Rail Guns don't recoil
![Canup, Robert E. (JSC-EV)[ESCG]'s profile photo](http://lh3.googleusercontent.com/a-/ALV-UjWPB3lGLd9GsBtBDsNcsKfQgxdGzspfF0vg5Dd38EWxzYJjCA=s40-c)
Canup, Robert E. (JSC-EV)[ESCG]
hold certainly push the boundaries of this newsgroup's prohibition
of "excessively speculative" material. However, there do seem to be
reasonably clearcut -- and possibly informative -- physics issues
here concerning momentum flux in systems with moving conductors.
Two very informative discussions on this topic are
http://en.wikipedia.org/wiki/Faraday_paradox
and sections 17-2 and 17-4 of volume II of the Feynman lectures.
-- jt]]
In a paper from the Naval Postgraduate School an attempt to measure
recoil forces in a rail gun failed to find any within experimental
error. In particular note what occurred when the paper's author clamped
the projectile to the rails of the device.
http://stinet.dtic.mil/cgi-bin/GetTRDoc?AD=ADA473387&Location=U2&doc=Get
TRDoc.pdf
There is very little room for skepticism about the paper. Large scale
tests performed by the US Navy of a prototype rail gun involved a 3.35
Kg projectile with a muzzle velocity of 2520 meters/sec. This gives a
momentum in excess of 8000 Kg-meters/sec, enough to send a 200 Kg rail
gun backward at over 40 meters per second. A conventional gun with
similar performance would require a massive and extensive recoil
absorption apparatus. There is none needed with a rail gun.
The attached photo shows a shot of the Naval prototype rail gun firing.
Note the thin pedestal mounts used to hold the rail gun apparatus.
http://en.wikipedia.org/wiki/Image:Railgun_usnavy_2008.jpg
Other pictures of prototype high performance rail guns at a development
facility show the rail gun sitting on wheeled carts while firing.
The lack of recoil in rail guns has disastrous consequences for physics;
it is a direct and unequivocal demonstration that the law of
conservation of momentum is incorrect.
I would like to make one prediction about rail guns: It will be found
that there is also no back e.m.f. generated by the devices; they not
only show conservation of momentum to be untrue, they violate
conservation of energy also.
Glen Herrmannsfeldt
> [[Mod. note -- Claims that energy and/or momentum conservation don't
> hold certainly push the boundaries of this newsgroup's prohibition
> of "excessively speculative" material. However, there do seem to be
> reasonably clearcut -- and possibly informative -- physics issues
> here concerning momentum flux in systems with moving conductors.
> Two very informative discussions on this topic are
> http://en.wikipedia.org/wiki/Faraday_paradox
> and sections 17-2 and 17-4 of volume II of the Feynman lectures.
> -- jt]]
This reminds me of problems in which ferromagnetic materials
are used to build electromagnets. Consider the disk in the
Faraday_paradox above, made of iron. It is even easier to form
paradoxes in this case.
-- glen
Oh No
Of course they do recoil, but because the force is applied over a much
longer time period than an explosive it is less.
Regards
--
Charles Francis
moderator sci.physics.foundations.
charles (dot) e (dot) h (dot) francis (at) googlemail.com (remove spaces and
braces)
David M. Palmer
<Robert....@nasa.gov> wrote:
> In a paper from the Naval Postgraduate School an attempt to measure
> recoil forces in a rail gun failed to find any within experimental
> error. In particular note what occurred when the paper's author clamped
> the projectile to the rails of the device.
>
> http://stinet.dtic.mil/cgi-bin/GetTRDoc?AD=ADA473387&Location=U2&doc=Get
> TRDoc.pdf
As I wrote when the subject came up earlier, a couple of decades ago:
<http://groups.google.com/group/sci.physics/msg/4c1be3d2451b9687>
the force on the rails themselves will be outwards, perpendicular to
the rails. If the rails are held rigidly with respect to each other, so
that they cannot fly apart, the pair of rails feels no net force.
The equal and opposite force to that which drives the projectile
forward is applied to the current carrying components that go from rail
to rail.
In the thesis experiment, most of the circuit that would be expected to
feel the recoil force was tied down to the table, so the force on it
would not be observed. However, the liquid contacts could move, and
in fact they did.
FIgure 35 of the thesis shows that the liquid splattered outward, which
is the direction you would expect based on orthodox E&M. This the
author blamed on arcing.
