# Where exactly is the Coriolis acceleration direction on my picture?

I solved the following problem in my homework:

A bullet is fired from a gun towards a target at distance of D=800 meters. The bullet speed is V=800 meters per second. The shot is taken at latitude θ=45°, towards the north: What is the deviation of the bullet from the target?meters. The bullet speed is V=800 meters per second. The shot is taken at latitude θ=45°, towards the north: What is the deviation of the bullet from the target?

This is the drawing of the problem:

I Solved it like so:

Time for bullet to reach target = 1 second (target at 800m, bullet speed is 800 m/s).

W(Earth's rotational speed) = 2π/(24*60*60) = 0.0000727.

a(coriolios acceleration) = -2W×V = -2*(0.0000727)*800*sin(90-45) = 0.082 (m/s^2)

And then the deviation is x=at^2 / 2 = 0.041 meters.

Where is the 0.041m deviation in this drawing exactly?

On your diagram it is difficult to show the deflection other than to say it is out of the screen or East on the Earth.
In you diagram $\vec W \times \vec V$ is into the screen and so $-\vec W \times \vec V$ is out of the screen.

• Out of the page is typically shown as a circle with a dot $\cdot$ in the middle. Into the page is a circle with an $\times$ in it. Jun 12 '18 at 22:58
• Oh because the earth rotates eastwards, then the bullet will continue in straight line while the earth rotates a bit eastwards making this gap? Jun 13 '18 at 17:07
• @TTnote When fired the bullet has a component of velocity eastwards equa to the speed of the Earth eastwards. As the bullet moves North its constant component of velocity eastwards becomes larger than the velocity of the Earth eastwards. Jun 13 '18 at 20:30

The answer in this question is incorrect.

First there are 2 deflections

1) In the Northern Hemisphere a deflection right (left in South) 2) Shooting due west the target rises as earth rotates so creates a low impact (downward deflection) and shooting due east the target falls so the impact is high. Shooting due north or south and there is no effect.

Second the formula is wrong. This can easily be seen by simply writing the units for each term and canceling out. If we want m as the result then we need a formula that looks like (say) m/sec * sec as the sec units cancel leaving m.

The correct formula for the lateral deflection is:

Horizontal_deflection = Omega * range * sin(Latitude) * time_of_flight

Where omega = 2 * pi / (24*60*60)

The vertical formula is more complex and the explanation of same can be found on pp113-115 here:

https://archive.org/details/AppliedBallisticsForLongRange/page/n123

And of course third that this is only true in a vacuum - the time of flight to 800 m for a 308/7.62 NATO projectile is about 1.5s due to the shedding of velocity from air resistance