# civil engineering

• Jan 25th 2006, 02:04 PM
mathAna1ys!5
civil engineering
highway curves.... proper banking angle, theta, for a car making a turn of radius, r feet at a velocity v feet per second is given by tan theta = v^2 /gr, where g is the acceleration due to gravity. assume g = 32 ft/s^2. an engineer is designing a curve with a radius of 1000 feet. if the speed limit on the curve will be 55mph, at what angle should the curve be banked?
• Jan 25th 2006, 07:00 PM
ThePerfectHacker
Quote:

Originally Posted by mathAna1ys!5
highway curves.... proper banking angle, theta, for a car making a turn of radius, r feet at a velocity v feet per second is given by tan theta = v^2 /gr, where g is the acceleration due to gravity. assume g = 32 ft/s^2. an engineer is designing a curve with a radius of 1000 feet. if the speed limit on the curve will be 55mph, at what angle should the curve be banked?

Substitute that infromation in to the equation (wonder if this equation is true :rolleyes: )
$\displaystyle \tan\theta=\frac{v^2}{gr}$
Thus,
$\displaystyle (g,r)=(32,1000)$
However,$\displaystyle v$ must be converted to feet per second,
$\displaystyle 55\frac{\mbox{miles}}{\mbox{hour}}=55\frac{\mbox{5 ,280 feet}}{\mbox{3600 seconds}}=80.\bar 6$
Thus,
$\displaystyle \tan\theta=\frac{(80.\bar 6)^2}{32\times 1000}\approx .203$
Thus, (use arctan)
$\displaystyle \theta \approx 11.5$
• Jan 26th 2006, 04:45 AM
mathAna1ys!5
ty
I had another problem like this and was able to solve it just fine. thanks for the help.
• Jan 26th 2006, 10:30 AM
ThePerfectHacker
Welcome, I can see you do not like trigonometry that much.