Barge Stability Calculation - Hydraulic Learning

Context: Naval ArchitectureThe field of engineering dealing with the design, construction, and maintenance of marine vessels..

This exercise will guide you through the fundamental principles of buoyancy and stability, which are critical in hydraulic engineering and ship design. We will analyze a simple rectangular barge to determine its draft (immersed depth) and its initial stability by calculating the metacentric height.

Pedagogical Note: This exercise teaches the direct application of Archimedes' Principle to find the draft and introduces the concept of metacentric height (\(GM\)), a critical factor in determining the initial stability of any floating object.

Learning Objectives

Apply Archimedes' Principle to find the draft (\(T\)) of a floating rectangular object.
Calculate the location of the Center of Buoyancy (\(KB\)).
Determine the transverse metacentric radius (\(BM_{\text{T}}\)) using the waterplane's moment of inertia.
Calculate the transverse metacentric height (\(GM_{\text{T}}\)) and assess the barge's initial stability.

Study Data

A rectangular barge is floating upright in a calm, fresh water lake. We need to determine its draft and initial stability.

System Specifications

Characteristic	Value
Barge Shape	Rectangular Prism
Fluid	Fresh Water
Specific Weight of Water (\(\gamma\))	62.4 lb/ft³

Barge Cross-Section and Forces

Parameter	Symbol	Value	Unit
Length	\(L\)	100	ft
Beam (Width)	\(B\)	30	ft
Total Barge Weight	\(W\)	500,000	lbs
Height of G from Keel	\(KG\)	4.0	ft

Questions to Address

Calculate the draft (\(T\)) of the barge.
Determine the height of the Center of Buoyancy (\(KB\)) from the keel.
Calculate the transverse metacentric radius (\(BM_{\text{T}}\)).
Calculate the height of the transverse metacenter from the keel (\(KM_{\text{T}}\)).
Determine the transverse metacentric height (\(GM_{\text{T}}\)) and state if the barge is stable.

Fundamentals of Buoyancy and Stability

To solve this exercise, we rely on two core principles of naval architecture and fluid mechanics.

1. Archimedes' Principle (Buoyancy)
A floating object is in equilibrium when its total weight (\(W\)) is equal to the buoyant force (\(F_{\text{B}}\)). The buoyant force is the weight of the fluid displaced by the object's submerged volume (\(V_{\text{sub}}\)). The specific weight of the fluid is \(\gamma\). \[ W = F_{\text{B}} = \gamma \times V_{\text{sub}} \]

2. Metacentric Height (Stability)
The initial stability of a vessel is determined by its metacentric height (\(GM\)). It is the vertical distance between the center of gravity (\(G\)) and the metacenter (\(M\)). \[ GM_{\text{T}} = KM_{\text{T}} - KG \] Where \(KG\) is the (given) center of gravity, and \(KM_{\text{T}} = KB + BM_{\text{T}}\). For the vessel to be stable, \(GM_{\text{T}} > 0\).

Solution: Barge Stability Calculation

Question 1: Calculate the draft (\(T\)) of the barge.

Principle

We apply Archimedes' Principle. For the barge to float, its total weight (\(W\)) must be exactly balanced by the upward buoyant force (\(F_{\text{B}}\)), which is the weight of the water it displaces.

Mini-Lesson

The submerged volume (\(V_{\text{sub}}\)) of a rectangular barge is its length (\(L\)) times its beam (\(B\)) times its unknown draft (\(T\)). So, \(V_{\text{sub}} = L \times B \times T\). We will use this to solve for \(T\).

Pedagogical Note

Think of this as a simple balance. The barge "pushes down" with 500,000 lbs. The water must "push up" with 500,000 lbs. Our job is to find the depth (\(T\)) needed to create that 500,000 lb upward force.

Norms

This calculation is based on first principles (Archimedes' Principle), which is a fundamental law of physics rather than a specific building code or maritime standard. The standards (like those from the US Coast Guard) apply to the *results* of stability (like required \(GM_{\text{T}}\)), not to the calculation of buoyancy itself.