They "were unable to continue the test for two or more seconds due to
the erratic behavior of the fluid. While the results were somewhat
satisfactory, until we were confident that we would be able to predict
the behavior of the fluid we did not feel as though the design had the
desired characteristics of being repeatable, predictable, and safe. "
See my sci.physics post, linked to above, for an ASCII diagram and more
details.
> There is very little room for skepticism about the paper. Large scale
> tests performed by the US Navy of a prototype rail gun involved a 3.35
> Kg projectile with a muzzle velocity of 2520 meters/sec. This gives a
> momentum in excess of 8000 Kg-meters/sec, enough to send a 200 Kg rail
> gun backward at over 40 meters per second. A conventional gun with
> similar performance would require a massive and extensive recoil
> absorption apparatus. There is none needed with a rail gun.
>
> The attached photo shows a shot of the Naval prototype rail gun firing.
> Note the thin pedestal mounts used to hold the rail gun apparatus.
>
> http://en.wikipedia.org/wiki/Image:Railgun_usnavy_2008.jpg
>
> Other pictures of prototype high performance rail guns at a development
> facility show the rail gun sitting on wheeled carts while firing.
>
> The lack of recoil in rail guns has disastrous consequences for physics;
> it is a direct and unequivocal demonstration that the law of
> conservation of momentum is incorrect.
>
> I would like to make one prediction about rail guns: It will be found
> that there is also no back e.m.f. generated by the devices; they not
> only show conservation of momentum to be untrue, they violate
> conservation of energy also.
--
David M. Palmer dmpa...@email.com (formerly @clark.net, @ematic.com)
Richard D. Saam
Energy and Momentum are conserved in the total system.
The other part of the system not addressed is the massive mechanical
rotating homopolar rotor generator below decks.
Specification from some of the supporting work
done at the University of Texas about 15 years ago:
Maximum energy storage capacity 6.7 megajoule
Peak discharge current 1,000,000 amperes
Maximum open circuit voltage 50 volts DC
Maximum rotor speed 6,500 RPM
Effective capacity 5,360 farads
Maximum slip ring speed 229 m/sec
Approximate dimensions 84 cm diameter x 64 cm long
Approximate weight 1,600 kg
motoring time to maximum rotor speed 2 minutes
cycle rate Full discharge every 5 minutes.
Discharge torque 51,000 N-m (60,000 ft-lb)
Contact is made and the mechanical rotating energy/momentum is converted
into projectile energy/momentum via electrical current with some waste heat.
A very large system but no new physics here.
Richard D. Saam
Tom Roberts
> The attached photo shows a shot of the Naval prototype rail gun firing.
> Note the thin pedestal mounts used to hold the rail gun apparatus.
> http://en.wikipedia.org/wiki/Image:Railgun_usnavy_2008.jpg
I'm not going to read that 93-page thesis. But I note that if one
instead fired a rocket along the rails of a railgun, one would not
expect to measure any significant reaction force on the rails. That
picture looks indistinguishable from a rocket doing just that.
So one must ask: how much momentum is contained in all that recoiling
fire and gas? How much is in the high-power electromagnetic radiation
involved? And where does all that fire and smoke come from?
Tom Roberts
Dirk Bruere at NeoPax
> In article <mt2.1-19645...@argon.astro.indiana.edu>, JSC-EV
> <Robert....@nasa.gov> wrote:
>
>> In a paper from the Naval Postgraduate School an attempt to measure
>> recoil forces in a rail gun failed to find any within experimental
>> error. In particular note what occurred when the paper's author clamped
>> the projectile to the rails of the device.
>>
>> http://stinet.dtic.mil/cgi-bin/GetTRDoc?AD=ADA473387&Location=U2&doc=Get
>> TRDoc.pdf
>
> As I wrote when the subject came up earlier, a couple of decades ago:
> <http://groups.google.com/group/sci.physics/msg/4c1be3d2451b9687>
> the force on the rails themselves will be outwards, perpendicular to
> the rails. If the rails are held rigidly with respect to each other, so
> that they cannot fly apart, the pair of rails feels no net force.
>
In which case conservation of momentum still fails.
--
Dirk
http://www.transcendence.me.uk/ - Transcendence UK
http://www.theconsensus.org/ - A UK political party
J. J. Lodder
That's great.
Catch the projectile, convert its kinetic energy to electricity,
and you have invented the perpetuum mobile.
Since stopping the projectile by conventional means
should satisfy momentum conservation
you will also have invented the reactionless space propulsion.
You have three guesses about this being right,
Jan
Dirk Bruere at NeoPax
The whole thesis is pretty easy to test.
