The Joule and the Impact of the hit with heavier BBs

[Barf]

For those who were at the last sniper clinic, I went over the difference between momentum and kinetic energy.

Momentum = mass * velocity = kg * m/s

Kinetic Energy = 0.5 * mass * velocity² = kg * (m/s)²

A small car (assume 1 metric tonne) moving at 1 m/s has the same momentum as 1 kg block at 1000 m/s, but the block has 1000 times the kinetic energy.

For bb impact, kinetic energy is the important factor, ie your body is not trying to stop the bb (momentum), the bb is imparting/transferring it's energy to you (kinetic energy).

[WhatTheWho]

Originally Posted by mcguyver

Remember the gyroscopic effect of hop-up on a BB. The heavier it is, the more energy it has, hence the less affected by gravity. It will "appear" to the naked eye to fall to the ground slower, all the while travelling straighter.

It's not gryoscopic. It's the Magnus effect which explains why hop-up works.

Gravity is constant not matter what mass object; it does not have less of an effect because a mass is greater or lesser than another. You calibrate hop-up for a straight trajectory where in effect, you are adding a force with an acceleration of 9.81m/s^2 to counteract that of gravity's.

The only reason why a heavier BB "could" retain it's hop-up effect better than a lighter BB, and therefore counteract gravity better, is due to it's mass being greater such that the force of friction from the surrounding air is not able to impart enough force to decelerate its spin when compared to a lighter BB.

I.E. Backspin is reduced faster on a BB with lighter mass when compared to a BB with heavier mass.


This explanation itself holds true to why a heavier BB retains more kinetic energy than a lighter BB when both energy outputs are equal; air resistance has less effect on an object with more mass, therefore it decelerates less than an object with less mass.

Less deceleration means it retains more of it's velocity in a given period of times than a lighter BB, and so it will retain more energy, and that dictates impact power.

It's not due to the momentum or anything like that, those are only relative to the reason and definition as to why more mass retains more energy. It's due to air resistance that a heavier projectile will retain more energy than a lighter projectile.

Another thing to note is that a sphere traveling faster experiences more drag (air resistance) than a slower traveling sphere; that would also be indicative of why a heavier BB will keep its velocity, therefore its kinetic energy, better.

An equation to relate why velocity and mass dictate air resistance, and how that relates to the retention of energy:

a = (Cd * r * V^2 * A / 2) / m
a = deceleration due to air resistance
Cd = coefficient of drag for a smooth sphere, being 0.5
r = density of air, being 1.229 kg/m^3 at sea level
V = velocity, being variable
A = cross-sectional area of the ball, being 2.826e-5 for a 6mm BB
m = mass of BB, being variable

Simplifying the equation yields:

a = (8.68e-6 * V^2) / m (at sea level)

This shows the relationship of velocity and mass to air resistance, and that an increase in velocity or a decrease in mass will increase air resistance. That causes more deceleration, and more loss of energy.
(This equation is for an instantaneous calc. of deceleration with a given velocity, as the velocity decreases for the next calc. of deceleration since air resistance is continuous and therefore velocity changes continuously)

So even if two different mass BB's were fired with the same Joules and with their corresponding velocities, the heavier BB will retain more energy and have more "impact power".

[m102404]

My head's starting to hurt reading this thread. Cudos and appreciation to those who have obviously put a lot of time and thought into their posts. Not being as adept as others at physics/math/etc... I'd opt for the empirical test route. Definitely think this thread should be cleaned and stickied.

Go out to a game.

Load up a couple of heavier bbs in one mag and some lighter ones in another. Put your goggles on, give your AEG and mags to a buddy you trust who won't shoot you in the nuts or throat/teeth.

Pace off a good distance (50yrds?) and have him shoot you, semi auto, center mass. Wear kit or not. See for yourself. Have him switch up the mags and shoot you again. Go a bit further out and have him shoot you some more.

For determining range, keep going until he can't hit you reliably any more (i.e. 6 hits out of 10 or whatever you think is a good average). Have him switch mags and see if the effective range is further or less.

Throughout, don't have him tell you which mag/bb weight he's using (i.e. blind test so you're observations aren't tainted by preconceived notions). Don't switch guns, brands of bbs, hop-up settings, etc... Have him take notes and you do the same.

