Practice Problem #1

[John QPE]

OK, so Practice Problems will not be like NCEES problems, but more Lindeburg style, one question that tests several concepts. These may take you 10-20 minutes, but again, only testing concepts.

Given a Horizontal Curve with degree of curvature = 6* and Delta = 60*

1) Using arc definition, calculate: L (Length); R (Radius); T (Tangent); C (Long Chord); E (PI to curve); M (Middle Ordinate)

2) Using chord definition, calculate: L; R; T; C; E; M

3) Find the new radius R, if the forward tangent is moved 10 feet inward (towards the PI)

 

[ptatohed]

Come on people, it's Exam month.....

 

[John QPE]

I'm convinced no one is taking this exam other than Matt.

 

[matt267 PE]

What? When?

 

[Beej]

Thanks for posting these questions.Just started studying this week for the 2016 Spring Exam. This is what I got. Couldn't visualize the 3rd question. Possible to get a diagram?

Arc Definition:

L = 999.97 ft

R = 954.9 ft

T = 551.31 ft

C = 954.9 ft

E = 147.72 ft

M = 127.93 ft

Chord Definition:

L = 1000 ft

R = 955.4 ft

T = 551.6 ft

C = 955.4 ft

E = 147.8 ft

M = 127.99 ft

 

[John QPE]

Thanks for posting these questions.Just started studying this week for the 2016 Spring Exam. This is what I got. Couldn't visualize the 3rd question. Possible to get a diagram?

Arc Definition:

L = 999.97 ft

R = 954.9 ft

T = 551.31 ft

C = 954.9 ft

E = 147.72 ft

M = 127.93 ft

Chord Definition:

L = 1000 ft

R = 955.4 ft

T = 551.6 ft

C = 955.4 ft

E = 147.8 ft

M = 127.99 ft

You are correct.

As for problem #3, this is a very classical transportation geometrics problem. You need to know how to do this.

If you shoot me a PM with your email I can send you a visual aid.

But you are basically just taking the forward tangent and moving it in 10 feet .... draw it .... I think you'll get what I'm getting at right away.

 

[Beej]

You are correct.

As for problem #3, this is a very classical transportation geometrics problem. You need to know how to do this.

If you shoot me a PM with your email I can send you a visual aid.

But you are basically just taking the forward tangent and moving it in 10 feet .... draw it .... I think you'll get what I'm getting at right away.

Finally got around to doing this one.

1. Subtract 10 feet from T = 551.31 > Tnew = 541.31

2. Back Solve for New R using Tnew = Rnew x tan (D/2)

3. I got Rnew = 937.5 ft

 

[John QPE]

I don't have my solution on hand, but I can tell you right now, if you simply subtracted 10' from Told your solution is incorrect. That is the whole trick to the problem.

Draw it, and rethink.

Here is a hint:

The forward tangent shifts parallel in position, it does not slide along the back tangent.

 

[ptatohed]

I don't have my solution on hand, but I can tell you right now, if you simply subtracted 10' from Told your solution is incorrect. That is the whole trick to the problem.

Draw it, and rethink.

Here is a hint:

The forward tangent shifts parallel in position, it does not slide along the back tangent.

So, you mean the Fwd Tan is 'offset'?

Also, what do you mean by " ... if the forward tangent is moved 10 feet inward (towards the PI)"?  The fwd (and back) tan makes up the PI, so how can it be moved toward the PI?  I assume you mean offset inward toward the curve?

 

[ptatohed]

Also, to solve part 3, do we use the arc def radius or the chord def radius?

 

[ptatohed]

Ok, so I went for it. 

1.) 

R = 954.93',  L = 1000.00',  T = 551.33',  C = 954.93',  E = 147.73', M = 127.94'

2.) 

R = 955.37',  L = 1000.46',  T = 551.58',  C = 955.37',  E = 147.80', M = 128.00'

3.)

(Assuming arc definition, DAo = 6o, ROrig = 954.93')  RNew = 934.93'

Am I close? 

 

[ptatohed]

Finally got around to doing this one.

1. Subtract 10 feet from T = 551.31 > Tnew = 541.31

2. Back Solve for New R using Tnew = Rnew x tan (D/2)

3. I got Rnew = 937.5 ft

What about answers for parts 1 and 2?

 

[P-E]

PE Rule #6.9  if the problem involves back solving or iteration skip it.  

 

[John QPE]

Ok, so I went for it. 

1.) 

R = 954.93',  L = 1000.00',  T = 551.33',  C = 954.93',  E = 147.73', M = 127.94'

2.) 

R = 955.37',  L = 1000.46',  T = 551.58',  C = 955.37',  E = 147.80', M = 128.00'

3.)

(Assuming arc definition, DAo = 6o, ROrig = 954.93')  RNew = 934.93'

Am I close? 

You're good.

Yeah, use arc definition.

View attachment Practice Problem-1.pdf

 

[Beej]

I don't have my solution on hand, but I can tell you right now, if you simply subtracted 10' from Told your solution is incorrect. That is the whole trick to the problem.

Draw it, and rethink.

Here is a hint:

The forward tangent shifts parallel in position, it does not slide along the back tangent.

ahhhh, I did exactly that! Thanks will rework the problem.

 

[Beej]

What about answers for parts 1 and 2?

The answers for Parts 1 and 2 are in post #5. Thanks for the diagram.