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I second jijir83's opinions. Dr. Ibrahim made sure we understood the material and got proficient in answering questions on the exam. When I started working on the actual exam problems, I realized that I had solved a bunch of them similar to the practice problems and practice tests. Whether I can...

That is the best strategy... work on the ones that you feel confident and then attack the rest.  Good luck!

I have taken the Seismic a week ago. I am an EET student, and I sure am glad that I was.

I am too old to even remember what my PE exam was like!!! Now I am trying to study for the California special exams, so I am reliving those "student" days. Anyway, success usually sides with the one who is prepared. Some people may have studied very hard in their college days and that...

Congratulations and thank you for sharing your experience. It is very helpful for future exam takers.

I can see the pictures better on the desktop now... the brothers seem to be proud of their little sister... little do they know that she is going to wrap them around her fingers and make them her puppets  :)

congratulations... 

Got that email and already emailed them asking them to upgrade my account using their magic software. 

This is the reason why I believe that there should be ONE national license. For people who work in multiple states, it will save them money and more importantly, there will be uniformity. 

Thanks for the various alternate solutions. It helps the readers to get different perspectives and we all learn new techniques.

I am getting R=434'. Do you mind sharing your method? See these image below for my designations. Theta1=Theta2=I, as tangents are paralled. It is given that R1=R2. BV1=EV2=T and T1=T2=T=R*Tan(I/2). V1C=V2C=T=R*Tan(I/2). Also, V1C+V2C=2R*Tan(I/2). V1V2 Sin(I)= S = 120ft. also...

I came up with this problem based on problems that I have seen in workbooks. I believe a similar problem can be expected in the exam. The wording maybe a little confusing, but sketch it out and it should be pretty easy.  The center of two parallel roads is 120 feet apart. They are to be...

This post from someone with just 1 post is not passing my smell test... 

Please see attached. I hope it makes sense. There is a formula given in http://www.ce.memphis.edu/3161/pdf files/GEOMETRIC DESIGN.pdf which is what Reza is using. However, I believe, it is better to use first principles to come up with the answer. However, for exam purposes, a formula like...

I have an old version of the book and there is no problem 60 for chapter 6. It ends at problem no. 54. If you can scan in and post the problem, I can try to help.  Thank you.

When I got my PE back many many years ago in a state that did not require seismic and survey, I got around a decent raise. I don't remember the exact numbers but it was somewhere between 5 and 10%.  Now that I spend time preparing cost proposals, here is my take on how to calculate a fair...

Not talking about this particular seminar, but I prefer the webinar format. It helps me to save time on commute, because every minute saved is a minute gained that you can either rest or study... or as in my case spend time with family.  Good luck.

I think you should contact LAPELS directly and get their opinion. In case you don't like their finding, then you can have a meeting with them and discuss. Otherwise, in case you do some work that according to LAPELS is beyond the scope of your license, you risk censure and bad publicity.  Good...

Thank you. Once you figure out that you need to draw those perpendicular lines, the problems become easier to solve. 

I am getting a different answer for part 1. Draw two lines from B, one perpendicular to DE (Call it BQ) and one perpendicular to FE (Call it  BN). Angle QBE = Angle NBE, as BQ=BN=8, BE is common, and BQE=BNE=90^. QBE+NBE=360-(90+90+10^30'+(180-67^10'))=56^40'. QBE=NBE=28^20'. Azimuth of...