This document summarizes key aspects of railway engineering related to curves and banking. It discusses vertical and horizontal curves, including formulas for calculating curve radius, banking, and elevation changes. It also covers superelevation, the relationship between curve radius and speed, and formulas for calculating centrifugal force on circular horizontal curves.

1. Railway Engineering
Chapter (2)
Eng: Mohamed Ismail Kotb

2. ‫اﻟرﺣﯾم‬ ‫اﻟرﺣﻣن‬ ‫ﷲ‬ ‫ﺑﺳم‬
( ‫اﻟﻣﻧﺣﻧﯾﺎت‬ ) ‫اﻟﺛﺎﻧﻰ‬ ‫اﻟﺑﺎب‬
‫إﻟﻰ‬ ‫اﻟﻣﻧﺣﻧﯾﺎت‬ ‫ﺗﻧﻘﺳم‬:
1(‫اﻟرأﺳﯾﺔ‬ ‫اﻟﻣﻧﺣﻧﯾﺎت‬2(‫اﻷﻓﻘﯾﺔ‬ ‫اﻟﻣﻧﺣﻧﯾﺎت‬
‫اﻟرأﺳﻰ‬ ‫اﻻﺳﺗدارة‬ ‫ﻟﻣﻧﺣﻧﻰ‬ ‫ﺑﮫ‬ ‫اﻟﺗﺧطﯾط‬ ‫ﯾﻣﻛن‬ ‫ﻗطر‬ ‫ﻧﺻف‬ ‫أﻗل‬
‫اﻟﻣرﻛزﯾﺔ‬ ‫اﻟطﺎرد‬ ‫اﻟﻘوة‬FC
= = ∝ ∝ = 2%
‫ﺑﺎﻟﻤﺘﺮ‬ ‫أﺳﻲ‬‫ﺮ‬‫اﻟ‬ ‫اﻻﺳﺘﺪارة‬ ‫ﻣﻨﺤﻨﻰ‬ ‫ﻗﻄﺮ‬ ‫ﻧﺼﻒ‬:Rv
: V‫اﻟﺴﺮﻋﺔ‬(km/h)
‫اﻟﺳﻠس‬ ‫اﻟﺳﯾر‬ ‫ﻟﺗﺣﻘﯾق‬ ‫ﻟﻠﻣﻧﺣﻧﻰ‬ ‫ﻗطر‬ ‫ﻧﺻف‬ ‫أﻧﺳب‬Rv =0.4 V2
‫ﯾﻘل‬ ‫أﻻ‬ ‫ﯾﺟب‬ ‫و‬Rv‫ﻋن‬)(‫أو‬2000‫أﻛﺑر‬ ‫أﯾﮭﻣﺎ‬ ‫ﻣﺗر‬.
≥‫ﻛﺑﯾرة‬ ‫ﺑدرﺟﺔ‬ ‫اﻟﺳﻛﺔ‬ ‫ﻋﻠﻰ‬ ‫اﻟﺿﻐط‬ ‫ﯾزﯾد‬ ‫ﻻ‬ ‫ﺣﺗﻰ‬.
2000 m≥Rv‫ﻣروﻧﺗﮭﺎ‬ ‫ﺣدود‬ ‫ﻋن‬ ‫اﻟﻔﺧذ‬ ‫ﻛﻣرة‬ ‫ﻓﻰ‬ ‫اﻹﺿﺎﻓﯾﺔ‬ ‫اﻟﺿﻐوط‬ ‫ﺗزﯾد‬ ‫ﻻ‬ ‫ﺣﺗﻰ‬ ‫ﻣﺗر‬.
‫اﻟرأﺳﯾﺔ‬ ‫اﻟﻣﻧﺣﻧﯾﺎت‬
FC
w
FC
w
w

