Amit & Deepali Sinha Foundation

The Triangle Behind a Southeastern Wind

Irodov 1.6DifficultyUniqueness

Problem Irodov 1.6

A ship sails due east along the equator at a constant velocity v0=30v_0 = 30 km/h. A wind, measured relative to the ground, blows from the southeast at a velocity v=15v = 15 km/h, its direction making an angle φ=60∘\varphi = 60^\circ with the equator. Find the velocity v′v' of the wind and the angle φ′\varphi' it makes with the equator, both as observed in the reference frame fixed to the moving ship.

The two given ground-frame velocities drawn from a common point A, with a compass rose at the right fixing the bearings. v_0 is the ship's velocity, due east; v is the wind's, at the given angle φ to the dashed equator.

The two given ground-frame velocities drawn from a common point AA, with a compass rose at the right fixing the bearings. v0\vect v_0 is the ship’s velocity, due east; v\vect v is the wind’s, at the given angle φ\varphi to the dashed equator.

Given
  • v0=30v_0 = 30 km/h
  • v=15v = 15 km/h
  • φ=60∘\varphi = 60^\circ
Find
v′v' and φ′\varphi', the wind’s speed and bearing relative to the ship

Solution

Step 1Set up the two ground-frame velocities

Take east as the xx axis and north as yy. The ship’s velocity is v0=(v0,0)\vect v_0 = (v_0,0). A southeastern wind blows from the southeast toward the northwest, so it carries a westward and a northward component; measuring its angle φ\varphi from the equator (the line it is blowing across) on the upwind side,

v=(−vcos⁡φ, vsin⁡φ).(1)\vect v = (-v\cos\varphi,\ v\sin\varphi). \tag{1}
Step 2Subtract to get the velocity relative to the ship
v′=v−v0=(−vcos⁡φ−v0, vsin⁡φ).(2)\vect v' = \vect v-\vect v_0 = (-v\cos\varphi-v_0,\ v\sin\varphi). \tag{2}
The velocity triangle this subtraction closes. Side AB is v_0 and side AC is v, so the included angle at A is 180°-φ; the closing arrow from B to C is v', making the angle φ' at B.

The velocity triangle this subtraction closes. Side ABAB is v0\vect v_0 and side ACAC is v\vect v, so the included angle at AA is 180∘−φ180^\circ-\varphi; the closing arrow from BB to CC is v′\vect v', making the angle φ′\varphi' at BB.

Step 3Its magnitude, by the law of cosines on the triangle
v′2=(vcos⁡φ+v0)2+(vsin⁡φ)2=v2+v02+2v0vcos⁡φ.(3)v'^2 = (v\cos\varphi+v_0)^2+(v\sin\varphi)^2 = v^2+v_0^2+2v_0v\cos\varphi. \tag{3}

Taking the square root and substituting v0=30v_0=30 km/h, v=15v=15 km/h, φ=60∘\varphi=60^\circ,

v′=v02+v2+2v0vcos⁡φ=302+152+2×30×15×0.5 km/h.v' = \sqrt{v_0^2+v^2+2v_0v\cos\varphi} = \sqrt{30^2+15^2+2\times30\times15\times0.5}\ \text{km/h}.
v′=v02+v2+2v0vcos⁡φ≈40 km/hv' = \sqrt{v_0^2+v^2+2v_0v\cos\varphi} \approx 40\text{ km/h}
(4)
Step 4Its bearing relative to the equator

Since vx′=−(v0+vcos⁡φ)<0v'_x=-(v_0+v\cos\varphi)<0 and vy′=vsin⁡φ>0v'_y=v\sin\varphi>0 (Step 2), v′\vect v' points into the second quadrant, west of north; the acute angle φ′\varphi' it makes with the equator, measured from due west, is the angle at vertex BB of the triangle, the arctangent of the ratio of its component magnitudes.

tan⁡φ′=∣vy′∣∣vx′∣=vsin⁡φv0+vcos⁡φ\tan\varphi' = \frac{|v'_y|}{|v'_x|} = \frac{v\sin\varphi}{v_0+v\cos\varphi}
tan⁡φ′=vsin⁡φv0+vcos⁡φ\tan\varphi' = \dfrac{v\sin\varphi}{v_0+v\cos\varphi}
(5)
φ′=arctan⁡15×sin⁡60∘30+15×cos⁡60∘≈19∘\varphi' = \arctan\dfrac{15\times\sin60^\circ}{30+15\times\cos60^\circ} \approx 19^\circ
(6)
Check

If the wind died out (v→0v\to0) the triangle collapses to v′=−v0\vect v'=-\vect v_0: the wind would appear, relative to the ship, to blow straight out of the east at speed v0v_0, i.e. φ′→0\varphi'\to0, and (5) indeed gives tan⁡φ′→0\tan\varphi'\to0 in that limit.

Remark

This is exactly the ”apparent wind” every sailor reckons with: the wind felt aboard a moving vessel is never the true, ground-frame wind but its vector difference with the ship’s own velocity, so a following or beam wind can appear to blow from a substantially different bearing, and even a day with no true wind at all still delivers a stiff headwind if the ship runs fast enough.