10- كيفية حل المسائل الحسابية (1).pdf1. اﻟﻤﺴﺎﺣﺔ ﻋﻠﻢ ﻓﻰ اﻹﻟﻴﻜﺘﺮوﻧﻴﺔ اﻟﻤﺤﺎﺿﺮات ﺳﻠﺴﻠﺔ
اﻟﻤﺴﺎﺣﺔ ﻋﻠﻢ ﻓﻰ اﻹﻟﻴﻜﺘﺮوﻧﻴﺔ اﻟﻤﺤﺎﺿﺮات ﺳﻠﺴﻠﺔ
E
E –
– Learning courses
Learning courses
اﻟﻤﺴﺎﺣﻴﺔ اﻟﻤﺴﺎﺋﻞ ﺣﻞ آﻴﻔﻴﺔ
اﻟﻤﺴﺎﺣﻴﺔ اﻟﻤﺴﺎﺋﻞ ﺣﻞ آﻴﻔﻴﺔ
How to solve
How to solve
Surveying Problems
Surveying Problems
أ
أ
.
.
د
د
/
/
اﻟﻤﻐﺮﺑﻰ ﺳﻌﻴﺪ
اﻟﻤﻐﺮﺑﻰ ﺳﻌﻴﺪ
ﻣﺪﻧﻰ ﻗﺴﻢ
ﻣﺪﻧﻰ ﻗﺴﻢ
-
-
اﻷزهﺮ هﻨﺪﺳﺔ
اﻷزهﺮ هﻨﺪﺳﺔ
2. ٢
٢
Prof. Dr. Eng. Said Elmaghraby
Prof. Dr. Eng. Said Elmaghraby
E_Learning courses in Engineering Surveying
E_Learning courses in Engineering Surveying
اﻟﻤﺤﺘﻮﻳﺎت
اﻟﻤﺤﺘﻮﻳﺎت
اﻷول اﻟﺒﺎب
اﻷول اﻟﺒﺎب
:
:
اﻟﺘﺮاﻓﻴﺮﺳﺎت ﺑﺎب ﻋﻠﻰ ﻣﺤﻠﻮﻟﺔ أﻣﺜﻠﺔ
اﻟﺘﺮاﻓﻴﺮﺳﺎت ﺑﺎب ﻋﻠﻰ ﻣﺤﻠﻮﻟﺔ أﻣﺜﻠﺔ
اﻟﺜﺎﻧﻰ اﻟﺒﺎب
اﻟﺜﺎﻧﻰ اﻟﺒﺎب
:
:
اﻟﻤﻴﺰاﻧﻴﺔ ﺑﺎب ﻋﻠﻰ ﻣﺤﻠﻮﻟﺔ أﻣﺜﻠﺔ
اﻟﻤﻴﺰاﻧﻴﺔ ﺑﺎب ﻋﻠﻰ ﻣﺤﻠﻮﻟﺔ أﻣﺜﻠﺔ
اﻟﺜﺎﻟﺚ اﻟﺒﺎب
اﻟﺜﺎﻟﺚ اﻟﺒﺎب
:
:
واﻟﺮدم اﻟﺤﻔﺮ ﻣﻜﻌﺒﺎت ﺣﺴﺎب ﻋﻠﻰ ﻣﺤﻠﻮﻟﺔ أﻣﺜﻠﺔ
واﻟﺮدم اﻟﺤﻔﺮ ﻣﻜﻌﺒﺎت ﺣﺴﺎب ﻋﻠﻰ ﻣﺤﻠﻮﻟﺔ أﻣﺜﻠﺔ
اﻟﺮاﺑﻊ اﻟﺒﺎب
اﻟﺮاﺑﻊ اﻟﺒﺎب
:
:
اﻷراﺿﻰ ﺗﺴﻮﻳﺔ ﻋﻠﻰ ﻣﺤﻠﻮﻟﺔ أﻣﺜﻠﺔ
اﻷراﺿﻰ ﺗﺴﻮﻳﺔ ﻋﻠﻰ ﻣﺤﻠﻮﻟﺔ أﻣﺜﻠﺔ
3. ٣
٣
Prof. Dr. Eng. Said Elmaghraby
Prof. Dr. Eng. Said Elmaghraby
E_Learning courses in Engineering Surveying
E_Learning courses in Engineering Surveying
اﻷول اﻟﺒﺎب
اﻷول اﻟﺒﺎب
ﻣﺤﻠﻮﻟﺔ أﻣﺜﻠﺔ
ﻣﺤﻠﻮﻟﺔ أﻣﺜﻠﺔ
ﻋﻠﻰ
ﻋﻠﻰ
اﻟﺘﺮاﻓﻴﺮﺳﺎت ﺑﺎب
اﻟﺘﺮاﻓﻴﺮﺳﺎت ﺑﺎب
4. ٤
٤
Prof. Dr. Eng. Said Elmaghraby
Prof. Dr. Eng. Said Elmaghraby
E_Learning courses in Engineering Surveying
E_Learning courses in Engineering Surveying
ﻣﺜﺎل
ﻣﺜﺎل
١
١
)
)
اﻟﺪاﺧﻠﻴﺔ واﻟﺰواﻳﺎ اﻟﻘﻔﻞ ﺧﻄﺄ ﺣﺴﺎب ﻋﻠﻰ
اﻟﺪاﺧﻠﻴﺔ واﻟﺰواﻳﺎ اﻟﻘﻔﻞ ﺧﻄﺄ ﺣﺴﺎب ﻋﻠﻰ
:(
:(
ﻳﻠ آﻤﺎ ﻓﻜﺎﻧﺖ د ﺟـ ب ا اﻟﻤﻘﻔﻞ اﻟﺘﺮاﻓﻴﺮس ﻓﻰ اﻟﺪاﺧﻠﻴﺔ اﻟﺰواﻳﺎ ﻗﻴﺴﺖ
ﻳﻠ آﻤﺎ ﻓﻜﺎﻧﺖ د ﺟـ ب ا اﻟﻤﻘﻔﻞ اﻟﺘﺮاﻓﻴﺮس ﻓﻰ اﻟﺪاﺧﻠﻴﺔ اﻟﺰواﻳﺎ ﻗﻴﺴﺖ
ﻰ
ﻰ
:
:
اﻟﺰواﻳﺎ ﻗﻴﻢ وﺣﺴﺎب اﻟﺰاوى اﻟﻘﻔﻞ ﺧﻄﺄ ﻗﻴﻤﺔ ﺣﺴﺎب واﻟﻤﻄﻠﻮب
اﻟﺰواﻳﺎ ﻗﻴﻢ وﺣﺴﺎب اﻟﺰاوى اﻟﻘﻔﻞ ﺧﻄﺄ ﻗﻴﻤﺔ ﺣﺴﺎب واﻟﻤﻄﻠﻮب
اﻟﻤﺼﺤﺤﺔ اﻟﺪاﺧﻠﻴﺔ
اﻟﻤﺼﺤﺤﺔ اﻟﺪاﺧﻠﻴﺔ
.
.
اﻟﺜﺎﻧﻴﺔ
اﻟﺪﻗﻴﻘﺔ
اﻟﺪرﺟﺔ
ا
22
46
99
ب
47
50
122
ﺟـ
15
26
72
د
32
56
64
اﻟﻨﻘﻄﺔ
اﻟﺪاﺧﻠﻴﺔ ااﻟﺰاوﻳﺔ
5. ٥
٥
Prof. Dr. Eng. Said Elmaghraby
Prof. Dr. Eng. Said Elmaghraby
E_Learning courses in Engineering Surveying
E_Learning courses in Engineering Surveying
ﻣﺜﺎل ﺣﻞ
ﻣﺜﺎل ﺣﻞ
١
١
:
:
ﻣﺠﻤﻮع ﺣﺴﺎب ﻳﺘﻢ
ﻣﺠﻤﻮع ﺣﺴﺎب ﻳﺘﻢ
اﻟﺪاﺧﻠﻴﺔ اﻟﺰواﻳﺎ
اﻟﺪاﺧﻠﻴﺔ اﻟﺰواﻳﺎ
اﻟﺰاوى اﻟﻘﻔﻞ ﺧﻄﺄ
اﻟﺰاوى اﻟﻘﻔﻞ ﺧﻄﺄ
=
=
٠٠
٠٠
٠٠
٠٠
٣٦٠
٣٦٠
–
–
٥٦
٥٦
٥٩
٥٩
٣٥٩
٣٥٩
=
=
٤
٤
ﺛﻮاﻧﻰ
ﺛﻮاﻧﻰ
ﻧﻮزع ﺣﻴﺚ
ﻧﻮزع ﺣﻴﺚ
١
١
زاوﻳﺔ آﻞ ﻋﻠﻰ ﺛﺎﻧﻴﺔ
زاوﻳﺔ آﻞ ﻋﻠﻰ ﺛﺎﻧﻴﺔ
.
.
اﻟﺜﺎﻧﻴﺔ
اﻟﺪﻗﻴﻘﺔ
اﻟﺪرﺟﺔ
اﻟﺜﺎﻧﻴﺔ
اﻟﺪﻗﻴﻘﺔ
اﻟﺪرﺟﺔ
ا
22
46
99
23
46
99
ب
47
50
122
48
50
122
ﺟـ
15
26
72
16
26
72
د
32
56
64
33
56
64
اﻟﻤﺠﻤﻮع
116
178
357
120
178
357
56
59
359
0
0
360
اﻟﻨﻘﻄﺔ
اﻟﺪاﺧﻠﻴﺔ ااﻟﺰاوﻳﺔ
اﻟﻤﺼﺤﺤﺔ اﻟﺪاﺧﻠﻴﺔ اﻟﺰاوﻳﺔ
6. ٦
٦
Prof. Dr. Eng. Said Elmaghraby
Prof. Dr. Eng. Said Elmaghraby
E_Learning courses in Engineering Surveying
E_Learning courses in Engineering Surveying
ﻣﺜﺎل
ﻣﺜﺎل
٢
٢
)
)
اﻻﻧﺤﺮاف ﺣﺴﺎب آﻴﻔﻴﺔ ﻋﻠﻰ
اﻻﻧﺤﺮاف ﺣﺴﺎب آﻴﻔﻴﺔ ﻋﻠﻰ
:(
:(
اﻟﺨﻂ آﺎن إذا اﻟﺴﺎﺑﻖ اﻟﻤﺜﺎل ﻓﻰ
اﻟﺨﻂ آﺎن إذا اﻟﺴﺎﺑﻖ اﻟﻤﺜﺎل ﻓﻰ
ﻓﺎﺣﺴﺐ اﻟﺸﺮق ﻧﺤﻮ ﻳﺘﺠﻪ ب ا
ﻓﺎﺣﺴﺐ اﻟﺸﺮق ﻧﺤﻮ ﻳﺘﺠﻪ ب ا
اﻟﺘﺮاﻓﻴﺮس ﻷﺿﻼع اﻟﺪاﺋﺮى اﻻﻧﺤﺮاف
اﻟﺘﺮاﻓﻴﺮس ﻷﺿﻼع اﻟﺪاﺋﺮى اﻻﻧﺤﺮاف
.
.
ﻣﺜﺎل ﺣﻞ
ﻣﺜﺎل ﺣﻞ
٢
٢
:
:
اﻟﻤﺼﺤﺤﺔ اﻟﺪاﺧﻠﻴﺔ اﻟﺰواﻳﺎ ﻗﻴﻢ آﺎﻧﺖ اﻟﺴﺎﺑﻖ اﻟﻤﺜﺎل ﻣﻦ
اﻟﻤﺼﺤﺤﺔ اﻟﺪاﺧﻠﻴﺔ اﻟﺰواﻳﺎ ﻗﻴﻢ آﺎﻧﺖ اﻟﺴﺎﺑﻖ اﻟﻤﺜﺎل ﻣﻦ
)
)
ﺑﻌﺪ
ﺑﻌﺪ
اﻟﺰاوى اﻟﻘﻔﻞ ﺧﻄﺄ ﺗﺼﺤﻴﺢ
اﻟﺰاوى اﻟﻘﻔﻞ ﺧﻄﺄ ﺗﺼﺤﻴﺢ
(
(
ﻳﻠﻰ آﻤﺎ
ﻳﻠﻰ آﻤﺎ
:
:
اﻟﺜﺎﻧﻴﺔ
اﻟﺪﻗﻴﻘﺔ
اﻟﺪرﺟﺔ
ا
23
46
99
ب
48
50
122
ﺟـ
16
26
72
د
33
56
64
اﻟﻨﻘﻄﺔ
اﻟﺪاﺧﻠﻴﺔ اﻟﺰاوﻳﺔ
7. ٧
٧
Prof. Dr. Eng. Said Elmaghraby
Prof. Dr. Eng. Said Elmaghraby
E_Learning courses in Engineering Surveying
E_Learning courses in Engineering Surveying
”
”
ﺗﺎﺑﻊ
ﺗﺎﺑﻊ
“
“
ﻣﺜﺎل ﺣﻞ
ﻣﺜﺎل ﺣﻞ
٢
٢
:
:
آﺮوآﻰ رﺳﻢ ﻳﺘﻢ
آﺮوآﻰ رﺳﻢ ﻳﺘﻢ
ﻟﻠﺘﺮاﻓﻴﺮس
ﻟﻠﺘﺮاﻓﻴﺮس
اﻻﻧﺤﺮاف ﻋﻠﻴﻪ ﻣﻮﺿﺢ
اﻻﻧﺤﺮاف ﻋﻠﻴﻪ ﻣﻮﺿﺢ
اﻟﺪاﺧﻠﻴﺔ اﻟﺰواﻳﺎ وﻗﻴﻢ ب ا ﻟﻠﺨﻂ اﻟﺪاﺋﺮى
اﻟﺪاﺧﻠﻴﺔ اﻟﺰواﻳﺎ وﻗﻴﻢ ب ا ﻟﻠﺨﻂ اﻟﺪاﺋﺮى
ﻟﻠﺨﻂ اﻷﻣﺎﻣﻰ اﻻﻧﺤﺮاف
ﻟﻠﺨﻂ اﻷﻣﺎﻣﻰ اﻻﻧﺤﺮاف
=
=
اﻟﺴﺎﺑﻖ ﻟﻠﺨﻂ اﻷﻣﺎﻣﻰ اﻻﻧﺤﺮاف
اﻟﺴﺎﺑﻖ ﻟﻠﺨﻂ اﻷﻣﺎﻣﻰ اﻻﻧﺤﺮاف
±
±
١٨٠
١٨٠
–
–
اﻟﺪاﺧﻠﻴﺔ اﻟﺰاوﻳﺔ
اﻟﺪاﺧﻠﻴﺔ اﻟﺰاوﻳﺔ
ا ب
ﺟـ
د
اﻟﻘﺎﻋﺪة ﻧﻄﺒﻖ ﺛﻢ
اﻟﻘﺎﻋﺪة ﻧﻄﺒﻖ ﺛﻢ
:
:
8. ٨
٨
Prof. Dr. Eng. Said Elmaghraby
Prof. Dr. Eng. Said Elmaghraby
E_Learning courses in Engineering Surveying
E_Learning courses in Engineering Surveying
ا ب
ﺟـ
د
”
”
ﺗﺎﺑﻊ
ﺗﺎﺑﻊ
“
“
ﻣﺜﺎل ﺣﻞ
ﻣﺜﺎل ﺣﻞ
٢
٢
:
:
9. ٩
٩
Prof. Dr. Eng. Said Elmaghraby
Prof. Dr. Eng. Said Elmaghraby
E_Learning courses in Engineering Surveying
E_Learning courses in Engineering Surveying
ﻣﺜﺎل
ﻣﺜﺎل
٣
٣
)
)
اﻟﻤﺮآﺒﺎت ﺣﺴﺎب ﻋﻠﻰ
اﻟﻤﺮآﺒﺎت ﺣﺴﺎب ﻋﻠﻰ
:(
:(
أﻃﻮال آﺎﻧﺖ إذا اﻟﺴﺎﺑﻖ اﻟﻤﺜﺎل ﻓﻰ
أﻃﻮال آﺎﻧﺖ إذا اﻟﺴﺎﺑﻖ اﻟﻤﺜﺎل ﻓﻰ
ﺛﻢ اﻟﻀﻠﻌﻰ اﻟﺨﻄﺄ ﻗﻴﻤﺔ ﻓﺎﺣﺴﺐ ﺑﺎﻟﺸﺮﻳﻂ ﻗﻴﺴﺖ اﻟﺨﻄﻮط
ﺛﻢ اﻟﻀﻠﻌﻰ اﻟﺨﻄﺄ ﻗﻴﻤﺔ ﻓﺎﺣﺴﺐ ﺑﺎﻟﺸﺮﻳﻂ ﻗﻴﺴﺖ اﻟﺨﻄﻮط
اﻟﻤﺮآﺒﺎت اﺣﺴﺐ
اﻟﻤﺮآﺒﺎت اﺣﺴﺐ
اﻟﻤﺼﺤﺤﺔ واﻟﺮأﺳﻴﺔ اﻷﻓﻘﻴﺔ
اﻟﻤﺼﺤﺤﺔ واﻟﺮأﺳﻴﺔ اﻷﻓﻘﻴﺔ
ﺑﺎﺳﺘﺨﺪام اﻟﺘﺮاﻓﻴﺮس ﻷﺿﻼع
ﺑﺎﺳﺘﺨﺪام اﻟﺘﺮاﻓﻴﺮس ﻷﺿﻼع
ا
ا
(
(
ﺑﺎﺳﺘﺨﺪام اﻷﻃﻮال ﺑﻨﺴﺒﺔ اﻟﺘﺼﺤﻴﺢ ﻃﺮﻳﻘﺔ
ﺑﺎﺳﺘﺨﺪام اﻷﻃﻮال ﺑﻨﺴﺒﺔ اﻟﺘﺼﺤﻴﺢ ﻃﺮﻳﻘﺔ
Bowditch method
Bowditch method
ب
ب
(
(
ﺑﺎﺳﺘﺨﺪام اﻟﻤﺮآﺒﺎت ﺑﻨﺴﺒﺔ اﻟﺘﺼﺤﻴﺢ ﻃﺮﻳﻘﺔ
ﺑﺎﺳﺘﺨﺪام اﻟﻤﺮآﺒﺎت ﺑﻨﺴﺒﺔ اﻟﺘﺼﺤﻴﺢ ﻃﺮﻳﻘﺔ
Transit method
Transit method
10. ١٠
١٠
Prof. Dr. Eng. Said Elmaghraby
Prof. Dr. Eng. Said Elmaghraby
E_Learning courses in Engineering Surveying
E_Learning courses in Engineering Surveying
ﻣﺜﺎل ﺣﻞ
ﻣﺜﺎل ﺣﻞ
٣
٣
)
)
ا
ا
:(
:(
ﺑﺎﺳﺘﺨﺪام اﻷﻃﻮال ﺑﻨﺴﺒﺔ اﻟﺘﺼﺤﻴﺢ ﻃﺮﻳﻘﺔ
ﺑﺎﺳﺘﺨﺪام اﻷﻃﻮال ﺑﻨﺴﺒﺔ اﻟﺘﺼﺤﻴﺢ ﻃﺮﻳﻘﺔ
Bowditch method
Bowditch method
ا
ب ا
61.31
0
0
90
61.310
0.000
61.299
0.025
ب
ﺟـ ب
71.85
12
9
147
38.971
-60.363
38.958
-60.334
ﺟـ
د ﺟـ
120.31
56
42
254
-116.055
-31.715
-116.076
-31.666
د
هـ د
93.29
23
46
9
15.836
91.936
15.819
91.974
ا
اﻟﻤﺠﻤﻮع
346.76
0.062
-0.142
0.000
0.000
اﻟﺨﻂ
اﻟﻄﻮل
اﻟﺮأﺳﻴﺔ
اﻟﻤﺼﺤﺤﺔ
اﻷﻓﻘﻴﺔ
اﻟﻤﺼﺤﺤﺔ
اﻟﻨﻘﻄﺔ
اﻷﻣﺎﻣﻰ اﻹﻧﺤﺮاف
اﻷﻓﻘﻴﺔ
اﻟﻐﻴﺮ
ﻣﺼﺤﺤﺔ
اﻟﺮأﺳﻴﺔ
اﻟﻐﻴﺮ
ﻣﺼﺤﺤﺔ
اﻟﺜﺎﻧﻴﺔ
اﻟﺪﻗﻴﻘﺔ
اﻟﺪرﺟﺔ
11. ١١
١١
Prof. Dr. Eng. Said Elmaghraby
Prof. Dr. Eng. Said Elmaghraby
E_Learning courses in Engineering Surveying
E_Learning courses in Engineering Surveying
ﻣﺜﺎل ﺣﻞ
ﻣﺜﺎل ﺣﻞ
٣
٣
)
)
ب
ب
:(
:(
ﺑﺎﺳﺘﺨﺪام اﻟﻤﺮآﺒﺎت ﺑﻨﺴﺒﺔ اﻟﺘﺼﺤﻴﺢ ﻃﺮﻳﻘﺔ
