Learn: Specific gravity
Specific gravity compares a material’s density with a reference water density. It tells you how dense the material is relative to water without carrying a physical unit.
What the formula is saying
Divide material density ρs by water density ρw. If a material is 2.65 times as dense as the chosen water reference, its specific gravity is 2.65.
Read the symbols in plain language
- ρs
- Material density
Mass per unit volume for the stated material and condition. This is density, not weight per volume.
kg/m³Use kg/m³ as the base unit shown here. Both densities use kg/m³ or any identical density unit. The units cancel, leaving a dimensionless ratio. Do not label specific gravity kg/m³ or kN/m³.
- ρw
- Reference density
Mass per unit volume for the stated material and condition. This is density, not weight per volume.
kg/m³Use kg/m³ as the base unit shown here. Both densities use kg/m³ or any identical density unit. The units cancel, leaving a dimensionless ratio. Do not label specific gravity kg/m³ or kN/m³.
- Gs
- Result to find
Specific gravity. Matching density units cancel when material density is divided by reference water density.
ratio / no unit
Sort out the units first
Both densities use kg/m³ or any identical density unit. The units cancel, leaving a dimensionless ratio. Do not label specific gravity kg/m³ or kN/m³.
Assumptions before calculating
Assume the material density and reference water density are defined at suitable stated conditions. For soil-particle specific gravity, use solid-particle density rather than bulk soil density including voids.
Let’s solve one together
Read the given values, follow each operation, then check what the result means.
Read the supplied values as one complete study case. Find Gs and explain the result in the stated output unit.
- ρs · Material density
- 2650 kg/m³
- ρw · Reference density
- 1000 kg/m³
Identify the stated material density
Choose the material or particle density that matches the definition being studied.
(2650) = 2650 kg/m³Compare with the water reference
Matching density units cancel when material density is divided by reference water density.
(2650) ÷ (1000) = 2.65
Does this worked answer make sense?
Equal densities give a ratio of 1. A denser material gives a ratio greater than 1, while a less dense one gives a value between 0 and 1.
A second worked example — different values
A second case uses different data. Predict which way the answer will change, then calculate it without reusing the first answer.
- ρs · Material density
- 2400 kg/m³
- ρw · Reference density
- 1000 kg/m³
Identify the stated material density
Choose the material or particle density that matches the definition being studied.
(2400) = 2400 kg/m³Compare with the water reference
Matching density units cancel when material density is divided by reference water density.
(2400) ÷ (1000) = 2.4
Now try your own values
Change a value or its unit. The same method will show your calculation, step by step.
Results update only when you calculate. The lesson example above stays unchanged.
Your turn — check your understanding
Solve this separate case yourself. Use only the values below; the two worked examples use different data. Give the requested result in the selected unit.
- ρs · Material density
- 7850 kg/m³
- ρw · Reference density
- 998 kg/m³
Find: Learn: Specific gravity
A hint, not the answer
Divide material density ρs by water density ρw. If a material is 2.65 times as dense as the chosen water reference, its specific gravity is 2.65.
Both densities use kg/m³ or any identical density unit. The units cancel, leaving a dimensionless ratio. Do not label specific gravity kg/m³ or kN/m³.
Show the full practice solution
Compare the steps with your work; revealing a solution does not mark the lesson complete.
Identify the stated material density
Choose the material or particle density that matches the definition being studied.
(7850) = 7850 kg/m³Compare with the water reference
Matching density units cancel when material density is divided by reference water density.
(7850) ÷ (998) ≈ 7.865731463
Avoid the common trap
Do not use mass in the numerator and density in the denominator. Do not reverse the ratio or substitute soil unit weight without a consistent conversion.
When this method applies — and when it does not
Water density depends on temperature and the reference convention. The ratio does not by itself determine a porous body’s flotation, strength or durability; its bulk condition and entrained air can matter.
For study and understanding, not approval of a real structure, site operation or design. Apply the correct standard, National Annex and professional review to actual engineering work.
One idea understood. Keep going.
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Sources & further reading
References open in a new tab and explain the underlying principles. The teaching text and examples here are SimpleFlick’s own; the source organisations have not endorsed this calculator.
Reading focus: Specific gravity. Read the relevant soil phase, seepage, earth-pressure, settlement or foundation topic. Effective stress, drainage and idealized geometry determine whether the relationship applies.
Lesson updated: · Both examples and the separate practice case are checked against an independent high-precision numerical implementation. This verifies arithmetic for the stated model, not engineering certification.