Bullet rise.... Does it exist? Physics and Trigonometry content inside! That has been a topic of discussion lately on here and I want to clarify a few things for those who are dead set on claiming that a bullet cannot rise after it has left the muzzle. 1) First, we must understand that everything is relative in this world; and the same goes for establishing contextual data for physics and trigonometry calculations. For instance, while I am driving in my car at 60mph on the freeway, my body is considered to be stationary with regards to the car, but traveling at a velocity of 60mph with regards to ground speed. This same necessity for establishing a proper context before wrapping your mind around a concept is applicable when pondering projectile trajectory (bullet flight path in this case) too. In the case of projectile flight paths we need to define what is down and what is up. Now, most people define down as the direction in which things fall due to gravity. From this point out, the axis upon which things fall due to gravity will be referred to as the Y axis (vertical axis). The axis that is exactly 90* perpendicular to this, and perfectly horizontal to the ground, will be referred to as the X axis. Keep in mind that this X axis DOES NOT correlate to the bore axis! If we define “down” as any drop from the bore axis, then the bullet will never “rise” above the bore axis. This is important, because what I will be referring to from here on out will be pertaining to the horizontal X axis, or Earth’s surface. 2) In our example we will be using a standard telescopically sighted rifle that is shooting a bullet traveling at 3,000fps (914 meters per second) at a target that is as high off of the ground as the line of sight of the crosshairs of the scope are. Another thing that a person needs to understand is that the bore axis of a rifle and the sight axis of the scope are NOT perfectly parallel! If the sight axis is held perfectly horizontal, then the bore axis actually slopes upwards at an ever so slight angle compared to this. This angle difference is what allows the bullet to intersect the sight axis even when gravity is constantly trying to pull the bullet in a downward direction. For the purpose of this example, we will assume an axis orientation difference of 2* (the barrel is pointed 2* above the horizontal X axis). 3) With any physics problem involving force vectors, trajectory, etc the best way to break things down is into their X and Y components. What I mean is that if I throw a baseball to my son who is 20 meters away from me, and the ball reaches a maximum height of 10 meters before falling back down, then the ball has attained a maximum Y direction height of 10 meters and has traveled a total of 20 meters along the X axis. So, in this case, the X component of the baseball trajectory is 20 meters and the Y component is 10 meters. This X and Y component way of doing things also works the same with acceleration and velocity too. Going back to the baseball example, IGNORING ALL DRAG RESISTANCE, if I threw the ball at a 25* angle with regard to the ground (horizontal axis), and it was initially traveling at 20 meters per second, then I could use simple SINE and COSINE trigonometry functions in my calculator to figure for its initial X and Y component velocity when it left my hand. For the X component, the calculation would be 20cos(25*) = 18.1 meters per second initial velocity in the X direction. For the Y component, the calc would be 20sin(25*) = 8.5 meters per second initial velocity in the Y direction. So, now that I know to calculate what my X and Y initial velocity components are, we can apply this to the rifle, remembering that the departure angle example was 2* and the initial velocity was 914 meters per second (m/s). For the X direction we get 914cos(2*) = 913.4 m/s initial X component velocity. For the Y component we get 914sin(2*) = 31.9 m/s initial Y component velocity in the positive (upward) direction. Now, keep in mind that were are talking about VELOCITY and NOT ACCELERATION! They are two vastly different things! Velocity is the speed at which an object is traveling. Acceleration is the rate of change in that object’s velocity. 4) We honestly do not need to move past this point with our calculations, because if we use common logic, we can determine the general characteristics of the bullet flight path. So, we know it is traveling with an initial horizontal (X axis) velocity of 913.5 m/s and, more importantly, an initial vertical (Y axis) velocity of 31.9 m/s in the upward direction. So, if something is traveling with an initial velocity of 31.9 meters per second in the upward direction, then it will move upwards. However, I get the impression that logic is lost on some folks, so I will trudge on with further computation and elaboration. 5) The constant force of Earth’s gravitational pull is 9.8 meters per second squared when standing on Earth’s surface. That is an ACCELERATION force, NOT a VELOCITY. Since we based our Y axis on how objects fall due to gravity in section 1), then we know that when dealing with acceleration due to gravity, it will only be used for travel in the Y direction (vertical). A few useful physics equations for figuring for trajectory path, travel time, initial velocity, final velocity, x displacement, and acceleration are as follows: Vf = V+AT X=VT+1/2AT^2 Vf^2=V^2+2AX X=1/2(V+Vf)T Vf= Final Velocity, V=Initial Velocity, A=Acceleration, T=Time, X=Travel distance/displacement. Let’s start with figuring just how long the bullet will be in the air for first using the Y direction component. NOTE: We are not accounting for air drag resistance, since this would overcomplicate things beyond the realm of a simple FB explanation. Using the first equation, lets plug what we know into it so far regarding the Y direction velocity. V= 31.9 m/s, Final Y direction velocity at the bullets highest point will be 0 m/s, so Vf=0, and the acceleration, or A, will be A=-9.8m/s^2 since gravity is decelerating the bullet’s upwards travel velocity. We don’t know time of travel yet. After being plugged in, our equation looks like this: 0 = 31.9+(-9.8)(T) ------> So from this we get that T=3.26 seconds. That figure is just upwards travel time; and what goes up must come down. So, it will take the same time to come down the same distance as it took to go up. So, we multiply 3.26 seconds by 2 to get a TOTAL travel time of 6.52 seconds. Now let’s move on to figuring just how high the bullet will reach above the sight line (our “zero height” point). We know that Vf=0, V=31.9, T=3.26, and A=-9.8. Using the last equation of X=1/2(V+Vf)T we can figure for Y direction displacement (aka max height of the bullet). Our data plugged in looks like this: Y=1/2(31.9+0)3.26 ------> So from this we get that Y=52 Meters What that means is that is a bullet is shot from a gun that is sitting on the ground, the bbl angle is 2* above horizontal (sight line), and initial projectile velocity is 914 m/s, then the absolute maximum height that bullet will achieve will be 52 meters, not taking air drag into account. We can take this one step further and calculate for X direction displacement (total horizontal travel distance) by using the X components. Initial X component velocity was 913.5 m/s and the bullet traveled for a total of 6.52 seconds before touching down. So VT=X --------> From this we get a total horizontal travel distance of 5,956 meters. Closing: So even though acceleration due to Earth’s gravity is ALWAYS pulling the bullet towards the Earth at a rate of 9.8 meters per second squared, the initial upward travel velocity of the bullet is enough to counteract that for enough time so as to rise above the sight line (NOT bore axis!). In our example, disregarding air resistance, the bullet will fly for a total of 6.52 seconds before hitting the ground if it was fired from a rifle that was at ground level, it will have traveled to a maximum Y direction apogee displacement of 52 meters before coming back down, and it will travel a maximum horizontal distance of 5,956 meters. This has been a LONG and complicated instructional that I assume few will read. Hopefully it educates the few that do not understand the key differences between VELOCITY and ACCELERATION, and how they play into explaining how bullet trajectory can indeed “rise” above the sight line/X axis. In short, when speaking with regard to the horizontal X travel axis/sight axis, the bullet does indeed rise above it for a period of time. When speaking with regard to bore axis, the bullet does NOT rise above it for a period of time.
Posted on: Tue, 28 Oct 2014 16:26:28 +0000
Recently Viewed Topics
© 2015