Issue with Volume in Elevation Table

[Anonymous]

Hi everyone,

Please see attached DWG file. As you can see, the volume does not match up with the elevation.  For example the volume between elevation = -0.2m to elevation = 0 is 4,786.70 m3. The 3D surface area for the elevation range is 8,990.75 m2.

Even if we assume that all cut depth for it is 0.2m, then the maximum volume would be 8,990.75*0.2 = 1,798.15 m3 not 4,786.70 m3!!!

Any idea why the shown volume is not in line with what it has to be logically?! Please correct me if I am wrong.

Please note that if you sum all shown volume in the table, it will be 14,935.86 which is exactly the same that you can see in the cut/fill table as per below. Thanks.

[Anonymous]

Hi everyone,

Please see attached DWG file. As you can see, the volume does not match up with the elevation.  For example the volume between elevation = -0.2m to elevation = 0 is 4,786.70 m3. The 3D surface area for the elevation range is 8,990.75 m2.

Even if we assume that all cut depth for it is 0.2m, then the maximum volume would be 8,990.75*0.2 = 1,798.15 m3 not 4,786.70 m3!!!

Any idea why the shown volume is not in line with what it has to be logically?! Please correct me if I am wrong.

Please note that if you sum all shown volume in the table, it will be 14,935.86 which is exactly the same that you can see in the cut/fill table as per below. Thanks.

[ChrisRS · Accepted solution]

I don't know why you PMed me, but I'll take the challenge!

Good question.

This topic does not seem to be very well documented.

It is a good idea to verify that we correctly understand how the software we use works, and that is working properly. 

The areas listed in the table are the surface area within each elevation range. It is the area between 2 contours,  not the area within a specific contour. 

For the -0.2 to 0.0 range the green shaded area = 8872.55. This is the area between the -0.2 and 0.0 contours.

To do volume calculations, the areas within contours are more useful.

For the volume between -0.2 and 0.0 can be estimated using:

     the area within the -0.2 contour = 19945 (Red+Red-orange+Orange+Yellow-orange+Yellow) 

     and the area within the 0.0 contour = 28818 (Red+Red-orange+Orange+Yellow-orange+Yellow+Green)

Using the conic method the interval volume = (19945 + 28818 + SQRT(19945*28818))*0.2/3 = 4849

This is in good agreement with the volume of 4786.70 in the table.

 

The fact that the manually computed values are not exact matches to the table values is to be expected. 

The table appears to create ad hoc TIN Volume Surfaces for each interval. This accounts for minor variations in the surfaces. This method is more accurate than "end area" estimating methods. I am somewhat surprised that they are this close. 

FYI:

Good luck.

[ChrisRS · Accepted solution]

I don't know why you PMed me, but I'll take the challenge!

Good question.

This topic does not seem to be very well documented.

It is a good idea to verify that we correctly understand how the software we use works, and that is working properly. 

The areas listed in the table are the surface area within each elevation range. It is the area between 2 contours,  not the area within a specific contour. 

For the -0.2 to 0.0 range the green shaded area = 8872.55. This is the area between the -0.2 and 0.0 contours.

To do volume calculations, the areas within contours are more useful.

For the volume between -0.2 and 0.0 can be estimated using:

     the area within the -0.2 contour = 19945 (Red+Red-orange+Orange+Yellow-orange+Yellow) 

     and the area within the 0.0 contour = 28818 (Red+Red-orange+Orange+Yellow-orange+Yellow+Green)

Using the conic method the interval volume = (19945 + 28818 + SQRT(19945*28818))*0.2/3 = 4849

This is in good agreement with the volume of 4786.70 in the table.

 

The fact that the manually computed values are not exact matches to the table values is to be expected. 

The table appears to create ad hoc TIN Volume Surfaces for each interval. This accounts for minor variations in the surfaces. This method is more accurate than "end area" estimating methods. I am somewhat surprised that they are this close. 

FYI:

Good luck.

[Anonymous]

Hi Chris,

Thank you so much for your message.

It clarified most of my questions.

Just another question; I need to know what is 2D/3D surface area WITHIN (not between) elevation=0 to elevation=-0.2 in the TIN volume surface.

In another word, I need to know what will be the total cut area which its depth is <=0.2m.

Could you please assist me with that? Many thanks.

K1

[Anonymous]

Hi Chris,

Thank you so much for your message.

It clarified most of my questions.

Just another question; I need to know what is 2D/3D surface area WITHIN (not between) elevation=0 to elevation=-0.2 in the TIN volume surface.

