻄﺘﻴﻦ ) ،(a , bﻓﻰ ﺣﺎﻟﺔ ﻋﺪم وﺟﻮد ﺟﺎذﺑﻴﺔ ﻣﺤﻠﻴﺔ ﻓﺈن اﻟﻔﺮق ﺑ ﻴﻦ اﻹﻧﺤ ﺮاف اﻟﺨﻠﻔ ﻰ واﻷﻣ ﺎﻣﻰ ﻟﻠﺨ ﻂ ﻳ ﺴﺎوى ،180أﻣ ﺎ ﻓ ﻰ ﺣﺎﻟ ﺔ وﺟ ﻮد ﺟﺎذﺑﻴ ﺔ ﻣﺤﻠﻴ ﺔ آﻤ ﺎ ﺑﺎﻟﺸﻜﻞ:
F2 + δb – 180 = F1 + δa )………… (3
)δb - δa = 180 – (F2 – F1
وﻹﻳﺠﺎد ﻗﻴﻢ ) (δa , δb , δc , …….ﻧﺘﺒﻊ اﻟﺨﻄﻮات اﻵﺗﻴﺔ: -1ﻧﻴﻌﻦ اﻟﻔﺮق ﺑﻴﻦ اﻹﻧﺤﺮاف اﻟﺨﻠﻔﻰ واﻷﻣﺎﻣﻰ ﻟﻜﻞ ﺧﻂ )ﺑﻌﺪ اﻟﺘﺼﺤﻴﺢ اﻟﺰاوى( -2ﻧﻄﺮح اﻟﻔﺮق )ﺑﺪون إﺷﺎرة( ﻣﻦ 180ﻓﻨﺤﺼﻞ ﻋﻠﻰ ﻋﺪد ﻣﻦ اﻟﻤﻌﺎدﻻت ﻋﻠﻰ اﻟﺼﻮرة:
δb - δa = k1 δc - δb = k2 δd - δc = k3 .. ………
وﻳﺠﺐ أن ﻳﻜﻮن ﻣﺠﻤﻮع هﺬﻩ اﻟﻤﻌﺎدﻻت = ﺻﻔﺮ ،أى أن:
k1 + k2 + k3 + ………. = 0.0
10
Prepared By:/ Amr A. El-Sayed, Civil Eng Dept., El-Minia Univ., Eg.
ﻣﺴﺎﺋﻞ ﻓﻰ اﻟﻤﺴﺎﺣﺔ اﻟﻤﺴﺘﻮﻳﺔ
Line
Corrected Front Bearing
Corrected Back Bearing
Difference Back - Front
180 - d
AB
64o 18`
244o 18`
180
0.0
BC
128o 19`
307o 49`
179o 30`
0o 30`
CD
201o 05`
22o 34`
-178o 31`
1o 29`
DE
288o 48`
107o 57`
-180o 51`
-00 51`
EA
324o 18`
144o 10`
-180o 8`
-00 8`
δb - δa = 0.0 δc - δb = 0o 30` δd - δc = -1o 29` δe - δd = 0o 51` δa - δe = 00 8`
for check 0o 30` - 1o 29` + 0o 51` + 00 8` = 0.0 δb = δa = 0.0 δc = 00 30` δd = -1o 29` + δc = -1o 29` + 0o 30` = -0o 59` δe = 0o 51` + δd = 0o 51` - 0o 59` = -0o 8`
the last equation is used for check δa - δe = 0.0 – (-0o 8`) = 0o 8`
o.k.
Line
δfront
δBack
AB
0.0
0.0
BC
0.0
0o 30`
CD
0o 30`
-0o 59`
DE
-0o 59`
-0o 8`
EA
-0o 8`
0.0
Corrected Front Bearing
64o 18` + 0.0 = 64o 18` 128o 19` + 0.0 = 128o 19` 201o 05` + 00 30` = 201o 35` 288o 48` - 0o 59` = 287o 49` 324o 18` - 0o 8` = 324o 10`
Corrected Back Bearing
244o 18` + 0.0 = 244o 18` 307o 49` + 0o 30` = 308o 19` 22o 34` - 0o 59` = 21o 35` 107o 57` - 0o 8` = 107o 49` 144o 10` + 0.0 = 144o 10`
Prepared By:/ Amr A. El-Sayed, Civil Eng Dept., El-Minia Univ., Eg.
11
ﻣﺴﺎﺋﻞ ﻓﻰ اﻟﻤﺴﺎﺣﺔ اﻟﻤﺴﺘﻮﻳﺔ :(ﺗﺼﺤﻴﺢ ﺧﻄﺄ اﻟﻘﻔﻞ )ﻋﻦ ﻃﺮﻳﻖ اﻟﻤﺮآﺒﺎت
_ + + East
West
+
_
_ _ +
N.E (North-East) N.W (North-West) S.E (South-East) S.W (South-West)
: وﺑﺎﻟﺘﺎﻟﻰ ﻧﻌﺮف ﻣﻮﻗﻌﻪ،ﻧﺤﺪد اﻹﻧﺤﺮاف اﻟﻤﺨﺘﺼﺮ ﻟﻜﻞ ﺿﻠﻊ the Vertical component is (+) and Horizontal component is (+) the Vertical component is (+) and Horizontal component is (-) the Vertical component is (-) and Horizontal component is (+) the Vertical component is (-) and Horizontal component is (-)
Then, The vertical component (V) = L * cos (θ) The horizontal component (H) = L * sin (θ), θ
Where
is the reduced bearing ()اﻹﻧﺤﺮاف اﻟﻤﺨﺘﺼﺮ
Line
Length
AB BC CD DE EA
96.23 121.15 79.16 102.77 94.75 494.06
Length of error = =
Corrected Front Bearing
Reduced Bearing
64o 18` 128o 19` 201o 35` 287o 49` 324o 10`
N.E 64o 18` S.E 51o 41` S.W 21o 35` N.W 72o 11` N.W 35o 50`
Vertical Horizontal Component Component +41.731 -75.114 -73.61 +31.445 +76.816 +1.2684
∑ (Vertical Components ) + ∑ (Horizontal Components ) 2
+86.711 +95.054 -29.119 -97.841 -55.469 -0.6655 2
(1.2684) 2 + (−0.6655) 2 = 1.4324
Prepared By:/ Amr A. El-Sayed, Civil Eng Dept., El-Minia Univ., Eg.
12
ﻣﺴﺎﺋﻞ ﻓﻰ اﻟﻤﺴﺎﺣﺔ اﻟﻤﺴﺘﻮﻳﺔ
Length of error 1.4324 = = 2.8992*10-3. Total length of traverse lines 494.06
= % of error
ﺗﻨﺺ اﻟﻤﻮاﺻﻔﺎت ﻋﻠﻰ أﻻ ﻳﺘﺠﺎوز ﻧﺴﺒﺔ ﺧﻄ ﺄ اﻟﻘﻔ ﻞ ) (1:100ﻓ ﻰ اﻷراﺿ ﻰ اﻟ ﻮﻋﺮة ذات اﻟﻄﺒﻮﻏﺮاﻓﻴ ﺔ اﻟ ﺸﺪﻳﺪة ،ﻣ ﻊ اﻟﻘﻴﺎس ﺑﺎﻟﺠﻨﺰﻳﺮ ،وأﻻ ﻳﺘﺠﺎوز ) (1:500ﻓﻰ اﻟﻤﺪن. ﻳﺘﻢ ﺗﺼﺤﻴﺢ اﻟﻤﺮآﺒﺔ اﻟﺮأﺳﻴﺔ ﻟﻜﻞ ﺧﻂ آﺎﻵﺗﻰ: ﻃﻮل اﻟﺨﻂ ﻣﺠﻤﻮع أﻃﻮال ﺧﻄﻮط اﻟﻤﻀﻠﻊ
ﺗ ﺼﺤﻴﺢ اﻟﻤﺮآﺒ ﺔ اﻟﺮأﺳ ﻴﺔ ﻟﻠﺨﻂ =
× اﻟﻤﺮآﺒﺔ اﻟﺮأﺳﻴﺔ ﻟﺨﻄﺄ اﻟﻘﻔﻞ
وﻳﺘﻢ ﺗﺼﺤﻴﺢ اﻟﻤﺮآﺒﺔ اﻷﻓﻘﻴﺔ ﻟﻜﻞ ﺧﻂ آﺎﻵﺗﻰ: ﻃﻮل اﻟﺨﻂ ﻣﺠﻤﻮع أﻃﻮال ﺧﻄﻮط اﻟﻤﻀﻠﻊ
ﺗﺼﺤﻴﺢ اﻟﻤﺮآﺒﺔ اﻷﻓﻘﻴ ﺔ ﻟﻠﺨ ﻂ =
× اﻟﻤﺮآﺒﺔ اﻷﻓﻘﻴﺔ ﻟﺨﻄﺄ اﻟﻘﻔﻞ
وﻧﻼﺣ ﻆ أﻧ ﻪ إذا آﺎﻧ ﺖ اﻟﻤﺮآﺒ ﺔ اﻟﺮأﺳ ﻴﺔ ﻟﺨﻄ ﺄ اﻟﻘﻔ ﻞ ﻣﻮﺟﺒ ﺔ ،ﻓ ﺈن ﺗ ﺼﺤﻴﺢ اﻟﻤﺮآﺒ ﺎت اﻟﺮأﺳ ﻴﺔ ﻟﺨﻄ ﻮط اﻟﻤﺮﺑ ﻊ ﺗﻜ ﻮن ﺳﺎﻟﺒﺔ ،وهﻜﺬا ﻓﺈن إﺷﺎرة اﻟﺘﺼﺤﻴﺢ ﺗﻜﻮن ﻋﻜﺲ إﺷﺎرة اﻟﺨﻄﺄ.
Corrected Corrected Vl. Comp. Hl. Comp. 41.484 86.841 -75.425 95.217 -73.813 -29.012 31.181 -97.703 76.573 -55.341 0.0 0.0
Vl. Corr. -0.2471 -0.3110 -0.2032 -0.2638 -0.2433
Hl. Corr. +0.1296 +0.1632 +0.1066 +0.1384 +0.1276
Vertical Horizontal Component Component +41.731 +86.711 -75.114 +95.054 -73.61 -29.119 +31.445 -97.841 +76.816 -55.469 +1.2684 -0.6655
Length
Line
96.23 121.15 79.16 102.77 94.75 494.06
AB BC CD DE EA
ﺗﺼﺤﻴﺢ اﻟﺨﻄﺄ ﺑﺎﻟﻄﺮﻳﻘﺔ اﻟﺒﻴﺎﻧﻴﺔ:
D
C `A
A
A
B
`B
D `D `C
13
B
A
C
Prepared By:/ Amr A. El-Sayed, Civil Eng Dept., El-Minia Univ., Eg.
ﻣﺴﺎﺋﻞ ﻓﻰ اﻟﻤﺴﺎﺣﺔ اﻟﻤﺴﺘﻮﻳﺔ
Area: 1- It is required to find the area of the closed traverse, if the coordinates of its corners are as follows: Point XCoordinate YCoordinate
A
B
C
D
E
F
10
45
87
138
104
36
40
77
65
83
3
14
B
D C
A F
E
X
Area = 0.50 { (X1.Y2 + X2.Y3 + X3.Y4 + X4.Y5 + X5.Y6 + X6.Y1) – (Y1.X2 + Y2.X3 + Y3.X4 + Y4.X5 + Y5.X6 + Y6.X1) } = 0.50 { (10 * 77 + 45 * 65 + 87 * 83 + 138 * 3 + 104 * 14 + 36 * 40) – (40 * 45 + 77 * 87 + 65 * 138 + 83 * 104 + 3 * 36 + 14 * 10) = 0.50 * 12123 = 6061.50
Prepared By:/ Amr A. El-Sayed, Civil Eng Dept., El-Minia Univ., Eg.
14
ﻣﺴﺎﺋﻞ ﻓﻰ اﻟﻤﺴﺎﺣﺔ اﻟﻤﺴﺘﻮﻳﺔ
2- The area by the lines’ components: A
E
B
C
D
Line AB BC CD DE EA
X component 150 175 -65 -160 -100
Y component -275 -135 -130 300 240
Vertical Y1 = -275 2Y1+Y2 = -685 2(Y1+Y2)+Y3 = -950 2(Y1+Y2+Y3)+Y4 = -780 2(Y1+Y2+Y3+Y4)+Y5 = -240
2*Area -41250 -119875 61750 124800 24000 49425
Area = 0.50 * 49425 = 24712.50
Prepared By:/ Amr A. El-Sayed, Civil Eng Dept., El-Minia Univ., Eg.
15
ﻣﺴﺎﺋﻞ ﻓﻰ اﻟﻤﺴﺎﺣﺔ اﻟﻤﺴﺘﻮﻳﺔ
اﻟﻤﺴﺎﺣﺔ اﻟﺘﺎآﻴﻮﻣﺘﺮﻳﺔ هﻰ اﻟﻤﺴﺎﺣﺔ اﻟ ﺴﺮﻳﻌﺔ ،وﺗ ﺘﻢ ﺑﺎﺳ ﺘﺨﺪام ﺗﻴﻮدﻳﻠﻴ ﺖ ﻣﺠﻬ ﺰ ﺑ ﺒﻌﺾ اﻹﺿ ﺎﻓﺎت اﻟﺨﺎﺻ ﺔ .وﺑﻮاﺳ ﻄﺔ ه ﺬﻩ اﻹﺿ ﺎﻓﺎت وﻣ ﻊ اﺳﺘﺨﺪام ﺑﻌﺾ اﻷﺟﻬﺰة اﻟﻤﺴﺎﻋﺪة ﻳﻤﻜﻦ إﻳﺠﺎد اﻟﻤﺴﺎﻓﺎت اﻷﻓﻘﻴﺔ ،واﻟﺮأﺳﻴﺔ ﺑﺴﺮﻋﺔ وﺑﺪﻗﺔ ﻣﻌﻘﻮﻟﺔ.