You don't need a vast megajoule railgun. Something much smaller that
doesn't generate fire and smoke will suffice. Put the whole arrangement
on a trolley and shoot a 1kg weight at around 5m/s. Any recoil will be
very apparent.
David M. Palmer
Bruere at NeoPax <dirk....@gmail.com> wrote:
> David M. Palmer wrote:
> > As I wrote when the subject came up earlier, a couple of decades ago:
> > <http://groups.google.com/group/sci.physics/msg/4c1be3d2451b9687>
> > the force on the rails themselves will be outwards, perpendicular to
> > the rails. If the rails are held rigidly with respect to each other, so
> > that they cannot fly apart, the pair of rails feels no net force.
> >
>
> In which case conservation of momentum still fails.
No it doesn't. The force on the projectile is balanced by the equal
and opposite force on the rest of the circuit, the part that completes
the circuit between the rails.
That part, including the batteries, switches and cables, is bolted to
the table and is not attached to the force-measuring equipment.
Rich L.
I'm afraid the authors of that study did not understand how the recoil
force would be applied to the gun. They correctly calculate the
armature force, but there is an equal and opposite force at the
"breach" of the gun where the power supply connects. By clamping
their liquid metal power supply pool to the table, they effectively
transfered this recoil force directly to the table, and presumably the
ground.
Note in Figure 35 the "splatter" of the liquid metal. This is the
recoil force (only part of it actually) that the liquid metal was
responding to. This experiment would have worked out better if they
had suspended the entire power supply, including the car batteries, on
the pendulum. If they had done this, they would have found no
disagreement with conservation of energy or momentum.
Rich L.
Dirk Bruere at NeoPax
Well, let me summarise my beliefs on the matter.
The projectile leaves the railgun with +mv of momentum. The rails and
conductors recoil with -mv of momentum. It all balances out correctly
according to standard Newtonian physics.
--
Dirk
http://www.transcendence.me.uk/ - Transcendence UK
http://www.theconsensus.org/ - A UK political party
http://www.onetribe.me.uk/wordpress/?cat=5 - Our podcasts on weird stuff
Neil B.
news:mt2.1-11417...@argon.astro.indiana.edu...
carried in the "exhaust" analog of the rail gun: electromagnetic waves.
After all, that's the point of Feynman's charged disk paradox (from
_Lectures_ Vol. II). But I don't think there's enough in EM waves to
carry off the many kg*m/s shot out by a rail gun (Anymore than in the
case of me pressing two magnets together and then letting them go, etc.)
It takes a lot of EM power to have significant (linear) momentum. Note
that we need about 300 MW of power from a photon rocket to get a mere
one newton of thrust!
Also of course, statements about momentum being directed out to the
sides don't solve anything - the conservation law is a vector identity!
It seems to me the rail magnets, in near proximity to the projectile,
should have nearly complete reaction forces right against them instead
of needing to appeal to equipment below. So this empirical result
(correct I suppose?) is still perplexing.
Actually, I am skeptical of the claim of negligible reaction momentum
since I remember seeing proposals to use the reaction from rail guns to
apply thrust to asteroids! (Mine ferrous slugs for projectiles and maybe
nuclear plants for the power, make the asteroid into a "spaceship" etc.)
OTOH, that was likely a theoretical calculation, perhaps simplistically
assuming action-reaction: what if actual results are contradictory?
There has been a paradox going around about classical derivation of spin
from forces between neighboring charges, it really is weird. See for
example about the curious work and claims of Anders O. Wistrom and V. M.
Khachatourian, at the link
http://www.aip.org/tip/INPHFA/vol-9/iss-4/p8.html from _The Industrial
Physicist_, August/September 2003 (Volume 9, Issue 4):
"Spin and energy�free?
Most physicists would not expect startling new theoretical conclusions
to emerge from electrostatics, whose basic mathematical structure was
completed 150 years ago. Yet two researchers at the University of
California, Riverside, arrived at conclusions that, if true, would be
revolutionary. In a forthcoming paper ( J. Physics A: Math. Gen. 2003,
36, 6495), Anders O. Wistrom and V. M. Khachatourian say they have
proven mathematically that electrostatic forces among three charged,
perfectly conducting spheres will cause them to start spinning. This
conclusion, which the authors have derived from Coulomb�s law,
contradicts long-held assumptions about how electrical fields behave.
Equally striking, it implies that, in theory, an arrangement of three
such spheres could transfer unlimited amounts of energy into the
spinning spheres�a violation of conservation of energy."