In another word, I need to know what will be the total cut area which its depth is <=0.2m.

Could you please assist me with that? Many thanks.

K1

[Anonymous]

Hi Chris,

Thank you so much for your message.

It clarified most of my questions.

Just another question; I need to know what is 2D/3D surface area WITHIN (not between) elevation=0 to elevation=-0.2 in the TIN volume surface.

In another word, I need to know what will be the total cut area which its depth is <=0.2m.

Could you please assist me with that? Many thanks.

K1

[Anonymous]

Hi Chris,

Thank you so much for your message.

It clarified most of my questions.

Just another question; I need to know what is 2D/3D surface area WITHIN (not between) elevation=0 to elevation=-0.2 in the TIN volume surface.

In another word, I need to know what will be the total cut area which its depth is <=0.2m.

Could you please assist me with that? Many thanks.

K1

[ChrisRS · Accepted solution]

I am not sure that I fully understand your question, but I will try to answer.

Using your example/table:

The total cut area WITHIN the 0.0 to -0.2 range is 8873 m2 (2D) - the green area

The total cut area BETWEEN the 0.0 contour and the -0.2 contour is also 8873 m2 (2D) - the green area.

I suggest that you ignore the 3D area of Tin Surface volumes.

To me being within a range or being between contours is the same thing.

Being within a contour is confusing.

I would interpret the cut area within -1.0 to be the sum of the red + red-orange + orange + yellow-orange areas. You start cutting at the highest peaks.

I would interpret the fill area within 1.0 to be the sum of the black + dark-blue + blue + medium-blue areas. You start filling at the lowest holes.

Regarding the 2D/3D area issue, refer to the illustration below.

The green area represents the 2D planimetric area. That is the area projected to a flat plane. In you 0.0 --0.2 example the elevation of the plane would be 0.0 feet of cut.

The red area represents the 3D "surface area" of the tin volume surface.* The 3D area reflects the slope and ups and down of the surface. As you see in the cross sections on the left side of the illustrations, the red lines are longer that the green lines, the 3D area will always be larger than the 2D area.

* It is important to realize that the slope of a Tin Volume surface is not necessarily representative of the slope of the finished grade surface. I see little use for the 3D area calulations for tin volume surfaces.
With steep slopes of the finished grade, the 2D vs 3D area difference can be substantial. The surface area of a 1:1 slope (100%) is 41% greater that a flat slope. If you are taking off quantities for turf or rip-rap this can be significant. Flater slopes are less critical. The surface area of a 5:1 slope (20%) is 2% greater that a  flat slope. 

I hope this answers your questions.

[ChrisRS · Accepted solution]

I am not sure that I fully understand your question, but I will try to answer.

Using your example/table:

The total cut area WITHIN the 0.0 to -0.2 range is 8873 m2 (2D) - the green area

The total cut area BETWEEN the 0.0 contour and the -0.2 contour is also 8873 m2 (2D) - the green area.

I suggest that you ignore the 3D area of Tin Surface volumes.

To me being within a range or being between contours is the same thing.

Being within a contour is confusing.

I would interpret the cut area within -1.0 to be the sum of the red + red-orange + orange + yellow-orange areas. You start cutting at the highest peaks.

I would interpret the fill area within 1.0 to be the sum of the black + dark-blue + blue + medium-blue areas. You start filling at the lowest holes.

Regarding the 2D/3D area issue, refer to the illustration below.

The green area represents the 2D planimetric area. That is the area projected to a flat plane. In you 0.0 --0.2 example the elevation of the plane would be 0.0 feet of cut.

The red area represents the 3D "surface area" of the tin volume surface.* The 3D area reflects the slope and ups and down of the surface. As you see in the cross sections on the left side of the illustrations, the red lines are longer that the green lines, the 3D area will always be larger than the 2D area.

* It is important to realize that the slope of a Tin Volume surface is not necessarily representative of the slope of the finished grade surface. I see little use for the 3D area calulations for tin volume surfaces.
With steep slopes of the finished grade, the 2D vs 3D area difference can be substantial. The surface area of a 1:1 slope (100%) is 41% greater that a flat slope. If you are taking off quantities for turf or rip-rap this can be significant. Flater slopes are less critical. The surface area of a 5:1 slope (20%) is 2% greater that a  flat slope. 

I hope this answers your questions.

[Anonymous]

Hello Chris,

Thank you for your prompt reply.

The purpose of my post from beginning was to calculate the extra cut volume on top of total cut volume(=14,935.86 m3) that we need to do on site for Topsoil.

