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boyakko [2]
2 years ago
10

A +7.5% grade meets a horizontal grade on a section of a rural mountainous highway. If the length of the crest vertical curve fo

rmed in that section is 300 ft long, determine the safe operating speed on the highway.
Engineering
1 answer:
Rasek [7]2 years ago
4 0

Using the formula of comfort condition it is possible to determine the safe operating speed on the highway.

The comfort condition formula is given by: L=2(\frac{NV^{3} }{C} )^{\frac{1}{2} }

Where, L is the length of the vertical curve, N is the grade, V is teh safe speed operation and assuming the standart value for C (0,6 m/s3) it is possible to calculate:

C = 0,6 m/s^{3} = 1,968 ft/s^{3}\\ N= N1 - N2 = 7,5-0 = 7,5% = 0,075

Changing the values in the formula:

L=2(\frac{NV^{3} }{C}){\frac{1}{2} }  \\300 = 2(\frac{0,075.V^{3} }{1,968})^{\frac{1}{2} } \\V = 83,89 ft/s

So, the safe operating speed on the highway is 83,89 ft/s.

See more about engineering at: brainly.com/question/12812715

#SPJ1

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Answer:

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Explanation:

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solution

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SSD = 1.47 ut + \frac{u^2}{30(\frac{a}{g}\pm G)}     ..............1

here u is speed and t is reaction time i.e 2.5 second and a is here deceleration rate i.e 11.2 ft/s² and g is gravitational force i.e 32.2 ft/s² and G is gradient i.e 0 here

so put here value and we get

SSD = 1.47 × 65 ×2.5  + \frac{65^2}{30(\frac{11.2}{32.2}\pm 0)}

solve it we get

SSD = 644 ft  

so here minimum distance clear from the inside edge of the inside lane is

Ms = Rv ( 1  - cos (\frac{28.65 SSD}{Rv}) )        .....................2

here Rv is = R - one lane width

Rv = 2300 - 6 = 2294 ft

put value in equation 2 we get

Ms = 2294 ( 1  - cos (\frac{28.65 \times 664}{2294})  )  

solve it we get

Ms = 22.57 ft

and

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R  = \frac{u^2}{15(e+f)}  ..................3

here f is coefficient of friction that is 0.10

put here value and we get e

2300 = \frac{65^2}{15(e+0.10)}

solve it we get

e = 2%

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Required

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The pseudocode and program makes use of a 1 dimensional array to accept input for the 10 numbers;

The largest of the 10 numbers is then saved in variable Largest and printed afterwards.

Pseudocode (Number lines are used for indentation to illustrate the program flow)

1. Start:

2. Declare Number as 1 dimensional array of 10 integers

3. Initialize: counter = 0

4. Do:

4.1 Display “Enter Number ”+(counter + 1)

4.2 Accept input for Number[counter]

4.3 While counter < 10

5. Initialize: Largest = Number[0]

6. Loop: i = 0 to 10

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6.3 End Loop:

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C# Program (Console)

Comments are used for explanatory purpose

using System;

namespace ConsoleApplication1

{

   class Program

   {

       static void Main(string[] args)

       {

           int[] Number = new int[10];  // Declare array of 10 elements

           //Accept Input

           int counter = 0;

           while(counter<10)

           {

               Console.WriteLine("Enter Number " + (counter + 1)+": ");

               string var = Console.ReadLine();

               Number[counter] = Convert.ToInt32(var);

               counter++;                  

           }

           //Initialize largest to first element of the array

           int Largest = Number[0];

           //Determine Largest

           for(int i=0;i<10;i++)

           {

               if(Largest < Number[i])

               {

                   Largest = Number[i];

               }

           }

           //Print Largest

           Console.WriteLine("The largest input is "+ Largest);

           Console.ReadLine();

       }

   }

}

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