|MadSci Network: Physics|
Dear David: Thank you for your question. The problem you are addressing is an old one and it looks deceivingly simple. We need to start mentioning Newton's Second Law. In the following link there is a nice explanation about this subject: ballistic trajectories This page reads in part: "In this lesson we will experiment with computing and visualizing ballistic trajectories. A ballistic trajectory is the path followed by an object which, after it is given some initial velocity, travels only under the influence of gravity. For the purposes of this lesson we will ignore the effects of air resistance." Well, ignoring air resistance, or, more generally, the medium resistance, is a huge simplification. Nevertheless, we will go along for the moment. The Ballistic Trajectory Problem One of the first problems studied in an introductory physics class is the ballistic trajectory problem. Let's assume that we are standing in a flat field and that we throw a ball so that it starts out moving at some velocity V > 0 and at some angle theta radians with the ground. At any given time t, the ball's horizontal distance is given by Vt cos (theta) and its vertical height is given by Vt sin(theta) - (1/2)gt^2 where g is the earth's gravitational constant, or 9.8 meters/sec/sec. This assumes that the ball's initial horizontal and vertical positions are both 0. It will prove convenient to define functions for distance and height that take initial velocity, initial angle, and time as arguments and return the distance and height of the ball. distance := (V, theta, t) -> V*t*cos(theta) height := (V, theta, t) -> V*t*sin(theta) - .5*9.8*t^2; So far, so good. It is clear that if we shot a weapon horizontally (i.e., theta = 0), after certain time, the bullet will hit the floor. This is the so called range of the weapon. A nice link presenting interactive graphics is the following: http://www.mcasco.com/p1anlm.html A more formal approach could be find in: http://www.phys.uidaho.edu/~pbickers/Course s/310/Notes/part2/node20.html Now, since in real life we could not ignore the drag due to the air (or water) resistance, we need to refine our ideas. It is evident that the drag is a force opposing the movement of the bullet. How big is this force?. In http: //www.i3solutions.com/vehiclesimulation/dragequations.htm we find: Air (water) Friction: Fair = 1/2 Cd A đ V2 Rolling Resistance: Rolling = (RRConst + Velocity * RRXcoef) * Weight * Cos(slope)/100 Grade: FGrade = Weight*Sin(slope * (22/7)/180) Total Losses: FTotal = Fair + FRolling + FGrade where Cd is coefficient of drag, A is the cross sectional area, V is the velocity, đ is the density of the medium. It is evident that the bigger the density the shortest distance traveled by the bullet. Finally, in http://www.las cruces.com/~jbm/ballistics/secdens.html we find: Sectional Density and Ballistic Coefficient The BC, or ballistic coefficient is defined as: BC = w / [i d2] where the diameter is specified in inches and the weight in pounds and the form factor is found using: i = CD / CDG The sectional density is defined as: SD = w / d2 making the ballistic coefficient BC = SD / i So this means that the ballistic coefficient is proportional to the weight of the bullet and inversely proportional to the diameter squared. (Keep in mind that the ballistic coefficient is also inversely proportional to the form factor which depends on the shape of the bullet!) Calculation of the sectional density is straight forward. For a 300 grain, .338 caliber bullet, the sectional density is: SD = [ 300 gr / (7000 gr/lb) ] / [ 0.338 in ]2 = 0.375 lb/in2 NOTE: With the common definition of the sectional density, the units have to be converted when used with drag functions, velocity, etc, to convert the in2 to ft2 resulting in a factor of 144. Variables d bullet diameter w bullet weight SD sectional density BC ballistic coefficient i form factor G "G" function CD drag coefficient CDG drag coefficient of the standard bullet The numerical answer, of course, depends of the particular values of all this parameters. There is a nice book devoted to such problems. It could be found in: http://www.border- barrels.com/book.htm I hope this helps Regards Jaime Valencia
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