Topsoil is the cutting whole site regardless if it is cut or fill. Topsoil depth is 0.2m. Assume we need to export whole topsoil from site.

My method to calculate extra volume for topsoil is as below. Please advise if it is not correct.

To calculate topsoil volume in fill area, I have to multiply 2D total area of fill (=35,915.18) * 0.2m = 7,183.036 m3

So, for the fill area, first we need to cut 7,183.036 m3 which is topsoil and then fill 7,183.036 m3+25635.09m3 = 32,818.126 m3. So, 32818.126 m3 is the total volume of fill without considering any inflation or compaction ratio.

With cut area, the total volume of topsoil is 28,818.04*0.2m = 5,763.608 m3. But majority of this volume is already within total cut volume(=14,935.86 m3). No need to add anything for cut area >0.2m depth cut.

But, For the cut area <=0.2m, Am I right to say that the total topsoil volume is 8872.55*0.2 = 1,774.51 m3? 

If yes, this volume (assuming whole area has been cut for 0.2m not less) is less than table volume(=4,786.70m3)!! So, something is wrong. How can I calculate the extra cut volume for topsoil for cut depth<=0.2m?

Regardless of my above assumption, could you please help me to find what I need to add to total cut volume (=14,935.86 m3) to make sure that Topsoil volume is allowed for cut depth<=0.2m?

Best regards,

K1

[Anonymous]

Hello Chris,

Thank you for your prompt reply.

The purpose of my post from beginning was to calculate the extra cut volume on top of total cut volume(=14,935.86 m3) that we need to do on site for Topsoil.

Topsoil is the cutting whole site regardless if it is cut or fill. Topsoil depth is 0.2m. Assume we need to export whole topsoil from site.

My method to calculate extra volume for topsoil is as below. Please advise if it is not correct.

To calculate topsoil volume in fill area, I have to multiply 2D total area of fill (=35,915.18) * 0.2m = 7,183.036 m3

So, for the fill area, first we need to cut 7,183.036 m3 which is topsoil and then fill 7,183.036 m3+25635.09m3 = 32,818.126 m3. So, 32818.126 m3 is the total volume of fill without considering any inflation or compaction ratio.

With cut area, the total volume of topsoil is 28,818.04*0.2m = 5,763.608 m3. But majority of this volume is already within total cut volume(=14,935.86 m3). No need to add anything for cut area >0.2m depth cut.

But, For the cut area <=0.2m, Am I right to say that the total topsoil volume is 8872.55*0.2 = 1,774.51 m3? 

If yes, this volume (assuming whole area has been cut for 0.2m not less) is less than table volume(=4,786.70m3)!! So, something is wrong. How can I calculate the extra cut volume for topsoil for cut depth<=0.2m?

Regardless of my above assumption, could you please help me to find what I need to add to total cut volume (=14,935.86 m3) to make sure that Topsoil volume is allowed for cut depth<=0.2m?

Best regards,

K1

[ChrisRS · Accepted solution]

I am on deadline so this will be quick.

Your explanation of the fill area methodology seems correct. I did not check the numbers.

Regarding the fill areas: 

 0.0 .           .

      \         / Your volume accounts for this area.

-0.2   \_______/

 0.0 .          .

     |\        /| You need to account for the approximately 

-0.2 | \      / | triangular areas around the perimeter.

A good approximation is: (2D Area between 0.0 and -0.2) x H / 2

In this case H = (0.0) - (-.02) = 0.2

(reasonableness check: is this a small percentage of the overall cut? if so you are good)

A slightly different work flow may be easier to follow and more accurate:

1. You already have EG and FG.
2. Create a new surface: CLEAR
   1. Paste EG into CLEAR
   2. Edit Clear to lower 0.2 m
3. Create a new Tin Volume surface: STRIP
   Base = EG; Comparison = CLEAR;
   Volume = Material to be cleared and hauled out. 
4. Create a new Tin Volume surface: CUT-FILL
   Base = CLEAR; Comparison = FG;
   Cut Volume = Extra Material to be cut below the cleared top soil.
   Fill Volume = Material to be filled for the cleared top soil up to FG.

[ChrisRS · Accepted solution]

I am on deadline so this will be quick.

Your explanation of the fill area methodology seems correct. I did not check the numbers.

Regarding the fill areas: 

 0.0 .           .

      \         / Your volume accounts for this area.

-0.2   \_______/

 0.0 .          .

     |\        /| You need to account for the approximately 

-0.2 | \      / | triangular areas around the perimeter.