Thank him for not shooting you in the nuts, then compare your observations with him and report back to the forum. Let us know what you observed.

Personally, I've noticed that, reduced or increased range or not, the cluster/groupings of shots at the outermost extent of the range of my AEGs tend to be tighter with 0.25bbs (vs. 0.20 bbs). Less spread and fewer fliers. To me, that means more hits on target. I think everything else is theoretical, if you can't hit what you're aiming at, a little bit more kinetic energy/momentum/terminal velocity, etc... won't make a bit of difference. I'm of the opinion that, you've got to know your limits and play effectively within them. If someone doesn't seem to feel the first hit or two, give them a burst of full auto and they'll hopefully notice all the BBs bouncing off them. If they are still too "caught up in the moment" to notice your shots, empty the mag at them, swear/bitch/cry (whatever makes you feel better), change out to a fresh mag and keep playing.

If after you make your observations, you want to increase power, increase range, tighten up groupings, increase rate of fire, etc.... search through this forum. Do a thorough search. It's filled with great advice and good people willing to take the time to help you out.

Just my $0.02.

Have fun, it's just a game. But the theory debate is neat.

[Flatlander]

Originally Posted by WhatTheWho

It's not gryoscopic. It's the Magnus effect which explains why hop-up works.

Gravity is constant not matter what mass object; it does not have less of an effect because a mass is greater or lesser than another. You calibrate hop-up for a straight trajectory where in effect, you are adding a force with an acceleration of 9.81m/s^2 to counteract that of gravity's.

WhatTheWho's got it pretty close, IMO.
The magnus effect is where you create a pressure differential between the top and bottom of the bb. The backspin creates higher air velocity (velocity of the air moving over the surface of the bb) on the top of the bb and slower air velocity on the bottom of the bb. This velocity difference results in a low pressure on top and a high pressure on the bottom (like how an airplane wing works). The high pressure wants to "push up" to the low pressure to obtain an equilibrium, thus creating "lift" (a force upwards counteracting the weight force down). This is why if you have too much backspin (hopup) you will actually get the bb's to rise as the "lift" generated is greater than the weight force.

Originally Posted by WhatTheWho

The only reason why a heavier BB "could" retain it's hop-up effect better than a lighter BB, and therefore counteract gravity better, is due to it's mass being greater such that the force of friction from the surrounding air is not able to impart enough force to decelerate its spin when compared to a lighter BB.

I.E. Backspin is reduced faster on a BB with lighter mass when compared to a BB with heavier mass.

This is due to momentum! Momentum is loosely defined as the difficulty to stop an object. So we're both correct in a way, I believe.

Originally Posted by WhatTheWho

This explanation itself holds true to why a heavier BB retains more kinetic energy than a lighter BB when both energy outputs are equal; air resistance has less effect on an object with more mass, therefore it decelerates less than an object with less mass.

Less deceleration means it retains more of it's velocity in a given period of times than a lighter BB, and so it will retain more energy, and that dictates impact power.

This is also true but again, its due to it's greater momentum that it retains it's velocity (or energy) better, for lack of a better term.

It appears to me you haven't studied mechanics/dynamics extensively as you say impact POWER. Power and energy are, again, very different! Units of energy are in joules, units of power are joules/second. Not meaning to cut you down; maybe just a mental lapse...

Originally Posted by WhatTheWho

Another thing to note is that a sphere traveling faster experiences more drag (air resistance) than a slower traveling sphere; that would also be indicative of why a heavier BB will keep its velocity, therefore its kinetic energy, better.

Close...if you fired bb's of different masses at the SAME velocity, the one with the greater mass will have it's velocity decrease at a slower rate.

Originally Posted by WhatTheWho

An equation to relate why velocity and mass dictate air resistance, and how that relates to the retention of energy:

a = (Cd * r * V^2 * A / 2) / m
a = deceleration due to air resistance
Cd = coefficient of drag for a smooth sphere, being 0.5
r = density of air, being 1.229 kg/m^3 at sea level
V = velocity, being variable
A = cross-sectional area of the ball, being 2.826e-5 for a 6mm BB
m = mass of BB, being variable

Simplifying the equation yields:

a = (8.68e-6 * V^2) / m (at sea level)

This shows the relationship of velocity and mass to air resistance, and that an increase in velocity or a decrease in mass will increase air resistance. That causes more deceleration, and more loss of energy.
(This equation is for an instantaneous calc. of deceleration with a given velocity, as the velocity decreases for the next calc. of deceleration since air resistance is continuous and therefore velocity changes continuously)

So even if two different mass BB's were fired with the same Joules and with their corresponding velocities, the heavier BB will retain more energy and have more "impact power".