-1ﻃﺮﻳﻘﺔ ﺷﻌﺮات اﻹﺳﺘﺎدﻳﺎ: وﺗﺘﻢ ﺑﺎﺳﺘﺨﺪام ﺗﻴﻮدﻳﻠﻴﺖ ﻳﺰود دﻟﻴﻠﻪ ﺑﺸﻌﺮﺗﻴﻦ أﻓﻘﻴﺘﻴﻦ إﺿﺎﻓﻴﺘﻴﻦ أﻋﻠﻰ وأﺳﻔﻞ اﻟﺸﻌﺮة اﻟﺮﺋﻴﺴﻴﺔ: X1
a1 X2
X
a
H
b1
o
b
m
c
t k
c1
D
اﻟﻤﺮآﺰ اﻟﺒﺼﺮى ﻟﻠﻌﺪﺳﺔ اﻟﺸﻴﺌﻴﺔ ﺷﻌﺮﺗﺎ اﻹﺳﺘﺎدﻳﺎ اﻟﺸﻌﺮة اﻷﻓﻘﻴﺔ اﻟﻮﺳﻄﻰ ﺗﻘﺎﻃﻊ ﺷﻌﺮﺗﻰ اﻹﺳﺘﺎدﻳﺎ ،واﻟﺸﻌﺮة اﻟﻮﺳﻄﻰ ﻣﻊ اﻟﻘﺎﻣﺔ اﻟﺒﻌﺪ اﻟﺒﺆرى ﻟﻠﻌﺪﺳﺔ اﻟﺸﻴﺌﻴﺔ اﻟﻤﺴﺎﻓﺔ اﻷﻓﻘﻴﺔ ﺑﻴﻦ اﻟﻘﺎﻣﺔ واﻟﻤﺮآﺰ اﻟﺒﺼﺮى ﻟﻠﻌﺪﺳﺔ اﻟﺸﻴﺌﻴﺔ اﻟﺒﻌﺪ اﻷﻓﻘﻰ ﺑﻴﻦ ﻣﺮآﺰ اﻟﻌﺪﺳﺔ اﻟﺸﻴﺌﻴﺔ ،وﻣﺴﺘﻮى ﺣﺎﻣﻞ اﻟﺸﻌﺮات اﻟﺒﻌﺪ اﻷﻓﻘﻰ ﺑﻴﻦ اﻟﻤﺮآﺰ اﻟﺒﺼﺮى ﻟﻠﻌﺪﺳﺔ اﻟﺸﻴﺌﻴﺔ ،واﻟﻤﺤﻮر اﻟﺮأﺳ ﻰ ﻟ ﺪوران اﻟﻤﻨﻈﺎر. اﻟﻤﺴﺎﻓﺔ اﻟﻤﻘﻄﻮﻋﺔ ﻋﻠﻰ اﻟﻘﺎﻣﺔ ﺑﻴﻦ ﺷﻌﺮﺗﻰ اﻹﺳﺘﺎدﻳﺎ a1.b1 = ، ﺑﺆرة اﻟﻌﺪﺳﺔ اﻟﺸﻴﺌﻴﺔ
m a,c b a1 , b1 , c1 X X1 X2 t H o X )+ ( X + t ab
ﺣﻴﺚ ) = (abﻃﻮل ﺣﺎﻣﻞ اﻟﺸﻌﺮات X = K1 ab
ﻳ ﺴﻤﻰ اﻟﺜﺎﺑ ﺖ اﻟﺘ ﺎآﻴﻮﻣﺘﺮى
=
اﻟﺒﻌﺪ اﻟﺒﺆرى ﻟﻠﻌﺪﺳﺔ اﻟﺸﻴﺌﻴﺔ ﻃﻮل ﺣﺎﻣﻞ اﻟﺸﻌﺮات
وﺗﺘﺮاوح ﻗﻴﻤﺔ K1ﺑﻴﻦ 50 , 100 , 200 16
Prepared By:/ Amr A. El-Sayed, Civil Eng Dept., El-Minia Univ., Eg.
D = H.
ﻣﺴﺎﺋﻞ ﻓﻰ اﻟﻤﺴﺎﺣﺔ اﻟﻤﺴﺘﻮﻳﺔ (X + t ) = Kﻳﺴﻤﻰ اﻟﺜﺎﺑﺖ اﻹﺿﺎﻓﻰ = اﻟﺒﻌﺪ اﻟﺒﺆرى ﻟﻠﻌﺪﺳﺔ اﻟﺸﻴﺌﻴﺔ +اﻟﻤﺴﺎﻓﺔ ﻣﻦ ﻣﺴﺘﻮى اﻟﻌﺪﺳ ﺔ اﻟ ﺸﻴﺌﻴﺔ إﻟ ﻰ اﻟﻤﺤﻮر اﻟﺮأﺳﻰ وﻳﺘﺮاوح ﻗﻴﻤﺔ اﻟﺜﺎﺑﺖ اﻹﺿﺎﻓﻰ ﺑﻴﻦ 60 ، 30ﺳﻢ.
D = H . K1 + K
( ……….
اﻟﻤﺴﺎﻓﺔ اﻷﻓﻘﻴﺔ ﺑﻴﻦ ﻧﻘﻄ ﺔ اﻟﺮﺻ ﺪ إﻟ ﻰ اﻟﻘﺎﻣ ﺔ = ﻓ ﺮق ﻗ ﺮاءة اﻟ ﺸﻌﺮﺗﻴﻦ اﻟﻌﻠﻴ ﺎ واﻟ ﺴﻔﻠﻰ × اﻟﺜﺎﺑ ﺖ اﻟﺘ ﺎآﻴﻮﻣﺘﺮى +اﻟﺜﺎﺑ ﺖ اﻹﺿﺎﻓﻰ. وﻓﻰ ﺣﺎﻟﺔ أﺧﺬ رﺻﺪة ﻣﺎﺋﻠﺔ )زاوﻳﺔ اﻟﻨﻈﺮ ﻣﺮﺗﻔﻌﺔ(: a1
H
X
b1
Y
c1
X1
Y1 θ
X2 m a
k
t
b c
D
X . cos 2 θ + ( X + t ). cosθ ab
( ……….
D = H . K1 . cos2 θ + K . cos θ
وﻳﻤﻜﻦ إﻳﺠﺎد إرﺗﻔﺎع أو إﻧﺨﻔﺎض اﻟﻨﻘﻄﺔ اﻟﻤﺮﺻﻮدة ﻋﻦ اﻟﻤﺴﺘﻮى اﻻﻓﻘﻰ اﻟﻤﺎر ﺑﺎﻟﺠﻬﺎز )(Y ( ……….
17
D = H.
Y = D . tan θ
Prepared By:/ Amr A. El-Sayed, Civil Eng Dept., El-Minia Univ., Eg.
ﻣﺴﺎﺋﻞ ﻓﻰ اﻟﻤﺴﺎﺣﺔ اﻟﻤﺴﺘﻮﻳﺔ
ﻣﺜﺎل: رﺻﺪت ﻧﻘﻄﺔ ﺑﺠﻬﺎز ﺗﻴﻮدﻟﻴﺖ ﻣﺰود ﺑﺸﻌﺮات اﻹﺳﺘﺎدﻳﺎ ،وآﺎﻧﺖ زاوﻳﺔ اﻟﻨﻈﺮ أﻓﻘﻴ ﺔ ،ﻓﻜﺎﻧ ﺖ ﻗ ﺮاءة اﻟ ﺸﻌﺮات – 2.613 -1 ﻼ ﺑﺰاوﻳ ﺔ 7درﺟ ﺎت ،4.106ﺛ ﻢ ﻗﻴ ﺴﺖ اﻟﻤ ﺴﺎﻓﺔ ﺑﻮاﺳ ﻄﺔ ﺷ ﺮﻳﻂ ﻓﻜﺎﻧ ﺖ 150م .رﺻ ﺪت ﻧﻘﻄ ﺔ أﺧ ﺮى ،وآ ﺎن ﺧ ﻂ اﻟﻨﻈ ﺮ ﻣ ﺎﺋ ً وآﺎﻧ ﺖ ﻗ ﺮاءة اﻟ ﺸﻌﺮات 3.154 – 1.146وﻗﻴ ﺴﺖ اﻟﻤ ﺴﺎﻓﺔ ﺑ ﺸﺮﻳﻂ ﻓﻜﺎﻧ ﺖ 200ﻣﺘ ﺮ ،أوﺟ ﺪ اﻟﺜﺎﺑ ﺖ اﻟﺘ ﺎآﻴﻮﻣﺘﺮى ،واﻟﺜﺎﺑ ﺖ اﻹﺿﺎﻓﻰ. Solution
ﻓﻰ ﺣﺎﻟﺔ ﺧﻂ اﻟﻨﻈﺮ أﻓﻘﻰ: D = H . K1 + K 150 = (4.106-2.613) . K1 + K
)………. (1 ﻓﻰ ﺣﺎﻟﺔ ﺧﻂ اﻟﻨﻈﺮ ﻳﻤﻴﻞ ﺑﺰاوﻳﺔ 7درﺟﺎت: )………. (2
2
D = H . K1 . cos θ + K . cos θ ) 200 = (3.154 – 1.146) . K1.cos2 ( 7 ) + K . cos ( 7
ﺑﺤﻞ اﻟﻤﻌﺎدﻟﺘﻴﻦ 2 ،1ﺁﻧﻴًﺎ ﻳﻨﺘﺞ أن: )اﻟﺜﺎﺑﺖ اﻟﺘﺎآﻴﻮﻣﺘﺮى( )اﻟﺜﺎﺑﺖ اﻹﺿﺎﻓﻰ(
K1 = 103 K = 3.785
-2ﻓﻰ ﺟﻬﺎز ﺗﻴﻮدﻟﻴﺖ ﻣﺰود ﺑﺤﺎﻣ ﻞ ﺷ ﻌﺮات آﺎﻧ ﺖ اﻟﻤ ﺴﺎﻓﺔ ﺑ ﻴﻦ آ ﻞ زوج ﻣ ﻦ اﻟ ﺸﻌﺮات 40/1ﺑﻮﺻ ﺔ ،واﻟﺒﻌ ﺪ اﻟﺒ ﺆرى ﻟﻠﻌﺪﺳ ﺔ اﻟﺸﻴﺌﻴﺔ ) 9 = (Xﺑﻮﺻﺔ ،واﻟﻤﺴﺎﻓﺔ ﺑﻴﻦ اﻟﻤﺤﻮر اﻟﺮأﺳﻰ ﻟﻠﺠﻬﺎز إﻟﻰ ﻣﺴﺘﻮى اﻟﻌﺪﺳﺔ اﻟﺸﻴﺌﻴﺔ ) 4.5 = (tﺑﻮﺻﺔ .رﺻﺪت ﻧﻘﻄﺔ ﺑﻮاﺳ ﻄﺔ ه ﺬا اﻟﺠﻬ ﺎز وآ ﺎن ﺧ ﻂ اﻟﻨﻈ ﺮ ﻳﻤﻴ ﻞ ﺑﺰاوﻳ ﺔ 9درﺟ ﺎت ،وآﺎﻧ ﺖ ﻗ ﺮاءة اﻟ ﺸﻌﺮات 2.65 - 2.14 – 1.63ﻋﻠ ﻰ ﻗﺎﻣ ﺔ ﻣﻘﺴﻤﺔ ﺑﺎﻟﻘﺪم واﻟﺒﻮﺻﺔ .أوﺟﺪ اﻟﻤﺴﺎﻓﺔ اﻷﻓﻘﻴﺔ ﺑﻴﻦ اﻟﺠﻬﺎز واﻟﻨﻘﻄﺔ اﻟﻤﺮﺻﻮدة ،وإذا آﺎن ﻣﻨﺴﻮب اﻟﻨﻘﻄﺔ اﻟﻤﺮﺻﻮدة = 80ﻗﺪم ،وإرﺗﻔﺎع ﺧﻂ اﻟﻨﻈﺮ 4.50ﻗﺪم ،أوﺟﺪ ﻣﻨﺴﻮب ﻧﻘﻄﺔ اﻟﺮﺻﺪ )اﻟﻨﻘﻄﺔ اﻟﺘﻰ ﺑﻬﺎ اﻟﺠﻬﺎز(. Solution
inch X )= 9.0 * 20.0 = 180 (dimensionless value ab
Ft
= , then K1
inch = 13.50 / 12.0 = 1.125
inch
X = 9.0
K = X + t = 9.0 + 4.50 = 13.50 Ft
Ft
1 1 = 40 20
* ab = 2
H = 2.65 – 1.63 = 1.02
D = H . K1 . cos2 θ + K . cos θ = 1.02 * 180 * cos2 (9) + 1.125 * cos (9) = 182.451
اﻟﻤﺴﺎﻓﺔ اﻟﺮأﺳﻴﺔ ﺑﻴﻦ ﺧﻂ اﻟﻨﻈﺮ ،وﺑﻴﻦ اﻟﻨﻘﻄﺔ اﻟﻤﺮﺻﻮدة : Ft
18
Y = D . tan θ = 182.451 * tan ( 9 ) = 28.879
Prepared By:/ Amr A. El-Sayed, Civil Eng Dept., El-Minia Univ., Eg.
ﻣﺴﺎﺋﻞ ﻓﻰ اﻟﻤﺴﺎﺣﺔ اﻟﻤﺴﺘﻮﻳﺔ a1
H
b1
Y1 Y θ
c1
a b c
D V . ﻓﺮق اﻟﻤﻨﺴﻮب ﺑﻴﻦ ﻧﻘﻄﺔ اﻟﺮﺻﺪ واﻟﻨﻘﻄﺔ اﻟﻤﺮﺻﻮدة: Ft ﻣﻨﺴﻮب ﻧﻘﻄﺔ اﻟﺮﺻﺪ:
= Y – Y1 + V = 28.879 – 2.14 + 4.5 = 31.239 Ft
19
= 80 – 31.239 = 48.761
Prepared By:/ Amr A. El-Sayed, Civil Eng Dept., El-Minia Univ., Eg.
