Canon shooting area in polar coordinates

In summary, the conversation discusses the calculation of the surface equation of cannon shots, taking into account the direction of the cannon and gravity. The suggested method is to use equations of motion, specifically dv=vvt-gt2/2, to find the horizontal and vertical distance (r) and ultimately the desired equation. The goal is to find the surface where the cannon ball would land, rather than a specific point. Useful resources for equations of motion are provided.
  • #1
prehisto
115
0

Homework Statement


hi,guys.
The directions of shooting e=cos[itex]\alpha[/itex]cos[itex]\varphi[/itex]i+cos[itex]\alpha[/itex]sin[itex]\varphi[/itex]j+sin[itex]\alpha[/itex]k
0<[itex]\varphi[/itex]<=2π;[itex]\varphi[/itex] -horizontally
[itex]\alpha[/itex][0,π];[itex]\alpha[/itex] is vertically
initial speed=v0

I need to calculate the surface equation of canon shots (where it hits).
In other words equations of the surface which is made from canon hits.

Homework Equations


The Attempt at a Solution



My view is that i need to gain horizontal and vertical distance (r),and then i can get the desired equation.

So horizontal distance is r=t|vxy|=tv0|ex+ey|=tv0(cos[itex]\alpha[/itex]cos[itex]\varphi[/itex]i+cos[itex]\alpha[/itex]sin[itex]\varphi[/itex]j)1/2=tv0cos([itex]\alpha[/itex])

So now i need height or in other words the distance in vertical direction(z -direction) or all together?.. I have hard time recalling use of equations of motion.
some help,please?
 
Last edited by a moderator:
Physics news on Phys.org
  • #2
Is there a question in the problem statement ?
You'll find equations of motion here, or otherwise ##\vec F = m\vec a## is a good one too.

List what you need under 2 and use less shorthand than in
So horizontal distance is r=t|vxy|=tv0|e_x+e_y|=tv0(cos[itex]\alpha[/itex]cos[itex]\varphi[/itex]i+cos[itex]\alpha[/itex]sin[itex]\varphi[/itex]j)1/2=tv0cos([itex]\alpha[/itex])
 
  • #3
BvU said:
Is there a question in the problem statement ?
You'll find equations of motion here, or otherwise ##\vec F = m\vec a## is a good one too.

List what you need under 2 and use less shorthand than in

Im sorry,i edited the original post,tried to be more specific.
 
  • #4
Let me try to translate: you want to calculate where a cannon ball lands that is shot from a cannon that is placed on the x-y plane in an empty flatland with no atmosphere, and gravity acceleration g in the -z direction, as a function of ##\phi, \theta## v0 and g.
That right ? Something like http://www.rabidgeek.net/physics-applets/projectile-motion/ ?

Did you find something useful among the equations of motion? Or here or here ?
 
  • #5
BvU said:
Let me try to translate: you want to calculate where a cannon ball lands that is shot from a cannon that is placed on the x-y plane
Yes,and the direction of cannon is described by [itex]\vec{e}[/itex]=cos[itex]\alpha[/itex]cos[itex]\varphi[/itex]i+cos[itex]\alpha[/itex]sin[itex]\varphi[/itex]j+sin[itex]\alpha[/itex]k

Because 0<[itex]\varphi[/itex]<2pi and [itex]\alpha[/itex][0,pi].Result is not one spot where it would land,but a surface which I am looking for.

Gravity g=-g[itex]\vec{k}[/itex]

BvU said:
Did you find something useful among the equations of motion? Or here or here ?

Yes,I found something useful,I think I could use dv=vvt-gt2/2
 
  • #6
Yes, so dv=0 has two solutions, one trivial at shooting off and one at "landing".
Bingo.
 
  • Like
Likes 1 person

Related to Canon shooting area in polar coordinates

1. What is a Canon shooting area in polar coordinates?

A Canon shooting area in polar coordinates is a mathematical representation of the area that a Canon camera can capture in a photograph. It is based on the camera's focal length, sensor size, and aperture setting.

2. How is the Canon shooting area in polar coordinates calculated?

The Canon shooting area in polar coordinates is calculated using trigonometric functions based on the camera's focal length, sensor size, and aperture setting. It is typically represented as a circular area on a polar coordinate system, with the camera at the center.

3. How does the Canon shooting area in polar coordinates differ from the shooting area in rectangular coordinates?

The Canon shooting area in polar coordinates takes into account the camera's focal length, which affects the angle of view and the distortion of the image. In rectangular coordinates, the shooting area is represented as a rectangular frame, without taking into account these factors.

4. Can the Canon shooting area in polar coordinates be adjusted?

Yes, the Canon shooting area in polar coordinates can be adjusted by changing the camera's focal length, sensor size, and aperture setting. These adjustments will change the size and shape of the shooting area on the polar coordinate system.

5. How can knowing the Canon shooting area in polar coordinates be useful for photographers?

Knowing the Canon shooting area in polar coordinates can help photographers visualize the coverage of their camera and plan their shots accordingly. It can also be useful for understanding the limitations and capabilities of their camera, and for comparing it to other cameras.

Similar threads

  • Advanced Physics Homework Help
Replies
2
Views
2K
  • Advanced Physics Homework Help
Replies
7
Views
2K
  • Advanced Physics Homework Help
Replies
1
Views
5K
  • Advanced Physics Homework Help
Replies
3
Views
5K
  • General Math
Replies
4
Views
1K
  • Advanced Physics Homework Help
Replies
9
Views
5K
  • Advanced Physics Homework Help
Replies
3
Views
3K
Replies
13
Views
2K
  • Programming and Computer Science
Replies
5
Views
2K
  • STEM Educators and Teaching
Replies
2
Views
1K
Back
Top