Coriolis Force - North/South
So, recall from the great circle explanation that when the globe isn't spinning, something shot North or South will continue to go that way, since all longitude lines are great circles. However, once you start spinning the globe, you add movement in the direction of spin as well, and how much there is depends on how far you are from the poles. At the poles, there's no "sideways" speed, and at the equator you have the maximum speed. So something moving North or South will find that its spin-based speed will soon no longer match the spin speed of the ground under it.
 We have two mecha (yes, I know their names, but I'm trying to avoid too much trademark-mentioning here) on a small planetoid with a bumpy orange surface, a rapid spin and enough gravity that their shots will follow a great circle. The one on the equator takes the first shot. In the time the yellow shot takes to reach the latitude of its target, both mecha will have traveled in the direction of the blue arrow to the next black longitude line. The only real force on the yellow projectile is gravity. In other words, there's a force directed to the center of the planetoid, which will keep the shots from traveling in a straight line. Our combatants are aware of this force and are taking it into account...but they're not taking rotation into account.     Notice that the green mech only travels 7 units to reach the next longitude, while the blue and gray mech travels 9 units in the same amount of time. That means that the yellow projectile has a speed of 9 to the right as well, in addition to the speed at which it's fired North. To find where the projectile is after the two mecha have rotated around to their new positions can be approximated by moving it North to the green mech's latitude and then 9 units along a great circle to the East (an exact location requires more math, but it won't be too far from the approximated spot). We've taken the Northward velocity, added it to the Eastward velocity, and then had gravity bend it into an arc around the planet (we're not worrying about altitude right now, assume it's going just fast enough to stay at the same height above the ground).     The yellow dots drawn in show the path that the firer sees his shot take. As the shot moves North, it's going East faster than the ground below it is, so seems to curve right. It's also a bit South, because of the whole great circle effect, but the important point is that it's passing East of the target.     If the blue and grey mech had wanted to hit his target, he would have had to aim a little West of it to compensate for the fact his shot starts off moving Eastward faster than the target is. Now the green mech returns fire, but he hasn't learned the lesson of compensating for spin yet, and he simply aims due South. Again, in the time it takes the yellow shot to move down to the equator, both mecha will rotate to the next black longitude line. The green one will move 7 units, the blue and gray one will move 9 units. Again, we can approximate the final location of the shot by moving it to the correct latitude and then East 7 units along a great circle. There's no North/South oddity in the approximation this time, since the equator is a great circle, but a more exact solution would put it a little South since it had Eastward motion while North of the equator.     The shot drifts to the right again, missing its target by 2 units to the West. Since the green mecha isn't moving East as quickly as his foe, he'd need to "lead" his target by aiming a bit to the left (East).

Just like in the 2-D case, there isn't actually a force pulling either shot to the right. Instead, it's just a case of the ground moving under the shot, either faster or slower depending on whether you're shooting South or North.

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