You are used to the question giving the difference between TAS and G/S as a wind component. This wind component is simply added to or subtracted from the TAS to find the groundspeed. This is rather a simplification of the issue, to reduce the complexity of the question.
In this example the wind component is not given. Instead you are given wind direction and speed and the desired track to fly outbound. Obviously for a PSR the return track must be the reciprocal of the outbound track.
In this case to calculate the groundspeed, instead of just adding or subtracting a given component you need to work on the CRP-5, the vector-calculator side. All instructions assume you use the wind-down method.
1. Place a wind mark (turn the wind direction to the "True Heading" arrow, count down from the centre bug by the wind speed and place a small cross)
2. Place the centre bug on the TAS
3. Turn the outbound track to the "True Heading" arrow - note this is only an approximate heading as it is the required track.
4. Find the drift under the wind cross.
5. Turn the track (towards the wind cross for the first turn) until the track sits against the drift on the drift ark at the top corresponding to the drift under the wind cross.
6. Check the drift under the wind cross, and repeat step 5 and this step if required.
7. Read the groundspeed under the wind cross
Note this groundspeed as "O", the outbound G/S.
Repeat steps 3 to 7 using the return track, note the groundspeed as "H", the homebound groundspeed.
Use O and H in the PSR formula.
Notice that with any crosswind component the difference between groundspeed and TAS is not the same on reciprocal tracks, i.e. the wind components are not the same but negative as they so often are on exam questions.
For example, imagine a tailwind, not a direct tailwind but a tailwind from the left, on the outbound leg. The aircraft would be heading left of track to account for drift, reducing the tailwind component. Reversing track the wind is a headwind from the right. The aircraft would be turned a little right to account for drift, increasing the headwind component. Therefore any headwind component will always be greater than the tailwind component on a reciprocal track, unless the wind is straight down the track when they are equal.
Try this example (from the JAA question bank):
Wind 045º/50kts
Track 090º out (so 270º home)
TAS 180 kts
How far can you go out and return within 1 hour?
Find groundspeeds, use the PSR formula with an endurance of 1 hour.
You should find groundspeeds of about 141 kts out, 212 kts home (within 1 kt) and a distance of 85 nm (within 1 nm).
Send Clowns
Gen Nav
BCFT
P.S. Although we can offer the brush up too, I can recommend Steve as a good bloke to get help through.