Wikipedia:Reference desk/Archives/Science/2019 July 4

= July 4 =

Bullet Energy and Height
My friend and I are developing a homebrew pen and paper combat game where each player moves miniatures around on a board and shoots at the other team. To make things more interesting, we added vertical elevation into the mix. This lead to the following debate: on the one side, it feels like - for a fixed horizontal distance - that adding elevation should make the bullet have more energy when it arrives at the target because of gravity; however, increasing elevation also increases the distance from the target, which feels like it should reduce the bullets energy when it arrives at the target. So, how does adding elevation affect the energy of the bullet upon reaching the target - does terminal velocity factor in? I find the question a little more fascinating than the rules of the game (will probably just go with whatever feels best to us), it is the actual science of the thing that seems neat here, I'd love to know what all variables are involved and what the broadest possible result is. Thank you for any help:-)24.3.61.185 (talk) 09:36, 4 July 2019 (UTC)


 * There's a well-known effect that elevation relative to the target increases artillery range (although most of that is just due to there not being any ground to hit when it loses elevation), but I'm not sure if the same would apply to bullets, since air resistance is more of a factor there. Ideally, the gravitational potential energy would transform into kinetic energy closer to the end, so that the increased velocity would occur then, not near the start, where increased velocity would also increase aerodynamic drag. SinisterLefty (talk) 09:57, 4 July 2019 (UTC)
 * Doing some really rough calculations. A 7.62x51mm Nato bullet is 10g, dropping it from 35 meters in height should 3.43 joules at impact. A 7.62x39mm bullet will have 1119 joules of energy at 100 yards and 813 joules at 200 yards. So, supposing the target was a horizontal distance of 120 meters and that linearly interpolate energy from 100 to 200 yards (which isn't accurate, but good enough for a rough-ish estimate), that's 1056 joules at point of impact. If we raise the shooter 35 meters into the air, that changes the distance to 125 meters, giving 1043 joules + 3 joules from gravity = 1046 joules (the 39 is lighter, but 3 seems decent). So, going by that, it looks like there would be little difference, but favouring the non-elevated shooter. Raising the elevation to 70 meters drops the joules at impact to around 1005ish, more so in favour of the ground based shooter. Of course, I'm sure something is being neglected, here, since these are very rough methods, but it does seem to favour the ground level shooter. Though, since most people aren't going to be shooting from elevations exceeding 30 meters, it seems like there wouldn't be any appreciable difference - for larger horizontal ranges, the gap should close even more, eventually winning out for the elevated shooter, but not by a lot and at distances that would severely reduce accuracy. All that said, I feel like I must be missing something, physics is not my strong suit.24.3.61.185 (talk) 10:33, 4 July 2019 (UTC)


 * A quantitative answer involving air resistance, velocity calculations etc. is hard to give, but a qualitative answer is fairly easy to come by.
 * The question can be converted into whether a gun located at $$x=0, z=z_0>0$$ can hit a target located at $$x=x_t, z=0$$, assuming the same gun can hit the same target when located at $$x=0, z=0$$. In the higher position, if the bullet is fired with the same angle, it with pass through the point $$x=x_t, z=z_0$$ and continue, falling down somewhere beyond the target. On the other hand, it is trivial to shoot the bullet somewhere closer to the shooter than the target is, say, towards $$x=0, z=0$$. Because the position of the intersection of the bullet trajectory and the ground is a continuous function of all parameters at hand (this seems obvious, but it is not true in general; proving it is true in the case at hand would involve lots of math that I am not sure to be able to do), the intermediate value theorem ensures there is some settings for which the bullet reaches the target.
 * As for whether the energy hitting the target would be higher or lower, both can happen. If the bullet speed just outside the rifle is gigantic (far above free fall terminal speed), and the target is very close on the x-axis, then the dominant effect is friction dissipating the initial kinetic energy, so a point-blank shot would do more damage. On the other hand, if the target is very far, the bullet speed is low and the height difference is huge, the dominant effect will be bullet acceleration due to gravity. Not sure how to put that into a nice adimensional number, though... Tigraan Click here to contact me 10:28, 4 July 2019 (UTC)


 * Practical shooting teach that you have two situations:
 * If the trajectory is basically a straight line (as opposed to: significantly parabolic), that is, if the target is close enough in respect to the speed of the bullet), so that you don't need Iron_sights: you just don't care about height and gravity
 * If not, just don't shoot, you'll miss unless you ARE a sharpshooter
 * This is not an energy problem, rather an aiming problem.
 * Gem fr (talk) 23:12, 4 July 2019 (UTC)

External_ballistics deals mathematically with the trajectory of a bullet in flight. Gravity can add to the bullet's energy on hitting the target only if the shooter is at higher elevation than the target, and then by only by a tiny amount corresponding to its fall during the short time of travel. A Terminal_velocity is the limit to the downward velocity of a free-falling object where atmospheric drag equals its weight so that it cannot be further accelerated. For a Parachute terminal velocity is slow and is quickly reached but a bullet is normally stopped by a target well before its rate of fall reaches terminal velocity. DroneB (talk) 09:01, 5 July 2019 (UTC)

Particle/ projectile emitted from neutron star or white dwarf
If someone throws a rock straight up on the moon or some other airless planet, I can estimate the changes in velocity with v=v1-at and v^2-v1^2=2a(y-y1), with a=gravity acceleration and y height. For a projectile that starts faster i think i would have to use change in kinetic energy =k(1/y -1/y1), since the force of gravity falls off as y^-2. What about a projectile from a white dwarf(if it were rotating slowly or not at all for simplicity)? Would relativistic effect change the formulas? Thanks! Rich (talk) 18:39, 4 July 2019 (UTC)
 * Not that much. Ruslik_ Zero 19:47, 4 July 2019 (UTC)