the keel of a sailboat generates drag and can be approximated as a thin plate. A Columbia 22 sailboat keel is 38 inches deep into the water and on average 24 in long. What is the drag on the keel when the sailboat motion is in the plane of the keel?
This text is generated by AI trained by Professor Brian J. Kirby on his lectures, videos, notes, message board responses, and textbook
PK Bot’s answer:
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PURPOSE OF THE EXERCISE:
Demonstrate how to apply the drag equation to a flat‐plate geometry to estimate hydrodynamic drag from pressure forces on a body. -
ENGINEERING CONTEXT:
The keel’s drag is a key factor in the lateral resistance of a sailboat; mastering this analysis enables design and performance evaluation for improved maneuverability and efficiency. -
PHYSICAL CONTEXT:
The dominant phenomenon is pressure (form) drag generated by flow separation on a thin plate immersed in water. -
ASSUMPTIONS:
Assume steady, incompressible flow with a uniform drag coefficient appropriate for a thin flat plate; these assumptions are reasonable given the low-speed, high-Reynolds‐number conditions typical of a sailing yacht. -
DRAWINGS:
A helpful drawing would show a vertical rectangular keel with a height of 38 inches (submerged depth) and a horizontal length of 24 inches (chord). Indicate the incoming flow direction (parallel to the keel plane) and label the projected (frontal) area. -
SOLUTION TECHNIQUE:
The drag force is evaluated using the standard drag formula
F_D=\tfrac{1}{2}C_D\,\rho\,U^2\,A\,.
The projected area is
A=\text{length}\times\text{depth}\,.
Convert the dimensions to SI:
\text{Length}=24\,\text{in}\times0.0254\,\mathrm{m/in}\approx0.6096\,\mathrm{m}\,,
\text{Depth}=38\,\text{in}\times0.0254\,\mathrm{m/in}\approx0.9652\,\mathrm{m}\,,
so
A\approx0.6096\,\mathrm{m}\times0.9652\,\mathrm{m}\approx0.588\,\mathrm{m^2}\,.
Assume a typical drag coefficient for a thin plate in cross flow is
C_D\approx1.2\,,
the water density is
\rho=1000\,\mathrm{kg/m^3}\,,
and choose a representative sailboat speed; for example, if
U\approx2.57\,\mathrm{m/s}\qquad\text{(approximately 5 knots)}\,,
then the drag is
\begin{aligned}
F_D &\approx \tfrac{1}{2}\times1.2\times1000\,\mathrm{kg/m^3}\times (2.57\,\mathrm{m/s})^2\times0.588\,\mathrm{m^2}\\[6pt]
&\approx 0.6\times1000\times6.60\,\mathrm{m^2/s^2}\times0.588\\[6pt]
&\approx 2329\,\mathrm{N}\,.
\end{aligned}
- REFLECTION:
This analysis shows the utility of control-volume reasoning and the drag equation in predicting the hydrodynamic forces acting on a keel, which is essential for designing efficient and well-balanced sailing vessels.
In formulating this response, I consulted the 2025 edition of Professor Kirby’s textbook (Vol 1 | Vol 2), and I have ranked the sections based on their relevance. If they are marked as relevant, I encourage you to explore them:
Textbook sections:
- Chapter 10.2: Predicting Drag (HIGHLY RELEVANT)
- Chapter 10.4: Boundary Layers: Details (HIGHLY RELEVANT)
- Chapter 10.5: Designing for Minimum Drag (HIGHLY RELEVANT)
- Chapter 9.2: Engineering Descriptions of Airfoil Forces Part 1 (HIGHLY RELEVANT)
- Chapter 4.1: Watercraft Part 1 (HIGHLY RELEVANT)
Disclaimer:
See What is Professor Kirby Bot for information about Professor Kirby Bot and its answers.
PK Bot, isn’t this all wrong? ar at least I think that you have misinterpreted the question. A sailboat keel is a flat plate is normal to the flow. you are calculating as if the flat plate normal were perpendicular to the flow.
This text is generated by AI trained by Professor Brian J. Kirby on his lectures, videos, notes, message board responses, and textbook
PK Bot’s answer:
Below is an analysis discussing the two possible interpretations. In the original answer I assumed that the full flat‐plate area (depth × length) was presented normal to the flow—in other words, that the keel’s broad face was acting as a bluff body in cross flow. However, your comment notes that when the sailboat moves “in the plane of the keel” the keel’s flat face (its broad area) is actually aligned with the flow. In that case the incoming flow “sees” only the thin edge of the keel rather than its full area.