See also commentary and responses at
http://www.aip.org/tip/INPHFA/vol-9/iss-6/p4.html.
I'm surprised not to hear more about this work anymore - anyone have
scoop?
(Note about rocket thrust: Take f = v(dm/dt) where dm/dt is the rate of
mass exiting the rocket. That's how we can derive the thrust from a
photon rocket, since we have f = c(dm/dt) equivalent = (dE/dt)/c.
However, there is a pseudo-paradox with rockets, since technically the
momentum of the whole rocket itself seemingly should have a term dp/dt =
v(dm/dt) where the same term refers to the changing "mass of the
rocket." However this is spurious because it is a changing redefinition
of the mass being considered; for any one particle the usual f = ma =
dp/dt still holds. (Yet many students and even text-book writers still
get confused.)
Richard D. Saam
m*v^2 = V*I*d/v
Putting some approximate numbers to it based on previously posted
homopolar generator.
projectile mass(m) = 10 kg
Voltage(V) = 20 volt
Amperage(I) = 1,000,000 amp
Rail launch distance(d)= 10 meter
projectile velocity(v) = 271 m/sec
A little low but the idea is presented
It is assumed that switch, cable, contact etc are impedance matched for
delivering maximum energy to the projectile.
Perhaps a fluid system would be analogous consisting of a pump, conduit
and fluid:
-------------------
|*****************| F
|*****************| |
|*****************| conduit V
|**PUMP***********--------------------
|*********************************************************** ->
|*****************--------------------
|*****************| ^
|*****************| |
|*****************| F
-------------------
Assuming zero fluid frictional losses in the conduit there are no back
reaction forces to the conduit in transmitting the fluid.
pump pressure(P) is converted into fluid velocity(v) P = v^2/(2g)
There would be orthogonal forces on conduit as the fluid accelerates or
decelerates.
Could this be likened to the projectile not imparting a back force to
the two rails which would never the less have a tremendous force between
them during projectile launch.
In the real world, minor losses (electrical contact losses) would result
in measureable back reaction but much less than projectile mv.
System momentum and energy are maintained.
Richard D. Saam
Oh No
>Also of course, statements about momentum being directed out to the
>sides don't solve anything - the conservation law is a vector identity!
>It seems to me the rail magnets, in near proximity to the projectile,
>should have nearly complete reaction forces right against them instead
>of needing to appeal to equipment below.
I am sure this is right.
> So this empirical result (correct I suppose?) is still perplexing.
The original statement confused different ideas. The comment was that
the rails do not need anywhere near the strength of material required to
control the kickback of an explosive weapon. But that is natural. In
rough terms, the peak force applied in an impulse is inversely
proportional to the duration over which the impulse is applied. I should
think there is a difference in duration for applying the accelerating
force, or impulse, of two orders of magnitude or more compared to an
explosive weapon. That means there can be just as much reaction
momentum, but it might require less than one percent of the force to
control it, hence one percent of the strength of material.
Regards
--
Charles Francis
moderator sci.physics.foundations.
charles (dot) e (dot) h (dot) francis (at) googlemail.com (remove spaces and
braces)
David M. Palmer
will talk about the Wistrom & Kachatourin paradox.
In article <_t6dncyYGK3lEqfU...@posted.widowmaker>, Neil
B. <neil_...@caloricmail.com> wrote:
> "Tom Roberts" <tjrobe...@sbcglobal.net> wrote in message
> news:mt2.1-11417...@argon.astro.indiana.edu...
> > So one must ask: how much momentum is contained in all that recoiling
> > fire and gas? How much is in the high-power electromagnetic radiation
> > involved? And where does all that fire and smoke come from?
> >
> Tom, I'm glad someone finally brought up the issue of momentum being
> carried in the "exhaust" analog of the rail gun: electromagnetic waves.
> After all, that's the point of Feynman's charged disk paradox (from
> _Lectures_ Vol. II). But I don't think there's enough in EM waves to
> carry off the many kg*m/s shot out by a rail gun (Anymore than in the
> case of me pressing two magnets together and then letting them go, etc.)
> It takes a lot of EM power to have significant (linear) momentum. Note
> that we need about 300 MW of power from a photon rocket to get a mere
> one newton of thrust!
A good analog is this:
The lower the mass, the higher the energy required to get the same
momentum. For massless photons it requires a lot of energy to get a
little momentum.
