SOH-CAH-TOA excludes projectile's inertia, projectile surface area and drag (aerodynamic coefficient x air density). If you are planning to put howitzer shell down the chimney 6km away on a hot windy day, it get bit more complex
You are assuming that the moment the projectile enters flight it instantly has the same lateral velocity of the wind. I doesn't work that way at all! F=ma! The wind drag equations can be solved for force and used to find an acceleration. You would then have to use kinematics to find final distance traveled. You are very wrong.
These equations are not needed to counter wind. Use this: distance to target in yards divided by 100 times wind value in mph divided by MOA wind constant . Example 175 SMK 308 at 2610 shooting to 600 yards in 10 mph full value wind is 600 divided by 100 times 10 mph divided by a constant of 10 ( proper constant for bullet’s external ballistics) equals 6 MOA of wind drift or 36 inches of favor. The way it is shown here the wind would change many times before a calculation could be finished.
This approach to calculating wind drift is way over-simplified and results in gross over-prediction of bullet's wind drift. It ignores the fact that at t=0 when the bullet leaves the muzzle at high velocity in x-direction, it's velocity relative to the crosswind (y-direction) is zero. During the time-of-flight t=T, the bullet accelerates in y-direction from zero to some velocity smaller or equal to crosswind velocity if T is long enough. The distance traveled in y-direction will depend on acceleration in y-direction which, in turn, depends on bullet's drag force in y-direction and its Magnus force (due to spin).
Show calculation, let`s do this. Yes, wind is never mathematically uniform. Bottom line, it will push the projectile, ship or aircraft sideways. Yes, surface area matters, mass matters, things will still drift sideways. My 5 minute calculation is not meant to substitute 4 years of college, or qualify anyone in aerodynamic engineering. It is to point out that winds and currents are reasonably calculable with simple math skills for fun.
If wind speed south direction 6m/s, velocity of arrow 195km/h and distance of hitting target is 72meter, arrow weight 34gm,then please tell me where I aiming to shoot in 10 ring?
Wind deflection on objects have to take into account Bc, mass, shape, as well as any localized motion on the object being pushed(rotational). It's a dynamic relationship, so I do see how you have isolated one singular variable, but unfortunately not very useful for ballistic calculation.
That's the problem. Mr. Here tries to apply simply algebra to calculate ballistic behavior. He doesn't seem to understand the affects of arrow dynamics and how the wind is affecting the projectile. He instantly assumes that the bullet travels at wind speed which is ...
