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Xiolablu3
05-15-2007, 06:41 PM
In Predator, one of the soldiers is carrying a small minigun which iis very effective in the film.

I was just wondering if this was a real handeld gun, or is it complete fantasy?

I would think the main problem about this minigin would be weught and the weight of ammo that you would have to fire.


Also the inoitial 'spin up' of the gun may also be a problem. WHen you are likely to meet the enemy you dont want to have to wait a second after you pull the trigger.


EDIT: It seems it is real! As I was searching for a picture I came across this page! I gather this is real?>

http://www.kitsune.addr.com/Firearms/Machine-Guns/GE_XM214_Minigun.htm

Ernst_Rohr
05-15-2007, 06:57 PM
Actually, there were a couple of developers who did indeed build a "man portable" version of this gun.

It still weighs a ton, and to actually have it function effectively, they had to seriously tone down the rate of fire on the minigun to make it managable.

It still isnt.

While the did indeed succeed in making it, as the article notes, its completely impractical. You cant carry sufficent ammo, batteries are still heavy, and the recoil, even at reduced RoF is still brutal.

There are still some man portable mass saturation heavy weapon projects toying with the idea, like the derivitves of the Metalstorm system out of Australia, but none of them are field functional.

Viper2005_
05-15-2007, 08:00 PM
Recoil is rate of change of momentum.

R = M*V

Going entirely from wikipedia numbers, the XM-214 fires a 5.56 round at a muzzle velocity of 3250 fps. That's 990.6 m/s.

Your standard NATO 5.56 round weighs in at 62 grains. That's 0.00401753242 kg.

MV is therefore 3.97976762 kg*m/s per round.

That implies an average recoil force of 3.98 N per round per second or per 60 rpm if you prefer. So 1000 rpm would give us 66 N or so, which is actually only about 6.8 kgf even if you round up. Not too terrifying.

And the recoil force scales directly with ROF, not geometrically. So if you were feeling crazy and you ran the poor thing flat out at 6000 rpm you'd produce a recoil force of 398 N or so, which is "only" about 40.6 kgf. Not exactly beyond the capabilities of a reasonably fit soldier.

Of course, if you're making a movie then you only need to fire blanks, so the recoil force will be rather smaller, which means that you can use a higher ROF for any given recoil force and therefore produce a more impressive muzzle flash.

Of course IRL you'd want to try to control the flash since as soon as the enemy work out where that hail of lead is coming from you can bet that they'll return the favour!

As has already been mentioned, the trouble with this sort of weapon is that the power and ammunition required to keep it operational for any length of time is considerable. For a lower logistical effort you could instead run 6 conventional gas operated guns at 1000 rpm for the same length of time, and go after multiple targets.

High ROF is really about 3 things

1) Making sure that the enemy can't sneak through your line of fire between rounds; at even 600 rpm there is only a 0.1s window between rounds so the target will have to either be pretty small or pretty fast to get through. Imagine shooting at a man from the side. Let's have him run at the predicted limiting olympic sprint speed - 100 m in 9.48 s, or 10.54 m/s

http://condellpark.com/kd/sprintlogistic.htm

Now we can look at some dimensions:

http://www.geocities.com/sketchup_lessons/human_dimensions.jpg

From this I shall assume a target width of 1'1" in profile (diagram one up from the bottom on the far right).

That's 0.3302 m.

Therefore it would take our sprinter 0.0313282732 s to pass through the line of fire

Now it follows that to always hit such a sprinter, we would need a cyclic rate of fire of 1915 rpm. There is simply no point in adopting a higher ROF unless you're trying to kill a significantly smaller or faster moving target, or assume that a single round is unlikely to do the job.

Since people like round numbers, it follows that a cyclic rate of 2000 rpm is the highest rate one might expect to require for tackling human targets.

With such a cyclic rate, you can be utterly certain that your stream of fire is an effective barrier to human passage. This means that if you can set up a pair of guns like this:

..EEEEE..
.........
.........
.........
G.......G

You may now walk the fire of those guns inwards, safe in the knowledge that the enemy cannot escape from the crossfire, other than perhaps by digging very quickly... http://forums.ubi.com/images/smilies/784.gif

2) Making lots of noise to keep the enemy's heads down. Pretty simple really - if you're shooting and he's forced to keep his head down then he's not shooting back at you! And of course your friends can use the fact that he's tied up to move into a more useful position.

3) Clearing out a large volume very quickly. With a sufficiently high cyclic rate, as has already been explained, the enemy cannot sneak through your line of fire unscathed. The other side of that coin is that if you sweep your line of fire through an arc then with a sufficiently high cyclic rate the enemy again cannot escape.

Consider enemy soldiers arranged in a perfect circular arc around our hero. Let's have a worst case scenario and arrange lots of them at long range. For the sake of argument let's assume a rather optimistic maximum effective range of 800 m.

Now from the human dimensions already referenced, our enemy soldiers are likely to be 1'8" or so from shoulder to shoulder.

That's 0.508 m.

A circle of radius 800 m has a circumference of 1600*pi m, or 5026.55 m. We may therefore arrange about 9895 men along it.

To put them out of action, assuming a single round is up to the task at that range, we therefore need 9895 rounds. Call it an even 10,000 for the sake of argument.

With our cyclic rate of 2000 rpm we could do the job in about 5 seconds flat!

But that isn't the key point.

The key point is that nobody in that circle could escape taking at least one round assuming that we limit our rate of turn to 72º/s, unless they are right at the edge of that range and turn sideways on or otherwise reduce their cross-section.

Since a circle of radius 800 m has an area of 2010619.3 m^2, it follows that if we assume that our target is facing us head-on, we can clear an area of 402123.86m^2/s at a cyclic rate of 2000 rpm.

In the general case therefore, the rate of area clearance ~ 200 m^2/s/rpm

Clearly in this instance cyclic rate pays. But I suspect that turning round at a steady rate of more than 72º/s would likely be something of a challenge, especially when carrying a heavy weapon and other kit.

As such we may again reasonably suppose that there is little to be gained from achieving cyclic rates in excess of 2000 rpm with a personal firearm, especially when one considers the accuracy required in elevation to consistently hit targets at 800 m whilst turning.

***

Given that you "only" need about 2000 rpm from this sort of weapon, there isn't a very strong case for lugging 6 barrels around with you; the job could adequately be done with a pair, saving a good deal of weight and expense.

Post exam insomnia is annoying...

BRASSTURTLE
05-15-2007, 08:38 PM
WOW.

Dude, put down the blow & step away from the keyboard.

Hope you passed your exams.

Try to sleep.

Sturmtrooper
05-15-2007, 08:57 PM
Posted Tue May 15 2007 19:38 Hide Post
WOW.

Dude, put down the blow & step away from the keyboard.

Hope you passed your exams.

Try to sleep

http://media.ubi.com/us/forum_images/gf-glomp.gif

http://forums.ubi.com/images/smilies/88.gif

DooDaH2007
05-15-2007, 09:07 PM
"
The General Electric Minigun used by Blain in the film has become a running gag in the books of Robert Rankin; after the words: "Fires up to 6000 rounds per minute. 7.62 x 51 mm shells. 1.36 kg recoil adapters. Six muzzle velocity of 869 m/s." were uttered by Elvis in They Came And Ate Us. (In reality, the recoil from such a gun is so powerful that no one would be able to stay upright and shoot it. It must be vehicle-mounted, and if carried by a person it would run out of ammunition in seconds. Also, being electrically-driven, the user would have to carry a massive battery-pack in addition to the gun and ammunition, making an impossibly heavy load. The gun used in the movie fired blanks, which produce minimal recoil, and was powered by a remote generator, with the cord concealed beneath Blain's pantleg.)
"

http://en.wikipedia.org/wiki/Predator_1

-

ake109
05-15-2007, 09:18 PM
As an analogy, they have tried mounting the damn thing on a Humvee and the associated weight and space of the mount, weapon and ammo was considered restrictive..

So if its considered restrictive for a freaking Humvee, it would be rather crazy for a human to try it.

As other posters pointed out, the movie version was firing blanks and had a generator off screen to power it.

Korolov1986
05-15-2007, 10:42 PM
M240B or M2HB is all you need for ground support.

Save M134 for the helicopters.

Capt.LoneRanger
05-16-2007, 12:37 AM
Nice formulas. You're related to raaaid in some way?

As he usually does, you forgot some very basic things, though.

1. The minugun has a good portion of it's own mass rotating around a small axis at high speed. As with any gyroscope, any force applied, will give it a momentum in another direction. So, if the gun is fired by hand, even at lower ROF, the recoil won't just push it back or up, but depending on length of the salvo, lead to an uncontrollable drift, which is higher than the recoil itself.
2. You're equation misses the fact, that not all energy from the charge is transformed into velocity of the bullet.
3. You made a little miscalculation http://forums.ubi.com/images/smilies/16x16_smiley-wink.gif
3.98kg*m/s is not 3.98N/s, but 39.8N/s.

If recoil per round and second is roughly 4kg x m/s x round, you have a recoil of 4.4kg*m/s with 66RPM. The Gun fired with 166 RPM in the movie, which would give you 11kg*m/s of recoil. The SpecialForces test version had a setting of 300prm, which would give a recoil of 20kg*m/s.
6000 RPM means 100 times 4kg*m/s 400kg of recoil.

Additionally the gun alone weights 12,3 kg and the complete setup with "only" 3000 rounds is about 39kg.


Besides that, the pure ammount of energy delivered to the target is not only useull to keep the enemy down, but also to shredder the targets cover. The gun was not adopted by the special forces, because it was unreliable in field operation and the high ammount of bullets by far increased the chance of ricochets going wild.

Panzerknacker02
05-16-2007, 01:57 AM
It'd be a damn sweet hedge trimmer though http://forums.ubi.com/images/smilies/784.gif If Wehrmacht machine gunners found it hard to keep up with the ammunition demands of their MG42s the minigun would be insane...

Viper2005_
05-16-2007, 06:26 AM
Originally posted by Capt.LoneRanger:
Nice formulas. You're related to raaaid in some way? Nope.


Originally posted by Capt.LoneRanger:As he usually does, you forgot some very basic things, though.

1. The minugun has a good portion of it's own mass rotating around a small axis at high speed. As with any gyroscope, any force applied, will give it a momentum in another direction. So, if the gun is fired by hand, even at lower ROF, the recoil won't just push it back or up, but depending on length of the salvo, lead to an uncontrollable drift, which is higher than the recoil itself.
[quote]

Gyroscopic effects are rather beyond the scope of this discussion. However, if the weapon has no tendency to pitch up then there will be no precession and therefore no problem. If OTOH the weapon is pitching up when it is fired then it's inaccurate anyway. Exactly in what direction your rounds deviate from the aiming point is somewhat academic...

Of course there will be a gyroscopic effect if the gun is turned whilst it is being fired, but if you turn slowly then this effect will be small.

[QUOTE]Originally posted by Capt.LoneRanger:
2. You're equation misses the fact, that not all energy from the charge is transformed into velocity of the bullet.


No. I just looked up the muzzle velocity from the wikipedia page on the weapon, and the mass of the bullet from the wikipedia page on the NATO 5.56 round. I then calculated the kinetic energy of the bullet from e = 0.5*m*v^2.

http://en.wikipedia.org/wiki/XM214_Microgun
http://en.wikipedia.org/wiki/5.56_x_45_mm_NATO

I make no claim as to the accuracy of this data. http://forums.ubi.com/groupee_common/emoticons/icon_wink.gif

I actually did this as a first step towards calculating the recoil associated with a blank-firing version of the weapon for film use, but since the recoil of the live firing version isn't all that terrifying at 2000 rpm I didn't bother...

(The method would be to assume that the propellant gas has the same muzzle velocity as the round, therefore calculate its kinetic energy and add this to that of the round. Then consider the blank firing version and simply assume that this kinetic energy is given to the propellant mass. Hence calculate velocity, and hence calculate momentum; the result will be smaller than for the live firing version, which explains a lot about films!)


Originally posted by Capt.LoneRanger:
3. You made a little miscalculation http://forums.ubi.com/images/smilies/16x16_smiley-wink.gif
3.98kg*m/s is not 3.98N/s, but 39.8N/s.

If recoil per round and second is roughly 4kg x m/s x round, you have a recoil of 4.4kg*m/s with 66RPM. The Gun fired with 166 RPM in the movie, which would give you 11kg*m/s of recoil. The SpecialForces test version had a setting of 300prm, which would give a recoil of 20kg*m/s.
6000 RPM means 100 times 4kg*m/s 400kg of recoil.

Additionally the gun alone weights 12,3 kg and the complete setup with "only" 3000 rounds is about 39kg.


Nope. The Newton is the SI unit of force. The kilogram is the SI unit of mass. You have confused them.


Originally posted by Capt.LoneRanger:Besides that, the pure ammount of energy delivered to the target is not only useull to keep the enemy down, but also to shredder the targets cover. The gun was not adopted by the special forces, because it was unreliable in field operation and the high ammount of bullets by far increased the chance of ricochets going wild. Well, shredding cover is again beyond the scope of this rather simplistic analysis. However, it's quite easy to see that rates of fire significantly greater than 2000 rpm aren't going to be very useful since the the second round will be so close behind the first round that it will hit the debris shed when the first round hits whatever cover the enemy is hiding behind, and therefore accuracy will deteriorate significantly.

This doesn't matter at short range, but then at short range you're going to saturate the target anyway...

BSS_Goat
05-16-2007, 07:59 AM
VIPER FTW!!!


http://forums.ubi.com/images/smilies/11.gif

K_Freddie
05-16-2007, 11:57 AM
Well if you want muzzle velocity, rate of fire, and gyroscopic gyration effects, maybe you should try these Predators.. http://forums.ubi.com/images/smilies/heart.gif

Hey how do you make the pic smaller ??
http://i8.photobucket.com/albums/a6/talonarms/TalonsGirls914.jpg

T_O_A_D
05-16-2007, 02:44 PM
Trying to figure it out again.
K_Freddie

http://www.practicalecommerce.com/articles/135/Encoding-Images-Using-HTML/

I_KG100_Prien
05-16-2007, 03:23 PM
Heh, just by itself a 5.56mm round out of an M-16 doesn't have much in the way of recoil.. But slap that '16 into auto and lay on the trigger... Even it can be hard to keep the barrel down. Screw trying to control something firing at 3,000rpm.

Ditto on the M240. I love the 240. Though the Ma Deuce still has a special place in my heart. Especially when there are two side by side in a twin mount.. WHEEEEEEEEEEEEEEEEEEEE.

Siwarrior
05-16-2007, 09:30 PM
whoa this thread is pretty interesting
say they had this sorta tech back in the '40s, i wonder if you be feasable to mount this onto a fighter or maybe bomber?

also if the mg42 had a high ROF 1200+? why wasnt it implemented into 109s or 190s, unless it is an d im mistaken http://forums.ubi.com/groupee_common/emoticons/icon_smile.gif

LEBillfish
05-16-2007, 11:21 PM
You all are forgetting....It's recoil, ammo, batteries might be too much for the average person to bear....Yet it was Jesse Ventura the mad dog wrestling governor from Minnisota!! http://forums.ubi.com/images/smilies/16x16_smiley-surprised.gif

and yes, that's right all you non Americans...He's a skinney midget compared to most here, and all American women must be "Playboy" quality or we're are sent elsewhere.

