9mm travel distance - are my calculations correct?

D = 1/2AT^2

Acceleration due to gravity is 32 feet per second per second. Dropping an object from a height of six feet, the formula becomes

6 = 1/2 x 32/T x T^2
6 = 16 T
T = 6/16 = .375 seconds

For a muzzle velocity of 1200 feet-per-second, the distance traveled (ignoring deceleration due to drag) is:

1200 x .375 = 450 feet


I find it interesting (and confounding) that so many of us are using what we think is the same formula but we're all getting a different answer. I tried to look at that and see what's wrong. For example, Throttleup said the drop takes 0.61 seconds. Looking at that, I had to conclude that's not possible. The formula for velocity is:

V = AT (velocity equals acceleration x time)

For an object starting from velocity zero and accelerating due to gravity at 32 feet per second per second, look at an elapsed time of exactly one second. The velocity would be:

V = 32
 
Aguila Blanca said:
D = 1/2AT^2

Acceleration due to gravity is 32 feet per second per second. Dropping an object from a height of six feet, the formula becomes

6 = 1/2 x 32/T x T^2
6 = 16 T
T = 6/16 = .375 seconds

For a muzzle velocity of 1200 feet-per-second, the distance traveled (ignoring deceleration due to drag) is:

1200 x .375 = 450 feet


I find it interesting (and confounding) that so many of us are using what we think is the same formula but we're all getting a different answer.
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What happens to t^2 in your calculation further on?

If you take your initial equation and manipulate it to isolate for t, you get what I have above in post #4.

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Guys you are way overintellectualizing this issue.

Firstly,.. most projectiles are going to impact some obstacle way before they reach their maximum travel distance.

Secondly.. a 9mm fired from approximately 6 feet (horizontally) can very easily travel the distance of a football field(300 feet) and I am not using any math equation to figure that out.. it just will, i have seen it.

Just using plain ole guesstimation.. I would say that near 600 feet or so before the bullet hits the dirt all by itself.
 
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The OP asked a specific question. It was answered a while ago. Math and physics are things. Some of us even enjoy them.

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What I think several of the folks posting are forgetting about the other factors involved when they shoot a handgun.

1. The sights and barrel are usually not parallel, so the bullet is not leaving the gun in a path parallel to the sights or the ground.

2. When you shoot a handgun recoil affects the barrel alignment with the ground. Heavier recoil equals sights that are further out of alignment with the barrel.

3. The bullet is traveling upward as it leaves the barrel. Therefore it crosses line of sight 2 times, once close to the gun and again far from the gun. Every gun, load, and shooter will have various results at different distances.

4. The math equation used to calculate the OP question is for a barrel anchored level with the ground, no recoil, no sights used.
 
big al hunter said:
What I think several of the folks posting are forgetting about the other factors involved when they shoot a handgun.



1. The sights and barrel are usually not parallel, so the bullet is not leaving the gun in a path parallel to the sights or the ground.



2. When you shoot a handgun recoil affects the barrel alignment with the ground. Heavier recoil equals sights that are further out of alignment with the barrel.



3. The bullet is traveling upward as it leaves the barrel. Therefore it crosses line of sight 2 times, once close to the gun and again far from the gun. Every gun, load, and shooter will have various results at different distances.



4. The math equation used to calculate the OP question is for a barrel anchored level with the ground, no recoil, no sights used.
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If you want to add an angle of elevation to the shot there are equations for that as well. There are also equations to take into account the ballistic coefficient of the bullet. You have to start somewhere and some simplistic assumptions will get you at least a pretty good initial guess if you don't have the practical experience to draw from.

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big al hunter said:
What I think several of the folks posting are forgetting about the other factors involved when they shoot a handgun.

1. The sights and barrel are usually not parallel, so the bullet is not leaving the gun in a path parallel to the sights or the ground.

2. When you shoot a handgun recoil affects the barrel alignment with the ground. Heavier recoil equals sights that are further out of alignment with the barrel.

3. The bullet is traveling upward as it leaves the barrel. Therefore it crosses line of sight 2 times, once close to the gun and again far from the gun. Every gun, load, and shooter will have various results at different distances.

4. The math equation used to calculate the OP question is for a barrel anchored level with the ground, no recoil, no sights used.
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Which is exactly what was specified in the original question -- bullet fired horizontally, over level ground.
 
Which is exactly what was specified in the original question -- bullet fired horizontally, over level ground.
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Yes, which is why someone shooting at a target at any given distance can't compare it to the OP.... Sometimes real life doesn't compare directly to the math of a hypothetical question.

.If you want to add an angle of elevation to the shot there are equations for that as well. There are also equations to take into account the ballistic coefficient of the bullet.
Click to expand...

Math is fun, isn't it. Gives us a way to determine what should happen, if all the variables are accounted for.
 
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