Hi Nick,
You have provided me with as much information as anyone has. For that, I am truly appreciative.
But, for the pleasure of pushing the envelope, or pushing buttons, the following is offered.
Your 13-feet per second represents a crosswind of 9-miles per hour. Drag is based on the square of the wind's velocity, therefor a crosswind of 25-miles per hour will exert a force of 68 pounds.
Assuming that the tail-rotor is capable of delivering 4 pounds of thrust per horsepower, then the required power is
( 68 lbs / 4lbs ) = 17 hp.
The craft's gross weight to power is
(11,700 lbs / 1600 hp) = 7.3 lbs per hp, therefor the weight loss is
(17 hp * 7.3 lbs) = 124 lbs.
The craft's useful load is 4627 lbs., therefor the loss of payload is
(124 lbs / 4627 lbs) = 2.7%.
Margin of error; +/- 2000%.
Sources of information; ~
Sikorsky. S76 ~
Drag.
Perhaps the advantage may not be significant for a single rotor helicopter; particularly if part of this VS is necessary to 'balance' the drag at front of the craft. For a helicopter, with dual laterally mounted rotors, which are in close proximity to each other, it might be a big advantage. This will be particularly true if the yaw is by opposed longitudinal cyclic.