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miloc

Left-Turn length question

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How would you answer this question:

post-18619-0-76014600-1332282546_thumb.j

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How would you answer this question:

Easiest way is to use formula give in Traffic Engineering Handbook, Pg: 332

L = VK25(1+p)/Nc

v= peak 15-min flow rate

K = constant to reflact random arrival of vehicle, usually 2

p = percent trucks

Nc = no. of cycle per hour

= 300*2*25 (1+0.1)/40

Answer is B = 413 ft

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How would you answer this question:

Easiest way is to use formula give in Traffic Engineering Handbook, Pg: 332

L = VK25(1+p)/Nc

v= peak 15-min flow rate

K = constant to reflact random arrival of vehicle, usually 2

p = percent trucks

Nc = no. of cycle per hour

= 300*2*25 (1+0.1)/40

Answer is B = 413 ft

You are in the right direction. I have the answer but it does not make sense (so far) for me. How did you know about that formula? I looked in the index but couldnt find it. The solution also involves deceleration length. I didn't include the answer bc I want to know if it is possible to get it with the HCM or Green Book instead.

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I would hope that even if the formula answer gave 1000' that ncees wouldn't have that as an answer or even a question, cause there aren't any 1000' left turn lanes around....

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I'm curious to the solution myself.

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I came across similar problem in 6 min. solutions and they had referred to same formula in their solution.

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Not sure if I am helping or not. This is just a quick and dirty rule of thumb but, using it, I get answer B.

X vph = X feet of Storage. Transition length usually = 90' for lower speeds, 120' for higher speeds.

Storage + Transition = Total left turn length

300 + say,100 = 400'

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I would hope that even if the formula answer gave 1000' that ncees wouldn't have that as an answer or even a question, cause there aren't any 1000' left turn lanes around....

We have some left turn lanes around that are two or three wide and over a quarter mile long...

Traffic engineers here have a lot of challenges, and I think that they have to throw away the book. With the 2nd worst traffic in the country here in the Metro DC area, often the answer to the question above would be "how much room do we have?" The politics and cause and effect of everything must be weighed. Relieving congestion at one intersection often just makes it worse at the next one.

For example in the question above, it may need to be longer if the next street has a high amount of left turners, therefore resulting in a blockage of the left turn lane at this intersection as people stack up to turn left at the next street...

On a similar topic, do traffic models use vehicles that during slow-downs, will merge into the on-ramp of a highway just to pass a few cars, therefore slowing down traffic more, causing more people to get into the on-ramp to cut back in?

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On a similar topic, do traffic models use vehicles that during slow-downs, will merge into the on-ramp of a highway just to pass a few cars, therefore slowing down traffic more, causing more people to get into the on-ramp to cut back in?

Unfortunately, there is no model for idiocy - there is a dire need for such models in the DC area

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Thanks for all your input guys. I did ask to the senior traffic engineer in my office about why and how to find the equation and table in the IE traffic engineering handbook, the answer "It's something that you have to know", as simple as that. P.S. there's a little mistake in the solution regarding the table (or isn't it?) the figure mentioned is for RIGHT-TURN deceleration bay, the question asks for LEFT-TURN.

p.s. I couldnt find the option to upload an image. I'm pasting the solution:

See Figure 10-15 in Traffic Engineering Handbook by ITE

STEP 1 - First determine the required deceleration length (d2+d3) using Table 10-7A

of ITE Traffic Engineering Handbook

Deceleration length for 50 mph speed = 565 ft

STEP 2 - Compute storage length (d4) using left-turn storage length equation

25(L =VK 1+ p) / Nc = 300*2*25(1+0.1)/40 = 412.50 ft

STEP 3 - Total left-turn bay length = 565+412.5 = 977.50 ft

Answer C

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I was going to send exact the same explanation as Jaa046 and few more guys which describe the solution in the same way. One more thing is puzzle me. There is 10% trucks and there is a table in (attached) saying s=30 (not 25). If i use data from the table my answer wouldn't be correct. why this data and table are given?

the answer is 413= C

I was trying to attache the table, but can't find how to do it here.

trucks avarage vech. length

<4% - 25

5% 27

10% 30

15% 35

Help me with attachment button, please

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Thanks for all your input guys. I did ask to the senior traffic engineer in my office about why and how to find the equation and table in the IE traffic engineering handbook, the answer "It's something that you have to know", as simple as that. P.S. there's a little mistake in the solution regarding the table (or isn't it?) the figure mentioned is for RIGHT-TURN deceleration bay, the question asks for LEFT-TURN.

p.s. I couldnt find the option to upload an image. I'm pasting the solution:

See Figure 10-15 in Traffic Engineering Handbook by ITE

STEP 1 - First determine the required deceleration length (d2+d3) using Table 10-7A

of ITE Traffic Engineering Handbook

Deceleration length for 50 mph speed = 565 ft

STEP 2 - Compute storage length (d4) using left-turn storage length equation

25(L =VK 1+ p) / Nc = 300*2*25(1+0.1)/40 = 412.50 ft

STEP 3 - Total left-turn bay length = 565+412.5 = 977.50 ft

Answer C

miloc,

I have the 5th edition (not the 6th) but on page 329 of the TEH, under the "Left-Turn Deceleration Lane" heading, last sentence of the second to last paragraph, it states that Fig 10-15 and Table 10-7 can be used for left turns (even though the figure is titled as right turn).

Thanks for posting the answer.

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