jasonmcbride
New member
- Joined
- Mar 9, 2020
- Messages
- 10
so maybe around 5.6 cubic metres then? top and bottom are flat, no curvesIf we increase the front left to 500 mm, we get an overestimate, which would be a triangular prism. If we decrease the front right to 200 mm, we get an underestimate. So the volume is probably between those, and maybe close to their average. So you might try that. But without knowing how the bottom surface curves, we can't be sure.
I'm curious how you got that.
You're making an assumption about the shape, namely that the cross-sections in your chosen direction are triangles, requiring that the surface of the soil be a ruled surface with straight lines in that direction. You could instead assume that other cross-sections of the surface are straight, or that they are any other curve. There are many ways to warp a plane.I would use Calculus. Taking the origin of a 3D coordinate system at the lower left front side, the lower front line is x= t, y= 0, z= 0, t going from 0 to 6. The upper front line goes from (0, 0, .2) to (6, 0, .5) so is given by x= t, y= 0, z= 0.2+ 0.05t, t going from 0 to 6 . The back edge is x= t, y= 4, z= 0, t going from 0 to 6.
If we "slice" the object perpendicular to the x-axis, the cut faces are triangles with height 0.2+ 0.05 t and width 4 so area (1/2)(0.2+ 0.05t)(4)= 0.4+ 0.1t. Taking the thickness to be "dt", the volume of each slice is (0.4+ 0.1t)dt so the volume of the object is \(\displaystyle \int_0^6 (0.4+ 0.1t)dt= \left[0.4t+ 0.05t^2\right]_0^6= 0.4(6)+ 0.05(36)= 2.4+ 1.8= 4.2\) cubic meters as Jasonmcbride said in post #4.