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§ 25.535
Auxiliary float loads.
(a) General. Auxiliary floats and their attachments and supporting structures must be designed for the conditions prescribed in this section. In the cases specified in paragraphs (b) through (e) of this section, the prescribed water loads may be distributed over the float bottom to avoid excessive local loads, using bottom pressures not less than those prescribed in paragraph (g) of this section.
(b) Step loading. The resultant water load must be applied in the plane of symmetry of the float at a point three-fourths of the distance from the bow to the step and must be perpendicular to the keel. The resultant limit load is computed as follows, except that the value of L need not exceed three times the weight of the displaced water when the float is completely submerged:
\[ L = \frac{C_5\ V_{S_0^2}\ W^{\frac{2}{3}}}{tan^{\frac{2}{3}} \beta_s \left( 1 + r_y^2 \right)^{\frac{2}{3}}} \]
where—
L = limit load (lbs.);
C5 = 0.0053;
VS0 = seaplane stalling speed (knots) with landing flaps extended in the appropriate position and with no slipstream effect;
W = seaplane design landing weight in pounds;
βS = angle of dead rise at a station 34 of the distance from the bow to the step, but need not be less than 15 degrees; and
ry = ratio of the lateral distance between the center of gravity and the plane of symmetry of the float to the radius of gyration in roll.
(c) Bow loading. The resultant limit load must be applied in the plane of symmetry of the float at a point one-fourth of the distance from the bow to the step and must be perpendicular to the tangent to the keel line at that point. The magnitude of the resultant load is that specified in paragraph (b) of this section.
(d) Unsymmetrical step loading. The resultant water load consists of a component equal to 0.75 times the load specified in paragraph (a) of this section and a side component equal to 3.25 tan β times the load specified in paragraph (b) of this section. The side load must be applied perpendicularly to the plane of symmetry of the float at a point midway between the keel and the chine.
(e) Unsymmetrical bow loading. The resultant water load consists of a component equal to 0.75 times the load specified in paragraph (b) of this section and a side component equal to 0.25 tan β times the load specified in paragraph (c) of this section. The side load must be applied perpendicularly to the plane of symmetry at a point midway between the keel and the chine.
(f) Immersed float condition. The resultant load must be applied at the centroid of the cross section of the float at a point one-third of the distance from the bow to the step. The limit load components are as follows:
\[ \text{vertical} = _{\rho g}V \\ \text{aft} = C_{x}\ _{2}^{\rho}V^{\frac{2}{3}} \left( K\ V_{S_0} \right)^2 \\ \text{side} = C_{y}\ _{2}^{\rho}V^{\frac{2}{3}} \left( K\ V_{S_0} \right)^2\]
where—
ρ = mass density of water (slugs/ft.2);
V = volume of float (ft.2);
Cx = coefficient of drag force, equal to 0.133;
Cy = coefficient of side force, equal to 0.106;
K = 0.8, except that lower values may be used if it is shown that the floats are incapable of submerging at a speed of 0.8 VS0 in normal operations;
VS0 = seaplane stalling speed (knots) with landing flaps extended in the appropriate position and with no slipstream effect; and
g = acceleration due to gravity (ft./sec.2).
(g) Float bottom pressures. The float bottom pressures must be established under § 25.533, except that the value of K2 in the formulae may be taken as 1.0. The angle of dead rise to be used in determining the float bottom pressures is set forth in paragraph (b) of this section.
[Doc. No. 5066, 29 FR 18291, Dec. 24, 1964, as amended by Amdt. 25-23, 35 FR 5673, Apr. 8, 1970]