## Cross wind component INSTANT ANSWER

• greg
• Topic Author

### greg created the topic: Cross wind component INSTANT ANSWER

Ok this is a little tip that has given me confidence with airfield wind conditions that is instant and it will always make you do this calc in flight and especially on final with the wind sock in site.
Ok, here is the SIMPLE genius with how you think with this important technique.
Remember, simple and quick head calculations are everything when flying and after I explain this trick ill tell you why!
Get a pen and paper to draw the lines as the explanation is difficult to follow with words.

For simplicity, assume your flying heading 000deg,M ,wind is 090M deg 20kts ....
Without thinking, the crosswind component is 20 kts from your right and 0 kts headwind and 0 kts tailwind.
Next scenario.
Now your flying 000deg M and the wind is 045deg M 20 kts.
Again without thinking to much the crosswind is 10kts from your right and headwind is 10 kts on your nose.
Simple, however, an interesting relationship with wind direction and wind strength has just schematically developed with your DI (directional indicator)
Look upon your DI now and put yourself in the pilots seat.
Heading 000 deg wind 045 20kts
Imagine the whole outer circle to be 20 kts ( the forecast or calculated wind)
Now drop an imaginary line straight down from your heading bug until it intersects the base line, ( the base line stretches from 270deg to 090 deg passing thru the centre point of the DI)

Where the imagined line intersects the base line from the centre point of the DI is exactly half the length to the outer edge of the DI from the centre (ie, 50% of the 20 kts that we call the outer circle)

Similarly , from the heading bug imagine a line intersecting the vertical line from centre of the DI to the top outer edge .
This point would be 1/2 of the total length of the line from centre to top OR 1/2of 20 being 10 kts

If the centre is deemed 0 kts and the edge of the circle is deemed the full strength of the wind ie: 20 kts then the half way point would be 10 kts.or crosswind 10 kts from the right and 10 kts of headwind.

Ok
If we were heading 000deg and the wind was 022.5deg 20 kts then if you drop a imaginary line vertically down from the bug once again to meet the base line it would intersect at exactly 1/4 of the total length of the line from the centre, to the edge being 1/4 of 20 kts
Similarly, imagining a horizontal line from the heading bug intersecting the centre line it would measure 3/4 of the length from centre to top being 15 kts.
Ie: crosswind 5 kts and headwind 15 ( being 20 kts minus 5 kts)

Ok play with any example you want calling the wind 30 kts or 50 kts and saying the wind is 015deg or 035 deg and you'll find the VISUAL formula infallible. Use wind as 130 deg and you'll get a tail wind or 240 and the same infallible quick answers just show themselves to you instantly.

Get used to it and it is so easy and takes no thinking.

Now, here's the kicker .
If your on final for say runway 04 .
IFR
Waiting to break visual.
Late at night unmanned aerodrome.
No current forecast, rain squalls, heart rate up wondering the conditions.
Then, bang , your visual and all you see is the runway lights and wind sock.
Then all you do is superimpose the windsock direction and strength of wind on your DI and you will have an instant answer as your mechanical wind calculator you paid 80 bucks for sails past your head.
Therefore, continuing, you don't even have to know the hdg or the wind direction, just an idea of strength and you will instantly know if you can ,after all the getting there, LAND.......
And..... If your in a Tomahawk with like 12 kts MAX crosswind component, then you'll be glad you read this!

All the best

Greg