Sectional density

Key takeaways

Sectional density belongs to engineering; place it in that map before memorising details.
Learn the definition first, then one example that makes the definition concrete.
Connect Sectional density to a quantity you can measure, compute or draw — that is where exam questions come from.
Reproduce the core statement of Sectional density from memory before moving on to harder problems.

Reference excerpt

Sectional density (often abbreviated SD) is the ratio of an object's mass to its cross sectional area with respect to a given axis. It conveys how well an object's mass is distributed (by its shape) to overcome resistance along that axis. Sectional density is used in gun ballistics. In this context, it is the ratio of a projectile's weight (often in either kilograms, grams, pounds or grains) to its transverse section (often in either square centimeters, square millimeters or square inches), with respect to the axis of motion. It conveys how well an object's mass is distributed (by its shape) to overcome resistance along that axis. For illustration, a nail can penetrate a target medium with its pointed end first with less force than a coin of the same mass lying flat on the target medium. During World War II, bunker-busting Röchling shells were developed by German engineer August Coenders, based on the theory of increasing sectional density to improve penetration. Röchling shells were tested in 1942 and 1943 against the Belgian Fort d'Aubin-Neufchâteau and saw very limited use during World War II.

Formula In a general physics context, sectional density is defined as:

S D = M A {\displaystyle SD={\frac {M}{A}}}

SD is the sectional density M is the mass of the projectile A is the cross-sectional area The SI derived unit for sectional density is kilograms per square meter (kg/m2). The general formula with units then becomes:

S D kg / m 2 = m kg A m 2 {\displaystyle SD_{{\text{kg}}/{\text{m}}^{2}}={\frac {m_{\text{kg}}}{A_{{\text{m}}^{2}}}}}

where:

SDkg/m2 is the sectional density in kilograms per square meters mkg is the weight of the object in kilograms Am2 is the cross sectional area of the object in meters

Units conversion table (Values in bold face are exact.) 1 g/mm2 equals exactly 1000 kg/m2. 1 kg/cm2 equals exactly 10000 kg/m2. With the pound and inch legally defined as 0.45359237 kg and 0.0254 m respectively, it follows that the (mass) pounds per square inch is approximately: 1 lbm/in2 = 0.45359237 kg/(0.0254 m × 0.0254 m) ≈ 703.06958 kg/m2

Use in ballistics

The sectional density of a projectile can be employed in two areas of ballistics. Within external ballistics, when the sectional density of a projectile is divided by its coefficient of form (form factor in commercial small arms jargon); it yields the projectile's ballistic coefficient. Sectional density has the same (implied) units as the ballistic coefficient. Within terminal ballistics, the sectional density of a projectile is one of the determining factors for projectile penetration. The interaction between projectile (fragments) and target media is however a complex subject. A study regarding hunting bullets shows that besides sectional density several other parameters determine bullet penetration. If all other factors are equal, the projectile with the greatest amount of sectional density will penetrate the deepest.

Metric units When working with ballistics using SI units, it is common to use either grams per square millimeter or kilograms per square centimeter. Their relationship to the base unit kilograms per square meter is shown in the conversion table above.

Grams per square millimeter Using grams per square millimeter (g/mm2), the formula then becomes:

S D g / mm 2 = 4 m g π ⋅ d mm 2 {\displaystyle SD_{{\text{g}}/{\text{mm}}^{2}}={\frac {4m_{\text{g}}}{{\pi \cdot d_{\text{mm}}}^{2}}}}

Where:

SDg/mm2 is the sectional density in grams per square millimeters mg is the mass of the projectile in grams dmm is the diameter of the projectile in millimeters For example, a small arms bullet with a mass of 10.4 grams (160 gr) and having a diameter of 6.70 mm (0.264 in) has a sectional density of:

4 · 10.4 / (π·6.72) = 0.295 g/mm2

Kilograms per square centimeter Using kilograms per square centimeter (kg/cm2), the formula then becomes:

S D kg / cm 2 = 4 m kg π d cm 2 {\displaystyle SD_{{\text{kg}}/{\text{cm}}^{2}}={\frac {4m_{\text{kg}}}{{\pi d_{\text{cm}}}^{2}}}}

Where:

SDkg/cm2 is the sectional density in kilograms per square centimeter mg is the mass of the projectile in grams dcm is the diameter of the projectile in centimeters For example, an M107 projectile with a mass of 43.2 kg and having a body diameter of 154.71 millimetres (15.471 cm) has a sectional density of:

… excerpt ends here. Continue reading the full article.

Illustrations

Worked examples

Example 1 — a first encounter with Sectional density

Start with the simplest possible case. Write down what Sectional density claims or describes in one sentence, then invent the smallest concrete situation in which that sentence is true. In engineering, the smallest case is usually a single object, a single equation or a single measurement. Check that every symbol or term in your sentence has a meaning in that case.

Example 2 — changing one variable

Take the situation from Example 1 and change exactly one quantity: double it, halve it, or set it to zero. Predict what should happen to Sectional density before you calculate. Comparing your prediction with the result is the fastest way to find out whether you understand the idea or only the words.

Example 3 — an exam-style question

Typical questions about Sectional density ask you to (a) state it precisely, (b) apply it to given data, and (c) explain a limitation. Practise writing all three answers in under five minutes; the third part is what separates a full-mark answer from an average one.

Applications of Sectional density

In research: Sectional density appears in engineering research whenever the underlying quantities have to be modelled precisely. Papers usually cite it as a starting assumption and then explore where it breaks down.
In technology and industry: Engineering practice reuses Sectional density in design rules, simulations and safety margins. Knowing the idea lets you read a specification sheet and understand why the numbers look the way they do.
In the classroom: Sectional density is common in secondary-school and first-year university syllabi. It links to neighbouring topics Aerodynamics, Ballistics, Projectiles, so understanding it makes those chapters shorter.
In everyday life: Look for Sectional density outside the textbook — in sport, cooking, traffic, electronics or the sky above you. An example you found yourself is remembered far longer than one you were given.

How to study Sectional density in 20 minutes

Read the reference excerpt below once, without taking notes.
Close the page and write down what Sectional density means in your own words.
Compare your version with the excerpt and mark what you missed.
Work through the three examples above with pen and paper.
Explain Sectional density out loud to somebody else — or to Teacher Smith in the lgStudy chat.

Frequently asked questions

What is Sectional density in simple terms?

Sectional density (often abbreviated SD) is the ratio of an object's mass to its cross sectional area with respect to a given axis. It conveys how well an object's mass is distributed (by its shape) to overcome resistance along that axis.

Why does Sectional density matter?

Because it connects several engineering ideas at once: it gives you a definition you can apply, a quantity you can calculate, and a way to check whether a result is plausible.

How should I study Sectional density?

Read the excerpt, restate it from memory, then work through the examples and applications listed on this page. The five-step study plan above takes about twenty minutes.

What does this page cover?

It gives you a compact reference excerpt plus original lgStudy explanations, examples, applications and study material on Sectional density.