Consider a gun barrel without a breech block, just a tube open at both
ends. Put a bullet and a small amount of gunpowder in it and set it
off. The burnt gunpowder gas and smoke flies out the back and the
bullet goes out the front. If the mass of the gunpowder is much less
than the bullet, most of the energy goes into the kinetic energy of the
gunpowder, and not much into the bullet.
Now add a breech block to close the tube at one end. This time, the
burnt gunpowder is trapped in the tube and presses against breech and
the bullet in opposite directions. The total momentum at the end is
again zero. The kinetic energy in this case is mostly in the bullet, a
little in the more massive breechblock plus the barrel (assuming that
they are joined together) and only a little in the gunpowder, which
doesn't have much of the momentum. This is a much more effective gun
than the open tube.
The open tube case is the photon-rocket equivalent.
The breech block case is the railgun, where the breech block analog is
the conductors, power source, etc. that go between rails. This
circuitry traps the magnetic field and prevents it from blowing out the
back of the railgun.
And that's why you don't need 300 MW to apply 1 Newton to the
projectile.
The burnt powder applies no direct force on the barrel in each case.
However, the breech block in a practical gun is firmly attached to the
barrel, and it pulls back on it so that the whole breech block and
barrel recoil as a unit. If they were not attached, the breech block
would fly backwards, but the barrel would remain stationary.
Likewise, in the case of the thesis experiment where the rails are not
attached to the 'breech block' they have no net force.
A railgun does not violate conservation of momentum. A well-designed
railgun does not balance the momentum by putting significant momentum
into a radiated EM field.
--
Dirk Bruere at NeoPax
> Thus spake Neil B. <neil_...@caloricmail.com>
>> Also of course, statements about momentum being directed out to the
>> sides don't solve anything - the conservation law is a vector identity!
>> It seems to me the rail magnets, in near proximity to the projectile,
>> should have nearly complete reaction forces right against them instead
>> of needing to appeal to equipment below.
>
> I am sure this is right.
>
>> So this empirical result (correct I suppose?) is still perplexing.
>
> The original statement confused different ideas. The comment was that
> the rails do not need anywhere near the strength of material required to
> control the kickback of an explosive weapon. But that is natural. In
> rough terms, the peak force applied in an impulse is inversely
> proportional to the duration over which the impulse is applied. I should
> think there is a difference in duration for applying the accelerating
> force, or impulse, of two orders of magnitude or more compared to an
> explosive weapon. That means there can be just as much reaction
> momentum, but it might require less than one percent of the force to
> control it, hence one percent of the strength of material.
This cannot be true.
A projectile launched by gas pressure is accelerating all the way down
the tube. This is also the case with a railgun projectile. For the same
length and same final momentum the forces are going to be approximately
the same, spread over the same duration.
Glen Herrmannsfeldt
> Oh No wrote:
(snip)
>> The original statement confused different ideas. The comment was that
>> the rails do not need anywhere near the strength of material required to
>> control the kickback of an explosive weapon. But that is natural. In
>> rough terms, the peak force applied in an impulse is inversely
>> proportional to the duration over which the impulse is applied. I should
>> think there is a difference in duration for applying the accelerating
>> force, or impulse, of two orders of magnitude or more compared to an
>> explosive weapon. That means there can be just as much reaction
>> momentum, but it might require less than one percent of the force to
>> control it, hence one percent of the strength of material.
> This cannot be true.
> A projectile launched by gas pressure is accelerating all the way down
> the tube. This is also the case with a railgun projectile. For the same
> length and same final momentum the forces are going to be approximately
> the same, spread over the same duration.
Assuming the comparison is to an ordinary rifle or handgun,
it is certainly possible to design a system where the explosive
material burns at the appropriate speed to keep a constant pressure
in the barrel through the travel time of the bullet, but I don't
know that is usually done.
For larger guns it would seem more important and more likely
that it is done that way.
http://community.seattletimes.nwsource.com/archive/?date=19900412&slug=1066024
Is a story from some years ago about Iraq trying to build the
worlds longest gun. It seems 131 feet and 40 inch bore.
To make that work, it would seem you would want the pressure
somewhat constant during the time the payload is in the
barrel.
Otherwise, it seems that traditional cannon barrels have
varying wall thickness, presumably because the pressure isn't
constant.
Oh No
acceleration at the first moment when they are fired. The peak force
will be orders of magnitude greater than a rail gun which applies
acceleration uniformly.
John J
> Assuming the comparison is to an ordinary rifle or handgun,
> it is certainly possible to design a system where the explosive
> material burns at the appropriate speed to keep a constant pressure
> in the barrel through the travel time of the bullet, but I don't
> know that is usually done.