Glad to be an American http://forums.ubi.com/images/smilies/16x16_smiley-wink.gif



http://forums.ubi.com/images/smilies/16x16_smiley-very-happy.gif

Freelancer-1
05-16-2007, 11:45 PM
I feel one of those rare LB photo postings coming on http://forums.ubi.com/images/smilies/16x16_smiley-wink.gif

leitmotiv
05-16-2007, 11:58 PM
Billfish is right. We deport all our non-Playboy quality women and men who are whimpier than Jesse, so watch out. When we are not mass producing uber-wenches and uber-studs, we are committing to memory entire volumes of physics while bench pressing Abrams tanks. Argh.

Vidar_1
05-17-2007, 12:37 AM
Originally posted by Viper2005_:
Recoil is rate of change of momentum.

R = M*V

Going entirely from wikipedia numbers, the XM-214 fires a 5.56 round at a muzzle velocity of 3250 fps. That's 990.6 m/s.

Your standard NATO 5.56 round weighs in at 62 grains. That's 0.00401753242 kg.

MV is therefore 3.97976762 kg*m/s per round.

That implies an average recoil force of 3.98 N per round per second or per 60 rpm if you prefer. So 1000 rpm would give us 66 N or so, which is actually only about 6.8 kgf even if you round up. Not too terrifying.


Actually, you do not get the recoil force by multiplying mass by velocity. You can see that by looking at the sort: Force in Newton is kg*m/s**2 not kg*m/s

In your equation R = m*v you are missing a "delta time" on the left side.

Should be R* deltaT= m *deltaV (this is the impulse equation useful for missile calculations etc)

Solving R= m* deltaV/deltaT

This equation gives the correct sort for force in Newton: kg*m/s**2

CD_kp84yb
05-17-2007, 12:41 AM
HI,

I only have some figures for the GAU-2 B/A 7.62 (minigun)
Caliber: 7.62mm
Lenght: 84.83 Cm
Weight: Gun: 15.86Kg
electric Engine : 3.4Kg
Recoil absorber: 1.36 Kg.

Now this is without the ammo or batteries (the setup of this gun was in gunships)
Rate of fire max 6000 rpm
max sustained fire 2000 rounds at rof 6000rpm
Shots until misfire 25000
Lifespan of the gun 100.000 shots
overhaul some parts after 15.000 shots

Here it comes :
normal recoil at 6000 rpm 135.9 Kg
max recoil at rof 6000 is 271.8 Kg
measured not calculated.

recoil travel at rpm 6000 is 3.2mm max
forward travel at sam rof is 1.6 mm max

Engine for rof 40000 28V DC 70Ah
engine for rof 6000 22V DC 130Ah

this comes from a data book from the army.
Attention this is not the 5.56mm

Cheers

Vidar_1
05-17-2007, 02:04 AM
The mingun is another product of the American obsession with firepower: If force is not solving your problem you are not using enough. A M16 won't do the job? Go get the minigun! GET SOME!!!!! YEAH!!!!!

Allt this bravado and overkill mentality: So totally different from the stylish European way of solving things http://forums.ubi.com/images/smilies/16x16_smiley-wink.gif

Read the book Unknown Soldier by Vin Linna and see the movie: Quit men just doing the job on skis with a mauser rifle slung over their shoulder. Rounding up the enemy in "motti". No inane Jarhead bravado with miniguns. That's waging war with style and brains!

"hakkaa plle pohjan poika"
("hugg in nordens sner")!

The-Pizza-Man
05-17-2007, 03:06 AM
Originally posted by Vidar_1:
<BLOCKQUOTE class="ip-ubbcode-quote"><div class="ip-ubbcode-quote-title">quote:</div><div class="ip-ubbcode-quote-content">Originally posted by Viper2005_:
Recoil is rate of change of momentum.

R = M*V

Going entirely from wikipedia numbers, the XM-214 fires a 5.56 round at a muzzle velocity of 3250 fps. That's 990.6 m/s.

Your standard NATO 5.56 round weighs in at 62 grains. That's 0.00401753242 kg.

MV is therefore 3.97976762 kg*m/s per round.

That implies an average recoil force of 3.98 N per round per second or per 60 rpm if you prefer. So 1000 rpm would give us 66 N or so, which is actually only about 6.8 kgf even if you round up. Not too terrifying.


Actually, you do not get the recoil force by multiplying mass by velocity. You can see that by looking at the sort: Force in Newton is kg*m/s**2 not kg*m/s

In your equation R = m*v you are missing a "delta time" on the left side.

Should be R* deltaT= m *deltaV (this is the impulse equation useful for missile calculations etc)

Solving R= m* deltaV/deltaT

This equation gives the correct sort for force in Newton: kg*m/s**2 </div></BLOCKQUOTE>

If you care to check vipers calculations you'll see he's done it correctly.

Vidar_1
05-17-2007, 04:44 AM
Originally posted by The-Pizza-Man:
<BLOCKQUOTE class="ip-ubbcode-quote"><div class="ip-ubbcode-quote-title">quote:</div><div class="ip-ubbcode-quote-content">Originally posted by Vidar_1:
<BLOCKQUOTE class="ip-ubbcode-quote"><div class="ip-ubbcode-quote-title">quote:</div><div class="ip-ubbcode-quote-content">Originally posted by Viper2005_:
Recoil is rate of change of momentum.

R = M*V

Going entirely from wikipedia numbers, the XM-214 fires a 5.56 round at a muzzle velocity of 3250 fps. That's 990.6 m/s.

Your standard NATO 5.56 round weighs in at 62 grains. That's 0.00401753242 kg.

MV is therefore 3.97976762 kg*m/s per round.

That implies an average recoil force of 3.98 N per round per second or per 60 rpm if you prefer. So 1000 rpm would give us 66 N or so, which is actually only about 6.8 kgf even if you round up. Not too terrifying.


Actually, you do not get the recoil force by multiplying mass by velocity. You can see that by looking at the sort: Force in Newton is kg*m/s**2 not kg*m/s

In your equation R = m*v you are missing a "delta time" on the left side.

Should be R* deltaT= m *deltaV (this is the impulse equation useful for missile calculations etc)

Solving R= m* deltaV/deltaT

This equation gives the correct sort for force in Newton: kg*m/s**2 </div></BLOCKQUOTE>

If you care to check vipers calculations you'll see he's done it correctly. </div></BLOCKQUOTE>

OK, If that's so, can you be kind enough to point out where I got it wrong Pizza-man? http://forums.ubi.com/groupee_common/emoticons/icon_smile.gif

Vidar_1
05-17-2007, 05:39 AM
Actually, I would claim that the recoil is not coupled to the muzzle velocity at all but to the speed of the gas mass flow leaving the barrel AFTER the bullet has LEFT the barrel. This also explains the difference in the recoil force calculted by Viper and the actual recoil posted by CD_kp84yb above (even if the difference in calibre is considered)

Consider a rifle firing a standard round: Anyone with any military experience will tell you that a short carbine firing that same round kicks worse than a "long" like an FN. Why? It's not primarily because of the weight difference, it's because the recoil comes from the gases leaving the muzzle acting like a rocket if you like. Before the bullet leaves the barrel, it's all internal forces, bullet accelerating down the barrel? Yes, so does this this start a recoil? No: the same bullet is causing an opposed force acting on the rifling in the barrel pushing the rifle forward. If you want to get a grip on the recoil, you need to calculate the mass flow and speed of the jet leaving the barrel. That is the source of the recoil. This is why cannon like the Mk108 have low muzzle velocity but a comparatively high recoil. The same round fired in a longer barrel Mk108 would have HIGHER muzzle velocity but LOWER recoil.

So recoil is not directly coupled to the muzzle velocity but rather to the rocket effect of the cartridge gases.

Viper2005_
05-17-2007, 08:11 AM
Originally posted by Vidar_1:
<BLOCKQUOTE class="ip-ubbcode-quote"><div class="ip-ubbcode-quote-title">quote:</div><div class="ip-ubbcode-quote-content">Originally posted by Viper2005_:
Recoil is rate of change of momentum.

R = M*V

Going entirely from wikipedia numbers, the XM-214 fires a 5.56 round at a muzzle velocity of 3250 fps. That's 990.6 m/s.

Your standard NATO 5.56 round weighs in at 62 grains. That's 0.00401753242 kg.

MV is therefore 3.97976762 kg*m/s per round.

That implies an average recoil force of 3.98 N per round per second or per 60 rpm if you prefer. So 1000 rpm would give us 66 N or so, which is actually only about 6.8 kgf even if you round up. Not too terrifying.


Actually, you do not get the recoil force by multiplying mass by velocity. You can see that by looking at the sort: Force in Newton is kg*m/s**2 not kg*m/s

In your equation R = m*v you are missing a "delta time" on the left side.

Should be R* deltaT= m *deltaV (this is the impulse equation useful for missile calculations etc)

Solving R= m* deltaV/deltaT

This equation gives the correct sort for force in Newton: kg*m/s**2 </div></BLOCKQUOTE>

I just set delta t to 1 second because I'm lazy. As a result the answer should be correct.

Viper2005_
05-17-2007, 08:45 AM
Originally posted by Vidar_1:
So recoil is not directly coupled to the muzzle velocity but rather to the rocket effect of the cartridge gases.

http://www.youtube.com/watch?v=LIBcQaZzr4w

Note that the barrel starts to recoil before the round and propellant gasses have left the muzzle. The propellant can of course contribute to recoil, but the momentum of the round is the primary source in most cases.

Of course, a well designed muzzle brake can make use of the momentum of the propellant gas to reduce the overall recoil.

Philipscdrw
05-17-2007, 09:10 AM
Originally posted by LEBillfish:
You all are forgetting....It's recoil, ammo, batteries might be too much for the average person to bear....Yet it was Jesse Ventura the mad dog wrestling governor from Minnisota!! http://forums.ubi.com/images/smilies/16x16_smiley-surprised.gif

and yes, that's right all you non Americans...He's a skinney midget compared to most here, and all American women must be "Playboy" quality or we're are sent elsewhere.

Glad to be an American http://forums.ubi.com/images/smilies/16x16_smiley-wink.gif



http://forums.ubi.com/images/smilies/16x16_smiley-very-happy.gif

Schwarzenegger is skinny compared to most Americans? OK, I can understand that... although Britain is moving in that direction too unfortunately. http://forums.ubi.com/images/smilies/16x16_smiley-very-happy.gif

Vidar_1
05-17-2007, 10:55 AM
Originally posted by Viper2005_:
<BLOCKQUOTE class="ip-ubbcode-quote"><div class="ip-ubbcode-quote-title">quote:</div><div class="ip-ubbcode-quote-content">Originally posted by Vidar_1:
So recoil is not directly coupled to the muzzle velocity but rather to the rocket effect of the cartridge gases.

http://www.youtube.com/watch?v=LIBcQaZzr4w

Note that the barrel starts to recoil before the round and propellant gasses have left the muzzle. The propellant can of course contribute to recoil, but the momentum of the round is the primary source in most cases.

Of course, a well designed muzzle brake can make use of the momentum of the propellant gas to reduce the overall recoil. </div></BLOCKQUOTE>

I don't agree. Show me a rifle or gun video or article that supports your theory that recoil is more due to the momentum of the bullet than gases.

If the camera would stay to show the barrel after the missile has left in the video you linked, my guess is you will see a more significant recoil AFTER the missile has left the barrel.

How do you explain that a carbine firing the SAME round in a shorter barrel with LOWER muzzle velocity has a higher recoil than a long rifle where the recoil is lower but the muzzle velocity is HIGHER. It completely contradicts your theory.

Dagnabit
05-17-2007, 11:03 AM
Originally posted by I_KG100_Prien:
Heh, just by itself a 5.56mm round out of an M-16 doesn't have much in the way of recoil.. But slap that '16 into auto and lay on the trigger... Even it can be hard to keep the barrel down. Screw trying to control something firing at 3,000rpm.

Ditto on the M240. I love the 240. Though the Ma Deuce still has a special place in my heart. Especially when there are two side by side in a twin mount.. WHEEEEEEEEEEEEEEEEEEEE.

.223/5.56mm on full auto is a very easy to fire weapon, very controlable. My wife and daughter have fired M16A2, as well as the XM177 (my brother has a class3 permit) And they also like my AR15s. But they were surprised at how easy they were to control, and how accurate they are.
Also to even consider toting around a minigun in .308/7.62X51mm would not be practical for a number of reasons, the recoil being the least of them. The weight of the unit itself would offset much of the recoil, but vibration, the complexities of the piece, and the ammo consumption problem would make it a no go in any case.
I wasent aware that they had chambered the minigun in .223/5.56mm. http://forums.ubi.com/images/smilies/16x16_smiley-surprised.gif
Regards
Dag

Vidar_1
05-17-2007, 11:10 AM
I got another video for ya. Where would the bullet go if the recoil this lass is getting in her hand occurs BEFORE the bullet leaves the gun?

http://www.youtube.com/watch?v=gLv4A5UO4Ow

Freelancer-1
05-17-2007, 11:21 AM
Here's some great recoil examples and some rippin' fast guns:

http://www.youtube.com/watch?v=FVms5xxwpEE&mode=related&search=

Check out the Glock 18 http://forums.ubi.com/groupee_common/emoticons/icon_eek.gif

Dagnabit
05-17-2007, 11:45 AM
Check em out recoil???

http://www.liveleak.com/view?i=ba4_1172300448

http://www.youtube.com/watch?v=ILUXYVUUZrY

Vidar, actually the way that girl is handling that pistol, there is no way to say for sure where the bullet will go. She appears to be a first timer, and has no idea at all how to handle a firearm safely. Her Dad should have fully explained exactly how that gun operates, and had her dry fire (on snap caps) before she ever touched a gun with live ammo. She made a number of safety errors and was lucky if she wasent injured. You just NEVER turn someone loose with a loaded gun without KNOWING for sure that they can handle the situation.
Regards
Dag

HayateAce
05-17-2007, 12:03 PM
Yikes, except for the blondie those girlz are a bit homely. I take it Talon is working with a third-world type budget.

LStarosta
05-17-2007, 12:07 PM
Originally posted by Vidar_1:
The mingun is another product of the American obsession with firepower: If force is not solving your problem you are not using enough. A M16 won't do the job? Go get the minigun! GET SOME!!!!! YEAH!!!!!

Allt this bravado and overkill mentality: So totally different from the stylish European way of solving things http://forums.ubi.com/images/smilies/16x16_smiley-wink.gif


Yeah, those Zyclon-B showers and supersize ovens sure blend in well with the decor. You go, girlfriend!

JG7_Rall
05-17-2007, 12:17 PM
Originally posted by LStarosta:
<BLOCKQUOTE class="ip-ubbcode-quote"><div class="ip-ubbcode-quote-title">quote:</div><div class="ip-ubbcode-quote-content">Originally posted by Vidar_1:
The mingun is another product of the American obsession with firepower: If force is not solving your problem you are not using enough. A M16 won't do the job? Go get the minigun! GET SOME!!!!! YEAH!!!!!

Allt this bravado and overkill mentality: So totally different from the stylish European way of solving things http://forums.ubi.com/images/smilies/16x16_smiley-wink.gif


Yeah, those Zyclon-B showers and supersize ovens sure blend in well with the decor. You go, girlfriend! </div></BLOCKQUOTE>

OWNED.

