Hello everyone,
I have a quick question about volume calculation. I’m a bit stuck at the moment.
A slope is to be set back by 2.50 m (8 feet) at both the top and bottom edges over a length of 35 m (115 feet). The slope angle should remain the same as it is now. The slope ratio is 1:1.5. The length from the bottom edge to the top edge is approximately 5.50 m (18 feet).
If this area is excavated as described, how many cubic meters (cubic yards) or tons of soil would be removed?
Do you need any additional information for the calculation?
I appreciate any help.
Thank you!
Best regards
I have a quick question about volume calculation. I’m a bit stuck at the moment.
A slope is to be set back by 2.50 m (8 feet) at both the top and bottom edges over a length of 35 m (115 feet). The slope angle should remain the same as it is now. The slope ratio is 1:1.5. The length from the bottom edge to the top edge is approximately 5.50 m (18 feet).
If this area is excavated as described, how many cubic meters (cubic yards) or tons of soil would be removed?
Do you need any additional information for the calculation?
I appreciate any help.
Thank you!
Best regards
If the same slope angle is to be recreated, I would virtually remove a rectangular block measuring 2.5m x 35m x 3.05m (8ft 2in x 115ft x 10ft) from the back (which will need to be disposed of) and shift the slope 2.5m (8ft 2in) backward.
For better visualization, I could lend a few building blocks from my child.
So, I don’t think the 755.9m³ (26,679ft³) figure is correct.
For better visualization, I could lend a few building blocks from my child.
So, I don’t think the 755.9m³ (26,679ft³) figure is correct.
B
Bieber081530 Nov 2016 16:58Musketier schrieb:
move the slope back by 2.5 m (8 feet 2 inches). This works great with building blocks, but in reality, a new slope has to be excavated. So, practically speaking:
- Excavate the old slope (volume of the triangular cross-section multiplied by 35 m (115 feet) length)
- Excavate the blocks (2.5 m (8 feet 2 inches) offset multiplied by height and length)
- Excavate the new slope (volume same as 1.)
I don’t want to recalculate again... it should be correct :P
Musketier schrieb:
If the same slope angle is to be maintained, I would virtually remove a cuboid measuring 2.5 m x 35 m x 3.05 m (8.2 ft x 115 ft x 10 ft) from the back (which must be disposed of) and shift the slope 2.5 m (8.2 ft) backward. I could lend a few building blocks from my child to help visualize it better.
So I don’t think the 755.9 m³ (26,679 ft³) is correct.Actually, it is correct. At first, you only create a hole where the slope can fit – but the slope itself is still in its original position, meaning it must be excavated and rebuilt on your remaining retaining wall.
One has to be careful to consider what volume is truly being moved – the total volume, when you excavate at the front and then reconstruct at the back, ends up being the same; which, of course, does not mean that the moved volume is zero.
Best regards,
Dirk Grafe
P
Painkiller1 Dec 2016 10:33Well, he’s not entirely wrong...
If you assume that you only excavate what needs to be removed, you would have a parallelogram with edge lengths of 2.5 m (8.2 ft) and 5.5 m (18 ft). To calculate the area, you need the height → 3.05 m (10 ft). That would be the rectangular volume taken out at the back.
That adds up to the mentioned 262.5 m³ (9,271 ft³).
You don’t need to excavate the entire slope with 4.5 m (15 ft) because it is only shifted backward by 2.5 m (8.2 ft).
If you assume that you only excavate what needs to be removed, you would have a parallelogram with edge lengths of 2.5 m (8.2 ft) and 5.5 m (18 ft). To calculate the area, you need the height → 3.05 m (10 ft). That would be the rectangular volume taken out at the back.
That adds up to the mentioned 262.5 m³ (9,271 ft³).
You don’t need to excavate the entire slope with 4.5 m (15 ft) because it is only shifted backward by 2.5 m (8.2 ft).
B
Bieber08151 Dec 2016 11:37Painkiller schrieb:
Parallelogram with edge lengths of 2.5 m (8.2 ft) and 5.5 m (18 ft) The base of the slope is 4.58 m (15 ft) long, plus an offset of 2.5 m (8.2 ft). Therefore, the parallelogram has a baseline of (4.58 + 2.5) m = 7.08 m (23.2 ft).
A sketch would help illustrate this...
P
Painkiller1 Dec 2016 12:07I think the original poster means something different because:
If he simply wants to move the ridge and base point 2.5 m (8 feet) backward, the baseline would also be only 2.5 m (8 feet)...
R3dDevil schrieb:
Top and bottom edges to be offset by 2.50 m (8 feet) over a length of 35 m (115 feet).
If he simply wants to move the ridge and base point 2.5 m (8 feet) backward, the baseline would also be only 2.5 m (8 feet)...
Similar topics