ﺑﺎﺳﺘﺨﺪام اﻟﻤﺮآﺒﺎت ﺑﻨﺴﺒﺔ اﻟﺘﺼﺤﻴﺢ ﻃﺮﻳﻘﺔ
Transit method
Transit method
ا
ب ا
61.31
0
0
90
61.310
0.000
61.294
0.000
ب
ﺟـ ب
71.85
12
9
147
38.971
-60.363
38.961
-60.316
ﺟـ
د ﺟـ
120.31
56
42
254
-116.055
-31.715
-116.086
-31.691
د
هـ د
93.29
23
46
9
15.836
91.936
15.831
92.007
ا
اﻟﺠﺒﺮى ﻣﺠـ
346.76
0.062
-0.142
0.000
0.000
اﻟﻌﺪدى ﻣﺠـ
232.171
184.014
اﻟﺨﻂ
اﻟﻄﻮل
اﻟﺮأﺳﻴﺔ
اﻟﻤﺼﺤﺤﺔ
اﻷﻓﻘﻴﺔ
اﻟﻤﺼﺤﺤﺔ
اﻟﻨﻘﻄﺔ
اﻷﻣﺎﻣﻰ اﻹﻧﺤﺮاف
اﻷﻓﻘﻴﺔ
اﻟﻐﻴﺮ
ﻣﺼﺤﺤﺔ
اﻟﺮأﺳﻴﺔ
اﻟﻐﻴﺮ
ﻣﺼﺤﺤﺔ
اﻟﺜﺎﻧﻴﺔ
اﻟﺪﻗﻴﻘﺔ
اﻟﺪرﺟﺔ
12. ١٢
١٢
Prof. Dr. Eng. Said Elmaghraby
Prof. Dr. Eng. Said Elmaghraby
E_Learning courses in Engineering Surveying
E_Learning courses in Engineering Surveying
ﻣﺜﺎل
ﻣﺜﺎل
٤
٤
)
)
اﻻﺣﺪاﺛﻴﺎت ﺣﺴﺎب ﻋﻠﻰ
اﻻﺣﺪاﺛﻴﺎت ﺣﺴﺎب ﻋﻠﻰ
:(
:(
ﻧﻘﻄ إﺣﺪاﺛﻴﺎت آﺎﻧﺖ إذا اﻟﻤﻀﻠﻊ ﻧﻘﺎط إﺣﺪاﺛﻴﺎت اﺣﺴﺐ اﻟﺴﺎﺑﻖ اﻟﻤﺜﺎل ﻓﻰ
ﻧﻘﻄ إﺣﺪاﺛﻴﺎت آﺎﻧﺖ إذا اﻟﻤﻀﻠﻊ ﻧﻘﺎط إﺣﺪاﺛﻴﺎت اﺣﺴﺐ اﻟﺴﺎﺑﻖ اﻟﻤﺜﺎل ﻓﻰ
ﺔ
ﺔ
ا
ا
)
)
٥٠٠
٥٠٠
ﺷﺮﻗﺎ
ﺷﺮﻗﺎ
،
،
١٠٠٠
١٠٠٠
ﺷﻤﺎﻻ
ﺷﻤﺎﻻ
(
(
ﺑﺎﻟﻄﺮﻳﻘﺘﻴﻦ اﻟﺘﺼﺤﻴﺢ ﺣﺎﻟﺔ ﻓﻰ وذﻟﻚ
ﺑﺎﻟﻄﺮﻳﻘﺘﻴﻦ اﻟﺘﺼﺤﻴﺢ ﺣﺎﻟﺔ ﻓﻰ وذﻟﻚ
:
:
ا
ا
(
(
ﺑﺎﺳﺘﺨﺪام اﻷﻃﻮال ﺑﻨﺴﺒﺔ اﻟﺘﺼﺤﻴﺢ ﻃﺮﻳﻘﺔ
ﺑﺎﺳﺘﺨﺪام اﻷﻃﻮال ﺑﻨﺴﺒﺔ اﻟﺘﺼﺤﻴﺢ ﻃﺮﻳﻘﺔ
Bowditch method
Bowditch method
ب
ب
(
(
ﺑﺎﺳﺘﺨﺪام اﻟﻤﺮآﺒﺎت ﺑﻨﺴﺒﺔ اﻟﺘﺼﺤﻴﺢ ﻃﺮﻳﻘﺔ
ﺑﺎﺳﺘﺨﺪام اﻟﻤﺮآﺒﺎت ﺑﻨﺴﺒﺔ اﻟﺘﺼﺤﻴﺢ ﻃﺮﻳﻘﺔ
Transit method
Transit method
13. ١٣
١٣
Prof. Dr. Eng. Said Elmaghraby
Prof. Dr. Eng. Said Elmaghraby
E_Learning courses in Engineering Surveying
E_Learning courses in Engineering Surveying
ﻣﺜﺎل ﺣﻞ
ﻣﺜﺎل ﺣﻞ
٤
٤
)
)
ا
ا
: (
: (
ﺑﺎﺳﺘﺨﺪام اﻷﻃﻮال ﺑﻨﺴﺒﺔ اﻟﺘﺼﺤﻴﺢ ﻃﺮﻳﻘﺔ
ﺑﺎﺳﺘﺨﺪام اﻷﻃﻮال ﺑﻨﺴﺒﺔ اﻟﺘﺼﺤﻴﺢ ﻃﺮﻳﻘﺔ
Bowditch method
Bowditch method
ا
500.000
1000.000
ب ا
61.31
61.299
0.025
ب
561.299
1000.025
ﺟـ ب
71.85
38.958
-60.334
ﺟـ
600.257
939.691
د ﺟـ
120.31
-116.076
-31.666
د
484.181
908.026
هـ د
93.29
15.819
91.974
ا
500.000
1000.000
اﻟﻤﺠﻤﻮع
346.76
0.000
0.000
اﻟﻨﻘﻄﺔ
اﻟﺨﻂ
اﻻﺣﺪاﺛﻰ
اﻷﻓﻘﻰ
اﻻﺣﺪاﺛﻰ
اﻟﺮأﺳﻰ
اﻟﻄﻮل
اﻟﺮأﺳﻴﺔ
اﻟﻤﺼﺤﺤﺔ
اﻷﻓﻘﻴﺔ
اﻟﻤﺼﺤﺤﺔ
14. ١٤
١٤
Prof. Dr. Eng. Said Elmaghraby
Prof. Dr. Eng. Said Elmaghraby
E_Learning courses in Engineering Surveying
E_Learning courses in Engineering Surveying
ﻣﺜﺎل ﺣﻞ
ﻣﺜﺎل ﺣﻞ
٤
٤
)
)
ب
ب
: (
: (
ب
ب
(
(
ﺑﺎﺳﺘﺨﺪام اﻟﻤﺮآﺒﺎت ﺑﻨﺴﺒﺔ اﻟﺘﺼﺤﻴﺢ ﻃﺮﻳﻘﺔ
ﺑﺎﺳﺘﺨﺪام اﻟﻤﺮآﺒﺎت ﺑﻨﺴﺒﺔ اﻟﺘﺼﺤﻴﺢ ﻃﺮﻳﻘﺔ
Transit method
Transit method
ا
500.000
1000.000
ب ا
61.31
61.294
0.000
ب
561.294
1000.000
ﺟـ ب
71.85
38.961
-60.316
ﺟـ
600.254
939.684
د ﺟـ
120.31
-116.086
-31.691
د
484.169
907.993
هـ د
93.29
15.831
92.007
ا
500.000
1000.000
اﻟﺠﺒﺮى ﻣﺠـ
346.76
0.000
0.000
اﻟﻌﺪدى ﻣﺠـ
اﻻﺣﺪاﺛﻰ
اﻷﻓﻘﻰ
اﻻﺣﺪاﺛﻰ
اﻟﺮأﺳﻰ
اﻟﻄﻮل
اﻟﺮأﺳﻴﺔ
اﻟﻤﺼﺤﺤﺔ
اﻷﻓﻘﻴﺔ
اﻟﻤﺼﺤﺤﺔ
اﻟﻨﻘﻄﺔ
اﻟﺨﻂ
15. ١٥
١٥
Prof. Dr. Eng. Said Elmaghraby
Prof. Dr. Eng. Said Elmaghraby
E_Learning courses in Engineering Surveying
E_Learning courses in Engineering Surveying
ب ا
107.12
0
20
85
ﺟـ ب
63.18
10
46
163
د ﺟـ
126.84
40
46
262
هـ د
27.99
50
30
348
ا هـ
40.04
10
29
6
اﻷﻣﺎﻣﻰ اﻹﻧﺤﺮاف
اﻟﺜﺎﻧﻴﺔ
اﻟﺪﻗﻴﻘﺔ
اﻟﺪرﺟﺔ
اﻟﺨﻂ
اﻟﻄﻮل
ﻣﺜﺎل
ﻣﺜﺎل
٥
٥
)
)
اﻟﻤﺼﺤﺤﺔ اﻟﻤﺮآﺒﺎت ﺣﺴﺎب ﻋﻠﻰ
اﻟﻤﺼﺤﺤﺔ اﻟﻤﺮآﺒﺎت ﺣﺴﺎب ﻋﻠﻰ
:(
:(
ﻷﺿﻼع اﻟﻤﺼﺤﺤﺔ واﻟﺮأﺳﻴﺔ اﻷﻓﻘﻴﺔ اﻟﻤﺮآﺒﺎت اﺣﺴﺐ
ﻷﺿﻼع اﻟﻤﺼﺤﺤﺔ واﻟﺮأﺳﻴﺔ اﻷﻓﻘﻴﺔ اﻟﻤﺮآﺒﺎت اﺣﺴﺐ
ﺗﺮاﻓﻴﺮس
ﺗﺮاﻓﻴﺮس
اﻟﺘﻴﻮدوﻟﻴﺖ
اﻟﺘﻴﻮدوﻟﻴﺖ
اﻟﺘﺎﻟﻰ ﺑﺎﻟﺠﺪول ﻣﻮﺿﺤﺔ اﻷرﺻﺎد ﺑﺄن ﻋﻠﻤﺎ هـ د ﺟـ ب ا اﻟﻤﻘﻔﻞ
اﻟﺘﺎﻟﻰ ﺑﺎﻟﺠﺪول ﻣﻮﺿﺤﺔ اﻷرﺻﺎد ﺑﺄن ﻋﻠﻤﺎ هـ د ﺟـ ب ا اﻟﻤﻘﻔﻞ
:
:
16. ١٦
١٦
Prof. Dr. Eng. Said Elmaghraby
Prof. Dr. Eng. Said Elmaghraby
E_Learning courses in Engineering Surveying
E_Learning courses in Engineering Surveying
ا
ب ا
107.12
0
20
85
106.765
8.715
107.774
8.754
ب
ﺟـ ب
63.18
10
46
163
17.659
-60.662
17.826
-60.392
ﺟـ
د ﺟـ
126.84
40
46
262
-125.834
-15.946
-124.644
-15.875
د
هـ د
27.99
50
30
348
-5.574
27.429
-5.521
27.552
هـ
ا هـ
40.04
10
29
6
4.523
39.784
4.566
39.961
ا
اﻟﺠﺒﺮى ﻣﺠـ
-2.460
-0.680
0.000
0.000
اﻟﻌﺪدى ﻣﺠـ
260.354
152.536
اﻟﻨﻘﻄﺔ
اﻷﻣﺎﻣﻰ اﻹﻧﺤﺮاف
اﻷﻓﻘﻴﺔ
اﻟﻐﻴﺮ
ﻣﺼﺤﺤﺔ
اﻟﺮأﺳﻴﺔ
اﻟﻐﻴﺮ
ﻣﺼﺤﺤﺔ
اﻟﺜﺎﻧﻴﺔ
اﻟﺪﻗﻴﻘﺔ
اﻟﺪرﺟﺔ
اﻟﺨﻂ
اﻟﻄﻮل
اﻟﺮأﺳﻴﺔ
اﻟﻤﺼﺤﺤﺔ
اﻷﻓﻘﻴﺔ
اﻟﻤﺼﺤﺤﺔ
ﻣﺜﺎل ﺣﻞ
ﻣﺜﺎل ﺣﻞ
٥
٥
:
:
آﺎﻵﺗﻰ ﺟﺪول ﻓﻰ اﻟﺤﺴﺎب ﺛﻢ اﻷرﺻﺎد وﺿﻊ ﻳﺘﻢ
آﺎﻵﺗﻰ ﺟﺪول ﻓﻰ اﻟﺤﺴﺎب ﺛﻢ اﻷرﺻﺎد وﺿﻊ ﻳﺘﻢ
17. ١٧
١٧
Prof. Dr. Eng. Said Elmaghraby
Prof. Dr. Eng. Said Elmaghraby
E_Learning courses in Engineering Surveying
E_Learning courses in Engineering Surveying
ب ا
363.900
363.900
ﺟـ ب
341.360
0.000
د ﺟـ
0.000
-324.150
هـ د
-231.180
-400.420
و هـ
-334.180
193.220
ا و
-138.980
167.200
اﻷﻓﻘﻴﺔ
اﻟﻐﻴﺮ
ﻣﺼﺤﺤﺔ
اﻟﺮأﺳﻴﺔ
اﻟﻐﻴﺮ
ﻣﺼﺤﺤﺔ
اﻟﺨﻂ
ﻣﺜﺎل
ﻣﺜﺎل
٦
٦
)
)
اﻻﺣﺪاﺛﻴﺎت ﺣﺴﺎب ﻋﻠﻰ
اﻻﺣﺪاﺛﻴﺎت ﺣﺴﺎب ﻋﻠﻰ
:(
:(
ﻓﻰ
ﻓﻰ
ﺗﺮاﻓﻴﺮس
ﺗﺮاﻓﻴﺮس
اﻟﺘﻴﻮدوﻟﻴﺖ
اﻟﺘﻴﻮدوﻟﻴﺖ
اﻷﻓﻘﻴﺔ اﻟﻤﺮآﺒﺎت ﻗﻴﻤﺔ ﻣﻌﻠﻮم و هـ د ﺟـ ب ا اﻟﻤﻘﻔﻞ
اﻷﻓﻘﻴﺔ اﻟﻤﺮآﺒﺎت ﻗﻴﻤﺔ ﻣﻌﻠﻮم و هـ د ﺟـ ب ا اﻟﻤﻘﻔﻞ
آﺎﻧﺖ إذا اﻟﻤﻀﻠﻊ ﻧﻘﺎط اﺣﺪاﺛﻴﺎت ﺣﺴﺎب وﻣﻄﻠﻮب ﻣﺼﺤﺤﺔ اﻟﻐﻴﺮ واﻟﺮأﺳﻴﺔ
آﺎﻧﺖ إذا اﻟﻤﻀﻠﻊ ﻧﻘﺎط اﺣﺪاﺛﻴﺎت ﺣﺴﺎب وﻣﻄﻠﻮب ﻣﺼﺤﺤﺔ اﻟﻐﻴﺮ واﻟﺮأﺳﻴﺔ
ا اﺣﺪاﺛﻴﺎت
ا اﺣﺪاﺛﻴﺎت
)
)
ﺷﺮﻗﺎ ﺻﻔﺮ
ﺷﺮﻗﺎ ﺻﻔﺮ
،
،
ﺷﻤﺎﻻ ﺻﻔﺮ
ﺷﻤﺎﻻ ﺻﻔﺮ
(
(
18. ١٨
١٨
Prof. Dr. Eng. Said Elmaghraby
Prof. Dr. Eng. Said Elmaghraby
E_Learning courses in Engineering Surveying
E_Learning courses in Engineering Surveying
ا
0.000
0.000
ب ا
363.900
363.900
363.662
363.963
ب
363.662
363.963
ﺟـ ب
341.360
0.000
341.137
0.000
ﺟـ
704.800
363.963
د ﺟـ
0.000
-324.150
0.000
-324.094
د
704.800
39.869
هـ د
-231.180
-400.420
-231.331
-400.351
هـ
473.469
-360.482
و هـ
-334.180
193.220
-334.398
193.253
و
139.071
-167.229
ا و
-138.980
167.200
-139.071
167.229
ا
0.000
0.000
ﺟﺒﺮى ﻣﺠـ
0.920
-0.250
0.000
0.000
ﻋﺪدى ﻣﺠـ
1409.600
1448.890
اﻟﻨﻘﻄﺔ
اﻷﻓﻘﻴﺔ
اﻟﻐﻴﺮ
ﻣﺼﺤﺤﺔ
اﻟﺮأﺳﻴﺔ
اﻟﻐﻴﺮ
ﻣﺼﺤﺤﺔ
اﻟﺨﻂ
اﻻﺣﺪاﺛﻰ
اﻷﻓﻘﻲ
اﻻﺣﺪاﺛﻰ
اﻟﺮأﺳﻲ
اﻟﺮأﺳﻴﺔ
اﻟﻤﺼﺤﺤﺔ
اﻷﻓﻘﻴﺔ
اﻟﻤﺼﺤﺤﺔ
ﻣﺜﺎل ﺣﻞ
ﻣﺜﺎل ﺣﻞ
٦
٦
:
:
آﺎﻵﺗﻰ ﺟﺪول ﻓﻰ اﻟﺤﺴﺎب ﺛﻢ اﻷرﺻﺎد وﺿﻊ ﻳﺘﻢ
آﺎﻵﺗﻰ ﺟﺪول ﻓﻰ اﻟﺤﺴﺎب ﺛﻢ اﻷرﺻﺎد وﺿﻊ ﻳﺘﻢ
19. ١٩
١٩
Prof. Dr. Eng. Said Elmaghraby
Prof. Dr. Eng. Said Elmaghraby
E_Learning courses in Engineering Surveying
E_Learning courses in Engineering Surveying
ﻣﺜﺎل
ﻣﺜﺎل
٧
٧
)
)
اﻟﺘﺮاﻓﻴﺮس ﻣﺴﺎﺣﺔ ﺣﺴﺎب ﻋﻠﻰ
اﻟﺘﺮاﻓﻴﺮس ﻣﺴﺎﺣﺔ ﺣﺴﺎب ﻋﻠﻰ
:(
:(
اﻟﺘﺮاﻓﻴﺮس ﻣﺴﺎﺣﺔ اﺣﺴﺐ
اﻟﺘﺮاﻓﻴﺮس ﻣﺴﺎﺣﺔ اﺣﺴﺐ
اﻟﺴﺎﺑﻖ اﻟﻤﺜﺎل ﻓﻰ
اﻟﺴﺎﺑﻖ اﻟﻤﺜﺎل ﻓﻰ
.
.