ﻣﺴﺎﺋﻞ ﻓﻰ اﻟﻤﺴﺎﺣﺔ اﻟﻤﺴﺘﻮﻳﺔ
-2ﻃﺮﻳﻘﺔ اﻟﻈﻼل: وﺗﺮﺟﻊ ﺟﺬور هﺬﻩ اﻟﻄﺮﻳﻘ ﺔ إﻟ ﻰ اﻟﻌ ﺎم ،1850وﺑﻮاﺳ ﻄﺔ ه ﺬﻩ اﻟﻄﺮﻳﻘ ﺔ ﻳﻤﻜ ﻦ ﺗﻌﻴ ﻴﻦ اﻟﻤ ﺴﺎﻓﺔ اﻷﻓﻘﻴ ﺔ واﻟﺒﻌ ﺪ اﻟﺮأﺳ ﻰ ﺑﺎﺳﺘﺨﺪام ﺗﻴﻮدﻟﻴﺖ ﻋﺎدى ،وﺗﻮﺟﺪ ﺣﺎﻟﺘﺎن ﻟﻠﺮﺻﺪ : أ -إذا آﺎﻧﺖ ﻃﺒﻴﻌﺔ اﻷرض ﺗﺴﻤﺢ ﺑﺄﺧﺬ رﺻﺪة أﻓﻘﻴﺔ: ﻧ ﻀﻊ اﻟﺘﻴﻮدﻟﻴ ﺖ ﻋﻨ ﺪ ﻧﻘﻄ ﺔ اﻟﺮﺻ ﺪ )ﻳﻄﻠ ﻖ ﻋﻠﻴﻬ ﺎ ﻋ ﺎدة اﻟﻨﻘﻄ ﺔ اﻟﻤﺤﺘﻠ ﺔ( ،واﻟﻘﺎﻣ ﺔ ﻋﻨ ﺪ اﻟﻨﻘﻄ ﺔ اﻟﻤﺮﺻ ﻮدة ،ﺛ ﻢ ﻧﺄﺣ ﺬ ﻗﺮاءة اﻟﻘﺎﻣﺔ ﻋﻨﺪﻣﺎ ﻳﻜﻮن ﺧﻂ اﻟﻨﻈﺮ أﻓﻘﻴﺎً ،وﻟﺘﻜﻦ ) .(bﺛﻢ ﻧﻤﻴﻞ اﻟﻤﻨﻈﺎر ﺑﺰاوﻳﺔ ﻣﻌﻴﻨ ﺔ )إﻟ ﻰ أﻋﻠ ﻰ أو إﻟ ﻰ أﺳ ﻔﻞ( وﻧﻌ ﻴﻦ ﻗ ﺮاءة اﻟﻘﺎﻣﺔ ﻓﻰ اﻟﺤﺎﻟﺔ اﻟﺜﺎﻧﻴﺔ ):(a
D a H θ
b
H=a–b H =D tan θ ب -ﻋﻨﺪﻣﺎ ﻻ ﺗﺴﻤﺢ ﻃﺒﻴﻌﺔ اﻷرض ﺑﺄﺧﺬ ﻧﻈﺮات أﻓﻘﻴﺔ: ﻧﻮﺟ ﻪ اﻟﻤﻨﻈ ﺎر إﻟ ﻰ اﻟﻘﺎﻣ ﺔ ﺑﺰاوﻳ ﺔ ﻣﻴ ﻞ ﻣﻌﻴﻨﺔ ) (θ1وﻧﺄﺧﺬ اﻟﻘﺮاءة ﻋﻠﻰ اﻟﻘﺎﻣﺔ ) ،(bﺛﻢ ﻧﻐﻴ ﺮ زاوﻳ ﺔ ﻣﻴ ﻞ اﻟﻤﻨﻈ ﺎر وﻟ ﺘﻜﻦ ) ،(φوﻧﺄﺧ ﺬ اﻟﻘ ﺮاءة اﻟﺜﺎﻧﻴﺔ ﻋﻠﻰ اﻟﻘﺎﻣﺔ ).(a
a H θ
φ
b
c
D
20
Prepared By:/ Amr A. El-Sayed, Civil Eng Dept., El-Minia Univ., Eg.
ﻣﺴﺎﺋﻞ ﻓﻰ اﻟﻤﺴﺎﺣﺔ اﻟﻤﺴﺘﻮﻳﺔ
ac – bc = H
bc = D . tan θ
H tan φ − tan θ
=D
ac = D . tan φ, )H = D (tan φ - tan θ
-3ﻃﺮﻳﻘﺔ ﻗﻀﻴﺐ اﻟﻤﺴﺎﻓﺎت:
a
θ
c
b D وﺗﺘﻢ ﻋﻦ ﻃﺮﻳﻖ ﻗﻴﺎس زاوﻳﺔ أﻓﻘﻴﺔ ﻋﻠﻰ ﻃﺮﻓﻰ ﻗﻀﻴﺐ ﻣﻌﺮوف ﻃﻮﻟﻪ ﺑﺪﻗﺔ ﺷﺪﻳﺪة ،وﺑﺎﻟﺘﺎﻟﻰ ﻳﻤﻜﻦ ﻣﻌﺮﻓﺔ اﻟﻤﺴﺎﻓﺔ اﻷﻓﻘﻴﺔ ﺑﻴﻦ اﻟﻨﻘﻄﺔ اﻟﻤﺤﺘﻠﺔ واﻟﻨﻘﻄﺔ اﻟﻤﺮﺻﻮدة آﺎﻵﺗﻰ: 1 θ .ab. tan 2 2
21
Prepared By:/ Amr A. El-Sayed, Civil Eng Dept., El-Minia Univ., Eg.
=D
ﻣﺴﺎﺋﻞ ﻓﻰ اﻟﻤﺴﺎﺣﺔ اﻟﻤﺴﺘﻮﻳﺔ
ﻣﺴﺎﺋﻞ ﻋﻠﻰ اﻟﻘﻴﺎس اﻟﺘﺎآﻴﻮﻣﺘﺮى -1أﺧ ﺬت اﻟﻘ ﺮاءات اﻵﺗﻴ ﺔ ﻋﻠ ﻰ ﻗﺎﻣ ﺔ رأﺳ ﻴﺔ ﻣﻮﺿ ﻮﻋﺔ ﻋﻨ ﺪ ﻧﻘﻄﺘ ﻴﻦ ﺑﻮاﺳ ﻄﺔ ﺟﻬ ﺎز ﺗ ﺎآﻴﻮﻣﺘﺮى ﺑﻐ ﺮض ﺗﻌﻴ ﻴﻦ اﻟﺜﺎﺑ ﺖ اﻟﺘﺎآﻴﻮﻣﺘﺮى ،واﻹﺿﺎﻓﻰ .ﻋﻴﻦ اﻟﺜﺎﺑﺘﻴﻦ: اﻟﻤﺴﺎﻓﺔ زاوﻳﺔ ﻗﺮاءات اﻟﻘﺎﻣﺔ اﻷﻓﻘﻴﺔ اﻹرﺗﻔﺎع 150 ﺻﻔﺮ 4.106 3.359 2.613 200 7درﺟﺎت 3.154 2.150 1.146
1 -2ﺗﺎآﻮﻣﺘﺮ ﻣﺰود ﺑﺜﻼث ﺷﻌﺮات اﻟﻤﺴﺎﻓﺔ ﺑﻴﻦ آﻞ زوج ﻣﻨﻬﺎ 40 1 ﻣ ﻦ اﻟ ﺸﻴﺌﻴﺔ ﻟﻠﻤﺤ ﻮر اﻟﺮأﺳ ﻰ 4ﺑﻮﺻ ﺔ .وﺿ ﻌﺖ ﻗﺎﻣ ﺔ رأﺳ ﻴﺔ ﻋﻨ ﺪ ﻧﻘﻄ ﺔ ﻣﻨ ﺴﻮﺑﻬﺎ 80.0ﻗ ﺪم ،أﻣﻴ ﻞ اﻟﻤﻨﻈ ﺎر 9.00ﻋﻠ ﻰ 2 ﻣﻦ اﻟﺒﻮﺻﺔ ،واﻟﺒﻌ ﺪ اﻟﺒ ﺆرى ﻟﻠ ﺸﻴﺌﻴﺔ 9.0ﺑﻮﺻ ﺔ واﻟﻤ ﺴﺎﻓﺔ
اﻷﻓﻘﻰ ،وآﺎﻧﺖ ﻗﺮاءات اﻟﻘﺎﻣﺔ : (1.63 – 2.14 – 2.65 ) feet أوﺟﺪ اﻟﻤﺴﺎﻓﺔ اﻷﻓﻘﻴﺔ ﺑﻴﻦ اﻟﺠﻬﺎز واﻟﻘﺎﻣﺔ ،وآﺬﻟﻚ ﻣﻨﺴﻮب ﺧﻂ اﻟﻨﻈﺮ ﻋﻠﻤ ًﺎ ﺑﺄن ارﺗﻔﺎع اﻟﺠﻬﺎز 4.50ﻗﺪم.
-3ﻋﻴﻦ ﻣﻌﺪل اﻹﻧﺤﺪار ﺑﻴﻦ ﻧﻘﻄﺘﻴﻦ ) (A, Bﻣﻦ اﻷرﺻﺎد اﻵﺗﻴﺔ اﻟﻤﺄﺧﻮذة ﺑﺘﺎآﻴﻮﻣﺘﺮ ﻣﺠﻬﺰ ﺑﻌﺪﺳﺔ ﺗﺤﻠﻴﻠﻴﺔ )اﻟﺜﺎﺑﺖ اﻹﺿ ﺎﻓﻰ ﻟ ﻪ = ﺻﻔﺮ( ،وﺛﺎﺑﺘﻪ اﻟﺘﺎآﻴﻮﻣﺘﺮى .100.0 اﻟﺰاوﻳﺔ اﻟﻘﺮاءات اﻹﻧﺤﺮاف ﻣﻦ اﻟﺮأﺳﻴﺔ اﻟﺠﻬﺎز `+10 o 12 2.80 2.00 1.20 75o إﻟﻰ A o o `- 4 48 1.30 2.10 2.90 345 إﻟﻰ B
-4ﻗﻴﺲ ﺧﻂ ) (ABﺑﺎﺳﺘﺨﺪام ﻗﻀﻴﺐ اﻹﻧﻔﺎر ) ، (Subtense Barوﺧﻂ ﻗﺎﻋﺪة ﻣﺴﺎﻋﺪ ﻋﻤﻮدى ﻋﻠﻰ ﺟﺎﻧﺐ واﺣﺪ ﻣ ﻦ )،(AB وﻓﻰ ﻣﻨﺘ ﺼﻔﻪ ﺗﻤﺎﻣ ﺎً ،ﻓ ﺈذا آ ﺎن ﻃ ﻮل اﻟﺨ ﻂ ) (ABه ﻮ 684.0ﻣﺘ ﺮ ،وﻃ ﻮل اﻟﻘﺎﻋ ﺪة اﻟﻤ ﺴﺎﻋﺪ ه ﻮ 28.0ﻣﺘ ﺮ ،ﻓﻌ ﻴﻦ اﻟﺰاوﻳ ﺔ اﻷﻓﻘﻴﺔ ﺑﻴﻦ ﻧﻬﺎﻳﺘﻰ اﻟﻘﻀﻴﺐ )زاوﻳﺔ اﻟﺒﺮاﻟﻜﺲ( ،واﻟﺰاوﻳﺔ اﻟﻤﻮﺟﻮدة ﻋﻨﺪ ﻧﻬﺎﻳﺔ اﻟﺨﻂ. وإذا آﺎن ﺧﻂ اﻟﻘﺎﻋﺪة اﻟﻤﺴﺎﻋﺪ ﻳﻨﺼﻒ ) ،(ABوﻋﻠﻰ ﺟﺎﻧﺒﻴﻪ ،ﻓﻤﺎ هﻰ زاوﻳﺔ اﻟﺒﺮاﻟﻜﺲ؟ واﻟﺰاوﻳﺔ اﻟﻤﺮﺻﻮدة ﻋﻨﺪ ﻧﻬﺎﻳﺔ اﻟﺨﻂ. -5ﻗﻤﺔ ﺗﻞ ﻣﻌﻠﻮم إرﺗﻔﺎﻋﻬﺎ ﺑﺄﻧﻪ 2055ﻣﺘﺮ ﻓﻮق ﺳﻄﺢ اﻟﻤﻴﺎﻩ ﻓﻰ ﺑﺤﻴﺮة .رﺻﺪت هﺬﻩ اﻟﻘﻤﺔ ﻣﻦ اﻟﺠﺎﻧﺐ اﻵﺧﺮ ﻟﻠﺒﺤﻴﺮة ،وآﺎﻧﺖ زاوﻳﺔ إرﺗﻔﺎﻋﻬﺎ `` ، 5o 10ﻓﺈذا آﺎﻧﺖ زاوﻳﺔ إﻧﺨﻔﺎض ﺻﻮرة اﻟﻘﻤ ﺔ ﻓ ﻰ ﻣﻴ ﺎﻩ اﻟﺒﺤﻴ ﺮة ،8o 40`` :أوﺟ ﺪ اﻟﻤ ﺴﺎﻓﺔ اﻷﻓﻘﻴ ﺔ ﻣ ﻦ اﻟﺠﻬﺎز إﻟﻰ اﻟﺘﻞ ،وأوﺟﺪ آﺬﻟﻚ اﻟﻔﺮق ﺑﻴﻦ ﻣﻨﺴﻮﺑﻰ اﻟﻨﻘﻄﺘﻴﻦ )اﻋﺘﺒﺮ أن ﻣﻌﺎﻣﻞ اﻹﻧﻜﺴﺎر ﻟﻠﻤﺎء هﻮ ﻧﻔﺴﻪ ﻟﻠﻬﻮاء(. -6ﻹﻳﺠﺎد ﻣﻨﺴﻮب اﻟﻨﻘﻄﺔ ) (Aﻣﻦ اﻟﻨﻘﻄﺔ ) (Bاﻟﻤﻌﻠﻮم ﻣﻨﺴﻮﺑﻬﺎ وﺿﻊ اﻟﺘﻴﻮدوﻟﻴ ﺖ ﻓ ﻮق ﻧﻘﻄ ﺔ ﺟﺪﻳ ﺪة ) (Cوأﺧ ﺬت اﻟﻘ ﺮاءات اﻵﺗﻴﺔ ﻋﻠﻰ اﻟﻘﺎﻣﺘﻴﻦ اﻟﻤﻮﺿﻮﻋﺘﻴﻦ رأﺳﻴ ًﺎ ﻓﻮق ) (A, Bﻓﻜﺎﻧﺖ: اﻟﻘﺎﻣﺔ A B
اﻟﺰاوﻳﺔ اﻟﺮأﺳﻴﺔ ``-7o 35` 30 ``+10o 50` 30
ﻗﺮاءات اﻟﺸﻌﺮات )ﻣﺘﺮ( 0.950 1.502 2.055 2.98 2.0 1.018
ﻓﺈذا ﻋﻠﻢ أن اﻟﺠﻬﺎز ﺑﻪ ﻋﺪﺳﺔ ﺗﺤﻠﻴﻠﻴﺔ ،واﻟﺜﺎﺑ ﺖ اﻟﺘ ﺎآﻴﻮﻣﺘﺮى = 100وأن ﻣﻨ ﺴﻮب ﻧﻘﻄ ﺔ ) -42.03 = (Bﻣﺘ ﺮ .أوﺟ ﺪ ﻣﻨ ﺴﻮب ﻧﻘﻄﺔ ).(A 22
Prepared By:/ Amr A. El-Sayed, Civil Eng Dept., El-Minia Univ., Eg.
ﻣﺴﺎﺋﻞ ﻓﻰ اﻟﻤﺴﺎﺣﺔ اﻟﻤﺴﺘﻮﻳﺔ -7أوﺟﺪ إﻟﻰ أى زاوﻳﺔ رأﺳﻴﺔ ﻳﻤﻜﻦ اﻋﺘﺒﺎر اﻟﻤﺴﺎﻓﺔ اﻟﻤﺎﺋﻠﺔ ﺗﺴﺎوى اﻟﻤﺴﺎﻓﺔ اﻷﻓﻘﻴﺔ ﻓ ﻰ اﻟﻘﻴ ﺎس اﻟﺘ ﺎآﻴﻮﻣﺘﺮ ﺑﺤﻴ ﺚ أن اﻟﺨﻄ ﺄ ﻻ
1 ﻳﺰﻳﺪ ﻋﻦ 300
،وﻳﻮﺟﺪ ﺑﺎﻟﺠﻬﺎز ﻋﺪﺳﺔ ﺗﺤﻠﻴﻠﻴﺔ )أى أن اﻟﺜﺎﺑﺖ اﻹﺿﺎﻓﻰ = ﺻﻔﺮ(.
