**Student Land Surveyors UNITE!** This hub is for surveying in schools and student surveyor support on Land Surveyors United. Join this group to share resources and tips for students of land surveying. In this Hub for Young Surveyors you can ask questions to both your fellow student surveyors as well as Educators inside the Land Surveyors United Community. Start a Discussion and Introduce yourself.

Tell us what you are learning and keep an on-going log of everything you learned!

**Things You Can Do in Student Surveyors Hub:**

- Share and Compare Class Notes
- Record and Stream Classroom Sessions
- Discuss What Other Students are Learning around the world
- Share Photos from Surveying Class
- Collaborate on Projects
- The Sky is the Limit!

Professors will soon have the ability to host and administer their entire classroom inside a hub of their own. Stay tuned!

## Replies

abdur,

Perhaps the terminology is at fault, but I'm not sure what you are suggesting different. Each of the areas is calculated using the area of each section. The term 'average' refers to interpolating that the area changes from one section to the next in a way that we can only assume is linear because we don't have any intermediate information (so we can only average the two sections together over that station length).

The average end area method can produce significant errors particularly where the cross sections transition from 0. (eg pyramidal volume).

My favourite volume calculation is:-

V = distance*(A1 + sqrt(A1*A2) +A2)/3

From 0m to 40m in Rick's solution above.

the volumes become:-

00-20m V=20(0 + sqrt(26.32*0) +26.32)/3 = 175.47

20-40m V=20(26.32 + sqrt(26.32*75.62) +75.62)/3 = 977.02

Volume 00-40m = 1152.49m^2

In this case the formula treats volume 0-20 as a pyramid d(0+A2)/3 which is probably understating the volume.

But the section 20-40 is treated as a truncated pyramid and is likely closer to the volume calculated by Simpsons Rule.

This method gives similar results to Simpsons Rule in general calculation and is particularly useful when calculating volumes of truncated irregular prisms like the volume between contours.

Dear All,

Average method is ok for the cross section quantity but as i strongly believe that if we make a close figer of cross section its will produce the more accurate result for the quantity.

Malik,

Rich Maher said:

I'm in agreement with Rich.

Looking at the problem, you need to get all the relevant data. Any reliable calc is based on good data. To do this with any method you need the station intervals (0, 20, 40 ect), the offsets of your levels (I see you used fixed offsets - or at least the distance between the shots(10m shots)).

Using average ends method is the simplest, as Rich mentioned, you need to plot (do this by hand whilst your a student - this way you grasp the extent of what your trying to achieve) the ground line on the design. Calc your side slopes and draw them in. Do this for every section.

You will use the sections in pairs. First 0 and 20. What I did was to get the section area by coordinating using excel using a cartecian layout. Using the area by coorinates method you determine the area cut and area fill for every section.

Now the cut volume from Ch0 to Ch20 = (Cut area @0 + Cut area @20) / 2.0

If there was fill from Ch0 to Ch20 = (Fill area @0 + Fill area @20) / 2.0

The next pair will be from 20 to 40 is (Cut area @20 + Cut area @40) / 2.0

And the same for the fill.

This process is repeated for all the section pairs.

The sum of the cut values Sigma n=(the be number of sections - 1) is the total value for the cutting.

The tricky part is the area by coordinates for the sections. Be carefull - there can be multiple cut and fill areas per section in more complex designs.

Always plot your sections (Even if you do it with some sort of software) and here is the important part - LOOK AT THEM. Be sure that everything went right with the ground line and design line. Once you have checked this, run the volume calc and report the volume.

It appears that Kendall is willing to run your project or homework through his earthwork software ... definitely a valid way to get to an answer. Most of us in our day to day business do it this way ... having tested our software, found it accurate, and trust it with minimal questions but occasional checks.

I think it is interesting to, when able, test our software with some "hand" methods. I always ask my employees if things pass the reasonability test (aka stupid test) and use that as an opportunity to teach the underlying math behind our programs or some solid traditional methods.

