I really appreciate his teaching; he brings exceptional clarity to every concept. His concise and to-the-point method is truly effective. I believe he deserves more recognition
Homework Solution: Question number 4 Given: Nw=30, Tw=3 minutes Na=160, Ta=4 minutes Stationary time(t) = 15 minutes. Length of the route from X to Y = 2.5Km To be found: Number of vehicles observed by a stationary observer (N) Solution: qw=Nw/Tw = 600 veh/hr (Flow along the traffic stream) qa=Na/Ta = 2400 veh/hr (Flow against the traffic stream) Average speed of moving observer a) along the stream = (Vo)w=Distance/Time = 2.5×(3/60) = 50 km/hr b) against the stream = (Vo)a = Distance/Time = 2.5×(4/60)=37.5 kmph 2400=K(Vs+37.5) (Flow against the traffic stream) 600=K(Vs-50) (Flow along the traffic stream) By solving the above equation K=20.57 Vs=79.17 Traffic stream flow q=KVs=1628.5 veh/hr q=N/t Therefore N=408
In Floating car method the observer vehicle first travels along the direction of the stream and then against the direction of the stream with velocity (v0) with the help of this method observer can calculate stream velocity(vs= avg. vel of the vehicle ) flow of the stream, the flow of stream along and against he direction of traffic. it is also used to calculate the no. of vehicles overtaking the observer veh or no. of vehicles which the observer vehicle overtook we can also find the density of stream (k) Eqns use in moving car method 1. q(along )= k(vs-v0) 2. q(along)=N1-N2/tw 3. tw= L/vo we can use the above 3 equations when the test vehicle moves in the direction of the stream When the observer vehicle moves against the stream following result can be used 1.q(against)= k(vs+v0) 2. q(against)= N/tw 3. tw=L/vo To calculate the flow of stream (q=kvs) when the observer is stationary then also we can use the above equation N1 = No. of vehicles overtaking test vehicle N2 = No. of vehicle overtaken by test vehicle tw= time of observation L= Distance N= no of the vehicle crossing the observed vehicle in the opposite lane The velocity of observer can be different in moving with the traffic and against the traffic.
Homework Solution: Question number 3 Given: qw=50 veh/hr qa=200 veh/hr (Vo)w = 20 kmph (Vo)a = 30 kmph To be found: K veh/Km Solution: 200=K(Vs+30) ( Flow against the traffic stream) 50=K(Vs-20) (Flow along the traffic stream) By solving the above two equation K = 3 veh/km
Homework Numerical Answer 1. Difference in No. of vehicle overtaking and overtaken by test vehicle =40 2. No. Vehicle Counted By Adam = 40 3. Density of Traffic = 3 veh/km 4. No. of vehicle stationary observer count in 15 minutes= 407.14=408
Homework solution: Question number 2 Given: qw = 200 veh/hr qa = 1200 veh/hr (Vo)w = 30 kmph (Vo)a = 20 kmph Stationary period (t) : 3 minutes To be found: Number of vehicles overtaking Adam's car when the car was stationary (N) Solution: 1200=K(Vs+20) (Flow against the traffic stream) 200=K(Vs-30) (Flow along the traffic stream) By solving the above two equations. K=20 Vs=40 Traffic flow q = KVs = 800 veg/hr But q=N/t Therefore N=40 vehicles. When Adam's car was stationary 40 vehicles overtook him.
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Homework Numericals:- 1. Stream Flow = 800=K*Vs. qa=1200=K(Vs+20). Solving the above two equations, we get K = 20, & Vs= 40. For moving along the flow, tw=12/60=0.2 hour. qw=nw/0.2=K(Vs-30). Putting values of K & V, we get the difference (nw)=40. 2. qw=200=K(Vs-30). qa=1200=K(Vs+20). Solving the above two equations, we get K = 20, & Vs= 40. Stream Flow=20*40=800. No. of vehicles counted by Adam = 800*(3/60)=40. 3. qw=50=K(Vs-20), qa=200=k(Vs+30), Solving the above two equations, we get density of traffic stream (K) = 3 vehicles/km. 4. For moving with the flow, Vo=50. qw=600=K(Vs-50) For moving against the flow, Vo=37.5. qa=2400=K(Vs-37.5) Solving the above two equations, we get K = 20.57, & Vs= 79.12. Stream Flow = 20.57*79.12 = 1627.5. No. of vehicles observed by stationary vehicle in 15 mins = 1627.5*(15/60)=406.875 =407.
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It is the first time I have understood this Floating Car Method completely well. Requesting you to please take a session on Rotary Design as well.
I really appreciate his teaching; he brings exceptional clarity to every concept. His concise and to-the-point method is truly effective. I believe he deserves more recognition
4:51 Traffic flow , Density and Stream Velocity
10:57 Moving with traffic
27:02 Moving against traffic
33:59 Numericals
Homework Solution: Question number 4
Given:
Nw=30, Tw=3 minutes
Na=160, Ta=4 minutes
Stationary time(t) = 15 minutes.