3. ‫اﻟرأﺳﯾﺔ‬ ‫اﻟﻣﻧﺣﻧﯾﺎت‬ ‫ﻗواﻧﯾن‬
1(‫ﻷﻋﻠﻰ‬ ‫رأﺳﻰ‬ ‫ﻣﻧﺣﻧﻰ‬
= 0.4
∆ =
| − |
1000
.
= ∗ ∆
‫ﻧﻘﻄﺔ‬ ‫ﻣﻨﺴﻮب‬a:
= ˋ −
2
×
| |
1000
‫ﻧﻘﻄﺔ‬ ‫ﻣﻨﺴﻮب‬b:
= ˋ −
2
×
| |
1000
‫اﻟﻣﻧﺣﻧﻰ‬ ‫ﻣﻧﺗﺻف‬ ‫ﻣﻧﺳوب‬c:
=
2
= ˋ −
cˋ
g 1 %o
g 2 %o

4. 2(‫ﻷﺳﻔل‬ ‫رأﺳﻰ‬ ‫ﻣﻧﺣﻧﻰ‬
= 0.4
∆ =
| − |
1000
.
= ∗ ∆
‫ﻧﻘﻄﺔ‬ ‫ﻣﻨﺴﻮب‬a:
= ˋ +
2
×
| |
1000
‫ﻧﻘﻄﺔ‬ ‫ﻣﻨﺴﻮب‬b:
= ˋ +
2
×
| |
1000
‫اﻟﻣﻧﺣﻧﻰ‬ ‫ﻣﻧﺗﺻف‬ ‫ﻣﻧﺳوب‬c:
=
2
= ˋ +
LV
RV
x Y
y
a b
c
-6‰ +5‰
Dg
bˋ
g 1 %o
g 2 %o

5. ‫اﻟﺑطن‬ ‫ﻋن‬ ‫اﻟظﮭر‬ ‫ارﺗﻔﺎع‬ ‫ﻣﻌﺎدﻟﺔ‬
‫اﻟﺒﻄﻦ‬ ‫ﻋﻦ‬ ‫اﻟﻈﻬﺮ‬ ‫ﺗﻔﺎع‬‫ر‬‫ا‬ ‫ﻗﻴﻤﺔ‬ ‫اﺳﺘﻨﺒﺎط‬ ‫ﻳﻤﻜﻦ‬E)‫ﻣﻢ‬(‫اﻟﻘﻄﺮ‬ ‫ﻧﺼﻒ‬ ‫ﺑﺪﻻﻟﺔ‬R)‫ﻣﺘﺮ‬(‫واﻟﺴﺮﻋﺔ‬V)‫ﻛﻢ‬/‫اﻟﺴﺎﻋﺔ‬(‫اﻟﺘﺎﻟﻲ‬ ‫اﻟﻮﺟﻪ‬ ‫ﻋﻠﻰ‬.
FC× Cos = W×sin + DFC
sin =
G
E
, Cos 1
FC = W × + DFC
×
= w × + m × a
×
= m × g × + m × a
a = −
×
‫اﻷﻓﻘﯾﺔ‬ ‫اﻟداﺋرﯾﺔ‬ ‫اﻟﻣﻧﺣﻧﯾﺎت‬
ν ‫م‬/‫ث‬
‫ق‬‫ر‬
=
‫ﻛـ‬.‫س‬2
‫ر‬
R
G
E
h



FC
FC Cos
FC sinw
w Cos
wsin
‫اﻟﻌﺮﺑﺎت‬ ‫ﺛﻘﻞ‬ ‫ﻣﺮﻛﺰ‬
)+
)-(
‫ﻋﺠﻠﺔ‬ ‫اﺗﺠﺎه‬
‫اﻟﻄﺮد‬
‫اﻟﻤﺮﻛﺰﯾﺔ‬
FC =
R
Vm 2


