Transportation Practice Problems

[owillis28]

A horizontal curve is going to be constructed for a paved county road in northern Michigan. The

pavement is 20 ft wide with 4 ft shoulders. The legs of the proposed curve are perpendicular. A design

speed of 60 mph is desired. What is most nearly the minimum radius needed to provide a design

where there is no reliance on a lateral friction force?

Two reverse-horizontal curves have external angles Δ1 = 30° and Δ2 = 60°. The distance between

PI and PI is 1000 ft. The maximum allowable side friction factor is 0.06, and the maximum

super-elevation rate is 0.06. What is most nearly the maximum velocity (design speed) that can be

provided while allowing for a 3 sec tangent distance between curves?

The baseline of a ramp curves to the left at a radius of 3000 ft and must clear a pier corner by 50 ft.

The pier corner is located 27 ft to the left of the curve back tangent at sta 108 + 50. What is the

station of the PC?

A new highway with a design speed of 50 mph is to be connected to an existing highway by a pair of

reverse 1° curves. The new highway is parallel to the existing highway, offset 30 ft to the right. What

is the length of the reverse curve transition, from PC to PT?

A horizontal curve is to be constructed for a paved county road in northern Michigan. The pavement is

20 ft wide with 4 ft shoulders. The legs of the proposed curve are perpendicular. A design speed of 60

mph is desired. What is most nearly the minimum length of the curve?

A ramp is to tie into an existing road at elevation 1207.44 ft, sta 16 + 00 on a 3% downgrade. The

vertical curve must clear a brick arch sewer by 4.5 ft. The sewer is 125 ft back from the tie point, and

the top of the brick arch is at elevation 1204.69 ft. The grade leading into the vertical curve is + 6%.

What is the length of the vertical curve?

The design of a high-speed train curve transition spirals calls for a maximum rate of change in lateral

acceleration of 1 ft/sec3. A curve has a radius of 10,000 ft and is to be designed for a speed of 150

mph. What is the length of spiral transition required?

 

[tall]

A horizontal curve is going to be constructed for a paved county road in northern Michigan. Thepavement is 20 ft wide with 4 ft shoulders. The legs of the proposed curve are perpendicular. A design

speed of 60 mph is desired. What is most nearly the minimum radius needed to provide a design

where there is no reliance on a lateral friction force?

Two reverse-horizontal curves have external angles Δ1 = 30° and Δ2 = 60°. The distance between

PI and PI is 1000 ft. The maximum allowable side friction factor is 0.06, and the maximum

super-elevation rate is 0.06. What is most nearly the maximum velocity (design speed) that can be

provided while allowing for a 3 sec tangent distance between curves?

The baseline of a ramp curves to the left at a radius of 3000 ft and must clear a pier corner by 50 ft.

The pier corner is located 27 ft to the left of the curve back tangent at sta 108 + 50. What is the

station of the PC?

A new highway with a design speed of 50 mph is to be connected to an existing highway by a pair of

reverse 1° curves. The new highway is parallel to the existing highway, offset 30 ft to the right. What

is the length of the reverse curve transition, from PC to PT?

A horizontal curve is to be constructed for a paved county road in northern Michigan. The pavement is

20 ft wide with 4 ft shoulders. The legs of the proposed curve are perpendicular. A design speed of 60

mph is desired. What is most nearly the minimum length of the curve?

A ramp is to tie into an existing road at elevation 1207.44 ft, sta 16 + 00 on a 3% downgrade. The

vertical curve must clear a brick arch sewer by 4.5 ft. The sewer is 125 ft back from the tie point, and

the top of the brick arch is at elevation 1204.69 ft. The grade leading into the vertical curve is + 6%.

What is the length of the vertical curve?

The design of a high-speed train curve transition spirals calls for a maximum rate of change in lateral

acceleration of 1 ft/sec3. A curve has a radius of 10,000 ft and is to be designed for a speed of 150

mph. What is the length of spiral transition required?

Would you please post the solutions for the problems

 

[owillis28]

Would you please post the solutions for the problems

I screwed up on one of the solutions. Let me work on it and hopefully have the solutions posted by the end of the day tomorrow. Sorry for the delay.

Were any of them giving you problems?

Matt

 

[owillis28]

I screwed up on one of the solutions. Let me work on it and hopefully have the solutions posted by the end of the day tomorrow. Sorry for the delay.
Were any of them giving you problems?

Matt

Here are my solutions. Not sure if they are correct but most of them seemed to make sense.

I could really use some help on the second problem. I am completely stumped. I asked a couple of PE's in the office but they didn't have an answer.

