Ah yes, aracer – good physics.

He's right, and that's also the reason why a side wind whilst driving will have a negative impact on fuel economy (I think).

I 'spose the nearest you can get this scenario to a conveyor belt is a geostationary satellite. It's moving really fast but it's not…freaky!!

Doesn't a bullet decrease in mass once fired? That would mean the stationary bullet would accelerate quicker and hit the ground first.

aracer – Member

Horizontal motion does not have an effect, fluid dynamics or not.

Well yes it does – for the reason I gave, that the drag force is non linear (proportional to the square of the speed). It doesn't need to generate lift, just drag.

To give some numbers, this means that if the bullet is going at 10m/s vertically (after just over a second of falling), then the vertical drag force on the dropped bullet is 100N, where N is a constant.

If the bullet is doing 100m/s horizontally (it will be going far faster when fired, I'm trying to make numbers simpler), then when it is also doing 10m/s vertically, the total speed is 100.5m/s. This results in a total drag force of 10100N in the opposite direction to the direction the bullet is moving in, the vertical component of which is 1005N. So 10 times the vertical drag force.

Am not sure how you got 1005N out of that… just remember similar triangles and all that…

If the horizontal velocity is 100m/s and the vertical velocity is 10m/s then you are correct that the overall velocity of the bullet is 100.5

If the drag works on the square of velocity, then you can do the calcs on it's overall velocity, but you can also do it on the vertical and horizontal components, and it will come to the same answer.

So, if you square the overall velocity to get your overall drag, so 10100N. You can also do it by squaring the horizontal velocity and adding it to the square of the vertical… (10000 + 100)N

It doesn't change the vertical component at all, which stays the same as the bullet that has been dropped.

shaggy… tut tut.. didn't your physics teacher teach you anything? Irrespective of mass, both bullets would fall at the same rate… I think Galileo proved this a few years ago…

:o)

Ah yes, aracer – good physics.

8)

Am not sure how you got 1005N out of that… just remember similar triangles and all that…

Easy. Total drag force is 10100N. This acts opposite to the direction of travel of the bullet. You can make a triangle with sides 100, 10 and 100.5 to represent the horizintal, vertical and resultant velocities – the force is along the hypotenuse, length 100.5, you resolve this into the component directions, hence the vertical component is 10100N/100.5*10=1005N (horiz component is 10100N/100.5*100=10050N).

It doesn't change the vertical component at all, which stays the same as the bullet that has been dropped.

Quite clearly from the maths above yes it does – because the drag force is non linear.

If it was a lead shot they would hit the ground time but surly the rifling of the bullet would have an affect?

Nick: I was hoping to push this in to relitivity 😮

What force does the bullet exert on the Earth? 🙂

I think my demo uses low velocities so we can ignore air resistance

Well that's where you're going wrong – if you use a real gun with normal muzzle velocity you can't ignore air resistance

My demo does use low velocities thats why it works, I don't use a gun do I. Yes I can ignore air resistance in my demo

aracer… you know when you have one of those d'oh! moments…

Hmmm…. so for drag does this mean you can't work it out using component vectors?

Ahhh, now in my hunt for this I found the equations that describe the drag induced on moving objects… they are quite complicated aren't they…

:o)

shaggy… surely the same as the earth exerts on it!

:o)

Forward motion alone won't be enough for a bullet to generate lift.
I was talking to Bez

Er. I assume you just read my second post (about forward motion of an aeroplane) and assumed it was a reply to the OP? An aeroplane generates lift (asumming the wings are still attached of course), a bullet doesn't.

As the above discussion might hint, fluid dynamics is hideously complex, which is why you leave it out of a nice simple thing like this 😉 Never mind just not using component vectors, you've got some tricky stuff going on with a body spinning extremely quickly and airflow suddenly separating and going turbulent behind it. I'm not accepting any reasoning from anyone on the aerodynamics of a bullet unless they come armed with a decent CFD model 🙂

Hmmm…. so for drag does this mean you can't work it out using component vectors?

You can't treat the orthogonal vectors independently no – that's only possible if everything is linear.

I found the equations that describe the drag induced on moving objects… they are quite complicated aren't they…

Yes, but proportionality to the square of the speed is a pretty good approximation.

As the above discussion might hint, fluid dynamics is hideously complex, which is why you leave it out of a nice simple thing like this Never mind just not using component vectors…

Fine, but that means the answer you get is wrong. Not being able to break it down into orthogonal vectors doesn't cause that many problems with the maths as I demonstrated above, and it's far from impossible to determine the approximate effect the fluid dynamics has by using a simplification. You can just ignore all the more complex effects and focus on the basic aero drag as I showed – a perfectly valid simplification given that the more complex stuff won't decrease the vertical force on the fired bullet to less than that of the dropped one.

Fine, but that means the answer you get is wrong.

In fairness the original scenario never mentioned performing the test in a viscous fluid, so it's not necessarily wrong 😉

By the way, this bit…

the total speed is 100.5m/s. This results in a total drag force of 10100N

First statement, agreed. What's the working to get to the second?

In fairness the original scenario never mentioned performing the test in a viscous fluid

Yeah, but it seems a not unreasonable assumption – has anybody ever fired a gun in a vacuum?

"the total speed is 100.5m/s. This results in a total drag force of 10100N"

First statement, agreed. What's the working to get to the second?
I used dodgy symbols – really should have been anything other than "N" to signify an arbitrary constant (it was late, I wasn't thinking straight). Just working on the principle that drag force is approximately proportional to the square of the velocity.

BTW whilst I was googling to see if I could find any info on firing a gun in a vacuum I found http://www.nennstiel-ruprecht.de/bullfly/faq.htm#Q9

I used dodgy symbols – really should have been anything other than "N"

Ah! Yeah 🙂

But isn't that comparison based on an assumption that the coefficient of drag will be the same for a spinning bullet fired forwards as for a bullet dropped on its side…?

