How much will this object be moved by this force?

This is all hypothetical. I am writing a paper about rifles and need to consider the effect of air density and wind speed on the displacement of a bullet. For the purposes of the calculation, I'm using 7.62 x 51 mm NATO bullet, which weighs approximately 9.5 grams. I am also going to hypothetically assume that the bullet is a perfect cylinder, in which case it would have a surface area of 1312.09 mm^2. In an assumed scenario of sea level at 20 degrees Celsius, the force exerted by a 1 meter per hour wind would approximately be 1.21528 x 10^-10 Newtons.





First of all, someone tell me if that's correct.





Next, can anyone tell me how I can determine what distance the wind will push the bullet?How much will this object be moved by this force?
I have never actually seen a calculation for the deflection of a bullet assuming that the bullet is a cylinder. Normally, the calculation is performed using a parameter called the ballistic coefficient (BC). See reference 1 below.





There are two ways that I have seen people use to model bullet motion in wind:





a) relative motion argument (see reference 2 for more details)


b) gyroscopic argument (see reference 3 for more details)





Both methods give similar answers, but the relative motion argument is conceptually simpler. For a nice summary of all the physical effects on a bullet, see reference 4.





Now let's look at your solution. Reference 1 gives examples of the typical wind deflections for bullets. Let's do an order of magnitude calculation. A 7.62 NATO round will have approximately 0.67 inches of deflection at 100 yards for a 10 mile per hour wind (assuming one minute of angle [MOA] equals one inch at 100 yards). Since this round has muzzle velocity of 2500 feet per second ( 762 meters/second), the round will take 0.12 seconds to hit the target. Assuming a constant force on a bullet with constant mass, the acceleration of the bullet can be calculated as follows.





d = 1/2*a*t^2 =%26gt; a = 2.364*m/sec^2





fWind= ma = 0.022N, which is much greater than the force you list. Of course, I am assuming a 10 mile per hour wind. A 1 meter per hour wind is very low (did you mean 1 meter per second?). So I do not believe that your answer is correct.





To calculate the force on a bullet, I would use one of the references below (like Reference 2) to estimate the effective crosswind induced angle or velocity. Then I would compute the effective force using the method I used above.





Wish I had a better solution for you.How much will this object be moved by this force?
I agree with you that intuitively the effect of wind, and air density would affect the trajectory of a bullet. Drag force opposes motion, and would be higher with higher density fluid. For example shooting a bullet into water would affect the path of the bullet more than it would in air.





I would use fluid mechanics to estimate the drag on the bullet. The general formula for drag is:





Drag Force = Cd * Area * fluid density * velocity^2 / 2





If the wind has a component opposing forward motion (i.e. across the target line), this drag force will be imbalanced and cause acceleration which will push the bullet offline. The higher the air density, the higher the drag and larger the offline force.