 

[tall]

Here are my solutions. Not sure if they are correct but most of them seemed to make sense.
I could really use some help on the second problem. I am completely stumped. I asked a couple of PE's in the office but they didn't have an answer.

If you try to find the S/L the length between the two tangents & divide to the time 3 sec will give the speed.

Talll

 

[owillis28]

If you try to find the S/L the length between the two tangents & divide to the time 3 sec will give the speed.
Talll

 

[paul]

owillis28,

Do you know why the tangent distance for a 3 sec transition is calculated by "1.47*t*v"? Why is it not "t*v"?

Thanks!

 

[paul]

Sorry for the question, it is just 1mph=1.47 ft/s

 

[Undertaker]

Thanks owillis28

 

[passman]

Thanks owillis28

Is there anyway you could post the solutions again to these helpful problems?

Thnaks!

 

[owillis28]

Is there anyway you could post the solutions again to these helpful problems?
Thnaks!

[attachment=1201:67032.pdf

 

[riz]

on this quastion ""Two reverse-horizontal curves have external angles Δ1 = 30° and Δ2 = 60°. The distance between

PI and PI is 1000 ft. The maximum allowable side friction factor is 0.06, and the maximum

super-elevation rate is 0.06. What is most nearly the maximum velocity (design speed) that can be

provided while allowing for a 3 sec tangent distance between curves? ""

I don't know if you have any luck getting the answer.

Well here is my Clue

let speed = V V^2 = ((15* (0.06+0.06))R

V = square root (1.8 R)

V = Dist / time

Distance to allow 3 sec tangent Dist = V x time (3 sec)

then 1000'=Tangent curve 1 + Dist + Tangent Curve 2

1000= R x tan (30/2) + 3 x Speed + R tan (60/2)

then 0.22679 + 3 V + 0.577 R - 1000 = 0

0.8453 R + 3 (square root (1.8 R) - 1000 = 0

Solve for R R= 1171.7

Base on R = V^2 / 15 (0.06+0.06) V= 46 ft/sec

:screwloose: I believe the Methodology is right but some how I don't get the right answer

please if you see any mistake or miscalculation just post it

Good luck to all of you , next week this time every thing is over

 

[riz]

oops actually the result is 46 mph

I believe I did extra step, this

1 - find R = V^2/15*(0.06+0.06)

R = 0.555 V^2 (v is in mph)

2- Distance between Tangent = 3sec / 3600 sec per hour x V

Dist = 0.0008 V (v is mph)

3- T1 = R tan 30/2

= 0.2679 x0.555 V^2 = 0.148 V^2

4- T2 = R Tan 60/2

= 0.577 x 0.555 V^2 = 0.32 V^2

5- 1000= (0.32+0.148) V^2 + 0.0008 V

Solve for V = 46 Mph

 

[riz]

for previous quastion there was a unit convertion mistake

2- Distance between Tangent = 3sec / 3600 sec per hour x V

Dist = 0.0008 V (v is mph)

the right calc. for this step is (t=1.47 t t v=( 1.47 )( 3 sec )v=4.41v )

I just got this from a friend , copy paste it for you

hope is not too late

Problem 3: Transportation

Two reverse-horizontal curves have external angles Δ1 = 30° and Δ2 = 60°. The distance between PI1and PI2is 1000 ft. The maximum allowable side friction factor is 0.06, and the maximum super-elevation rate is 0.06. What is most nearly the maximum velocity (design speed) that can be provided while allowing for a 3 sec tangent distance between curves?

32 mph 36 mph 42 mph 47 mph

Solution:

The minimum radius and the velocity are related by

R= v 2 15( e+f ) = v 2 ( 15 )( 0.06+0.06 ) = v 2 1.8

The tangent legs of the curves are given by

T 1 =Rtan Δ 1 2 =( v 2 1.8 )tan 30° 2 =.149 v 2 T 2 =Rtan Δ 2 2 =( v 2 1.8 )tan 60° 2 =0.32 v 2

The tangent distance for a 3 sec transition between curves can be calculated from

t=1.47 t t v=( 1.47 )( 3 sec )v=4.41v

The total available distance is 1000 ft.

1000 ft=t+ T 1 + T 2

1000 ft=4.41 v+.149 v 2 +0.32 v 2

v 2 +9.39 v−2129=0

v= −B± B 2 −4AC 2A

= − 9.39 ± ( 9.39 ) 2 −( 4 )( 1 )( −2129 ) ( 2 )( 1 )

=41.7 mi / hr  ( 42 mph )