I've never known of such a weapon. There are devices that carry a
propellant inside the bullet so that the bullet is propelled outside the
barrel IOW, the bullet is a rocket and the barrel an alignment device.
> For larger guns it would seem more important and more likely
> that it is done that way.
>
> http://community.seattletimes.nwsource.com/archive/?date=19900412&slug=1066024
No such cannon exists that has constant pressure.
> Is a story from some years ago about Iraq trying to build the
> worlds longest gun. It seems 131 feet and 40 inch bore.
> To make that work, it would seem you would want the pressure
> somewhat constant during the time the payload is in the
> barrel.
Very large cannons use a low-velocity, high expansion fuel (charge)
similar to black powder and 'Pyrodex' fuels. And they recoil, of course.
> Otherwise, it seems that traditional cannon barrels have
> varying wall thickness, presumably because the pressure isn't
> constant.
Conventional cannon barrels have not had significantly varying wall
thickness since the old iron days, and in that case it was done because
casting processes and the making of iron was crude, not well known, and
the iron barrels were fragile. The cannons one sees from, for example,
the 1600s are those that did not explode, or were never used.
Back to the rail gun - speaking of the device where the magnetically
susceptible projectile is FAPP suspended above a rail in which
electrical pulses along the rail are used to forward the projectile -
the force is efficiently moved from the rail's emission to the
projectile so that the projectile behaves more like a rocket in an
open-ended tube - whereas a conventional rifle acts like the rocket
itself with the projectile being part of the overall charge - it is
expelled by the rifle.
See recoil-less weapons - they are open at each end. The projectile is a
rocket.
John J
> The breech block case is the railgun, where the breech block analog is
> the conductors, power source, etc. that go between rails. This
> circuitry traps the magnetic field and prevents it from blowing out the
> back of the railgun.
A rail gun does not require a barrel. A round barrel is often used
because it is a coil and allows a shorter device than a pair of rails of
the same length as the coiled medium.
Force is exerted upon the rails. A pair of rails would move dramatically
if not secured properly. They would move sideways and slightly to the
opposite direction of the projectile. Keep in mind that the projectile
is impelled at intervals (via pulses, change or creation of polarity)
along the rails. The 'pressure' against the projectile is relatively
constant, and always greater than the inertial resistance of the
projectile, hence acceleration.
Dirk Bruere at NeoPax
>>
> Explosive weapons and even airguns will apply almost all the
> acceleration at the first moment when they are fired. The peak force
> will be orders of magnitude greater than a rail gun which applies
> acceleration uniformly.
Are claiming that for most guns the projectile ceases to accelerate some
considerable distance before it exits the barrel? In which case, why
have the extra (redundant) length of barrel? A reasonable approximation
is that acceleration is inversely proportional to the distance of the
bullet from the breech, and is zero when the gas pressure (which falls
linearly proportional to length) can no longer accelerate the bullet
against the barrel friction. At which point the bullet begins to lose
velocity.
In reality, sawing off 50% of the length of a rifle barrel will severely
impact the exit velocity of the bullet.
Dirk Bruere at NeoPax
>> Explosive weapons and even airguns will apply almost all the
>> acceleration at the first moment when they are fired. The peak force
>> will be orders of magnitude greater than a rail gun which applies
>> acceleration uniformly.
>
> Are claiming that for most guns the projectile ceases to accelerate some
> considerable distance before it exits the barrel? In which case, why
> have the extra (redundant) length of barrel? A reasonable approximation
> is that acceleration is inversely proportional to the distance of the
> bullet from the breech, and is zero when the gas pressure (which falls
> linearly proportional to length) can no longer accelerate the bullet
> against the barrel friction. At which point the bullet begins to lose
> velocity.
>
> In reality, sawing off 50% of the length of a rifle barrel will severely
> impact the exit velocity of the bullet.
>
I should add that quite often there is still propellant burning as the
bullet exits the muzzle, so the approximation of assuming it all turns
to gas before the bullet starts to move is overly simplistic.
John J
>> Explosive weapons and even airguns will apply almost all the
>> acceleration at the first moment when they are fired. The peak force
>> will be orders of magnitude greater than a rail gun which applies
>> acceleration uniformly.
>
> Are claiming that for most guns the projectile ceases to accelerate some
> considerable distance before it exits the barrel?
