I've been doing using stereographic paper to chart thermal winds recently, it's time consuming when done in mass quantities...so I wanted to find a geometric calculation to shorten the calculation time. Please Register or Log in to view the hidden image! Two vectors with the properties horizontal angle and wind speed. To find the resultant component I did this. \(V_a = [\theta_a, A]\) \(V_b = [\theta_b, B]\) Note these angles could be for example, 210 and 230 degrees...making the actual angle 20. Resultant vector should be. \(C = |(sqrt([Asin(\theta_a)+Bsin(\theta_b)]^2 + [Acos(\theta_a)+Bcos(\theta_b)]^2)|\) Angle A-C should be... \(\frac{sin(\phi_{A-C})}{B} = \frac{sin(|\theta_a - \theta_b|)}{C}\) Then solve for \(\phi_{A-C}\) Then you need to solve for \(\phi_{A-D}\) assuming \(\phi_{A-D} = 180-\phi_{A-C} - 90\) The resultant \(D\) length should be... \(asin(\phi_{A-D}) = \frac{D}{A}\) Then solving for D.... the orange vector...ugh...Q...should then be the length \(Q = sqrt(A^2 - D^2)\) Which by the way, is the rate of vertical advection resulting from thermal wind between 2 pressures. Does anyone see mistake?