(Meh – anyway – the point is purely explaining that forces don't affect perpendicular motion, it's not supposed to be a fluid dynamicist's problem :))

Ignoring drag, if the bullet being dropped is released at the exact instant the gun is fired then it will hit the ground slightly before the one fired from the gun. This is due to the extra time taken for the fired round to accelerate the travel horizontally through the guns barrel before starting it's acceleration downwards in the vertical plane. It will only be a small fraction of a second difference, much smaller than the margin of error you could realistically obtain when trying to fire the gun and drop the round at the same time.

aracer – so a bullet fired into the wind would hit the ground later than it would on a still day?

But isn't that comparison based on an assumption that the coefficient of drag will be the same for a spinning bullet fired forwards as for a bullet dropped on its side…?

As I alluded previously, any difference in cD is nowhere near sufficient to overcome the huge difference in the term in the drag equation due to the velocity (to a first approximation we're talking a factor of 30 difference for a bullet fired with a slowish muzzle velocity of 300m/s after 1s of falling – with larger factors beforehand).

anyway – the point is purely explaining that forces don't affect perpendicular motion

Bah – I'm just busy enjoying proving that's not the case!

so a bullet fired into the wind would hit the ground later than it would on a still day?

Interesting question. Based on my simplification, yes, but the difference is so small that it's likely to be overwhelmed by all the things I'm ignoring.

what if a squirrel fired the gun?

Sorry,i have not read all the previous posts so if this has already been stated then i apologise.I have done quite a lot of work with regards to internal and external ballistics in the past and it will all depend on a number of factors.

1.muzzle velocity
2.bullet ballistic co-efficient.(can be calculated on G1 OR G7 scale)
3.bullet weight
4.sectional density of the bullet.
5.rifling twist which will effect rotational speed depending on the bullets weight,length and bearing surface.

Some examples. You can have two bullets of the same weight but with a different ballistic co-efficient,fired at the same speed the higher b.c on e will fly flatter and for longer before hitting the ground.

Muzzle velocity,you could shoot a 15 grain weight bullet from an air rifle at 600 feet per second or you could shoot a 15 grain bullet from a centerfire rifle at over 400 feet per second.

the rate of twist with regards to rifling will affect how stable the bullet is in flight and therefore how long before drag starts to have and effect on the bullet slowing it down and causing it to fall.

I also noticed in a post early on that someone mentioned that the bullet will fly in an ark when it leaves the muzzle,it wont.As soon as the bullet leaves the muzzle it starts to drop,the reason people think they fly in an ark is because they cross the line of sight of your telescopic sight or open sights.

sorry for boring you all ridged.

AAHH I see well that explains it then. Stupendous!!!

Nevertheless, if you did this on the moon (assuming no atmosphere so no drag), the fired bullet and the dropped bullet would hit the surface at the same time?

: P

kaesae, that's not the way to answer a theoretical physics problem.

: P

Pierre

Do you have any evidence and can you prove it's not the right way to answer?

Apart from kaesae being a complete cant I agree with *same time*, what on earthe are all those words ^^^^^^^for?

Sorry,i have not read all the previous posts so if this has already been stated then i apologise.I have done quite a lot of work with regards to internal and external ballistics in the past and it will all depend on a number of factors.

Yes, but the question was which will hit the ground first, a dropped bullet or a fired one. Surely none of your points apply to the dropped bullet, and none of them will make the fired bullet fall faster than the dropped bullet, you're just suggesting that calculating how much slower it falls compared to the dropped bullet is rather difficult.

AAHH I see well that explains it then. Stupendous!!!

*sigh* – I ought to point out, FWIW, that the post I answered said something along the lines of "There are too many factors so the question is null and void!!!" which kaesae then edited to the above italics.

Dick.

: P

Head

😆

To clear things up a little i have no doubt that out of the millions of cartridge case,bullet and powder options out there that you can find a combo that will allow the fired bullet to hit the ground before the dropped bullet or you could change things around ballistically so that the dropped bullet will hit the ground first etc.

There are so many factors that it would be a total nightmare to control all of them at once.If for instance you found an ideal load for the rifle or pistol that would let the dropped bullet and fired bullet hit the ground at the exact same time at sea level.If you then moved the test up a mountain then you would get a different result.

If you changed the bullet for another bullet of exact same weight fired at the exact same muzzle velocity but with a higher ballistic co-efficient then the dropped bullet would hit the ground first.

Cheers aracer, taught me something. 🙂

I have no doubt that out of the millions of cartridge case,bullet and powder options out there that you can find a combo that will allow the fired bullet to hit the ground before the dropped bullet

I might not be an expert in ballistics, but I know my physics, and I reckon your lack of doubt simply means you're not thinking it through properly. I have no doubt that there's no ballistic combination which will result in less aero drag in a vertical direction than the bullet which is dropped, simply for the reasons I've explained above – there's simply no way to reduce the drag coefficient low enough to make up for the huge difference in velocity (it needs to be less than 1% of the drag coefficient of the dropped bullet).

Here's another theoretical question for you to solve.

Why did the chicken cross the road?

I hate to admit it, but I have just spent 15 minutes of my life that I'm not going to get back putting a few numbers on this to prove aracer wrong. And the sad fact is, he's right! 😳

Fully agree, and the result surprised me (I have a background in aerodynamics!).

Becky said… my excuse is I'm a lazy chemist and not a physicist

So's Mrs Higgo and, at times like these, I sometimes suggests she goes back to looking at the pretty colours and putting elements in rows depending on how much they fizz.