A good approximation is: (2D Area between 0.0 and -0.2) x H / 2

In this case H = (0.0) - (-.02) = 0.2

(reasonableness check: is this a small percentage of the overall cut? if so you are good)

A slightly different work flow may be easier to follow and more accurate:

1. You already have EG and FG.
2. Create a new surface: CLEAR
   1. Paste EG into CLEAR
   2. Edit Clear to lower 0.2 m
3. Create a new Tin Volume surface: STRIP
   Base = EG; Comparison = CLEAR;
   Volume = Material to be cleared and hauled out. 
4. Create a new Tin Volume surface: CUT-FILL
   Base = CLEAR; Comparison = FG;
   Cut Volume = Extra Material to be cut below the cleared top soil.
   Fill Volume = Material to be filled for the cleared top soil up to FG.

[kostial.mk]

Hello @ChrisRS,

I came across this topic while wondering about the same "inaccuracy" in the elevation table as kkhcivil. I want to confirm, that I understand it right.

What I expected (and I understand kkhcivil expected the same) was the volume for case B in picure bellow, i.e. approximately the area between contours x average elevation of contours. If I understand what you wrote, in the table is really listed the volume of case A, i.e. the volume between two plates with elevations of contours. Correct?

If so, I'd like to ask, if there is an easy way of calculating the volume of case B, without the necessity of slicing the surface manualy and calculating volume of each of those partial surfaces.

Thank you in advance!

[kostial.mk]

Hello @ChrisRS,

I came across this topic while wondering about the same "inaccuracy" in the elevation table as kkhcivil. I want to confirm, that I understand it right.

What I expected (and I understand kkhcivil expected the same) was the volume for case B in picure bellow, i.e. approximately the area between contours x average elevation of contours. If I understand what you wrote, in the table is really listed the volume of case A, i.e. the volume between two plates with elevations of contours. Correct?

If so, I'd like to ask, if there is an easy way of calculating the volume of case B, without the necessity of slicing the surface manualy and calculating volume of each of those partial surfaces.

Thank you in advance!

[ChrisRS]

Yes, the shows "A".

If I understand you illustration, for "B",. you want the ring volume from datum (bottom of Red) to surface (Blue).

It is not simple, nut can be done. Bounding surgacesesurgaces maye work. I haveno time oww, but will check over the weekend.

[ChrisRS]

Yes, the shows "A".

If I understand you illustration, for "B",. you want the ring volume from datum (bottom of Red) to surface (Blue).

It is not simple, nut can be done. Bounding surgacesesurgaces maye work. I haveno time oww, but will check over the weekend.

[Anonymous]

My comment on the mismatch of the color area of the layer according to the depth specified in the table with the size of each layer in the Civil 3D

My analysis of this is that; the Volume is correct, but the areas of colors are incorrect according to the actual area that should be displayed (except for the inner color only), as the Civil 3D program reads the color areas, which is naturally visible as a Horizontal Projection, which should be understood as follows:

It is that the inner color is only the color that includes the real area, then the real area of ​​the next color includes (the Area of the same color + the Area of the previous color).

Where all colors cannot be represented on a Horizontal Projection, it must be a three-dimensional projection.

To make sure of this, take color area of last layer by the value next to the color of the penultimate layer, and lower the surface to the required depth you will find that the quantity is exactly the same as the table quantity and you will find also the real area has become logical.

Or in another way, specify in the table only the beginning and end of the required layer depth, and you will find the same result.

And therefore; The problem is that it is difficult to show the required layer color area on a Horizontal Projection and not the problem with the quantities.

[Anonymous]

My comment on the mismatch of the color area of the layer according to the depth specified in the table with the size of each layer in the Civil 3D

My analysis of this is that; the Volume is correct, but the areas of colors are incorrect according to the actual area that should be displayed (except for the inner color only), as the Civil 3D program reads the color areas, which is naturally visible as a Horizontal Projection, which should be understood as follows:

It is that the inner color is only the color that includes the real area, then the real area of ​​the next color includes (the Area of the same color + the Area of the previous color).

Where all colors cannot be represented on a Horizontal Projection, it must be a three-dimensional projection.

To make sure of this, take color area of last layer by the value next to the color of the penultimate layer, and lower the surface to the required depth you will find that the quantity is exactly the same as the table quantity and you will find also the real area has become logical.

Or in another way, specify in the table only the beginning and end of the required layer depth, and you will find the same result.

And therefore; The problem is that it is difficult to show the required layer color area on a Horizontal Projection and not the problem with the quantities.