Calculation of volume under x,y coordinates 2

[musilj]

Is there a method to calculate volume of an irregular poly formed by crushed ore stockpile. I plotted the rough poly with excel, I intend to do it manually and checking accuracy with a software program. i have calculated the area under the coordinates. how do i calculate the volume? i have the heights of the coordinates, sorry i am definitely not a Pro.

 

[IRstuff]

So why can't you simply multiply the heights by the patch sizes? That would be the simplest numerical integration.



TTFN

 

[musilj]

I haven't got the patch sizes. all I have calculated is the area covered by the poly coordinates, which to me appears like the plan/base area(from top), without taking into consideration the slope. the shape formed looks like a frustum but not exactly. Simpson's might work but is there any other method i can use which base its calculations on coordinates and incremental heights. By the way can I calculate the patch sizes from the coordinates?

 

[IRstuff]

That's what I was assuming you had already done.

Each patch boundary is simply half the distance to the adjacent data point.

TTFN

 

[CarlB]

I don't follow what you have coordinates for. Just the outline of a pile at the base level? ... intermediate or top elevations? If you can't approximate the pile as some geometric shape, or have enough points to draw multiple sections, either horizontal or vertical, then you probably need to put the points into survey software to model the surface. (and what is a "patch"?)

 

[IRstuff]

If the points are irregularly spaced, then differential area associated with each data point could be called a "patch" of area.

TTFN

 

[Denial]

Sounds to me like you need to apply some sort of triangulation process to the irregularly located (x,y) data points, with each point being a vertex for several adjoining triangles. Then calculate the volume of each triangle (above your z-datum)as
A * (z₁ + z₂ + z₃)/3
where A is the plan area of the triangle.

There are heaps of programs that will do this sort of thing, and all you'd have to do is import your data into one of these.

 

[musilj]

I have the coordinates as shown, i have calculated the total area which i realise might be the plan area. I need to calculate the volume of the pile.
Northing Easting RL (metres)

1838.9 2222.3 191.8

1849.2 2221 192

1858.3 2222.3 191.8

1865.6 2231.5 190.6

1862 2239.8 189.9

1857.5 2243.6 189.5

1849.1 2245 189.3

1839 2240.9 189.8

1833.2 2232.4 190.5

1849.9 2228.5 201.3

1854 2231.9 201

1848.7 2234 200.4

 

[IRstuff]

ARe those ALL the points? Do they represent the boundary points? Are there only 3 interior points?



TTFN

 

[musilj]

YES, those are ALL the points representing boundary points, and YES there are only 3 interior points.

 

[IRstuff]

OK, so the simplest thing you can do, barring any additional information, is to assume that the surface is comprised of triangles whose vertices are the points on your list. Calculate the area of each triangle, multiply by the lowest of the three vertices to get the volume of the flat-topped prism. Then, you calculate the volume of the chunk above that.

The remainder's volumes should be extensions of the 1/3 base_area*height equation from solid geometry. The general case involves finding the one side of the solid that is a quadrilateral and using that as the base. The altitude is the distance from the point that doesn't reside in the quadrilateral.

TTFN

 

[musilj]

I have been trying to interpret your reply, but probably not smart enough, I don't understand whether I will need to calculate the sides of the triangles? Do you mean joining three points (3 sets Eastings, Northings, and RL) as given in the data to form the triangle vertices? I tried some form of triangulation joining the triangles to one vertex 1865.6(N) 2231.5(E) 190.6(RL), it looks like it didnt come out good.

 

[IRstuff]

Sorry I couldn't make it clearer. Consider each triplet of points as forming the top vertices of a triangular prism with a flat bottom and a tilted top. You can break the prism into two solids, one with flat top and bottom and the top part which is a 5-sided solid with a triangular base.

The volume of the bottom part is relatively trivial, i.e., calculate the area of the base and multiply by the height.

The left over solid is more complicated. The general case is that one of the three vertical sides will be a quadrilateral, while the other two vertical sides, the base, and the top are triangular.

I'm guessing the simplest solution is to calculate the area of the quadrilateral and multiply by 1/3 the altitude of the solid referenced to the plane of the quadrilateral.

TTFN

 

[TDAA]

Enter into CAD, get cross sections, get area of the cross section, use average end areas and distance between.

 

[musilj]

Yep i intended to use CAD as a check for the manual calculations. thanks all, i got a good estimate.

 

[MEM1]

2318.7 cubic meters