Question about the Coriolis effect

Click For Summary

Homework Help Overview

The discussion revolves around the Coriolis effect as it pertains to the deviation of a bullet fired towards a target on Earth's surface. The original poster presents a scenario involving a bullet fired at a distance of 800 meters with a speed of 800 meters per second from a latitude of 45° and seeks to understand the implications of the calculated deviation.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants explore the meaning of the calculated deviation of 0.041 meters and seek clarification on its representation in relation to the target. There is a focus on the vector nature of Coriolis acceleration and its directional implications.

Discussion Status

The discussion is ongoing, with participants providing insights into the vectorial aspects of the Coriolis effect and suggesting ways to visualize the deviation. Some participants express a desire for a visual representation to better understand the deviation's location relative to the target.

Contextual Notes

Participants mention the complexity of the Coriolis effect and its dependence on angular velocity and the velocity of the bullet. There is an acknowledgment of the challenges in visualizing the deviation in the context of the problem's setup.

Foruer
Messages
7
Reaction score
0

Homework Statement


It's a question about the deviation of a bullet fired on Earth's surface:
**I solved the question, but I can't figure out what exactly this deviation means, where is it on the following question's picture:

XTc8Gya.jpg

The problem:
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?

Homework Equations


ac = -2W×V

The Attempt at a Solution


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
ac = -2W×V = -2*(0.0000727)*800*sin(45) = 0.082 (m/s2)

And then the deviation is x=at2 / 2 = 0.041 meters.

Where exactly is this deviation?
 

Attachments

  • XTc8Gya.jpg
    XTc8Gya.jpg
    8.5 KB · Views: 1,375
Physics news on Phys.org
Foruer said:

Homework Statement


It's a question about the deviation of a bullet fired on Earth's surface:
**I solved the question, but I can't figure out what exactly this deviation means, where is it on the following question's picture:

View attachment 226866
The problem:
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?

Homework Equations


ac = -2W×V

The Attempt at a Solution


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
ac = -2W×V = -2*(0.0000727)*800*sin(45) = 0.082 (m/s2)

And then the deviation is x=at2 / 2 = 0.041 meters.

Where exactly is this deviation?
The Coriolis force and Coriolis acceleration are vectors, so they have direction, depending on the directions of the angular velocity ##\vec Ω ## and ##\vec v ##, the velocity of the body in the rotating frame of reference . https://hu.wikipedia.org/wiki/Coriolis-erő
 
  • Like
Likes   Reactions: Foruer
Thank you, I do know that, but in this specific problem I can't point to where this 0.041(m) is. I'd like someone to draw a line or any other way to show where exactly is this deviation. it's 0.041(m) in relation to what exactly?
 
Foruer said:
Thank you, I do know that, but in this specific problem I can't point to where this 0.041(m) is. I'd like someone to draw a line or any other way to show where exactly is this deviation. it's 0.041(m) in relation to what exactly?
The Coriolis acceleration is proportional to the vector product of the angular velocity and the velocity of the object. For that you can figure out the direction of the acceleration. Without the Coriolis force, the body would travel in the direction of the target, along the initial velocity. In what direction does it deviate from it because of the Coriolis force?
 
You can also forget the vectorial product if use spherical coordinates. Then all the accelerations appears in a analitical way from the Chirstoffel Symbols in spherical coordinates.

All signs are referred to the coordinate system. You can see that in the next link, last post (in the link), EXAMPLE 2.
(note that in the link appears all accelerations not only coriolis, but there are explanation for all terms, coriolis also an their sings and direction)

https://www.physicsforums.com/threa...-aceleration-from-christoffel-symbols.923208/

I'm sorry because I can not quote only EXAMPLE 2, and need to post entire link.
Edit: nothe that in the link theta is the angle between the z axis an position vector R.
in your case this are not problem because your latitude is 45º and then sin45º=cos45º.
(note in example 2 that, the cos in equation is equivalent to your sin in vectorial product)
(If you see your own picture the angle between V and W is 90-theta)
 
Last edited:
Yes, sorry it's 90-theta, but since it's 45 I didn't write it (because it's the same). About the deviation , I still don't know, would like someone to draw on my picture where exactly is it so I can understand :)
 
Yes, I see that the link is hard to you forget it.

Then we go back to the vectorial product: Ac=-2·WxV
Look in your draw the direction of W and V, and then remember the rule of the thumb of the right hand for the vectorial product. (And the Ac has the oposite direction because the - sign)
And in the link you see Ac is in the ##\varphi## direction... in your case is to the East or to the West...
 

Similar threads

  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 25 ·
Replies
25
Views
2K
  • · Replies 9 ·
Replies
9
Views
8K
  • · Replies 3 ·
Replies
3
Views
5K
  • · Replies 4 ·
Replies
4
Views
8K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
10K