Can't completely comment on this except you have to be careful. You're finding its acceleration (or deceleration in this case) which deals with FORCES (F=ma) and then talking about energy then mistake the term "impact power" which should be (I think) impract energy as they would be completely different.

I'm not completely sold that impact energy (kinetic energy) is the definitive way of saying what will hurt more. I could be wrong and will look into it as I'm quite curious....

I remembered we did a test one day out at the field to see what hurts more, .20's or .25's. We had mixed results to say the least. I think the test must be done point blank (or close as possible) so that they (in theory) have relatively the same impact energy (ie. little to none is reduced by drag)

Just look at the Mythbuster's "frozen vs thaw" chicken test. They thought that the penetration should be equal as they will have the same kinetic energy when they impact the windshield. This was obviously not the case as they saw the frozen had MUCH greater penetration. True, they do have the same "impact energy", but they transfer this energy completely different! Not quite a good analogy for our BB case but it gets your mind around energy being the "be all, end all"....

A lot of terminology is very confusing and gets thrown around loosely. I find it tough for me to keep it all straight at time. FYI/summary for those interested:

-Mass isn't the same as weight! Mass is essentially "how much stuff is in something". You have the same mass on earth as on the moon, however your WEIGHT will change! (weight=mass*acceleration due to gravity). Weight is also a FORCE. Note that imperial units (pounds,etc) are weights (forces) and metric units (kilograms,etc) are masses. Gets tricky dealing with both!

-Momentum and energy are NOT the same. Have the same parameters but are quite different. Note that work is also energy (ie. you must do X amount of work to obtain X amount of kinetic energy)

This is a light-hearted debate (on my mind) and enjoy getting the brain going even though it is summer time! This is a huge portion of what I study at school so it quite interests me. Time to get something done at work now...

[Jayhad]

M102404, I am with you on this one. I enjoy studying physics and do so in my spare time but nothing beats practical testing.

[Flatlander]

I'm hoping someone out there will have some test results and post them! That way I can see if my theory is flawed or not.

I'm also very curious to see the velocity results from an AEG when shooting various BB weights (masses!) and see if their kinetic energies are all the same or close (ignoring rotational energy).

So much for getting any work done...

[WhatTheWho]

Originally Posted by Flatlander

This is due to momentum! Momentum is loosely defined as the difficulty to stop an object. So we're both correct in a way, I believe.

Actually, it's not due to momentum.

I could have a 1g sphere traveling at 10m/s compared to a 10g sphere traveling at 1m/s; momentum would be the equal. But, the 1g sphere will decelerate faster than the 10g sphere.

Originally Posted by Flatlander

It appears to me you haven't studied mechanics/dynamics extensively as you say impact POWER. Power and energy are, again, very different! Units of energy are in joules, units of power are joules/second. Not meaning to cut you down; maybe just a mental lapse...

Just a lapse. I forgot to throw quotation marks around "impact power" as it was the term everyone was throwing around in this thread to define the amount of "hurt" a person feels. I even quoted it in my last remarks of my previous post.

Originally Posted by Flatlander

Close...if you fired bb's of different masses at the SAME velocity, the one with the greater mass will have it's velocity decrease at a slower rate.

What you say is true, but the concept of higher velocity creating for air resistance is what I'm trying to indicate.

My statement holds true as I am wording it to keep in context of the debate. The question that I gathered from this thread was, given the same gun which would fire off different mass BB's at consistent kinetic energies, the lighter BB would have a higher FPS than the heaver BB.

Originally Posted by Flatlander

Can't completely comment on this except you have to be careful. You're finding its acceleration (or deceleration in this case) which deals with FORCES (F=ma) and then talking about energy then mistake the term "impact power" which should be (I think) impract energy as they would be completely different.

As I mentioned, I forgot to throw quotation marks around "impact power."
Follow through the equation itself with a set of numbers for a greater mass BB and a smaller mass BB, ensuring that the velocities of both will produce the same kinetic energy. The lighter BB will have more air resistance.

