Calculating the volume of a sloped wedge?

Are you implying that the back end is just a line, with zero height all along it, but the front is a trapezoid?

Then the top is not flat (planar), and in order to determine the volume, we would need to know its exact shape. The left and right top edges are not parallel.
 
Ok, yes, the back end is 0mm at both corners. Just trying to determine how much soil I will need to fill a retaining wall on sloped ground (top will be flat, bottom won't) what more info would I need. The ground slopes left to right and back to front.
Is it even possible to calculate this, or will I just have to estimate.
Cheers
Jason
 
If 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.
 
Dr.Peterson said:
If 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.
Click to expand...
so maybe around 5.6 cubic metres then? top and bottom are flat, no curves
 
Again, they can't be literally flat, but sort of warped, though it might not be noticeable! But for the overestimate, I got [MATH]\frac{4\times 0.5\times 6}{2} = 6\text{ m}^3[/MATH], and for the underestimate [MATH]\frac{4\times 0.2\times 6}{2} = 2.4\text{ m}^3[/MATH], for an average of 4.2 m3, which happens to be what you suggested before. I'm curious how you got that.

I could try other formulas that take more work, with specific assumptions, but before doing that I'd want to look at your actual location and do some extra measurements, which I wouldn't even know how to describe without being there. Just decide how much more than 4.2 m3 you need to be on the safe side.
 
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 ∫06(0.4+0.1t)dt=[0.4t+0.05t2]06=0.4(6)+0.05(36)=2.4+1.8=4.2\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.
 
HallsofIvy said:
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 ∫06(0.4+0.1t)dt=[0.4t+0.05t2]06=0.4(6)+0.05(36)=2.4+1.8=4.2\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.
Click to expand...
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.

But this does seem to be a reasonable guess.
 
How did it turn out? Was the answer correct in the real world? Don't keep us in suspense, mate!!
 
You must log in or register to reply here.
Top