Freelancer-1
05-17-2007, 12:28 PM
Originally posted by LStarosta:
<BLOCKQUOTE class="ip-ubbcode-quote"><div class="ip-ubbcode-quote-title">quote:</div><div class="ip-ubbcode-quote-content">Originally posted by Vidar_1:
The mingun is another product of the American obsession with firepower: If force is not solving your problem you are not using enough. A M16 won't do the job? Go get the minigun! GET SOME!!!!! YEAH!!!!!

Allt this bravado and overkill mentality: So totally different from the stylish European way of solving things http://forums.ubi.com/images/smilies/16x16_smiley-wink.gif


Yeah, those Zyclon-B showers and supersize ovens sure blend in well with the decor. You go, girlfriend! </div></BLOCKQUOTE>

Nothing like the righteous indignation of the American gun lover http://forums.ubi.com/images/smilies/16x16_smiley-mad.gif

M_Gunz
05-17-2007, 01:17 PM
Originally posted by Viper2005_:
I then calculated the kinetic energy of the bullet from e = 0.5*m*v^2.

I thought you were trying to find the recoil, not the kinetic energy.

Recoil is momentum, mass x velocity only. Funny how many mix and match potential with kinetic
when they are measured so differently. Same mass, same velocity, the two differ often widely.

EDIT: I've been out of school for a long time now even by my measure but IIRC MOMENTUM IS
CONSERVED any time masses seperate or collide. I go and look at the science experiments on
the web (how homework assignments get passed out it seems) concerning momentum and kinetic
and they still think the same is true, momentum is conserved. Mass x Velocity of bullet plus
Mass x Velocity of gasses is balanced by Mass x Velocity of the weapon and whatever holds it.

And btw, kick is not momentum just as penetration is not knock-down. Go read up on serious
rifle sites. When science students can't tell PE from KE then we might have trouble of it.

Vidar_1
05-17-2007, 01:21 PM
Originally posted by JG7_Rall:
<BLOCKQUOTE class="ip-ubbcode-quote"><div class="ip-ubbcode-quote-title">quote:</div><div class="ip-ubbcode-quote-content">Originally posted by LStarosta:
<BLOCKQUOTE class="ip-ubbcode-quote"><div class="ip-ubbcode-quote-title">quote:</div><div class="ip-ubbcode-quote-content">Originally posted by Vidar_1:
The mingun is another product of the American obsession with firepower: If force is not solving your problem you are not using enough. A M16 won't do the job? Go get the minigun! GET SOME!!!!! YEAH!!!!!

Allt this bravado and overkill mentality: So totally different from the stylish European way of solving things http://forums.ubi.com/images/smilies/16x16_smiley-wink.gif


Yeah, those Zyclon-B showers and supersize ovens sure blend in well with the decor. You go, girlfriend! </div></BLOCKQUOTE>

OWNED. </div></BLOCKQUOTE>

STRIKE! And it's a BIG one! Bzzzzzzzzzzzzzzzzzzz..... look at that reel smoke!!

M_Gunz
05-17-2007, 01:46 PM
Originally posted by Freelancer-1:
Nothing like the righteous indignation of the American gun lover http://forums.ubi.com/images/smilies/16x16_smiley-mad.gif

Until the opposite that may be non-American or a gun hater comes out and parades around.
You can tell easy enough how much hate those types have and who they blame for the world.
Someone who goes out and shoots target or hunts, yeah those are the ones that drive the
greed that shapes the conflicts. Right. It is not about people putting pressure for
always more that makes the politics, crime and war but ONLY the ones with the guns. And
then you have washed your hands and are innocent. But once you've "had it", you will call
for others to carry guns for your own "righteous indignation", I am sure.

Get your bashing in, it only makes your personal history tighter for the moment, your
decision of who should suffer more sure. You know who the enemy is, don't you? Do you
think the people who shoot paper also know to hate you as well?

Vidar_1
05-17-2007, 01:53 PM
Seriously, I think the US is great: East and west coast liberals, Barak Obama, Hilary Clinton, Michael Moore, Paul Auster, Kurt Vonnegut. I could go on and on...

What I do not like though, are ignorant rednecks spreading democracy through the barrel of a gun.

Zyclon-B, ovens and showers? Shame on you.....

M_Gunz
05-17-2007, 01:54 PM
Originally posted by Vidar_1:
I got another video for ya. Where would the bullet go if the recoil this lass is getting in her hand occurs BEFORE the bullet leaves the gun?

http://www.youtube.com/watch?v=gLv4A5UO4Ow

Recoil could be said to begin when the bullet begins moving. Center of mass of the weapon is
changing after all.

Why don't you find how long the bullet spends traversing the barrel and stop talking about being
able to see such effects as bullet in the barrel on aim? That is lunacy.

How many grains/milligrams of powder converted to gasses (conservation of mass, mass of powder
is equal to mass of gasses and burned solids though some stays inside the barrel) at what
pressure (Western and Remington may have figures) release makes how much recoil from gas?
Fire a blank after firing a bullet some time and notice the difference... even blanks do
fire plug material that amounts to more than the gasses alone!

Vidar_1
05-17-2007, 02:04 PM
Originally posted by M_Gunz:
<BLOCKQUOTE class="ip-ubbcode-quote"><div class="ip-ubbcode-quote-title">quote:</div><div class="ip-ubbcode-quote-content">Originally posted by Vidar_1:
I got another video for ya. Where would the bullet go if the recoil this lass is getting in her hand occurs BEFORE the bullet leaves the gun?

http://www.youtube.com/watch?v=gLv4A5UO4Ow

Recoil could be said to begin when the bullet begins moving. Center of mass of the weapon is
changing after all.

Why don't you find how long the bullet spends traversing the barrel and stop talking about being
able to see such effects as bullet in the barrel on aim? That is lunacy.

How many grains/milligrams of powder converted to gasses (conservation of mass, mass of powder
is equal to mass of gasses and burned solids though some stays inside the barrel) at what
pressure (Western and Remington may have figures) release makes how much recoil from gas?
Fire a blank after firing a bullet some time and notice the difference... even blanks do
fire plug material that amounts to more than the gasses alone! </div></BLOCKQUOTE>

Have you ever fired a pistol? I've shot quite a lot of 9 mm parabellum. Strangely I do hit the target roughly where I'm aiming. The bullet does not go 45 degrees in a ballistic trajectory. My point has been: the MAJOR and DOMINANT force that you experience as recoil is due to the rocket motor effect of the cartridge gases not the bullet momemtum as it travels down the barrel. Do you think you would hit jacksh!t if this was not so?

Viper2005_
05-17-2007, 02:09 PM
Originally posted by M_Gunz:
<BLOCKQUOTE class="ip-ubbcode-quote"><div class="ip-ubbcode-quote-title">quote:</div><div class="ip-ubbcode-quote-content">Originally posted by Viper2005_:
I then calculated the kinetic energy of the bullet from e = 0.5*m*v^2.

I thought you were trying to find the recoil, not the kinetic energy.

Recoil is momentum, mass x velocity only. Funny how many mix and match potential with kinetic
when they are measured so differently. Same mass, same velocity, the two differ often widely.

EDIT: I've been out of school for a long time now even by my measure but IIRC MOMENTUM IS
CONSERVED any time masses seperate or collide. I go and look at the science experiments on
the web (how homework assignments get passed out it seems) concerning momentum and kinetic
and they still think the same is true, momentum is conserved. Mass x Velocity of bullet plus
Mass x Velocity of gasses is balanced by Mass x Velocity of the weapon and whatever holds it.

And btw, kick is not momentum just as penetration is not knock-down. Go read up on serious
rifle sites. When science students can't tell PE from KE then we might have trouble of it. </div></BLOCKQUOTE>

Obviously you didn't take the time to read my post. Had you done so you would have seen that the only reason for calculating KE was as a first step in calculating the recoil associated with blank firing, the methodology for which I explained. This calculation was unnecessary since the mean recoil thrust associated with live firing was quite reasonable, even at high cyclic rates.

Please take the time to read what I write before lecturing me. You may find that I'm not quite as stupid as you think. http://forums.ubi.com/images/smilies/16x16_smiley-wink.gif

Vidar_1
05-17-2007, 02:15 PM
M_Gunz: Trick Question: Does this Desert Eagle shooter experience most recoil before or after the bullet left the barrel?

http://www.youtube.com/watch?v=__JkOUheyVk

Vortex_79
05-17-2007, 02:16 PM
Originally posted by Vidar_1:
The mingun is another product of the American obsession with firepower: If force is not solving your problem you are not using enough. A M16 won't do the job? Go get the minigun! GET SOME!!!!! YEAH!!!!!

Allt this bravado and overkill mentality: So totally different from the stylish European way of solving things http://forums.ubi.com/images/smilies/16x16_smiley-wink.gif

Well, I'm European and this is me using the M134 last week... Link (http://smg.photobucket.com/albums/v331/_79_Vortex/?action=view&current=MOV00002.flv)

Have I used it for real? Nope, not yet but I may be soon though. My colleagues have though and they can't rate it enough...

_79_Vortex http://img.photobucket.com/albums/v331/_79_Vortex/sc-rambo.gif

Freelancer-1
05-17-2007, 02:19 PM
Originally posted by M_Gunz:
<BLOCKQUOTE class="ip-ubbcode-quote"><div class="ip-ubbcode-quote-title">quote:</div><div class="ip-ubbcode-quote-content">Originally posted by Freelancer-1:
Nothing like the righteous indignation of the American gun lover http://forums.ubi.com/images/smilies/16x16_smiley-mad.gif

Until the opposite that may be non-American or a gun hater comes out and parades around.
You can tell easy enough how much hate those types have and who they blame for the world.
Someone who goes out and shoots target or hunts, yeah those are the ones that drive the
greed that shapes the conflicts. Right. It is not about people putting pressure for
always more that makes the politics, crime and war but ONLY the ones with the guns. And
then you have washed your hands and are innocent. But once you've "had it", you will call
for others to carry guns for your own "righteous indignation", I am sure.

Get your bashing in, it only makes your personal history tighter for the moment, your
decision of who should suffer more sure. You know who the enemy is, don't you? Do you
think the people who shoot paper also know to hate you as well? </div></BLOCKQUOTE>

Easy Gunz. http://forums.ubi.com/images/smilies/touche.gif

I admit to a poor choice of words and for that I apologize.

I don't bash those of other nationality's as a rule.

I do however bash people who trivialize the holocaust like Luke did in that pathetic retort of his. This kind of sad attempt to justify a point of view just chaps my ***. And it serves no purpose but to be hurtful at the expense of belittling an entire important historical event.

Got my hackles up and I typed a quick response without thinking.

Again I apologize to any I offended. http://forums.ubi.com/images/smilies/16x16_smiley-indifferent.gif

Vidar_1
05-17-2007, 02:21 PM
Originally posted by Viper2005_:
<BLOCKQUOTE class="ip-ubbcode-quote"><div class="ip-ubbcode-quote-title">quote:</div><div class="ip-ubbcode-quote-content">Originally posted by M_Gunz:
<BLOCKQUOTE class="ip-ubbcode-quote"><div class="ip-ubbcode-quote-title">quote:</div><div class="ip-ubbcode-quote-content">Originally posted by Viper2005_:
I then calculated the kinetic energy of the bullet from e = 0.5*m*v^2.

I thought you were trying to find the recoil, not the kinetic energy.

Recoil is momentum, mass x velocity only. Funny how many mix and match potential with kinetic
when they are measured so differently. Same mass, same velocity, the two differ often widely.

EDIT: I've been out of school for a long time now even by my measure but IIRC MOMENTUM IS
CONSERVED any time masses seperate or collide. I go and look at the science experiments on
the web (how homework assignments get passed out it seems) concerning momentum and kinetic
and they still think the same is true, momentum is conserved. Mass x Velocity of bullet plus
Mass x Velocity of gasses is balanced by Mass x Velocity of the weapon and whatever holds it.

And btw, kick is not momentum just as penetration is not knock-down. Go read up on serious
rifle sites. When science students can't tell PE from KE then we might have trouble of it. </div></BLOCKQUOTE>

Obviously you didn't take the time to read my post. Had you done so you would have seen that the only reason for calculating KE was as a first step in calculating the recoil associated with blank firing, the methodology for which I explained. This calculation was unnecessary since the mean recoil thrust associated with live firing was quite reasonable, even at high cyclic rates.

Please take the time to read what I write before lecturing me. You may find that I'm not quite as stupid as you think. http://forums.ubi.com/images/smilies/16x16_smiley-wink.gif </div></BLOCKQUOTE>

Yeah, I think I was a bit quit on the trigger when I saw your first post Viper. I got stuck on the sort Kg*m/s part but I saw you integrated the force over 1 sec in the text to get Newton so I'm with you there.

Still think most of the recoil is due to the rocket motor effect though http://forums.ubi.com/images/smilies/16x16_smiley-wink.gif

M_Gunz
05-17-2007, 02:23 PM
Originally posted by Vidar_1:
Consider a rifle firing a standard round: Anyone with any military experience will tell you that a short carbine firing that same round kicks worse than a "long" like an FN. Why?

Try comparing the mass of the carbine vs the mass of that FN. Because the lighter the rifle,
the heavier the kick from the same round. Also add in effect of recoil absorbers in the two weapons when using your shoulder to measure the effect.

Don't feel bad, we have a trained machinegunner here who has stated categorically that .50
cal bullets speed up after leaving the muzzle and are really going faster at 300 meters than
at muzzle -- all from that good military knowledge that MY trainers seemed to have forgotten.
Oh well, when you have 'authority' and don't know something then just make it as you go!


It's not primarily because of the weight difference, it's because the recoil comes from the gases leaving the muzzle acting like a rocket if you like.

Hand load some blanks with only a tiny bit of tissue holding the powder and feel the great
recoil from those. I can tell you, it ain't much.


This is why cannon like the Mk108 have low muzzle velocity but a comparatively high recoil. The same round fired in a longer barrel Mk108 would have HIGHER muzzle velocity but LOWER recoil.

You are saying that the Mk 103 had LESS recoil per shot than the Mk 108? http://forums.ubi.com/images/smilies/16x16_smiley-very-happy.gif

Perhaps you should get with Jokf. He is re-creating whatever science gets in his way too!

Vidar_1
05-17-2007, 02:33 PM
Originally posted by Vortex_79:
<BLOCKQUOTE class="ip-ubbcode-quote"><div class="ip-ubbcode-quote-title">quote:</div><div class="ip-ubbcode-quote-content">Originally posted by Vidar_1:
The mingun is another product of the American obsession with firepower: If force is not solving your problem you are not using enough. A M16 won't do the job? Go get the minigun! GET SOME!!!!! YEAH!!!!!

Allt this bravado and overkill mentality: So totally different from the stylish European way of solving things http://forums.ubi.com/images/smilies/16x16_smiley-wink.gif

Well, I'm European and this is me using the M134 last week... Link (http://smg.photobucket.com/albums/v331/_79_Vortex/?action=view&current=MOV00002.flv)

Have I used it for real? Nope, not yet but I may be soon though. My colleagues have though and they can't rate it enough...

_79_Vortex http://img.photobucket.com/albums/v331/_79_Vortex/sc-rambo.gif </div></BLOCKQUOTE>

Reminds me of a scene in Kubrik's full metal jacket. The grunt shooting up farmers in a field from the chopper. Why? Because he was in a free fire zone!

