Safe Design speed Help!

[MaryJ]

I just worked a problem and came up with a Radius of 5729.578 and a HSO value of 23.32.

The problem basically ask for the safe design speed of a highway, given D=(9 degrees, 30 minutes), lanes are 12 feet wide and a continuous tree hedge is located along the inside of the curve, exactly 29.32 ft from center of the highway. PI= 16+00

my solution:

R= 5729.578/9.5= 603.11 (radius of roadway centerline)

Determine the radius of the centerline of the inside lane (603.11 - 6')= 597.113'

Find actual HSO distance: 29.32 - 6' = 23.32'

Now the problem continue by stating that I should plug R and HSO values into CERM Eq 79.44 (CERM 10-th edition) to get a "S" value of 335'

Is there an error in the solution of the problem as I have plug these values into the equation a couple of times and cannot come up with this answer.

formula:

S = ( R/28.65 )( arccos (( R-M )/R )

 

[sac_engineer]

I just worked a problem and came up with a Radius of 5729.578 and a HSO value of 23.32.
The problem basically ask for the safe design speed of a highway, given D=(9 degrees, 30 minutes), lanes are 12 feet wide and a continuous tree hedge is located along the inside of the curve, exactly 29.32 ft from center of the highway. PI= 16+00

my solution:

R= 5729.578/9.5= 603.11 (radius of roadway centerline)

Determine the radius of the centerline of the inside lane (603.11 - 6')= 597.113'

Find actual HSO distance: 29.32 - 6' = 23.32'

Now the problem continue by stating that I should plug R and HSO values into CERM Eq 79.44 (CERM 10-th edition) to get a "S" value of 335'

Is there an error in the solution of the problem as I have plug these values into the equation a couple of times and cannot come up with this answer.

formula:

S = ( R/28.65 )( arccos (( R-M )/R )

Your approach is correct but you've only calculated the required stopping distance based on the geometry and sight obstruction properties of the curve. You need to do one more calc to compute the design speed for the previously calculated stopping distance.

SSD = 1.47xSpeed(in mph) + (speed)^2 / (30*((acc/32.2) +/- G))

where acc = 11.2 ft/s^2 (Default deceleration rate) and G = grade (use zero as default, i.e. level grade)

Use the quadratic equation to solve for Speed.

Thus when SSD = 335, Speed is approx. 42 mph.

 

[sac_engineer]

CORRECTION:

formula is SSD = 1.47xSpeed(in mph) x tr + (speed)^2 / (30*((acc/32.2) +/- G))

where tr = 2.5 sec (default), acc = 11.2 ft/s^2 (Default deceleration rate) and G = grade (use zero as default, i.e. level grade)