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Land & plots

Soil fill volume and delivery loads calculator

Separate compacted target volume from loose delivery volume and apply entered truck-body and payload limits.

Your numbers

Plain decimal, up to 6 decimal places.
Plain decimal, up to 6 decimal places.
A supplied loose/compacted volume ratio, not a percentage or recommended soil factor.
Plain decimal, up to 6 decimal places.
Plain decimal, up to 6 decimal places.
Plain decimal, up to 6 decimal places.
Plain decimal, up to 6 decimal places.

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Your result

Ready when you are

Enter your numbers and select Calculate.

Try it, understand it

Explore the result

Keep the fill target and truck inputs fixed, then change only the entered loose density to see which capacity limits the equivalent load.

Worked example comparison
ExampleOutputWhat to notice
Volume and weight limits agreeModeled delivery trips7At 1,600 kg/m³, the 5 m³ body holds exactly the entered 8 tonnes. The loose quantity requires 7 modeled trips.
The same body, denser materialModeled delivery trips9At 2,000 kg/m³, the 8 tonne payload limits the equivalent load to 4 m³. The same loose volume requires 9 trips.
No target depthModeled delivery trips0Zero compacted depth gives zero target/delivery volume, zero modeled trips and zero haul cost.

Check your understanding

Does a truck body of 5 m³ guarantee it can carry that volume of every material?

No. At a supplied density of 2,000 kg/m³, 5 m³ would weigh 10 tonnes. With the entered 8 tonne payload assumption the equivalent capacity is only 4 m³.

Is a compacted/loose factor of 0.8 entered as 0.8 here?

No. This field requests loose per compacted volume. If your actual supplied relationship is compacted/loose = 0.8, its reciprocal for this field is 1.25. No suitable material factor is inferred.

Behind the numbers

How this calculation works

First calculate the compacted target from plan area and depth. Apply the explicitly entered loose/compacted volume factor, then determine whether body volume or payload mass limits an equivalent full load. Round trips upward only after checking both limits.

THE FORMULATrips = ceil(max(loose volume/body volume, loose mass/payload mass)); loose volume = compacted volume × factor

A worked example

1,000 sq ft with 0.3 m compacted depth gives 27.870912 m³ compacted. A factor of 1.25 gives 34.83864 m³ loose. With density 1,600 kg/m³, a 5 m³ body and 8 tonne payload, 7 modeled trips cost ₹17,500 at ₹2,500 each.

Good to know

Uniform target depth and a supplied loose/compacted factor only. No factor, density or truck limit is recommended or certified. No stripping, cut/fill surface model, compaction suitability, road legality, safe loading, material price or waiting charge is inferred. Haul cost uses the same entered flat rate for every trip.

Reference: FHWA: earthwork material-state factors and geotechnical assumptions (opens in a new tab) ↗

Check the land-unit definitions and sources

The standard factors use the international foot and an acre of 43,560 sq ft. Cent means 1/100 acre; gunta means 1/40 acre; the selected Chennai ground means 2,400 sq ft.

The Assam report defines bigha as 14,400 sq ft, katha as 2,880 sq ft and lecha as 144 sq ft. The cited West Bengal court order mentions a 720-sq-ft katha convention. These are explicit presets, not an automatic definition for every record or every place using the same name. Match your record to the stated convention, or use its verified custom factor.

The acre relationships are converted using exact base factors. Rounded or unrelated regional values in a reference table are not substituted for these definitions.

Name the material state before using a volume factor

For 1,000 sq ft and a compacted target depth of 0.3 m, the target is 27.870912 m³. If your supplied factor says 1.25 loose cubic metres are needed per compacted cubic metre, the delivery quantity is 34.83864 m³. The factor’s direction is part of its meaning.

A record stating compacted/loose = 0.8 describes the reciprocal relationship: loose/compacted = 1/0.8 = 1.25. Entering 0.8 directly into this tool’s loose-per-compacted field would reverse the adjustment. The example factor is not a tested value for a soil type.

Measured plot area can provide a horizontal area for a known shape. A uniform fill depth is a separate assumption. If existing and target levels vary across the plot, area times one depth is not a substitute for a surveyed cut-and-fill surface model.

A vehicle has both volume and mass constraints

At an entered loose density of 1,600 kg/m³, a 5 m³ body would contain 8 tonnes. An entered 8 tonne payload therefore matches the body limit. The example 34.83864 m³ requires seven modeled trips, with an equivalent last volume of 4.83864 m³ after six full equivalent loads.

Change only the supplied density to 2,000 kg/m³ and the same 8 tonne payload limits an equivalent full load to 4 m³. The delivery quantity then needs nine trips, even though the body is still described as 5 m³. The density is applied to the loose material state.

The density calculator can check a compatible mass/volume observation. This tool does not verify a truck rating, safe loading practice or legal road limit. Its cost is only a flat entered haul rate times modeled trips; material price, preparation, waiting and other charges are not included.

A few questions, answered

Is a sand unit or tractor load assigned a standard volume?

No. Enter actual usable body volume and payload assumptions. Vehicle names and local shorthand do not define a universal capacity here.

Is the last trip charged partially?

No. The flat price applies to every trip, including the equivalent partial last load. Different commercial terms require a different price model.

What if the required depth is zero?

Both target and delivery volumes are zero and the modeled trip count and haul cost are zero.

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