Loads change shape, and supports close the load path#
Imagine pressure on one face exceeding the other by 10 Pa. On a small 0.10 m² patch, a uniform difference would push with 1 N perpendicular to the cloth. This is an illustrative load, not measured sail pressure. Air friction can also exert force along the patch.
From air motion to surface force#
A normal points perpendicular to a surface; a tangent runs along it. Let \(\mathbf{n}\) be a unit normal, a dimensionless vector of length one. Label the face it points toward “plus” and the other “minus.” These labels are not automatically high- and low-pressure faces.
Use \(p\) for air pressure (Pa), \(\mathbf{u}\) for air velocity (m/s), and \(\eta\) for dynamic viscosity (Pa s), as in Chapter 3. Choose \(\eta=1.8\times10^{-5}\) Pa s only for illustrative calculations.
A stress tensor, written \(\boldsymbol{\sigma}\), is a table of nine numbers organizing force per area in different directions on differently oriented surfaces. It has units Pa. The table times a unit normal gives a force-per-area vector.
Before its equation, \(\mathbf{I}\) is the identity table, which leaves a vector unchanged when multiplied by it. \(\nabla\mathbf{u}\) is the table of spatial rates of velocity change (1/s). A superscript \(T\) means transpose, swapping rows and columns. Adding the table to its transpose selects deformation, rather than a simple turning of the whole air parcel. For incompressible Newtonian air,
The first term is compressive pressure; the second is viscous stress. Their derivation and the force-per-area rule are in MIT’s stress and viscosity notes, equations (338)–(345). Our formula includes pressure as well as viscosity.
Write \(\boldsymbol{\sigma}^{+},\boldsymbol{\sigma}^{-}\) for stress on the two faces and \(p^{+},p^{-}\) for their pressures. Superscripts here label faces, not powers. Let \(\mathbf{q}\) be the net air force per current cloth area (N/m²). The air on the plus face acts through the direction \(\mathbf{n}\); the minus-face direction is its negative. Summing these forces gives
Without viscous stress, this reduces to pressure alone:
Higher minus-face pressure pushes toward the plus face, as in our example. Reversing the labels reverses the normal too, leaving the physical force unchanged. This sign follows from the two face forces; it is not an extra empirical law.
Summing loads over the cloth#
Let \(S\) be the current cloth surface and \(dA\) an arbitrarily small patch of its current area (m²). Unlike Chapter 1’s original area \(dA_0\), this area can change when cloth stretches. Let \(\mathbf{F}\) be the whole sail’s air force (N). An integral adds patch forces:
For a uniform flat patch this is just force per area times area. For a curved sail the force direction varies, so the vector contributions must be added, not just their positive magnitudes.
What accelerates a cloth patch?#
Inertia is the resistance of mass to acceleration. Newton’s law is force equals mass times acceleration, not force equals velocity. To account for cloth mass, use original area as in Chapter 1.
Before the next equation, define:
\(\mu_0\): mass per original area, 0.22 kg/m² in our example.
\(\mathbf{r}\): a cloth point’s current position (m). \(d^2\mathbf{r}/dt^2\) is its acceleration (m/s²), following the same point; it means take its time derivative twice.
\(J=dA/dA_0\): dimensionless local area ratio. A patch changing from 0.10 to 0.11 m² has \(J=1.10\).
\(\mathbf{f}\): internal elastic force per original area (N/m²), transmitted by neighboring cloth.
\(\mathbf{f}_d\): force per original area associated with physical energy loss (N/m²); the subscript names dissipation, conversion of mechanical energy to heat or other untracked forms.
\(\mathbf{g}\): gravity, choose (0, −9.81, 0) m/s².
Then the patch balance is
Every term is force per original area, in N/m². The factor \(J\) converts air force per current area to original-area accounting; it does not create a new force. Dissipation requires measured material behavior, not a convenient arbitrary damping value.
The relation is Newton’s law applied to a material patch; see OpenStax’s discussion of Newton’s second law. The area conversion follows from equality of the force on the same patch.
Where does the internal elastic force come from?#
Let \(E=E_m+E_b\) be the sum of stretching/shearing and bending energy from Chapter 1 (J). An elastic force acts in the direction of decreasing energy. In a one-direction example, let \(s\) be displacement (m) and \(f\) the corresponding elastic force (N). Then
The minus sign means that extending a spring pulls back toward smaller extension. For a whole sheet this same principle uses a small displacement \(\delta\mathbf{r}\) at each point and the resulting energy change \(\delta E\), measured in joules; the trial displacement is in metres. Here \(\delta\) means an infinitesimal trial change, not a new time step. As in Chapter 1, \(S_0\) is the original sheet and \(dA_0\) is a tiny original-area patch (m²). With attachments fixed,
The dot selects force along the trial displacement. Force per area times displacement times area has units J. This is the continuous counterpart of the spring rule, expressing work and stored energy; it is the principle used in MIT’s variational plate derivation, Section 3.1.2. Here “variational” means testing arbitrarily small shape changes. One still needs the chosen material energy and attachment conditions.
Why the supports matter#
Static equilibrium means a time-independent shape with no acceleration. Let \(M\) be total cloth mass (kg), and \(\mathbf{R}\) the total mast/boom support reaction on it (N). Internal forces transmit load to the supports but do not hold up the whole sheet without an external reaction. At equilibrium,
For example, an illustrative 100 N sideways air load requires an opposite sideways support reaction as well as support for weight. Loads travel through cloth into mast and boom, then into the ground.
A moment, or torque, is a force’s turning effect about a point (N m). Moments must also balance; force balance alone cannot fix where a force acts. See OpenStax on equilibrium conditions. Fixed supports transmit force without doing work because they do not move. During cloth motion, external forces change total momentum rather than necessarily adding to zero. Slow motion under artificial damping is not proof of physical equilibrium.
Lift and drag are a way of describing the same force#
Let \(\mathbf{W}\) be incoming velocity (m/s) and \(U\) its nonzero speed. Define the unit wind direction \(\mathbf{e}\), drag \(D\) (N), and remaining transverse air force \(\mathbf{L}\) (N). “Transverse” means perpendicular to incoming wind, not necessarily horizontal. The dot product projects onto a direction:
This is geometric decomposition, not two additional force mechanisms. With wind along the boom and an illustrative air force (10, 3, 40) N, drag is 10 N and transverse force is (0, 3, 40) N. The norm \(\|\mathbf{L}\|\) means its magnitude: approximately 40.11 N, without a sign. A signed horizontal component is different. Weight and reactions are not included in aerodynamic force \(\mathbf{F}\). At zero wind there is no unique wind direction for this decomposition.