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
B
Bieber081529 Nov 2016 13:31The cross-section of the slope is a right triangle. The hypotenuse (the part of the slope visible from the outside) measures 5.5 m (18 feet). We also know that the lengths of the legs are in a ratio of 1:1.5 (this is how I interpret your statement).
1.) a² + b² = c² with c = 5.5 m (18 feet)
2.) a : b = 1 : 1.5
Two equations, two unknowns. This results in a = 3.48 m (11.4 feet), b = 4.26 m (14 feet), assuming I haven’t made a mistake.
The slope volume follows from the triangular area and the length of 35 m (115 feet), totaling 259.4 m³ (9,160 ft³).
Setting back by 2.5 m (8.2 feet) means additional earth movement behind the slope, so 35 m (115 feet) * 2.5 m (8.2 feet) * 3.48 m (11.4 feet) = 304.5 m³ (10,760 ft³) (assuming the height of the slope is 3.48 m (11.4 feet), depending on how you define your ratio).
No guarantee.
1.) a² + b² = c² with c = 5.5 m (18 feet)
2.) a : b = 1 : 1.5
Two equations, two unknowns. This results in a = 3.48 m (11.4 feet), b = 4.26 m (14 feet), assuming I haven’t made a mistake.
The slope volume follows from the triangular area and the length of 35 m (115 feet), totaling 259.4 m³ (9,160 ft³).
Setting back by 2.5 m (8.2 feet) means additional earth movement behind the slope, so 35 m (115 feet) * 2.5 m (8.2 feet) * 3.48 m (11.4 feet) = 304.5 m³ (10,760 ft³) (assuming the height of the slope is 3.48 m (11.4 feet), depending on how you define your ratio).
No guarantee.
P
Painkiller30 Nov 2016 12:55Since the slope is meant to be restored at the end, I would use a parallelogram for the calculation.
If the slope has a ratio of 1:1.5 (a to b), then 3.48 and 4.26 are incorrect, as that would correspond to a ratio of 1:1.22.
I calculate the height of the slope to be approximately 3m (10 feet).
The entire structure is planned to be set back by 2.5m (8 feet 2 inches).
This results in a cross-sectional area of 7.5m² (81 square feet).
Over a length of 35m (115 feet), the volume would then be 262.5m³ (9,270 cubic feet).
If the slope has a ratio of 1:1.5 (a to b), then 3.48 and 4.26 are incorrect, as that would correspond to a ratio of 1:1.22.
I calculate the height of the slope to be approximately 3m (10 feet).
The entire structure is planned to be set back by 2.5m (8 feet 2 inches).
This results in a cross-sectional area of 7.5m² (81 square feet).
Over a length of 35m (115 feet), the volume would then be 262.5m³ (9,270 cubic feet).
B
Bieber081530 Nov 2016 14:11You are absolutely right! I clearly forgot to square the 1.5 in the calculation. The height of 3.05 m (10 feet) is correct for a slope ratio of 1:1.5. The depth of the slope is then 4.58 m (15 feet).
The 7.5 m² (81 ft²) apparently results from multiplying the height by 2.5 m (8 feet). However, the volume that needs to be moved includes not only this rectangular cross-section but also the slope itself.
The above method (slope plus backfill area, rectangular) with the corrected figures gives 244.5 m³ (8,630 ft³) + 266.9 m³ (9,420 ft³) = 511.4 m³ (18,050 ft³). It is correct that the slope must be counted twice: 511.4 + 244.5 = 755.9 m³ (26,670 ft³).
The parallelogram (good idea) has a cross-sectional area of 21.6 m² (233 ft²) (slope depth of 4.6 m (15 feet) plus 2.5 m (8 feet) offset), resulting in a total volume of 756 m³ (26,680 ft³).
The 7.5 m² (81 ft²) apparently results from multiplying the height by 2.5 m (8 feet). However, the volume that needs to be moved includes not only this rectangular cross-section but also the slope itself.
The above method (slope plus backfill area, rectangular) with the corrected figures gives 244.5 m³ (8,630 ft³) + 266.9 m³ (9,420 ft³) = 511.4 m³ (18,050 ft³). It is correct that the slope must be counted twice: 511.4 + 244.5 = 755.9 m³ (26,670 ft³).
The parallelogram (good idea) has a cross-sectional area of 21.6 m² (233 ft²) (slope depth of 4.6 m (15 feet) plus 2.5 m (8 feet) offset), resulting in a total volume of 756 m³ (26,680 ft³).
P
Painkiller30 Nov 2016 14:52@Bieber0815: That's right, I overlooked the slope and only considered the rectangular part. So your 755.9 m³ (27,672 ft³) are correct.
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