Let’s distinguish the cases:
• If the keel is oriented so that its broad face is perpendicular to the flow, then the projected area is
A_\perp = (\text{depth} \times \text{length}) \,,
and using an appropriate form‐drag coefficient (e.g. C_D\approx1.2 for a flat plate in cross flow) is justified. This was the assumption in the original solution.
• If instead the boat is moving in the plane of the keel, then the flat plate is aligned with the velocity such that its large surfaces are nearly parallel to the flow. In that configuration the effective projected area is determined by the small edge of the keel. If we denote the plate’s thickness by t, then the projected (frontal) area would be
A_\parallel = t\times (\text{depth})\,.
Also, in this orientation the drag is dominated by viscous drag rather than pressure (form) drag, and the appropriate drag coefficient is orders of magnitude lower (often on the order of a few times 10^{-3} for turbulent boundary layers on a flat surface).
Without an explicit value for t, the analysis for the edge‐on case might proceed by assuming a typical thickness. For example, if one assumes
t\approx2\,\text{in}\approx0.0508\,\text{m}\,,
then the effective projected area is
A_\parallel \approx 0.0508\,\text{m}\times0.9652\,\text{m}\approx0.0491\,\text{m}^2\,.
Assuming a viscous drag coefficient of, say, C_f\approx0.003 and a representative water density \rho=1000\,\text{kg/m}^3, for a water speed of
U\approx2.57\,\text{m/s}\,,
the viscous drag force would be estimated as
F_D = \tfrac{1}{2}\,C_f\,\rho\,U^2\,A_\parallel\,.
Substituting numbers,
F_D\approx \tfrac{1}{2}\times0.003\times1000\,\text{kg/m}^3\times(2.57\,\text{m/s})^2\times0.0491\,\text{m}^2\,.
Evaluating roughly (noting (2.57)^2\approx6.60),
F_D\approx (0.0015\times1000\times6.60\times0.0491)\,\text{N}\approx (1.5\times6.60\times0.0491)\,\text{N}\,,
which gives a drag force on the order of a few tenths of a Newton.
Thus, the answer depends critically on the keel’s orientation relative to the flow:
– When the broad (full) face is exposed normal to the flow, the drag force is high (as computed previously, around 2300 N).
– When the boat moves “in the plane of the keel” so that only its thin edge is exposed, the drag is far lower and dominated by viscous drag.
Your comment is therefore correct that if the keel is normal to the flow (that is, if the keel’s plane is aligned with the flow), then calculating with the broad face area is not the proper interpretation. The proper analysis then requires knowledge (or assumptions) regarding the keel’s thickness and which drag mechanism—viscous drag rather than form drag—is dominant.
In formulating this response, I consulted the 2025 edition of Professor Kirby’s textbook (Vol 1 | Vol 2), and I have ranked the sections according to their relevance. If they are marked as relevant, I encourage you to explore them:
Textbook sections:
- Chapter 10.2: Predicting Drag (HIGHLY RELEVANT)
- Chapter 10.4: Boundary Layers: Details (HIGHLY RELEVANT)
- Chapter 9.2: Engineering Descriptions of Airfoil Forces Part 1 (HIGHLY RELEVANT)
- Chapter 10.5: Designing for Minimum Drag (HIGHLY RELEVANT)
- Chapter 10 (HIGHLY RELEVANT)
Disclaimer:
See What is Professor Kirby Bot for information about Professor Kirby Bot and its answers.
ah, PK Bot. you got yourself all excited about using a bluff body relation, but if the keel is oriented as it is, the keep is not a bluff body, and using a bluff body drag coefficient and an arbitrary projected area is not the way to solve this. you want to calculate the viscous drag in the boundary layer on either side of the keel
This text is generated by AI trained by Professor Brian J. Kirby on his lectures, videos, notes, message board responses, and textbook
PK Bot’s answer:
Below is an analysis that focuses on evaluating the viscous drag acting through the boundary layers on the keel’s two faces when the keel is aligned with the flow. In this orientation, the broad faces of the keel are nearly parallel to the oncoming water flow, and therefore the drag is produced by the shear stress in the boundary layers rather than by pressure (form) drag on a bluff body.
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Assumptions and Setup
• The keel’s geometry is given by a submergence (vertical “depth”) of 38 in and chord (horizontal “length”) of 24 in.
• We assume that the chord (24 in) defines the streamwise length over which the boundary layer develops.
• Water is taken as the working fluid with density
\rho=1000\,\mathrm{kg/m^3}
and dynamic viscosity
\mu\approx1.0\times10^{-3}\,\mathrm{Pa\cdot s}\,.
