Muzzle velocity given test range, time of flight and ballistic properties.

In summary, this isn't really homework, but I figure this is the most appropriate place to post this... Please do say if you think there is a more likely place it will get answered - I'm new!
  • #1
Carpet_Diver
2
0
This isn't really homework, but I figure this is the most appropriate place to post this... Please do say if you think there is a more likely place it will get answered - I'm new!

Homework Statement
I am trying to calculate the muzzle velocity of an air rifle. I can find the time of flight over a known horizontal range. I know the mass and quadratic velocity damping constant due to air drag of the pellet. I also know that over a reasonable range (20m), air drag cannot be ignored, so [itex]\large \dot{x}_0 = \frac{r}{T}[/itex] is not true.

I will assume the trajectory of the pellet is a parabola, because the trajectory is so flat a parabola is a good enough approximation to real life. I will treat the motion decay problem as one dimensional (along the parabola arc length), so gravity can be ignored from here on.

Relevant equations
X, parabola arc length is known, [itex]\large X = T \sqrt{(\frac{r}{T})^{2}+(\frac{g T}{2})^{2}}[/itex], where r is horizontal range, and g is gravity.
T, time of flight is known
C, quadratic velocity damping constant is known
M, pellet mass is known
Acceleration due to quadratic velocity damping is given by [itex]\Large \ddot{x} = \frac{C \dot{x}^{2}}{M}[/itex]
[itex]\dot{x}_{0}[/itex], the initial velocity, is unknown.

The attempt at a solution so far
I have already worked out the arc length, as seen above. I have come up with the following expression for velocity as a function of time, which may, or may not be useful;

[itex]\LARGE \dot{x}(t) = \frac{\dot{x}_{0}}{1 + \frac{t C \dot{x}_{0}}{M}}[/itex]

and an expression for distance, as a function of time;

[itex]\LARGE x(t) = \frac{M}{C} ln(1 + \frac{t C \dot{x}_{0}}{M})[/itex]

I am not really sure what to do next, to get what I want; an expression for [itex]\dot{x}_{0}[/itex] as a function of X, T, C and M.
 
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  • #2
Carpet_Diver said:
This isn't really homework, but I figure this is the most appropriate place to post this... Please do say if you think there is a more likely place it will get answered - I'm new!

Homework Statement
I am trying to calculate the muzzle velocity of an air rifle. I can find the time of flight over a known horizontal range. I know the mass and quadratic velocity damping constant due to air drag of the pellet. I also know that over a reasonable range (20m), air drag cannot be ignored, so [itex]\large \dot{x}_0 = \frac{r}{T}[/itex] is not true.

I will assume the trajectory of the pellet is a parabola, because the trajectory is so flat a parabola is a good enough approximation to real life. I will treat the motion decay problem as one dimensional (along the parabola arc length), so gravity can be ignored from here on.

Relevant equations
X, parabola arc length is known, [itex]\large X = T \sqrt{(\frac{r}{T})^{2}+(\frac{g T}{2})^{2}}[/itex], where r is horizontal range, and g is gravity.
T, time of flight is known
C, quadratic velocity damping constant is known
M, pellet mass is known
Acceleration due to quadratic velocity damping is given by [itex]\Large \ddot{x} = \frac{C \dot{x}^{2}}{M}[/itex]
[itex]\dot{x}_{0}[/itex], the initial velocity, is unknown.

The attempt at a solution so far
I have already worked out the arc length, as seen above. I have come up with the following expression for velocity as a function of time, which may, or may not be useful;

[itex]\LARGE \dot{x}(t) = \frac{\dot{x}_{0}}{1 + \frac{t C \dot{x}_{0}}{M}}[/itex]

and an expression for distance, as a function of time;

[itex]\LARGE x(t) = \frac{M}{C} ln(1 + \frac{t C \dot{x}_{0}}{M})[/itex]

I am not really sure what to do next, to get what I want; an expression for [itex]\dot{x}_{0}[/itex] as a function of X, T, C and M.

Assuming that all of your math is correct (I haven't gone through the steps), could you not just take your last expression and invert it for the initial velocity by taking the exponential of both sides?
 
  • #3
Wow, thanks, I am obviously in need of sleep!
 

What is muzzle velocity?

Muzzle velocity is the speed at which a projectile leaves the barrel of a firearm or other projectile weapon. It is measured in feet per second (fps) or meters per second (m/s).

How is muzzle velocity calculated?

Muzzle velocity is calculated by dividing the distance traveled (test range) by the time it takes for the projectile to reach that distance (time of flight). This calculation takes into account the ballistic properties of the projectile, such as weight and shape. It can also be determined through the use of specialized equipment such as a chronograph.

Why is muzzle velocity important?

Muzzle velocity is important because it affects the accuracy and range of a projectile. It also plays a role in the amount of force and damage a projectile can inflict on its target. Knowing the muzzle velocity can help with making adjustments for better accuracy and predicting the trajectory of the projectile.

What factors can affect muzzle velocity?

There are several factors that can affect muzzle velocity, including the type and weight of the projectile, the type and condition of the firearm, and environmental factors such as air temperature, humidity, and altitude. Other factors such as the type and amount of gunpowder used and the length of the barrel can also impact muzzle velocity.

How does muzzle velocity differ from muzzle energy?

Muzzle velocity and muzzle energy are related but different concepts. Muzzle velocity refers to the speed of the projectile, while muzzle energy refers to the amount of kinetic energy the projectile possesses. Muzzle energy takes into account the weight and speed of the projectile, while muzzle velocity only considers the speed.

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