# At what speed would a wind affect a bullet? [closed]

Firing a gun loaded with the fastest bullet (.220 Swift 1,422m/s or any bullet that is super fast and excellent aero dynamics) in a close range (2cm) from the tip of an air blower. What would be the speed of the air coming out of the air blower to be able to deflect the bullet off course 90 degrees?

• Depends upon time given for pushing the bullet Commented Dec 20, 2021 at 15:16

For the bullet traveling directly at the air blower, to stop the bullet within $$2cm$$ needs, from the equations of motion a deceleration of $$5\times10^{7}m/s^{2}$$

Air resistance is $$F=\frac{1}{2}\rho ACv^2$$ see for example this website

For air $$\rho = 1.2$$, $$C=0.2$$ (estimate) and $$A=\pi r^2$$ with $$r=2.8\times 10^{-3}m$$

so from $$F=ma$$ with a mass of the bullet of $$2g$$

$$0.377r^2v^2 = 2\times 10^{-3}\times5\times10^{7}$$

$$v^2 = 3.4\times10^{10}$$

$$v = 184,000m/s$$

We could subtract the speed of the bullet from this, but it doesn't make much difference.

• So what you're saying is that, to stop the world fastest bullet with a wind blower, the blower has to blow air at 184km/s? Commented Dec 21, 2021 at 1:05
• @Hither Joe Yes, if the calculations are correct Commented Dec 21, 2021 at 8:10

Remember that the bullet does not know what the wind speed is. The bullet only knows to travel in its given medium. So if the wind was blowing from the side at the same 1,422m/s, then the bullet would travel sideways in that medium at the same rate that it is travelling forwards. In this case, in one second it would travel forward 1,422 metres and sideways 1,422 metres, so it would travel at a 45 degree angle. This might be the answer you are looking for. About 5,000 kph.

Otherwise, to travel in a complete 90 degree angle, it would have to have no forward travel at all, and complete sideways travel from the moment it leaves the gun. So if you said that in the first millisecond after leaving the gun, it travelled directly sideways, then the wind speed would have to be 1,000 x 1,422 = 1,422,000 metres per second. Darned fast.

But in general, any wind speed at all changes the path of a bullet. This is a key aspect of being a target shooter, and is the reason why they set up ribbons along the rifle range, to show the wind speed at different points.

• I think you might clarify your first paragraph. The bullet will leave the barrel in the same direction the barrel is pointed. Then the bullet will curve in the direction of the wind as it travels. So in other words, the bullet won't be going in a straight line. I think you are saying that it moves at a 45 degree angle in the moving reference frame of the wind, but that's only true at the instant it leaves the barrel. Commented Dec 20, 2021 at 19:02
• @James I had made the assumption of a constant speed for the bullet. It would travel in a straight line at a 45 degree angle. it would travel one foot to the side for each foot forward, or 45 degrees. It would not form a curve until the bullet started to slow down, which of course does happen immediately. Commented Dec 20, 2021 at 19:50
• For that to be true, the bullet would have to instantaneously change from 0 m/s side motion to 1422 m/s side motion, which would be an infinite acceleration. Instead, the bullet is pushed by the wind, and the bullet will be accelerated sideways. The side velocity of the bullet will depend on things such as the bullet cross-section, the wind speed, and the amount of time it's been pushed. Commented Dec 20, 2021 at 19:57
• I'm trying to take this seriously. You would have to have a very very fast moving air blower, or a very very (world record) long one blowing incredibly fast and its anybody's guess how long it would take to deflected even an average bullet (in a circular arc). Commented Dec 20, 2021 at 20:49
• Saying that an object will instantly move at the velocity of the medium it is in is just not true. If it were true, then the bullet shot from gun on a day with no wind would just instantly stop... and float I guess. Commented Dec 20, 2021 at 21:37