That depends upon the charge behind the bullet, but I must say that I
know of very few weapons that do not maintain acceleration in the
barrel, however many have barrels longer than necessary - the extra
length of the barrel serves only two purposes - to enhance sighting
along the barrel and to be legal. Most modern shotguns have very little
ballistic advantages with a barrel longer than 12 inches, but that's too
short to satisfy legalities, and (as mentioned) such short barrels have
major ergonomic/sighting deficiencies.
> In which case, why
> have the extra (redundant) length of barrel?
To be legal and to assist in balance of the weapon and in sighting it.
> [...]
> In reality, sawing off 50% of the length of a rifle barrel will severely
> impact the exit velocity of the bullet.
If we take the nominal/norm barrel to be, say, 30" then that is not true
at all with any consumer firearm. I mean, anything less than the 50
calibre Browning (machine gun) round.
Dirk Bruere at NeoPax
>> [...]
>> In reality, sawing off 50% of the length of a rifle barrel will
>> severely impact the exit velocity of the bullet.
>
> If we take the nominal/norm barrel to be, say, 30" then that is not true
> at all with any consumer firearm. I mean, anything less than the 50
> calibre Browning (machine gun) round.
Barrel lengths seldom exceed 24". More common is 20" and at a push 16"
for compact assault rifles.
Richard D. Saam
>> Explosive weapons and even airguns will apply almost all the
>> acceleration at the first moment when they are fired. The peak force
>> will be orders of magnitude greater than a rail gun which applies
>> acceleration uniformly.
>
> Are claiming that for most guns the projectile ceases to accelerate some
> considerable distance before it exits the barrel? In which case, why
> have the extra (redundant) length of barrel? A reasonable approximation
> is that acceleration is inversely proportional to the distance of the
> bullet from the breech, and is zero when the gas pressure (which falls
> linearly proportional to length) can no longer accelerate the bullet
> against the barrel friction. At which point the bullet begins to lose
> velocity.
>
> In reality, sawing off 50% of the length of a rifle barrel will severely
> impact the exit velocity of the bullet.
>
the reaction force at the gun barrel surface
adjacent to the traveling projectile
is negligible (projectile-barrel frictional forces)
and is in the same direction as the traveling projectile.
The primary gun recoil force is:
Force = Gun barrel gas pressure/Gun barrel breech Area pi D^2 / 4
where D is gun barrel breech internal diameter.
In the case of the rail gun, the recoil
is at the energy storage capacitor, flywheel or battery.
Richard D. Saam
Dirk Bruere at NeoPax
>
> In the case of the rail gun, the recoil
> is at the energy storage capacitor, flywheel or battery.
>
> Richard D. Saam
Let me get this right.
You are claiming that if the power supply for a railgun is (for
arguments sake) a couple of miles away from the gun itself then when the
railgun is fired the whole generator (a couple of miles away at the end
of a cable) recoils at 180deg to the direction the gun is pointing? That
somehow the generator knows what direction the gun fired in and kicks
accordingly?
John J
> I should add that quite often there is still propellant burning as the
> bullet exits the muzzle, so the approximation of assuming it all turns
> to gas before the bullet starts to move is overly simplistic.
Indeed, you are correct.
Presuming inner-barrel pressure is null (or equal to atmospheric
pressure) is simply incorrect. The only percussion (sound) one would
experience is that of the projectile moving through the air.
All the propellant does not turn to gas before the projectile begins to
move. Ballistics people call that an explosion, or detonation, and it
usually results in a catastrophic failure of the firearm. The propellant
burns as the projectile moves and sometimes for the entire length of the
barrel. (Muzzle-flash is name of the visual effect.)
Richard D. Saam
>>
>> In the case of the rail gun, the recoil
>> is at the energy storage capacitor, flywheel or battery.
>>
>> Richard D. Saam
>
> Let me get this right.
> You are claiming that if the power supply for a railgun is (for
> arguments sake) a couple of miles away from the gun itself then when the
> railgun is fired the whole generator (a couple of miles away at the end
> of a cable) recoils at 180deg to the direction the gun is pointing? That
> somehow the generator knows what direction the gun fired in and kicks
> accordingly?
>
I am saying that for example
if the power source for the rail gun in Phoenix, Arizona
is a generator with coupled hydroturbine at the Hoover dam
then at the time of projectile launch
the Hoover dam generator hydroturbine
will have a corresponding change in momentum (m*v/t).
The impulse m*v/t or Q*rho*v/t is at the penstock turbine interface
affecting all the other power requirements
such as someone in Phoenix
using their computer of turning on the lights.