PaulV2007
05-17-2007, 02:41 PM
Originally posted by Vidar_1:
Have you ever fired a pistol? I've shot quite a lot of 9 mm parabellum. Strangely I do hit the target roughly where I'm aiming. The bullet does not go 45 degrees in a ballistic trajectory. My point has been: the MAJOR and DOMINANT force that you experience as recoil is due to the rocket motor effect of the cartridge gases not the bullet momemtum as it travels down the barrel. Do you think you would hit jacksh!t if this was not so?

Thank you Sir Isaac Newton. http://forums.ubi.com/groupee_common/emoticons/icon_wink.gif

For every action there's an equal and opposite reaction. The force of that action is equal to the mass being moved times the acceleration. Assuming a fairly typical 9mm load of a 115 grain bullet over 5 grains of powder and that both experience roughly the same acceleration, which one is exerting the greater force on the barrel? I'm even giving you credit here for converting the entire 5 grains of powder into gas exiting the barrel. Were that it only so then I could get rid of all those expensive and smelly gun cleaning products.

Vidar_1
05-17-2007, 02:51 PM
Originally posted by M_Gunz:
<BLOCKQUOTE class="ip-ubbcode-quote"><div class="ip-ubbcode-quote-title">quote:</div><div class="ip-ubbcode-quote-content">Originally posted by Vidar_1:
Consider a rifle firing a standard round: Anyone with any military experience will tell you that a short carbine firing that same round kicks worse than a "long" like an FN. Why?

Try comparing the mass of the carbine vs the mass of that FN. Because the lighter the rifle,
the heavier the kick from the same round. Also add in effect of recoil absorbers in the two weapons when using your shoulder to measure the effect.

Don't feel bad, we have a trained machinegunner here who has stated categorically that .50
cal bullets speed up after leaving the muzzle and are really going faster at 300 meters than
at muzzle -- all from that good military knowledge that MY trainers seemed to have forgotten.
Oh well, when you have 'authority' and don't know something then just make it as you go!


It's not primarily because of the weight difference, it's because the recoil comes from the gases leaving the muzzle acting like a rocket if you like.

Hand load some blanks with only a tiny bit of tissue holding the powder and feel the great
recoil from those. I can tell you, it ain't much.


This is why cannon like the Mk108 have low muzzle velocity but a comparatively high recoil. The same round fired in a longer barrel Mk108 would have HIGHER muzzle velocity but LOWER recoil.

You are saying that the Mk 103 had LESS recoil per shot than the Mk 108? http://forums.ubi.com/images/smilies/16x16_smiley-very-happy.gif

Perhaps you should get with Jokf. He is re-creating whatever science gets in his way too! </div></BLOCKQUOTE>

ROTFLMAO: Thank God your views are solely your own! You are so wrong! Where do I begin.....

1) Stripping away the ranting about the carbine and the "long", the carbine would kick worse even when ballasted to match the rifle weight. Why? the gas pressure drops down the barrel and there will be more pressure to propel the gases at carbine muzzle length than at rifle length,

2)Blanks will not build up as much pressure. There is nothing keeping the pressure up during the burn. Try sprinkling some powder on a plate and set fire. See how slow it burns?

3)The best for last: If you knew anything about the Mk108 and Mk103 you would know that they use quite a different round. The Mk108 cartridge is puny compared to the Mk103 http://forums.ubi.com/images/smilies/88.gif

I really enjoyed your science lesson M_Gunz! Have not had so much fun in years! Right now I have to leave, but I will return tomorrow to check up on this thread and read more funny stories!

The Mk108 and Mk103 comparison......priceless!

K_Freddie
05-17-2007, 03:02 PM
I like the hardware on the left. Her muzzle velocity does it for me. She makes me go ballistic.... http://forums.ubi.com/groupee_common/emoticons/icon_biggrin.gif

M_Gunz
05-17-2007, 03:05 PM
Originally posted by Vidar_1:
Have you ever fired a pistol? I've shot quite a lot of 9 mm parabellum. Strangely I do hit the target roughly where I'm aiming. The bullet does not go 45 degrees in a ballistic trajectory. My point has been: the MAJOR and DOMINANT force that you experience as recoil is due to the rocket motor effect of the cartridge gases not the bullet momemtum as it travels down the barrel. Do you think you would hit jacksh!t if this was not so?

Only a few hundred times, I was much more into rifles back in the day.
The bullet leaves your 9mm barrel in something under 1/1000th of a second, your pressure on
the trigger has more time to screw your aim. Remember the delta-T?

Really you should read up more on internal and external ballistics. There are texts that do
go into gas release which until the bullet leaves the barrel unless the weapon is gas operated
are at very high pressures -- but you take the volume of your pistol barrel at 3000 psi and
at release there is a millisecond puff is all. Something measurable but not the main event.

When the bullet leaves the barrel that is the time when recoil is measured by people with
actual measuring devices and not slow senses.

Look at the mass of powder compared to the mass of bullet and subtract from the powder the
energy imparted to the bullet, what is left is your rocket gasses from the muzzle.

Of course you do not write about firing blanks at all. Even when it is noted that blanks
are used to lower the recoil of the hollywood stunt and make it possible to fire high-rate.

Maj.Kaos
05-17-2007, 03:26 PM
Originally posted by Xiolablu3:
In Predator, one of the soldiers is carrying a small minigun which iis very effective in the film.

I was just wondering if this was a real handeld gun, or is it complete fantasy?

I would think the main problem about this minigin would be weught and the weight of ammo that you would have to fire.


Also the inoitial 'spin up' of the gun may also be a problem. WHen you are likely to meet the enemy you dont want to have to wait a second after you pull the trigger.


EDIT: It seems it is real! As I was searching for a picture I came across this page! I gather this is real?>

http://www.kitsune.addr.com/Firearms/Machine-Guns/GE_XM214_Minigun.htm

Fantasy. If it were real I wouldn't carry it in the field, and I highly doubt anyone else would want to. At 23 lbs the M60 plus ammo was heavy enough to make a long day in the field. When fired "Rambo" style it was a ridiculous waste of ammo and fire efficiency, thus came the smaller and lighter SAW.

Spraying rounds toward target is only intimidating, and more dangerous to the person standing up in the open and pulling the trigger. One the other hand, applying rounds on target, however slowly, is morale breaking if not often fatal. One round per second on target will keep heads down as effectively and more efficiently than 166 rounds per second.

K_Freddie
05-17-2007, 03:27 PM
Originally posted by M_Gunz:
Of course you do not write about firing blanks at all. Even when it is noted that blanks
are used to lower the recoil of the hollywood stunt and make it possible to fire high-rate.

Sorry - you're wrong here...
Recoil effect force is reliant on the object(mass) it has to push forward (eg. the bullet head).
The heavier the mass, the more the recoil(back pressure) effect, thus a quicker reload time and fire.

With a Blank there is no mass to push forward, so the back pressure is reliant on the diameter of the barrel and the powder mixture. For the same powder mixture as a 'non blank', the recoil effect will definitely be less leading to a lower fire rate.

http://forums.ubi.com/groupee_common/emoticons/icon_wink.gif

M_Gunz
05-17-2007, 03:27 PM
Originally posted by Vidar_1:
1) Stripping away the ranting about the carbine and the "long", the carbine would kick worse even when ballasted to match the rifle weight. Why? the gas pressure drops down the barrel and there will be more pressure to propel the gases at carbine muzzle length than at rifle length,


The carbine does not kick worse ballasted unless the recoil mechanism in the FN is much more.

The things you say about the gas being the largest source of recoil, you don't back up at all.

Modern smokeless powder burns slower with pressure unlike black powder that burns faster with
pressure. It is a propery of propellant vs explosive and carefully worked out. Your idea of
how fast the burn propagates with powder poured onto a plate is completely different from
powder ignited by cap in a confined space at the closed end of a barrel. Does the powder on
the plate make a bang? The blank when fired does! It is fired quicker and under pressure
from the primer against the pressure of air in the barrel, and the little plug. It does try
to leave the barrel faster than air permits which is not an issue on the plate.

5 or 6 GRAINS of powder in the 9mm and bullets are how much mass? 120 to 150 grains?
YOU do the math how much faster the gasses would have to leave the barrel to make that the
major force.

M_Gunz
05-17-2007, 04:48 PM
Originally posted by K_Freddie:
<BLOCKQUOTE class="ip-ubbcode-quote"><div class="ip-ubbcode-quote-title">quote:</div><div class="ip-ubbcode-quote-content">Originally posted by M_Gunz:
Of course you do not write about firing blanks at all. Even when it is noted that blanks
are used to lower the recoil of the hollywood stunt and make it possible to fire high-rate.

Sorry - you're wrong here...
Recoil effect force is reliant on the object(mass) it has to push forward (eg. the bullet head).
The heavier the mass, the more the recoil(back pressure) effect, thus a quicker reload time and fire.

With a Blank there is no mass to push forward, so the back pressure is reliant on the diameter of the barrel and the powder mixture. For the same powder mixture as a 'non blank', the recoil effect will definitely be less leading to a lower fire rate.

http://forums.ubi.com/groupee_common/emoticons/icon_wink.gif </div></BLOCKQUOTE>

There is the mass of air in the barrel which does block movement and is why blanks go "bang!".
That bang is supersonic air leaving the barrel. At the speeds of the reaction, air is such.
So the blank does push air and burned powder.

Firing rate? Only when you leave the blank adapter on the gas-operated weapon do you get a
fir rate with blanks. Otherwise you have to keep cocking the thing.

I agree with you that it is mass times velocity that makes the recoil, conservation of momentum
and Newton's Second Law of Motion at work (ala Viper and others).

Compare the mass of gasses from powder to mass from bullet, like 6 grains powder for a 124
grain bullet in 9mm pistol and HOW FAST would those gasses have to be to make as much recoil
as the bullet? My point is that it's DAFT to think that the bullet makes less!

Put the weapon on one ballistic pendulum and have it fire into another 10 meters away.
I would like to see that less than 99% of the momentum is transferred to the target.
At 10 meters I expect that not much of the magic rocket exhaust will be causing the target
pendulum to move.

Some info on RECOIL including PERCEIVED vs MEASURED. (http://www.chuckhawks.com/rifle_recoil.htm)

* .300 WSM (180 grain, MV 2950 fps), 6.5 lb. rifle = 30.8 ft. lbs. / 17.5 fps
* .300 WSM (180 grain, MV 2950 fps), 8.5 lb. rifle = 23.6 ft. lbs. / 13.4 fps

* .45-70 (300 grain, MV 1900 fps), 7.0 lb. rifle = 26.6 ft. lbs. / 15.6 fps
* .45-70 (300 grain, MV 1900 fps), 8.5 lb. rifle = 21.9 ft. lbs. / 12.9 fps

Let's compare the weight of the carbine to the FN? Oh no! That makes no difference!

" Muzzle blast subjectively seems to increase recoil, although it has no actual bearing on the free recoil."

Gee, from people who actually DO put rifles onto pendulums. Takes all the fun out sitting
back over beers and applying Aristotle-Thinking that so many "important" people feel is their
right, which is consistent with Aristotle's methods.

This page just for fun, see how much velocity drops at 200 yards. (http://www.chuckhawks.com/rifle_ballistics_table.htm)

And last of all;

Here on the Recoil Table Page (http://www.chuckhawks.com/recoil_table.htm)

The recoil energy and recoil velocity figures are taken from various sources including the recoil nomograph in the Handloader's Digest 8th Edition, various online recoil calculators, or calculated from the formula given in the Lyman Reloading Handbook, 43rd Edition. The formula is:

E = 1/2 (Wr / 32) (Wb x MV + 4700 x Wp / 7000 x Wr)squared.

Where E = recoil Energy in ft. lbs., Wr = Weight of rifle in pounds, Wb = Weight of bullet in grains, MV = Muzzle Velocity of bullet in feet-per-second, Wp = Weight of powder in grains.

Yeah, the WEIGHT OF POWDER is in there times 4700 divided by 7000, as in 67% of powder mass
to 100% (EDIT: times bullet muzzle velocity which powder weight does not get, powder weight
only gets times 4700/7000) of bullet mass in the formula. Bullet mass x velocity vs powder
mass x 67% as factors. Hmmmm. Gonna take a lot of beer to forget that!

Still want to say the rocket exhaust is the big factor?

HeibgesU999
05-17-2007, 05:20 PM
Worse mistake the US Army mistake was going to the SAW. 5.56mm is terrible machine gun cartdridge.

NO PLUNGING FIRE. So no reverse slope defense. D'uh. What? And infantryman thought this was a good idea?

Can't take out armored vehicles. With the M-60 you could take out a BMP. Itty Bitty little 5.56mm doesn't have the punch of 7.62mm

The SAW was about as well thought out as the Bradley.

M2 is too heavy for light infantry unit really. Unless they are lugging it around in a humvee or susvee.

M_Gunz
05-17-2007, 05:30 PM
Plunging fire? Isn't that why there's grenades and launchers?
Surely not with an M-60! They hardly drop much at 800 meters.

But I do go with ya on the power, M-60 is as light as I'd want.

Besides, if the enemy is well fortified then you need a good FO!
Sit back and let the gun bunnies have em! Or the Jr. Birdmen....

Vidar_1
05-18-2007, 01:22 PM
Originally posted by M_Gunz:
<BLOCKQUOTE class="ip-ubbcode-quote"><div class="ip-ubbcode-quote-title">quote:</div><div class="ip-ubbcode-quote-content">Originally posted by Vidar_1:
Have you ever fired a pistol? I've shot quite a lot of 9 mm parabellum. Strangely I do hit the target roughly where I'm aiming. The bullet does not go 45 degrees in a ballistic trajectory. My point has been: the MAJOR and DOMINANT force that you experience as recoil is due to the rocket motor effect of the cartridge gases not the bullet momemtum as it travels down the barrel. Do you think you would hit jacksh!t if this was not so?

Only a few hundred times, I was much more into rifles back in the day.
The bullet leaves your 9mm barrel in something under 1/1000th of a second, your pressure on
the trigger has more time to screw your aim. Remember the delta-T?

Really you should read up more on internal and external ballistics. There are texts that do
go into gas release which until the bullet leaves the barrel unless the weapon is gas operated
are at very high pressures -- but you take the volume of your pistol barrel at 3000 psi and
at release there is a millisecond puff is all. Something measurable but not the main event.

When the bullet leaves the barrel that is the time when recoil is measured by people with
actual measuring devices and not slow senses.

Look at the mass of powder compared to the mass of bullet and subtract from the powder the
energy imparted to the bullet, what is left is your rocket gasses from the muzzle.

Of course you do not write about firing blanks at all. Even when it is noted that blanks
are used to lower the recoil of the hollywood stunt and make it possible to fire high-rate. </div></BLOCKQUOTE>

M_Gunz: You really don't get this do you? I'll try it one more time and do it a bit more stepwize (as in slower) this time.

I belive K_Freddie set you straight on the blanks issue so I don't have to do that anyway.

By your own words above: "The bullet leaves your 9mm barrel in something under 1/1000th of a second, your pressure on the trigger has more time to screw your aim."

Glad we agree. This means that at the time the bullet leaves the barrel the muzzle is still pointing at the target. If not convinced go back to see the Desert Eagle pulverizing the melons and the effect it has on the shooter AFTER the bullet left the barrel. Link reposted below for your convenience.

