Calculating Max Height & Time of Rifle Trajectory at 300 Yards

In summary: He is asking if he can use calculus to find the maximum height of the round at a specific point in its flight path. This is a difficult problem to solve analytically, and so it is not often explicitly proven. However, if you assume the bullet is subject to uniform gravity and that drag is not a factor, then the equation for the maximum height of the round can be solved using a derivative function. In this case, the function would be h(t) = 4t^2+48t+3. The time t at which the maximum height is reached would be found by solving for it.
  • #1
surferbarney0729
32
0
I am a formal instructor and one of the classes I teach revolves
around the ballistics of a bullet. I am currently in a Calculus class
and will be starting as an Economics major In the fall when I finish
my time in the Marines. I am hoping to see if there is any way,
through calculus that I can prove the max height of the round through
actual mathematical proof. Many of the lesson plans and outlines
simply indentify what I have attached below and do not mention proof
of this. I am simply curious if I can prove the max height of the
round and at what time during its flight path does it reach this
height. Any help would be greatly appreciated. Below is a simple
diagram of a rifle trajectory at 300 yards. I attached the image to
give a preliminary view of what I am referencing.



Maximum Ordinate - The highest point in the trajectory of the round
on its route to the target
The event I would like to know if I can prove is how high the maximum
ordinate should be. I am new to calculus and especially the
application of calculus to real world events. Can I use a Derivative
function formula to find the max?

If so, what variables do I need for this, the rounds are all constant,
and with that the velocity of the round at specific ranges is also
fixed (in theory).
Would I use a function like h(t) = 4t^2+48t+3 to find the maximum h
and the corresponding time t. The conventional knowledge says that
roughly 2/3rds of the way to the target the projectile reaches its
maximum height. It is said to be roughly 7 inches above the line of
sight at its highest moment, which is supposedly 2/3rd of the way to 300 yards. This is what I am trying to prove or disprove throughcalculus.
All in all, any help or nudges to this type of problem would be
greatly appreciated.
 
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  • #2
This seems like an appropriate problem for calculus, but I'm unsure of what assumptions are necessary to get the correct result. In other words, it's usually an introductory physics problem to treat the trajectory of an object under the influence of uniform gravity without drag; in that scenario, the maximum height always occurs at halfway through the trajectory. I suspect drag is needed for this problem, however.

In other words, you'll have to solve a vector differential equation:

[tex]\frac{d\vec p}{dt} = \vec F_{\text{drag}} + \vec F_{\text{gravity}}[/tex]

The gravitational force is just going to be downward, but the drag will generally be in the opposite direction of the velocity. This is probably a tough problem to attack analytically, to be honest, which is why (I suspect) it's not often explicitly proven.
 
  • #3
Thanks for your response. I am going to have to do some research and reading to figure out a way to solve this. I have come across many issues with some of the material I teach that should be able to be proven mathematically.

You mention that max height always occurs halfway through trajectory, but in this case all the literature teaches that 2/3rds of the way to the target a bullet will reach its max ordinate, or highest point.
 
  • #4
No, he said that without drag "the maximum height always occurs at halfway through the trajectory". With drag, which reduces the horizontal speed of the bullet more than the vertical speed, the trajectory is steeper later than before and so the maximum height is farther on the trajectory. What formula for "drag" is assumed in the literature that gives 2/3? Both "proportional to the speed" and "proportional to the square of the speed" are often used and will give different results.
 
  • #5
I appreciate the input. As i might have not mentioned, I am new to the world of calculus, and especially new to the world of attempting to apply it to a real life scenario. I teach marksmanship, and the pubs are simply what our organization gives us for lesson plans. The lesson plan says 2/3rds of the way to the target at 300 yards, the round reaches roughly 7 inches above the line of sight.

After the maximum ordinate (height) is taught and preached to be what I mentioned above, the lesson plan does indeed talk about drag and gravity.

How would you guys, as the expert apply write a formula to prove or disprove when exactly over the 300 yards from end of barrel of rifle to impact of target that the round is at its highest over the line of sight? I have a table of the velocity of the round at specific distances. This is strictly for enjoyment of math and again I apologize if much of this is elementary, but I came to this sight to get input from those who have applied and mastered math in this form.

Can anyone create a formula or throw a bone.

Range is 300 yards

the max ordinate is supposed to be at 200 yards (im assuming that is because of drag and external forces)

The round weighs 62 gr and has velocity of 3,100 ft/s and is being shot out of a rifle with a muzzle velocity of 2900 ft per second.

Am i missing any other important data? Do you need the ballistic coefficient of the round?
 

1. What is the formula for calculating the maximum height of a rifle trajectory at 300 yards?

The formula for calculating the maximum height of a rifle trajectory at 300 yards is:
h = (v2sin2θ) / 2g
where h is the maximum height, v is the initial velocity of the bullet, θ is the angle of elevation, and g is the acceleration due to gravity.

2. How do I determine the initial velocity of the bullet?

The initial velocity of the bullet can be determined by using a chronograph. A chronograph is a device that measures the speed of a bullet as it passes through two sensors. Alternatively, you can also use the manufacturer's stated muzzle velocity for the specific type of bullet you are using.

3. What is the angle of elevation for a rifle shot at 300 yards?

The angle of elevation for a rifle shot at 300 yards will depend on the height of the shooter's eye above the bore of the rifle. Generally, a shooter's eye will be around 1.5 inches above the bore of the rifle. Using this as an estimation, the angle of elevation would be approximately 0.3 degrees.

4. How accurate is the calculation for max height and time of rifle trajectory at 300 yards?

The accuracy of the calculation will depend on several factors such as the consistency of the shooter, the accuracy of the rifle, and external factors such as wind and air resistance. However, the formula provides a good estimation for the maximum height and time of the rifle trajectory at 300 yards.

5. Can this formula be used for other distances besides 300 yards?

Yes, this formula can be used for other distances as well. However, it is important to note that the initial velocity and angle of elevation may vary for different distances, so the results may not be as accurate. It is recommended to use a ballistic calculator or consult a ballistics expert for more precise calculations at different distances.

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