You've asked for a solution, not an answer, so I'll give you an idea. Let me know if this makes sense and works for you. If Kendall will run this through his software, it would be nice to get a comparative answer ... otherwise I'll do it for you; after you throw up an answer the old way.

I understand your premise this way: You have provided an existing surface, collected as station, outs, and elevations. Along this route you have a template to be cut that includes side slopes.

You are looking for the cut/fill volumes along this route applying the proposed template to the existing surface. I think all the information needed for the calculation is provided.

By hand I would do use the Average End Area Method. You have sections of the existing surface and you can create sections of the proposed surface. Since your existing surface is only at 20 meter intervals, there isn't really anything much gained by evaluating the math at a smaller interval. So we only have 5 sections we need to work with.

The volume you are looking for is generated by multiplying your cut area drawn (m^2) by length (m) to get the volume (m^3). You need to get the area of each of the 5 sections. Since the area changes from section to section, we assume that the change is linear and we use the average area for the reach we are evaluating (the genesis of the name of the method).

Draw Section 0 (existing and proposed) and calculate the area of cut.

Draw Section 20 (existing and proposed) and calculate the area of cut.

You then average the area at each end (add them and divide by two). This is the average cross section area of the 20 meter reach that you multiply by 20m to get the volume.

Repeat this action for 20-40, 40-60, and 60-80. The sum of the cut volumes is your answer for the entire reach.

To calculate the areas you draw ... draw cross-sections by hand and use a planimeter, draw sections in CAD and use the area command, turn the sections into coordinates and use the Area Coordinate method.

So, Noor ... let's have your hand answer and one of us will give you a verification the easy way.

Rich

and what i understand is that only volume of excavation is needed in all offsets , so back filling is not necessary

I am really new in this and the way you explaining is quite difficult for me

I dont want software solution also

step by step guide will be really appreciated

Thanks

How are you doing? Have you given some of our suggestions a shot?

I'll give you a head start if you haven't ... I'm using a CAD program to calculate my area. As I spoke of earlier, you can use a planimeter, calculate it rigorously, or with some graph paper, even estimate it by counting squares and estimating partial squares. I'll discuss the volume here, but we can branch off and discuss the area calculation if necessary.

Attached you'll find a sketch showing the sections at 0, 20, and 40 meters including the area I inquired using an "area" command in CAD. The sections are drawn 1:1 vertically to horizontally.

For the average end area method, we assume the transition of cut area from section to section occurs in a linear fashion and if we use the average, we can take the average section area and drag it the distance between sections to get a volume.

Area at 0 (0m^2) + Area at 20 (26.32m^2) all divided by 2 (average) is 13.16m^2

13.16m^2 over a reach of 20m is 263.2m^3 (cubic meters)

Area at 20 (26.32m^2) + Area at 40 (75.62m^2) all divided by 2 (average) is 50.97m^2

50.97m^2 over a reach of 20m is 1,019.4m^3 (cubic meters)

So from station 0 to station 40 we have a cut volume of 1,282.6 cubic meters

Two more sections to get the cut areas, two more average sections to multiply the area by the length, and then you add those to the sum and there is the answer.

When you get done, do some basic checks and make sure there aren't any blunders ... a reasonability check.

Rich

Road : lsu

------ XS area ------ ------ Volume ------ ---- Accumulated ----

Statn Cut Fill Cut Fill Cut Fill

Meters Sq m Sq m Cu m Cu m Cu m Cu m

0.000 0.00 0.00 0.00 0.00 0.00 0.00

10.000 13.33 0.00 66.67 0.00 66.67 0.00

20.000 26.32 0.00 198.29 0.00 264.96 0.00

30.000 50.21 0.00 382.67 0.00 647.63 0.00

40.000 75.62 0.00 629.12 0.00 1276.75 0.00

50.000 96.37 0.00 859.91 0.00 2136.66 0.00

60.000 118.68 0.00 1075.24 0.00 3211.89 0.00

70.000 140.31 0.00 1294.96 0.00 4506.86 0.00

80.000 162.25 0.00 1512.78 0.00 6019.64 0.00