-8اﻟﺒﻌﺪ اﻟﺒﺆرى ﻟﻠﻌﺪﺳﺔ اﻟﺸﻴﺌﻴﺔ ﻓ ﻰ ﻣﻨﻈ ﺎر ه ﻮ 12.0ﺑﻮﺻ ﺔ ،واﻟﻤﺤ ﻮر اﻟﺮأﺳ ﻰ ﻟﻠ ﺪوران ﻓ ﻰ ﻣﻨﺘ ﺼﻒ اﻟﻤ ﺴﺎﻓﺔ ﺑ ﻴﻦ اﻟﺒ ﺆرة، واﻟﻌﺪﺳ ﺔ اﻟ ﺸﻴﺌﻴﺔ .وﺿ ﻌﺖ ﻗﺎﻣ ﺔ ﻋﻠ ﻰ ﺑﻌ ﺪ 301.50ﻗ ﺪم ﻣ ﻦ اﻟﻤﺤ ﻮر اﻟﺮأﺳ ﻰ ﻟﻠﺠﻬ ﺎز ،وآ ﺎن اﻟﺠ ﺰء اﻟﻤﻘﻄ ﻮع ﺑ ﻴﻦ ﺷ ﻌﺮﺗﻰ اﻻﺳﺘﺎدﻳﺎ ﻋﻠﻰ اﻟﻘﺎﻣﺔ = 3.0ﻗﺪم. ﻣﺎ هﻰ اﻟﻤﺴﺎﻓﺔ ﺑﻴﻦ ﺷﻌﺮﺗﻰ اﻟﺴﺘﺎدﻳﺎ ﻋﻠﻰ ﺣﺎﻣﻞ اﻟﺸﻌﺮات ﻓﻰ اﻟﺠﻬﺎز؟ -10وﺿ ﻌﺖ ﻗﺎﻣ ﺔ رأﺳ ﻴﺔ ورﺻ ﺪت ﺑﺘﻴﻮدوﻟﻴ ﺖ ﻋ ﺎدى ،ورﺻ ﺪت اﻟﺰواﻳ ﺎ اﻟﺮأﺳ ﻴﺔ ﻟﻬ ﺪﻓﻴﻦ ﻋﻠ ﻰ اﻟﻘﺎﻣ ﺔ ،ﻓ ﺈذا آﺎﻧ ﺖ اﻟﻤ ﺴﺎﻓﺔ اﻟﺮأﺳ ﻴﺔ ﺑﻴﻨﻬﻤ ﺎ 4.26ﻣﺘ ﺮ ،واﻟﻔ ﺮق ﺑ ﻴﻦ ﻇﻠ ﻰ زاوﻳﺘ ﻰ اﻹرﺗﻔ ﺎع = ، 0.044ﻣ ﺎهﻮ ﻣﻨ ﺴﻮب ﻧﻘﻄ ﺔ اﻟﻘﺎﻣ ﺔ إذا آ ﺎن ﻇ ﻞ زاوﻳ ﺔ اﻟﻬ ﺪف اﻟ ﺴﻔﻠﻰ = 0.161واﻹرﺗﻔ ﺎع ﻣ ﻦ اﻷرض ﻟﻠﻬ ﺪف اﻟﻌﻠ ﻮى = 1.75ﻣﺘ ﺮ ،وﻣﻨ ﺴﻮب ﺳ ﻄﺢ اﻟﺠﻬ ﺎز ﺗﺤ ﺖ ﺳ ﻄﺢ اﻟﺒﺤ ﺮ ﺑﻤﻘﺪار 4.0ﻣﺘﺮ. -12ﻓﻰ ﺗﺮاﻓﺮس ،أرﻳﺪ إﻳﺠﺎد اﻟﻤﺴﺎﻓﺔ ﺑﻴﻦ اﻟﻨﻘﻄﺘﻴﻦ ) (A, Bاﻟﻨﻘﻄﺔ Aآﺎﻧ ﺖ ﻇ ﺎهﺮة ﻣ ﻦ ﻧﻘﻄ ﺔ Xاﺣ ﺪى ﻧﻘ ﻂ اﻟﺘﺮاﻓ ﺮس ،أﻣ ﺎ ﻧﻘﻄﺔ Bﻓﻜﺎﻧﺖ ﻇﺎهﺮة ﻣﻦ ﻧﻘﻄﺔ أﺧﺮى . Yأﺧﺬت أرﺻ ﺎد ﺗﺎآﻴﻮﻣﺘﺮﻳ ﺔ ﻣ ﻦ اﻟﻨﻘﻄﺘ ﻴﻦ ) (X, Yﻋﻠ ﻰ ﻗ ﺎﺋﻤﺘﻴﻦ ﻣﻮﺿ ﻮﻋﺘﻴﻦ ﻓ ﻮق اﻟﻨﻘﻄﺘﻴﻦ ) (A, Bوآﺎﻧﺖ اﻷرﺻﺎد آﺎﻵﺗﻰ: Vertical Hair angle Readings 09o 22` 1.50 2.11 2.72 12o 18` 1.80 2.25 2.70
Traverse Bearing Point A `326o 42 B `10o 27
Occupied Co-ordinates Point Horizontal Vertical X 3800 E 1000 N Y 3600 W 740 N
اﻟﺠﻬﺎز ﻣﺰود ﺑﻌﺪﺳﺔ ﺗﺤﻠﻴﻠﻴﺔ ،واﻟﺜﺎﺑﺖ اﻟﺘﺎآﻴﻮﻣﺘﺮ = . 100اﺣﺴﺐ اﻟﻤﺴﺎﻓﺔ ،ABواﻧﺤﺮاف اﻟﺨﻂ .AB -13أﺧﺬت ﻗﺮاءات ﻣﻦ ﺟﻬﺎزى ﺗﺎآﻴﻮﻣﺘﺮ ﻋﻨﺪ ﻧﻘﻄﺔ ،Aاﻟﺘﻰ ﻣﻨﺴﻮﺑﻬﺎ 15.05ﻣﺘﺮ ،إﻟﻰ ﻗﺎﻣﺔ ﻣﻮﺿﻮﻋﺔ ﻓﻮق :B اﻟﺠﻬﺎز اﻷول ) : ( ρﺛﺎﺑﺘﻪ اﻟﺘﺎآﻴﻮﻣﺘﺮى = ، 100وﺛﺎﺑﺘﻪ اﻹﺿﺎﻓﻰ = 14.40ﺑﻮﺻﺔ. اﻟﺠﻬﺎز اﻟﺜﺎﻧﻰ ) : ( φﺛﺎﺑﺘﻪ اﻟﺘﺎآﻴﻮﻣﺘﺮى = ، 95وﺛﺎﺑﺘﻪ اﻹﺿﺎﻓﻰ = 15.0ﺑﻮﺻﺔ. وآﺎن ارﺗﻔﺎع اﻟﺠﻬﺎز ) ( ρﻋﻨﺪ ﻧﻘﻄﺔ 1.45 = Aﻣﺘﺮ ،وزاوﻳﺔ اﻹرﺗﻔﺎع
23
Prepared By:/ Amr A. El-Sayed, Civil Eng Dept., El-Minia Univ., Eg.
ﻣﺴﺎﺋﻞ ﻓﻰ اﻟﻤﺴﺎﺣﺔ اﻟﻤﺴﺘﻮﻳﺔ
Theory of Errors آﻠﻤﺎ وﺟﺪت أرﺻﺎد ،آﻠﻤﺎ وﺟﺪت أﺧﻄﺎء ،وﺗﺨﺘﻠﻒ ﻃﺒﻴﻌﺔ هﺬﻩ اﻷﺧﻄﺎء ،وﻗﺪ ﺗﻌﻮد إﻟ ﻰ اﻟ ﺸﺨﺺ اﻟﺮاﺻ ﺪ أو اﻵﻟ ﺔ اﻟﺘ ﻰ ﻳﺮﺻﺪ ﺑﻬﺎ أو إﻟﻰ ﻋﻮاﻣﻞ ﻃﺒﻴﻌﻴﺔ ﺧﺎرﺟﻴﺔ: 1- Types of Errors: 1-1- According to the sources:
a- Personal Errors b- Instrumental Errors c- Natural Errors 2-1 According to kind
a- Systematic Error b- Un-systematic (random) Error c- Gross error 2- propagation of Errors ﻋﻨﺪ ﻗﻴﺎس ﺧﻂ ﻃﻮﻟﻪ ﻳﺰﻳﺪ ﻋﻦ ﻃﻮل اﻟﺸﺮﻳﻂ ﻋﺪة ﻣﺮات ،ﻓﺈن ﻋﺪد ﻣﺮات اﻟﻘﻴﺎس ) ،(nوإذا وﺟﺪ ﺧﻄﺄ ﻓﻰ اﻟﻘﻴﺎس ﻓﺈﻧ ﻪ ﻳﺘﻜﺮر ﺑﻨﻔﺲ اﻟﻌﺪد ﻣﻦ اﻟﻤﺮات ،ﻓﺈذا آﺎﻧﺖ ﻗﻴﻤﺔ هﺬا اﻟﺨﻄﺄ ﻓﻰ اﻟﻤﺮة اﻟﻮاﺣﺪة ) ،(Eﻓﺈن اﻟﺨﻄﺄ اﻟﻨﺎﺗﺞ ﻓﻰ اﻟﺨﻂ آﻠﻪ = ).(n.E وﻧﻼﺣﻆ أن اﻟﺨﻄﺄ ﻳﺘﺠﻤﻊ وﻻ ﺗﺘﻐﻴﺮ ﻧﺴﺒﺘﻪ ﺑﺰﻳﺎدة ﻃﻮل اﻟﺨﻂ. 3- Random Errors: اﻷﺧﻄﺎء ﻟﻬﺎ ﻣﺼﺎدر ﻣﺘﻌﺪدة ،وﻹرﺟﺎع اﻟﺨﻄﺄ إﻟﻰ ﻣﺼﺪرﻩ ﺳﻮاء آﺎن ﺷﺨﺼﻰ أو ﻣﻦ اﻟﺠﻬﺎز اﻟﻤﺴﺘﺨﺪم ﻳﺘﻢ اﻵﺗﻰ: ﻋﻨﺪ ﻋﻤﻞ ﻣﺠﻤﻮﻋﺔ ﻣﻦ اﻷﺷﺨﺎص ﻋﻠﻰ ﻧﻔﺲ اﻟﺠﻬﺎز ﻓﺈن ﻣﺼﺪر اﻷﺧﻄﺎء ﻳﻜﻮن ﺷﺨﺼﻰ. ﻋﻨﺪ اﺳﺘﺨﺪام ﺷﺨﺺ واﺣﺪ ﻣﺠﻤﻮﻋﺔ ﻣﻦ اﻷﺟﻬﺰة ،ﻓﺈن ﻣﺼﺪر اﻟﺨﻄﺄ ﻳﻜﻮن ﻣﻦ اﻟﺠﻬﺎز،ﺑﻌﺪ اﻟﺘﺨﻠﺺ ﻣﻦ اﻷﺧﻄﺎء اﻟﻤﻨﺘﻈﻤﺔ ،واﻟﺜﺎﺑﺘﺔ ﺗﺒﻘﻰ اﻷﺧﻄﺎء اﻟﻌﺎرﺿﺔ ،وﺗﺘﻤﻴﺰ ﺑﺄﻧﻬﺎ ﺻﻐﻴﺮة ﺟﺪًا ،وﻏﻴﺮ ﺛﺎﺑﺘﺔ اﻹﺷﺎرة أو اﻟﻘﻴﻤﺔ ،وﻻ ﻳﻤﻜﻦ ﻣﻌﺮﻓﺔ أﺳﺒﺎﺑﻬﺎ أو اﻟﺘﺨﻠﺺ ﻣﻨﻬﺎ.
ﺑﻌﺾ اﻟﺘﻌﺎرﻳﻒ اﻟﻬﺎﻣﺔ Direct Measure -1اﻟﻘﻴﺎس اﻟﻤﺒﺎﺷﺮ: هﻮ اﻟﻘﻴﺎس اﻟﺬى ﻳﺘﻢ ﺑﻮاﺳﻄﺘﻪ ﺗﺤﺪﻳﺪ اﻟﻜﻤﻴﺔ اﻟﻤﻘﺎﺳﺔ ﻣﺒﺎﺷﺮة ،ﻣﺜﻞ ﻗﻴﺎس ﻃﻮل ﺧﻂ ﺑﻮاﺳ ﻄﺔ اﻟ ﺸﺮﻳﻂ ،أو ﻗﻴ ﺎس ﻗﻴﻤ ﺔ زاوﻳﺔ ﺑﻮاﺳﻄﺔ اﻟﺘﻴﻮدوﻟﻴﺖ. Indirect Measure -2اﻟﺮﺻﺪ أو اﻟﻘﻴﺎس ﻏﻴﺮ اﻟﻤﺒﺎﺷﺮ: اﻟﻘﻴﺎس اﻟﻐﻴﺮ ﻣﺒﺎﺷﺮ ﻳﻌﻨ ﻰ اﻟﺤ ﺼﻮل ﻋﻠ ﻰ آﻤﻴ ﺔ ،ﻟ ﻢ ﻳ ﺘﻢ ﻋﻤ ﻞ ﻗﻴ ﺎس ﻣﺒﺎﺷ ﺮ ﻟﻬ ﺎ .ﻣﺜ ﻞ ﺣ ﺴﺎب ﻃ ﻮل ﺿ ﻠﻊ ﻓ ﻰ ﻣﺜﻠ ﺚ ﺑﻤﻌﻠﻮﻣﻴﺔ اﻟﻀﻠﻌﻴﻦ اﻵﺧﺮﻳﻦ ،واﻟﺰاوﻳﺔ ﺑﻴﻨﻬﻤﺎ ،ﻓﺎﻟﻀﻠﻊ اﻟﺬى ﺗﻢ اﻟﺤﺼﻮل ﻋﻠﻰ ﻗﻴﻤﺔ ﻗﻴﺎﺳﻪ ﻟﻢ ﻳﺴﺘﺨﺪم اﻟﺸﺮﻳﻂ ﻓﻰ ﻗﻴﺎﺳﻪ ،وإﻧﻤﺎ اﺳﺘﺨﺪﻣﺖ ﻋﻼﻗﺔ ﺧﺎﺻﺔ ﺗﺮﺑﻂ أﺿﻼع اﻟﻤﺜﻠﺚ. -3اﻟﻘﻴﻤﺔ اﻷﻛﺜﺮ اﺣﺘﻤﺎﻻ ً:
The most probable value
ﻳﺘﻌﺬر اﻟﺘﺨﻠﺺ ﺗﻤﺎﻣ ًﺎ ﻣﻦ اﻷﺧﻄﺎء اﻟﻌﺎدﻳﺔ واﻟﻤﻨﺘﻈﻤﺔ ،آﻤﺎ ﻳﺘﻌﺬر أﻳﻀ ًﺎ أﺧﺬ ﻋﺪد ﻻ ﻧﻬ ﺎﺋﻰ ﻣ ﻦ اﻷرﺻ ﺎد ﻟﻠﺤ ﺼﻮل ﻋﻠ ﻰ اﻟﻘﻴﻤﺔ اﻟﺤﻘﻴﻘﻴﺔ ﻟﻠﻜﻤﻴﺔ اﻟﻤﻘﺎﺳﺔ .وﺑﻔﺮض أن اﻷﺧﻄﺎء اﻟﻌﺎدﻳﺔ واﻟﻤﻨﺘﻈﻤﺔ ﺗﻼﺷﺖ إﻟﻰ أﻗﺼﻰ ﺣﺪ ﻣﻤﻜﻦ ﺑﺤﻴﺚ ﻳﻤﻜﻦ اهﻤﺎﻟﻬﺎ ،ﻓﺈن أﻓﻀﻞ ﻗﻴﻤﺔ اﺣﺘﻤﺎﻟﻴﺔ ﺗﻜﻮن هﻰ اﻟﻘﻴﻤﺔ اﻟﺘﻰ ﻳﺤﺘﻤﻞ ﺻﺤﺘﻬﺎ أآﺜﺮ ﻣﻦ ﻏﻴﺮهﺎ ﺗﺒﻌ ًﺎ ﻟﻨﻈﺮﻳﺔ أﻗﻞ ﻣﺠﻤﻮع ﻟﻤﺮﺑﻌﺎت اﻟﺘﺼﺤﻴﺢ. 24
Prepared By:/ Amr A. El-Sayed, Civil Eng Dept., El-Minia Univ., Eg.