6. ‫ﺣﻴﺚ‬a=‫ﻛﺰﻳﺔ‬‫اﻟﻤﺮ‬ ‫اﻟﻄﺮد‬ ‫ﻋﺠﻠﺔ‬)‫داﺧﻠﻴﺔ‬ ‫أو‬ ‫ﺧﺎرﺟﻴﺔ‬(‫م‬/‫ث‬2
‫داﺋﺮي‬ ‫ﺧﻂ‬ ‫ﻋﻠﻰ‬ ‫ﺑﺴﺮﻋﺔ‬ ‫ﺗﺘﺤﺮك‬ ‫ﻣﺎ‬ ‫ﻛﺘﻠﺔ‬‫ﻋﻠﻰ‬ ‫ﺗﺆﺛﺮ‬ ‫اﻟﺘﻲ‬ ‫اﻟﻘﻮة‬ ‫ﻣﻘﺪار‬ ‫ﻋﻠﻰ‬ ‫ﺗﺪل‬ ‫وﻫﻰ‬ ،
‫ﻗﻄﺮ‬ ‫ﻧﺼﻒ‬ ‫ذو‬R.
‫وﻟﻜﻦ‬
6.3
)/( hrkmV
= ν(m/s)،G=1500‫اﻟﻌﺎدي‬ ‫اﻻﺗﺴﺎع‬ ‫ﺣﺎﻟﺔ‬ ‫ﻓﻲ‬ ‫ﻣﻢ‬
a =
12.96
−
9.81
1500
a =
12.96
−
153
/ (1)
OR
∴ =
11.8
− 153 (2)
‫ﻟﻠﻤﻌﺎدﻟﺔ‬ ً‫ﺎ‬‫وﺗﺒﻌ‬)1(‫ﻛﺎﻧﺖ‬‫إذا‬a‫اﻟﺪاﺧﻞ‬ ‫إﻟﻰ‬ ‫ﺗﺆﺛﺮ‬ ‫اﻟﻌﺠﻠﺔ‬ ‫أن‬ ‫ﻋﻠﻰ‬ ‫ﺗﺪل‬ ‫ﻓﺈﻧﻬﺎ‬ ‫ﺳﺎﻟﺒﺔ‬ ‫ﻛﺎﻧﺖ‬‫وإذا‬ ‫اﻟﺨﺎرج‬ ‫إﻟﻰ‬ ‫ﺗﺆﺛﺮ‬ ‫اﻟﻌﺠﻠﺔ‬ ‫أن‬ ‫ﻋﻠﻰ‬ ‫ﺗﺪل‬ ‫ﻓﺈﻧﻬﺎ‬ ‫ﻣﻮﺟﺒﺔ‬.
=
. ×
‫ﺣﺎﻻت‬ ‫ﻓﻰ‬ ‫ﺗﺳﺗﺧدم‬:
، ‫ﺿواﺣﻰ‬ ، ‫ﻣﺗرو‬ ‫ﺧط‬
‫ﺳرﻋﺔ‬‫ﺛﺎﺑﺗﺔ‬.
=
×
=
. × ∑ ×
× ∑
‫ﻗطﺎرات‬ ‫ﺳﯾر‬ ‫ﺣﺎل‬ ‫ﻓﻰ‬ ‫ﺗﺳﺗﺧدم‬
‫أ‬ ‫و‬ ‫ﻣﺧﺗﻠﻔﺔ‬ ‫ﺑﺳرﻋﺎت‬‫ﻣﺧﺗﻠﻔﺔ‬ ‫وزان‬.
=
‫اﻟﺣﺻول‬ ‫ﺣﺎﻟﺔ‬ ‫ﻓﻰ‬ ‫ﺗﺳﺗﺧدم‬
‫ﯾﻣﻛن‬ ‫ﺳرﻋﺔ‬ ‫أﻋﻠﻰ‬ ‫ﻋﻠﻰ‬
‫ﺑﮭﺎ‬ ‫ﯾﺳﯾر‬ ‫أن‬ ‫ﻟﻠﻘطﺎر‬.
‫ﺣﺎﻟﺔ‬ ‫وﻓﻰ‬Emax < E
=
. ×
−
‫ﺣﺎﻟﺔ‬ ‫ﻓﻰ‬ ‫ﺗﺳﺗﺧدم‬Emin > E