Formula(s)

The core formulas for this step.

Equilibrium Equation

W = F_{\text{B}} = \gamma \times V_{\text{sub}}

Solving for Draft (T)

W = \gamma \times (L \times B \times T) \implies T = \frac{W}{\gamma \times L \times B}

Hypotheses

We make the following standard assumptions.

The barge is floating in calm, still water.
The barge is floating upright (no initial heel or trim).
The barge itself is a perfect rectangular prism.

Data

We pull the required values from the problem statement.

Parameter	Symbol	Value	Unit
Barge Weight	\(W\)	500,000	lbs
Specific Weight (Water)	\(\gamma\)	62.4	lb/ft³
Length	\(L\)	100	ft
Beam	\(B\)	30	ft

Tips

The most common error is a unit mismatch. Here, all units are in pounds (lbs) and feet (ft), so we are safe. \(T\) will come out in feet, which is what we want.

Schematic (Before Calculations)

This schematic shows the cross-section of the barge. The draft \(T\) is the unknown value we are solving for.

Barge Cross-Section (Solving for T)

Calculation(s)

We will first calculate the denominator (weight per foot of draft), then solve for \(T\).

Step 1: Calculate the Buoyant Force per Foot of Draft

This is the weight of water displaced for every 1 ft of draft.

\begin{aligned} \text{Force/Draft} &= \gamma \times L \times B \\ &= 62.4 \text{ lb/ft}^3 \times 100 \text{ ft} \times 30 \text{ ft} \\ &= 187,200 \text{ lbs/ft} \end{aligned}

This means every foot of draft supports 187,200 lbs of weight.

Step 2: Calculate the Draft (T)

We now find how many "feet of draft" are needed to support the total weight.

\begin{aligned} T &= \frac{W}{\text{Force/Draft}} \\ T &= \frac{500,000 \text{ lbs}}{187,200 \text{ lbs/ft}} \\ T &= 2.6709... \text{ ft} \end{aligned}

We will round this to 2.67 ft.

Schematic (After Calculations)

The barge is submerged 2.67 ft into the water.

Barge with Calculated Draft

Analysis

The draft, or immersed depth, of the barge is 2.67 ft. This is the first step required for any stability analysis.

Cautionary Points

Always use the specific weight for the correct fluid. If this were salt water (≈ 64 lb/ft³), the draft would be slightly less.

Key Takeaways

Equilibrium: \(Weight = Buoyant Force\).
\(V_{\text{sub}}\) for a box is \(L \times B \times T\).

Did You Know?

Large ships have "Plimsoll lines" painted on their hulls to show the maximum legal draft for different water types (like fresh, salt, and tropical) to prevent overloading.

FAQ

Common questions about this step.

Final Result

The draft (\(T\)) of the barge is 2.67 ft.

Your Turn

If the barge's weight was 600,000 lbs instead, what would be the new draft in ft?

Memo Card

Question 1 Summary:

Key Concept: Archimedes' Principle
Essential Formula: \(T = W / (\gamma \times L \times B)\)
Key Data: \(W = 500,000 \text{ lbs}\), \(L \times B = 3000 \text{ ft}^2\)

Question 2: Determine the height of the Center of Buoyancy (\(KB\)) from the keel.

Principle

The Center of Buoyancy (\(B\)) is the geometric center, or centroid, of the *submerged volume* (\(V_{\text{sub}}\)). It is the single point where the total buoyant force (\(F_{\text{B}}\)) can be considered to act.

Mini-Lesson

For a simple, upright rectangular prism, the centroid of its volume is at its geometric center. Vertically, this is simply at half the height of the submerged shape, which is the draft (\(T\)). Therefore, its distance from the Keel (K) is \(T/2\).

Pedagogical Note

This is a common-sense check. The buoyant force "pushes up" from the average location of all the displaced water. For a simple box shape, that average location is exactly halfway up the submerged part.