ا
0.000
0.000
ب
363.662
363.963
704.800
256520.865
ﺟـ
704.800
363.963
341.137
124161.249
د
704.800
39.869
-231.331
-9222.866
هـ
473.469
-360.482
-565.729
203935.225
و
139.071
-167.229
-473.469
79177.645
ا
0.000
0.000
224.592
0.000
ب
363.662
363.963
اﻟﻤﺴﺎﺣﺔ ﺿﻌﻒ
654572.118
اﻟﻤﺴﺎﺣﺔ
327286.059
اﻟﻨﻘﻄﺔ
ﻓﺮق
اﻟﻤﺮآﺒﺎت
اﻷﻓﻘﻴﺔ
* اﻟﺮأﺳﻴﺔ
(اﻟﻤﺮآﺒﺎت )ﻓﺮق
اﻟﻤﺮآﺒﺔ
اﻷﻓﻘﻴﺔ
اﻟﻤﺮآﺒﺔ
اﻟﺮأﺳﻴﺔ
20. ٢٠
٢٠
Prof. Dr. Eng. Said Elmaghraby
Prof. Dr. Eng. Said Elmaghraby
E_Learning courses in Engineering Surveying
E_Learning courses in Engineering Surveying
ﻣﺜﺎل
ﻣﺜﺎل
٨
٨
)
)
اﻟﻨﺎﻗﺼﺔ اﻷرﺻﺎد
اﻟﻨﺎﻗﺼﺔ اﻷرﺻﺎد
:
:
واﻧﺤﺮاﻓﻪ ﺧﻂ ﻃﻮل
واﻧﺤﺮاﻓﻪ ﺧﻂ ﻃﻮل
: (
: (
ﺑﺎﻟﺠﺪول اﻟﻤﻮﺿﺤﺔ اﻷرﺻﺎد أﺧﺬت
ﺑﺎﻟﺠﺪول اﻟﻤﻮﺿﺤﺔ اﻷرﺻﺎد أﺧﺬت
ﻟﻠﺘﺮاﻓﻴﺮس
ﻟﻠﺘﺮاﻓﻴﺮس
ﺣﺴﺎب واﻟﻤﻄﻠﻮب د ﺟـ ب ا اﻟﻤﻘﻔﻞ
ﺣﺴﺎب واﻟﻤﻄﻠﻮب د ﺟـ ب ا اﻟﻤﻘﻔﻞ
ب ا اﻟﺨﻂ واﻧﺤﺮاف ﻃﻮل
ب ا اﻟﺨﻂ واﻧﺤﺮاف ﻃﻮل
.
.
اﻟﺜﺎﻧﻴﺔ
اﻟﺪﻗﻴﻘﺔ
اﻟﺪرﺟﺔ
ﺟـ ا
130.75
0
0
0
د ﺟـ
210.75
0
39
54
ب د
65.00
0
57
328
ا ب
L
α
اﻷﻣﺎﻣﻰ اﻹﻧﺤﺮاف
اﻟﺨﻂ
اﻟﻄﻮل
21. ٢١
٢١
Prof. Dr. Eng. Said Elmaghraby
Prof. Dr. Eng. Said Elmaghraby
E_Learning courses in Engineering Surveying
E_Learning courses in Engineering Surveying
ﻣﺜﺎل ﺣﻞ
ﻣﺜﺎل ﺣﻞ
٨
٨
:
:
آﺎﻵﺗﻰ ﺟﺪول ﻓﻰ اﻟﺤﺴﺎب ﺛﻢ اﻷرﺻﺎد وﺿﻊ ﻳﺘﻢ
آﺎﻵﺗﻰ ﺟﺪول ﻓﻰ اﻟﺤﺴﺎب ﺛﻢ اﻷرﺻﺎد وﺿﻊ ﻳﺘﻢ
:
:
اﻟﺜﺎﻧﻴﺔ
اﻟﺪﻗﻴﻘﺔ
اﻟﺪرﺟﺔ
ا
ﺟـ ا
130.75
0
0
0
0.000
130.750
ﺟـ
د ﺟـ
210.75
0
39
54
171.895
121.934
د
ب د
65.00
0
57
328
-33.526
55.687
ب
ا ب
L
L sin α
L cos α
ا
اﻟﻤﺠﻤﻮع
138.369
308.370
L sin α
L cos α
α
اﻟﻨﻘﻄﺔ
اﻟﻤﺮآﺒﺔ
اﻟﺮأﺳﻴﺔ
اﻷﻣﺎﻣﻰ اﻹﻧﺤﺮاف
اﻟﺨﻂ
اﻟﻤﺮآﺒﺔ
اﻷﻓﻘﻴﺔ
اﻟﻄﻮل
22. ٢٢
٢٢
Prof. Dr. Eng. Said Elmaghraby
Prof. Dr. Eng. Said Elmaghraby
E_Learning courses in Engineering Surveying
E_Learning courses in Engineering Surveying
”
”
ﺗﺎﺑﻊ
ﺗﺎﺑﻊ
“
“
ﻣﺜﺎل ﺣﻞ
ﻣﺜﺎل ﺣﻞ
٨
٨
:
:
اﻷﻓﻘﻴﺔ اﻟﻤﺮآﺒﺎت ﻣﺠـ
اﻷﻓﻘﻴﺔ اﻟﻤﺮآﺒﺎت ﻣﺠـ
=
=
ل
ل
ﺟﺎ
ﺟﺎ
α
α
+
+
١٣٨٫٣٦٩
١٣٨٫٣٦٩
=
=
ﺻﻔﺮ
ﺻﻔﺮ
ل
ل
ﺟﺎ
ﺟﺎ
α
α
=
=
-
-
١٣٨٫٣٦٩
١٣٨٫٣٦٩
اﻟﺮأﺳﻴﺔ اﻟﻤﺮآﺒﺎت ﻣﺠـ
اﻟﺮأﺳﻴﺔ اﻟﻤﺮآﺒﺎت ﻣﺠـ
=
=
ل
ل
ﺟﺘﺎ
ﺟﺘﺎ
α
α
+
+
٣٠٨٫٣٧
٣٠٨٫٣٧
=
=
ﺻﻔﺮ
ﺻﻔﺮ
ل
ل
ﺟﺘﺎ
ﺟﺘﺎ
α
α
=
=
-
-
٣٠٨٫٣٧
٣٠٨٫٣٧
اﻟﻤﻌﺎدﻟﺘﻴﻦ ﺑﻘﺴﻤﺔ
اﻟﻤﻌﺎدﻟﺘﻴﻦ ﺑﻘﺴﻤﺔ
ﻇﺎ
ﻇﺎ
α
α
=
=
٠٫٤٤٨٧١
٠٫٤٤٨٧١
إذا
إذا
α
α
=
=
٥٩
٥٩
ث
ث
٩
٩
ق
ق
٢٤
٢٤
د
د
ﻓﻰ اﻟﺨﻂ ﻓﻴﻜﻮن ﺳﺎﻟﺒﺔ واﻟﺮأﺳﻴﺔ اﻷﻓﻘﻴﺔ اﻟﻤﺮآﺒﺔ ﻣﻦ آﻼ ﻷن وﻧﻈﺮا
ﻓﻰ اﻟﺨﻂ ﻓﻴﻜﻮن ﺳﺎﻟﺒﺔ واﻟﺮأﺳﻴﺔ اﻷﻓﻘﻴﺔ اﻟﻤﺮآﺒﺔ ﻣﻦ آﻼ ﻷن وﻧﻈﺮا
اﻟﺜﺎﻟﺚ اﻟﺮﺑﻊ
اﻟﺜﺎﻟﺚ اﻟﺮﺑﻊ
اﻻﻧﺤﺮاف
اﻻﻧﺤﺮاف
=
=
٥٩
٥٩
ث
ث
٩
٩
ق
ق
٢٤
٢٤
د
د
+
+
١٨٠
١٨٠
د
د
=
=
٥٩
٥٩
ث
ث
٩
٩
ق
ق
٢٠٤
٢٠٤
د
د
ل اﻟﻄﻮل ﻳﻜﻮن اﻟﻤﻌﺎدﻟﺘﻴﻦ اﺣﺪى ﻓﻰ وﺑﺎﻟﺘﻌﻮﻳﺾ
ل اﻟﻄﻮل ﻳﻜﻮن اﻟﻤﻌﺎدﻟﺘﻴﻦ اﺣﺪى ﻓﻰ وﺑﺎﻟﺘﻌﻮﻳﺾ
=
=
٣٧٧٫٩٩
٣٧٧٫٩٩
ﻣﺘﺮ
ﻣﺘﺮ
23. ٢٣
٢٣
Prof. Dr. Eng. Said Elmaghraby
Prof. Dr. Eng. Said Elmaghraby
E_Learning courses in Engineering Surveying
E_Learning courses in Engineering Surveying
ﻣﺜﺎل ﻓﻰ اﻟﻨﺘﺎﺋﺞ ﻣﻦ اﻟﺘﺤﻘﻖ
ﻣﺜﺎل ﻓﻰ اﻟﻨﺘﺎﺋﺞ ﻣﻦ اﻟﺘﺤﻘﻖ
٨
٨
:
:
اﻷرﺻﺎد ﻣﻮﺿﻊ اﻟﻨﺘﺎﺋﺞ وﺿﻊ ﻳﺘﻢ
اﻷرﺻﺎد ﻣﻮﺿﻊ اﻟﻨﺘﺎﺋﺞ وﺿﻊ ﻳﺘﻢ
اﻟﺘﺤﻘﻴﻖ وﻳﻜﻮن آﺎﻵﺗﻰ ﺟﺪول ﻓﻰ اﻟﺤﺴﺎب إﻋﺎدة ﺛﻢ اﻟﻨﺎﻗﺼﺔ
اﻟﺘﺤﻘﻴﻖ وﻳﻜﻮن آﺎﻵﺗﻰ ﺟﺪول ﻓﻰ اﻟﺤﺴﺎب إﻋﺎدة ﺛﻢ اﻟﻨﺎﻗﺼﺔ
ﻟﻠﺼﻔﺮ ﻣﺴﺎوﻳﺎ واﻟﺮأﺳﻴﺔ اﻷﻓﻘﻴﺔ اﻟﻤﺮآﺒﺎت ﺑﻤﺠﻤﻮع
ﻟﻠﺼﻔﺮ ﻣﺴﺎوﻳﺎ واﻟﺮأﺳﻴﺔ اﻷﻓﻘﻴﺔ اﻟﻤﺮآﺒﺎت ﺑﻤﺠﻤﻮع
:
:
اﻟﺜﺎﻧﻴﺔ
اﻟﺪﻗﻴﻘﺔ
اﻟﺪرﺟﺔ
ا
ﺟـ ا
130.75
0
0
0
0.000
130.750
ﺟـ
د ﺟـ
210.75
0
39
54
171.895
121.934
د
ب د
65.00
0
57
328
-33.526
55.687
ب
ا ب
337.99
59
9
204
-138.369
-308.369
ا
اﻟﻤﺠﻤﻮع
0.000
0.000
اﻟﻨﻘﻄﺔ
اﻟﻤﺮآﺒﺔ
اﻟﺮأﺳﻴﺔ
اﻷﻣﺎﻣﻰ اﻹﻧﺤﺮاف
اﻟﺨﻂ
اﻟﻤﺮآﺒﺔ
اﻷﻓﻘﻴﺔ
اﻟﻄﻮل
24. ٢٤
٢٤
Prof. Dr. Eng. Said Elmaghraby
Prof. Dr. Eng. Said Elmaghraby
E_Learning courses in Engineering Surveying
E_Learning courses in Engineering Surveying
اﻟﺜﺎﻧﻴﺔ
اﻟﺪﻗﻴﻘﺔ
اﻟﺪرﺟﺔ
ب ا
500.0
0
0
0
ﺟـ ب
L
0
0
45
د ﺟـ
854.0
0
33
110
هـ د
1019.8
0
41
168
و هـ
1118.0
ا و
656.8
0
54
305
α
اﻟﻄﻮل
اﻷﻣﺎﻣﻰ اﻹﻧﺤﺮاف
اﻟﺨﻂ
ﻣﺜﺎل
ﻣﺜﺎل
٩
٩
)
)
اﻟﻨﺎﻗﺼﺔ اﻷرﺻﺎد
اﻟﻨﺎﻗﺼﺔ اﻷرﺻﺎد
:
:
ﺁﺧﺮ ﺧﻂ واﻧﺤﺮاف ﺧﻂ ﻃﻮل
ﺁﺧﺮ ﺧﻂ واﻧﺤﺮاف ﺧﻂ ﻃﻮل
: (
: (
و هـ د ﺟـ ب ا اﻟﻤﻘﻔﻞ اﻟﺘﺮاﻓﻴﺮس ﻓﻰ اﻟﻨﺎﻗﺼﺔ اﻷرﺻﺎد اﺣﺴﺐ
و هـ د ﺟـ ب ا اﻟﻤﻘﻔﻞ اﻟﺘﺮاﻓﻴﺮس ﻓﻰ اﻟﻨﺎﻗﺼﺔ اﻷرﺻﺎد اﺣﺴﺐ
.
.
25. ٢٥
٢٥
Prof. Dr. Eng. Said Elmaghraby
Prof. Dr. Eng. Said Elmaghraby
E_Learning courses in Engineering Surveying
E_Learning courses in Engineering Surveying
اﻟﺜﺎﻧﻴﺔ
اﻟﺪﻗﻴﻘﺔ
اﻟﺪرﺟﺔ
ا
ب ا
500.00
0
0
0
0.000
500.000
ب
ﺟـ ب
L
0
0
45
0.707 L
0.707 L
ﺟـ
د ﺟـ
854.00
0
33
110
799.657
-299.775
د
هـ د
1019.80
0
41
168
200.117
-999.973
هـ
و هـ
1118.00
1118 sin α
1118 cos α
و
ا و
656.80
0
54
305
-532.035
385.129
ا
اﻟﻤﺠﻤﻮع
467.738
-414.618
0.707 L
0.707 L
1118 sin α
1118 cos α
اﻟﻤﺮآﺒﺔ
اﻟﺮأﺳﻴﺔ
اﻷﻣﺎﻣﻰ اﻹﻧﺤﺮاف
اﻟﺨﻂ
α
اﻟﻤﺮآﺒﺔ
اﻷﻓﻘﻴﺔ
اﻟﻄﻮل اﻟﻨﻘﻄﺔ
ﻣﺜﺎل ﺣﻞ
ﻣﺜﺎل ﺣﻞ
٩
٩
:
:
آﺎﻵﺗﻰ ﺟﺪول ﻓﻰ اﻟﺤﺴﺎب ﺛﻢ اﻷرﺻﺎد وﺿﻊ ﻳﺘﻢ
آﺎﻵﺗﻰ ﺟﺪول ﻓﻰ اﻟﺤﺴﺎب ﺛﻢ اﻷرﺻﺎد وﺿﻊ ﻳﺘﻢ
:
:
26. ٢٦
٢٦
Prof. Dr. Eng. Said Elmaghraby
Prof. Dr. Eng. Said Elmaghraby
E_Learning courses in Engineering Surveying
E_Learning courses in Engineering Surveying
”
”
ﺗﺎﺑﻊ
ﺗﺎﺑﻊ
“
“
ﻣﺜﺎل ﺣﻞ
ﻣﺜﺎل ﺣﻞ
٩
٩
:
:
اﻷﻓﻘﻴﺔ اﻟﻤﺮآﺒﺎت ﻣﺠـ
اﻷﻓﻘﻴﺔ اﻟﻤﺮآﺒﺎت ﻣﺠـ
=
=
٤٦٧٫٧٣٨
٤٦٧٫٧٣٨
+
+
٠٫٧٠٧
٠٫٧٠٧
ل
ل
+
+
١١١٨
١١١٨
ﺟﺎ
ﺟﺎ
α
α
=
=
ﺻﻔﺮ
ﺻﻔﺮ
اﻟﺮأﺳﻴﺔ اﻟﻤﺮآﺒﺎت ﻣﺠـ
اﻟﺮأﺳﻴﺔ اﻟﻤﺮآﺒﺎت ﻣﺠـ
=
=
-
-
٤١٤٫٦١٨
٤١٤٫٦١٨
+
+
٠٫٧٠٧
٠٫٧٠٧
ل
ل
+
+
١١١٨
١١١٨
ﺟﺘﺎ
ﺟﺘﺎ
α
α
=
=
ﺻﻔﺮ
ﺻﻔﺮ
اﻟﻤﻌﺎدﻟﺘﻴﻦ ﺑﻄﺮح
اﻟﻤﻌﺎدﻟﺘﻴﻦ ﺑﻄﺮح
)
)
ﺟﺘﺎ
ﺟﺘﺎ
α
α
–
–
ﺟﺎ
ﺟﺎ
α
α
= (
= (
٠٫٧٨٩٢٢٧
٠٫٧٨٩٢٢٧
٢
٢
ﺟﺘﺎ
ﺟﺘﺎ
٢
٢
)
)
٤٥
٤٥
+
+
α
α
= (
= (
٠٫٧٨٩٢٢٧
٠٫٧٨٩٢٢٧
اذا
اذا
α
α
=
=
٣٩
٣٩
ث
ث
٤
٤
ق
ق
١١
١١
د
د
و هـ اﻟﺨﻂ اﻧﺤﺮاف وﻳﻜﻮن
و هـ اﻟﺨﻂ اﻧﺤﺮاف وﻳﻜﻮن
)=
)=
٠٠
٠٠
ث
ث
٠٠
٠٠
ق
ق
٢٧٠
٢٧٠
د
د
(
(
–
–
)
)
٣٩
٣٩
ث
ث
٤
٤
ق
ق
١١
١١
د
د
= (
= (
٢١
٢١
ث
ث
٥٥
٥٥
ق
ق
٢٥٨
٢٥٨
د
د
اﻟﻤﻌﺎدﻟﺘﻴﻦ إﺣﺪى ﻓﻰ وﺑﺎﻟﺘﻌﻮﻳﺾ
اﻟﻤﻌﺎدﻟﺘﻴﻦ إﺣﺪى ﻓﻰ وﺑﺎﻟﺘﻌﻮﻳﺾ
ﺟـ ب اﻟﻀﻠﻊ ﻃﻮل ﻗﻴﻤﺔ ﻋﻠﻰ ﻧﺤﺼﻞ
ﺟـ ب اﻟﻀﻠﻊ ﻃﻮل ﻗﻴﻤﺔ ﻋﻠﻰ ﻧﺤﺼﻞ
)
)
ل
ل
= (
= (
٨٩٠٫١٥٩
٨٩٠٫١٥٩
م
م
27. ٢٧
٢٧
Prof. Dr. Eng. Said Elmaghraby
Prof. Dr. Eng. Said Elmaghraby
E_Learning courses in Engineering Surveying
E_Learning courses in Engineering Surveying
اﻟﺜﺎﻧﻴﺔ
اﻟﺪﻗﻴﻘﺔ
اﻟﺪرﺟﺔ
ا
ب ا
500.00
0
0
0
0.000
500.000
ب
ﺟـ ب
890.16
0
0
45
629.437
629.437
ﺟـ
د ﺟـ
854.00
0
33
110
799.657
-299.775
د
هـ د
1019.80
0
41
168
200.117
-999.973
هـ
و هـ
1118.00
21
55
258
-1097.170
-214.809
و
ا و
656.80
0
54
305
-532.035
385.129
ا
اﻟﻤﺠﻤﻮع
0.000
0.000
اﻟﻤﺮآﺒﺔ
اﻷﻓﻘﻴﺔ
اﻟﻄﻮل اﻟﻨﻘﻄﺔ
اﻟﻤﺮآﺒﺔ
اﻟﺮأﺳﻴﺔ
اﻷﻣﺎﻣﻰ اﻹﻧﺤﺮاف
اﻟﺨﻂ
ﻣﺜﺎل ﺣﻞ ﻣﻦ اﻟﺘﺤﻘﻖ
ﻣﺜﺎل ﺣﻞ ﻣﻦ اﻟﺘﺤﻘﻖ
٩
٩
:
:
ﻓﻰ اﻟﺤﺴﺎب ﺛﻢ اﻟﻨﺎﻗﺼﺔ اﻷرﺻﺎد وﺿﻊ ﻳﺘﻢ
ﻓﻰ اﻟﺤﺴﺎب ﺛﻢ اﻟﻨﺎﻗﺼﺔ اﻷرﺻﺎد وﺿﻊ ﻳﺘﻢ
آﺎﻵﺗﻰ ﺟﺪول
آﺎﻵﺗﻰ ﺟﺪول
:
:
28. ٢٨
٢٨
Prof. Dr. Eng. Said Elmaghraby
Prof. Dr. Eng. Said Elmaghraby
E_Learning courses in Engineering Surveying
E_Learning courses in Engineering Surveying
ﻣﺜﺎل
ﻣﺜﺎل
١٠
١٠
)
)
اﻟﻨﺎﻗﺼﺔ اﻷرﺻﺎد
اﻟﻨﺎﻗﺼﺔ اﻷرﺻﺎد
:
:
ﺿﻠﻌﻴﻦ ﻃﻮﻻ
ﺿﻠﻌﻴﻦ ﻃﻮﻻ
: (
: (
وﻗ ﻟﻸﺿﻼع اﻷﻣﺎﻣﻰ اﻻﻧﺤﺮاف ﻗﻴﺎس ﺗﻢ د ﺟـ ب ا اﻟﻤﻘﻔﻞ اﻟﺘﺮاﻓﻴﺮس ﻓﻰ
وﻗ ﻟﻸﺿﻼع اﻷﻣﺎﻣﻰ اﻻﻧﺤﺮاف ﻗﻴﺎس ﺗﻢ د ﺟـ ب ا اﻟﻤﻘﻔﻞ اﻟﺘﺮاﻓﻴﺮس ﻓﻰ
ﻃﻮل ﻴﺎس
ﻃﻮل ﻴﺎس
ب ا اﻟﻀﻠﻌﻴﻦ ﻣﻦ آﻞ
ب ا اﻟﻀﻠﻌﻴﻦ ﻣﻦ آﻞ
،
،
د ﺟـ اﻟﻀﻠﻌﻴﻦ ﻃﻮل ﻗﻴﺎس وﺗﻌﺬر ﺟـ ب
د ﺟـ اﻟﻀﻠﻌﻴﻦ ﻃﻮل ﻗﻴﺎس وﺗﻌﺬر ﺟـ ب
،
،
ا د
ا د
.
