al applied

Suppose that the plane is steered in the direction with an acute angle α to AB (counted in opposite sense to θ) so that the plane's resultant velocity in perp. to AB is zero, i.e.

w sin θ = v sin α ... (1)

Then in forward journey, the resultant velocity is:

v cos α + w cos θ along AB

In backward journey, the resultant velocity is:

v cos α + w cos θ along BA

Thus:

t1 = d/(v cos α + w cos θ)

t2 = d/(v cos α - w cos θ)

t1t2 = d2/(v2 cos2 α - w cos2 θ)

= d2/(v2 - v2 sin2 α - w2 + w2 sin2 θ)

= d2/(v2 - w2) by (1)

t1 + t2 = d/(v cos α + w cos θ) + d/(v cos α - w cos θ)

= 2dv cos α/(v2 cos2 α - w cos2 θ)

= 2d√(v2 - v2 sin2 α)/(v2 cos2 α - w cos2 θ)

= 2d√(v2 - w sin2 θ)/(v2 cos2 α - w cos2 θ) by (1)

2011-08-02 08:40:34 補充：
In backward journey, the resultant velocity should be v cos α - w cos θ along BA

2011-08-02 08:41:59 補充：
And for t1 + t2, the denominator part would be v^2 - w^2 after simplification.