Q is water flowrate (cubic feet/second)
rho is water density
t is water resident time in the turbine.
v is change in water velocity at the turbine.
Richard D. Saam
Dirk Bruere at NeoPax
As here :-)
http://beard.freaksho.net/data/etc/people/flash.jpg
Which is one reason people do not saw too much off of their rifle
barrels, and why ammunition has to be tailored to the weapon (or vice
versa).
Dirk Bruere at NeoPax
I can quite see how the energy stored as angular momentum in the
generator rotor will change, and hence the angular momentum of the rotor
will change, when the gun is fired.
However, the recoil of the slug fired from the railgum has *linear*
momentum, and I do not see how that is transferred over wires to be
absorbed miles away.
Tom Roberts
> the recoil of the slug fired from the railgum has *linear*
> momentum, and I do not see how that is transferred over wires to be
> absorbed miles away.
Throughout this discussion the physical configuration has been
imprecisely specified, leading to confusion. And different contributors
have used different configurations.
A railgun is topologically an electric circuit in the shape of a "U",
with a movable crossbar sliding "upward" along the parallel rails (upper
2/3 of the letter "U"); there is a power generator somewhere in the
circuit "below" the place where the crossbar starts.
All relationships in "quotes" apply to the "U", and must
be re-oriented to apply to any real railgun.
Regardless of where the generator is located, the momentum recoil will
apply to the conductor at the bottom of the "U", balancing the linear
momentum carried off by the projectile sliding "upward" along the rails.
This is precisely the same mechanism that tries to force the rails apart
(except they are rigidly mounted to prevent that). If you put the
generator at the "bottom" of the "U", then the generator must absorb
this momentum. If you extend the bottom of the "U" with wires to a
generator far away, then the projection of those wires and generator
onto the "bottom" of the "U" must absorb the momentum. Momentum is a
vector quantity, and in this case it is headed "downward", so the
"up/down" portion of wires receive no momentum kick, but all
"left/right" portions do; all wires have large forces on them, so all
but the crossbar must be rigidly mounted to maximize the energy
transferred to the projectile/crossbar -- this makes the earth absorb
the recoil.
I'm ignoring any momentum carried by the EM fields,
and any momentum from the enormous clouds of smoke and
fire shown in the picture referenced earlier in this
thread (which was indistinguishable from a rocket).
Contrary to the Subject line, railguns DO recoil. But existing ones
transfer it to the earth where it is not noticeable.
Tom Roberts
Dirk Bruere at NeoPax
> Contrary to the Subject line, railguns DO recoil. But existing ones
> transfer it to the earth where it is not noticeable.
As do most field guns
http://www.military.ie/army/equipment/weapons/arty/25pdr/25pound.htm
Richard D. Saam
A dimensional correction from above: Q*rho*v/g versus Q*rho*v/t = m*v/t
The impulse F = m*v/t or Q*rho*v/g is at the penstock turbine interface
affecting all system power requirements
such as someone in Phoenix
using their computer of turning on the lights
or operating a rail gun.
Excuse a relapse into English dimensional units
Q is water flowrate ft^3 /sec
rho is water specific density 62.4 lb/ft^3
g is gravity acceleration 32.2 ft/sec^2
t is water resident time in the turbine sec
v is change in water velocity at the turbine ft/sec
In answer to your question,
the water traveling at the Hoover dam penstock has linear momentum
as it hits the turbine which is coupled to the generator rotor.
Linear momentum is changed into rotational momentum
and by work energy principle Fds
a electromagnetic pressure ~(E^2 + B^2)
is produced in the Hoover Power system electrical grid.
What is flowrate part 'Q' of the Hoover dam total flowrate is required
to power a 1 Horsepower motor in Phoenix
assuming the water velocity change in the turbine is 20 ft/sec
Fluid power = Q * rho * v^2 / g
1 HP = 550 ft lb / sec = Q rho v^2 / g
Q = .7 ft^3 /sec
Now carry this water analogy a bit further.
Put yourself at the end of a fire hose with no nozzle.
Their would be very little 'back pressure' on you
as you held the hose as water streamed out the end.
Now let another person place a rock in the streaming water at some
distance from the end of the hose. Momentum would be transfered from
the streaming water to the rock and the rock would move in the direction
of the streaming water. There would be no additional back reaction to
you holding the hose as the rock is placed in the streaming water.
Maybe something like this is happening in the rail gun system
wherein the projectile momentum is only partially projected into the
launch rails(the major portion of momentum exists in the E & B fields
traceable back to the generator).
Richard D. Saam