Now you claim the recoil is due to the bullet? But the bullet has already left the barrel. So when your arm gets thrown up after this event there is some force causing that. This force is obviously at work AFTER the bullet has left the barrel. Please explain the nature of that force and from where it originates and how it is that it causes this effect on your arm as the bullet is downrange speeding towards the target. http://forums.ubi.com/images/smilies/88.gif

If you have problems working this out and you are still unhappy about cartridge gas jet effect then consult your machine gun pal you mentioned in your earlier post. I'm sure you two together can come up with an explanation for how the bullet that just left your barrel mysteriously reaches back to throw your arm up into the air after the bullet has left the barrel. You two seem to be on the same level of scientific understanding.

BTW: As to your style of posting I agree with Viper. It's annoying when someone lectures you without grasping the contents of your post.

http://www.youtube.com/watch?v=__JkOUheyVk

M_Gunz
05-18-2007, 01:44 PM
Here ya go AGAIN, look at the formula and tell me that your idea of gas from the barrel
being most of the recoil or even much at all ... tell me PLEASE what a bunch of idiots
they are at Lyman or throw out another minor excuse to ignore what the industry has
determined through deep study. Unless of course you have other real information.

Weight of powder is in there as a factor in recoil. It counts for LITTLE but it IS
in there IF you can read formula otherwise find someone that can to help you.


Originally posted by M_Gunz:
Here on the Recoil Table Page (http://www.chuckhawks.com/recoil_table.htm)

The recoil energy and recoil velocity figures are taken from various sources including the recoil nomograph in the Handloader's Digest 8th Edition, various online recoil calculators, or calculated from the formula given in the Lyman Reloading Handbook, 43rd Edition. The formula is:

E = 1/2 (Wr / 32) (Wb x MV + 4700 x Wp / 7000 x Wr)squared.

Where E = recoil Energy in ft. lbs., Wr = Weight of rifle in pounds, Wb = Weight of bullet in grains, MV = Muzzle Velocity of bullet in feet-per-second, Wp = Weight of powder in grains.

Yeah, the WEIGHT OF POWDER is in there times 4700 divided by 7000, as in 67% of powder mass
to 100% (EDIT: times bullet muzzle velocity which powder weight does not get, powder weight
only gets times 4700/7000) of bullet mass in the formula. Bullet mass x velocity vs powder
mass x 67% as factors. Hmmmm. Gonna take a lot of beer to forget that!

Still want to say the rocket exhaust is the big factor?

Viper2005_
05-18-2007, 01:47 PM
Take one gun.

Fire a round.

The bullet whizzes off at high speed - for the sake of argument let's say it has a mass of 0.01 kg and is moving at 500 m/s.

It has a momentum of 5 kg*m/s.

Momentum is conserved. Let's say the gun has a mass of 1 kg.

Therefore when the bullet leaves the barrel the gun is recoiling at about 5 m/s.

If the barrel is 15 cm long and the acceleration is constant then the bullet will take 0.0006 seconds to leave the barrel, at which point the barrel will have recoiled 0.0015 m, or 1.5 mm if you prefer.

Despite the fact that in this case the bullet "supplies" all the momentum the gun won't have moved very far by the time the round leaves the barrel and there will therefore be very little aiming error associated with the recoil of a single round.

It takes quite a long time for the human body to respond to and compensate for the recoil of the gun, and therefore the observer will see the majority of the recoil movement taking place long after the bullet has left the barrel, and indeed probably after it has hit its target.

It is also worth pointing out that since f=ma and v=u+at, quite large forces would actually be associated with this recoil action - to get the gun up to 5 m/s in the time available implies a recoil force of over 3 kN, which might well be quite painful. Therefore most modern weapons use some form of damping system to spread the recoil impulse out over a longer period of time.

This would affect the result of any serious high-speed testing. However, the video you post only slows down to 1/4 speed, and therefore the dominant reason for the recoil seeming to kick in after the flash is simply the poor resolution of the recording. The barrel is recoiling from the moment the bullet starts to move - but it doesn't have time to move very far before the bullet leaves and you see the muzzle flash.

M_Gunz
05-18-2007, 02:55 PM
Originally posted by Vidar_1:
Glad we agree. This means that at the time the bullet leaves the barrel the muzzle is still pointing at the target. If not convinced go back to see the Desert Eagle pulverizing the melons and the effect it has on the shooter AFTER the bullet left the barrel. Link reposted below for your convenience.

45k dialup here. I don't do video downloads. I am very aware since the mid-60's about pistols
and rifles jerking upwards after firing. I learned about that when I was a kid.


Now you claim the recoil is due to the bullet? But the bullet has already left the barrel. So when your arm gets thrown up after this event there is some force causing that. This force is obviously at work AFTER the bullet has left the barrel. Please explain the nature of that force and from where it originates and how it is that it causes this effect on your arm as the bullet is downrange speeding towards the target. http://forums.ubi.com/images/smilies/88.gif

Your hand and arm not being solid as metal take some time to absorb the energy of recoil of
the weapon, it first moves back and only then does the barrel lift if at all. Your bones and
muscles and reactions are also part of the effect. The gasses are gone long before the barrel
lifts as well.

In case you didn't know, all aspects of shooting is a past science which is why I am so happy
to have found the formula showing the factors precisely and taken this out of SUPPOSITION and
SUBJECTIVE GUESS-LAND.


If you have problems working this out and you are still unhappy about cartridge gas jet effect

Come back with something solid about that, which no you have not.


then consult your machine gun pal you mentioned in your earlier post. I'm sure you two together can come up with an explanation for how the bullet that just left your barrel mysteriously reaches back to throw your arm up into the air after the bullet has left the barrel. You two seem to be on the same level of scientific understanding.

One; I am the machine pal. Two; my level of science is well beyond yours and still nothing to
brag about. No mystery when you quit treating your arm as a rigid attached device. Try getting
someone to explain to you about non-elastic collisions or better yet, fire a 30-06 when you
have the butt 1" from your shoulder and see how fast the butt and your shoulder meet.

If I attach the weapon into a ballistic pendulum and fire it then the bullet will leave the
barrel and at that precise moment the weapon will START to swing upwards. It keeps swinging
up even after the bullet is gone and that is not due to rocket gasses, it is due to the initial
impulse of bullet and weapon separating and Newton's 2nd Law. There is by Lyman some effect
due to the weight of the powder which is where those rocket gasses comes from and THAT has been
qualified-quantified and counts for very little unless of course you use one hell of a LOT
of powder -- ie you made a rocket.

If you still can't figure it out then go READ about recoil (why bother, you know it all from
just shooting!) at the link I gave. I can give you a few more that explain what you seem to
be having trouble with but that one alone explains your PERCEPTIONS very well.

Of course since it's not video that you can re-interpret how you want then it must be useless.


BTW: As to your style of posting I agree with Viper. It's annoying when someone lectures you without grasping the contents of your post.

Do tell. You keep practicing at it. I've fully grasped your errors and provided answers that
you don't seem to notice let alone grasp and yet you come back with what you have.

Get this through your head -- the people I quote have put the weapons in test setups that do
not include humans and do measure all aspects. And you have noticed how the barrel lifts
after the bullet leaves the barrel. Never mind that the gasses in the barrel are GONE long
before the barrel lifts else how do you think that 1000+ ROF single barrel MG's work?
Chew on it, the gas pressure is gone before you can blink and yet, the barrel still lifts --
maybe you should invoke ghosts!

Vidar_1
05-19-2007, 02:50 PM
Originally posted by Viper2005_:
Take one gun.

Fire a round.

The bullet whizzes off at high speed - for the sake of argument let's say it has a mass of 0.01 kg and is moving at 500 m/s.

It has a momentum of 5 kg*m/s.

Momentum is conserved. Let's say the gun has a mass of 1 kg.

Therefore when the bullet leaves the barrel the gun is recoiling at about 5 m/s.

If the barrel is 15 cm long and the acceleration is constant then the bullet will take 0.0006 seconds to leave the barrel, at which point the barrel will have recoiled 0.0015 m, or 1.5 mm if you prefer.

Despite the fact that in this case the bullet "supplies" all the momentum the gun won't have moved very far by the time the round leaves the barrel and there will therefore be very little aiming error associated with the recoil of a single round.

It takes quite a long time for the human body to respond to and compensate for the recoil of the gun, and therefore the observer will see the majority of the recoil movement taking place long after the bullet has left the barrel, and indeed probably after it has hit its target.

It is also worth pointing out that since f=ma and v=u+at, quite large forces would actually be associated with this recoil action - to get the gun up to 5 m/s in the time available implies a recoil force of over 3 kN, which might well be quite painful. Therefore most modern weapons use some form of damping system to spread the recoil impulse out over a longer period of time.

This would affect the result of any serious high-speed testing. However, the video you post only slows down to 1/4 speed, and therefore the dominant reason for the recoil seeming to kick in after the flash is simply the poor resolution of the recording. The barrel is recoiling from the moment the bullet starts to move - but it doesn't have time to move very far before the bullet leaves and you see the muzzle flash.

That's a nice explanation and I buy it in general terms. However, the reason I made a post in this thread was to make a point about the recoil forces. You have only include the bullet momentum part here as you did in the initial post. My point all along has been that this is to much of a simplification and that you need to consider the gas forces since they are not negligible but quite the contrary.

Going back you made the followin calculation:
5.56 NATO, 990.6 m/s muzzle velocity, 0.004018 grams.

Using this you calculated the average force 3.98 N. This was then used to calculate the 6000 rpm average recoil force 398 N, or 40.57 Kgf as you put it.

Then CD-kp84yb posted recoil forces for the 7.62mm NATO chambered minigun from an army handbook: 135.9 Kgf "normal" and 271.8 Kgf "max".

Now these are not comparable so I calculated the recoil for the 7.62 mm NATO (From Wikipedia: 840 m/s, 9.5 g bullet) using your method:

840 x 0.0095 x 6000/60 = 798 N or 81.35 Kgf

Comparing this to the army figures:

81.35/135.9 gives roughly 60%

81.35/271.8 gives roughly 30%

So using only the bullet momentum way we end up between 40-70% short in recoil. My conclusion from this exercise is that you cannot exclude the cartridge gas jet effect or you will underestimate recoil.

M_Gunz
05-19-2007, 03:58 PM
And we know that Wikipedia numbers are the same bullet and muzzle velocity as the Army used.

I do believe that even Vidar can agree with this:


The principle factors that must be considered to calculate recoil are bullet weight (mass), bullet velocity, powder charge, and rifle weight (mass). The mass times the velocity of everything ejected from the muzzle of a rifle (principally the bullet and power gasses) will be equaled by the mass times the velocity of the recoiling rifle.


the formula given in the Lyman Reloading Handbook, 43rd Edition which includes the powder:

E = 1/2 (Wr / 32) (Wb x MV + 4700 x Wp / 7000 x Wr)squared.

Where E = recoil Energy in ft. lbs., Wr = Weight of rifle in pounds, Wb = Weight of bullet in grains, MV = Muzzle Velocity of bullet in feet-per-second, Wp = Weight of powder in grains.

The free recoil velocity is how fast the gun comes back at the shooter. The faster a gun comes back at you the more it hurts. This is because your body has less time to give with the recoil. You have probably heard about the "long, slow push" touted by some big bore fans as opposed to the "sharp rap" supposedly delivered by high velocity cartridges. Recoil velocity is a real, measurable effect. But the "long slow push" appears to be a myth. Both my own personal experiences with a reasonable range of rifle calibers from .22 LR to .458 Win. Mag. and a quick perusal of the "Rifle Recoil Table" will show that the recoil velocity tends to increase as the recoil energy increases. The following examples of recoil energy and velocity are all measured in 8 pound rifles. (Caliber [bullet weight, muzzle velocity] = free recoil energy & free recoil velocity.)

* 6mm Rem. (100 grain, MV 3100 fps) = 10.0 ft. lbs. & 9.0 fps
* .270 Win. (140 grain, MV 3000 fps) = 17.1 ft. lbs. & 11.7 fps
* .30-06 (180 grain, MV 2700 fps) = 20.3 ft. lbs. & 12.8 fps
* .35 Whelen (250 grain, MV 2400 fps) = 26.1 ft. lbs. & 14.5 fps
* .450 Marlin (350 grain, MV 2100 fps) = 35.7 ft. lbs. & 17.0 fps
* .458 Win. Mag. (500 grain, MV 2050 fps) = 68.9 ft. lbs. & 23.5 fps

In the typical examples above, as the bullet weight goes up the muzzle velocity (MV) goes down; but the recoil energy and recoil velocity both continue to go up. The heavy bullets at relatively low velocity do not deliver a "long slow push," they deliver a progressively harder and faster blow to the shooter. Note that the high velocity .270 with its 140 grain bullet has a recoil velocity of only 11.7 fps, while the relatively low velocity .450 Marlin with its 350 grain bullet at 2100 fps has a recoil velocity of 17 fps!

Rifle weight plays an important role in determining the amount of recoil delivered to the shooter's shoulder. For any given caliber and load, a lighter rifle kicks more than a heavier rifle. Which is why I avoid ultra-light hunting rifles. Here are a couple of examples showing the effect rifle weight has on recoil energy and velocity when shooting the exact same load.

* .300 WSM (180 grain, MV 2950 fps), 6.5 lb. rifle = 30.8 ft. lbs. / 17.5 fps
* .300 WSM (180 grain, MV 2950 fps), 8.5 lb. rifle = 23.6 ft. lbs. / 13.4 fps

* .45-70 (300 grain, MV 1900 fps), 7.0 lb. rifle = 26.6 ft. lbs. / 15.6 fps
* .45-70 (300 grain, MV 1900 fps), 8.5 lb. rifle = 21.9 ft. lbs. / 12.9 fps

This is all well and good for theoretical models, but in reality the shooter's perception of recoil is influenced by the free recoil energy and the free recoil velocity his or her body must absorb and the shooter's pain threshold plus other factors. Some of these include stock fit, the size and shape and consistency of the rifle's recoil pad or butt plate, muzzle brake (if any), and even action type in the case of a gas operated autoloader. Muzzle blast subjectively seems to increase recoil, although it has no actual bearing on the free recoil. If the shooter is only wearing a light shirt the kick will feel worse than if he is wearing a heavy jacket or padded shooting vest.

Vidar_1
05-19-2007, 03:58 PM
Originally posted by M_Gunz:
Here ya go AGAIN, look at the formula and tell me that your idea of gas from the barrel
being most of the recoil or even much at all ... tell me PLEASE what a bunch of idiots
they are at Lyman or throw out another minor excuse to ignore what the industry has
determined through deep study. Unless of course you have other real information.

Weight of powder is in there as a factor in recoil. It counts for LITTLE but it IS
in there IF you can read formula otherwise find someone that can to help you.

<BLOCKQUOTE class="ip-ubbcode-quote"><div class="ip-ubbcode-quote-title">quote:</div><div class="ip-ubbcode-quote-content">Originally posted by M_Gunz:
Here on the Recoil Table Page (http://www.chuckhawks.com/recoil_table.htm)

The recoil energy and recoil velocity figures are taken from various sources including the recoil nomograph in the Handloader's Digest 8th Edition, various online recoil calculators, or calculated from the formula given in the Lyman Reloading Handbook, 43rd Edition. The formula is:

E = 1/2 (Wr / 32) (Wb x MV + 4700 x Wp / 7000 x Wr)squared.