ﻣﺴﺎﺋﻞ ﻓﻰ اﻟﻤﺴﺎﺣﺔ اﻟﻤﺴﺘﻮﻳﺔ ﻟﺬﻟﻚ ﻓﺈن اﻟﻘﻴﻤﺔ اﻟﻤﺘﻮﺳﻄﺔ ﻟﻸرﺻﺎد اﻟﻌﺪﻳﺪة ﻟﻠﻜﻤﻴﺔ اﻟﻤﻄﻠﻮﺑ ﺔ ﻳﻄﻠ ﻖ ﻋﻠﻴﻬ ﺎ اﻟﻘﻴﻤ ﺔ اﻷآﺜ ﺮ اﺣﺘﻤ ﺎﻻً ،وﻳﻘ ﺼﺪ ﺑ ﺬﻟﻚ أﻧﻬ ﺎ أﻗ ﺮب ﻣ ﺎ ﻳﻤﻜﻦ ﻟﻠﻜﻤﻴﺔ اﻟﺤﻘﻴﻘﻴﺔ. -4اﻟﺨﻄﺄ اﻟﺤﻘﻴﻘﻰ:
True error
هﻮ اﻟﻔﺮق ﺑﻴﻦ اﻟﻘﻴﻤﺔ اﻟﺤﻘﻴﻘﻴﺔ ﻟﻠﻜﻤﻴﺔ اﻟﻤﺮﺻﻮدة )وهﻰ ﻳﺘﻌﺬر او ﻳﺴﺘﺤﻴﻞ اﻟﺤ ﺼﻮل ﻋﻠﻴﻬ ﺎ( وﺑ ﻴﻦ اﻟﻘﻴﻤ ﺔ اﻟﻤﺮﺻ ﻮدة ﻟﻬ ﺎ .وﻟﻤ ﺎ آﺎن اﻟﺤﺼﻮل ﻋﻠﻰ اﻟﻘﻴﻤﺔ اﻟﺤﻘﻴﻘﻴﺔ ﻣﺴﺘﺤﻴﻞ ،ﻓﺈن اﻟﺤﺼﻮل ﻋﻠﻰ اﻟﺨﻄﺄ اﻟﺤﻘﻴﻘﻰ ﻣﺴﺘﺤﻴﻞ أﻳﻀ ًﺎ. اﻟﺨﻄﺄ اﻟﺤﻘﻴﻘﻰ = اﻟﻘﻴﻤﺔ اﻟﻤﺮﺻﻮدة – اﻟﻘﻴﻤﺔ اﻟﺤﻘﻴﻘﻴﺔ -5اﻷﺧﻄﺎء اﻟﺒﺎﻗﻴﺔ:
Residual error
هﻮ اﻟﻔﺮق ﺑﻴﻦ اﻟﻘﻴﻤﺔ اﻟﻤﺮﺻﻮدة ،وﺑﻴﻦ اﻟﻘﻴﻤﺔ اﻷآﺜﺮ اﺣﺘﻤﺎ ًﻻ ﻟﻬﺎ اﻟﺨﻄﺄ اﻟﺒﺎﻗﻰ = اﻟﻘﻴﻤﺔ اﻟﻤﺮﺻﻮدة – اﻟﻘﻴﻤﺔ اﻷآﺜﺮ اﺣﺘﻤﺎ ًﻻ
-6اﻟﻤﺘﻮﺳﻂ اﻟﺤﺴﺎﺑﻰArithmetic mean : ﻋﻨﺪﻣﺎ ﻳﻘﻮم اﻟﺮاﺻﺪ ﺑﺮﺻﺪ آﻤﻴﺔ ﻣﺎ ﻋﺪد ﻣ ﻦ اﻟﻤ ﺮات = ،nﻓﺈﻧ ﻪ ﻳﺤ ﺼﻞ ﻋﻠ ﻰ ﻧﺘ ﺎﺋﺞ ﻣﺨﺘﻠﻔ ﺔ ،وﻣﺘﻮﺳ ﻂ ه ﺬﻩ اﻷرﺻ ﺎد ﻳﺴﺎوى اﻟﻤﺠﻤﻮع اﻟﻜﻠﻰ ﻟﻸرﺻﺎد ÷ ﻋﺪد ﻣﺮات اﻟﺮﺻﺪ. وﻟﻠﻤﺘﻮﺳﻂ اﻟﺤﺴﺎﺑﻰ ﻋﺪة ﺧﻮاص أهﻤﻬﺎ: أ -ﻣﺠﻤﻮع اﻷﺧﻄﺎء اﻟﺒﺎﻗﻴﺔ ﻳﺴﺎوى ﺻﻔﺮ )ﻷن اﻟﺨﻄﺄ اﻟﺒ ﺎﻗﻰ = اﻟﻘﻴﻤ ﺔ اﻟﻤﺮﺻ ﻮدة – اﻟﻘﻴﻤ ﺔ اﻷآﺜ ﺮ اﺣﺘﻤ ﺎ ًﻻ( ،وﻟﻤ ﺎ آﺎﻧ ﺖ اﻟﻘﻴﻤ ﺔ اﻷآﺜﺮ اﺣﺘﻤﺎ ًﻻ ﺗﻤﺜﻞ اﻟﻤﺘﻮﺳﻂ اﻟﺤﺴﺎﺑﻰ ﻓﺈن ﻣﺠﻤﻮع اﻷﺧﻄﺎء اﻟﺒﺎﻗﻴﺔ ﻟﻸرﺻﺎد = ﺻﻔﺮ. ب -ﻣﺠﻤﻮع ﻣﺮﺑﻌﺎت اﻷﺧﻄﺎء اﻟﺒﺎﻗﻴﺔ أﻗﻞ ﻣﺎ ﻳﻤﻜﻦ -7اﻟﺨﻄﺄ اﻟﺘﺮﺑﻴﻌﻰ اﻟﻤﺘﻮﺳﻂ:
)Mean square error (MSE
ﻟﺘﻘﺪﻳﺮ دﻗﺔ اﻷرﺻﺎد ﻟﻜﻤﻴﺔ ﻣﺎ ﻣﻌﻠﻮم ﻗﻴﻤﺘﻬ ﺎ اﻟﺤﻘﻴﻘﻴ ﺔ ﻳ ﺴﺘﺨﺪم اﻹﻧﺤ ﺮاف اﻟﻤﻌﻴ ﺎرى أو ﻣ ﺎ ﻳ ﺴﻤﻰ ﻓ ﻰ ﻧﻈﺮﻳ ﺔ اﻷﺧﻄ ﺎء ﺑﺎﻟﺨﻄﺄ اﻟﺘﺮﺑﻴﻌﻰ اﻟﻤﺘﻮﺳﻂ ).Mean Square Error (M.S.E -اﻟﺨﻄﺄ اﻟﺘﺮﺑﻴﻌﻰ اﻟﻤﺘﻮﺳﻂ ﻟﻠﺮﺻﺪة اﻟﻮاﺣﺪة:
MSE of a single observation
) ∑ (X i − X n
2
i =1
n −1
=σ
اﻟﺨﻄﺄ اﻟﺘﺮﺑﻴﻌﻰ اﻟﻤﺘﻮﺳﻂ ﻟﻠﻤﺘﻮﺳﻂ اﻟﺤﺴﺎﺑﻰMSE of the arithmetic mean :ﻣﻦ اﻟﻤﻌﻠﻮم أن دﻗﺔ ﺗﻌﻴﻴﻦ اﻟﻜﻤﻴﺔ اﻟﻤﺠﻬﻮﻟﺔ ﺑﺎﺳﺘﺨﺪام اﻟﻤﺘﻮﺳﻂ اﻟﺤﺴﺎﺑﻰ أآﺒﺮ ﻣﻦ ﺗﻌﻴﻴﻨﻬﺎ ﺑﺎﺳﺘﺨﺪام أى رﺻﺪة أﺧ ﺮى. وﺑﺎﻟﺘﺎﻟﻰ ﻓﺈن اﻟﺨﻄﺄ اﻟﺘﺮﺑﻴﻌﻰ اﻟﻤﺘﻮﺳ ﻂ ﻟﻠﻤﺘﻮﺳ ﻂ اﻟﺤ ﺴﺎﺑﻰ أﻗ ﻞ ﻣ ﻦ اﻟﺨﻄ ﺄ اﻟﺘﺮﺑﻴﻌ ﻰ اﻟﻤﺘﻮﺳ ﻂ ﻟﻠﺮﺻ ﺪة اﻟﻮاﺣ ﺪة ،واﻟﻌﻼﻗ ﺔ ﺑ ﻴﻦ اﻟﺨﻄﺄ اﻟﺘﺮﺑﻴﻌﻰ اﻟﻤﺘﻮﺳﻂ ﻟﻠﺮﺻﺪة اﻟﻮاﺣﺪة ،وﺑﻴﻦ اﻟﺨﻄﺄ اﻟﺘﺮﺑﻴﻌﻰ اﻟﻤﺘﻮﺳﻂ ﻟﻠﻤﺘﻮﺳﻂ اﻟﺤﺴﺎﺑﻰ هﻰ:
σ2 n the mean square error of the arithmetic mean
where σo
σ n
25
Prepared By:/ Amr A. El-Sayed, Civil Eng Dept., El-Minia Univ., Eg.
= σ o2
= σo
ﻣﺴﺎﺋﻞ ﻓﻰ اﻟﻤﺴﺎﺣﺔ اﻟﻤﺴﺘﻮﻳﺔ
∑ (X i − X ) n
σo =
2
i =1
n.(n − 1) Relative error
: اﻟﺨﻄﺄ اﻟﻨﺴﺒﻰ-8
ﻓﺈﻧ ﻪ ﻣ ﻦ،ﻼ ﻃ ﻮل وزاوﻳ ﺔ( واﻟﻤﻄﻠ ﻮب ﻣﻌﺮﻓ ﺔ أﻳﻬﻤ ﺎ أدق ﻓ ﻰ اﻟﻘﻴ ﺎس ً ﻋﻨﺪ اﺧﺘﻼف اﻟﻜﻤﻴﺎت اﻟﻤﻘﺎﺳﺔ ﻓﻰ اﻟﻮﺣﺪات )ﻣﺜ . ﻟﺬﻟﻚ ﺗﻌﻴﻦ آﻤﻴﺔ ﺗﺴﻤﻰ اﻟﺨﻄﺄ اﻟﻨﺴﺒﻰ ﻟﻠﻤﻘﺎرﻧﺔ ﻟﺘﺤﺪﻳﺪ أى اﻷرﺻﺎد أدق،اﻟﺨﻄﺄ ﻣﻘﺎرﻧﺔ اﻷﺧﻄﺎء ﻓﻰ اﻟﺤﺎﻟﺘﻴﻦ
Re lative Error =
σ X Mean error
: اﻟﺨﻄﺄ اﻟﻤﺘﻮﺳﻂ-9
: وﻗﺪ أﻣﻜﻦ اﺛﺒﺎت أن،(η) وﻳﺮﻣﺰ ﻟﻪ ﺑﺎﻟﺮﻣﺰ،وهﻮ ﻳﺴﺎوى اﻟﻤﺘﻮﺳﻂ اﻟﺤﺴﺎﺑﻰ ﻟﻸﺧﻄﺎء ﻣﺠﺮدة ﻣﻦ اﻹﺷﺎرة
η=
2
π
.σ = 0.7979 σ
Probable error
: اﻟﺨﻄﺄ اﻟﻤﺤﺘﻤﻞ-10
ﻟ ﺬﻟﻚ ﻓﻬ ﻮ ﻳﻤﺜ ﻞ اﻟ ﺮﻗﻢ اﻷوﺳ ﻂ ﻓ ﻰ ﺗﺮﺗﻴ ﺐ، ﻣﻦ وﻗﻮع ﺟﻤﻴ ﻊ اﻷﺧﻄ ﺎء50% = هﻮ اﻟﺨﻄﺄ اﻟﺬى ﻳﻜﻮن اﺣﺘﻤﺎل وﻗﻮﻋﻪ : وﻗﺪ أﻣﻜﻦ اﺛﺒﺎت أن اﻟﺨﻄﺄ اﻟﻤﺤﺘﻤﻞ ﻳﻤﻜﻦ ﺗﻌﻴﻨﻪ ﻣﻦ،اﻷﺧﻄﺎء اﻟﺒﺎﻗﻴﺔ r = 0.6745 σ example: 3- A distance was measured eight times and the observations were as follows: 118.167 m 118.750 m 118.273 m 118.266 m 118.165 m 118.167 m 118.760 m 118.280 m a- calculate the mean value ( X ) of this distance, and the mean square error (MSE) of the observations (σx) of a single measurement. b- Calculate the (MSE) of the arithmatic mean (σx). c- Calculate the relative error. d- Find the mean error (a) and the probable error (r) from the above observations, then check their relations with (σ) e- Supposing that if the true value of that distance was 118.248 m, what would you comment upon these observations?
Prepared By:/ Amr A. El-Sayed, Civil Eng Dept., El-Minia Univ., Eg.
26
ﻣﺴﺎﺋﻞ ﻓﻰ اﻟﻤﺴﺎﺣﺔ اﻟﻤﺴﺘﻮﻳﺔ Solution No.