Norms

This calculation is based on the geometric principles of finding a centroid. It is a fundamental concept in physics and engineering. Specific maritime standards do not dictate *how* to find a centroid, but they rely on its correct calculation as an input for the stability criteria.

Formula(s)

For a rectangular prism floating upright:

KB = \frac{T}{2}

Hypotheses

We assume the barge is a simple rectangular prism and is floating upright (no heel or trim).

Data

We use the calculated draft from Question 1.

Parameter	Symbol	Value	Unit
Draft (from Q1)	\(T\)	2.67	ft

Tips

Do not confuse \(KB\) (Center of Buoyancy) with \(KG\) (Center of Gravity). \(KB\) only depends on the *shape of the submerged volume*, not on how the barge's weight is distributed.

Schematic (Before Calculations)

We need to find the vertical position of point \(B\), which is the centroid of the shaded submerged volume.

Finding Center of Buoyancy (B)

Calculation(s)

We apply the simple formula using our calculated draft.

\begin{aligned} KB &= \frac{T}{2} \\ KB &= \frac{2.67 \text{ ft}}{2} \\ KB &= 1.335 \text{ ft} \end{aligned}

This places the Center of Buoyancy 1.335 ft above the keel.

Schematic (After Calculations)

The point \(B\) is now located 1.335 ft above the keel.

Center of Buoyancy Located

Analysis

The center of the buoyant force (\(F_{\text{B}}\)) is located 1.335 ft above the keel. This is the second piece of the stability puzzle, after \(T\).

Cautionary Points

This \(T/2\) formula is *only* for rectangular (box) shapes. A V-shaped hull, a circular pontoon, or a real ship hull would have a different, more complex formula for its centroid.

Key Takeaways

\(KB\) is the centroid of the *submerged volume*.
For a box, \(KB = T/2\), measured from the keel (K).

Did You Know?

For a submerged triangular (V-shaped) hull, the centroid \(KB\) would be at \( (2/3) \times T \). For a semi-circle, it would be at \( (4 \times R) / (3 \times \pi) \).

FAQ

Common questions about this step.

Final Result

The Center of Buoyancy (\(KB\)) is 1.335 ft above the keel (K).

Your Turn

If a different barge has a draft (\(T\)) of 5.0 ft, what is its \(KB\)?

Memo Card

Question 2 Summary:

Key Concept: Centroid of Submerged Volume
Essential Formula: \(KB = T / 2\) (for a box)
Key Data: \(T = 2.67 \text{ ft}\)

Question 3: Calculate the transverse metacentric radius (\(BM_{\text{T}}\)).

Principle

The metacentric radius (\(BM_{\text{T}}\)) is a measure of stability that comes purely from the *shape of the waterplane* (the 2D area of the barge at the waterline). It is the ratio of the waterplane's transverse moment of inertia (\(I_{\text{T}}\)) to the total submerged volume (\(V_{\text{sub}}\)).

Mini-Lesson

The Moment of Inertia (\(I_{\text{T}}\)) measures how "spread out" an area is, which in this case, resists rolling. A wider barge (larger \(B\)) has a *much* larger \(I_{\text{T}}\) (since it's \(B^3\)), making it more stable. A larger submerged volume (\(V_{\text{sub}}\)) makes it harder to rotate, so it *decreases* \(BM_{\text{T}}\).

Pedagogical Note

Think of \(BM_{\text{T}}\) as a "shape stability" factor. A wide, flat log (\(I_{\text{T}}\) is large, \(V_{\text{sub}}\) is small) is very stable and hard to roll. A round log (\(I_{\text{T}}\) is small) is very tippy. We are calculating this shape-based stability factor.

Norms

The formula for the Moment of Inertia of a rectangle is a standard, fundamental equation from engineering mechanics and calculus. Maritime regulations (like 46 CFR) do not define this formula, but they mandate its use in stability calculations.