Where E = recoil Energy in ft. lbs., Wr = Weight of rifle in pounds, Wb = Weight of bullet in grains, MV = Muzzle Velocity of bullet in feet-per-second, Wp = Weight of powder in grains.

Yeah, the WEIGHT OF POWDER is in there times 4700 divided by 7000, as in 67% of powder mass
to 100% (EDIT: times bullet muzzle velocity which powder weight does not get, powder weight
only gets times 4700/7000) of bullet mass in the formula. Bullet mass x velocity vs powder
mass x 67% as factors. Hmmmm. Gonna take a lot of beer to forget that!

Still want to say the rocket exhaust is the big factor? </div></BLOCKQUOTE>

Once again a rambling post. This one putting words in my mouth as well. Have I ever questioned Lyman or the equation above?

The way you set up this post is classic: the way you formulated it is if I question you then I question Lyman and call them a bunch of morons. Well guess what? I do not question the formula or Lyman. I only question the way you hide behind Lyman to boost your credibility and intrepret the equation to suit your ends. I only see one moron here and it's not Lyman.

First of all we have a disagreement on the applicability of this eqation. I really see no problem with it. My guess is it is an empirical or semiempirical equation that provides guidance to gauge the TOTAL recoil force when tinkering with bullet and powder loads.

You on the other hand seem to think that this equation can be used to see how much of the recoil is due to bullet momentum and how much is due to the powder. Frankly. I find the way you dissicate the equation to support your case ridiculus. Source please?

Try contacting Lyman and ask them about how the eqation is meant to be used. Ask them if the recoil E calculated can be broken down to show recoil due to powder and recoil due to bullet momemtum according to your analysis. They could probably do with a laugh.

Now let's get back to the issue at hand:

You keep on claiming that the gas jet is not a factor and that only the bullet momemtum needs to be considered to gauge recoil:

"Weight of powder is in there as a factor in recoil. It counts for LITTLE but it IS
in there IF you can read formula otherwise find someone that can to help you."

"There is by Lyman some effect
due to the weight of the powder which is where those rocket gasses comes from and THAT has been
qualified-quantified and counts for very little unless of course you use one hell of a LOT
of powder -- ie you made a rocket."

So according to you the powder gases account for a recoil force on the scale very little to LITTLE.

Well apparently Weatherby do not agree with you. They even went so far as to come up with a solution to ease the recoil on your shoulder. You can open up the link below to see for yourself but here are a few examples:

.270 Wby: 25.2 Ftlbs without muzzle brake and 11.8 Ftlbs with: reduction 53%

.375 Wby: 46.1 Ftlbs without muzzle brake and 21.7 Ftlbs with: reduction 53%

So the Weatherby Accubrake reduces recoil by around 53%. That's right. the muzzle brake get's rid of more than half the recoil. There are muzzle brakes that give less peformance but we are not concerned with muzzle brake performance here but how much of the recoil force is due to the gases. Now consider that a muzzle brake can never be 100% efficient. Even if a lot of the gas pressure is vented at right angles by necessity there will be gases behind the bullet as it leaves the barrel. This means that a larger reduction than 53% would be possible by an ideal muzzle brake and that consequently the cartridge gas jet effect is the DOMINANT recoil force at large here.

So in your book, a 53% reduction in recoil is very LITTLE is it M_Gunz?
http://www.weatherby.com/products/performance/accubrake.asp

M_Gunz
05-19-2007, 04:11 PM
Calcualting recoil, trajectory, etc.

Q: I shoot ________ (make and model) gun. What is the recoil? What is my trajectory like and for what distance should I set the sights?

A: Please do not ask me to compute your recoil or trajectory for you; these are things that you can figure out for yourself. GUNS AND SHOOTING ONLINE includes information to help you do this. Take a look at the relevant lists and tables on the "Tables, Charts and Lists Page," they are there for your benefit.


Since I don't have a membership I can only refer to the Lyman formula given by the same people
rather than give the expanded table of all types.

These are the people who say "The mass times the velocity of everything ejected from the muzzle of a rifle (principally the bullet and power gasses) will be equaled by the mass times the velocity of the recoiling rifle." and that is the formula for when you don't have a
recoil pendulum or other measuring rig.

Here is a section of the HANDGUN RECOIL TABLE they did put up, sorry the column formatting
gets lost between their site and UBI post window:

The following are taken from various sources, including online recoil calculators, or calculated from the formula given in the Lyman Reloading Handbook, 43rd Edition. The formula is:

E = 1/2 (Wg / 32) (Wb x MV + 4700 x Wp / 7000 x Wg)squared

Where E = recoil Energy in ft. lbs., Wg = Weight of gun in pounds, Wb = Weight of bullet in grains, M = Muzzle, V = Velocity in feet-per-second, Wp = Weight of powder in grains

Note: All bullets in table below are jacketed unless otherwise noted

Cartridge (Wb@MV) Pistol Wt. (lbs.) Recoil E. (ft. lbs.) Recoil V. (fps)
.25 ACP (50 at 800) 0.75 0.9 8.7
.30 Carbine (110 at 1400) 3.0 4.9 10.2
.32 ACP (71 at 910) 1.0 1.7 10.5
.32 H&R Mag. (100 at 1100) 2.0 2.7 9.4
.380 ACP (90 at 1000) 1.5 2.5 10.4
9mm Makarov (95 at 1025) 1.5 3.0 11.2
9x19 (124 at 1125) 1.5 6.0 16.0
.38 Super (125 at 1250) 2.25 4.9 11.9
.357 SIG (125 at 1350) 1.75 7.4 16.6
.38 Spec. (125 at 850) 1.0 5.6 18.9
.38 Spec. +P (125 at 975) 2.25 2.9 9.2
.357 Mag. (158 at 1250) 2.75 8.7 14.3


With APOLOGIES to Viper, yes I can see that they do indeed use MV^2/2 for energy units.
The difference between the two 9mm's given makes that obvious as does the formula given.
But... knockdown is still by momentum http://forums.ubi.com/images/smilies/16x16_smiley-happy.gif

CD_kp84yb
05-19-2007, 04:40 PM
This is what i have found for the 5.56 xm124

caliber 5.56mm
mv 3250fps
rof 400- 6000spm
time-to-fire .5 s
time-to-stop .25s
recoil average 100LBS @4000spm
recoil max NA
barrel life 100.000shots (only at lower rof)
gun life 500 k
Reliability (MRBF) 200k
Drive (electric) .75hp to 3.2Hp
weight gun only 27Lbs

took some time to get it.Lots of paper piles

edit : Ow the 7.62 is listed in these papers as 300 Lbs @ 6000spm and max as 600Lbs

M_Gunz
05-19-2007, 05:23 PM
Originally posted by Vidar_1:
.270 Wby: 25.2 Ftlbs without muzzle brake and 11.8 Ftlbs with: reduction 53%

.375 Wby: 46.1 Ftlbs without muzzle brake and 21.7 Ftlbs with: reduction 53%

So the Weatherby Accubrake reduces recoil by around 53%. That's right. the muzzle brake get's rid of more than half the recoil. There are muzzle brakes that give less peformance but we are not concerned with muzzle brake performance here but how much of the recoil force is due to the gases. Now consider that a muzzle brake can never be 100% efficient. Even if a lot of the gas pressure is vented at right angles by necessity there will be gases behind the bullet as it leaves the barrel. This means that a larger reduction than 53% would be possible by an ideal muzzle brake and that consequently the cartridge gas jet effect is the DOMINANT recoil force at large here.

So in your book, a 53% reduction in recoil is very LITTLE is it M_Gunz?
http://www.weatherby.com/products/performance/accubrake.asp

You can't read MATH so I guess this isn't going to mean much to ya.

The brake throws gasses back and adds mass to the rifle which also reduces recoil.

I can significantly reduce recoil just by weighting the gun up alone.

And you still haven't shown that the bullet leaving the barrel is not the cause of recoil
which I do believe you have stated more than once here, so maybe watch it with pointing out
small mistakes and calling names cause Vidar, you started with a WHOPPER!

Really we are BOTH WRONG, btw. From the guys that make a living at it:


There are devices that can be added to rifles to reduce recoil. Muzzle brakes are one such accessory, and they come standard on some hard kicking rifles. They reduce recoil by redirecting the escaping powder gas to the side and rearward. Everything that leaves the barrel adds to the recoil, which in the case of a rifle is the bullet and the gasses that propel it down the barrel. (A sabot, should one be used, also adds to the recoil.) The jet effect of the escaping powder gasses makes up about 25% to 50% of the total recoil of a modern rifle. An efficient muzzle brake can reduce the actual recoil by about 20% in some cases (the figures vary widely). See my article "Muzzle Brakes" for more on this subject.

Again: The jet effect of the escaping powder gasses makes up about 25% to 50% of the total recoil of a modern rifle.

And yes, I know that Weatherby shows much more but then those cartridges use inordinate
amounts of powder too. The common expression is a beer can necked down to .460 but I can't
give you a web source for that. Again, a huge amount of powder by the formula will make a
large part of the recoil.

My mistake was not taking the whole powder weight factor to include the weight of the rifle
which not you nor anyone has bothered to point out yet.

Muzzle Brakes article (http://www.chuckhawks.com/muzzle_brakes.htm)


A properly designed muzzle brake can significantly reduce recoil. The actual effectiveness depends to an extent on the cartridge for which the rifle is chambered. Ahlman's claims a 50% recoil reduction when their Recoil Reducer muzzle brake is used on large magnum rifles. Mag-na-port International claims recoil reduction of up to 45% for their Mag-na-brake. Browning, whose BOSS (Ballistic Optimizing Shooting System) is both a muzzle brake and an accuracy tuning device, claims a recoil reduction of up to 30%. Weatherby, who claims that their Accubrake is the most effective on the market, claims recoil reduction of up to 53%.

In their literature, Weatherby compares the effectiveness of the Accubrake to several other makes of muzzle brake. According to Weatherby, who used a .416 Weatherby Magnum rifle for testing, the Recoil Reducer reduced recoil by 39%, the KDF Slimline reduced recoil by 40.6%, and the KDF Regular reduced recoil by 49%. These devices reduced recoil by an average of 42.86%. I find that pretty impressive.

The Recoil Reducer, Magna-brake, Accubrake and BOSS muzzle brakes are screwed onto the end of the barrel. They are essentially a ventilated steel tube bored a little larger than the groove diameter of the barrel. Screwed firmly into place, they add a couple of inches of length to the end of the barrel, and are usually slightly fatter than the normal outside contour of the barrel. There is no loss of bullet velocity or change in the rifle's ballistic performance with these muzzle brakes.

Please get this REAL STRAIGHT.

Shoulder cannon that use LARGE AMOUNTS OF POWDER are NOT Your 9mm pistol.

When I wrote that the gasses from your 9mm pistol do not amount to much of the total recoil
(which I still maintain it's less than half as opposed to ALL of the recoil) that does NOT
include extreme rifle cartridges whatsoever.

So don't bother jacking me off like that just to make yourself feel better. The gas pressure
in the barrel is dropped before you perceive any movement of your arm whatsoever, the only
force that lifts your arm leaves the barrel before you twitch.


from Vidar:
Have you ever fired a pistol? I've shot quite a lot of 9 mm parabellum. Strangely I do hit the target roughly where I'm aiming. The bullet does not go 45 degrees in a ballistic trajectory. My point has been: the MAJOR and DOMINANT force that you experience as recoil is due to the rocket motor effect of the cartridge gases not the bullet momemtum as it travels down the barrel. Do you think you would hit jacksh!t if this was not so?

MAJOR and DOMINANT. You still maintain that?


from Vidar:
M_Gunz: Trick Question: Does this Desert Eagle shooter experience most recoil before or after the bullet left the barrel?

Trick question indeed. The shooter's "experience" is due to effects that occurred in about
one millisecond. So yes, his "experience" in dealing with the recoil is post-shot but the
CAUSE happens before the shooter "experiences" any effect at all.


from Vidar:
Now you claim the recoil is due to the bullet? But the bullet has already left the barrel. So when your arm gets thrown up after this event there is some force causing that. This force is obviously at work AFTER the bullet has left the barrel. Please explain the nature of that force and from where it originates and how it is that it causes this effect on your arm as the bullet is downrange speeding towards the target. Too Happy

http://forums.ubi.com/images/smilies/10.gif the science guys call it "Momentum" http://forums.ubi.com/images/smilies/10.gif

I suppose that if you jump up in the air that something must keep pushing you up after your
feet leave the ground? Maybe that's rocket gasses too?



Going back you made the followin calculation:
5.56 NATO, 990.6 m/s muzzle velocity, 0.004018 grams.

Using this you calculated the average force 3.98 N. This was then used to calculate the 6000 rpm average recoil force 398 N, or 40.57 Kgf as you put it.


Must be your week for confusion, *I* did not calculate that.

And if you want to play with automatic fire recoil then don't just use the starting ROF.
M-16A1's I had STARTED at 600 rpm but by 15-20 rounds (if it don't jam) is up to 900 rpm.
So feeding bad data into formulae is only going to get you low numbers.


from Vidar:
Try contacting Lyman and ask them about how the eqation is meant to be used. Ask them if the recoil E calculated can be broken down to show recoil due to powder and recoil due to bullet momemtum according to your analysis. They could probably do with a laugh.

It is very clear where each of the factors is, the formula is straightforward and labeled
clearly. The only laugh they'd get is from another clueless dweeb that has to ask!



You keep on claiming that the gas jet is not a factor and that only the bullet momemtum needs to be considered to gauge recoil:

Where did I say "not a factor"? Please show. I said SMALL and VERY SMALL. The last is not
correct.

Viper2005_
05-19-2007, 05:58 PM
Originally posted by Vidar_1:
That's a nice explanation and I buy it in general terms. However, the reason I made a post in this thread was to make a point about the recoil forces. You have only include the bullet momentum part here as you did in the initial post. My point all along has been that this is to much of a simplification and that you need to consider the gas forces since they are not negligible but quite the contrary.

Going back you made the followin calculation:
5.56 NATO, 990.6 m/s muzzle velocity, 0.004018 grams. I think you mean kg...


Using this you calculated the average force 3.98 N. This was then used to calculate the 6000 rpm average recoil force 398 N, or 40.57 Kgf as you put it.

Then CD-kp84yb posted recoil forces for the 7.62mm NATO chambered minigun from an army handbook: 135.9 Kgf "normal" and 271.8 Kgf "max".

Now these are not comparable so I calculated the recoil for the 7.62 mm NATO (From Wikipedia: 840 m/s, 9.5 g bullet) using your method:

840 x 0.0095 x 6000/60 = 798 N or 81.35 Kgf

Comparing this to the army figures:

81.35/135.9 gives roughly 60%

81.35/271.8 gives roughly 30%

So using only the bullet momentum way we end up between 40-70% short in recoil. My conclusion from this exercise is that you cannot exclude the cartridge gas jet effect or you will underestimate recoil.

The method that I have used, and that you have subsequently used in your analysis calculates the average force over a 1 second period. As such the average figure is more appropriate than the maximum figure. The max figure is more likely to be a function of the recoil damping system than it is to be a measure of the true maximum recoil force, which will likely (considerably) exceed 3 kN for a 7.62 mm round depending upon the propellant.