Observation
1 2 3 4 5 6 7 8
118.167 118.750 118.273 118.266 118.165 118.167 118.760 118.280 Σ = 946.828
Residual error
Xi − X -0.1865 0.3965 -0.0805 -0.0875 -0.1885 -0.1865 0.4065 -0.0735
error Sorting
True error
Xi − Xt
error Sorting
0.0735 0.0805 0.0875 0.1865 0.1865 0.1885 0.3965 0.4065
-0.081 0.502 0.025 0.018 -0.083 -0.081 0.512 0.032
0.018 0.025 0.032 0.081 0.081 0.083 0.502 0.512
n
∑Xi a- mean value =
i =1
n
=
946.828 = 118.3535 m 8
∑ (X i − X ) n
- mean square error of a single observation (σn-1) =
i =1
σo =
= 0.252725
n −1
∑ (X i − X ) n
b- MSE of the arithmetic mean =
2
2
i =1
n.(n − 1)
=
σ n−1 n
=
0.252725 = 0.08935 8
c- relative error : Re =
σ n−1 X
=
0.252725 = 0.002135 118.3535 n
∑ Xi − X d- Mean error (a) =
i =1
n
=
1.606 = 0.20075 8
check of mean error value with (σn-1) a = 0.7979 σ = 0.7979 * 0.2527 = 0.2016 d- Probable error: إذا آ ﺎن ﻋ ﺪد اﻷرﺻ ﺎد ﻓﺮدﻳ ﺔ ﻓ ﺈن اﻟﺨﻄ ﺄ رﻗ ﻢ، ﺑ ﺪون اﻟﻨﻈ ﺮ إﻟ ﻰ اﻹﺷ ﺎرة،ﻳﺘﻢ ﺗﺮﺗﻴﺐ اﻷﺧﻄﺎء ﻣﻦ اﻷﺻﻐﺮ إﻟ ﻰ اﻷآﺒ ﺮ .ﻳﻤﺜﻞ اﻟﺨﻄﺄ اﻷآﺜﺮ اﺣﺘﻤﺎ ًﻻ
n +1 2
n n + 1 ( وإذا آﺎﻧﺖ ﻋﺪد اﻷرﺻﺎد زوﺟﻴﺔ ﻓﺈن اﻟﺨﻄﺄ اﻷآﺜﺮ اﺣﺘﻤﺎ ًﻻ = ﻣﺘﻮﺳﻂ اﻟﺨﻄﺄﻳﻦ رﻗﻤﻰ 2 2
. ) ,
Prepared By:/ Amr A. El-Sayed, Civil Eng Dept., El-Minia Univ., Eg.
27
ﻣﺴﺎﺋﻞ ﻓﻰ اﻟﻤﺴﺎﺣﺔ اﻟﻤﺴﺘﻮﻳﺔ
0.1865 + 0.1865 = 0.1865 2
=r
r = 0.6745 σ = 0.6745 * 0.2527 = 0.1704 e- if the true value = 118.248, then, it can be concluded that both observations (118.750, and 118.760) contains random errors, and must be excluded. -11اﻷوزان: ﻼ اﻟﺸﺨﺺ اﻟﺪﻗﻴﻖ ﻟﻪ أرﺻ ﺎد ﻟﻬ ﺎ وزن ،وﺑﺎﻟﺘ ﺎﻟﻰ ﻓ ﺈن اﻷرﺻ ﺎد اﻟﺘ ﻰ ﻟﻬ ﺎ وزن وﻳﻘﺼﺪ ﺑﻬﺎ ﻣﺪى اﻹﻋﺘﺪاد ﺑﺎﻟﺮﺻﺪة ،ﻣﺜ ً ﻋﺒﺎرة ﻋﻦ أرﺻﺎد ﻟﻬﺎ ﺛﻘﺔ ﻧﺴﺒﻴﺔ.
1
σ2
= weight Where:
Standard deviation
σ
) ∑ (X i − X n
2
i =1
n −1
=σ
واﻹﻧﺤﺮاف اﻟﻤﻌﻴﺎرى ﻳﻤﺜﻞ إﻧﺤﺮاف اﻟﺮﺻﺪة اﻟﻮاﺣﺪة ﻋﻦ اﻟﻘﻴﻤﺔ اﻟﻤﺘﻮﺳﻄﺔ ﻟﻸرﺻﺎد آﻠﻬﺎ .وﺑﺎﻟﺘﺎﻟﻰ آﻠﻤﺎ ﻗﻞ اﻹﻧﺤﺮاف اﻟﻤﻌﻴﺎرى ﻟﻠﺮﺻﺪة آﻠﻤﺎ زداد وزﻧﻬﺎ. -12ﺧﻮاص ﻣﻨﺤﻨﻰ اﻷﺧﻄﺎء )ﻣﻨﺤﻨﻰ ﺟﺎوس(: ﻋﺒﺎرة ﻋﻦ ﻣﻨﺤﻨﻰ ﻳﻤﺜﻞ ﻓﻴﻪ اﻟﻤﺤﻮر اﻷﻓﻘﻰ اﻟﻘﻴﻤﺔ اﻟﻌﺪدﻳﺔ واﻟﺠﺒﺮﻳﺔ ﻟﻸﺧﻄﺎء ،وﻳﻤﺜﻞ اﻟﻤﺤﻮر اﻟﺮأﺳﻰ اﺣﺘﻤﺎل وﻗﻮع اﻟﺨﻄﺄ أو ﻋﺪد ﻣﺮات ﺣﺪوﺛﻪ. ﻼ إذا ﻗﺴﻨﺎ آﻤﻴﺔ ﻣﺎ ﻋﺪد آﺒﻴﺮ ﺟﺪًا ﻣﻦ اﻟﻤﺮات ،ﺛﻢ ﺣﺪدﻧﺎ اﻟﺨﻄﺄ اﻟﺒﺎﻗﻰ وﻳﺴﺎوى اﻟﻔﺮق ﺑﻴﻦ اﻟﻘﻴﻢ اﻟﻤﺮﺻﻮدة وﺑﻴﻦ اﻟﻘﻴﻤﺔ ﻓﻤﺜ ً اﻷآﺜﺮ اﺣﺘﻤﺎﻻً ،ﺛﻢ ﺗﻢ ﺗﺮﺗﻴﺐ اﻟﻨﺘﺎﺋﺞ ﻓﻰ ﺟﺪول آﺎﻟﺘﺎﻟﻰ:
28
Prepared By:/ Amr A. El-Sayed, Civil Eng Dept., El-Minia Univ., Eg.
ﻣﺴﺎﺋﻞ ﻓﻰ اﻟﻤﺴﺎﺣﺔ اﻟﻤﺴﺘﻮﻳﺔ
:ﻣﺜﺎل 1- the angle (ABC) was measured 25 times, with equally accuracy. By taking the arithmetic mean of the observations, the most probable value of the angle was (94o 04` 20.55``). The values of observations are listed in the following table: No.
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
Observations o
Residual Error
+V
-V
1.80 1.45 1.45 1.20 0.95 0.70 0.70 0.70 0.40 0.30 0.15 0.05
1.80 1.50 1.50 1.05 1.00 0.80 0.55 0.55 0.30 0.05 0.05 0.05 0.05
Xi − X
94 04` 20.85`` 94o 04` 20.50`` 94o 04` 20.60`` 94o 04` 20.50 94o 04` 20.70`` 94o 04` 19.05`` 94o 04` 20.50 94o 04` 22.00`` 94o 04` 19.55`` 94o 04` 19.75`` 94o 04` 20.50`` 94o 04` 19.05`` 94o 04` 18.75`` 94o 04` 20.00`` 94o 04` 20.00`` 94o 04` 21.25`` 94o 04` 21.25`` 94o 04` 22.00`` 94o 04` 21.50`` 94o 04` 19.50`` 94o 04` 21.75`` 94o 04` 22.35`` 94o 04` 20.95`` 94o 04` 20.35`` 94o 04` 21.25``
0.30 -0.05 0.05 -0.05 0.15 -1.50 -0.05 1.45 -1.0 -0.80 -0.05 -1.50 -1.80 -0.55 -0.55 0.70 0.70 1.45 0.95 -1.05 1.20 1.80 0.40 -0.20 0.70
:( ﺛﻢ ﻧﺤﺪد ﻋﺪد اﻷﺧﻄﺎء ﻓﻰ آﻞ ﻓﺘﺮة آﺎﻟﺘﺎﻟﻰ0.50``) ﻧﻘﺴﻢ اﻷﺧﻄﺎء إﻟﻰ ﻓﺘﺮات وﻟﺘﻜﻦ ﻗﻴﻤﺔ اﻟﻔﺘﺮة Period
2.0-1.5 1.5-1.0 1.0-0.5 0.5-0.0 0.0-(-0.5) -0.5-(-1.0) -1.0-(-1.5) -1.5-(-2.0)
No. of Errors 1 3 4 4 5 4 3 1
، وﻳﻤﺜ ﻞ ارﺗﻔﺎﻋ ﻪ ﻋ ﺪد ﻣ ﺮات ﺗﻜ ﺮار اﻟﺨﻄ ﺄ، ﺑﺤﻴ ﺚ ﺗﻤﺜ ﻞ ﻗﺎﻋ ﺪة اﻟﻤ ﺴﺘﻄﻴﻞ ﻓﺘ ﺮة اﻟﺨﻄ ﺄ،ﺛﻢ ﻳﺘﻢ رﺳﻢ ﻋ ﺪة ﻣ ﺴﺘﻄﻴﻼت ﻣﺘﺠ ﺎورة :وﻳﻼﺣﻆ ﻓﻰ هﺬا اﻟﻤﻨﺤﻨﻰ اﻟﺬى ﻳﺴﻤﻰ ﻣﻨﺤﻨﻰ اﻟﺘﺠﻤﻴﻊ اﻟﺘﻜﺮارى اﻵﺗﻰ . اﻷﺧﻄﺎء ذات اﻟﻘﻴﻤﺔ اﻟﻮاﺣﺪة ﻳﺘﺴﺎوى ﻋﺪد اﻟﻤﻮﺟﺐ ﻣﻨﻬﺎ ﻣﻊ ﻋﺪد اﻟﺴﺎﻟﺐ-1 Prepared By:/ Amr A. El-Sayed, Civil Eng Dept., El-Minia Univ., Eg.
29
ﻣﺴﺎﺋﻞ ﻓﻰ اﻟﻤﺴﺎﺣﺔ اﻟﻤﺴﺘﻮﻳﺔ -2ﻳﺰداد ﻋﺪد اﻷﺧﻄﺎء آﻠﻤﺎ ﺻﻐﺮت ،وﻳﻘﻞ ﻋﺪدهﺎ آﻠﻤﺎ آﺒﺮ اﻟﺨﻄﺄ. -3ﻳﺰداد اﻟﻤﺘﻮﺳﻂ اﻟﺤﺴﺎﺑﻰ ﻗﺮﺑ ًﺎ ﻣﻦ اﻟﻘﻴﻤﺔ اﻟﺤﻘﻴﻘﻴﺔ آﻠﻤﺎ ازداد ﻋﺪد اﻷرﺻﺎد. 6 5
3 2
Repetation of Error
4
1
)(1.5 to 2.0
)(1.0 to 1.5
)(0.5 to 1.0
)(0.0 to 0.5
)(-0.5 to 0.0
)(-1.0 to -0.5
)(-1.5 to -1.0
)(-2 to -1.5
0
Period of Error
وﻓﻰ اﻟﺸﻜﻞ اﻟﺴﺎﺑﻖ ،إذا ﻗﻠﺖ اﻟﻔﺘﺮة إﻟﻰ أﻗﻞ ﻣﺎ ﻳﻤﻜﻦ ،وزاد ﻋﺪد اﻷرﺻﺎد إﻟﻰ أآﺒﺮ ﻣﺎ ﻳﻤﻜﻦ ﻓﺈﻧﻨ ﺎ ﻧﺤ ﺼﻞ ﻋﻠ ﻰ ﻣﻨﺤﻨ ﻰ اﻷﺧﻄﺎء ،واﻟﺬى ﻳﻄﻠﻖ ﻋﻠﻴﻪ )ﻣﻨﺤﻨﻰ ﺟﺎوس( ،وهﻮ ﻳﺸﺒﺔ ﻣ ﻦ ﺟﻤﻴ ﻊ اﻟﻮﺟ ﻮﻩ ﻣﻨﺤﻨ ﻰ اﻹﺣﺘﻤ ﺎﻻت وذﻟ ﻚ ﺑﻌ ﺪ اﺳ ﺘﺒﺪال اﻹﺣ ﺪاﺛﻰ اﻟﺮأﺳﻰ ،ﻓﺒﺪ ًﻻ ﻣﻦ ) (nاﻟﺘﻰ ﺗﺪل ﻋﻠﻰ ﻋﺪد ﻣﺮات ﺣﺪوث اﻟﺨﻄﺄ ﻧﻀﻊ ) (yاﻟﺘ ﻰ ﺗ ﺪل ﻋﻠ ﻰ اﺣﺘﻤ ﺎل ﺣ ﺪوث ﻧﻔ ﺲ اﻟﺨﻄ ﺄ اﻟ ﺴﺎﺑﻖ، وﻳﻄﻠﻖ ﻋﻠﻴﻪ ﻓﻰ هﺬﻩ اﻟﺤﺎﻟﺔ )ﻣﻨﺤﻨﻰ اﺣﺘﻤﺎل ﺣﺪوث اﻟﺨﻄﺄ( ،وﺗﺘﻀﺢ ﻣﻦ هﺬا اﻟﻤﻨﺤﻨﻰ اﻟﺤﻘﺎﺋﻖ اﻟﺘﺎﻟﻴﺔ: -1اﺣﺘﻤﺎل ﺣﺪوث اﻷﺧﻄﺎء اﻟﻤﻮﺟﺒﺔ ﻣﺜﻞ اﺣﺘﻤﺎل ﺣﺪوث اﻷﺧﻄﺎء اﻟﺴﺎﻟﺒﺔ اﻟﺘﻰ ﻟﻬﺎ ﻧﻔﺲ اﻟﻘﻴﻤﺔ اﻟﻌﺪدﻳﺔ . -2اﻷﺧﻄﺎء اﻟﺼﻐﻴﺮة أآﺜﺮ ﺣﺪوﺛ ًﺎ ﻣﻦ اﻷﺧﻄﺎء اﻟﻜﺒﻴﺮة -أى ﻳﻘﻞ اﺣﺘﻤﺎل ﺣﺪوث اﻟﺨﻄﺄ آﻠﻤﺎ ازدادت ﻗﻴﻤﺘﻪ. -3اﻷﺧﻄﺎء اﻟﻜﺒﻴﺮة ﻧﺎدرة اﻟﺤﺪوث ،وﻟﻬﺬا ﻓﺈن اﻟﻤﻨﺤﻨﻰ ﻳﻘﺎﺑﻞ ﻣﺤﻮر ) (Xﻓﻰ ﻣﺎﻻﻧﻬﺎﻳﺔ. )G(e
Error
30
Prepared By:/ Amr A. El-Sayed, Civil Eng Dept., El-Minia Univ., Eg.
