I Coriolis acceleration of a projectile launched at the equator

[jk185]

I have a question about the Coriolis acceleration experienced by a projectile launched from the surface of a rotating body.

Say a ball is launched at 45 degree angle relative to the surface at some initial velocity v0. Let's further specify that the ball is launched due north from the equator (i.e. latitude = 0). I want to calculate the Coriolis acceleration (and its x, y, z components) that is experienced by the ball. Since the launch occurs at the equator, most would say that the Coriolis acceleration is zero, but is this true? Isn't the Coriolis acceleration only 0 at the equator since the launch angle is assumed to be 0 (i.e. the motion is parallel to the surface), meaning the velocity vector is parallel to the rotational vector? If a ball is launched at some angle relative the surface, the velocity vector is no longer parallel to the surface, meaning that there should actually be a Coriolis acceleration that is experienced by the ball.

Also, I am particularly interested in long-range ballistic travel.

Is there something that I am missing here? I have looked everywhere for an answer to this question but haven't found anything that specifically addresses this question.

Any help is appreciated here. Thanks!

 

[Bystander]

https://man.fas.org/dod-101/navy/do...cs is the branch,force imparted on the target.

 

[jk185]

https://man.fas.org/dod-101/navy/docs/swos/gunno/INFO6.html#:~:text=Exterior ballistics is the branch,force imparted on the target.

This is not exactly what I was looking for, but thank you anyways.

 

[Nugatory]

Isn't the Coriolis acceleration only 0 at the equator since the launch angle is assumed to be 0 (i.e. the motion is parallel to the surface), meaning the velocity vector is parallel to the rotational vector? If a ball is launched at some angle relative the surface, the velocity vector is no longer parallel to the surface, meaning that there should actually be a Coriolis acceleration that is experienced by the ball.

That seems to be what the standard formula for calculating the Coriolis acceleration $a_C=2\vec{\Omega}\times\vec{v}$ says, yes.

 

[jbriggs444]

Look at the Latitude 0 section and the columns for a target at 0 degrees (North) or 180 degrees (South). Look at the row for 30,000 yard range. A shot at those angles is deflected 42 yards left (west) when firing north and 42 yards right (west) when firing south. This is presumably due to Coriolis acting on the arching shot.