Calculations including propellant are inherently far more complex than calculations based solely upon the mass of the bullet. Since initially the propellant is constrained to leave the barrel at the same velocity as the bullet, the obvious 1st order of magnitude calculation may be based upon the assumption that the propellant mass is simply added to the mass of the bullet.

If you want to get cleverer than that you enter the territory of serious maths courtesy of Navier, Stokes and friends.

Anyway, I think I'm going to go flying now. The propellant masses should be available on wikipedia - I'm quite sure you can do the maths yourself. The bottom line is that unless the barrel is stupidly short for the cartridge you're very unlikely to find the propellant to be the primary source of recoil since that's fundamentally what the engineers designing the kit are trying to avoid.

Vidar_1
05-20-2007, 04:28 AM
Originally posted by M_Gunz:
<BLOCKQUOTE class="ip-ubbcode-quote"><div class="ip-ubbcode-quote-title">quote:</div><div class="ip-ubbcode-quote-content">Originally posted by Vidar_1:
.270 Wby: 25.2 Ftlbs without muzzle brake and 11.8 Ftlbs with: reduction 53%

.375 Wby: 46.1 Ftlbs without muzzle brake and 21.7 Ftlbs with: reduction 53%

So the Weatherby Accubrake reduces recoil by around 53%. That's right. the muzzle brake get's rid of more than half the recoil. There are muzzle brakes that give less peformance but we are not concerned with muzzle brake performance here but how much of the recoil force is due to the gases. Now consider that a muzzle brake can never be 100% efficient. Even if a lot of the gas pressure is vented at right angles by necessity there will be gases behind the bullet as it leaves the barrel. This means that a larger reduction than 53% would be possible by an ideal muzzle brake and that consequently the cartridge gas jet effect is the DOMINANT recoil force at large here.

So in your book, a 53% reduction in recoil is very LITTLE is it M_Gunz?
http://www.weatherby.com/products/performance/accubrake.asp

You can't read MATH so I guess this isn't going to mean much to ya.

The brake throws gasses back and adds mass to the rifle which also reduces recoil.

I can significantly reduce recoil just by weighting the gun up alone.

And you still haven't shown that the bullet leaving the barrel is not the cause of recoil
which I do believe you have stated more than once here, so maybe watch it with pointing out
small mistakes and calling names cause Vidar, you started with a WHOPPER!

Really we are BOTH WRONG, btw. From the guys that make a living at it:


There are devices that can be added to rifles to reduce recoil. Muzzle brakes are one such accessory, and they come standard on some hard kicking rifles. They reduce recoil by redirecting the escaping powder gas to the side and rearward. Everything that leaves the barrel adds to the recoil, which in the case of a rifle is the bullet and the gasses that propel it down the barrel. (A sabot, should one be used, also adds to the recoil.) The jet effect of the escaping powder gasses makes up about 25% to 50% of the total recoil of a modern rifle. An efficient muzzle brake can reduce the actual recoil by about 20% in some cases (the figures vary widely). See my article "Muzzle Brakes" for more on this subject.

Again: The jet effect of the escaping powder gasses makes up about 25% to 50% of the total recoil of a modern rifle.

And yes, I know that Weatherby shows much more but then those cartridges use inordinate
amounts of powder too. The common expression is a beer can necked down to .460 but I can't
give you a web source for that. Again, a huge amount of powder by the formula will make a
large part of the recoil.

My mistake was not taking the whole powder weight factor to include the weight of the rifle
which not you nor anyone has bothered to point out yet.

Muzzle Brakes article (http://www.chuckhawks.com/muzzle_brakes.htm)


A properly designed muzzle brake can significantly reduce recoil. The actual effectiveness depends to an extent on the cartridge for which the rifle is chambered. Ahlman's claims a 50% recoil reduction when their Recoil Reducer muzzle brake is used on large magnum rifles. Mag-na-port International claims recoil reduction of up to 45% for their Mag-na-brake. Browning, whose BOSS (Ballistic Optimizing Shooting System) is both a muzzle brake and an accuracy tuning device, claims a recoil reduction of up to 30%. Weatherby, who claims that their Accubrake is the most effective on the market, claims recoil reduction of up to 53%.

In their literature, Weatherby compares the effectiveness of the Accubrake to several other makes of muzzle brake. According to Weatherby, who used a .416 Weatherby Magnum rifle for testing, the Recoil Reducer reduced recoil by 39%, the KDF Slimline reduced recoil by 40.6%, and the KDF Regular reduced recoil by 49%. These devices reduced recoil by an average of 42.86%. I find that pretty impressive.

The Recoil Reducer, Magna-brake, Accubrake and BOSS muzzle brakes are screwed onto the end of the barrel. They are essentially a ventilated steel tube bored a little larger than the groove diameter of the barrel. Screwed firmly into place, they add a couple of inches of length to the end of the barrel, and are usually slightly fatter than the normal outside contour of the barrel. There is no loss of bullet velocity or change in the rifle's ballistic performance with these muzzle brakes.

Please get this REAL STRAIGHT.

Shoulder cannon that use LARGE AMOUNTS OF POWDER are NOT Your 9mm pistol.

When I wrote that the gasses from your 9mm pistol do not amount to much of the total recoil
(which I still maintain it's less than half as opposed to ALL of the recoil) that does NOT
include extreme rifle cartridges whatsoever.

So don't bother jacking me off like that just to make yourself feel better. The gas pressure
in the barrel is dropped before you perceive any movement of your arm whatsoever, the only
force that lifts your arm leaves the barrel before you twitch.


from Vidar:
Have you ever fired a pistol? I've shot quite a lot of 9 mm parabellum. Strangely I do hit the target roughly where I'm aiming. The bullet does not go 45 degrees in a ballistic trajectory. My point has been: the MAJOR and DOMINANT force that you experience as recoil is due to the rocket motor effect of the cartridge gases not the bullet momemtum as it travels down the barrel. Do you think you would hit jacksh!t if this was not so?

MAJOR and DOMINANT. You still maintain that?


from Vidar:
M_Gunz: Trick Question: Does this Desert Eagle shooter experience most recoil before or after the bullet left the barrel?

Trick question indeed. The shooter's "experience" is due to effects that occurred in about
one millisecond. So yes, his "experience" in dealing with the recoil is post-shot but the
CAUSE happens before the shooter "experiences" any effect at all.


from Vidar:
Now you claim the recoil is due to the bullet? But the bullet has already left the barrel. So when your arm gets thrown up after this event there is some force causing that. This force is obviously at work AFTER the bullet has left the barrel. Please explain the nature of that force and from where it originates and how it is that it causes this effect on your arm as the bullet is downrange speeding towards the target. Too Happy

http://forums.ubi.com/images/smilies/10.gif the science guys call it "Momentum" http://forums.ubi.com/images/smilies/10.gif

I suppose that if you jump up in the air that something must keep pushing you up after your
feet leave the ground? Maybe that's rocket gasses too?



Going back you made the followin calculation:
5.56 NATO, 990.6 m/s muzzle velocity, 0.004018 grams.

Using this you calculated the average force 3.98 N. This was then used to calculate the 6000 rpm average recoil force 398 N, or 40.57 Kgf as you put it.


Must be your week for confusion, *I* did not calculate that.

And if you want to play with automatic fire recoil then don't just use the starting ROF.
M-16A1's I had STARTED at 600 rpm but by 15-20 rounds (if it don't jam) is up to 900 rpm.
So feeding bad data into formulae is only going to get you low numbers.


from Vidar:
Try contacting Lyman and ask them about how the eqation is meant to be used. Ask them if the recoil E calculated can be broken down to show recoil due to powder and recoil due to bullet momemtum according to your analysis. They could probably do with a laugh.

It is very clear where each of the factors is, the formula is straightforward and labeled
clearly. The only laugh they'd get is from another clueless dweeb that has to ask!



You keep on claiming that the gas jet is not a factor and that only the bullet momemtum needs to be considered to gauge recoil:

Where did I say "not a factor"? Please show. I said SMALL and VERY SMALL. The last is not
correct. </div></BLOCKQUOTE>

I sifted through your rambling above and basically I find nothing new. I think those who are interested can read through from page one and form their own conclusion about how the cartridge gases factor into recoil.

As to week of confusion: If you go back you will see that I never claimed you did the calculation on the 5.56 mm NATO cartridge. That was in a response I made to Viper. http://forums.ubi.com/images/smilies/16x16_smiley-wink.gif

Vidar_1
05-20-2007, 04:33 AM
Originally posted by Viper2005_:
<BLOCKQUOTE class="ip-ubbcode-quote"><div class="ip-ubbcode-quote-title">quote:</div><div class="ip-ubbcode-quote-content">Originally posted by Vidar_1:
That's a nice explanation and I buy it in general terms. However, the reason I made a post in this thread was to make a point about the recoil forces. You have only include the bullet momentum part here as you did in the initial post. My point all along has been that this is to much of a simplification and that you need to consider the gas forces since they are not negligible but quite the contrary.

Going back you made the followin calculation:
5.56 NATO, 990.6 m/s muzzle velocity, 0.004018 grams. I think you mean kg...


Using this you calculated the average force 3.98 N. This was then used to calculate the 6000 rpm average recoil force 398 N, or 40.57 Kgf as you put it.

Then CD-kp84yb posted recoil forces for the 7.62mm NATO chambered minigun from an army handbook: 135.9 Kgf "normal" and 271.8 Kgf "max".

Now these are not comparable so I calculated the recoil for the 7.62 mm NATO (From Wikipedia: 840 m/s, 9.5 g bullet) using your method:

840 x 0.0095 x 6000/60 = 798 N or 81.35 Kgf

Comparing this to the army figures:

81.35/135.9 gives roughly 60%

81.35/271.8 gives roughly 30%

So using only the bullet momentum way we end up between 40-70% short in recoil. My conclusion from this exercise is that you cannot exclude the cartridge gas jet effect or you will underestimate recoil.

The method that I have used, and that you have subsequently used in your analysis calculates the average force over a 1 second period. As such the average figure is more appropriate than the maximum figure. The max figure is more likely to be a function of the recoil damping system than it is to be a measure of the true maximum recoil force, which will likely (considerably) exceed 3 kN for a 7.62 mm round depending upon the propellant.

Calculations including propellant are inherently far more complex than calculations based solely upon the mass of the bullet. Since initially the propellant is constrained to leave the barrel at the same velocity as the bullet, the obvious 1st order of magnitude calculation may be based upon the assumption that the propellant mass is simply added to the mass of the bullet.

If you want to get cleverer than that you enter the territory of serious maths courtesy of Navier, Stokes and friends.

Anyway, I think I'm going to go flying now. The propellant masses should be available on wikipedia - I'm quite sure you can do the maths yourself. The bottom line is that unless the barrel is stupidly short for the cartridge you're very unlikely to find the propellant to be the primary source of recoil since that's fundamentally what the engineers designing the kit are trying to avoid. </div></BLOCKQUOTE>

I never claimed it was easy to factor in the cartridge gases into the equation, just that they had to be included to get the true picture of recoil. Seems we agree on that now.

M_Gunz
05-20-2007, 05:07 AM
Originally posted by Vidar_1:
I sifted through your rambling above and basically I find nothing new. I think those who are interested can read through from page one and form their own conclusion about how the cartridge gases factor into recoil.

Okay Aristotle. You can just look and decide beats the people who have measured and quantified.

Of course there's nothing new and that includes you still admit none of your own errors.


As to week of confusion: If you go back you will see that I never claimed you did the calculation on the 5.56 mm NATO cartridge. That was in a response I made to Viper. http://forums.ubi.com/images/smilies/16x16_smiley-wink.gif

This must be your week. You can point out 3 actual errors I made and point out 10+ errors I
didn't make as errors then shrug off a whole STACK of errors YOU made but still catch that
and reply to the rest with only dismissal. I guess that must make YOU right.

M_Gunz
05-21-2007, 03:25 AM
Also while hunting up cartridge loads that matched recoil data, I hit close every time even
when having to interpolate and a fingernail thickness off on the 45 ACP that the powder loading
data matches instead of nearly matches the recoil data table which I get close on anyway.


CLICK HERE (http://home.earthlink.net/%7Ehwsportsman/RecoilMgt.html)

This is from the 48th edition whereas the formula on the other page is from the 43rd edition.
The newer formula uses all pounds and fps and it's on a little square image, not text chars.

48th edition: I took the liberty to replace the numbered terms with lettered terms....

FRE (free recoil energy) = ( Wb * Vb + 4700 * Wp )^2 / (64.348 * Wg )
Wb = weight of bullet in lbs (1 lb is 7000 grains), Vb is muzzle velocity in fps,
Wp is weight of powder in lbs and Wg is weight of the gun in lbs.

Data on all those factors with apples to apples between recoil and powder load data ----

powder grains data from reloading page subject:
.45 ACP
230 grain Hornady HP-XTP (.451") bullet, 8.4 grains HS6 powder, Win. WLP primer, Hornady case. MV 850 fps in 5" pistol barrel. (Based on the Hornady Handbook, Sixth Edition.)

recoil data for matching bullet mass and velocity above:
Cartridge (Wb@MV) Pistol Wt. (lbs.) Recoil E. (ft. lbs.) Recoil V. (fps).45 ACP (230 at 850) 2.25 7.9 15.0

Wb (weight of bullet) is 230 gr / 7000 gr/lb = 0.032857 lbs (to 6 plcs)
Wp (weight of powder) is 8.4 gr / 7000 gr/lb = 0.0012 lbs
Wg (weight of gun) is 2.25 lbs

FRE (free recoil energy) = ( Wb * Vb + 4700 * Wp )^2 / (64.348 * Wg )
FRE = ( .032857 * 850 + 4700 * .0012 )^2 / ( 64.348 * 2.25 lbs )
FRE = ( 27.92845 + 5.64 )^2 / 144.783 = 1126.840835 / 144.783 = 7.7829 ft-lbs

And the chart I checked against has 7.9 for a VERY WELL known cartridge.
Note that all the numbers are here for inspection, nothing was fudged or adjusted.

In the formula the top is a binomial of two factors, bullet momentum at muzzle and gasses.
The two factors are added and squared so its a REAL SIMPLE AND CLEAR procedure to show how
much the bullet weighed in and how much the rocket exhaust did.
If you look at the beginning of the last line;

FRE = ( 27.92845 + 5.64 )^2 / 144.783

The 27.9... is the factor of bullet recoil and the 5.64 is the factor of the powder gasses.
I note that since the weight of powder term is lbs, the 4700 must be fps to get a usable
ft-lbs unit to be added to the bullet MV ft-lbs so yes indeed the rocket gases are counted
as being moving out of the muzzle at 4700 fps, don't forget your ear plugs.

Now I compare 28 to 6 and that puts the powder at about 18% of the total even rounding it
up more than the bullet momentum. And that is the way the formula works so I hope you can
understand enough to see for yourself.

What surprised me is that the gases even amount to more than half that.

I almost matched a 9x19 between powder and recoil, bullet mass and speed I couldn't match
so I went a fraction high on the powder and ended maybe a third of a ft-lb off.

I wouldn't mind if the guys with less rust in physics look this over.