ﻣﺴﺎﺋﻞ ﻓﻰ اﻟﻤﺴﺎﺣﺔ اﻟﻤﺴﺘﻮﻳﺔ
اﻟﻤﻌﺎدﻟﺔ اﻟﻌﺎﻣﺔ ﻟﻤﻨﺤﻨﻰ اﻹﺣﺘﻤﺎل: أوﺟﺪ ﺟﺎوس اﻟﻌﻼﻗﺔ اﻵﺗﻴﺔ ﻟﻤﻨﺤﻨﻰ اﻹﺣﺘﻤﺎل: ) .X 2
2
.e ( − h
h
π
= ) G (ε Where:
)Probability of error X (for the Y-axis )natural logarithmic base (2.71828 the error
)G(ε e X
= is a constant (h
h
)the standard deviation (MSE
σ
1 ) σ. 2
واﻟﺨﻄﺄ اﻟﺘﺮﺑﻴﻌﻰ اﻟﻤﺘﻮﺳﻂ ﻣﻘﺪار ﺛﺎﺑ ﺖ ) (σﺑﺎﻟﻨ ﺴﺒﺔ ﻟﻠﻤﻨﺤﻨ ﻰ اﻟﻮاﺣ ﺪ ،وﻳﺘﻮﻗ ﻒ ﻗﻴﻤﺘ ﻪ ﻋﻠ ﻰ دﻗ ﺔ اﻷرﺻ ﺎد ،واﺧ ﺘﻼف ﻗﻴﻤﺘﻪ ﻳﺆﺛﺮ ﻓﻰ ﺷﻜﻞ اﻟﻤﻨﺤﻨﻰ آﻤﺎ هﻮ ﻣﻮﺿﺢ ﻓﻰ اﻟﺸﻜﻞ اﻟﺘﺎﻟﻰ: -آﻠﻤﺎ ازدادت دﻗﺔ اﻷرﺻﺎد ،آﻠﻤﺎ ﻗﻠﺖ ﻗﻴﻤﺔ ) ،(σوﺑﺎﻟﺘﺎﻟﻰ ازداد ﺗﺤﺪب اﻟﻤﻨﺤﻨﻰ.
)G(e أرﺻﺎد دﻗﻴﻘﺔ
أرﺻﺎد ﻏﻴﺮ دﻗﻴﻘﺔ
Error إن ﻣﻌﻨﻰ اﺣﺘﻤﺎل ﺣﺪوث ﺷﻲء ﻣﺎ هﻮ ﻧﺴﺒﺔ ﺣﺪوﺛﻪ وﻳﻌﺒﺮ ﻋﻨﻪ ﺑﻜﺴﺮ ﻋﺸﺮى ﻗﻴﻤﺘﻪ ﺑﻴﻦ اﻟﺼﻔﺮ واﻟﻮاﺣﺪ اﻟﺼﺤﻴﺢ ،وﺑﻌﺮض ﻓﻜﺮة ﻣﺒﺴﻄﺔ ﻋﻦ ﻧﻈﺮﻳﺔ اﻹﺣﺘﻤﺎﻻت ﻧﻌﺮض اﻟﻤﺜﺎل اﻟﺘﺎﻟﻰ:
إﺣﺘﻤﺎل ﺣﺪوث ﺷﻲء ﻣﺎ أو ﺧﻄﺄ ﻣﺎ =
31
ﻋﺪد ﻣﺮات ﺗﻜﺮار اﻟﺤﺪث اﻟﻌﺪد اﻟﻜﻠﻰ ﻟﻸﺣﺪاث
Prepared By:/ Amr A. El-Sayed, Civil Eng Dept., El-Minia Univ., Eg.
ﻣﺴﺎﺋﻞ ﻓﻰ اﻟﻤﺴﺎﺣﺔ اﻟﻤﺴﺘﻮﻳﺔ ﻼ اﺣﺘﻤﺎل إﺧﺮاج آﺮة ﺣﻤﺮاء ﻣﻦ ﺻﻨﺪوق ﺑﻪ 10آﺮات ﺣﻤﺮاء ،و 15آﺮة ﺳﻮداء ،و 25آﺮة ﺑﻴﻀﺎء هﻮ: ﻓﻤﺜ ً
10 1 = = 20% 10 + 25 + 15 5 وﺑﺎﻟﺘﺎﻟﻰ ﻓﺈن اﻟﻤﺴﺎﺣﺔ ﺗﺤﺖ ﻣﻨﺤﻨﻰ ﺟﺎوس = ،1ﻷﻧﻬﺎ ﺗﻤﺜﻞ ﻣﺠﻤﻮع اﺣﺘﻤﺎﻻت ﺣﺪوث ﺟﻤﻴﻊ اﻷﺧﻄﺎء.
.dx = 1
.X 2
2
.e −h
h
π
∞+
∫
= Area under the (Gauss) curve
∞−
وﻟﻴﺲ ﻣﻦ اﻟﻤﻌﺘﺎد أن ﻳﺤﺴﺐ وﻗﻮع ﺧﻄﺄ ﻣﻌﻴﻦ ،وإﻧﻤﺎ ﻳﺤﺴﺐ اﺣﺘﻤﺎل وﻗﻮع ﺧﻄﺄ ﺑﻴﻦ ﻗﻴﻤﺘﻴﻦ )(X1 , X2
)G(e
X1
X2 X2 X1
Error
X1 X2
اﺣﺘﻤﺎل وﻗﻮع ﺧﻄﺄ ﺑﻴﻦ ﻗﻴﻤﺘﻴﻦ )= (X1 , X2 .X 2
32
2
.e −h
Prepared By:/ Amr A. El-Sayed, Civil Eng Dept., El-Minia Univ., Eg.
h
π
X2
∫
X1
=
ﻣﺴﺎﺋﻞ ﻓﻰ اﻟﻤﺴﺎﺣﺔ اﻟﻤﺴﺘﻮﻳﺔ ﻣﻘﺎرﻧﺔ ﻣﻌﺎﻳﻴﺮ دﻗﺔ اﻷرﺻﺎد: ﻣﻌﻨﻰ اﻟﺨﻄﺄ )اﻟﻤﺤﺘﻤﻞ ﺣﺪوﺛﻪ أآﺜﺮ ( أﻧﻪ ﻓﻰ أى ﻣﺠﻤﻮﻋﺔ ﻣﻦ اﻷرﺻ ﺎد ﻳﻜ ﻮن ﻋ ﺪد اﻷرﺻ ﺎد اﻟﺘ ﻰ ﺑﻬ ﺎ أﺧﻄ ﺎء أﺻ ﻐﺮ ﻣﻦ اﻟﺨﻄﺄ اﻟﻤﺤﺘﻤﻞ ﺗﺴﺎوى ﻋﺪد اﻷرﺻﺎد اﻟﺘﻰ ﺑﻬﺎ أﺧﻄﺎء أآﺒﺮ ﻣﻨﻪ – آﻤﺎ وﺿﺤﻨﺎ ذﻟﻚ ﻓﻰ ﺗﻌﺮﻳﻒ ). (Probable Error
r = 0.6745 σ وآﺬﻟﻚ اﻟﺨﻄﺄ اﻟﻤﺘﻮﺳﻂ ):(η
η = 0.7979 σ
وﻳﺘﻤﻴﺰ آﻞ ﻣﻦ اﻟﺨﻄﺄ اﻟﺘﺮﺑﻴﻌﻰ اﻟﻤﺘﻮﺳﻂ ) ، (σواﻟﺨﻄﺄ اﻟﻤﺤﺘﻤﻞ ﺑﺎﻵﺗﻰ: -1ﻻ ﻋﻼﻗﺔ ﻟﻜﻞ ﻣﻨﻬﻤﺎ ﺑﺈﺷﺎرة اﻟﺨﻄﺄ ﻓﻴﻤﺎ ﻟﻮ آﺎﻧﺖ ﺳﺎﻟﺒﺔ أو ﻣﻮﺟﺒﺔ )ﻟﻮﺟﻮد اﻟﺘﺮﺑﻴﻊ(. ﻼ ﻣﻦ ) (σو ) (rﻓﻰ ﻗﻴﺎس دﻗ ﺔ اﻷرﺻ ﺎد -2ﻳﻈﻬﺮ ﻓﻴﻬﻤﺎ ﺗﺄﺛﻴﺮ اﻷﺧﻄﺎء ﻣﺒﻜﺮًا ،وذﻟﻚ ﻷن اﻟﺘﺮﺑﻴﻊ ﻳﺰﻳﺪ اﻟﻘﻴﻤﺔ ،ﻟﺬﻟﻚ ﻳﺴﺘﺨﺪم آ ً واﺳﺘﺒﻌﺎد اﻟﻐﻴﺮ ﺻﺎﻟﺢ ﻣﻨﻬﺎ وﻳﻔﻀﻞ اﺳﺘﺨﺪام اﻟﺨﻄﺄ اﻟﺘﺮﺑﻴﻌﻰ اﻟﻤﺘﻮﺳﻂ MSEﻋﻨﺪ رﻓﺾ ﺑﻌﺾ اﻷرﺻﺎد اﻟﻐﻴﺮ ﻣﺤﺘﻤﻠ ﺔ اﻟﻮﻗ ﻮع ﺗﺤﺖ ﻣﻨﺤﻨﻰ اﻷﺧﻄﺎء. ﺣﺪ اﻷﺧﻄﺎء اﻟﺘﻰ ﻳﻘﻊ ﻓﻴﻬﺎ ) (50%ﻣﻦ اﻷرﺻﺎد:
r = ± 0.6745 σ
)G(e
r
0.6745 σ
0.6745 σ
Error
r
وآﻤﺎ ﺳﺒﻖ أن أﺷﺮﻧﺎ أن اﺣﺘﻤﺎل وﻗﻮع ﺧﻄﺄ ﺑﻴﻦ ﻗﻴﻤﺘﻴﻦ )= (X1 , X2 .X 2
33
2
.e −h
Prepared By:/ Amr A. El-Sayed, Civil Eng Dept., El-Minia Univ., Eg.
h
π
X2
∫
X1
=
ﻣﺴﺎﺋﻞ ﻓﻰ اﻟﻤﺴﺎﺣﺔ اﻟﻤﺴﺘﻮﻳﺔ واﻟﺨﻄﺄ اﻷآﺜﺮ اﺣﺘﻤﺎﻷ ) ( rﻧﺴﺒﺔ ﺣﺪوﺛﻪ ):(50% .r 2
2
.e −h
r
h
∫
π
= 0.50
−r
وآﺬﻟﻚ ﻳﻤﻜﻦ إﺛﺒﺎت اﻟﺘﺎﻟﻰ: -1اﺣﺘﻤﺎل ﺣﺪوث ﺧﻄﺄ ﻳﻘﻊ ﺑﻴﻦ ) (-σ , +σﻳﺴﺎوى 0.683
= 0.683
.σ 2
2
.e −h
h
π
-2اﺣﺘﻤﺎل ﺣﺪوث ﺧﻄﺄ ﻳﻘﻊ ﺑﻴﻦ ) (-2σ , +2σﻳﺴﺎوى 0.953
= 0.953
.( 2σ ) 2
2
.e −h
h
π
= 0.997
.( 3σ ) 2
2
∫
−σ
+2σ
∫
− 2σ
-3اﺣﺘﻤﺎل ﺣﺪوث ﺧﻄﺄ ﻳﻘﻊ ﺑﻴﻦ ) (-3σ , +3σﻳﺴﺎوى 0.997
h
.e −h
+σ
π
+3σ
∫
−3σ
وﻣﻦ اﻟﻤﻌﺎدﻟﺔ اﻟﺴﺎﺑﻘﺔ ﻳﺘﻀﺢ أن اﺣﺘﻤﺎل ﺣﺪوث أﺧﻄﺎء ﺗﺰﻳﺪ ﻋ ﻦ ± 3σﺗ ﺴﺎوى ) (100-99.7أى ﺣ ﻮاﻟﻰ ،0.30 % وﻟ ﺬﻟﻚ ﻓ ﺈن وﺟ ﻮد أرﺻ ﺎد ﺗﺰﻳ ﺪ أﺧﻄﺎؤه ﺎ اﻟﺒﺎﻗﻴ ﺔ )اﻟﻘﻴﻤ ﺔ اﻟﻤﺮﺻ ﻮدة – اﻟﻘﻴﻤ ﺔ اﻷآﺜ ﺮ اﺣﺘﻤ ﺎ ًﻻ( ﻋ ﻦ ± 3σﻳﻠ ﺰم رﻓ ﻀﻬﺎ ﻣ ﻦ ﻣﺠﻤﻮﻋﺔ اﻷرﺻﺎد. Example: 2- A distance was measured 40 times from which the standard deviation was found to be ±12 mm. How many of these forty observations containing errors that lie between –15 mm, and ?+10 mm Solution
σ = 12.0 mm
= 0.683
.X 2
)e F ( x = .dx )F `( x 2
2
e −h . X = . π − 2h 2 . X h
وﺑﺎﻟﺘﻌﻮﻳﺾ ﻓﻰ اﻟﻤﻌﺎدﻟﺔ اﻟﺴﺎﺑﻘﺔ ﻋﻦ ﻗﻴﻤﺔ Xﺑـ ± 12 34
Prepared By:/ Amr A. El-Sayed, Civil Eng Dept., El-Minia Univ., Eg.
2
.e −h
)F ( x
h
π
∫e
−h2 . X 2
+σ
∫
−σ
but
∫e
h
π
ﻣﺴﺎﺋﻞ ﻓﻰ اﻟﻤﺴﺎﺣﺔ اﻟﻤﺴﺘﻮﻳﺔ
−1 0.683 = 2h π
⎛ e −h2 (12)2 e −h2 ( −12)2 ⎜ − ⎜ 12 − 12 ⎝
−1 0.683 = 2h π
⎛ e −144h2 ⎜ ⎜ 6 ⎝
⎞ ⎟ ⎟ ⎠
(
⎞ ⎟ ⎟ ⎠
h = 0.0688 * e −144 h
2
)
: ﺗﺴﺎوى± 12 ﻋﺪد اﻷرﺻﺎد اﻟﺘﻰ ﺑﻬﺎ ﺧﻄﺄ ﻳﺘﺮاوح ﺑﻴﻦ
= 0.683 * 40 = 27 observations
G(e)
A = 0.683
σ σ
σ σ
Prepared By:/ Amr A. El-Sayed, Civil Eng Dept., El-Minia Univ., Eg.
