However, your assertion that there would some sort of wind shear going north (or south) from the equator is a fallacy.

Assuming none of the air moves ever and every atom stays exactly where it is, as though it were a solid ... you'd still be wrong. We see shear forces even in the solid and liquid of a round earth. This is why the earth bulges at the equator according to RE.

Your mixing things up here. Yes, of of curse there are forces, and despite them adding up to only .034M/s/s, they have an effect. I am not unaware of them. However, that really has nothing to do with the rate at which one particle moves WRT it's neighbor. For wind shears to occur, the rate must be larger, considerably larger, than the dynamic noise (thermal, turbulence) but it's not, it's vanishingly small.

If the air moves North or South, that air has mass. And you need to apply a force to either speed it up or slow it down. This as mentioned is just simple Newtonian physics.

Yes, I agree, but I'm not sure why you are mentioning this as it does not apply to resolving the rotating atmosphere problem. But let's see if we can apply it nonetheless. Let's assume there was enough of a velocity gradient stemming solely from the rotational speed of the atmosphere along a given longitude to produce a minor disturbance, such as Helmhotz waves. This will, as you accurately state, require energy to deflect the flow of air north and south of the shear line. Energy that must come from the difference in velocity across the shear. Which is just not available from the rotation of the atmosphere...

The velocity of the atmosphere is given by the simple equation:

**v= V**_{o}sin(a), where

**a** is the angle from the axis of rotation and

**V**_{o} is the velocity at the equator.

Taking the derivative we get simply,

**dv/da = V**_{o}cos(a) Such that as the velocity approaches it's highest, at the equator, the velocity gradient, which is tiny at it's worst, approaches zero.

So your original assertion:

So, at the surface the air must be moving with the earth then at 1040mph ... at the equator. As I move north and south, that air needs to decelerate to a full stop (well but for what must be a huge cyclone at the poles).

Is just not correct.

You also, correctly, said

The air, according to your theory must be changing velocity to match location imperceptibly...

Which is exactly what we see.

In fact, at it's worst the velocity gradient due to the rotation of the atmosphere is about 0.25 kph per kilometer as one leaves a pole heading toward the equator. This diminishes rapidly to zero at the equator.