7. ‫ﺳﺮﻋﺘﻪ‬ ‫ﻛﺎﻧﺖ‬‫إذا‬ ‫اﻻﻧﻘﻼب‬ ‫وﺷﻚ‬ ‫ﻋﻠﻰ‬ ‫ﻳﻜﻮن‬ ‫أﻧﻪ‬ ‫ﻓﺄﺛﺒﺖ‬ ، ‫ﺑﻄﻦ‬ ‫ﻋﻦ‬ ‫ﻇﻬﺮ‬ ‫ﺗﻔﺎع‬‫ر‬‫ا‬ ‫ﺑﺪون‬ ‫ﻣﻨﻔﺬ‬ ‫أﻓﻘﻰ‬ ‫ﻣﻨﺤﻨﻰ‬ ‫ﻋﻠﻰ‬ ‫ﻗﻄﺎر‬ ‫ﺳﺎر‬ ‫إذا‬:
≅ 8
∙
ℎ
‫ﺣﻴﺚ‬ G :‫اﻟﺴﻜﺔ‬ ‫اﺗﺴﺎع‬)‫اﻟﻘﻀﺒﺎن‬ ‫ﻣﺤﺎور‬ ‫ﺑﻴﻦ‬ ‫اﻟﻤﺴﺎﻓﺔ‬-‫ﻣﺘﺮ‬(
R :‫اﻟﻤﻨﺤﻨﻰ‬ ‫ﻗﻄﺮ‬ ‫ﻧﺼﻒ‬)‫ﻣﺘﺮ‬(
h :‫اﻟﻘﻀﺒﺎن‬ ‫ﺳﻄﺢ‬ ‫ﻋﻦ‬ ‫اﻟﺜﻘﻞ‬ ‫ﻛﺰ‬‫ﻣﺮ‬ ‫ﺗﻔﺎع‬‫ر‬‫إ‬)‫ﻣﺘﺮ‬(
‫اﻟﺣ‬‫ــــ‬‫ل‬
‫اﻟﻮزن‬ ‫ﻧﺘﻴﺠﺔ‬ ‫اﻟﻔﻌﻞ‬ ‫رد‬Re
2
w
=
‫ﻛﺰ‬‫اﻟﻤﺮ‬ ‫اﻟﻄﺎردة‬ ‫اﻟﻘﻮة‬ ‫ﻧﺘﻴﺠﺔ‬ ‫اﻟﻔﻌﻞ‬ ‫رد‬‫ﻳﺔ‬DRe
×
=
‫ـﺪاﺧﻠﻲ‬‫ـ‬‫ﻟ‬‫ا‬ ‫ـﻴﺐ‬‫ـ‬‫ﻀ‬‫اﻟﻘ‬ ‫ـﻰ‬‫ـ‬‫ﻠ‬‫ﻋ‬ ‫ـﻞ‬‫ـ‬‫ﻌ‬‫اﻟﻔ‬ ‫رد‬ ‫ـﺒﺢ‬‫ـ‬‫ﺻ‬‫أ‬ ‫إذا‬ ‫ـﻼب‬‫ـ‬‫ﻘ‬‫اﻻﻧ‬ ‫ـﻚ‬‫ـ‬‫ﺷ‬‫و‬ ‫ـﻰ‬‫ـ‬‫ﻠ‬‫ﻋ‬ ‫ـﺎر‬‫ـ‬‫ﻄ‬‫اﻟﻘ‬ ‫ـﺒﺢ‬‫ـ‬‫ﺼ‬‫ﻳ‬
‫ﺻﻔﺮ‬ ‫ﻳﺴﺎوي‬
Re = DRe
×
=
× ×
. × . × ×
=
. × ×
=
≅
×
G
h
FE
w
ReRe
DReDRe

8. ‫ارﺗﻔﺎع‬ ‫ﻣﻧﺣدر‬‫اﻻﻧﺗﻘﺎل‬ ‫وﻣﻧﺣﻧﻰ‬ ‫اﻟﺑطن‬ ‫ﻋن‬ ‫اﻟظﮭر‬
 =
 =
×
= 150
 =
×
/ℎ
 L =
1000
10 )()( ff EV 
 Sh =
R
LS
24
2
‫اﻟ‬‫ﻣﻧﺣ‬‫اﻟﻣرﻛب‬ ‫ﻧﻰ‬
 IF (R2 > R1 )
 =
×
= 150
 =
×
/ℎ ( v is the future speed)
 =
×