Length of the route from X to Y = 2.5Km
To be found:
Number of vehicles observed by a stationary observer (N)
Solution:
qw=Nw/Tw = 600 veh/hr (Flow along the traffic stream)
qa=Na/Ta = 2400 veh/hr (Flow against the traffic stream)
Average speed of moving observer
a) along the stream = (Vo)w=Distance/Time
= 2.5×(3/60) = 50 km/hr
b) against the stream = (Vo)a = Distance/Time
= 2.5×(4/60)=37.5 kmph
2400=K(Vs+37.5) (Flow against the traffic stream)
600=K(Vs-50) (Flow along the traffic stream)
By solving the above equation
K=20.57 Vs=79.17
Traffic stream flow q=KVs=1628.5 veh/hr
q=N/t
Therefore N=408
Correct Solution
Sir ,, aap mst samjhate ho sach me great full
I understood this method for the first time
Excellent sir
GOOD LECTURE SIR ,,,,,,,,TQQQQQQQQQ
Thanks a lot prakash
Welcome. Keep attending the lectures.
Thank you sir 😊
In Floating car method the observer vehicle first travels along the direction of the stream and then against the direction of the stream with velocity (v0) with the help of this method observer can calculate stream velocity(vs= avg. vel of the vehicle ) flow of the stream, the flow of stream along and against he direction of traffic. it is also used to calculate the no. of vehicles overtaking the observer veh or no. of vehicles which the observer vehicle overtook we can also find the density of stream (k)
Eqns use in moving car method
1. q(along )= k(vs-v0)
2. q(along)=N1-N2/tw
3. tw= L/vo
we can use the above 3 equations when the test vehicle moves in the direction of the stream
When the observer vehicle moves against the stream following result can be used
1.q(against)= k(vs+v0)
2. q(against)= N/tw
3. tw=L/vo
To calculate the flow of stream (q=kvs)
when the observer is stationary then also we can use the above equation
N1 = No. of vehicles overtaking test vehicle
N2 = No. of vehicle overtaken by test vehicle
tw= time of observation
L= Distance
N= no of the vehicle crossing the observed vehicle in the opposite lane
The velocity of observer can be different in moving with the traffic and against the traffic.
Very nicely compiled
Thank You Sir 🙏🙏
Thank you so much. It was very well explained.
Sir your lecture very excellent.plaese put a video about rotary design
Sure and thanks for such compliment
Will soon come up with this topic
Homework Solution: Question number 3
Given:
qw=50 veh/hr
qa=200 veh/hr
(Vo)w = 20 kmph
(Vo)a = 30 kmph
To be found:
K veh/Km
Solution:
200=K(Vs+30) ( Flow against the traffic stream)
50=K(Vs-20) (Flow along the traffic stream)
By solving the above two equation
K = 3 veh/km
Correct Solution
Homework Numerical Answer
1. Difference in No. of vehicle overtaking and overtaken by test vehicle =40
2. No. Vehicle Counted By Adam = 40
3. Density of Traffic = 3 veh/km
4. No. of vehicle stationary observer count in 15 minutes= 407.14=408
Correct Answer . Bravo !!!!
Homework solution: Question number 2
Given:
qw = 200 veh/hr
qa = 1200 veh/hr
(Vo)w = 30 kmph
(Vo)a = 20 kmph
Stationary period (t) : 3 minutes
To be found:
Number of vehicles overtaking Adam's car when the car was stationary (N)
Solution:
1200=K(Vs+20) (Flow against the traffic stream)
200=K(Vs-30) (Flow along the traffic stream)
By solving the above two equations.
K=20 Vs=40
Traffic flow q = KVs = 800 veg/hr
But q=N/t
Therefore N=40 vehicles.
When Adam's car was stationary 40 vehicles overtook him.
Correct Solution
nice work!
Thank you! Cheers!
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Homework Numericals:-
1. Stream Flow = 800=K*Vs.
qa=1200=K(Vs+20).
Solving the above two equations, we get K = 20, & Vs= 40.
For moving along the flow, tw=12/60=0.2 hour.
qw=nw/0.2=K(Vs-30).
Putting values of K & V, we get the difference (nw)=40.
2. qw=200=K(Vs-30).
qa=1200=K(Vs+20).
Solving the above two equations, we get K = 20, & Vs= 40.
Stream Flow=20*40=800.
No. of vehicles counted by Adam = 800*(3/60)=40.
3. qw=50=K(Vs-20),
qa=200=k(Vs+30),
Solving the above two equations, we get density of traffic stream (K) = 3 vehicles/km.
4. For moving with the flow, Vo=50.
qw=600=K(Vs-50)
For moving against the flow, Vo=37.5.
qa=2400=K(Vs-37.5)
Solving the above two equations, we get K = 20.57, & Vs= 79.12.
Stream Flow = 20.57*79.12 = 1627.5.
No. of vehicles observed by stationary vehicle in 15 mins = 1627.5*(15/60)=406.875 =407.
Correct Answer . Bravo !!!!!
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Thank you for this wonderful session sir. Please take more such sessions.
Welcome . Keep attending
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