Mythbusters is notorious for setting up test conditions where the margin of error is many times larger than the thing being measured. If the bullet is fired from shoulder height, it should take about 0.6 seconds to drop. During that time, it will travel several hundred meters. Keep in mind we need to measure milliseconds here. I estimate that a fired bullet should hit the ground maybe one or two milliseconds later than a dropped bullet. In the last millisecond, the bullets are moving downward approximately 1.7 mm/msec. So, we need to set up two targets several hundred meters apart and make sure they are perfectly level with each other, within 3 mm. Good luck with that. Don’t forget the curvature of the Earth. At 200 meters, it’s 3.1 mm. At 400 meters, it’s 12.6 mm.
Oh, and if your rifle barrel was off level by just 0.001 degrees, that translates to an error of +/- 7 mm at a distance of 400 meters.
Umm… bullets spin so never have a ‘nose high’ trajectory.
I think you are thinking about how sights are aliened that makes the trajectory of the bullet higher than the sight line (the barrel of the gun is pointed ‘up’ compared to the site line. Thus allowing for bullet drop over long distances and hit where the site/s are pointing.
Bullets at distance (imagine a rifle) shoot in an arc. Not an issue with pistols as the drop in 10-20-50 feet is not an issue.
Wiki - Point Blank It has a couple of meanings (depending on usage), but ‘point blank’ can mean anything from 10 feet to a few hundred yards if that is the distance the weapon is sighted in on (meaning that you don’t have to aim high to hit the intended target)
A bullet has a (very slight) “nose high” trajectory.
In the book Understanding Firearm Ballistics by Robert A. Rinker, the author cites a paper on this. When the bullet is travelling down range (with no wind present) the bullet travels slightly nose up, thereby creating an airfoil like effect and thus giving it a slight aerodynamic lift that makes it fall slower than the bullet dropped at the same instant. The author says this is true even when the barrel is aimed parallel with the ground. He also says the difference is almost insignificant.
I have the book. If you want the exact quote, give me some time to find the book…
Enipla, spinning doesn’t preclude a non-zero angle of attack. American footballs are thrown this way all the time. They spin about their major axis, but that axis isn’t parallel to the direction of travel. It almost certainly generates a little lift. I’m sure this has been studied to death.
I have no strong opinion about bullets’ angle of attack, but Crafter Man, I do wonder what Rinker has to say about what would cause a nose-high attitude. The bullet would have to “know” which way is up, which implies that the attitude is somehow caused by gravity. Hmmm…
Huh. Never heard of such a thing. The only possible explanation that I can think might be that since the back of a bullet is heavier, it tends to fly ‘nose up’? I’m really doubtful though. What else could cause this ‘airfoil’ effect? I’ll do some Googling myself when I have some time.
Oh dear … how to say “horsefeathers” in the form of a question …
I’m confused with Chrono’s explanation in his post #2 … drag is proportional to the square of velocity … okay … let’s split the velocity into horizontal and vertical components … (the square of the horizontal) plus (the square of the vertical) equals (the square of the true vector magnitude) … I’m just not seeing how any part of the horizontal component of drag is being transferred to the vertical, which is required for the one object to hit the ground later than the other … at all times, the vertical velocities are the same, the squares are the same, therefore the vertical drag is the same …
I’m going to suggest that the assumption of a constant flow velocity field is fine for small caliber bullets … but let’s consider a hollow sphere of aluminum with mass equal to a 747 or Saturn V rocket … first thing we notice is that the trajectory of the sphere with an initial horizontal velocity is no where close to ballistic … so there’s “other things” going on here … so my question:
Would stress cause more vertical drag in the horizontal moving object, and thus have the object with strictly vertical motion hit the ground first?
If you look at it via the math, I laid out all of my steps. Is there some step that you’re unsure about? If each step works, then the whole derivation must work.
If you want to look at it without the math, then it’s because the fired bullet is hitting more air molecules, as I explained in post #6.
You can’t separate the component velocities and then square them to find drag components. You have to square the total velocity to find total drag, then separate the drag components.
Instead of simply dropping one of the two objects, try this:
Launch ball #1 straight down at 100 m/s. Launch ball #2 with a horizontal velocity of 100 m/s and a downward velocity of 100 m/s. IOW, ball #2 has a 45-degree downward trajectory at 141.4 m/s. Both balls have the same vertical velocity, 100 m/s.
Drag is proportional to square of velocity. So ball #1 has a drag of 10,000 in the vertical direction. Ball #2 has a drag of 20,000 directed 45 degrees from vertical. The vertical component of that drag is 14,142. This is more vertical drag than ball #1.