9. L =
10 ( ) ( )
1000 L =
10 ( ) ( )
1000
L = L − L
Sh =
1
2
24 1
Sh =
2
2
24 2 Sh = Sh − Sh
‫إ‬‫اﻟظﮭر‬ ‫رﺗﻔﺎع‬‫اﻟﺣﺎﻟﻲ‬ ‫اﻟﺑطن‬ ‫ﻋن‬
=
8 ×
=
×
(V is present speed )
‫اﻟﻌﻛﺳﻲ‬ ‫ﻟﻠﻣﻧﺣﻧﻰ‬ ‫اﻟﺑطن‬ ‫ﻋن‬ ‫اﻟظﮭر‬ ‫ارﺗﻔﺎع‬ ‫ﻣﻧﺣدر‬ ‫ﺗﻧﻔﯾذ‬ ‫طرق‬:
1(‫ﻣﺳﺎﻓﺔ‬ ‫ﺗرك‬ ‫ﻣﻊ‬ ‫اﻟﺧﺎرﺟﻲ‬ ‫اﻟﻘﺿﯾب‬ ‫رﻓﻊ‬‫اﻟﺛﺎﻧﻲ‬ ‫اﻟﻣﻧﺣدر‬ ‫وﺑداﯾﺔ‬ ‫اﻷول‬ ‫اﻟﻣﻧﺣدر‬ ‫ﻧﮭﺎﯾﺔ‬ ‫ﺑﯾن‬ ‫ﻣﺳﺗﻘﯾﻣﺔ‬.
.ً‫ا‬‫ﺟد‬ ‫ﻗﻠﯾل‬ ‫زﻣن‬ ‫ﻓﻲ‬ ‫اﻟﻣﺳﺗﻘﯾﻣﺔ‬ ‫اﻟﻣﺳﺎﻓﺔ‬ ‫وﻧﮭﺎﯾﺔ‬ ‫ﺑداﯾﺔ‬ ‫ﻋﻧد‬ ‫ﺻدﻣﺔ‬ ‫ﺣدوث‬ : ‫ﻋﯾﺑﮭﺎ‬
2(‫اﻟﺛﺎﻧﻲ‬ ‫اﻟﻣﻧﺣﻧﻰ‬ ‫ﻓﻲ‬ ‫اﻟداﺧﻠﻲ‬ ‫اﻟﻘﺿﯾب‬ ‫وﺧﻔض‬ ‫اﻷول‬ ‫اﻟﻣﻧﺣﻧﻰ‬ ‫ﻓﻲ‬ ‫اﻟﺧﺎرﺟﻲ‬ ‫اﻟﻘﺿﯾب‬ ‫رﻓﻊ‬.
: ‫ﻋﯾوﺑﮭﺎ‬
-‫اﻟﺻﻌب‬ ‫ﻣن‬ ‫ﻟﻛن‬ ‫اﻟﻘﺿﯾب‬ ‫ﻣﻧﺳوب‬ ‫رﻓﻊ‬ ‫اﻟﺳﮭل‬ ‫ﻣن‬‫ﻣﻧﺳوﺑﮫ‬ ‫ﺧﻔض‬ ً‫ا‬‫ﺟد‬.
-‫اﻟﻣﻧﺣﻧﻰ‬ ‫ﻓﻰ‬ ‫ﺗﻌطﻰ‬ ‫اﻟﻣﻘﺎرﻧﺔ‬ ‫ﻧﻘط‬ ‫ﻣﻧﺎﺳﯾب‬)2(‫إذ‬ ، ‫اﻟﺻﯾﺎﻧﺔ‬ ‫ﻟﻌﻣﺎل‬ ‫ﻣﻌﺗﺎد‬ ‫ھو‬ ‫ﻣﺎ‬ ‫ﺧﻼف‬ ‫ﻋﻠﻰ‬ ‫اﻟﺧﺎرﺟﻰ‬ ‫اﻟﻘﺿﯾب‬ ‫ﻋﻠﻰ‬
‫اﻟداﺧﻠﻰ‬ ‫اﻟﻘﺿﯾب‬ ‫ﻋﻠﻰ‬ ً‫ﺎ‬‫داﺋﻣ‬ ‫ﺗﻧﺳب‬ ‫أﻧﮭﺎ‬.
3(‫ﻓﻲ‬ ‫وﺿﻌﮭﻣﺎ‬ ‫ﯾﺗﺑدل‬ ‫ﺣﺗﻰ‬ ‫اﻟوﻗت‬ ‫ﻧﻔس‬ ‫ﻓﻲ‬ ‫اﻟداﺧﻠﻲ‬ ‫اﻟﻘﺿﯾب‬ ‫ﻣﻧﺳوب‬ ‫رﻓﻊ‬ ‫ﻣﻊ‬ ‫اﻟﺧﺎرﺟﻲ‬ ‫اﻟﻘﺿﯾب‬ ‫ﻣﻧﺳوب‬ ‫ﺗﺧﻔﯾض‬
‫اﻟﺛﺎﻧﻲ‬ ‫اﻟﻣﻧﺣﻧﻰ‬.
.ً‫ﺎ‬‫داﺋﻣ‬ ‫اﻟﻣﺗﺑﻌﺔ‬ ‫اﻟﺳﻠﯾﻣﺔ‬ ‫اﻟطرﯾﻘﺔ‬ ‫وھﻰ‬ ‫اﻟﺳﺎﺑﻘﺔ‬ ‫اﻟﻌﯾوب‬ ‫ﺗﺗﻼﻓﻰ‬ : ‫ﻣﻣﯾزاﺗﮭﺎ‬
‫اﻟﻌﻛﺳﻰ‬ ‫اﻟﻣﻧﺣﻧﻰ‬