.
واﻟﻤﻄﻠﻮب
واﻟﻤﻄﻠﻮب
اﻟﺘﺎﻟﻰ ﺑﺎﻟﺠﺪول ﻣﻮﺿﺤﺔ اﻷرﺻﺎد ﺑﺄن ﻋﻠﻤﺎ أﻃﻮاﻟﻬﻤﺎ ﺣﺴﺎب
اﻟﺘﺎﻟﻰ ﺑﺎﻟﺠﺪول ﻣﻮﺿﺤﺔ اﻷرﺻﺎد ﺑﺄن ﻋﻠﻤﺎ أﻃﻮاﻟﻬﻤﺎ ﺣﺴﺎب
:
:
اﻟﺜﺎﻧﻴﺔ
اﻟﺪﻗﻴﻘﺔ
اﻟﺪرﺟﺔ
ب ا
150.0
0
30
178
ﺟـ ب
224.0
0
20
102
د ﺟـ
L
0
45
31
ا د
M
0
20
282
اﻷﻣﺎﻣﻰ اﻹﻧﺤﺮاف
اﻟﺨﻂ
اﻟﻄﻮل
29. ٢٩
٢٩
Prof. Dr. Eng. Said Elmaghraby
Prof. Dr. Eng. Said Elmaghraby
E_Learning courses in Engineering Surveying
E_Learning courses in Engineering Surveying
اﻟﺜﺎﻧﻴﺔ
اﻟﺪﻗﻴﻘﺔ
اﻟﺪرﺟﺔ
ا
ب ا
150.0
0
30
178
3.927
-149.949
ب
ﺟـ ب
224.0
0
20
102
218.830
-47.846
ﺟـ
د ﺟـ
L
0
45
31
0.526L
0.850L
د
ا د
M
0
20
282
-0.977M
0.214M
ا
اﻟﻤﺠﻤﻮع
222.757
-197.795
0.526L
0.850L
-0.977M
0.214M
اﻟﻨﻘﻄﺔ
اﻟﻤﺮآﺒﺔ
اﻟﺮأﺳﻴﺔ
اﻷﻣﺎﻣﻰ اﻹﻧﺤﺮاف
اﻟﺨﻂ
اﻟﻤﺮآﺒﺔ
اﻷﻓﻘﻴﺔ
اﻟﻄﻮل
ﻣﺜﺎل ﺣﻞ
ﻣﺜﺎل ﺣﻞ
١٠
١٠
:
:
آﺎﻵﺗﻰ ﺟﺪول ﻓﻰ اﻟﺤﺴﺎب ﺛﻢ اﻷرﺻﺎد وﺿﻊ ﻳﺘﻢ
آﺎﻵﺗﻰ ﺟﺪول ﻓﻰ اﻟﺤﺴﺎب ﺛﻢ اﻷرﺻﺎد وﺿﻊ ﻳﺘﻢ
:
:
30. ٣٠
٣٠
Prof. Dr. Eng. Said Elmaghraby
Prof. Dr. Eng. Said Elmaghraby
E_Learning courses in Engineering Surveying
E_Learning courses in Engineering Surveying
”
”
ﺗﺎﺑﻊ
ﺗﺎﺑﻊ
“
“
ﻣﺜﺎل ﺣﻞ
ﻣﺜﺎل ﺣﻞ
١٠
١٠
:
:
اﻷﻓﻘﻴﺔ اﻟﻤﺮآﺒﺎت ﻣﺠـ
اﻷﻓﻘﻴﺔ اﻟﻤﺮآﺒﺎت ﻣﺠـ
=
=
٢٢٢٫٧٥٧
٢٢٢٫٧٥٧
+
+
٠٫٥٢٦
٠٫٥٢٦
ل
ل
-
-
٠٫٩٧٧
٠٫٩٧٧
م
م
=
=
ﺻﻔﺮ
ﺻﻔﺮ
ل
ل
=
=
-
-
٤٢٣٫٤٩٢
٤٢٣٫٤٩٢
+
+
١٫٨٥٧
١٫٨٥٧
م
م
اﻟﺮأﺳﻴﺔ اﻟﻤﺮآﺒﺎت ﻣﺠـ
اﻟﺮأﺳﻴﺔ اﻟﻤﺮآﺒﺎت ﻣﺠـ
=
=
-
-
١٩٧٫٧٩٥
١٩٧٫٧٩٥
+
+
٠٫٨٥٠
٠٫٨٥٠
ل
ل
+
+
٠٫٢١٤
٠٫٢١٤
م
م
=
=
ﺻﻔﺮ
ﺻﻔﺮ
ل
ل
=
=
٢٣٢٫٧
٢٣٢٫٧
–
–
٠٫٢٥٢
٠٫٢٥٢
م
م
ﻣﺠﻬﻮﻟﻴﻦ ﻓﻰ اﻟﻤﻌﺎدﻟﺘﻴﻦ ﺑﺤﻞ
ﻣﺠﻬﻮﻟﻴﻦ ﻓﻰ اﻟﻤﻌﺎدﻟﺘﻴﻦ ﺑﺤﻞ
م اﻟﻄﻮل ﻳﻜﻮن
م اﻟﻄﻮل ﻳﻜﻮن
=
=
٣١١٫١٨٣
٣١١٫١٨٣
ﻣﺘﺮ
ﻣﺘﺮ
ل واﻟﻄﻮل
ل واﻟﻄﻮل
=
=
١٥٤٫٣٧٥
١٥٤٫٣٧٥
ﻣﺘﺮ
ﻣﺘﺮ
31. ٣١
٣١
Prof. Dr. Eng. Said Elmaghraby
Prof. Dr. Eng. Said Elmaghraby
E_Learning courses in Engineering Surveying
E_Learning courses in Engineering Surveying
ﻣﺜﺎل ﻓﻰ اﻟﻨﺘﺎﺋﺞ ﻣﻦ اﻟﺘﺤﻘﻖ
ﻣﺜﺎل ﻓﻰ اﻟﻨﺘﺎﺋﺞ ﻣﻦ اﻟﺘﺤﻘﻖ
١٠
١٠
:
:
ﻣﻮﺿﻊ اﻟﻨﺘﺎﺋﺞ وﺿﻊ ﻳﺘﻢ
ﻣﻮﺿﻊ اﻟﻨﺘﺎﺋﺞ وﺿﻊ ﻳﺘﻢ
وﻳﻜﻮن آﺎﻵﺗﻰ ﺟﺪول ﻓﻰ اﻟﺤﺴﺎب إﻋﺎدة ﺛﻢ اﻟﻨﺎﻗﺼﺔ اﻷرﺻﺎد
وﻳﻜﻮن آﺎﻵﺗﻰ ﺟﺪول ﻓﻰ اﻟﺤﺴﺎب إﻋﺎدة ﺛﻢ اﻟﻨﺎﻗﺼﺔ اﻷرﺻﺎد
ﻟﻠﺼﻔﺮ ﻣﺴﺎوﻳﺎ واﻟﺮأﺳﻴﺔ اﻷﻓﻘﻴﺔ اﻟﻤﺮآﺒﺎت ﺑﻤﺠﻤﻮع اﻟﺘﺤﻘﻴﻖ
ﻟﻠﺼﻔﺮ ﻣﺴﺎوﻳﺎ واﻟﺮأﺳﻴﺔ اﻷﻓﻘﻴﺔ اﻟﻤﺮآﺒﺎت ﺑﻤﺠﻤﻮع اﻟﺘﺤﻘﻴﻖ
:
:
اﻟﺜﺎﻧﻴﺔ
اﻟﺪﻗﻴﻘﺔ
اﻟﺪرﺟﺔ
ا
ب ا
150.0
0
30
178
3.927
-149.949
ب
ﺟـ ب
224.0
0
20
102
218.830
-47.846
ﺟـ
د ﺟـ
154.4
0
45
31
81.234
131.273
د
ا د
311.2
0
20
282
-304.001
66.468
ا
اﻟﻤﺠﻤﻮع
0.000
0.000
اﻟﻨﻘﻄﺔ
اﻟﻤﺮآﺒﺔ
اﻟﺮأﺳﻴﺔ
اﻷﻣﺎﻣﻰ اﻹﻧﺤﺮاف
اﻟﺨﻂ
اﻟﻤﺮآﺒﺔ
اﻷﻓﻘﻴﺔ
اﻟﻄﻮل
32. ٣٢
٣٢
Prof. Dr. Eng. Said Elmaghraby
Prof. Dr. Eng. Said Elmaghraby
E_Learning courses in Engineering Surveying
E_Learning courses in Engineering Surveying
ﻣﺜﺎل
ﻣﺜﺎل
١١
١١
)
)
اﻟﻨﺎﻗﺼﺔ اﻷرﺻﺎد
اﻟﻨﺎﻗﺼﺔ اﻷرﺻﺎد
:
:
ﺿﻠﻌﻴﻦ اﻧﺤﺮاف
ﺿﻠﻌﻴﻦ اﻧﺤﺮاف
: (
: (
ﻓﻴﻪ ا د ﺟـ ب ا اﻟﻤﻀﻠﻊ
ﻓﻴﻪ ا د ﺟـ ب ا اﻟﻤﻀﻠﻊ
:
:
ب ا
200.00
ﺟـ ب
350.00
0
0
102
د ﺟـ
150.00
هـ د
400.00
0
0
270
اﻟﺨﻂ
اﻟﻄﻮل
α
β
اﻷﻣﺎﻣﻰ اﻹﻧﺤﺮاف
اﻟﺜﺎﻧﻴﺔ
اﻟﺪﻗﻴﻘﺔ
اﻟﺪرﺟﺔ
ب ا اﻟﻀﻠﻌﻴﻦ اﻧﺤﺮاف إﻳﺠﺎد واﻟﻤﻄﻠﻮب
ب ا اﻟﻀﻠﻌﻴﻦ اﻧﺤﺮاف إﻳﺠﺎد واﻟﻤﻄﻠﻮب
،
،
د ﺟـ
د ﺟـ
33. ٣٣
٣٣
Prof. Dr. Eng. Said Elmaghraby
Prof. Dr. Eng. Said Elmaghraby
E_Learning courses in Engineering Surveying
E_Learning courses in Engineering Surveying
ﻣﺜﺎل ﺣﻞ
ﻣﺜﺎل ﺣﻞ
١١
١١
:
:
آﺎﻵﺗﻰ ﺟﺪول ﻓﻰ اﻟﺤﺴﺎب ﺛﻢ اﻷرﺻﺎد وﺿﻊ ﻳﺘﻢ
آﺎﻵﺗﻰ ﺟﺪول ﻓﻰ اﻟﺤﺴﺎب ﺛﻢ اﻷرﺻﺎد وﺿﻊ ﻳﺘﻢ
:
:
ا
ب ا
200.00
αﺟﺎ200
αﺟﺘﺎ200
ب
ﺟـ ب
350.00
0
0
102
342.352
-72.769
ﺟـ
د ﺟـ
150.00
βﺟﺎ150
βﺟﺘﺎ150
د
هـ د
400.00
0
0
270
-400.000
0.000
ا
اﻟﻤﺠﻤﻮع
1100.00
-57.648
-72.769
αﺟﺎ200
αﺟﺘﺎ200
βﺟﺎ150
βﺟﺘﺎ150
اﻷﻓﻘﻴﺔ
اﻟﻐﻴﺮ
ﻣﺼﺤﺤﺔ
اﻟﺮأﺳﻴﺔ
اﻟﻐﻴﺮ
ﻣﺼﺤﺤﺔ
اﻟﺜﺎﻧﻴﺔ
اﻟﺪﻗﻴﻘﺔ
اﻟﺪرﺟﺔ
α
β
اﻟﻨﻘﻄﺔ
اﻷﻣﺎﻣﻰ اﻹﻧﺤﺮاف
اﻟﺨﻂ
اﻟﻄﻮل
34. ٣٤
٣٤
Prof. Dr. Eng. Said Elmaghraby
Prof. Dr. Eng. Said Elmaghraby
E_Learning courses in Engineering Surveying
E_Learning courses in Engineering Surveying
٢٠٠
٢٠٠
ﺟﺎ
ﺟﺎ
α
α
=
=
٥٧٫٦٤٨
٥٧٫٦٤٨
–
–
١٥٠
١٥٠
ﺟﺎ
ﺟﺎ
β
β
ﺟﺎ
ﺟﺎ
α
α
=
=
٠٫٢٨٨٣٤
٠٫٢٨٨٣٤
–
–
٠٫٧٥
٠٫٧٥
ﺟﺎ
ﺟﺎ
β
β
ﻣﻌﺎدﻟﺔ
ﻣﻌﺎدﻟﺔ
)
)
١
١
(
(
٢٠٠
٢٠٠
ﺟﺘﺎ
ﺟﺘﺎ
α
α
=
=
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٧٢٫٧٦٩
–
–
١٥٠
١٥٠
ﺟﺘﺎ
ﺟﺘﺎ
β
β
ﺟﺘﺎ
ﺟﺘﺎ
α
α
=
=
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٠٫٣٦٣٨
–
–
٠٫٧٥
٠٫٧٥
ﺟﺘﺎ
ﺟﺘﺎ
β
β
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ﻣﻌﺎدﻟﺔ
)
)
٢
٢
(
(
”
”
ﺗﺎﺑﻊ
ﺗﺎﺑﻊ
“
“
ﻣﺜﺎل ﺣﻞ
ﻣﺜﺎل ﺣﻞ
١١
١١
:
:
ﺟﺎ
ﺟﺎ
٢
٢
α
α
=
=
٠٫٠٨٣
٠٫٠٨٣
+
+
٠٫٥٦٢٥
٠٫٥٦٢٥
ﺟﺎ
ﺟﺎ
٢
٢
β
β
-
-
٢
٢
)
)
٠٫٢٨٨٢٤
٠٫٢٨٨٢٤
x
x
٠٫٧٥
٠٫٧٥
ﺟﺎ
ﺟﺎ
β
β
(
(
ﻣﻌﺎدﻟﺔ
ﻣﻌﺎدﻟﺔ
)
)
٣
٣
(
(
اﻟﻤﻌﺎدﻟﺘﻴﻦ ﺑﺘﺮﺑﻴﻊ
اﻟﻤﻌﺎدﻟﺘﻴﻦ ﺑﺘﺮﺑﻴﻊ
١
١
،
،
٢
٢
ﺟﺘﺎ
ﺟﺘﺎ
٢
٢
α
α
=
=
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٠٫١٣٢٣
+
+
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٠٫٥٦٢٥
ﺟﺘﺎ
ﺟﺘﺎ
٢
٢
β
β
-
-
٢
٢
)
)
٠٫٣٦٣٨
٠٫٣٦٣٨
x
x
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ﺟﺘﺎ
β
β
(
(
ﻣﻌﺎدﻟﺔ
ﻣﻌﺎدﻟﺔ
)
)
٤
٤
(
(
35. ٣٥
٣٥
Prof. Dr. Eng. Said Elmaghraby
Prof. Dr. Eng. Said Elmaghraby
E_Learning courses in Engineering Surveying
E_Learning courses in Engineering Surveying
ﺟﺘﺎ
ﺟﺘﺎ
)
)
β
β
-
-
δ
δ
) = (
) = (
-
-
٠٫٢٢٢٢
٠٫٢٢٢٢
(
(
))
))
س
س
٢
٢
+
+
ص
ص
٢
٢
(
(
½
½
(
(
)
)
β
β
-
-
δ
δ
