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C4 K2 VS0
P =
tanβ
where–
P :pressure(p.s.i.); C4 :Value of C4 =0.078C1 (with C1 computed under Sec. 23.527); K2 :hull station weighingfactor,determinedin accordance with .gure 2 of appendix I of this part;
VS0 :seaplanestalling speed(knots) withlanding .apsextended in the appropriate position and with no slipstream e.ect; and
β :angle of dead rise at appropriate station.
(2)
The unsymmetrical pressure distribution consists of the pressures prescribed inparagraph(c)(1) of thissectionononesideof thehull ormain .oatcenter-line and one-half of that pressure on the other side of the hull or main .oat centerline in accordance with .gure 3 of appendix I of this part.

(3)
Thesepressures are uniform andmustbe applied simultaneously overthe entire hullormain .oatbottom.Theloadsobtained mustbecarriedintothesidewall structure of the hull proper, but need not be transmitted in a fore and aft direction as shear and bending loads.]

Amdt. 23-45, E.. 09/07/93
FAR23.535:[Auxiliary .oatloads.]
(a)
[General. Auxiliary .oats and their attachments and supporting structures must be designedfor the conditionsprescribedin this section.In the cases speci.edinpara-graphs(b)through(e) ofthis section, theprescribed waterloads maybedistributed overthe .oatbottomtoavoid excessivelocalloads,usingbottompressuresnotless thanthoseprescribedinparagraph(g) of this section.

(b)
Step loading. The resultant water load must be applied in the plane of symmetry of the .oat at a point three-fourths of the distance from the bow to the step and mustbeperpendiculartothekeel.The resultantlimitloadis computed asfollows, except that the value of L need not exceed three times the weight of the displaced water when the .oat is completely submerged :

2
C5 VS0 W
L =
23 23

/ tanβS / 1+ry2、、
where–
L :limit load(lbs.);
C5 :Constant C5 =0.0053;

VS0 :seaplanestalling speed(knots) withlanding .apsextended
inthe appropriateposition and with no slipstream e.ect; W :seaplanedesignlanding weightinpounds; βS :angle of dead rise at a station主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 .oat to the radius of gyration in roll.
(c) Bow loading. The resultant limit load must be applied in the plane of symmetry of the .oat at a point one-fourth of the distance from the bow to the step and must
′
Elodie Roux. Septembre 2003
Subpart C : Structure
be perpendicular to the tangent to the keel line at that point. The magnitude of the resultantloadisthat speci.edinparagraph(b) of this section.
(d)
Unsymmetrical step loading. The resultant waterload consists of a component equal to0.75timestheload speci.edinparagraph(a) ofthis section and a side component equal to 0.25 tanβ times theload speci.edinparagraph(b) of this section.The sideload mustbeappliedperpendicularly totheplaneof symmetry of the .oatat a point midway between the keel and the chine.

(e)
Unsymmetrical bow loading. The resultant water load consists of a component equal to0.75timestheload speci.edinparagraph(b) ofthis section and a side component equalto0.25tanβ times theload speci.edinparagraph(c) ofthis section.The side load mustbe appliedperpendicularly totheplane of symmetry at apoint midway between the keel and the chine.

(f)
Immersed .oat condition. The resultant load must be applied at the centroid of the cross section of the .oat at a point one-third of the distance from the bow to the step. The limit load components are as follows :

vertical = ρgV
2
2
3
Cx ρV
(KVS0)
aft =
2
2
2
3
Cy ρV
(KVS0)
side =
　
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