10. ‫اﻟﻌﻛﺳﻰ‬ ‫اﻟﻣﻧﺣﻧﻰ‬ ‫ﺗﺻﻣﯾم‬
 IF (R2 > R1 )
 =
×
= 150
 =
×
/ℎ ( v is the future speed)
R1
R2
E1
LS
E2
LS
1:n
1:n
E1
LS1
E2
LS2
1:n 1:n
V
10

E1
LS
E2
LS
ym
yn
En
xn
‫ﻟﻠﻤﻨﺤﻨﯿﯿﻦ‬ ‫اﻷﻓﻘﻰ‬ ‫اﻟﻤﺴﻘﻂ‬
‫أ‬(‫ﻓﻰ‬ ‫ﻣﺴﺎﻓﺔ‬ ‫ﺗﺮك‬
‫ب‬(‫اﻟﻈﮭﺮ‬ ‫ﻋﻦ‬ ‫اﻟﺒﻄﻦ‬ ‫ﺗﺨﻔﯿﺾ‬‫ﻓﻰ‬
‫ﺟـ‬(‫اﻟﻌﻜﺴﻰ‬ ‫اﻟﻤﻨﺤﺪر‬
ER1
SR1 SR2
ER2

11.  =
×
L =
10 ( ) ( )
1000
L =
10 ( ) ( )
1000
= .5 (L + L )
≥. ( + ) < .5 ( + )
= .5 (L + L ) > .5 ( + )
L =
8 ( ) ( )
1000
L =
8 ( ) ( )
1000
L +L = 2X AB … … .1 = 2
1
………2 solve
<. ( + )
Reduce the speed
L +L = 2X AB
80 ( )
1000 1
+
80 ( )
1000 2
= 2X AB
ّ‫اﻟﻠﻲ‬‫اﻟذى‬‫واﺣد‬ ‫إطﺎر‬ ‫ﻓﻲ‬ ‫ﻣﺛﺑﺗﯾن‬ ‫ﻣﺣورﯾن‬ ‫ذات‬ ‫ﺑﺿﺎﻋﺔ‬ ‫ﻋرﺑﺔ‬ ‫ﻟﮭﺎ‬ ‫ﺗﺗﻌرض‬
‫ﺑﻣﻘدار‬ ‫اﻷﺧرﯾﺎت‬ ‫ﻋن‬ ‫ﺗﻧﺧﻔض‬ ‫أو‬ ‫اﻟﻌﺟﻼت‬ ‫إﺣدى‬ ‫ﺗرﺗﻔﻊ‬ ‫اﻟﻣﻧﺣدر‬ ‫ﻋﻠﻰ‬ ‫ﻣﺣورﯾن‬ ‫ﻣرور‬ ‫ﻧﺗﯾﺟﺔ‬ ، ‫اﻟﻣوﺿﺢ‬ ‫اﻟرﺳم‬ ‫ﻣن‬‫ﺔ‬‫زاوﯾ‬ ‫ﺗﺻﺑﺢ‬ ‫وﺑذﻟك‬
ّ‫اﻟﻠﻲ‬‫ﻛﺎﻵﺗﻲ‬:(rad) =
nG
d

