Why does the Cross Wind Component plus the Head Wind Component equal more Wind than what is present? Is the calculation invalid?

Question

(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?😊

Ronald Levy · Answer

@David RingsmuthIf you need a physics explanation, think back to Einstein – all motion is relative, not absolute.  Head/tail wind is relative to the airplane's x-axis.  Crosswind is relative to the y-axis.  Total wind is relative to the earth, not the airplane.

Randy Sheidler · Answer

@David RingsmuthThe math or trig function for a right triangle is: leg 1 squared plus leg 2 squared equals the hypotenuse squared.  Maybe this is what you are thinking when you are trying to add the legs together and compare that sum to the hypotenuse.

David Ringsmuth · Answer

@David RingsmuthActually, 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.

Ronald Levy · Answer

@David RingsmuthIt'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.

David Ringsmuth · Answer

@Ronald LevyRonald,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.Thanks!