Formula(s)

Metacentric Radius

BM_{\text{T}} = \frac{I_{\text{T}}}{V_{\text{sub}}}

Transverse Moment of Inertia (Rectangle)

I_{\text{T}} = \frac{L \times B^3}{12}

Hypotheses

The waterplane (the area at the waterline) is a simple rectangle (\(L \times B\)). The roll is transverse (side-to-side).

Data

We need dimensions and the submerged volume.

Parameter	Symbol	Value	Unit
Length	\(L\)	100	ft
Beam	\(B\)	30	ft
Submerged Volume	\(V_{\text{sub}}\)	8012.8	ft³

Tip: We can find \(V_{\text{sub}}\) two ways:
1. \(V_{\text{sub}} = L \times B \times T = 100 \times 30 \times 2.67 = 8010 \text{ ft}^3\)
2. \(V_{\text{sub}} = W / \gamma = 500,000 / 62.4 = 8012.8 \text{ ft}^3\)
Method 2 is more precise as it avoids using the rounded \(T\). Let's use 8012.8 ft³.

Tips

The *most* common error is using the wrong moment of inertia formula. For *transverse* (side-to-side) rolling, we are rolling about the longitudinal (length) axis, so the \(B\) (Beam) term is cubed: \( (L \times B^3) / 12 \).

Schematic (Before Calculations)

We are analyzing the top-down 2D shape of the barge at the waterline (the "waterplane").

Waterplane Area (Top View)

Calculation(s)

Step 1: Calculate \(I_{\text{T}}\)

We plug \(L\) and \(B\) into the moment of inertia formula.

\begin{aligned} I_{\text{T}} &= \frac{L \times B^3}{12} \\ I_{\text{T}} &= \frac{100 \text{ ft} \times (30 \text{ ft})^3}{12} \\ I_{\text{T}} &= \frac{100 \text{ ft} \times 27,000 \text{ ft}^3}{12} \\ I_{\text{T}} &= \frac{2,700,000 \text{ ft}^4}{12} \\ I_{\text{T}} &= 225,000 \text{ ft}^4 \end{aligned}

The transverse moment of inertia of the waterplane is 225,000 ft⁴.

Step 2: Calculate \(BM_{\text{T}}\)

We divide \(I_{\text{T}}\) by the precise submerged volume \(V_{\text{sub}}\).

\begin{aligned} BM_{\text{T}} &= \frac{I_{\text{T}}}{V_{\text{sub}}} \\ BM_{\text{T}} &= \frac{225,000 \text{ ft}^4}{8012.8 \text{ ft}^3} \\ BM_{\text{T}} &= 28.08 \text{ ft} \end{aligned}

This gives a very large metacentric radius of 28.08 ft, which is the source of the barge's high stability.

Schematic (After Calculations)

The value \(BM_{\text{T}}\) is the distance from \(B\) *up* to the Metacenter \(M_{\text{T}}\). We now visualize this on a clean vertical diagram.

Metacentric Radius (\(BM_{\text{T}}\))

Analysis

This large value (28.08 ft) is characteristic of wide, flat-bottomed vessels like barges. It indicates that the waterplane shape itself provides a massive amount of stability, as the center of buoyancy shifts significantly when rolled.

Cautionary Points

Remember: \(I_{\text{T}} = (L \times B^3) / 12\). If you were calculating *longitudinal* stability (pitching end-to-end), you would use \(I_{\text{L}} = (B \times L^3) / 12\), which would be a much larger number.

Key Takeaways

\(BM_{\text{T}} = I_{\text{T}} / V_{\text{sub}}\)
Stability is proportional to the *cube* of the Beam (\(B^3\)), making width a critical factor.

Did You Know?

Catamarans are so stable because they have two narrow hulls, but the *overall* \(I_{\text{T}}\) of their waterplane (spanning the full width across both hulls) is enormous relative to their small \(V_{\text{sub}}\).

FAQ

Common questions about this step.

Final Result

The transverse metacentric radius (\(BM_{\text{T}}\)) is 28.08 ft.