IF AB (given)
E
G
d
1:n
d

12. ‫اﻟﺣدﯾدﯾﺔ‬ ‫ﻟﻠﺳﻛك‬ ‫اﻟدوﻟﻲ‬ ‫اﻻﺗﺣﺎد‬ ‫ﻟﻣواﺻﻔﺎت‬ ً‫ﺎ‬‫طﺑﻘ‬
‫اﻟﺧﺷن‬ ‫اﻟﺳﯾر‬ ‫ﺣﺎﻟﺔ‬ ‫ﻓﻲ‬
nmin = 8V
8V = 400 V = 50 km/hr
‫اﻟﺳﻠس‬ ‫اﻟﺳﯾر‬ ‫ﺣﺎﻟﺔ‬ ‫ﻓﻲ‬
nst = 10V
10V = 400 V = 40 km/hr
‫اﻟرأﺳﻰ‬ ‫اﻟﻣﻧﺣﻧﻰ‬ ‫ﻣﻊ‬ ‫اﻷﻓﻘﻰ‬ ‫اﻟﻣﻧﺣﻧﻰ‬
‫اﻟرأﺳﻰ‬ ‫اﻟﻣﻧﺣﻧﻰ‬ ‫أﻻﯾﺗداﺧل‬ ‫ﯾﺟب‬‫اﻻﻧﺗﻘﺎل‬ ‫ﻣﻧﺣﻧﻰ‬ ‫ﻣﻊ‬
 AB = .5X (LS + LV)
≥. ( + ) < .5 ( + )
= .5 (L + L ) > .5 ( + )
L =
8 ( ) ( )
1000
L =
L +L = 2X AB … … .1
<. ( + )
Reduce the speed
L +L = 2X AB
80 ( )
1000
+ ( )X θ = 2X AB
IF AB (given)

13. ‫ﺑطن‬ ‫ﻋن‬ ‫ظﮭر‬ ‫ارﺗﻔﺎع‬ ‫ﺑدون‬ ‫اﻟﻣﻧﺣﻧﯾﺎت‬
‫ﺣﺎﻟﺔ‬ ‫ﻓﻰ‬ ‫ﻟﻠﺧط‬ ‫اﻟﻘﺻوى‬ ‫واﻟﺳرﻋﺔ‬ ‫اﻷﻓﻘﯾﺔ‬ ‫اﻟداﺋرﯾﺔ‬ ‫اﻟﻣﻧﺣﻧﯾﺎت‬ ‫أﻗطﺎر‬ ‫أﻧﺻﺎف‬ ‫ﺑﯾن‬ ‫اﻟﻌﻼﻗﺔ‬‫اﺗﺻﺎل‬‫رة‬‫ﻣﺑﺎﺷ‬ ‫ﮭﺎ‬‫ﺑﺑﻌﺿ‬ ‫ﺎت‬‫اﻟﻣﻧﺣﻧﯾ‬ ‫ھذه‬
‫ﻣﻧﺣﻧﯾﺎت‬ ‫ﺑدون‬‫اﻧﺗﻘﺎل‬‫ﺗﻧﻔﯾذ‬ ‫أو‬‫ارﺗﻔﺎع‬‫ﺑطن‬ ‫ﻋن‬ ‫ظﮭر‬.
a =
R
V
96.12
2
-
153
E
‫ﺣﺎﻟﺔ‬ ‫وﻓﻲ‬E=‫ﺻﻔﺮ‬
a =
R
V
96.12
2
=
100096.12
2

