(Sine(30)+Sine(60)) > 1 by 25%.
Crosswind+Headwind > 1, except when one of the components is zero.
Please explain the physics that justify using this formula to calculate Wind Components?
Actually, the question's implications should be more easily obvious using Sine(45), which is 0.707…
It is impossible that total wind is 40% greater at a 45 degree angle.
What is really going on with the Wind Component formula?
Shouldn't the total Wind of both Headwind and Crosswind components equal 1?
Below CW is the degree value of Cross wind, and HW is the degree value of Head wind.
Sine(CW)- (1-(Sine(HW)+Sine(CW)) ) /2=True(CW)
Sine(HW)- (1-(Sine(HW)+Sine(CW)) ) /2=True(HW)
This would make the sum of wind components always = 1.
It's not physics, it's trigonometry. Think of the classic 3-4-5 right triangle. The sides (head/tail wind components) are 3 and 4, but the hypotenuse (wind velocity vector) is 5, not 7. Find a high school trig text for the full explanation.
My two equations are trigonometry results, and should add up to the total wind, if the equation fits the problem.
The question remains unanswered.
The result of the two wind components must not be greater than the available wind. Using the sine of the crosswind degrees, and the sine of the headwind degrees combined results in a total wind greater than the wind available.
The answer is a physics problem, because wind physics are what we are dealing with.
I’m sure you were trying to help.
Please retract your answer.
45° Crosswind is simultaneously 45° Headwind.
Sine of 45 degrees is .70
So the sum of crosswind and headwind is 140% of the available wind.
I can’t be the first person to see this?
This equation (sine of the wind angle) must not be the correct solution to finding the wind component.
You're assuming that head/tailwind component plus crosswind component cannot be more than the total wind velocity. I'm not sure where you got that but it's without basis. The magnitude of the vector sum of two vectors is always less than the sum of the individual vector magnitudes unless the vectors have the same direction. That's a basic trigonometric axiom.
I was memorizing crosswind values, researching presentations on solving for both Crosswind and Headwind. The graphics shown resolved to the two trig equations I used here. I believe they are wrong.
° - %
10 - 17 ~ 1/6
20 - 34 ~ 1/3
30 - 5 ~ 1/2
40 - 64 ~ 2/3
45 - 70
50 - 77 ~ 3/4
60 - 88 ~7/8
70 - 94
80 - 98
90 - 100
If the crosswind component is correctly calculated from the sine of the degree angle, than the headwind component must be the remainder of available wind.
If a 45 degree crosswind is a .7 of available wind, the headwind must be .3.
So the graphics which show the results of the two sines are wrong. This seems like the answer to the problem I have with the solutions presented on the web.