= (
= (
٤٢
٤٢
ث
ث
٣٦
٣٦
ق
ق
١٠٨
١٠٨
أو د
أو د
=
=
١٨
١٨
ث
ث
٢٣
٢٣
ق
ق
٢٥١
٢٥١
د
د
β
β
=
=
٠٥
٠٥
ث
ث
٠٠
٠٠
ق
ق
١٤٧
١٤٧
أو د
أو د
=
=
٤١
٤١
ث
ث
٤٦
٤٦
ق
ق
٢٨٩
٢٨٩
د
د
α
α
=
=
٤٢
٤٢
ث
ث
٠٥
٠٥
ق
ق
٣٥٣
٣٥٣
أو د
أو د
=
=
٠٩
٠٩
ث
ث
٤٣
٤٣
ق
ق
٨٣
٨٣
د
د
”
”
ﺗﺎﺑﻊ
ﺗﺎﺑﻊ
“
“
ﻣﺜﺎل ﺣﻞ
ﻣﺜﺎل ﺣﻞ
١١
١١
:
:
اﻟﻤﻌﺎدﻟﺘﻴﻦ ﺑﺠﻤﻊ
اﻟﻤﻌﺎدﻟﺘﻴﻦ ﺑﺠﻤﻊ
٣
٣
،
،
٤
٤
٠٫٤٣٢٣٦
٠٫٤٣٢٣٦
ﺟﺎ
ﺟﺎ
β
β
+
+
٠٫٥٤٥٧
٠٫٥٤٥٧
ﺟﺘﺎ
ﺟﺘﺎ
β
β
=
=
-
-
٠٫٢٢٢٢
٠٫٢٢٢٢
اﻟﺤﺪ أن وﺑﻔﺮض
اﻟﺤﺪ أن وﺑﻔﺮض
٠٫٤٣٢٣٦
٠٫٤٣٢٣٦
=
=
س
س
،
،
واﻟﺤﺪ
واﻟﺤﺪ
٠٫٥٤٥٧
٠٫٥٤٥٧
=
=
ﻣﻊ ص
ﻣﻊ ص
ﻣﺴﺎﻋﺪة ﺑﺰاوﻳﺔ اﻻﺳﺘﻌﺎﻧﺔ
ﻣﺴﺎﻋﺪة ﺑﺰاوﻳﺔ اﻻﺳﺘﻌﺎﻧﺔ
δ
δ
ﺣﻴﺚ
ﺣﻴﺚ
ﻇﺎ
ﻇﺎ
δ
δ
=
=
س
س
ص
ص
=
=
٠٫٧٩٢٣
٠٫٧٩٢٣
إذا
إذا
δ
δ
=
=
٢٣
٢٣
ث
ث
٢٣
٢٣
ق
ق
٣٨
٣٨
د
د
36. ٣٦
٣٦
Prof. Dr. Eng. Said Elmaghraby
Prof. Dr. Eng. Said Elmaghraby
E_Learning courses in Engineering Surveying
E_Learning courses in Engineering Surveying
ﻣﺜﺎل ﺣﻞ ﻣﻦ اﻟﺘﺤﻘﻖ
ﻣﺜﺎل ﺣﻞ ﻣﻦ اﻟﺘﺤﻘﻖ
١١
١١
)
)
اﻷول اﻟﺤﻞ
اﻷول اﻟﺤﻞ
:(
:(
اﻟ ﻣﻦ اﻟﺘﺄآﺪ آﻴﻔﻴﺔ ﻳﻮﺿﺢ اﻟﺘﺎﻟﻰ ﻓﺎﻟﺠﺪول ﺣﻼن ﻟﻬﺎ اﻟﻤﺴﺄﻟﺔ ﻷن ﻧﻈﺮا
اﻟ ﻣﻦ اﻟﺘﺄآﺪ آﻴﻔﻴﺔ ﻳﻮﺿﺢ اﻟﺘﺎﻟﻰ ﻓﺎﻟﺠﺪول ﺣﻼن ﻟﻬﺎ اﻟﻤﺴﺄﻟﺔ ﻷن ﻧﻈﺮا
اﻷول ﺑﺎﻟﺤﻞ ﻨﺘﺎﺋﺞ
اﻷول ﺑﺎﻟﺤﻞ ﻨﺘﺎﺋﺞ
ا
ب ا
200.00
42
5
353
-24.045
198.549
ب
ﺟـ ب
350.00
0
0
102
342.352
-72.769
ﺟـ
د ﺟـ
150.00
5
0
147
81.693
-125.803
د
هـ د
400.00
0
0
270
-400.000
0.000
ا
اﻟﻤﺠﻤﻮع
1100.00
0.000
0.000
اﻟﻨﻘﻄﺔ
اﻷﻣﺎﻣﻰ اﻹﻧﺤﺮاف
اﻷﻓﻘﻴﺔ
اﻟﻐﻴﺮ
ﻣﺼﺤﺤﺔ
اﻟﺮأﺳﻴﺔ
اﻟﻐﻴﺮ
ﻣﺼﺤﺤﺔ
اﻟﺜﺎﻧﻴﺔ
اﻟﺪﻗﻴﻘﺔ
اﻟﺪرﺟﺔ
اﻟﺨﻂ
اﻟﻄﻮل
37. ٣٧
٣٧
Prof. Dr. Eng. Said Elmaghraby
Prof. Dr. Eng. Said Elmaghraby
E_Learning courses in Engineering Surveying
E_Learning courses in Engineering Surveying
ا
ب ا
200.00
9
43
83
198.800
21.880
ب
ﺟـ ب
350.00
0
0
102
342.352
-72.769
ﺟـ
د ﺟـ
150.00
9
46
289
-141.159
50.735
د
هـ د
400.00
0
0
270
-400.000
0.000
ا
اﻟﻤﺠﻤﻮع
1100.00
0.000
0.000
اﻟﻨﻘﻄﺔ
اﻷﻣﺎﻣﻰ اﻹﻧﺤﺮاف
اﻷﻓﻘﻴﺔ
اﻟﻐﻴﺮ
ﻣﺼﺤﺤﺔ
اﻟﺮأﺳﻴﺔ
اﻟﻐﻴﺮ
ﻣﺼﺤﺤﺔ
اﻟﺜﺎﻧﻴﺔ
اﻟﺪﻗﻴﻘﺔ
اﻟﺪرﺟﺔ
اﻟﺨﻂ
اﻟﻄﻮل
ﻣﺜﺎل ﺣﻞ ﻣﻦ اﻟﺘﺤﻘﻖ
ﻣﺜﺎل ﺣﻞ ﻣﻦ اﻟﺘﺤﻘﻖ
١١
١١
)
)
اﻟﺜﺎﻧﻰ اﻟﺤﻞ
اﻟﺜﺎﻧﻰ اﻟﺤﻞ
:(
:(
اﻟﺜﺎﻧﻰ ﺑﺎﻟﺤﻞ اﻟﻨﺘﺎﺋﺞ ﻣﻦ اﻟﺘﺄآﺪ آﻴﻔﻴﺔ ﻳﻮﺿﺢ اﻟﺘﺎﻟﻰ اﻟﺠﺪول
اﻟﺜﺎﻧﻰ ﺑﺎﻟﺤﻞ اﻟﻨﺘﺎﺋﺞ ﻣﻦ اﻟﺘﺄآﺪ آﻴﻔﻴﺔ ﻳﻮﺿﺢ اﻟﺘﺎﻟﻰ اﻟﺠﺪول
38. ٣٨
٣٨
Prof. Dr. Eng. Said Elmaghraby
Prof. Dr. Eng. Said Elmaghraby
E_Learning courses in Engineering Surveying
E_Learning courses in Engineering Surveying
اﻟﺜﺎﻧﻰ اﻟﺒﺎب
اﻟﺜﺎﻧﻰ اﻟﺒﺎب
ﻣﺤﻠﻮﻟﺔ أﻣﺜﻠﺔ
ﻣﺤﻠﻮﻟﺔ أﻣﺜﻠﺔ
ﻋﻠﻰ
ﻋﻠﻰ
اﻟﻤﻴﺰاﻧﻴﺔ ﺑﺎب
اﻟﻤﻴﺰاﻧﻴﺔ ﺑﺎب
39. ٣٩
٣٩
Prof. Dr. Eng. Said Elmaghraby
Prof. Dr. Eng. Said Elmaghraby
E_Learning courses in Engineering Surveying
E_Learning courses in Engineering Surveying
ﻣﺜﺎل
ﻣﺜﺎل
١
١
:
:
ﻋﻠﻰ اﻟﻘﺮاءة ﻓﻜﺎﻧﺖ ب ا اﻟﻤﺴﺎﻓﺔ ﻣﻨﺘﺼﻒ ﻓﻰ ﻣﻴﺰان وﺿﻊ
ﻋﻠﻰ اﻟﻘﺮاءة ﻓﻜﺎﻧﺖ ب ا اﻟﻤﺴﺎﻓﺔ ﻣﻨﺘﺼﻒ ﻓﻰ ﻣﻴﺰان وﺿﻊ
ا اﻟﻘﺎﻣﺔ
ا اﻟﻘﺎﻣﺔ
)
)
١٫٦٨
١٫٦٨
م
م
(
(
ب اﻟﻘﺎﻣﺔ وﻋﻠﻰ
ب اﻟﻘﺎﻣﺔ وﻋﻠﻰ
)
)
١٫٩٥
١٫٩٥
م
م
(
(
اﻟﺠﻬﺎز ﻧﻘﻞ ﺛﻢ
اﻟﺠﻬﺎز ﻧﻘﻞ ﺛﻢ
ا اﻟﻘﺎﻣﺔ ﻋﻠﻰ اﻟﻘﺮاءة ﻓﻜﺎﻧﺖ ب اﻟﻨﻘﻄﺔ ﻣﻦ ﻗﺮﻳﺒﺎ ووﺿﻊ
ا اﻟﻘﺎﻣﺔ ﻋﻠﻰ اﻟﻘﺮاءة ﻓﻜﺎﻧﺖ ب اﻟﻨﻘﻄﺔ ﻣﻦ ﻗﺮﻳﺒﺎ ووﺿﻊ
)
)
١٫٣١
١٫٣١
م
م
(
(
ب اﻟﻘﺎﻣﺔ وﻋﻠﻰ
ب اﻟﻘﺎﻣﺔ وﻋﻠﻰ
)
)
١٫٧٤
١٫٧٤
م
م
.(
.(
ووﺿﺢ اﻟﻨﻈﺮ ﺧﻂ ﻳﻮﺿﺢ ﺷﻜﻼ ارﺳﻢ
ووﺿﺢ اﻟﻨﻈﺮ ﺧﻂ ﻳﻮﺿﺢ ﺷﻜﻼ ارﺳﻢ
ﻟﻤﻌﺎﻳﺮة اﻟﻼزم اﻟﺘﺼﺤﻴﺢ ﻗﻴﻤﺔ وﻣﺎ ﺑﺎﻟﻤﻴﺰان اﻟﺮﺻﺪ ﻳﻤﻜﻦ هﻞ
ﻟﻤﻌﺎﻳﺮة اﻟﻼزم اﻟﺘﺼﺤﻴﺢ ﻗﻴﻤﺔ وﻣﺎ ﺑﺎﻟﻤﻴﺰان اﻟﺮﺻﺪ ﻳﻤﻜﻦ هﻞ
اﻟﻤﻴﺰان
اﻟﻤﻴﺰان
.
.
40. ٤٠
٤٠
Prof. Dr. Eng. Said Elmaghraby
Prof. Dr. Eng. Said Elmaghraby
E_Learning courses in Engineering Surveying
E_Learning courses in Engineering Surveying
ب
ب
١٫٩٥
١٫٩٥ ١٫٦٨
١٫٦٨
ﻟﻠﻤﻴﺰان اﻟﺜﺎﻧﻰ اﻟﻮﺿﻊ
ﻟﻠﻤﻴﺰان اﻟﺜﺎﻧﻰ اﻟﻮﺿﻊ
ﻟﻠﻤﻴﺰان اﻷول اﻟﻮﺿﻊ
ﻟﻠﻤﻴﺰان اﻷول اﻟﻮﺿﻊ
١٫٣١
١٫٣١
ﻣﺜﺎل ﺣﻞ
ﻣﺜﺎل ﺣﻞ
١
١
:
:
اﻟﺤﺎﻟﺘﻴﻦ ﻓﻰ اﻟﺠﻬﺎز وﺿﻌﻰ رﺳﻢ ﻳﺘﻢ
اﻟﺤﺎﻟﺘﻴﻦ ﻓﻰ اﻟﺠﻬﺎز وﺿﻌﻰ رﺳﻢ ﻳﺘﻢ
اﻷول اﻟﻮﺿﻊ ﻓﻰ
اﻷول اﻟﻮﺿﻊ ﻓﻰ
:
:
ا ﺑﻴﻦ اﻟﺤﻘﻴﻘﻰ اﻟﻤﻨﺴﻮب ﻓﺮق
ا ﺑﻴﻦ اﻟﺤﻘﻴﻘﻰ اﻟﻤﻨﺴﻮب ﻓﺮق
،
،
ب
ب
=
=
١٫٩٥
١٫٩٥
-
-
١٫٦٨
١٫٦٨
=
=
٠٫٢٧
٠٫٢٧
م
م
ا
ا
ا
ا
ب
ب
١٫٧٤
١٫٧٤
اﻟﺜﺎﻧﻰ اﻟﻮﺿﻊ ﻓﻰ
اﻟﺜﺎﻧﻰ اﻟﻮﺿﻊ ﻓﻰ
:
:
ا ﺑﻴﻦ اﻟﻤﻨﺴﻮب ﻓﺮق ﻳﻜﻮن
ا ﺑﻴﻦ اﻟﻤﻨﺴﻮب ﻓﺮق ﻳﻜﻮن
،
،
ب
ب
=
=
١٫٧٤
١٫٧٤
-
-
١٫٣١
١٫٣١
=
=
٠٫٤٣
٠٫٤٣
م
م
اﻟﻤﻨﺴﻮب ﻓﺮق ﻻﺧﺘﻼف ﻧﻈﺮا
اﻟﻤﻨﺴﻮب ﻓﺮق ﻻﺧﺘﻼف ﻧﻈﺮا
اﻟﺮﺻﺪ ﻳﻤﻜﻦ ﻻ إذا اﻟﻮﺿﻌﻴﻦ ﻓﻰ
اﻟﺮﺻﺪ ﻳﻤﻜﻦ ﻻ إذا اﻟﻮﺿﻌﻴﻦ ﻓﻰ
ﻣﻌﺎﻳﺮة اﻟﻰ وﻳﺤﺘﺎج ﺑﺎﻟﻤﻴﺰان
ﻣﻌﺎﻳﺮة اﻟﻰ وﻳﺤﺘﺎج ﺑﺎﻟﻤﻴﺰان
.
.
ﻓﻰ ب ﻋﻨﺪ اﻟﺨﻄﺄ ان وﺑﺎﻋﺘﺒﺎر
ﻓﻰ ب ﻋﻨﺪ اﻟﺨﻄﺄ ان وﺑﺎﻋﺘﺒﺎر
ﺻﻔﺮ ﻳﺴﺎوى اﻟﺜﺎﻧﻰ اﻟﻮﺿﻊ
ﺻﻔﺮ ﻳﺴﺎوى اﻟﺜﺎﻧﻰ اﻟﻮﺿﻊ
،
،
اﻟﺘﺼﺤﻴﺢ ﻗﻴﻤﺔ ﻓﺘﻜﻮن
اﻟﺘﺼﺤﻴﺢ ﻗﻴﻤﺔ ﻓﺘﻜﻮن
:
:
١٫٧٤
١٫٧٤
=
=
٠٫٢٧
٠٫٢٧
+
+
١٫٣١
١٫٣١
-
-
س
س
س
س
=
=
٠٫١٦
٠٫١٦
م
م
س
س
41. ٤١
٤١
Prof. Dr. Eng. Said Elmaghraby
Prof. Dr. Eng. Said Elmaghraby
E_Learning courses in Engineering Surveying
E_Learning courses in Engineering Surveying
ﻣﺜﺎل
ﻣﺜﺎل
٢
٢
:
:
ﻟﻄﺮﻳﻖ ﻃﻮﻟﻰ ﻣﺤﻮر ﻋﻠﻰ ﺑﺎﻟﻤﻴﺰان اﻟﺘﺎﻟﻴﺔ اﻟﻘﺮاءات أﺧﺬت
ﻟﻄﺮﻳﻖ ﻃﻮﻟﻰ ﻣﺤﻮر ﻋﻠﻰ ﺑﺎﻟﻤﻴﺰان اﻟﺘﺎﻟﻴﺔ اﻟﻘﺮاءات أﺧﺬت
:
:
)
)
٢٫١٩
٢٫١٩
(
(
،
،
٢٫٥٠
٢٫٥٠
،
،
٢٫٣٢
٢٫٣٢
،
،
١٫٤٩
١٫٤٩
،
،
)
)
٣٫٠١
٣٫٠١
(
(
،
،
٢٫٥١
٢٫٥١
،
،
٢٫٨١
٢٫٨١
،
،
)
)
١٫٧٥
١٫٧٥
(
(
،
،
٣٫٨١
٣٫٨١
.
.
ﻧﻘﻄﺔ أول وﻣﻨﺴﻮب ﻣﺆﺧﺮات اﻷﻗﻮاس ﺑﻴﻦ اﻟﻘﺮاءات آﺎﻧﺖ ﻓﺈذا
ﻧﻘﻄﺔ أول وﻣﻨﺴﻮب ﻣﺆﺧﺮات اﻷﻗﻮاس ﺑﻴﻦ اﻟﻘﺮاءات آﺎﻧﺖ ﻓﺈذا
+)
+)
٣٠
٣٠
م
م
(
(
،
،
ﻣﺨﺘﻠﻔﺘﻴﻦ ﺑﻄﺮﻳﻘﺘﻴﻦ اﻟﻨﻘﻂ ﺑﺎﻗﻰ ﻣﻨﺎﺳﻴﺐ ﻓﺎﺣﺴﺐ
ﻣﺨﺘﻠﻔﺘﻴﻦ ﺑﻄﺮﻳﻘﺘﻴﻦ اﻟﻨﻘﻂ ﺑﺎﻗﻰ ﻣﻨﺎﺳﻴﺐ ﻓﺎﺣﺴﺐ
.
.
