I am a little confused about the force in psi applied to a bullet in a cross wind. The physics formula for the force applied to an object is .00256 x (speed in mph) squared which is exponential in result. The ballistic calculators double the drift if you double the MPH which is linear.
Maybe this will help get you going in the right direction.
You're talking about two different things:
1. The force applied to a bullet by a crosswind, which would use units of force (kg or lbf etc) and
2. Wind deflection, or "drift" which would use units of length (mm, inch, etc)
(psi is a unit of pressure, not force)
But you don't need to think in terms of force (or pressure) to calculate wind deflection. You can ignore it.
What I always remember for figuring wind deflection (and this helps me remember/understand ballistic coefficient) is this:
Wind Deflection = [Lag Time] x [Wind Speed] x [Any trigonometric correction factors for wind direction]
where
[Lag Time] = [Time of Flight] — [Time of Flight in a Vacuum]
and where
[Time of Flight in a Vacuum] = [how long it would take your bullet to cover x yards if its velocity at x yards was equal to, or the same as, its muzzle velocity ... in other words, if it didn't slow down at all]
The concept of "lag time" is what makes me remember that "the more streamlined a bullet is, the less drag it creates as it passes through the air, which means the less it slows down, compared to that same bullet flying through a vacuum -- or outer space -- where it doesn't have to push any air molecules out of its way." (A bullet flying through outer space will continue at the same speed forever, or until it hits something that has mass.)
One thing to remember is that in a strong wind (or
any wind, but it's easier to visualize if you consider a strong crosswind), a bullet's wind deflection increases exponentially, in much the same way that its acceleration downward, due to gravity, increases exponentially (10 meters/sec² or "32 feet per second per second"). In other words, wind deflection doesn't increase lineally, but exponentially. Another way to say this is that "the more
time that the crosswind acts upon the bullet, the
faster the bullet moves laterally" ... at least until the bullet's lateral speed "catches up to" the wind speed, at which point their lateral speeds will match. In other words, when the bullet exits the muzzle, it is not moving laterally (sideways) at all. But the longer the bullet is suspended in the air, the more closely it approaches the lateral speed of the air through which it flies, until eventually the bullet's lateral speed will match the wind speed, the same way a tumbling umbrella flies down the street faster and faster until eventually it matches the wind speed. In some ways, a bullet flying through the air is like a boat moving through the water...they're both subject to "set and drift."
Let's say you have a 40 mph (~59 ft/sec) crosswind at 90° (say 9 o'clock).
At a range of 0 yards, the bullet has a lateral speed of 0 ft/sec. It's not moving laterally at all...
yet.
At 100 yds, its lateral speed will be let's say 3 ft/sec (not real numbers -- just to illustrate). It's accelerating laterally.
At 200 yds, its lateral speed will be let's say 8 ft/sec. (Yardage doubled, but deflection
more than doubled.)
At 400 yds, its lateral speed will be let's say 20 ft/sec. (Yardage doubled again, but deflection more than doubled.)
At 800 yds, its lateral speed will be let's say 50 ft/sec.
In other words, the bullet's lateral speed due to the wind increases parabolically.
That means its wind deflection will
also increase parabolically, the more time that the wind acts upon it.
Bryan Litz's book "Applied Ballistics for Long Range Shooting" is where I read about all this stuff, and I hope that I'm not mangling the information in it too badly ... if I am, I hope Bryan or someone will set the record straight. (That book is a GREAT resource, by the way ... I learned a ton from it.)
F=m * dv/dt
I will trust my ballistics calculator, and the target.