Philipscdrw
05-21-2007, 06:07 AM
Originally posted by Vidar_1:
I sifted through your rambling above and basically I find nothing new. I think those who are interested can read through from page one and form their own conclusion about how the cartridge gases factor into recoil.

As to week of confusion: If you go back you will see that I never claimed you did the calculation on the 5.56 mm NATO cartridge. That was in a response I made to Viper. http://forums.ubi.com/images/smilies/16x16_smiley-wink.gif
OK, I've done that, I think I agree with M_Gunz. Recoil is principally caused by the momentum of the bullet, not by the gas, which has neligible effect unless you're using high-powered rifles. M_Gunz and Viper seem to know what they're talking about while you're being evasive.

Viper2005_
05-21-2007, 07:56 AM
Since we're still having the recoil debate I suppose I might as well go back and calculate the recoil of a blank firing 5.56 "round".

Let's say about 25 grains of propellant.

http://accurateshooter.net/Downloads/sierra223rembolt.pdf

That's about 1.62 grams.

Wikipedia quotes a muzzle energy of about 1785 J for the NATO 5.56 round, so let's now spread that over 1.62 grams of propellant.

e = 0.5*m*v^2

thus

v = (2e/m)^0.5

in this case v ~ 1484.5 m/s (1dp).

Momentum would therefore be 1484.5*1.62*10^-3 ~ 2.40 kg m/s (2dp)

Thus if we fire at 6000 rpm (ie 100 rounds/second), the recoil force will be 240 N.

If I rework the recoil calculation for the 62 grain NATO 5.56 bullet, this time using the muzzle energy of 1785 J, I get a muzzle velocity of 942 m/s and a momentum of 3.79 kg m/s. Thus at 6000 rpm a recoil force of 379 N.

Therefore, worst case if I were using the gun as a rocket the propellant would give about 63% of the recoil of the live round. However, this does not mean that when I fire a live round the propellant will produce this large recoil.

The muzzle velocity of the propellant is constrained to be the same as that of the bullet. Therefore, we may reasonably just add the propellant mass to the bullet mass. So with 25 grains of propellant and 62 grains of bullet we have a total mass of 87 grains. Since the momentum is distributed pro rata we may therefore conclude that the total recoil force will be about 531 N, of which around 28% will come from the propellant and the remaining 72% will come from the round. The truth will be somewhere between these two results since once the bullet leaves the barrel the remaining propellant gases are free to expand at a greater rate and may therefore reach a muzzle velocity higher than that of the round. However, I would suggest that despite this effect the chances are that the contribution of propellant gasses to recoil force will be much closer to 30% than to 60%.

BTW, the mass of the gun shouldn't change the recoil forces at all, since F=ma, which we can integrate with respect to time to p = mv.

http://isites.harvard.edu/fs/docs/icb.topic58975.files/ch3.pdf
(isn't the internet wonderful?)

It will change the distribution of energy between round and weapon however; let's assume our microgun weighs in at 40 kg. The round weighs in at about 4*10^-3 kg.
Thus there is a factor 10^4 difference in mass between the weapon and the round. It follows that for momentum to be conserved there must be a 10^4 difference in the velocity imparted to the weapon and the round. Thus the weapon will recoil at about 0.1 m/s each time a round is fired at about 10^3 m/s. The energy of the round will be 1*(10^4)^2/(10^4)*(10^-1)^2 times as large as the energy of the weapon, which comes out at 10^6. Of course, these are very much ball-park figures, but they serve to illustrate the point that because the weapon is much heavier than the round it fires the recoil wastes only a tiny amount of energy moving the weapon, even if the weapon is in free space. IRL it's attached to a person attached to the earth which is extremely massive...

Vidar_1
05-21-2007, 12:31 PM
Originally posted by Viper2005_:
Since we're still having the recoil debate I suppose I might as well go back and calculate the recoil of a blank firing 5.56 "round".


However, energy is not destroyed but converted. In this case, 30% to the bullet, the remaining 70% to other sources.
Let's say about 25 grains of propellant.

http://accurateshooter.net/Downloads/sierra223rembolt.pdf

That's about 1.62 grams.

Wikipedia quotes a muzzle energy of about 1785 J for the NATO 5.56 round, so let's now spread that over 1.62 grams of propellant.

e = 0.5*m*v^2

thus

v = (2e/m)^0.5

in this case v ~ 1484.5 m/s (1dp).

Momentum would therefore be 1484.5*1.62*10^-3 ~ 2.40 kg m/s (2dp)

Thus if we fire at 6000 rpm (ie 100 rounds/second), the recoil force will be 240 N.

If I rework the recoil calculation for the 62 grain NATO 5.56 bullet, this time using the muzzle energy of 1785 J, I get a muzzle velocity of 942 m/s and a momentum of 3.79 kg m/s. Thus at 6000 rpm a recoil force of 379 N.

Therefore, worst case if I were using the gun as a rocket the propellant would give about 63% of the recoil of the live round. However, this does not mean that when I fire a live round the propellant will produce this large recoil.

The muzzle velocity of the propellant is constrained to be the same as that of the bullet. Therefore, we may reasonably just add the propellant mass to the bullet mass. So with 25 grains of propellant and 62 grains of bullet we have a total mass of 87 grains. Since the momentum is distributed pro rata we may therefore conclude that the total recoil force will be about 531 N, of which around 28% will come from the propellant and the remaining 72% will come from the round. The truth will be somewhere between these two results since once the bullet leaves the barrel the remaining propellant gases are free to expand at a greater rate and may therefore reach a muzzle velocity higher than that of the round. However, I would suggest that despite this effect the chances are that the contribution of propellant gasses to recoil force will be much closer to 30% than to 60%.

BTW, the mass of the gun shouldn't change the recoil forces at all, since F=ma, which we can integrate with respect to time to p = mv.

http://isites.harvard.edu/fs/docs/icb.topic58975.files/ch3.pdf
(isn't the internet wonderful?)

It will change the distribution of energy between round and weapon however; let's assume our microgun weighs in at 40 kg. The round weighs in at about 4*10^-3 kg.
Thus there is a factor 10^4 difference in mass between the weapon and the round. It follows that for momentum to be conserved there must be a 10^4 difference in the velocity imparted to the weapon and the round. Thus the weapon will recoil at about 0.1 m/s each time a round is fired at about 10^3 m/s. The energy of the round will be 1*(10^4)^2/(10^4)*(10^-1)^2 times as large as the energy of the weapon, which comes out at 10^6. Of course, these are very much ball-park figures, but they serve to illustrate the point that because the weapon is much heavier than the round it fires the recoil wastes only a tiny amount of energy moving the weapon, even if the weapon is in free space. IRL it's attached to a person attached to the earth which is extremely massive...

I do not agree with "spreading" the bullet muzzle energy over the powder as you have done. It should be the other way around. The powder imparting SOME of it's energy to the bullet.

I will go with your numbers: 25 grain powder weight.

The figure I found for powder energy content is 178 ftlb/grain. Conversion 1ftlb=1.356 J gives:

25 x 178 x 1.356 = 6034 J

Now by your own calculation, the muzzle energy for the bullet is 1785 J.

1785/6034 = 0.296 or 29.6 % if you like

Naturally, all the residual powder energy is not converted to jet thrust. There are friction losses, transfer of heat etc but fact remains only about 30% in this case get's converted to bullet energy.

So the unnacounted for 40-70% in recoil indicated by the figures posted by CD-kp84yb for the minigun above tab up pretty well with the powder energy content analysis above if you deduct some for heat transfer and friction losses.

M_Gunz
05-21-2007, 06:51 PM
Originally posted by Viper2005_:
The muzzle velocity of the propellant is constrained to be the same as that of the bullet.

No, it isn't. Much higher. Consider a moment

* the bullet is accelerating all the way up the barrel requires energy.

* the bullet is in close contact with the barrel in a sliding friction state that still has
friction enough to heat the barrel, unless said barrel is worn out, requires energy.

* the bullet is still being spun up by the rifling which requires energy.

The energy is gotten from kinetic energy of the gas molecules that are the real speed of the gas
and not how fast the bullet is moving. Those molecules must impinge on the bullet hard enough
to drive it through a tight fit narrow barrel accelerating forward and in rotation at the same
time takes way more energy than just keeping up!
Once the lid (bullet) comes off, those gas molecules are free and that is where Lyman gives
the 4700 figure to go with the weight of powder, 4700 fps x lbs of powder is the gas momentum
by their instrumented-test and years at it backed reckoning. It's way better than what *I*
can do alone!

if there was no bullet to absorb energy from the powder burned then the rocket drive from the
powder alone would be more efficient. More of the chemical energy of the powder would drive
gas out the muzzle instead of converting the energy into the heavy bullet. It might burn
slower lit with a match but that's not sparked by a mercury fulminate primer. Smokeless
powder is formulated to not burn faster with increased pressure but maybe you'd need to be
down in the right library to find books on the subject. I've spent time in some good ones
when I was younger.

M_Gunz
05-21-2007, 07:04 PM
BTW, the mass of the gun shouldn't change the recoil forces at all, since F=ma, which we can integrate with respect to time to p = mv.

http://isites.harvard.edu/fs/docs/icb.topic58975.files/ch3.pdf
(isn't the internet wonderful?)

It will change the distribution of energy between round and weapon however; let's assume our microgun weighs in at 40 kg. The round weighs in at about 4*10^-3 kg.
Thus there is a factor 10^4 difference in mass between the weapon and the round. It follows that for momentum to be conserved there must be a 10^4 difference in the velocity imparted to the weapon and the round. Thus the weapon will recoil at about 0.1 m/s each time a round is fired at about 10^3 m/s. The energy of the round will be 1*(10^4)^2/(10^4)*(10^-1)^2 times as large as the energy of the weapon, which comes out at 10^6. Of course, these are very much ball-park figures, but they serve to illustrate the point that because the weapon is much heavier than the round it fires the recoil wastes only a tiny amount of energy moving the weapon, even if the weapon is in free space. IRL it's attached to a person attached to the earth which is extremely massive...

Momentum is conserved. So a lighter gun will recoil faster than a heavier one. The gun guys
are concerned with how hard the butt goes for shoulder.

Note how Lymans starts with bullet and powder momentums, adds those, squares that and divides
by a constant that I can't say just what about times the mass of the gun. So you can seperate
the factors easily to compare how much each is.

Viper2005_
05-22-2007, 07:09 AM
Originally posted by M_Gunz:
<BLOCKQUOTE class="ip-ubbcode-quote"><div class="ip-ubbcode-quote-title">quote:</div><div class="ip-ubbcode-quote-content">Originally posted by Viper2005_:
The muzzle velocity of the propellant is constrained to be the same as that of the bullet.

No, it isn't. Much higher. Consider a moment

* the bullet is accelerating all the way up the barrel requires energy.

* the bullet is in close contact with the barrel in a sliding friction state that still has
friction enough to heat the barrel, unless said barrel is worn out, requires energy.

* the bullet is still being spun up by the rifling which requires energy.

The energy is gotten from kinetic energy of the gas molecules that are the real speed of the gas
and not how fast the bullet is moving. Those molecules must impinge on the bullet hard enough
to drive it through a tight fit narrow barrel accelerating forward and in rotation at the same
time takes way more energy than just keeping up!
Once the lid (bullet) comes off, those gas molecules are free and that is where Lyman gives
the 4700 figure to go with the weight of powder, 4700 fps x lbs of powder is the gas momentum
by their instrumented-test and years at it backed reckoning. It's way better than what *I*
can do alone!

if there was no bullet to absorb energy from the powder burned then the rocket drive from the
powder alone would be more efficient. More of the chemical energy of the powder would drive
gas out the muzzle instead of converting the energy into the heavy bullet. It might burn
slower lit with a match but that's not sparked by a mercury fulminate primer. Smokeless
powder is formulated to not burn faster with increased pressure but maybe you'd need to be
down in the right library to find books on the subject. I've spent time in some good ones
when I was younger. </div></BLOCKQUOTE>

The propellant gasses cannot escape the barrel before the bullet does. Therefore their muzzle velocity is constrained to be the same as the bullet during live firing.

Now, once the bullet leaves the muzzle, they may of course expand further and accelerate to some higher velocity.

However, this process is going to be limited because:

1) Gas immediately behind the bullet can't expand very much at all before it also leaves the muzzle. At that point its expansion will be uncontrolled and there will be no further increase in its momentum. It will exert some limited pressure thrust though since its pressure is greater than atmospheric pressure.

2) Expansion is limited by ambient atmospheric pressure. As such, some propellant gasses will have to remain in the barrel to balance the books. (If you were to throw Navier-Stokes at the problem, you'd probably see initial over-expansion followed by some reverse flow due to the inertia of the propellant gas flow. However, we're only really interested in the net recoil forces so we can legitimately time average this behaviour under the carpet!)

It may interest you to note that my figure of 1484.5 m/s for the blank firing case could alternatively be written as 4870 fps, which is within 4% of your expert's figure of 4700 fps.

If this figure is accurate for the live firing case then that implies equal energy distribution between propellant gasses and bullet, which I would find slightly surprising given the mass difference between the two.

***

Vidar_1, I think your figures relate to the heat of combustion as measured by a chemist, which assumes that the products of combustion are brought back to ambient conditions.

Doing mechanical work with that energy is generally a rather different question, and it isn't reasonable to expect to see all of the heat of combustion realised as kinetic energy.

If your figure is correct then the implied thermal efficiency of the gun is about 30% which is pretty reasonable.

M_Gunz
05-22-2007, 08:01 PM
The gas behind the bullet is not a single thing even if equations may treat it so.
The pressure of the gas on the bullet is the average energy of the gas molecules that once
the muzzle is open do move at their own velocity that as I pointed out MUST BE greater than
the bullet or there would be no pressure on the bullet going up the barrel. The gases being
under pressure, the gas molecules -are- moving faster than the bullet. Once the bullet is
out of the barrel those gasses will move out faster than the bullet until pressure within
the barrel drops, a very short time.

M_Gunz
05-22-2007, 09:10 PM
If 1 Calorie (IIRC = 1000 food calories, small c but it's been decades so...) heats 500g of
water 1 degree C,

then how much is taken to warm the mass of the barrel in one shot from cold barrel?

There is a lot of friction and some straight heat transfer from the powder gases, but how much
in under 1 msec? The friction is the most, twist on the bullet absorbs energy not measureable
through muzzle velocity that also creates more normal force on the rifling, higher friction
that is measurable in barrel heat change... if you insulate the barrel calorimeter style then
the measure could be really accurate.

And how much energy does the spin of the bullet carry off with NO effect on recoil?

M_Gunz
05-24-2007, 10:44 PM
Apparently bullets need to spin up greatly to stay stable over seconds of flight.
And too much spin is also bad.
.357 in an example ( 8.25" twist and 1235 MV ) showed at muzzle spin as just over 101k rpm.
The factory formula is RPM = 720 x MV / twist. A long twist is 16 and there was a nice
story about the M-14 and Navy rebarreled M1's that gave the final figure on one barrel's
optimum twist as 11... and both rifles used the same ammo, 30 cal US M2.

100k rpm being low and 300k rpm being high range rpm's, that's 100,000 to 300,000+ rpm it
certainly does take energy, as in 100x whatever it takes to turn 1000 to 3000 rpm --- from
zero to MV and spinning like H in under 1/1000th sec. The bullet doesn't keep that spin up
but it did leave home with a full breakfast, paid for.