42. ٤٢
٤٢
Prof. Dr. Eng. Said Elmaghraby
Prof. Dr. Eng. Said Elmaghraby
E_Learning courses in Engineering Surveying
E_Learning courses in Engineering Surveying
ﻣﺜﺎل ﺣﻞ
ﻣﺜﺎل ﺣﻞ
٢
٢
:
:
اﻟﻤﻴﺰان ﺳﻄﺢ ﻣﻨﺴﻮب ﻃﺮﻳﻘﺔ ﺑﺎﺳﺘﺨﺪام
اﻟﻤﻴﺰان ﺳﻄﺢ ﻣﻨﺴﻮب ﻃﺮﻳﻘﺔ ﺑﺎﺳﺘﺨﺪام
اﻟﻨﻘﻄﺔ
اﻟﻤﺆﺧﺮة
اﻟﻤﺘﻮﺳﻄﺔ
اﻟﻤﻘﺪﻣﺔ
ﺳﻄﺢ ﻣﻨﺴﻮب
اﻟﻤﻴﺰان
ﻣﻨﺴﻮب
اﻟﻨﻘﻄﺔ
ﻣﻼﺣﻈﺎت
ا
2.19
32.19
30.00
روﺑﻴﺮ
ب
2.50
29.69
ﺟـ
2.32
29.87
د
3.01
1.49
33.71
30.70
دوران
هـ
2.51
31.20
و
1.75
2.81
32.65
30.90
دوران
ز
3.81
28.84
اﻟﻤﺠﻤﻮع
6.95
8.11
اﻟﻔﺮق
1.16
1.16
اﻟﺘﺤﻘﻴﻖ
O.K
43. ٤٣
٤٣
Prof. Dr. Eng. Said Elmaghraby
Prof. Dr. Eng. Said Elmaghraby
E_Learning courses in Engineering Surveying
E_Learning courses in Engineering Surveying
ﻣﺜﺎل ﺣﻞ
ﻣﺜﺎل ﺣﻞ
٢
٢
:
:
واﻻﻧﺨﻔﺎض اﻻرﺗﻔﺎع ﻃﺮﻳﻘﺔ ﺑﺎﺳﺘﺨﺪام
واﻻﻧﺨﻔﺎض اﻻرﺗﻔﺎع ﻃﺮﻳﻘﺔ ﺑﺎﺳﺘﺨﺪام
اﻟﻨﻘﻄﺔ
اﻟﻤﺆﺧﺮة
اﻟﻤﺘﻮﺳﻄﺔ
اﻟﻤﻘﺪﻣﺔ
ارﺗﻔﺎع
(+)
اﻧﺨﻔﺎض
(-)
ﻣﻨﺴﻮب
اﻟﻨﻘﻄﺔ
ﻣﻼﺣﻈﺎت
ا
2.19
30.00
روﺑﻴﺮ
ب
2.50
0.31
29.69
ﺟـ
2.32
0.18
29.87
د
3.01
1.49
0.83
30.70
دوران
هـ
2.51
0.50
31.20
و
1.75
2.81
0.30
30.90
دوران
ز
3.81
2.06
28.84
اﻟﻤﺠﻤﻮع
6.95
8.11
اﻟﻔﺮق
1.16
1.16
اﻟﺘﺤﻘﻴﻖ
O.K
44. ٤٤
٤٤
Prof. Dr. Eng. Said Elmaghraby
Prof. Dr. Eng. Said Elmaghraby
E_Learning courses in Engineering Surveying
E_Learning courses in Engineering Surveying
ﻣﺜﺎل ﺣﻞ
ﻣﺜﺎل ﺣﻞ
٣
٣
:
:
اﻟﻤﻴﺰان ﺳﻄﺢ ﻣﻨﺴﻮب ﻃﺮﻳﻘﺔ ﺑﺎﺳﺘﺨﺪام
اﻟﻤﻴﺰان ﺳﻄﺢ ﻣﻨﺴﻮب ﻃﺮﻳﻘﺔ ﺑﺎﺳﺘﺨﺪام
اﻟﻨﻘﻄﺔ
اﻟﻤﺆﺧﺮة
اﻟﻤﺘﻮﺳﻄﺔ
اﻟﻤﻘﺪﻣﺔ
ﺳﻄﺢ ﻣﻨﺴﻮب
اﻟﻤﻴﺰان
ﻣﻨﺴﻮب
اﻟﻨﻘﻄﺔ
ﻣﻼﺣﻈﺎت
ا
2.19
26.35
24.16
ب
2.50
23.85
ﺟـ
2.32
24.03
د
3.01
1.49
27.87
24.86
دوران
هـ
2.51
25.36
و
1.75
2.81
26.81
25.06
دوران
ز
3.81
23.00
روﺑﻴﺮ
اﻟﻤﺠﻤﻮع
6.95
8.11
اﻟﻔﺮق
1.16
1.16
اﻟﺘﺤﻘﻴﻖ
O.K
ﻣﺜﺎل
ﻣﺜﺎل
٣
٣
:
:
ﺑﻄﺮﻳﻘﺘﻴﻦ اﻟﺴﺎﺑﻖ اﻟﻤﺜﺎل ﻓﻰ اﻟﻨﻘﻂ ﻣﻨﺎﺳﻴﺐ ﺣﺴﺎب اﻟﻤﻄﻠﻮب
ﺑﻄﺮﻳﻘﺘﻴﻦ اﻟﺴﺎﺑﻖ اﻟﻤﺜﺎل ﻓﻰ اﻟﻨﻘﻂ ﻣﻨﺎﺳﻴﺐ ﺣﺴﺎب اﻟﻤﻄﻠﻮب
ﻧﻘﻄﺔ ﺁﺧﺮ ﻣﻨﺴﻮب آﺎن إذا ﻣﺨﺘﻠﻔﺘﻴﻦ
ﻧﻘﻄﺔ ﺁﺧﺮ ﻣﻨﺴﻮب آﺎن إذا ﻣﺨﺘﻠﻔﺘﻴﻦ
+)
+)
٢٣
٢٣
م
م
.(
.(
45. ٤٥
٤٥
Prof. Dr. Eng. Said Elmaghraby
Prof. Dr. Eng. Said Elmaghraby
E_Learning courses in Engineering Surveying
E_Learning courses in Engineering Surveying
ﻣﺜﺎل ﺣﻞ
ﻣﺜﺎل ﺣﻞ
٣
٣
:
:
واﻻﻧﺨﻔﺎض اﻻرﺗﻔﺎع ﻃﺮﻳﻘﺔ ﺑﺎﺳﺘﺨﺪام
واﻻﻧﺨﻔﺎض اﻻرﺗﻔﺎع ﻃﺮﻳﻘﺔ ﺑﺎﺳﺘﺨﺪام
اﻟﻨﻘﻄﺔ
اﻟﻤﺆﺧﺮة
اﻟﻤﺘﻮﺳﻄﺔ
اﻟﻤﻘﺪﻣﺔ
ارﺗﻔﺎع
(+)
اﻧﺨﻔﺎض
(-)
ﻣﻨﺴﻮب
اﻟﻨﻘﻄﺔ
ﻣﻼﺣﻈﺎت
ا
2.19
24.16
روﺑﻴﺮ
ب
2.50
0.31
23.85
ﺟـ
2.32
0.18
24.03
د
3.01
1.49
0.83
24.86
دوران
هـ
2.51
0.50
25.36
و
1.75
2.81
0.30
25.06
دوران
ز
3.81
2.06
23.00
اﻟﻤﺠﻤﻮع
6.95
8.11
اﻟﻔﺮق
1.16
1.16
اﻟﺘﺤﻘﻴﻖ
O.K
46. ٤٦
٤٦
Prof. Dr. Eng. Said Elmaghraby
Prof. Dr. Eng. Said Elmaghraby
E_Learning courses in Engineering Surveying
E_Learning courses in Engineering Surveying
اﻟﺜﺎﻟﺚ اﻟﺒﺎب
اﻟﺜﺎﻟﺚ اﻟﺒﺎب
ﻣﺤﻠﻮﻟﺔ أﻣﺜﻠﺔ
ﻣﺤﻠﻮﻟﺔ أﻣﺜﻠﺔ
ﻋﻠﻰ
ﻋﻠﻰ
واﻟﺮدم اﻟﺤﻔﺮ ﻣﻜﻌﺒﺎت ﺣﺴﺎب
واﻟﺮدم اﻟﺤﻔﺮ ﻣﻜﻌﺒﺎت ﺣﺴﺎب
47. ٤٧
٤٧
Prof. Dr. Eng. Said Elmaghraby
Prof. Dr. Eng. Said Elmaghraby
E_Learning courses in Engineering Surveying
E_Learning courses in Engineering Surveying
رﻗﻢ ﻣﺜﺎل
رﻗﻢ ﻣﺜﺎل
١
١
:
:
واﻟﺮدم اﻟﺤﻔﺮ ﻣﻜﻌﺒﺎت ﺣﺴﺎب
واﻟﺮدم اﻟﺤﻔﺮ ﻣﻜﻌﺒﺎت ﺣﺴﺎب
ﻣﻦ
ﻣﻦ
اﻟﺸﺒﻜﻴﺔ اﻟﻤﻴﺰاﻧﻴﺔ
اﻟﺸﺒﻜﻴﺔ اﻟﻤﻴﺰاﻧﻴﺔ
-
-
اﻟﺤﻔﺮ ﻣﻜﻌﺒﺎت ﺣﺴﺎب ﻣﻄﻠﻮب
اﻟﺤﻔﺮ ﻣﻜﻌﺒﺎت ﺣﺴﺎب ﻣﻄﻠﻮب
اﻷرض ﻟﺘﺴﻮﻳﺔ اﻟﻼزﻣﺔ واﻟﺮدم
اﻷرض ﻟﺘﺴﻮﻳﺔ اﻟﻼزﻣﺔ واﻟﺮدم
ﻣﻨﺴﻮب ﻋﻠﻰ ﺑﺎﻟﺸﻜﻞ اﻟﻤﻮﺿﺤﺔ
ﻣﻨﺴﻮب ﻋﻠﻰ ﺑﺎﻟﺸﻜﻞ اﻟﻤﻮﺿﺤﺔ
+
+
٧٫٠٠
٧٫٠٠
ﻣﺮﺑﻌﺔ اﻟﻮﺣﺪة ﺑﺄن ﻋﻠﻤﺎ ﻣﺘﺮ
ﻣﺮﺑﻌﺔ اﻟﻮﺣﺪة ﺑﺄن ﻋﻠﻤﺎ ﻣﺘﺮ
وأﺑﻌﺎدهﺎ اﻟﺸﻜﻞ
وأﺑﻌﺎدهﺎ اﻟﺸﻜﻞ
٣٠
٣٠
x
x
٣٠
٣٠
م
م
.
.
٦٫٥
٦٫٥
٧٫٥
٧٫٥
٦٫٨
٦٫٨
٨٫١
٨٫١
٨٫٣
٨٫٣
٧٫٣
٧٫٣
٧٫٠
٧٫٠
٧٫٢
٧٫٢
٧٫٥
٧٫٥
٦٫١
٦٫١
٦٫٨
٦٫٨
٦٫٥
٦٫٥ ٦٫٦
٦٫٦ ٦٫٧
٦٫٧
٦٫٠
٦٫٠ ٥٫٨
٥٫٨ ٥٫٥
٥٫٥
٦٫٠
٦٫٠
٥٫٣
٥٫٣
٩٫١
٩٫١
٨٫٥
٨٫٥
٨٫٥
٨٫٥
48. ٤٨
٤٨
Prof. Dr. Eng. Said Elmaghraby
Prof. Dr. Eng. Said Elmaghraby
E_Learning courses in Engineering Surveying
E_Learning courses in Engineering Surveying
رﻗﻢ ﻣﺜﺎل ﺣﻞ
رﻗﻢ ﻣﺜﺎل ﺣﻞ
١
١
:
:
٦٫٥
٦٫٥
٧٫٥
٧٫٥
٦٫٨
٦٫٨
٨٫١
٨٫١
٨٫٣
٨٫٣
٧٫٣
٧٫٣
٧٫٠
٧٫٠
٧٫٢
٧٫٢
٧٫٥
٧٫٥
٦٫١
٦٫١
٦٫٨
٦٫٨
٦٫٥
٦٫٥ ٦٫٦
٦٫٦ ٦٫٧
٦٫٧
٦٫٠
٦٫٠ ٥٫٨
٥٫٨ ٥٫٥
٥٫٥
٦٫٠
٦٫٠
٥٫٣
٥٫٣
٩٫١
٩٫١
٨٫٥
٨٫٥
٨٫٥
٨٫٥
-
-
رﺳﻢ ﻳﺘﻢ
رﺳﻢ ﻳﺘﻢ
آﻨﺘﻮر
آﻨﺘﻮر
+
+
٧٫٠٠
٧٫٠٠
م
م
-
-
اﻷﻋﻠﻰ اﻟﻘﻴﻤﺔ ذات اﻷرآﺎن
اﻷﻋﻠﻰ اﻟﻘﻴﻤﺔ ذات اﻷرآﺎن
ﻣﻨﺴﻮب ﻣﻦ
ﻣﻨﺴﻮب ﻣﻦ
+
+
٧٫٠٠
٧٫٠٠
م
م
اﻷﻗﻞ واﻷرآﺎن ﺣﻔﺮ ﻣﻨﺎﻃﻖ
اﻷﻗﻞ واﻷرآﺎن ﺣﻔﺮ ﻣﻨﺎﻃﻖ
ردم ﻣﻨﺎﻃﻖ
ردم ﻣﻨﺎﻃﻖ
.
.
ﺣﻔﺮ
ﺣﻔﺮ
ردم
ردم
49. ٤٩
٤٩
Prof. Dr. Eng. Said Elmaghraby
Prof. Dr. Eng. Said Elmaghraby
E_Learning courses in Engineering Surveying
E_Learning courses in Engineering Surveying
”
”
ﺗﺎﺑﻊ
ﺗﺎﺑﻊ
“
“
رﻗﻢ ﻣﺜﺎل ﺣﻞ
رﻗﻢ ﻣﺜﺎل ﺣﻞ
١
١
:
:
٠٫٥
٠٫٥
٠٫٥
٠٫٥
٠٫٢
٠٫٢
١٫١
١٫١
١٫٣
١٫٣
٠٫٣
٠٫٣
٠٫٠
٠٫٠
٠٫٢
٠٫٢
٠٫٥
٠٫٥
٠٫٩
٠٫٩
٠٫٢
٠٫٢
٠٫٥
٠٫٥ ٠٫٤
٠٫٤ ٠٫٣
٠٫٣
١٫٠
١٫٠ ١٫٢
١٫٢ ١٫٥
١٫٥
١٫٠
١٫٠
١٫٧
١٫٧
٢٫١
٢٫١
١٫٥
١٫٥
١٫٥
١٫٥
-
-
ﻃﺮح ﻳﺘﻢ
ﻃﺮح ﻳﺘﻢ
+
+
٧٫٠٠
٧٫٠٠
ﻣﻦ م
ﻣﻦ م
وﻧﻀﻊ اﻷرآﺎن ﺟﻤﻴﻊ ﻣﻨﺎﺳﻴﺐ
وﻧﻀﻊ اﻷرآﺎن ﺟﻤﻴﻊ ﻣﻨﺎﺳﻴﺐ
أو ﺟﺪﻳﺪ رﺳﻢ ﻋﻠﻰ اﻟﻔﺮق
أو ﺟﺪﻳﺪ رﺳﻢ ﻋﻠﻰ اﻟﻔﺮق
اﻟﺮﺳﻢ ﻧﻔﺲ ﻋﻠﻰ ﺁﺧﺮ ﺑﻠﻮن
اﻟﺮﺳﻢ ﻧﻔﺲ ﻋﻠﻰ ﺁﺧﺮ ﺑﻠﻮن
.
.
ﺣﻔﺮ
ﺣﻔﺮ
ردم
ردم
50. ٥٠
٥٠
Prof. Dr. Eng. Said Elmaghraby
Prof. Dr. Eng. Said Elmaghraby
E_Learning courses in Engineering Surveying
E_Learning courses in Engineering Surveying
”
”
ﺗﺎﺑﻊ
ﺗﺎﺑﻊ
“
“
رﻗﻢ ﻣﺜﺎل ﺣﻞ
رﻗﻢ ﻣﺜﺎل ﺣﻞ
١
١
:
:
-
-
ﻣﻜﻌﺒﺎت ﺣﺠﻢ ﺣﺴﺎب ﻳﺘﻢ
ﻣﻜﻌﺒﺎت ﺣﺠﻢ ﺣﺴﺎب ﻳﺘﻢ
اﻟﺼﺤﻴﺤﺔ اﻟﺮدم
اﻟﺼﺤﻴﺤﺔ اﻟﺮدم
)
)
ﻋﺪد
ﻋﺪد
٦
٦
ﻣﺮﺑﻌﺎت
ﻣﺮﺑﻌﺎت
:(
:(
٠٫٥
٠٫٥
٠٫٥
٠٫٥
٠٫٢
٠٫٢
١٫١
١٫١
١٫٣
١٫٣
٠٫٣
٠٫٣
٠٫٠
٠٫٠
٠٫٢
٠٫٢
٠٫٥
٠٫٥
٠٫٩
٠٫٩
٠٫٢
٠٫٢
٠٫٥
٠٫٥ ٠٫٤
٠٫٤ ٠٫٣
٠٫٣
١٫٠
١٫٠ ١٫٢
١٫٢ ١٫٥
١٫٥
١٫٠
١٫٠
١٫٧
١٫٧
٢٫١
٢٫١
١٫٥
١٫٥
١٫٥
١٫٥
ﺣﻔﺮ
ﺣﻔﺮ
ردم
ردم
١ع
٢ع
٣ع
٤ع
1
1.2
0.3
1.7
1.5
0.4
0.9
1
0.5
0.2
0.2
0.5
4.1
4.6
0
0.7
اﻟﺤﺠﻢ
اﻟﺤﺠﻢ
)) =
)) =
٣٠
٣٠
x
x
٣٠
٣٠
(
(
٤
٤
(
(
x
x
)
)
١
١
x
x
٤٫١
٤٫١
+
+
٢
٢
x
x
٤٫٦
٤٫٦
+
+
٤
٤
x
x
٠٫٧
٠٫٧
= (
= (
٣٦٢٢٫٥
٣٦٢٢٫٥
م
م
٣
٣
51. ٥١
٥١
Prof. Dr. Eng. Said Elmaghraby
Prof. Dr. Eng. Said Elmaghraby
E_Learning courses in Engineering Surveying
E_Learning courses in Engineering Surveying
”
”
ﺗﺎﺑﻊ
ﺗﺎﺑﻊ
“
“
رﻗﻢ ﻣﺜﺎل ﺣﻞ
رﻗﻢ ﻣﺜﺎل ﺣﻞ
١
١
:
:
-
-
ﻣﻜﻌﺒﺎت ﺣﺠﻢ ﺣﺴﺎب ﻳﺘﻢ
ﻣﻜﻌﺒﺎت ﺣﺠﻢ ﺣﺴﺎب ﻳﺘﻢ
اﻟﺼﺤﻴﺤﺔ اﻟﻐﻴﺮ اﻟﺮدم
اﻟﺼﺤﻴﺤﺔ اﻟﻐﻴﺮ اﻟﺮدم
)
)
ﻋﺪد
ﻋﺪد
٣
٣
أﺟﺰاء
أﺟﺰاء
:(
:(
٠٫٥
٠٫٥
٠٫٥
٠٫٥
٠٫٢
٠٫٢
١٫١
١٫١
١٫٣
١٫٣
٠٫٣
٠٫٣
٠٫٠
٠٫٠
٠٫٢
٠٫٢
٠٫٥
٠٫٥
٠٫٩
٠٫٩
٠٫٢
٠٫٢
٠٫٥
٠٫٥ ٠٫٤
٠٫٤ ٠٫٣
٠٫٣
١٫٠
١٫٠ ١٫٢
١٫٢ ١٫٥
١٫٥
١٫٠
١٫٠
١٫٧
١٫٧
٢٫١
٢٫١
١٫٥
١٫٥
١٫٥
١٫٥
ﺣﻔﺮ
ﺣﻔﺮ
ردم
ردم
I
I II
II III
III
ﺣﺠﻢ
ﺣﺠﻢ
I
I
) =
) =
١٥
١٥
x
x
٣٠
٣٠
(
(
x
x
))
))
٠٫٥
٠٫٥
+
+
٠٫٢
٠٫٢
+
+
٠٫٠
٠٫٠
+
+
٠٫٠
٠٫٠
(
(
٤
٤
=(
=(
٧٨٫٧٥
٧٨٫٧٥
م
م
٣
٣
ﺣﺠﻢ
ﺣﺠﻢ
II
II
))=
))=
١٥
١٥
+
+
٣٠
٣٠
(
(
٢
٢
(
(
x
x
٣٠
٣٠
(
(
x
x
))
))
٠٫٢
٠٫٢
+
+
٠٫٢
٠٫٢
+
+
٠٫٠
٠٫٠
+
+
٠٫٠
٠٫٠
(
(
٤
٤
=(
=(
٦٧٫٥
٦٧٫٥
م
م
٣
٣
ﺣﺠﻢ
ﺣﺠﻢ
III
III
))=
))=
٢٢٫٥
٢٢٫٥
+
+
٣٠
٣٠