Your Turn

If the barge was 40 ft wide (Beam \(B=40\)) but had the same \(V_{\text{sub}}\) (8012.8 ft³), what would \(I_{\text{T}}\) be? (Formula: \((100 \times 40^3) / 12\))

Memo Card

Question 3 Summary:

Key Concept: Waterplane Inertia
Essential Formula: \(BM_{\text{T}} = I_{\text{T}} / V_{\text{sub}}\)
Key Data: \(I_{\text{T}} = 225,000 \text{ ft}^4\), \(V_{\text{sub}} = 8012.8 \text{ ft}^3\)

Question 4: Calculate the height of the transverse metacenter from the keel (\(KM_{\text{T}}\)).

Principle

The Transverse Metacenter (\(M_{\text{T}}\)) is the key point for stability. Its vertical position is found by adding the metacentric radius (\(BM_{\text{T}}\)) *above* the Center of Buoyancy (\(B\)). Since \(KB\) is the height of \(B\) from the keel, we simply "stack" these two vertical distances.

Mini-Lesson

All vertical stability measurements are made from the Keel (\(K\)). We have found the height of \(B\) (which is \(KB\)) and the distance from \(B\) to \(M\) (which is \(BM_{\text{T}}\)). The total height of \(M\) from \(K\) is the sum of these two segments.

Pedagogical Note

This is a simple vertical addition step. We are finding the total height of the "stability pivot point" (\(M_{\text{T}}\)) by adding its height *relative to \(B\)* to \(B\)'s height *relative to \(K\)*.

Norms

This is a definitional calculation. The concept of \(KM_{\text{T}}\) is a cornerstone of naval architecture, and its definition as the sum of \(KB\) and \(BM_{\text{T}}\) is universally accepted. Standards (like 46 CFR) focus on the *final* \(GM_{\text{T}}\) value, not the intermediate steps.

Formula(s)

The formula is a simple sum of the two components.

KM_{\text{T}} = KB + BM_{\text{T}}

Hypotheses

We assume all points (K, B, M) lie on the vessel's vertical centerline (for an upright vessel).

Data

We use the results from Q2 and Q3.

Parameter	Symbol	Value	Unit
Center of Buoyancy (Q2)	\(KB\)	1.335	ft
Metacentric Radius (Q3)	\(BM_{\text{T}}\)	28.08	ft

Tips

Always measure from the same datum (reference point), which is the Keel (\(K\)). \(KB\) is from the keel, \(BM_{\text{T}}\) is from \(B\), so adding them gives \(KM_{\text{T}}\), which is also from the keel.

Schematic (Before Calculations)

The schematic from Q3 showed the segments. Now we are formally adding them to find the total height of \(M_{\text{T}}\).

Stacking the Heights

Calculation(s)

We now sum the height of the Center of Buoyancy (\(KB\)) and the Metacentric Radius (\(BM_{\text{T}}\)) to find the total height of the Metacenter from the keel.

\begin{aligned} KM_{\text{T}} &= KB + BM_{\text{T}} \\ KM_{\text{T}} &= 1.335 \text{ ft} + 28.08 \text{ ft} \\ KM_{\text{T}} &= 29.415 \text{ ft} \end{aligned}

The total height of the geometric stability point (\(M_{\text{T}}\)) is 29.415 ft above the keel.

Schematic (After Calculations)

This schematic shows the locations of all key points, including the given \(KG\).

Stability Points (Cross-Section)

Analysis

The transverse metacenter (\(M_{\text{T}}\)) is 29.415 ft above the keel. This is the "pivot point" for buoyant stability. Now we must compare this to the location of the center of gravity (\(G\)).

Cautionary Points

Do not subtract! \(M_{\text{T}}\) is located *above* \(B\). This value, \(KM_{\text{T}}\), represents the total "geometric" stability of the hull shape at this draft, independent of how the weight is loaded.

Key Takeaways

\(KM_{\text{T}} = KB + BM_{\text{T}}\).
All stability heights (\(KB, KG, KM_{\text{T}}\)) are measured vertically from the keel (K).

Did You Know?

