Choose a Topic |
Volumes From Spot Heights - Grid LevellingPreviously levels have been located on cross sections and along a centre line. A convenient variation of this is the method of grid levelling. It is used extensively for site investigations, for irrigation, for drawing contours, to minimise computations of position and for the easy calculation of volumes. Rectangular Base MethodThe figure below shows a typical survey of a site using grid levels. At each and every intersection of the grid (which unfortunately must be established in the field) a level is obtained. The general problem is to calculate the volume of material above or below a certain reduced level (RL). It should be pointed out that this method will only calculate the net volume and not the gross volume (made up of cut and fill volumes). Generally this is a disadvantage as the RL to which the calculations are referred MUST NOT intersect the surface defined by the grid of levels. This problem is investigated later in this section to locate the line of 'zero cut and fill'.
In the figure above the grid interval is s in both directions, therefore the area of a grid element is s2. The reference RL is defined as RL0 and the problem is to calculate the volume of material between this horizontal plane and the surface defined by the grid levels. The height difference at each grid point is given by; hi = RLi - RL0
For the second and third elements the volumes are,
The total volume of the area covered by the entire grid of levels is, V = [V1 + V2 + V3 + ............. + Vn] and therefore the volume in general terms may be expressed as,
s2 is the area of the grid cell Ni is the Occurrence number, hi is the height difference at each point,. This method is based on the assumption that the level for each grid element is the mean of the four corner levels. This is, in the majority of cases, an invalid assumption. It will not lead to significant errors if the terrain is reasonably flat or if the grid spacing (s) is small with respect to the surface roughness. The problem can be avoided (but not totally eliminated) if each square element is considered to be a prism on a triangular base.
|
||||||||||||||