If you doubt the equation, it can be looked up on google or such.

Originally Posted by Flatlander

I'm not completely sold that impact energy (kinetic energy) is the definitive way of saying what will hurt more. I could be wrong and will look into it as I'm quite curious....

True, kinetic energy is not the definitive was way of saying what will hurt more.

I would think that Impulse would be quantities that would need to be calculated in terms of how much "hurt" a BB has, given that it is your body which needs to decelerate the BB to zero.

[Flatlander]

Sorry let me try to clarify my "position" on things. To me it looks like we both are "right" but have simply technical discrepancies.

-I believe an AEG will impart the same kinetic energy to each BB reguardless of it's mass (I'm hoping to do some emperical tests to prove this THEORY).

-The BB with the greater mass will decelerate ("lose it's velocity") at a slower rate due to it having the greater momentum (I'm thinking of momentum more as "inertia" rather than strickly m*v).

Quick calculations:

Kinetic energy (.20 @ 90m/s) = 810 joules -> From this I can calculate the theoretical velocity of .25's (neglecting rotational energy)

Velocity of .25's with same KE = 80.5 m/s

Momentum of .20's = 18
Momentum of .25's = 20.125

Deceleration due to drag = k*v^2/m, where k=constant=coefficient of drag*radius/2 ; v=velocity; m=mass

-You can see here that the deceleration will ALWAYS be greater for the lighter BB fired from the same AEG (theoretically!). Also, you will notice that the heavier BB will ALWAYS have the greater amount of momentum when fired from the same AEG. Hence, why I said the momentum had something to do with the drag and deceleration and straighter flight, etc.

-Of course drag has many parameters such as surface area, fluid viscosity, as well as mass and velocity so by me referring to it’s momentum being the key factor maybe a bit of an oversimplification for our current airsoft situation?

-Drag is also a force (or in the above equation an acceleration assuming it’s correct), and to turn that into negative work (energy “stolen”) you must multiply it by the distance traveled (work = force*displacement….units of energy). Basically technical mumbojumboo that is very confusing.

-This is all assuming that the BB is NOT SPINNING. Once it starts spinning it gets A LOT more complicated (ie. Magnus/Bernoulli effects, rotational energy, etc).

Topic of what to use for calculating how much “hurt”:

-Note that impulse and momentum go hand-in-hand (same units and used together to do impulse/momentum analysis). Impulse is the integral of force*d(time)….which is basically Force*time for simplicity sake (assuming the force does NOT change with time, which usually isn’t the case)

-So again, I have no idea what to use for calculating “hurt” but I’m leaning towards impulse/momentum.

[Mud Gunner]

I think physical testing would be simpler and less "thinky". "ouch" what that was .2s "WTF?" would be .25s or higher

[mcguyver]

They're both in your yard Mud. Ask them over to play.

[Nik12]

Originally Posted by Flatlander

(I'm a mechanical engineering student).
The moment of inertia of a sphere is 2/5*m*R^2.

1) My bad, I usually deal with circular cross setions of beams (Civil Eng Student).

Originally Posted by Barf

Momentum = mass/velocity = kg/(m/s)
A small car (assume 1 metric tonne) moving at 1 m/s has the same momentum as 1 kg block at 1000 m/s, but the block has 1000 times the kinetic energy.

2) ...Well, actually it's mass multiplied by velocity, but that's ok.

Exactly. And personally, I think that 1kg block moving at 1000m/s will hurt ALOT more than the 1 tonne car moving at 1m/s. Would you not agree (since that's nearly 3 times the speed of sound...with general conditions). This kind of proves that impact energy is the main basis for hurt. Remember this, I'll be bringing it up later.

Originally Posted by Flatlander

Kinetic energy (.20 @ 90m/s) = 810 joules -> From this I can calculate the theoretical velocity of .25's (neglecting rotational energy)

Velocity of .25's with same KE = 80.5 m/s

3) I realise you're that you're just doing a basic comparison and that the extra powers of ten would just disappear anyways, but the BB has .81 joules (.2g is .0002kg, which is nessecary for J).