A "stiff" ship has a very high \(KM_{\text{T}}\) (often from a large \(BM_{\text{T}}\)), while a "tender" ship (which rolls slowly and gently) has a low \(KM_{\text{T}}\).

FAQ

Common questions about this step.

Final Result

The height of the transverse metacenter (\(KM_{\text{T}}\)) is 29.415 ft above the keel.

Your Turn

If \(KB = 2.0\) ft and \(BM_{\text{T}} = 15.5\) ft, what is \(KM_{\text{T}}\)?

Memo Card

Question 4 Summary:

Key Concept: Metacenter Height
Essential Formula: \(KM_{\text{T}} = KB + BM_{\text{T}}\)
Key Data: \(KB = 1.335 \text{ ft}\), \(BM_{\text{T}} = 28.08 \text{ ft}\)

Question 5: Determine the transverse metacentric height (\(GM_{\text{T}}\)) and state if the barge is stable.

Principle

The metacentric height (\(GM_{\text{T}}\)) is the final and most important measure of initial stability. It is the vertical separation between the Metacenter (\(M_{\text{T}}\)) and the Center of Gravity (\(G\)). If \(M\) is *above* \(G\) (\(GM_{\text{T}}\) is positive), the vessel is stable and will right itself. If \(M\) is *below* \(G\) (\(GM_{\text{T}}\) is negative), it is unstable and will capsize.

Mini-Lesson

Stability is a "battle" between two forces. The barge's weight (\(W\)) acts *downward* through \(G\). The buoyant force (\(F_{\text{B}}\)) acts *upward* through \(B\). When the barge rolls, \(B\) shifts, and the buoyant force now acts upward along a line pointing to \(M\). If \(M\) is above \(G\), these two forces create a "righting couple" that rotates the barge back upright.

Pedagogical Note

Think of it this way: \(G\) is where the weight "pushes down." \(M_{\text{T}}\) is the effective point where buoyancy "pushes up" during a small roll. For stability, the "up" push must be higher than the "down" push. We need \(M_{\text{T}}\) to be above \(G\).

Norms

This is where standards are critical. For example, the U.S. Coast Guard (in 46 CFR Subchapter S) specifies minimum intact stability requirements. While specific values vary by vessel type, a common requirement for barges is to have a positive \(GM_{\text{T}}\) and a righting arm curve that meets certain criteria, often implying a minimum \(GM_{\text{T}}\) of around 1.0 to 3.0 ft to ensure a margin of safety.

Formula(s)

The final calculation compares the geometric stability (\(KM_{\text{T}}\)) to the weight distribution (\(KG\)).

GM_{\text{T}} = KM_{\text{T}} - KG

Hypotheses

The given \(KG\) value of 4.0 ft accurately represents the vertical center of the barge's total weight.

Data

We use the result from Q4 and the given \(KG\).

Parameter	Symbol	Value	Unit
Metacenter Height (Q4)	\(KM_{\text{T}}\)	29.415	ft
Gravity Height (Given)	\(KG\)	4.0	ft

Tips

This is the final step. We compare the "geometric stability" (\(KM_{\text{T}}\)) with the "weight stability" (\(KG\)). As long as \(KM_{\text{T}} > KG\), the vessel is stable.

Schematic (Before Calculations)

We are finding the distance between \(M_T\) and \(G\) on our stability diagram.

Finding Metacentric Height (\(GM_{\text{T}}\))

Calculation(s)

We find the final metacentric height by subtracting the height of the Center of Gravity (\(KG\)) from the height of the Metacenter (\(KM_{\text{T}}\)).

\begin{aligned} GM_{\text{T}} &= KM_{\text{T}} - KG \\ GM_{\text{T}} &= 29.415 \text{ ft} - 4.0 \text{ ft} \\ GM_{\text{T}} &= 25.415 \text{ ft} \end{aligned}

The result is a large, positive value, confirming the barge's stability.

Schematic (After Calculations)

The final diagram shows a large, positive \(GM_T\), indicating high stability.

Final Stability Diagram

Analysis

The calculated metacentric height is **+25.415 ft**.