Try launching ball #2 with the same vertical velocity but even more horizontal velocity, 200 m/s. Now it’s moving at 224 m/s, at an angle 26.6 degrees from horizontal. Total drag force is 50,000, vertical component of that drag force is 22,361. This is quite a bit more vertical drag than ball #1.
the discrepancy is reduced for lower vertical and/or horizontal velocities, but for any two objects released at the same time - one dropped or fired straight down, the other launched with the same initial vertical speed but also some horizontal speed - it seems clear that the one with some horizontal speed must experience greater vertical drag at any given moment, and therefore less vertical acceleration at any given moment, than the object that was dropped or fired straight down.
Also wonder what data Rinker has to support this. And why this phenomena would occur.
Sure, a football released from the QB’s hand at a certain angle giving the front a nose up attitude. A barrel of a gun has no such attribute. And looking at ballistic tables, there is plenty of information about bullet drop, but nothing about bullet elevation gain. I’ve been shooting for 45 years and have never heard of such a thing.
Let’s assume the bullet is fired perfectly horizontal. Since it spins along its axis, then it’s not “nose up” at the moment it leaves the barrel. BUT it immediately starts dropping due to gravity, following a parabolic arc, while the bullet’s spin axis remains horizontal, which means the bullet IS nose-up, relative to the parabolic arc of travel. Hence it’s nose-up relative to the wind drag, which means it probably does experience some aerodynamic lift. But the lift would only happen while the bullet is falling below horizontal, and the closer the trajectory returned to horizontal, the less the effect would be, diminishing towards zero. Therefore, there’s no way the lift could cause the trajectory to angle upwards or oven stay horizontal. All the lift could possibly accomplish is slow down the rate of descent.
It seems EdelweissPirate and enipla are both right. The bullet is nose-up relative to its trajectory (much like a football) but it doesn’t rise unless the barrel was angled above horizontal.
I asked the question about a bullet 10 years ago. It started with useful discussion of the drag, curvature of the earth, and spin of the bullet. Then it got completely sidetracked by a now-banned poster confused about simple physics.
On second thought, maybe don’t bother reading that thread.
I was taking “all is totally horizontal (ground and throw trajectory) and gravity is perfectly perpendicular”, in the OP, to mean (among other things) that we were still neglecting curvature of the Earth, and that the only complication we were adding was atmospheric drag.
Right. A bullet fired horizontal to the ground (which is what we are talking about) produces no lift and has no aerodynamic effect to produce lift. It simply moves forward and falls.
As an aside, I remembered that old thread too. That’s why I was able to give a quick answer in this thread, before I took the time to do (and type out) the detailed calculations: I had already done the calcs back in that other thread, and remembered the result.
Careful. It doesn’t produce lift while it’s traveling on a horizontal path (i.e. a path parallel to its spin axis), but that condition lasts for only a moment after it leaves the barrel. The spin axis should remain horizontal while it flies, but as soon as it begins falling, it’s got a velocity vector that is not parallel to its spin axis, and it will develop lift. Note that “lift” does not necessarily mean a force equal in magnitude to the weight of the object; “lift” is any force perpendicular to the direction of fluid motion. So for a bullet, the lift produced during its descent isn’t enough to offset its weight, but it may be enough to cause significant deviation from a truly parabolic/ballistic trajectory.
The spinning isn’t directly relevant to the lift (though it will result in a lateral force). The lift is due to the bullet’s orientation. Spinning just helps it to maintain that orientation.
Yes, I am talking about a gyroscopic effect due to the spin of the bullet: the effect is that the bullet’s spin will cause it to remain in a horizontal orientation even after its velocity is no longer purely horizontal.
No, the spin will not cause it to counteract gravity. But the spin will maintain its horizontal orientation, even while its velocity becomes decidedly non-horizontal (i.e. once it begins to fall). Its horizontal orientation, combined with its non-horizontal velocity (once it begins to fall), will produce a force perpendicular to the path of travel (in this case, approximately opposite the gravity vector). This is independent of the drag produced by forward motion; that drag, once the bullet begins to fall, will have a vertical component that is independent of (and additive with) the aforementioned lift.
It’s a bit like launching a plane off a carrier deck with zero angle of attack: as soon as it leaves the end of the deck, it will begin accelerating downward. The only difference is that airplane wings are so good at producing lift that as long as you’re above stall speed, you’ll settle into some stable descent path (again, this assumes the plane’s long axis remains perfectly level with respect to the ground). A bullet isn’t nearly as good at making lift. It’s like a plane flung off the deck at well below stall speed: it’ll make some lift, but not nearly enough to bear its entire weight. But that lift may be enough to cause meaningful deviation from a true ballistic trajectory.