1. ## Cut/Fill calcs (Non CAD)

Registered forum members do not see this ad.

Working on a surveying assignment doing cut fill calcs.

I have 5 ordinates (cross sections along 5 grid lines). Gridlines spaced 10m apart.

With areas of my ordinates calculated what is the more accurate method of calculating area.

Simpson's Rule or find the average area (sum of all areas/5) and times by 40m?

Thanks

2. "Survey assignment"? Is this for a class? I presume you are trying to calculate a volume, and not an area?

Did the instructor give you any specific instructions?

Simpson's Rule might give you a slightly better approximation in some cases, but how much better? Usually, at least in the real world, the difference isn't worth worrying about. So generally, average end area is used in this type of volume calculation, since it's simpler. And judging from the results I've seen, I believe the Corridor material volumes in Civil 3D are calculated using average end area.

3. Simpson's rule is more accurate than the average end area method , which is more accurate than averaging all 5 times 40. Averaging all 5 gives equal weight to the areas at the far ends.

4. Sinc - You're right I meant volume and this part of an assignment due in on Thursday.

I was skeptical about the accuracy of Simpson's Rule with only 5 ordinates though. Are you sure this is the case?

We can approach this however we like. However, I've tried both methods and my results are not close enough for me to feel confident about using either method due to the relatively low quantitiy of material involved.

Distance between the 5 ordinates = 10m
Simpson's
10/39((0.65+12.34) + 4(0.61+5.79) + 2(2.16)) = 143m cubed to 0 dp

Average:
40((0.65+0.61+2.16+5.79+12.34)/5) = 172m cubed to 0dp

edit - Just re-read your post CarlB, if what you say is correct then using average end method should give a result closer to the Simposons result rather than the average overall method. However using this method givess = 259.8m Cubed.

5. What do you mean, your "results are not close enough"? Do you mean that the various methods create results that are not what you consider to be close enough to each other?

If so, then the problem may be your data set. It's the old rule of Garbage In = Garbage Out. If you do not have adequate survey data, it doesn't matter which method you use. Meanwhile, as the number of sample points go up, the different calculation methods tend to converge on the same result.

You might also want to go back and check your math. From the data you've given in this thread, I come up with 151 cubic meters, using the average end area method.

6. The areas of my ordinates are correct and checked. I have used all the available data as per our practical grid survey.

I just assumed that the average end method was calculated like:

1/2(1st ordinate + 2nd ordinate) x 40

but due to your result, I guess not! If I had done it correctly then 151 cubic meters when compared with my results would follow what CarlB had suggested.

I'll stick with Simpsons.

7. They're simply different methods of computing the average area. Hopefully this should lay out the difference between the three methods...

The first method simply gets an area at each sample point, averages them, and multiplies by the total distance. In other words, we can come up with an average area like this:

average area = (a1 + a2 + a3 + ... + an) / n

We then take that average area and multiply by the total length, to come up with an approximate volume.

In this particular example, the average area method yields:

40 * (0.65 + 0.61 + 2.16 + 5.79 + 12.34) / 5 = 172 m^3

The problem with this method is that it puts too much weight on the two ends (e.g., the first and last sections). All of the other sections are both the start of one segment, and the end of another segment. And of course, this method also assumes that the length of each segment is identical, which is not always the case in the real world.

The Average End Area method corrects for this problem. In this method, we look at each segment individually, so that we don't apply too much weight to the first and last areas. Strictly speaking, we should calculate each segment individually. For example, the volume of the first section would be calculated by:

segment volume = [(start area + end area) / 2] * dist

If the distance between each section is the same (as in your example), then we can combine terms, and come up with a formula to calculate the average area that lets us easily see the difference between the previous method and this method:

average area = (a1/2 + a2 + a3 + ... + an/2) / (n - 1)

Once again, we can multiply this average area by the total length. In this particular case, the average end area method yields:

40 * ((0.65/2) + 0.61 + 2.16 + 5.79 + (12.34/2)) / 4 = 151 m^3

When we get to Simpson's Rule, we're actually taking advantage of the fact that, in situations like this, we generally have a smooth curve between our sample points. In other words, in real life, we don't have a series of angle points between 10m stations. So Simpson's Rule applies a different approximation, which partially accounts for the fact that there is probably a smooth curve between data points, yielding an even more accurate approximation.

In order to do this, Simpson's Rule needs to know the value at the midpoint of each station. It then weights the midpoint more than the edges, to come up with an approximation of the volume for a parabola. In this method, we're still doing nothing but computing a weighted average, but this time the formula looks more like this:

segment volume = [(start area + 4 * midpoint area + end area) / 6] * dist

In this particular example, we don't know the section area at the midpoints of each range. But since each range is the same size, we can combine adjacent ranges. The end result is we have half as many segments, but in most cases, using the parabolic area comes up with a more-accurate volume, simply because it is usually more reflective of real-world conditions than a series of angle points. This time, we end up with this:

average area = (a1 + (4 * a2)+ (2 * a3)+ (4 * a4) + a5) / 12

Once again, we can multiply this average area by the total length. In this particular case, the Simpson's Rule method yields:

40 * (0.65 + (4*0.61) + (2*2.16) + (4*5.79) + 12.34) / 12 = 143 m^3

8. Summed up very well. Thanks.

9. I use the average end area system, as that is the one often used by local area dirt contractors. Need to be consistant with their desires or this creates an ambiguity when payment for services is issued. Use what they use.

It also is easy to load up into some Excell program and run, then compare notes with a contractor to see who is right in their numbers days later on.

The Simpson system may be accurate, but is not exactly the same system that the construction firms use around here.

Wm.

10. Registered forum members do not see this ad.

Originally Posted by Coosbaylumber
The Simpson system may be accurate, but is not exactly the same system that the construction firms use around here.
It wouldn't be, would it? They (the construction firms) are looking to get maximum money. They have not got time for accuracy. The measurement of earth moving quantities is not an exact science, so any slight misdirection, in favour of extra money, is used until someone is brave enough to point out the fallacies.

#### Posting Permissions

• You may not post new threads
• You may not post replies
• You may not post attachments
• You may not edit your posts