Originally Posted by Flatlander

Topic of what to use for calculating how much “hurt”:

-So again, I have no idea what to use for calculating “hurt” but I’m leaning towards impulse/momentum.

4) Summary: 1kg Block moving at 1000m/s verses 1000kg car moving at 1m/s. Momentum is equal, kinetic energy has a 10^3 difference in magnitude. 1kg block with 1000m/s velocity would porbably hurt more. For full details, see quote reply number two.

Originally Posted by Flatlander

I'm not completely sold that impact energy (kinetic energy) is the definitive way of saying what will hurt more. I could be wrong and will look into it as I'm quite curious....

5) I would say so. What else could cause hurt? When something has momentum, it also has energy. The transfer of energy to the tissue of the body would cause pain. Why can't momentum do this alone? Think electricity. Electircity doesn't have momentum (don't give me any quantum electron movement theories, because the momentum of an electron is rediculously small, since it has such a tiny tiny tiny tiny mass and they don't even move that fast either...not as fast as most people might think anyways). For all intents and purposes electricity is just energy. Pure energy. Yet when you're connected to a live wire or two, it can really hurt. Now, I'm not entirely sure on what I just said, since the closest thing to an electricity course I've taken is a first year intro to electricity and magnetism and a really simple electricity and thermophysics course, but it seems to make sense to me.

Originally Posted by Flatlander

Just look at the Mythbuster's "frozen vs thaw" chicken test. They thought that the penetration should be equal as they will have the same kinetic energy when they impact the windshield. This was obviously not the case as they saw the frozen had MUCH greater penetration. True, they do have the same "impact energy", but they transfer this energy completely different! Not quite a good analogy for our BB case but it gets your mind around energy being the "be all, end all"....

6)But then collision thoery has to be accounted for. The thawed chicken is a softer medium so it will absorb some of the impact energy (which would probably cause it to explode into little bits). The frozen chicken is a harder medium so it will not absorb nearly as much of the impact energy, which will force it all onto the object it is impacting. This is definately a very bad analogy for our BB case, since the BBs are of nearly the same hardness, and would produce the same type of impact.

Originally Posted by Flatlander

This is a light-hearted debate (on my mind) and enjoy getting the brain going even though it is summer time! This is a huge portion of what I study at school so it quite interests me.

7)Ditto. I may not be in Mech and may not need to know a whole lot about the way things move (aside from structures), but I certainly do enjoy knowing it. It's good to get out and strech the old brain out during the off season.

[Barf]

Originally Posted by Nik12

2) ...Well, actually it's mass multiplied by velocity, but that's ok.

Corrected. It was a typo.

[Barf]

Mass (g).........Velocity (fps).......Kinetic Energy (j)
0.20....................280....................0.7 3
0.20....................360....................1.2 0
0.20....................400....................1.4 9
0.20....................450....................1.8 8
0.20....................500....................2.3 2
0.20....................550....................2.8 1
0.20....................600....................3.3 4

0.25....................280....................0.9 1
0.25....................360....................1.5 1
0.25....................400....................1.8 6
0.25....................450....................2.3 5
0.25....................500....................2.9 0
0.25....................550....................3.5 1
0.25....................600....................4.1 8

0.29....................280....................1.0 6
0.29....................360....................1.7 5
0.29....................400....................2.1 6
0.29....................450....................2.7 3
0.29....................500....................3.3 7
0.29....................550....................4.0 7
0.29....................600....................4.8 5

0.30....................280....................1.0 9
0.30....................360....................1.8 1
0.30....................400....................2.2 3
0.30....................450....................2.8 2
0.30....................500....................3.4 8
0.30....................550....................4.2 2
0.30....................600....................5.0 2

0.36....................280....................1.3 1
0.36....................360....................2.1 7
0.36....................400....................2.6 8
0.36....................450....................3.3 9
0.36....................500....................4.1 8
0.36....................550....................5.0 6
0.36....................600....................6.0 2

...and why we don't play with 0.43's

0.43....................280....................1.5 7
0.43....................360....................2.5 9
0.43....................400....................3.2 0
0.43....................450....................4.0 4
0.43....................500....................4.9 9
0.43....................550....................6.0 4
0.43....................600....................7.1 9

[Nik12]

Well done. *applause*

[CDN_Stalker]

This is a great example to show off my basic math skills:

Chairsofters > Airsofters