- METHOD OF ESTIMATING DROGUE SIZE FOR
- LARGE BOAT (60 FT. LOA, 60,000 LBS DP.)

The incidence of breaking wave capsize diminishes rapidly with increase in displacement. Larger yachts can, however, be accelerated to wave speed (surfing speed) by a breaking wave. The boat can reach a high velocity and may plunge into the preceding wave or may broach and roll down. This behaviour can be prevented by the use of a properly designed drogue.

To select a drogue size for a boat displacing 60,000 lbs. The following conditions were assumed:

- Breaking wave with a crest velocity of 39 ft./sec. corresponding to a wave length of 300 ft. This is representative of a wave height of approximately 40 feet.
- The boat has been accelerated to wave speed and is riding the crest at a slope of 0.5 (27 deg.).
- The hull drag is zero because the water in the crest is. moving at the wave speed with the boat.
- The wind drag is assumed to be 6000 lbs. (75 mph wind).
- The boat is in a steady state with no horizontal or vertical acceleration.
- The drag of the drogue must equal the horizontal component of the buoyancy force plus the wind force. Thus the boat will be held on the face of the wave until the crest dissipates.

Figure 1C shows the assumed conditions:

- FB = Horizontal component of buoyancy force = 60,000 x 0.5 = 30,000 lbs
- FA = Air drag = 6000 lbs
- FD = Drogue force = C
_{D}A*P*/2 V^{2 }= 30,000 + 6000 = 36,000 lbs - C
_{D}= Drag coefficient of drogue @ 1.0,*P*/2@ 1.0 - A = Drogue cross section area
- V = Velocity of drogue = boat velocity = 39 ft./sec.
- Required drogue area = 36,000/(39)
^{2}= 23.7 sq. ft. - Drogue diameter =
- Ö
__4A__= 5.5 ft. - P