Since \(GM_{\text{T}}\) is a large positive number, the barge is extremely stable. This is expected for a wide, flat barge (like a pontoon), which has a very large \(BM_{\text{T}}\) (from its \(B^3\) term) and a low center of gravity (\(KG\)).

Cautionary Points

A very large \(GM_{\text{T}}\) (like this one) is not always desirable. It makes the barge "stiff," meaning it will roll back and forth very quickly and violently, which is uncomfortable for people and bad for sensitive cargo. A "tender" ship (small \(GM_{\text{T}}\)) rolls slowly and gently.

Key Takeaways

\(GM_{\text{T}} = KM_{\text{T}} - KG\).
If \(GM_{\text{T}} > 0\), the vessel is stable.
If \(GM_{\text{T}} < 0\), the vessel is unstable and will capsize.

Did You Know?

Passenger ships are often designed with a smaller, more comfortable \(GM_{\text{T}}\) (e.g., 2-4 ft). A cargo ship's \(GM_{\text{T}}\) can change dramatically as it's loaded, so the crew must re-calculate it for every single voyage to ensure it's not too low (unstable) or too high (stiff).

FAQ

Common questions about this step.

Final Result

The transverse metacentric height (\(GM_{\text{T}}\)) is 25.415 ft. Since \(GM_{\text{T}} > 0\), the barge is stable.

Your Turn

If the \(KG\) was 30.0 ft (e.g., top-heavy cargo), what would \(GM_{\text{T}}\) be? (Use \(KM_{\text{T}}=29.415\)).

Memo Card

Question 5 Summary:

Key Concept: Metacentric Height
Essential Formula: \(GM_{\text{T}} = KM_{\text{T}} - KG\)
Key Data: \(KM_{\text{G}} = 29.415 \text{ ft}\), \(KG = 4.0 \text{ ft}\)

Interactive Tool: Stability Simulator

Use the sliders to see how changing the barge's Weight (which affects draft) and Center of Gravity (cargo height) impacts the final stability (\(GM_{\text{T}}\)).

Input Parameters

Barge Weight (\(W\)) (k-lbs) 500 k-lbs

Center of Gravity (\(KG\)) (ft) 4.0 ft

Key Requirements

Calculated Draft (\(T\)) (ft) -

Metacenter Height (\(KM_{\text{T}}\)) (ft) -

Final Stability (\(GM_{\text{T}}\)) (ft) -

Final Quiz: Test Your Knowledge

1. If the weight of a barge increases, what happens to its draft (\(T\))?

It decreases (floats higher)
It increases (sinks lower)
It stays the same

2. What is the formula for the transverse moment of inertia (\(I_{\text{T}}\)) of a rectangular waterplane?

\(I_{\text{T}} = (L \times B^2) / 12\)
\(I_{\text{L}} = (B \times L^3) / 12\)
\(I_{\text{T}} = (L \times B^3) / 12\)

3. A barge is initially stable if:

\(GM_{\text{T}} > 0\) (Metacenter is above Center of Gravity)
\(GM_{\text{T}} < 0\) (Center of Gravity is above Metacenter)
\(KB = KG\)

Glossary

Archimedes' Principle: The principle that a floating body displaces a weight of fluid equal to its own weight.
Draft (\(T\)): The vertical distance from the bottom of the hull (keel) to the waterline.
Keel (\(K\)): The bottom-most structural member of a vessel, used as the vertical reference point (0 ft).
Center of Buoyancy (\(B\)): The geometric center (centroid) of the *submerged volume* of the vessel.
Center of Gravity (\(G\)): The point representing the average location of the vessel's total weight (hull, cargo, etc.).
Metacenter (\(M\)): A theoretical point of intersection of the buoyant force vectors for successive small angles of heel (roll). Its height is critical for stability.
Metacentric Height (\(GM\)): The vertical distance between the Center of Gravity (\(G\)) and the Metacenter (\(M\)). A positive \(GM\) indicates initial stability.