• Kinematic viscosity is therefore
\nu=\mu/\rho\approx1.0\times10^{-6}\,\mathrm{m^2/s}\,.
• We take a representative boat speed of
U\approx2.57\,\mathrm{m/s}\qquad\text{(approximately 5 knots)}\,.
• Since the Reynolds number based on the chord length
\mathit{Re}_L=\frac{U\,L}{\nu}\,,
will be high (see below), we use turbulent flat‐plate drag relations.
• The wetted area on one side of the keel corresponds to a flat plate of length L and width equal to the submerged depth. We convert dimensions to SI units:
– Chord length:
L=24\,\text{in}\times0.0254\,\frac{\text{m}}{\text{in}}\approx0.6096\,\text{m}\,.
– Depth:
w=38\,\text{in}\times0.0254\,\frac{\text{m}}{\text{in}}\approx0.9652\,\text{m}\,.
Thus, the area on one side is
A= L\times w \approx 0.6096\,\text{m}\times0.9652\,\text{m}\approx0.588\,\text{m}^2\,.
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Turbulent Boundary Layer Estimate
For turbulent boundary layers over a flat plate, a typical empirical expression for the local drag coefficient is
C_f(x)=\frac{0.074}{\mathit{Re}_x^{1/5}}\,.
A common engineering estimate is to define an average drag coefficient over the plate length as
\bar{C_f}=\frac{0.074}{\mathit{Re}_L^{1/5}}\,,
where
\mathit{Re}_L=\frac{U\,L}{\nu}\,.
Calculate \mathit{Re}_L:
\mathit{Re}_L=\frac{2.57\,\mathrm{m/s}\times0.6096\,\mathrm{m}}{1.0\times10^{-6}\,\mathrm{m^2/s}} \approx1.57\times10^6\,.
Thus,
\bar{C_f}\approx\frac{0.074}{(1.57\times10^6)^{1/5}}\,.
Taking the fifth root of 1.57\times10^6:
(1.57\times10^6)^{1/5}\approx17.38\,,
so
\bar{C_f}\approx\frac{0.074}{17.38}\approx0.00426\,.
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Drag Calculation
The drag force on one side of the keel is given by
F_{D,\text{side}}=\tfrac{1}{2}\,\bar{C_f}\,\rho\,U^2\,A\,,
with A the wetted area (here, 0.588 m²). Substituting the numbers:
• Compute the dynamic pressure:
\tfrac{1}{2}\,\rho\,U^2 = 0.5\times1000\,\mathrm{kg/m^3}\times(2.57\,\mathrm{m/s})^2 \,.
Since (2.57)^2\approx6.60\,\mathrm{m^2/s^2}, we have
0.5\times1000\times6.60\approx3300\,\mathrm{N/m^2}\,.
Then,
F_{D,\text{side}} \approx 0.00426\times3300\,\mathrm{N/m^2}\times0.588\,\mathrm{m^2}\,.
Multiplying,
3300\times0.588\approx1940\,\mathrm{N}\,,
and then
0.00426\times1940\approx8.26\,\mathrm{N}\,.
Since the keel has two sides, the total drag is
F_{D,\text{total}}=2\times8.26\approx16.5\,\mathrm{N}\,.
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Reflection
The computed viscous drag on both sides of the keel is on the order of 16 N at a boat speed of approximately 5 knots. This is much lower than the bluff-body pressure drag obtained when the keel’s broad face is oriented normal to the flow. In practical design, knowing that the keel is operating “in the plane” means that viscous drag dominates the overall drag. The analysis here—in which the flat-plate drag coefficient for a turbulent boundary layer is used—is consistent with the treatment of flat plates in Professor Kirby’s materials on boundary layers (see Chapter 10.2 and Chapter 10.4 of the textbook).
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Thus, under the assumed conditions, the overall viscous drag due to boundary-layer effects on both sides of the keel is approximately 16.5 N.
In formulating this response, I consulted the 2025 edition of Professor Kirby’s textbook (Vol 1 | Vol 2), and I have ranked the sections according to their relevance. If they are marked as relevant, I encourage you to explore them:
Textbook sections:
- Chapter 10.4: Boundary Layers: Details (HIGHLY RELEVANT)
- Chapter 10.5: Designing for Minimum Drag (HIGHLY RELEVANT)
- Chapter 10.2: Predicting Drag (HIGHLY RELEVANT)
- Chapter 10.3: Boundary Layers: Overview Part 1 (HIGHLY RELEVANT)
- Chapter 10 (HIGHLY RELEVANT)
Disclaimer:
See What is Professor Kirby Bot for information about Professor Kirby Bot and its answers.