Numerical Integration and Differentiation - Assignment Example

A road needs to be built through hilly terrain at a sloping angle of not more than 10 degrees with minimal material wastage. The various measurements (height of hill and distance across have been provided.

Method

In order to determine the volume of material to be removed the area of land above the road must be obtained. By assuming the longitudinal section is a curve it becomes easier to obtain it because it helps map it into a graph. The following specifies the steps

1. Height and distance readings were taken and plotted using the Matlab plot function
2. In order to obtain a slope for the road, it was assumed the road would be a line of best fit. In this case, it would neither be high-sloping nor low sloping.
3. The equation for the line of best fit was obtained. This can be easily obtained in Matlab using the polyfit () function.
4. Using the road as a second line, its points of intersection with the road were obtained. The area of the hill above the road was the one to be removed.

diff(sign(s-y)) was used to find the points of intersection. Any non-zero result indicated the point of crossing.

find(d>0)  and find(d<0) were used to obtain the non-zero points and the result was used to obtain the exact x-axis points of intersection. This was confirmed by comparing the result with that of the plot.

5. The area of the land above the planned road was obtained using the integration method (trapezoid method) to find the total area under the curve and subtract the total area of the area of land under the planned road. This was achieved through the MatLab trap ()

            Area of land to be removed=area of land under curve-area of land below the road

6. The areas of the different sections were summed up and multiplied by the width of the road to get the total volume

Fig3  From the function p=polyfit(x,y,1) the equation for the line of best fit was obtained which is

0.0049*x + 534.8242

On using the d =diff(sign(y-s)) function where s was the equation for the best fit and y was for the terrain curve the non-zero points were obtained.-2 and 2 were the non-zero points which corresponded to respective x points. those x points are shown below

It was noted that those points corresponded with the x points where the two curves intersected. These points were displayed in pairs(the first one showing the x point for the start of the region and the second one showing the endpoint for the region)e.g

V=1900  4400   showed that

  fig4  sample region with x-co-ordinates of the first region.

Next, the program requests the user to enter the number of regions in the graph. In our case, they were 7. This was made in order to allow the user more control over the area calculation and reduce errors.

Sample

 >>how many hills?7

The program then prompts the user for the first and second x-coordinate

>>Enter fp 

     Enter sp

Using the first region as an example where 1900 and 4400 were the values

>>Enter fp  1900

   Enter sp 4400

The program calculates the area of the region under the curve subtracts the region under the best fit then sums it up. It then asks for the second point. Calculates the area then adds it to the total. At the end of the last region, it multiplies the area by 10(road width and gives the total volume.