Compare Problems

[Ambrug20]

I have 2 problems from NCEES. They both look almost the same. But the solutions are different. If you use solution from problem 2 on problem 1, you'll get a wrong answer. Does any one can figure out why. Actualy, why don't we use 15' (distance from RR) and 5' (driver distance from front bumper) in first problem.

The way I worked out second problem was:

Found Sb (braking dist), then Sr (precip-react. dist), add it together and added 15' and 5'. Came to right aswer, or was it a coincidence?

I was trying to work first problem with Creen Book formula Ex. 9-103, but it didn't work.

Where is the clench?!

Trasp._solution.pdf

 

[civilized_naah]

I have 2 problems from NCEES. They both look almost the same. But the solutions are different. If you use solution from problem 2 on problem 1, you'll get a wrong answer. Does any one can figure out why. Actualy, why don't we use 15' (distance from RR) and 5' (driver distance from front bumper) in first problem.

There are 2 issues here.

1. When the phrase 'sight triangle distance' is used, typically you draw the two travel paths to intersect and then calculate the leg of each triangle from the point of conflict. However, by setting the stop line back 15 ft, the 'point of conflict' is set back (car is only allowed to go to stop line and not all the way to nearside rail) and therefore, the sight triangle leg distance must be reduced 15 ft. From that point of view, the first version is correct. The second version adds the 15 ft to the calculated stopping sight distance (not the length of the sight triangle)

2. The first version gives the friction factor f = 0.38 (the OLD Green Book approach), whereas the second version asks you to assume the default decelaration rate a = 11.2 ft/s^2 (which corresponds to a 'f' value = 0.348), which makes the second term different in the 2 calcs.

So, there seems to be a little bit of ambiguity in the question (as NCEES has proved by providing 2 different versions in the 2 printings of the sample questions book). It it were up to me, I would interpret the sight triangle distance as the distance from the driver's position (the sighter) to the stop line.

 

[maximus808]

I was looking at this problems and although I didn't look in the Green Book, I have a transportation book that states the following equation for rail intersection sight distance.

dh=1.47vt + v^2/(30)(a/g) + D + de

It assumed the road to be flat, and I believe a/g represents f (the friction factor) correct me if I'm wrong someone.

the equation now is:

dh=1.47vt + v^2/(30)(f+G) +D + de

dh=1.47(45)(2.5) + 45^2/(30)(0.38-0.01) + 15 + 5

t=2.5s (assumed)

G=-.01 highway grade

D=distance from rail tracks to stop bar

de = distance from driver's eye to front bumper

The question ask what distance must the car stop at the stop bar so you basically subtract D & de from dh.

I got a result of 347'

 

[maximus808]

Sorry, I just realized that I think the second solution is correct in your attached pdf. 340' is the SSD which does not account for the 15' stop line from rail and 5' eye distance from bumper. The green book has constants such as B=1.075 but I would probably use this equation dh=avt+bv^2/a+D+de for the correct result. The driver must have a greater sight distance to stop at the stop bar and account for the distance he is from his front bumper than just the SSD or else he would collide with the train. Does this sound right guys? Thanks.

 

[Ambrug20]

Sorry, I just realized that I think the second solution is correct in your attached pdf. 340' is the SSD which does not account for the 15' stop line from rail and 5' eye distance from bumper. The green book has constants such as B=1.075 but I would probably use this equation dh=avt+bv^2/a+D+de for the correct result. The driver must have a greater sight distance to stop at the stop bar and account for the distance he is from his front bumper than just the SSD or else he would collide with the train. Does this sound right guys? Thanks.

Both solutions are correct in these calculations, but why in the first problem they didn't use 15'from stop line and 5' back from front bumper. In my opinion, doesn't matter what formula to use, the result should be the same. Why this data wasn't used?

 

[sac_engineer]

I agree that both solutions need to include the extra stopping distance (stop line + bumper distance). The SSD implies the reaction and deceleration in relation to the passing train; however, the stop line and bumper distances must be considered so that the vehicle stops that the stop line. Therefore, the SSD calculation would only compute the minimum distance that the driver travels from their body position without considering the stop line and bumper distance.

 

[ryan_mizzou]

First I just want to say thanks for taking the time to help people out, this site has been very useful for me. Although I've lurked until now I've picked up some good info here!

Rather than start a new topic, I thought I would bump this one....

On the 2011 practice exam, #518 is very similar to the problem in this thread. Again it asks the the distance for the vehicle to stop AT THE STOP LINE. In my solution it adds the 15' (between stop bar and track) and the 8' between driver and front of vehicle. Page 9-188 in 2011 GDHS shows dH extending to the track. #518 shows dH only to the stop bar.

My main question is why the 15' and 8' are added if it's asking for distance to the stop line.

Thanks for the help!

 

[ptatohed]

First I just want to say thanks for taking the time to help people out, this site has been very useful for me. Although I've lurked until now I've picked up some good info here!

Rather than start a new topic, I thought I would bump this one....

On the 2011 practice exam, #518 is very similar to the problem in this thread. Again it asks the the distance for the vehicle to stop AT THE STOP LINE. In my solution it adds the 15' (between stop bar and track) and the 8' between driver and front of vehicle. Page 9-188 in 2011 GDHS shows dH extending to the track. #518 shows dH only to the stop bar.

My main question is why the 15' and 8' are added if it's asking for distance to the stop line.

Thanks for the help!

r_m, dude you bumped a 3.5 year old thread!

You bring up a great point. I think the question and solution/answer for 2011 NCEES Transpo #518 is correct but the diagram is wrong. dH should go to the tracks, you're right. But note the question is not asking for the distance to the stop line. It's asking for "the sight triangle distance along the hwy" which is, by definition, dH.

This reminds me of #528 which, in the problem statement, states that the skid marks start 150' from the intersection yet the diagram shows them all the way to the car!