(
(
٢
٢
(
(
x
x
٣٠
٣٠
(
(
x
x
))
))
٠٫٩
٠٫٩
+
+
٠٫٢
٠٫٢
+
+
٠٫٠
٠٫٠
+
+
٠٫٠
٠٫٠
(
(
٤
٤
=(
=(
٢١٦٫٥٦
٢١٦٫٥٦
م
م
٣
٣
اﻟﺮدم ﺣﺠﻢ اﺟﻤﺎﻟﻰ
اﻟﺮدم ﺣﺠﻢ اﺟﻤﺎﻟﻰ
=
=
٣٦٢٢٫٥
٣٦٢٢٫٥
+
+
٧٨٫٧٥
٧٨٫٧٥
+
+
٦٧٫٥
٦٧٫٥
+
+
٢١٦٫٥٦
٢١٦٫٥٦
=
=
٣٩٨٥٫٣١
٣٩٨٥٫٣١
م
م
٣
٣
52. ٥٢
٥٢
Prof. Dr. Eng. Said Elmaghraby
Prof. Dr. Eng. Said Elmaghraby
E_Learning courses in Engineering Surveying
E_Learning courses in Engineering Surveying
”
”
ﺗﺎﺑﻊ
ﺗﺎﺑﻊ
“
“
رﻗﻢ ﻣﺜﺎل ﺣﻞ
رﻗﻢ ﻣﺜﺎل ﺣﻞ
١
١
:
:
-
-
ﻣﻜﻌﺒﺎت ﺣﺠﻢ ﺣﺴﺎب ﻳﺘﻢ
ﻣﻜﻌﺒﺎت ﺣﺠﻢ ﺣﺴﺎب ﻳﺘﻢ
اﻟﺼﺤﻴﺤﺔ اﻟﺤﻔﺮ
اﻟﺼﺤﻴﺤﺔ اﻟﺤﻔﺮ
)
)
ﻋﺪد
ﻋﺪد
٤
٤
ﻣﺮﺑﻌﺎت
ﻣﺮﺑﻌﺎت
:(
:(
٠٫٥
٠٫٥
٠٫٥
٠٫٥
٠٫٢
٠٫٢
١٫١
١٫١
١٫٣
١٫٣
٠٫٣
٠٫٣
٠٫٠
٠٫٠
٠٫٢
٠٫٢
٠٫٥
٠٫٥
٠٫٩
٠٫٩
٠٫٢
٠٫٢
٠٫٥
٠٫٥ ٠٫٤
٠٫٤ ٠٫٣
٠٫٣
١٫٠
١٫٠ ١٫٢
١٫٢ ١٫٥
١٫٥
١٫٠
١٫٠
١٫٧
١٫٧
٢٫١
٢٫١
١٫٥
١٫٥
١٫٥
١٫٥
ﺣﻔﺮ
ﺣﻔﺮ
ردم
ردم
١ع
٢ع
٣ع
٤ع
0.5
0.2
1.5
0
2.1
0.3
1.3
1.5
0.5
1.1
5.4
3.6
0
0
اﻟﺤﺠﻢ
اﻟﺤﺠﻢ
)) =
)) =
٣٠
٣٠
x
x
٣٠
٣٠
(
(
٤
٤
(
(
x
x
)
)
١
١
x
x
٥٫٤
٥٫٤
+
+
٢
٢
x
x
٣٫٦
٣٫٦
= (
= (
٢٨٣٥
٢٨٣٥
م
م
٣
٣
I
I II
II III
III
53. ٥٣
٥٣
Prof. Dr. Eng. Said Elmaghraby
Prof. Dr. Eng. Said Elmaghraby
E_Learning courses in Engineering Surveying
E_Learning courses in Engineering Surveying
”
”
ﺗﺎﺑﻊ
ﺗﺎﺑﻊ
“
“
رﻗﻢ ﻣﺜﺎل ﺣﻞ
رﻗﻢ ﻣﺜﺎل ﺣﻞ
١
١
:
:
-
-
ﻣﻜﻌﺒﺎت ﺣﺠﻢ ﺣﺴﺎب ﻳﺘﻢ
ﻣﻜﻌﺒﺎت ﺣﺠﻢ ﺣﺴﺎب ﻳﺘﻢ
اﻟﺼﺤﻴﺤﺔ اﻟﻐﻴﺮ اﻟﺤﻔﺮ
اﻟﺼﺤﻴﺤﺔ اﻟﻐﻴﺮ اﻟﺤﻔﺮ
)
)
ﻋﺪد
ﻋﺪد
٣
٣
أﺟﺰاء
أﺟﺰاء
:(
:(
٠٫٥
٠٫٥
٠٫٥
٠٫٥
٠٫٢
٠٫٢
١٫١
١٫١
١٫٣
١٫٣
٠٫٣
٠٫٣
٠٫٠
٠٫٠
٠٫٢
٠٫٢
٠٫٥
٠٫٥
٠٫٩
٠٫٩
٠٫٢
٠٫٢
٠٫٥
٠٫٥ ٠٫٤
٠٫٤ ٠٫٣
٠٫٣
١٫٠
١٫٠ ١٫٢
١٫٢ ١٫٥
١٫٥
١٫٠
١٫٠
١٫٧
١٫٧
٢٫١
٢٫١
١٫٥
١٫٥
١٫٥
١٫٥
ﺣﻔﺮ
ﺣﻔﺮ
ردم
ردم
I
I II
II III
III
ﺣﺠﻢ
ﺣﺠﻢ
IV
IV
) =
) =
١٥
١٥
x
x
٣٠
٣٠
(
(
x
x
))
))
٠٫٥
٠٫٥
+
+
٠٫٢
٠٫٢
+
+
٠٫٠
٠٫٠
+
+
٠٫٠
٠٫٠
(
(
٤
٤
=(
=(
٧٨٫٧٥
٧٨٫٧٥
م
م
٣
٣
ﺣﺠﻢ
ﺣﺠﻢ
V
V
)) =
)) =
١٥
١٥
x
x
٣٠
٣٠
(
(
٢
٢
(
(
x
x
))
))
٠٫٢
٠٫٢
+
+
٠٫٠
٠٫٠
+
+
٠٫٠
٠٫٠
(
(
٣
٣
=(
=(
١٥٫٠
١٥٫٠
م
م
٣
٣
ﺣﺠﻢ
ﺣﺠﻢ
VI
VI
)) =
)) =
٧٫٥
٧٫٥
x
x
٣٠
٣٠
(
(
٢
٢
(
(
x
x
))
))
٠٫٣
٠٫٣
+
+
٠٫٠
٠٫٠
+
+
٠٫٠
٠٫٠
(
(
٣
٣
=(
=(
١١٫٢٥
١١٫٢٥
م
م
٣
٣
اﻟﺮدم ﺣﺠﻢ اﺟﻤﺎﻟﻰ
اﻟﺮدم ﺣﺠﻢ اﺟﻤﺎﻟﻰ
=
=
٢٨٣٥
٢٨٣٥
+
+
٧٨٫٧٥
٧٨٫٧٥
+
+
١٥٫٠
١٥٫٠
+
+
١١٫٢٥
١١٫٢٥
=
=
٢٩٤٠٫٠
٢٩٤٠٫٠
م
م
٣
٣
IV
IV V
V
VI
VI
54. ٥٤
٥٤
Prof. Dr. Eng. Said Elmaghraby
Prof. Dr. Eng. Said Elmaghraby
E_Learning courses in Engineering Surveying
E_Learning courses in Engineering Surveying
ﻣﺜﺎل
ﻣﺜﺎل
٢
٢
:
:
أﺑﻌﺎدﻩ ﺳﺒﺎﺣﺔ ﺣﻮض ﻹﻧﺸﺎء
أﺑﻌﺎدﻩ ﺳﺒﺎﺣﺔ ﺣﻮض ﻹﻧﺸﺎء
٢٠
٢٠
x
x
٥٠
٥٠
ﻣﻴﺰاﻧﻴﺔ أﺟﺮﻳﺖ م
ﻣﻴﺰاﻧﻴﺔ أﺟﺮﻳﺖ م
ﻳﻠﻰ آﻤﺎ ﻓﻜﺎﻧﺖ اﻟﻄﻮﻟﻰ اﻟﻤﺤﻮر ﻋﻠﻰ ﻃﻮﻟﻴﺔ
ﻳﻠﻰ آﻤﺎ ﻓﻜﺎﻧﺖ اﻟﻄﻮﻟﻰ اﻟﻤﺤﻮر ﻋﻠﻰ ﻃﻮﻟﻴﺔ
:
:
اﻟﻤﺴﺎﻓﺔ
0
10
20
30
40
50
اﻷرض ﻣﻨﺴﻮب
اﻟﻄﺒﻴﻌﻴﺔ
39.30
40.00
41.50
39.70
39.50
37.70
اﻟﺤﻮض ﺑﺪاﻳﺔ ﻋﻨﺪ اﻹﻧﺸﺎء ﺧﻂ ﻣﻨﺴﻮب آﺎن ﻓﺈذا
اﻟﺤﻮض ﺑﺪاﻳﺔ ﻋﻨﺪ اﻹﻧﺸﺎء ﺧﻂ ﻣﻨﺴﻮب آﺎن ﻓﺈذا
٣٨٫٥
٣٨٫٥
م
م
اﻟﺤﻮض ﻗﺎع واﻧﺤﺪار
اﻟﺤﻮض ﻗﺎع واﻧﺤﺪار
٥
٥
%
%
اﻟﻰ ﺻﻔﺮ ﻣﺴﺎﻓﺔ ﻣﻦ ﻷﺳﻔﻞ
اﻟﻰ ﺻﻔﺮ ﻣﺴﺎﻓﺔ ﻣﻦ ﻷﺳﻔﻞ
٢٠
٢٠
م
م
ﺛﻢ
ﺛﻢ
٤
٤
%
%
ﻣﻦ ﻷﺳﻔﻞ
ﻣﻦ ﻷﺳﻔﻞ
٢٠
٢٠
اﻟﻰ
اﻟﻰ
٣٠
٣٠
ﻣﻦ أﻓﻘﻰ ﺛﻢ م
ﻣﻦ أﻓﻘﻰ ﺛﻢ م
٣٠
٣٠
اﻟﻰ
اﻟﻰ
٥٠
٥٠
م
م
.
.
واﻟﺮدم اﻟﺤﻔﺮ ﻣﻜﻌﺒﺎت وﺣﺴﺎب ﻃﻮﻟﻰ ﻗﻄﺎع رﺳﻢ واﻟﻤﻄﻠﻮب
واﻟﺮدم اﻟﺤﻔﺮ ﻣﻜﻌﺒﺎت وﺣﺴﺎب ﻃﻮﻟﻰ ﻗﻄﺎع رﺳﻢ واﻟﻤﻄﻠﻮب
.
.
55. ٥٥
٥٥
Prof. Dr. Eng. Said Elmaghraby
Prof. Dr. Eng. Said Elmaghraby
E_Learning courses in Engineering Surveying
E_Learning courses in Engineering Surveying
٣٧
٣٨
٣٩
٤٠
٤١
٣٦
٣٥
ءﺎﺸﻧﻹا ﻂﺧ
ﺔﻴﻌﻴﺒﻄﻟا ضرﻷا
اﻟﻤﺴﺎﻓﺔ
0
10
20
30
40
50
اﻟﻄﺒﻴﻌﻴﺔ اﻷرض ﻣﻨﺴﻮب
39.30
40.00
41.50
39.70
39.50
37.70
اﻹﻧﺸﺎء ﺧﻂ ﻣﻨﺴﻮب
38.50
38.00
37.50
37.10
37.10
37.10
اﻟﺤﻔﺮ ارﺗﻔﺎع
0.80
2.00
4.00
2.60
2.40
0.60
اﻟﺤﻔﺮ ﻣﺴﺎﺣﺔ
16.00
40.00
80.00
52.00
48.00
600.00
اﻟﺤﻔﺮ ﺣﺠﻢ
1120
2400
2640
2000
12960
اﻟﻜﻠﻰ اﻟﺤﺠﻢ
21120
:
:٢
٢ ﻣﺜﺎل ﺣﻞ
ﻣﺜﺎل ﺣﻞ
56. ٥٦
٥٦
Prof. Dr. Eng. Said Elmaghraby
Prof. Dr. Eng. Said Elmaghraby
E_Learning courses in Engineering Surveying
E_Learning courses in Engineering Surveying
ﻣﺜﺎل
ﻣﺜﺎل
٣
٣
:
:
آﻞ ﻃﻮﻟﻴﺔ ﻣﻴﺰاﻧﻴﺔ أﺟﺮﻳﺖ ﻃﺮﻳﻖ إﻧﺸﺎء ﺑﻬﺪف
آﻞ ﻃﻮﻟﻴﺔ ﻣﻴﺰاﻧﻴﺔ أﺟﺮﻳﺖ ﻃﺮﻳﻖ إﻧﺸﺎء ﺑﻬﺪف
١٠٠
١٠٠
ﻋﻠﻰ م
ﻋﻠﻰ م
ﻳﻠﻰ آﻤﺎ ﻓﻜﺎﻧﺖ اﻟﻄﻮﻟﻰ اﻟﻤﺤﻮر
ﻳﻠﻰ آﻤﺎ ﻓﻜﺎﻧﺖ اﻟﻄﻮﻟﻰ اﻟﻤﺤﻮر
:
:
اﻟﻤﺴﺎﻓﺔ
0
100
200
300
400
500
اﻷرض ﻣﻨﺴﻮب
اﻟﻄﺒﻴﻌﻴﺔ
26.00
27.50
25.00
23.00
23.00
24.00
واﺣﺪ ﻣﺘﺮ ﻳﺴﺎوى اﻟﻄﺮﻳﻖ ﺑﺪاﻳﺔ ﻋﻨﺪ اﻟﺤﻔﺮ ارﺗﻔﺎع آﺎن ﻓﺈذا
واﺣﺪ ﻣﺘﺮ ﻳﺴﺎوى اﻟﻄﺮﻳﻖ ﺑﺪاﻳﺔ ﻋﻨﺪ اﻟﺤﻔﺮ ارﺗﻔﺎع آﺎن ﻓﺈذا
،
،
واﻟﻄﺮﻳﻖ
واﻟﻄﺮﻳﻖ
ﺑﻨﺴﺒﺔ أﺳﻔﻞ اﻟﻰ ﻳﻨﺤﺪر
ﺑﻨﺴﺒﺔ أﺳﻔﻞ اﻟﻰ ﻳﻨﺤﺪر
٢٠٠:١
٢٠٠:١
ﻣﺴﺎﻓﺔ اﻟﻰ ﺻﻔﺮ ﻣﺴﺎﻓﺔ ﻣﻦ
ﻣﺴﺎﻓﺔ اﻟﻰ ﺻﻔﺮ ﻣﺴﺎﻓﺔ ﻣﻦ
٢٠٠
٢٠٠
م
م
،
،
ﻣﺴﺎﻓﺔ ﻣﻦ أﻓﻘﻰ ﺛﻢ
ﻣﺴﺎﻓﺔ ﻣﻦ أﻓﻘﻰ ﺛﻢ
٢٠٠
٢٠٠
اﻟﻰ
اﻟﻰ
٣٠٠
٣٠٠
م
م
،
،
ﺑﻨﺴﺒﺔ أﻋﻠﻰ اﻟﻰ ﺛﻢ
ﺑﻨﺴﺒﺔ أﻋﻠﻰ اﻟﻰ ﺛﻢ
١٠٠:١
١٠٠:١
اﻟﻄﺮﻳﻖ ﻧﻬﺎﻳﺔ ﺣﺘﻰ
اﻟﻄﺮﻳﻖ ﻧﻬﺎﻳﺔ ﺣﺘﻰ
.
.
اﻟﻄﺮﻳﻖ وﻋﺮض
اﻟﻄﺮﻳﻖ وﻋﺮض
١٥
١٥
وﻣﻴﻮﻟﻪ م
وﻣﻴﻮﻟﻪ م
اﻟﺠﺎﻧﺒﻴﺔ
اﻟﺠﺎﻧﺒﻴﺔ
٢:١
٢:١
.
.
واﻟﺮدم اﻟﺤﻔﺮ ﻣﻜﻌﺒﺎت وﺣﺴﺎب ﻟﻠﻄﺮﻳﻖ ﻃﻮﻟﻰ ﻗﻄﺎع رﺳﻢ واﻟﻤﻄﻠﻮب
واﻟﺮدم اﻟﺤﻔﺮ ﻣﻜﻌﺒﺎت وﺣﺴﺎب ﻟﻠﻄﺮﻳﻖ ﻃﻮﻟﻰ ﻗﻄﺎع رﺳﻢ واﻟﻤﻄﻠﻮب
.
.
57. ٥٧
٥٧
Prof. Dr. Eng. Said Elmaghraby
Prof. Dr. Eng. Said Elmaghraby
E_Learning courses in Engineering Surveying
E_Learning courses in Engineering Surveying
٢٤
٢٥
٢٦
٢٧
٢٨
٢٣
٢٢
ءﺎﺸﻧﻹا ﻂﺧ
ﺔﻴﻌﻴﺒﻄﻟا ض
رﻷا
:
:٣
٣ ﻣﺜﺎل ﺣﻞ
ﻣﺜﺎل ﺣﻞ
٢
١
١٥
م
واﻟﺮدم اﻟﺤﻔﺮ ﻓﻰ اﻟﻘﻄﺎع ﻣﺴﺎﺣﺔ ﻣﻌﺎدﻟﺔ
=
ع
x
)
١٥
+
١٥
+
٤
ع
(
٢
=
ع
x
)
١٥
+
٢
ع
(
ع
١٥
+
٤
ع
اﻟﺤﻔﺮ
اﻟﺮدم
اﻟﻔﺎﺻﻞ
اﻟﺤﺪ
58. ٥٨
٥٨
Prof. Dr. Eng. Said Elmaghraby
Prof. Dr. Eng. Said Elmaghraby
E_Learning courses in Engineering Surveying
E_Learning courses in Engineering Surveying
اﻟﻤﺴﺎﻓﺔ
0.0
100.0
200.0
300.0
400.0
500.0
اﻷرض ﻣﻨﺴﻮب
اﻟﻄﺒﻴﻌﻴﺔ
26.0
27.5
25.0
23.0
23.0
24.0
ﺧﻂ ﻣﻨﺴﻮب
اﻹﻧﺸﺎء
25.0
24.5
24.0
24.0
25.0
26.0
اﻟﺤﻔﺮ ارﺗﻔﺎع
1
3
1
اﻟﺮدم ارﺗﻔﺎع
1
2
2
اﻟﺤﻔﺮ ﻣﺴﺎﺣﺔ
17
63
17
0
اﻟﺮدم ﻣﺴﺎﺣﺔ
0
17
38
38
اﻟﺤﻔﺮ ﺣﺠﻢ
4000
4000
425
اﻟﺮدم ﺣﺠﻢ
425
2750
3800
اﻟﺤﻔﺮ إﺟﻤﺎﻟﻰ
اﻟﺮدم ﺣﺠﻢ
6975
8425
2
5
0
.0
:
:٣
٣ ﻣﺜﺎل ﺣﻞ
ﻣﺜﺎل ﺣﻞ
اﻟﻔﺎﺻﻞ
اﻟﺤﺪ
اﻟﻔﺎﺻﻞ
اﻟﺤﺪ
59. ٥٩
٥٩
Prof. Dr. Eng. Said Elmaghraby
Prof. Dr. Eng. Said Elmaghraby
E_Learning courses in Engineering Surveying
E_Learning courses in Engineering Surveying
ﻣﺜﺎل
ﻣﺜﺎل
٤
٤
:
:
ﻣﻘﺘﺮح ﻃﺮﻳﻖ ﻣﺤﻮر ﻋﻠﻰ واﻟﻤﻨﺎﺳﻴﺐ اﻟﻤﺴﺎﻓﺎت أﺧﺬت
ﻣﻘﺘﺮح ﻃﺮﻳﻖ ﻣﺤﻮر ﻋﻠﻰ واﻟﻤﻨﺎﺳﻴﺐ اﻟﻤﺴﺎﻓﺎت أﺧﺬت
ﻳﻠﻰ آﻤﺎ ﻓﻜﺎﻧﺖ
ﻳﻠﻰ آﻤﺎ ﻓﻜﺎﻧﺖ
:
:
اﻟﻤﺴﺎﻓﺔ
0
100
200
300
400
500
اﻷرض ﻣﻨﺴﻮب
اﻟﻄﺒﻴﻌﻴﺔ
23.50
24.50
25.70
23.90
21.60
21.50
ﺻﻔﺮ ﻣﺴﺎﻓﺔ ﻋﻨﺪ اﻹﻧﺸﺎء ﺧﻂ ﻣﻨﺴﻮب آﺎن ﻓﺈذا
ﺻﻔﺮ ﻣﺴﺎﻓﺔ ﻋﻨﺪ اﻹﻧﺸﺎء ﺧﻂ ﻣﻨﺴﻮب آﺎن ﻓﺈذا
٢٣٫٣٠
٢٣٫٣٠
م
م
،
،
وﻋﺮض
وﻋﺮض
اﻟﻄﺮﻳﻖ
اﻟﻄﺮﻳﻖ
١٠
١٠
اﻟﺠﺎﻧﺒﻴﺔ وﻣﻴﻮﻟﻪ م
اﻟﺠﺎﻧﺒﻴﺔ وﻣﻴﻮﻟﻪ م
١:١
١:١
أﻋﻠﻰ اﻟﻰ ﻳﻤﻴﻞ واﻟﻄﺮﻳﻖ
أﻋﻠﻰ اﻟﻰ ﻳﻤﻴﻞ واﻟﻄﺮﻳﻖ
ﺑﻨﺴﺒﺔ
ﺑﻨﺴﺒﺔ
٥٠٠:١
٥٠٠:١
ﻣﺴﺎﻓﺔ اﻟﻰ ﺻﻔﺮ ﻣﺴﺎﻓﺔ ﻣﻦ
ﻣﺴﺎﻓﺔ اﻟﻰ ﺻﻔﺮ ﻣﺴﺎﻓﺔ ﻣﻦ
٣٠٠
٣٠٠
م
م
،
،
اﻟﻰ ﻳﻤﻴﻞ ﺛﻢ
اﻟﻰ ﻳﻤﻴﻞ ﺛﻢ
ﺑﻨﺴﺒﺔ أﺳﻔﻞ
ﺑﻨﺴﺒﺔ أﺳﻔﻞ
٤٠٠:١
٤٠٠:١
اﻟﻄﺮﻳﻖ ﻧﻬﺎﻳﺔ ﺣﺘﻰ
اﻟﻄﺮﻳﻖ ﻧﻬﺎﻳﺔ ﺣﺘﻰ
.