Error
35
ﻣﺴﺎﺋﻞ ﻓﻰ اﻟﻤﺴﺎﺣﺔ اﻟﻤﺴﺘﻮﻳﺔ اﻟﺨﻮاص اﻟﻬﻨﺪﺳﻴﺔ ﻟﻤﻨﺤﻨﻰ )ﺟﺎوس(:
-1اﻟﻤﻨﺤﻨﻰ ﻋﺒﺎرة ﻋﻦ داﻟﺔ زوﺟﻴﺔ )ﻣﺘﻤﺎﺛﻠﺔ ﺣﻮل ﻣﺤﻮر ( Yوذﻟﻚ ﻷن:
)Y(x) = Y(-x -2أﻗﺼﻰ ﻗﻴﻤﺔ ﻟﻠﺪاﻟﺔ ﺗﺤﺪث ﻋﻨﺪ ):(X = 0.0
h
π
= .e 0
h
π
= ) G (ε
at X = 0.0,
-2اﻟﻤﻨﺤﻨﻰ ﻟﻪ ﻧﻘﻄﺘﻰ اﻧﻘ ﻼب )ﻧﻘﻄ ﺔ اﻹﻧﻘ ﻼب ه ﻰ اﻟﻨﻘﻄ ﺔ اﻟﺘ ﻰ ﻳﻐﻴ ﺮ ﻓﻴﻬ ﺎ اﻟﻤﻤ ﺎس اﺷ ﺎرﺗﻪ( ،وﻟﺘﺤﺪﻳ ﺪ ﻧﻘﻄ ﺔ اﻹﻧﻘ ﻼب ﻧﻔﺎﺿ ﻞ اﻟﺪاﻟﺔ ﻣﺮﺗﻴﻦ ﺑﺎﻟﻨ ﺴﺒﺔ ﻟ ـ Xوﻧ ﺴﺎوى اﻟﻨ ﺎﺗﺞ ﺑﺎﻟ ﺼﻔﺮ ﻓﻨﺤ ﺼﻞ ﻋﻠ ﻰ ﻗﻴﻤ ﺔ Xاﻟﺘ ﻰ ﻳﺤ ﺪث ﻋﻨ ﺪهﺎ اﻹﻧﻘ ﻼب ،ﺛ ﻢ ﻧﻌ ﻮض ﻋ ﻦ ﺗﻠ ﻚ اﻟﻘﻴﻤﺔ ﻓﻰ ﻣﻌﺎدﻟﺔ اﻟﻤﻨﺤﻨﻰ ،ﻓﻨﺤﺼﻞ ﻋﻠﻰ ﻗﻴﻤﺔ ) G(εاﻟﻤﻘﺎﺑﻠﺔ ﻟﻬﺎ .وﻟﺘﻌﻴﻴﻦ ﻧﻘﻄﺘﻰ اﻹﻧﻘﻼب رﻳﺎﺿﻴ ًﺎ ﻧﻔﺎﺿﻞ اﻟﻤﻌﺎدﻟﺔ ﻣﺮﺗﻴﻦ:
d d h −h2 . X 2 = )G ( x .e dx dx π
.X 2
2
. X .e −h
* 1) = 0.0
.X 2
− 2h 3
π 2
) .(−2h 2 X
=
* −2h 2 . X + e −h *1
.X 2
2
.X 2
2
.X 2
2
.e −h
.( X * e −h
* 2h 2 . X = e − h
h
= `G
π
− 2h 3
= ``G
π .X 2
2
X * e −h
2h 2 . X 2 = 1 1 h. 2 1 h. 2
−h 2 .
.e
h
π
= and that point is corresponding to Y −h 2
-4اﻟﻤﺴﺎﺣﺔ ﺗﺤﺖ اﻟﻤﻨﺤﻨﻰ ﻣﻦ ﻧﻘﻄﺔ )∞ (-إﻟﻰ ﻧﻘﻄﺔ )∞1= (+
36
=X
Prepared By:/ Amr A. El-Sayed, Civil Eng Dept., El-Minia Univ., Eg.
.e
h
π
=
ﻣﺴﺎﺋﻞ ﻓﻰ اﻟﻤﺴﺎﺣﺔ اﻟﻤﺴﺘﻮﻳﺔ
Precision and Accuracy ﻟﻨﻌﺮف اﻟﻔﺮق ﺑﻴﻦ ) (Precision & Accuracyﻧﺴﻮق اﻟﻤﺜﺎل اﻟﺘﺎﻟﻰ: أراد ﺷﺨﺺ أن ﻳﻘﺲ زاوﻳﺔ ﻗﻴﻤﺘﻬﺎ اﻷﻗﺮب اﺣﺘﻤﺎ ًﻻ آﺎﻧﺖ )`` (75o 12` 40ﻓﻜﺎﻧﺖ أرﺻﺎدﻩ آﺎﻟﺘﺎﻟﻰ:
``70o 12` 42 ``70o 12` 40
``70o 12` 39.5 ``70o 12` 41
``70o 12` 39 ``70o 12` 39.5
``70o 12` 40.5 ``70o 12` 41.5
``70o 12` 41 ``70o 12` 40
ﻓﺈذا أردﻧﺎ أن ﻧﻌﻠﻖ ﻋﻠﻰ هﺬﻩ اﻷرﺻﺎد ﻓﻤﺎذا ﻧﻘﻮل؟ ﻧﺠﺪ أن اﻷرﺻﺎد ﻣﺘﻘﺎرﺑﺔ ﺟﺪًا ﻣﻦ ﺑﻌﻀﻬﺎ اﻟﺒﻌﺾ ،وﻟﻜﻨﻬﺎ ﺑﻌﻴﺪة ﻋﻦ اﻟﻘﻴﻤﺔ اﻟﺤﻘﻴﻘﻴﺔ ﻟﻠﻜﻤﻴﺔ اﻟﻤﺮﺻﻮدة ،وﺑﺎﻟﺘﺎﻟﻰ ﻓﺈن هﺬﻩ اﻷرﺻﺎد ﻳﻘﺎل ﻋﻠﻴﻬﺎ ).(Precise but not Accurate Precision: وﻧﻘﻴﺲ ﺑﻬﺎ دﻗﺔ اﻷرﺻﺎد ﻣﻦ ﺣﻴﺚ إرﺗﺒﺎﻃﻬﺎ ﺑﺒﻌﻀﻬﺎ اﻟﺒﻌﺾ. Accuracy: وﻧﻘﻴﺲ ﺑﻬﺎ دﻗﺔ اﻷرﺻﺎد ﻣﻦ ﺣﻴﺚ إﻗﺘﺮاﺑﻬﺎ ﻣﻦ اﻟﻘﻴﻤﺔ اﻟﺤﻘﻴﻘﻴﺔ ﻟﻠﻜﻤﻴﺔ اﻟﻤﺮﺻﻮدة. وﻳﻤﻜﻦ ﺗﻌﺮﻳﻒ اﻷرﺻﺎد ﻣﻦ ﺣﻴﺚ اﻟﺪﻗﺔ ﻓﻰ أرﺑﻌﺔ ﻣﺠﻤﻮﻋﺎت آﺎﻟﺘﺎﻟﻰ : ﻧﺘﺨﻴﻞ ﻟﻮﺣﺔ رﻣﺎﻳﺔ ﺗﻤﺜﻞ اﻟﺪاﺋﺮة اﻟﺪاﺧﻠﻴﺔ ﻓﻴﻬﺎ اﻟﻘﻴﻤﺔ اﻟﺤﻘﻴﻘﻴﺔ ﻟﻠﻜﻤﻴﺔ اﻟﻤﺮﺻﻮدة ،وﺗﻜﻮن اﺣﺘﻤﺎﻻت اﻃﻼق اﻟﺮﺻﺎص ﻋﻠﻰ اﻟﻬﺪف: -1ﻃﻠﻘﺎت اﻟﺮﺻﺎص ﻣﺘﻘﺎرﺑﺔ ﻣﻦ ﺑﻌﻀﻬﺎ اﻟﺒﻌﺾ وﻟﻜﻨﻬﺎ ﺑﻌﻴﺪة ﻋﻦ اﻟﻬﺪف: Precise but not Accurate
-2ﻃﻠﻘﺎت اﻟﺮﺻﺎص ﻗﺮﻳﺒﺔ ﻣﻦ اﻟﻬﺪف وﻟﻜﻨﻬﺎ ﻣﺘﺒﺎﻋﺪة ﻋﻦ ﺑﻌﻀﻬﺎ اﻟﺒﻌﺾ: Accurate but not Precise
-3ﻃﻠﻘﺎت اﻟﺮﺻﺎص ﻗﺮﻳﺒﺔ ﻣﻦ ﺑﻌﻀﻬﺎ اﻟﺒﻌﺾ وﻗﺮﻳﺒﺔ ﻣﻦ اﻟﻬﺪف: Precise and Accurate
-4ﻃﻠﻘﺎت اﻟﺮﺻﺎص ﺑﻌﻴﺪة ﻋﻦ ﺑﻌﻀﻬﺎ اﻟﺒﻌﺾ وﺑﻌﻴﺪة ﻋﻦ اﻟﻬﺪف: Not Accurate and Not Precise
37
Prepared By:/ Amr A. El-Sayed, Civil Eng Dept., El-Minia Univ., Eg.
ﻣﺴﺎﺋﻞ ﻓﻰ اﻟﻤﺴﺎﺣﺔ اﻟﻤﺴﺘﻮﻳﺔ
Precise and Accurate
Precise but not Accurate
Not Accurate & Not Precise
Accurate not Precise
)Coefficient of Coronation (CC
∑ X .Y ∑ X 2 .∑ Y 2
= Cc
Variances and co-variances
وهﻰ ﻋﺒﺎرة ﻋﻦ ﻋﻼﻗﺔ اﻷرﺻﺎد اﻟﻐﻴﺮ ﻣﺘﺮاﺑﻄﺔ ،وﺗﺤﺪث ﻓﻰ ﺣﺎﻟﺔ وﺟﻮد أرﺻﺎد آﺜﻴﺮة ﺟﺪًا واﻟﻤﺠﻬﻮل آﻤﻴﺔ واﺣﺪة. Co-variance: هﻮ اﻟﺬى ﻳﻮﺿﺢ اﻟﻌﻼﻗﺔ ﺑﻴﻦ ﻧﻮﻋﻴﻦ ﻣﻦ اﻟﺮﺻﺪات ﺧﻮاص ﻣﺼﻔﻮﻓﺔ اﻟـ Variance -1اﻟﻤﺼﻔﻮﻓﺔ ﻗﻄﺮﻳﺔ وﻋﻨﺎﺻﺮ اﻟﻘﻄﺮ اﻟﺮﺋﻴﺴﻰ ﺗﺴﺎوى ) (σ12 , σ22 , σ32 , σ42 , …….واﻟﺒﺎﻗﻰ أﺻﻔﺎر. ﺧﻮاص ﻣﺼﻔﻮﻓﺔ اﻟـ Co-variance -1اﻟﻤﺼﻔﻮﻓﺔ ﻣﺮﺑﻌﺔ )اﻟﺼﻔﻮف ﻣﺴﺎوﻳﺔ ﻟﻸﻋﻤﺪة( -2اﻟﻤﺼﻔﻮﻓﺔ ﻟﻬﺎ ﻣﻌﻜﻮس ﺿﺮﺑﻰ أى أن ﻣﺤﺪدهﺎ ﻻ ﻳﺴﺎوى ﺻﻔﺮ. -3ﻋﻨﺎﺻﺮ اﻟﻘﻄﺮ اﻟﺮﺋﻴﺴﻰ ﻣﻮﺟﺒﺔ داﺋﻤﺎً ،وﺑﺎﻗﻰ اﻟﻌﻨﺎﺻﺮ ﻣﺘﻤﺎﺛﻠﺔ ﺣﻮل اﻟﻘﻄﺮ اﻟﺮﺋﻴﺴﻰ. 38
Prepared By:/ Amr A. El-Sayed, Civil Eng Dept., El-Minia Univ., Eg.
ﻣﺴﺎﺋﻞ ﻓﻰ اﻟﻤﺴﺎﺣﺔ اﻟﻤﺴﺘﻮﻳﺔ
ﺗﻄﺒﻴﻘﺎت ﻋﻠﻰ اﻟﺘﺒﺎﻳﻦ ):(Varianc ﻓﻰ ﺣﻠﺔ وﺟﻮد أرﺻﺎد ﻏﻴﺮ ﻣﺘﺮاﺑﻄﺔ ،واﻟﻤﻄﻠﻮب إﻳﺠﺎد آﻤﻴﺔ ﻣﺠﻬﻮﻟﺔ ﺑﺎﺳﺘﺨﺪام هﺬﻩ اﻟﺮﺻﺪات: -1ﻧﻌﻴﻦ اﻟﻤﻌﺎدﻟﺔ اﻟﻌﺎﻣﺔ اﻟﺘﻰ ﺗﺮﺑﻂ هﺬﻩ اﻟﻜﻤﻴﺔ اﻟﻤﺠﻬﻮﻟﺔ ﺑﺎﻷرﺻﺎد ﻣﺜﻞ ﺣﺴﺎب إرﺗﻔﺎع ﻣﺜﻠﺚ ﻣﻦ زاوﻳﺔ وﺿﻠﻊ آﺎﻟﻤﺜﺎل اﻟﺘﺎل: ) H = L . tan (θ
H θ ﺗﻢ ﻗﻴﺎس اﻟﺰاوﻳﺔ واﻟﻄﻮل ﻋﺪة ﻣﺮات ﺛﻢLآﺎﻧﺖ اﻟﻘﻴﻢ اﻣﺘﻮﺳﻄﺔ واﻹﻧﺤﺮاف اﻟﻤﻌﻴﺎرى آﺎﻟﺘﺎﻟﻰ: -2ﻳﺤﺴﺐ ﻗﻴﻤﺔ اﻟﻤﺠﻬﻮل ﺑﺎﺳﺘﺨﺪام اﻟﻘﻴﻢ اﻟﻤﺘﻮﺳﻄﺔ ﻓﻘﻂ ﺑﺪون σ
θ → θ ± σθ L → L ± σL ) H = L . tan (θ
-3اﻟﻤﻌﺎدﻟﺔ اﻟﻌﺎﻣﺔ ﻟﺤﺴﺎب : σ 2
2
⎞ ⎛ ∂H ⎞ ⎛ ∂H 2 2 ⎜= ⎜ ⎟ .(σ L ) + ) ⎟ .(σ θ ⎠ ⎝ ∂L ⎠ ⎝ ∂θ
) (σ H
2
ﺛﻢ ﻧﻔﺎﺿﻞ اﻟﻤﻌﺎدﻟﺔ اﻟﻌﺎﻣﺔ وﻧﻌﻮض ﺑﻘﻴﻢ اﻟﻤﺘﻮﺳﻄﺎت ﻓﻨﺤﺼﻞ ﻋﻠﻰ ﻣﻌﺎدﻟﺔ ﺗﺮﺑﻂ ﺑﻴﻦ: σH , σL , σθ -4ﺗﻮﺣﻴﺪ اﻟﻮﺣﺪات : Scaling
π 180
إذا آﺎﻧﺖ اﻟﻮﺣﺪات ﻣﺨﺘﻠﻔﺔ )أﻃﻮال وزواﻳﺎ( ﻧﺤﻮل آﻞ اﻟﺰواﻳﺎ إﻟﻰ اﻟﺘﻘﺪﻳﺮ اﻟﺪاﺋﺮى :
* σθ → σθ
ﻧﻌﻮض ﻓﻰ اﻟﻤﻌﺎدﻟﺔ ﻓﻨﺤﺼﻞ ﻋﻠﻰ (σH → H ± σH) σH
39
Prepared By:/ Amr A. El-Sayed, Civil Eng Dept., El-Minia Univ., Eg.