True wind angle from true wind direction and heading

Is it possible to find a closed form formula to calculate the TWA (true wind angle), in the range [-180, 180), from the TWD (true wind direction) and the heading angle, both relative to north and in the range [0, 360)?

• Welcome to outdoors.SE! I know nothing about sailing, but as a physicist I hear this as a vector addition problem, and it sounds to me like you could do this, but it would require that you know both the boat's speed relative to the water and the wind's speed relative to the boat. Perhaps there are approximations you could make in the limit where the boat's speed is low compared to the wind's speed. GPS could also probably fill in one of the necessary variables.
– user2169
Mar 6 at 1:27
• @BenCrowell GPS as you suggest would be good - better than water speed which would bring in yet another velocity vector if there's any current. However looking at the definitions, because the apparent wind (which includes the vessel's motion) doesn't come into the question, I think it's just manipulating angles, but don't have time to get pen and paper out Mar 6 at 12:20
• @BenCrowell it is a vector addition problem but you need the speed of the boat relative to the land rather than the water, which is a whole extra complication Mar 6 at 12:34
• My question is why you'd want to do this. You sail on apparent wind rather than true wind, so all you really need to know is course, speed, apparent wind direction. Anything else can be calculated from those, but isn't required for the most part. Mar 6 at 12:38
• The question is about wind and not about apparent wind, so speed does not come into it, surely? Mar 6 at 22:06

If
D is true wind direction
A is true wind angle
then
A = D - H
and if A > 180 then A = A - 360
or if A < -180 then A = A + 360

```Example 1
H
/   H =  020
/    D =  090
/     A =   70
/ A
/_ _ _ _ _ _ D

Example 2
D
/   H =  090
/    D =  020
/     A =  -70
/ A
/_ _ _ _ _ _ H

Example 3
H          D
\        /   H =  340
\      /    D =  020
\ A  /     A = -320 + 360
\  /        =   40
\/

Example 4
D          H
\        /   H =  020
\      /    D =  340
\ A  /     A =  320 - 360
\  /        =  -40
\/
```
• I am assuming you are only talking true directions, not apparent directions. Mar 6 at 22:44

By definition

TWD = True Wind Direction (relative to true north)

TWA = True Wind Angle (relative to the bow)

Then,

• TWD = Angle between wind and North
• TWA = Angle between wind and heading

=> TWA = TWD - Heading

Then, as all angles are expressed module 360 (in degrees), then any angle in degrees can be expressed as:

``````Angle ≡ Angle + k * 360
``````

(where k can be any integer, including 0 and negative numbers)