Trajectory of a bullet with drag

In summary, the conversation discussed a physics problem involving the trajectory of a bullet with drag. The drag force was found to be dependent on the velocity squared, which led to difficulty in solving the problem. The conversation also delved into the validity of using the approximation of drag being proportional to the first or second power of velocity. However, it was suggested that this approximation may not fully capture the complexity of real-life drag, especially at high speeds. The conversation concluded by emphasizing the importance of using correct mathematical equations and discussing theoretical versus experimental approaches to understanding fluid drag.
  • #1
mekrob
11
0
Trajectory of a bullet with drag

In most physics books, they pass over drag problems because of their difficulty. However, I had to do one for a mid-term for a physics class. When the velocity is low (tens of m/s or less), then the drag is proportional to the velocity and the math is quite simple (at least in comparison). However, we were asked about a bullet which travels nearly 1000 m/s and thus is dependent upon velocity squared. I was utterly confused with how to progress with the math.

Here's what I had:

Since the bullet is fired at an initially positive angle [tex]\theta[/tex], the drag can be broken into two components (2-D only): Fx and Fy.

Fx = F * cos[tex]\theta[/tex]
Fy = F * sin[tex]\theta[/tex]

FDrag = [tex]\frac{(Drag Coefficient)*(\rho)*V^{2}*(Area)}{2}[/tex]
FDrag = k*V2, where K = .5*Cd*[tex]\rho[/tex]*Area

Assume that density is constant throughout time.

Fx = m[tex]\frac{\partial vx}{\partial t}[/tex] = -k*V2

[tex]\frac{\partial vx}{\partial t}[/tex] = -k*Vx2/m

[tex]\frac{\partial vx}{V^2}[/tex] =-k/m * dt

Integrate to get:

-1/Vx = -kt/m + C

Vx = m/kt + K

However, at t = 0, Vx = muzzle velocity*cos(theta). But according to the equation I have, Vx would go to infinity as t -> 0. Right?

I run into the same problem when trying to figure out Vy which is even more complicated since it involves gravity. Can anyone tell me where I went wrong with this?

Thanks
 
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  • #2


mekrob said:
Fx = m[tex]\frac{\partial vx}{\partial t}[/tex] = -k*V2

[tex]\frac{\partial vx}{\partial t}[/tex] = -k*Vx2/m

We'll get to your question, but I couldn't help to notice a bit of "creative math" in these two lines.

[tex]\vec v^2 \neq v_x^2 [/tex]

So in order to get the vec's right:

[tex] m\frac{\partial \vec v}{\partial t}=-k \vec v^2 \hat v[/tex]

(^ means unit vector) projecting onto the x-axis:

[tex] m\frac{\partial v_x}{\partial t}=-k (v_x^2+v_y^2) \frac{v_x}{\sqrt{v_x^2+v_y^2}}[/tex]

so

[tex] m\frac{\partial v_x}{\partial t}=-k v_x\sqrt{v_x^2+v_y^2}[/tex]

and similar for y:

[tex] m\frac{\partial v_y}{\partial t}=-k v_y\sqrt{v_x^2+v_y^2}[/tex]

To get to your question, set vy = 0 for a case of one dimensional drag motion. Here:

[tex] m\frac{\partial v_x}{\partial t}=-k v_x^2[/tex]

Now you integrate to obtain
mekrob said:
-1/Vx = -kt/m + C

Vx = m/kt + K
Which is not a good (even wrong?) way to do it. Rather stick to:

[tex]v_x=\frac{1}{kt/m+C}[/tex]

Where is it clear that v goes to v0 for t going to 0 for a suitable pick of the constant C (= 1/v0) and goes to 0 for t going to infinity.

Now you can tackle the 2-dimensional problem on your own :)
 
  • #3


Wow, thanks. Not that complicated when you do it correctly.
 
  • #4
Check out these videos if you want your mind blow about drag:

http://web.mit.edu/hml/ncfmf.html".

They might make you rethink the approximation that drag is proportional to the first or second power of the velocity.
 
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  • #5
tankFan86 said:
Check out these videos if you want your mind blow about drag:

http://web.mit.edu/hml/ncfmf.html".

They might make you rethink the approximation that drag is proportional to the first or second power of the velocity.

Why? The gentleman in the videos drew nice parabolic drag/speed dependencies.

Even if it is an approximation (it isn't, at least in some specific cases) who says that it is an unsatisfactory approximation?
 
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  • #6


Well the assumption does have some validity, for sure. But I think it would be naive to think that drag can be completely understood in these terms.
 
  • #8


It's nice to find some physics where the theory is less developed than the experiment!
 
  • #9


tankFan86 said:
Well the assumption does have some validity, for sure. But I think it would be naive to think that drag can be completely understood in these terms.

First it was an approximation, now it is an assumption?? Please understand that it is neither. Just about any textbook on fluid mechanics will derive that. Of course there are limitations to the validity of the exact dependence, eg. in transition from sub-sonic to super-sonic. Such special cases will have to be tackled separately, just like is the case in just about all physics. But nobody has ever claimed otherwise.

tankFan86 said:
It's nice to find some physics where the theory is less developed than the experiment!

No. The fundamental theory of fluid mechanics (Euler (Navier-Stokes) equation + no slip BC) has proven to be remarkably robust, when solved correctly!

It's time to call enough of this nonsense. If tankFan86 wishes to discuss disagreements in theory and experiment with regards to fluid drag, may I suggest that he provides a peer reviewed publication on which to base this discussion, such that we can all agree on the premises. This "I think..."-reasoning is useless.
 
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1. How does drag affect the trajectory of a bullet?

Drag is a force that opposes the motion of an object through a fluid, such as air. When a bullet is fired, it experiences drag, which can cause it to slow down and change its trajectory.

2. What factors influence the drag on a bullet?

The amount of drag on a bullet is influenced by its shape, speed, and the density of the fluid it is traveling through. The bullet's surface roughness and the fluid's viscosity also play a role in determining drag.

3. How does air density affect the trajectory of a bullet?

Air density, which is affected by factors such as altitude and temperature, can impact the drag force acting on a bullet. As air density increases, so does the drag force, which can cause a bullet to slow down and deviate from its intended trajectory.

4. Can the trajectory of a bullet with drag be predicted?

Yes, the trajectory of a bullet with drag can be predicted using mathematical models. These models take into account factors such as the bullet's initial velocity, mass, and drag coefficient to calculate its trajectory.

5. How can the drag on a bullet be reduced?

The drag on a bullet can be reduced by altering its shape to make it more aerodynamic, or by using materials that decrease drag, such as smoother surfaces or coatings. Additionally, increasing the bullet's velocity can help overcome the effects of drag.

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