.
واﻟﺮدم اﻟﺤﻔﺮ ﻣﻜﻌﺒﺎت وﺣﺴﺎب ﻟﻠﻄﺮﻳﻖ ﻃﻮﻟﻰ ﻗﻄﺎع رﺳﻢ واﻟﻤﻄﻠﻮب
واﻟﺮدم اﻟﺤﻔﺮ ﻣﻜﻌﺒﺎت وﺣﺴﺎب ﻟﻠﻄﺮﻳﻖ ﻃﻮﻟﻰ ﻗﻄﺎع رﺳﻢ واﻟﻤﻄﻠﻮب
.
.
60. ٦٠
٦٠
Prof. Dr. Eng. Said Elmaghraby
Prof. Dr. Eng. Said Elmaghraby
E_Learning courses in Engineering Surveying
E_Learning courses in Engineering Surveying
٢٤
٢٥
٢٦
٢٧
٢١
٢٣
٢٢
ءﺸﺎﻹﻧا ﻂﺧ
ﺔﻴﻌﻴﺒﻄﻟا ضرﻷا
:
:٤
٤ ﻣﺜﺎل ﺣﻞ
ﻣﺜﺎل ﺣﻞ
١
١
١٠
م
واﻟﺮدم اﻟﺤﻔﺮ ﻓﻰ اﻟﻘﻄﺎع ﻣﺴﺎﺣﺔ ﻣﻌﺎدﻟﺔ
=
ع
x
)
١٠
+
١٠
+
٢
ع
(
٢
=
ع
x
)
١٠
+
ع
(
ع
١٠
+
٢
ع
اﻟﺤﻔﺮ
اﻟﺮدم
اﻟﻔﺎﺻﻞ
اﻟﺤﺪ
61. ٦١
٦١
Prof. Dr. Eng. Said Elmaghraby
Prof. Dr. Eng. Said Elmaghraby
E_Learning courses in Engineering Surveying
E_Learning courses in Engineering Surveying
اﻟﻤﺴﺎﻓﺔ
0.0
100.0
200.0
300.0
400.0
500.0
اﻷرض ﻣﻨﺴﻮب
اﻟﻄﺒﻴﻌﻴﺔ
23.50
24.50
25.70
23.90
21.60
21.50
ﺧﻂ ﻣﻨﺴﻮب
اﻹﻧﺸﺎء
23.30
23.50
23.70
23.90
23.65
23.40
اﻟﺤﻔﺮ ارﺗﻔﺎع
0.20
1.00
2.00
0.00
اﻟﺮدم ارﺗﻔﺎع
0.00
2.05
1.90
اﻟﺤﻔﺮ ﻣﺴﺎﺣﺔ
2.04
11.00
24.00
0.00
اﻟﺮدم ﻣﺴﺎﺣﺔ
0.00
24.70
22.61
اﻟﺤﻔﺮ ﺣﺠﻢ
652.0
1750.0
1200.0
اﻟﺮدم ﺣﺠﻢ
1235.13
2365.63
اﻟﺤﻔﺮ إﺟﻤﺎﻟﻰ
اﻟﺮدم ﺣﺠﻢ
3600.75
3602.00
:
:٤
٤ ﻣﺜﺎل ﺣﻞ
ﻣﺜﺎل ﺣﻞ
ﺪﺤﻟا
ﺪﺤﻟا
ﻞﺻ
ﺎﻔﻟا
ﻞﺻ
ﺎﻔﻟا
62. ٦٢
٦٢
Prof. Dr. Eng. Said Elmaghraby
Prof. Dr. Eng. Said Elmaghraby
E_Learning courses in Engineering Surveying
E_Learning courses in Engineering Surveying
اﻟﻜﻨﺘﻮر ﺧﻂ رﻗﻢ
اﻟﻤﺴﺎﺣﺔ
33
300
32
500
31
600
30
750
29
800
28
1000
27
1200
26
1720
ﻣﺜﺎل
ﻣﺜﺎل
٥
٥
:
:
ﺧﻄﻮط داﺧﻞ اﻟﻤﺤﺼﻮرة اﻟﻤﺴﺎﺣﺔ ﻳﻮﺿﺢ اﻟﺘﺎﻟﻰ اﻟﺠﺪول
ﺧﻄﻮط داﺧﻞ اﻟﻤﺤﺼﻮرة اﻟﻤﺴﺎﺣﺔ ﻳﻮﺿﺢ اﻟﺘﺎﻟﻰ اﻟﺠﺪول
اﻟﻜﻨﺘﻮر
اﻟﻜﻨﺘﻮر
واﻟﺮدم اﻟﺤﻔﺮ آﻤﻴﺎت ﺣﺴﺎب واﻟﻤﻄﻠﻮب ﺗﻞ ﺑﻬﺎ ﻟﻤﻨﻄﻘﺔ
واﻟﺮدم اﻟﺤﻔﺮ آﻤﻴﺎت ﺣﺴﺎب واﻟﻤﻄﻠﻮب ﺗﻞ ﺑﻬﺎ ﻟﻤﻨﻄﻘﺔ
ﻣﻨﺴﻮب ﻋﻠﻰ اﻷرض ﻟﺘﺴﻮﻳﺔ
ﻣﻨﺴﻮب ﻋﻠﻰ اﻷرض ﻟﺘﺴﻮﻳﺔ
٣٠
٣٠
م
م
.
.
63. ٦٣
٦٣
Prof. Dr. Eng. Said Elmaghraby
Prof. Dr. Eng. Said Elmaghraby
E_Learning courses in Engineering Surveying
E_Learning courses in Engineering Surveying
اﻟﻜﻨﺘﻮر ﺧﻂ رﻗﻢ
اﻟﻤﺴﺎﺣﺔ
اﻟﺤﺠﻢ ﻣﻌﺎدﻟﺔ
اﻟﻜﻤﻴﺔ
33
300
(٥٠٠+٣٠٠)١ * ٠٫٥= ح
400.00
32
500
(٦٠٠+٥٠٠)١ * ٠٫٥= ح
550.00
31
600
(٧٥٠+٦٠٠)١ * ٠٫٥ = ح
675.00
30
750
اﻟﺤﻔﺮ اﺟﻤﺎﻟﻰ
1625.00
30
750
3250.00
(٧٥٠-٨٠٠)١ * ٠٫٥ = ح
25.00
29
800
(٨٠٠ - ١٠٠٠) ١٫٥ = ح
300.00
28
1000
(١٠٠٠ - ١٢٠٠) ٢٫٥ = ح
500.00
27
1200
(١٢٠٠- ١٧٢٠) ٣٫٥ = ح
1820.00
26
1720
اﻟﺮدم اﺟﻤﺎﻟﻰ
2645.00
اﻟﺮدم ﻣﻜﻌﺒﺎت
اﻟﺤﻔﺮ ﻣﻜﻌﺒﺎت :
:٥
٥ ﻣﺜﺎل ﺣﻞ
ﻣﺜﺎل ﺣﻞ
64. ٦٤
٦٤
Prof. Dr. Eng. Said Elmaghraby
Prof. Dr. Eng. Said Elmaghraby
E_Learning courses in Engineering Surveying
E_Learning courses in Engineering Surveying
اﻟﻜﻨﺘﻮر ﺧﻂ رﻗﻢ
ﻣﺮﺑﻊ ﻣﺘﺮ اﻟﻤﺴﺎﺣﺔ
40
22080
45
56340
50
58060
55
61940
60
68060
65
70960
ﻣﺜﺎل
ﻣﺜﺎل
٦
٦
:
:
ﺻﺨﺮﻳﺔ ﺑﻤﻨﻄﻘﺔ اﻟﻤﻴﺎﻩ ﻟﺤﺠﺰ ﺧﺰان ﺑﻨﺎء ﺗﻢ
ﺻﺨﺮﻳﺔ ﺑﻤﻨﻄﻘﺔ اﻟﻤﻴﺎﻩ ﻟﺤﺠﺰ ﺧﺰان ﺑﻨﺎء ﺗﻢ
،
،
اﻟﺘﺎﻟﻰ واﻟﺠﺪول
اﻟﺘﺎﻟﻰ واﻟﺠﺪول
ﺧﻄﻮط وﻣﺴﺎﺣﺔ ﻣﻨﺴﻮب ﻳﻮﺿﺢ
ﺧﻄﻮط وﻣﺴﺎﺣﺔ ﻣﻨﺴﻮب ﻳﻮﺿﺢ
اﻟﻜﻨﺘﻮر
اﻟﻜﻨﺘﻮر
اﻟﻤﻨﻄﻘﺔ ﻟﻬﺬﻩ
اﻟﻤﻨﻄﻘﺔ ﻟﻬﺬﻩ
،
،
واﻟﻤﻄﻠﻮب
واﻟﻤﻄﻠﻮب
اﻟﺨﺰان ﺳﻌﺔ ﻣﻨﺤﻨﻰ رﺳﻢ
اﻟﺨﺰان ﺳﻌﺔ ﻣﻨﺤﻨﻰ رﺳﻢ
.
.
65. ٦٥
٦٥
Prof. Dr. Eng. Said Elmaghraby
Prof. Dr. Eng. Said Elmaghraby
E_Learning courses in Engineering Surveying
E_Learning courses in Engineering Surveying
اﻟﺤﻔﺮ ﻣﻜﻌﺒﺎت ﻣﺴﺎﺋﻞ ﻣﺜﻞ ﺗﺤﺴﺐ اﻟﺨﺰان ﺳﻌﺔ ﻣﺴﺎﺋﻞ
اﻟﺤﻔﺮ ﻣﻜﻌﺒﺎت ﻣﺴﺎﺋﻞ ﻣﺜﻞ ﺗﺤﺴﺐ اﻟﺨﺰان ﺳﻌﺔ ﻣﺴﺎﺋﻞ :
:٦
٦ ﻣﺜﺎل ﺣﻞ
ﻣﺜﺎل ﺣﻞ
اﻟﻜﻨﺘﻮر ﺧﻂ رﻗﻢ
اﻟﻤﺴﺎﺣﺔ
آﻨﺘﻮر ﺧﻄﻲ آﻞ ﺑﻴﻦ اﻟﻤﻴﺎﻩ ﺣﺠﻢ ﻣﻌﺎدﻟﺔ
اﻟﺠﺰﺋﻰ اﻟﺤﺠﻢ
اﻟﻜﻠﻰ اﻟﺤﺠﻢ
40
22080
(٥٦٣٤٠+٢٢٠٨٠)٥ * ٠٫٥= ح
196050
196050
45
56340
(٥٦٣٤٠+٥٨٠٦٠)٥ * ٠٫٥= ح
286000
482050
50
58060
(٥٨٠٦٠+٦١٩٤٠)٥ * ٠٫٥ = ح
300000
782050
55
61940
(٦١٩٤٠+٦٨٠٦٠)٥ * ٠٫٥ = ح
325000
1107050
60
68060
(٦٨٠٦٠+٧٠٩٦٠)٥ * ٠٫٥ = ح
347550
1454600
65
70960
اﻟﻜﻠﻰ اﻟﺤﺠﻢ
1454600
66. ٦٦
٦٦
Prof. Dr. Eng. Said Elmaghraby
Prof. Dr. Eng. Said Elmaghraby
E_Learning courses in Engineering Surveying
E_Learning courses in Engineering Surveying
٥٠
٥٥
٦٠
٦٥
٤٥
٤٠
اﻟﺴﻌﺔ
اﻟﻤﻜﻌﺐ ﺑﺎﻟﻤﺘﺮ
ﺻﻔﺮ
٢٥٠٠٠٠
٥٠٠٠٠٠
٧٥٠٠٠٠
١٠٠٠٠٠٠
١٢٥٠٠٠٠
١٥٠٠٠٠٠
اﻟﻤﻨﺴﻮب
ﺔﻌﺴﻟا ﻰﻨﺤﻨﻣ
**
**
ﺑﻴﻨﻰ ﻣﻨﺴﻮب أى ﻋﻨﺪ اﻟﻤﺨﺰﻧﺔ اﻟﻤﻴﺎﻩ آﻤﻴﺔ ﻟﺘﺤﺪﻳﺪ اﻟﺴﻌﺔ ﻣﻨﺤﻨﻰ ﻳﺴﺘﺨﺪم
ﺑﻴﻨﻰ ﻣﻨﺴﻮب أى ﻋﻨﺪ اﻟﻤﺨﺰﻧﺔ اﻟﻤﻴﺎﻩ آﻤﻴﺔ ﻟﺘﺤﺪﻳﺪ اﻟﺴﻌﺔ ﻣﻨﺤﻨﻰ ﻳﺴﺘﺨﺪم
ﻣﻨﺴﻮب ﻋﻨﺪ اﻟﻤﺨﺰﻧﺔ اﻟﻤﻴﺎﻩ آﻤﻴﺔ ﻳﻮﺿﺢ اﻷﺳﻮد اﻟﺴﻬﻢ اﻟﻤﺜﺎل ﺳﺒﻴﻞ ﻋﻠﻰ
ﻣﻨﺴﻮب ﻋﻨﺪ اﻟﻤﺨﺰﻧﺔ اﻟﻤﻴﺎﻩ آﻤﻴﺔ ﻳﻮﺿﺢ اﻷﺳﻮد اﻟﺴﻬﻢ اﻟﻤﺜﺎل ﺳﺒﻴﻞ ﻋﻠﻰ
٦٣
٦٣
م
م
67. ٦٧
٦٧
Prof. Dr. Eng. Said Elmaghraby
Prof. Dr. Eng. Said Elmaghraby
E_Learning courses in Engineering Surveying
E_Learning courses in Engineering Surveying
اﻟﻜﻨﺘﻮر ﺧﻂ رﻗﻢ
ﻣﺮﺑﻊ ﻣﺘﺮ اﻟﻤﺴﺎﺣﺔ
120
4600
125
7100
130
8900
135
11200
140
12300
145
14100
ﻣﺜﺎل
ﻣﺜﺎل
٧
٧
:
:
ﺧﻄﻮط داﺧﻞ ﺧﺮﻳﻄﺔ ﻋﻠﻰ اﻟﻤﺴﺎﺣﺎت ﻗﻴﺴﺖ
ﺧﻄﻮط داﺧﻞ ﺧﺮﻳﻄﺔ ﻋﻠﻰ اﻟﻤﺴﺎﺣﺎت ﻗﻴﺴﺖ
اﻟﻜﻨﺘﻮر
اﻟﻜﻨﺘﻮر
ﻟﻤﻮﻗﻊ
ﻟﻤﻮﻗﻊ
ﻓﻜﺎﻧﺖ اﻟﺨﺰاﻧﺎت أﺣﺪ ﻹﻧﺸﺎء ﻣﻘﺘﺮح
ﻓﻜﺎﻧﺖ اﻟﺨﺰاﻧﺎت أﺣﺪ ﻹﻧﺸﺎء ﻣﻘﺘﺮح
:
:
ﻣﻨﺴﻮب ﻋﻠﻰ ﻳﻘﻊ اﻟﺨﺰان ﻗﺎع أن ﻋﻠﻤﺖ ﻓﺈذا
ﻣﻨﺴﻮب ﻋﻠﻰ ﻳﻘﻊ اﻟﺨﺰان ﻗﺎع أن ﻋﻠﻤﺖ ﻓﺈذا
١٢٠
١٢٠
ﺑﺎﻟﺨﺰان اﻟﻤﻴﺎﻩ وﺳﻄﺢ م
ﺑﺎﻟﺨﺰان اﻟﻤﻴﺎﻩ وﺳﻄﺢ م
ﻣﻨﺴﻮب ﻋﻠﻰ
ﻣﻨﺴﻮب ﻋﻠﻰ
١٤٢
١٤٢
ﻓﺎﺣﺴﺐ م
ﻓﺎﺣﺴﺐ م
:
:
ا
ا
(
(
إرﺳﻢ
إرﺳﻢ
اﻟﺨﺰان ﺳﻌﺔ ﻣﻨﺤﻨﻰ
اﻟﺨﺰان ﺳﻌﺔ ﻣﻨﺤﻨﻰ
ب
ب
(
(
اﻟﺨﺰان داﺧﻞ اﻟﻤﻴﺎﻩ ﺣﺠﻢ
اﻟﺨﺰان داﺧﻞ اﻟﻤﻴﺎﻩ ﺣﺠﻢ
ﺟـ
ﺟـ
(
(
ﺳﻌﺔ ﻋﻨﺪ ﺑﺎﻟﺨﺰان اﻟﻤﻴﺎﻩ ﺳﻄﺢ ﻣﻨﺴﻮب
ﺳﻌﺔ ﻋﻨﺪ ﺑﺎﻟﺨﺰان اﻟﻤﻴﺎﻩ ﺳﻄﺢ ﻣﻨﺴﻮب
١٠٠٠٠
١٠٠٠٠
ﻣﻜﻌﺐ ﻣﺘﺮ
ﻣﻜﻌﺐ ﻣﺘﺮ
68. ٦٨
٦٨
Prof. Dr. Eng. Said Elmaghraby
Prof. Dr. Eng. Said Elmaghraby
E_Learning courses in Engineering Surveying
E_Learning courses in Engineering Surveying
اﻟﻜﻨﺘﻮر ﺧﻂ رﻗﻢ
اﻟﻤﺴﺎﺣﺔ
آﻨﺘﻮر ﺧﻄﻲ آﻞ ﺑﻴﻦ اﻟﻤﻴﺎﻩ ﺣﺠﻢ ﻣﻌﺎدﻟﺔ
اﻟﺠﺰﺋﻰ اﻟﺤﺠﻢ
اﻟﻜﻠﻰ اﻟﺤﺠﻢ
120
4600
(٧١٠٠+٤٦٠٠)٥ * ٠٫٥= ح
29250
29250
125
7100
(٨٩٠٠+٧١٠٠)٥ * ٠٫٥= ح
40000
69250
130
8900
(١١٢٠٠+٨٩٠٠)٥ * ٠٫٥ = ح
50250
119500
135
11200
(١٢٣٠٠+١١٢٠٠)٥ * ٠٫٥ = ح
58750
178250
140
12300
(١٤١٠٠+١٢٣٠٠)٥ * ٠٫٥ = ح
66000
244250
145
14100
اﻟﻜﻠﻰ اﻟﺤﺠﻢ
244250
:
:٧
٧ ﻣﺜﺎل ﺣﻞ
ﻣﺜﺎل ﺣﻞ