 KV
Δa =
12960
2
KV D
‫أن‬ ‫ﺑﺎﻟﺘﺠﺮﺑﺔ‬ ‫وﺟﺪ‬ ‫وﻗﺪ‬Δa‫ﻋﻦ‬ ‫زادت‬ ‫إذا‬ ‫ﺧﻄﺮة‬ ‫ﺗﺼﺒﺢ‬0.7‫م‬/‫ث‬2
ΔK  2
7.012960
V

 2
9000
V
‫ﻣﻧﺣﻧﯾﺎت‬ ‫ﺑدون‬ ‫ﺑﺑﻌﺿﮭﺎ‬ ‫اﻟﻌﻛﺳﯾﺔ‬ ‫اﻷﻓﻘﯾﺔ‬ ‫اﻟداﺋرﯾﺔ‬ ‫اﻟﻣﻧﺣﻧﯾﺎت‬ ‫اﺗﺻﺎل‬ ‫ﻋﻧد‬ ‫ﺗواﻓرھﻣﺎ‬ ‫اﻟواﺟب‬ ‫اﻟﺷرطﺎن‬‫ھﻣﺎ‬ ‫اﻧﺗﻘﺎل‬:
1) ΔK  2
9000
V
2) ΔK  10
1(‫ﻣرﻛب‬ ‫و‬ ‫ﺑﺳﯾط‬ ‫ﻣﻧﺣﻧﻰ‬
 a =
R
V
96.12
2
( a≯.65 )‫اﻟﻣرﻛزﯾﺔ‬ ‫اﻟطرد‬ ‫ﻋﺟﻠﺔ‬ ‫ﺷرط‬
= .
2(‫ﻋﻛﺳﻰ‬ ‫ﻣﻧﺣﻧﻰ‬
 a =
R
V
96.12
2
a≯.65
= .

14.  Δa =
12960
2
KV D
)‫اﻟﻣرﻛزﯾﺔ‬ ‫اﻟطرد‬ ‫ﻋﺟﻠﺔ‬ ‫ﻓﻲ‬ ‫اﻟﺗﻐﯾر‬ ‫ﺷرط‬Δa≮.7‫م‬/‫ث‬2
(
ΔK  2
7.012960
V

 2
9000
V
‫ﺣﺎﻟﺔ‬ ‫وﻓﻰ‬R‫اﻟﻌﻜﺴﻰ‬ ‫اﻟﻤﻨﺤﻨﻰ‬ ‫ﻓﻰ‬ ‫ﻣﺘﺴﺎوﻳﺔ‬
ΔK = 2
9000
V
=
.
‫ﺣﺎﻛﻢ‬ ‫اﻧﺤﺪار‬ ‫وﺟﻮد‬ ‫ﺣﺎﻟﺔ‬ ‫وﻓﻰ‬
‫اﻟﻣﻧﺳوب‬ ‫ﻓرق‬=‫اﻷﻓﻘﯿﺔ‬ ‫اﻟﻤﺴﺎﻓﺎت‬+‫اﻟﻤﻨﺤﻨﯿﺔ‬ ‫اﻟﻤﺴﺎﻓﺎت‬
‫اﻟﺛﻼﺛﺔ‬ ‫اﻟﺷروط‬ ‫ﺗﺣﻘﯾق‬ ‫ﻋﻠﻰ‬ ‫ﯾﻌﻣل‬ ‫ﻗطر‬ ‫ﻧﺻف‬ ‫أﻗل‬ ‫ﻋﻠﻰ‬ ‫اﻟﺣﺻول‬ ‫ﯾﺗم‬ ‫وﺑذﻟك‬:
1(‫اﻟﻣرﻛزﯾﺔ‬ ‫اﻟطرد‬ ‫ﻋﺟﻠﺔ‬ ‫ﺷرط‬a≯.65
2(‫اﻟﻣرﻛزﯾﺔ‬ ‫اﻟطرد‬ ‫ﻋﺟﻠﺔ‬ ‫ﻓﻲ‬ ‫اﻟﺗﻐﯾر‬ ‫ﺷرط‬Δa≮.7‫م‬/‫ث‬2
3(‫اﻟﺣﺎﻛم‬ ‫اﻻﻧﺣدار‬ ‫ﺷرط‬