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How does one estimate the cubic yards of earth that need to be moved in excavating a pond? I am talking about a rectangular dug pond here with no dike etc.

I.E if a pond was 80 by 40 feet and the slope was 1:1 or 2:1 how would the amount of earth to be removed be calculated? Obviously the feet would be converted to yards, but I am not sure how proceed after that and account for the slope etc. There must be a formula or formula's used.


If pigs could fly bacon would be harder to come by and there would be a lot of damaged trees.






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How deep, Cecil?

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Length X width X depth will give you the volume of the whole area. If you use feet, divide by 27 to convert to cubic yards.
To find the volume of the sloped area use the formula: the square of c = the square of a + the square of b (where c is the length of the slope, a is the depth. Solve for b.
Then use (40 + 80) X (a X b)divided by 27. This figure will be the volume of the sloped area in cubic yards that must be subtracted from the volume of the whole area. The result of all that will be the answer to your question.
The problem can also be solved using trigonometric functions if you'd prefer that approach.

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Incidentally, Cecil, if you use a 1 to 1 slope, a and b will be = . If you use a 2 to 1 slope, b will be 2 x a. (remember, "a" is the depth).
So to calculate the volume of the sloped area with a 2 to 1 slope that would have to be subtracted from the total volume (measurements in feet) you could use (40 + 80) x (a x 2a). Divide that figure by 27 to get cubic yards.
If you decide on a 1 to 1 slope, use (40 + 80) x
(a x a). Divide that figure by 27 to get the amount to be subtracted.

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Dudley,

Max depth would be 7 feet. When you say "square of" do you mean square root of or squared?

40 feet X 80 feet X 7 feet = 22,400 cubic feet divided by 27 = 829 cubic yds. Of course it would be less than that due to the slope.

I was assuming 1:1 or 2:1 because the slope will be as steep as possible.

How do I proceed again?


If pigs could fly bacon would be harder to come by and there would be a lot of damaged trees.






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Treat the sloping sides as half of a solid
rectangle and subtract their volumn from
you 829 cu yds. ie, for 2:1 slope on sides :

Total yards :
829 - (14 * 80 * 7)/27 = 539 cu yds

Note: the volumn of one sloping side is
1/2 of the (14 * 80 * 7)

If you slope the ends also, then :
Total yards :
539 - (14 * 40 * 7 )/27 + 4*Vc = 410 cu yds

where Vc is the inclusive yardage in the
intersecting corners. This value can be
approximated in your this case at about
4 cu yrds.

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If 1:1 slope, 40ft + 80ft x 7ft x 7ft = 120ft x 49sq ft = 5880cu ft. Divide this figure by 27cu ft/cu yd. That would = 218 cu yd. Subtract this from 829cu yd and that would be the total cu yd removed (611cu yd)

If a 2:1 slope, 120ft x 7ft x 14ft = 11760cu ft. Dividedby 27 = 435cu yd. 829cu yd - 435cu yd = 393cu yd removed.

These calculations will put you within about 67cu yd of reality, Cecil, and should be close enough for an estimate. Because of the geometry of the corners, absolute (to the cu yd) calculations are much more exotic and would have to be done by a surveyor or an engineer ($$$).

Hope this helps.

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Dudley, I'm glad you pointed that out about the corners being irregular, I was worried someone would slip up on that. But Cecil I agree with Dudley, that should be close enough for a rough estimate. Plus most likely your corners will be rounded and not be the type of thing you get when two trangles intersect. Still a good estimate.

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Cecil, if you have figured it out from the previous posts, good. But I thought that I would splain it another way.

The formula that they are referring to is Pythagoreans theorem. Look it up on the web for more answers.

But, I think that there is an easier way.

If you are doing a 1 to 1 drop, imagine a square 10 feet by 10 feet, if it had a 1 to 1 drop from one corner to the other, the bisecting line would cut the square exactly in half. So, knowing that a 10 X 10 square is 100 square feet, half would be 50 feet. This can easily be converted to cubic feet.

In the same way, a 2 to 1 drop would take a 20 X 10 box and cut it in half, so the 200 square feet would be bisected into two 100 square foot sections.

So in your application, if you dig down 7 feet, in a 1 to 1 scenario, it would take 7 horizontal feet to get down 7 vertical feet. (Pythagorenans theorem says that the length of that slope would be 9.89 feet.) A 7 X 7 slope, running 10 feet long would be 7 x 7 x 10 = 490 cubic feet, the half you are removing is 245 cubic feet of dirt.

In the 2 to 1 scenario, it would take 14 horizontal feet to get down 7 vertical feet. (Pythagorean says the slope is 15.65 feet long.) A 14 x 7 x 10 cube = 980 cubic feet, taking out half; you are moving 490 cubic feet of dirt. But the flat bottom portion of the pond will be smaller. (By 7 feet on each side) so you have to calculate in a savings there.


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Thanks guys! I get it now! I do remember some of this vaguely from geometry in high school!

Thanks!


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Cecil, I've been thinking about this problem for the past month, searching for a simpler approach to calculating excavated yardage involving square and rectangular ponds. How about this approach:
Consider a point midway the sloped edges. At that point, the volume of dirt removed from the upper half will be very nearly the same as the volume of dirt remaining in the lower half of the sloped area. (because of the corners, the volumes won't be exactly equal, but in the interest of simplicity and because we're estimating, we'll ignore that minor difference).
In the case of the 1:1 slope, with a seven foot pond depth, the dimensions of the pond at mid-slope would be seven feet less than the full pond dimensions (3 1/2ft on each side). A 40ft x 80ft pond would measure 33ft x 73ft at mid-slope. If we use those dimensions as the basis for our estimate, we would have 33ft x 73ft x 7ft = 16863cu ft. Dividing this figure by 27 cu ft/cu yd will give us 625 cu yd to be removed.
With a 2:1 slope, the dimensions of the pond at mid-slope would be 14ft less than surface dimensions. 26ft x 66ft x 7ft = 12012cu ft. Dividing by 27 = 445cu yd.
This approach seems accurate to me, Cecil, and certainly easier to work with than all that squaring stuff.
I hope that Squeeky, Tritonvt, and Nick will offer their opinions.
As a reminder, if you're loading into trucks, the volume of the dirt can increase by up to 1/3 because of decompaction.

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I forgot to give you the working formulas, Cecil.
Where L = the length of the pond, W = width, and D = depth (all measurements in feet)
For a 1:1 slope: (L - D)x(W - D)xD divided by 27 = volume in cu yds to be removed.
For a 2:1 slope: (L - 2D)x(W - 2D)xD divided by 27 = cu yds to be removed.
For a 1.5:1 slope: (L - 1.5D)x(W - 1.5D)xD divided by 27 = cu yds to be removed.
For a 3:1 slope: (L - 3D)x(W - 3D)xD divided by 27 = cu yds to be removed.
I'm sure you get the picture. Now ain't that simpler than all that other stuff?
Best regards.

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Dudley you're my do right hero! \:D

Thanks for the info. Next time an excavator gives me an estimate I can show him that not only do I know what a core trench is (many around here don't have a clue), but I can dazzle him by already knowing how many cubic yards of earth need to be removed! \:\)


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Yes Dudley, you are actually computing the
volumn of a rectangular solid whose sides
extend through the mid-slope points of the
excavation. Ignoring the geometry of the corners,
You have come up with a niffty way to approximate the excavation volumn.

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Cecil,

Here is a website I found with volume calculators. If they don't have what you need, an internet search should be able to find others.

volume calculator

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Ah, Kevin. That is a fine site. With a tear and a smile, let us gratefully and eagerly embrace the new.


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