DREAD weapon system, power requirements

  • Thread starter Thread starter SK
  • Start date Start date
  • Tags Tags
    Power System
Click For Summary
The discussion centers on the DREAD weapon system, a centrifuge-based design that has raised skepticism due to its unrealistic assumptions, such as recoillessness. Calculations indicate that to achieve a bullet velocity of 800 m/s, the system requires an average power output of 0.81 Megawatts, which is considered excessively high. Concerns are raised about the mechanical stresses involved, particularly the extreme accelerations of 21,000 gravities, which may render the design impractical. The feasibility of sustained fire is questioned, as it would necessitate powerful motors and robust materials capable of handling significant forces. Overall, while the concept is intriguing, its practical implementation faces substantial engineering challenges.
  • #31
Antiphon said:
Don't confuse physics with engineering.

When he says no recoil, he means no appreciable net recoil on the
user/vehicle. He doesn't mean he's violating conservation of momentum.

You can envision the simultaneous release of a forward ball from the right
side with a backward ball from the left. No net recoil at the time of release.

(Of course you have to figure out how not to frag your buddy who's behind
you, but of course this IS in the realm of engineering and not physics.)

Nah. An engineer might say something like "It wouold have (mumble,mumble) the recoil of a conventional weapon firing at the same rate, because there are no exhaust gasses to contribute to the recoil.

To make statements like "it has no recoil" requires a salesman without any technical knowledge of his own. Usually said salesman is trying to sell something (that's what salesmen do). He may have the engineers try and design the totally unrelistic/impossible gadget he sold if/when he ever cons someone out of their money to actually buy the non-existant product.
 
Engineering news on Phys.org
  • #32
Antiphon said:
Don't confuse physics with engineering.

When he says no recoil, he means no appreciable net recoil on the
user/vehicle. He doesn't mean he's violating conservation of momentum.
Well it seems the author of the article confuses wishful thinking with engineering, like pervect said. I´d accept the "no appreciable net recoil on the vehicle" part (at least for slow ROFs), but they also write:
5) Satellite Defense/Space Defense Systems. Since the centrifuge weapon system doesn't generate any recoil impulse or gyroscopic effect, it can be fired from a satellite at various types of targets without throwing the satellite off it's orbital path. Theoretically, the centrifuge weapon system could be mounted on and employed by space vehicles as well, for the same purpose.
http://www.defensereview.com/modules.php?name=News&file=article&sid=678
Of course you can cancel recoil with firing two projectiles in exactly opposing directions. But that´s possible with any type of mass accelerator, be it DREAD, a chemical energy gun, railgun or whatever.


You can envision the simultaneous release of a forward ball from the right
side with a backward ball from the left. No net recoil at the time of release.
Sure you can, but neither illustrations nor patent show anything like that. It might would be possible to release a countermass that expands rapidly after release, thereby increasing its drag and generating a backblast zone behind the weapon of some (10+) meters. That´s the way some handheld AT weapons work.
Then again, how complicated is this thing going to be? So far we need a 1000+ HP engine, extremely expensive materials and as soon as there´s a malfunction you can write off the surrounding people and machinery. All this for accelerating some balls to fly a unaccurate trajectory with very low range compared to current military ammunitions.
 
Last edited by a moderator:
  • #33
Hello,
First I have to tell that I am french, and I don't speak english well. So it will be difficult to me to explain what I mean. Anyway, I read this subject with great interest, and I agree on all points to tell that this weapon cannot work : gyroscopic effect, energy needed, etc...
But I don't agree on the recoil point : If a conventional weapon have recoil effect, it is because there is a reaction from the acceleration of the bullet with two forces opposed. In the case of the dread system, why would it be a reaction, when the balls already go at constant speed ? imagine a plane flying at 300 m/s on a circular path, if it carries a bomb and throw it, the bomb will fly straight at 300 m/s, but there will be no recoil applied to the plane, no ?
I am not a mathematician, so thank you for your clear and simple explanation.
 
Last edited:
  • #34
Knowing that momentum is conserved, we can confidently say that anything that shoots bullets out in only one direction must recoil, even without working out the forces in detail.
 
  • #35
OK, but my (our?) goal here is to understand why it is or it is not...
What about my image of the plane ? do you agree at least that there is a big difference in the strenght of the reaction when you SHOOT a Ball and when you just LET it FALL down from your hand ?
 
  • #36
When you unbalance a rapidly spinning object, it's going to be a lot different than letting something fall from your hand.

Part of the problem constructing the device will be to deal with the resulting forces on an unbalanced rapidly rotating object. With strong enough materials and special construction it's theoretically possible, though I still don't know if there are any actual physical materials strong enough to allow the device as specified to be built. One way of constructing the device (at least in principle) is found by copying the desing of the household washing machine - the motor and rotor are connected together rigidly, but the whole assembly is suspended from cables, allowing them to move. Without some such special arrangement, the connection between the motor and rotor will break.

An analysis of the forces is still a complicated job. For starters,c onsider the following digram



left---------------right^


With "up" towards the top of the page, and "down" towards the bottom of the page, consider the sense of the rotation such that the right side is moving "up".

The clearest way to analyze the problem is to consider a non-rotating frame of reference, IMO - that has the added advantage of being the standard approach.

We can then ask the quesiton, with no other forces applied, what is the resulting motion when we detach a piece of the rotor on the right side?

The answer is that the detached piece will continue to move upwards. When we sum the total momentum of the rotating piece in the up down direction, we see that the rotating piece moves downwards. This momentum must remain conserved. The general motion of any rigid body will be a sum of rotation around it's center of mass, plus translation. So the motion of the main rotor will be that it will continue to rotate around its new center of mass (which will be displaced from the old center of mass), but it will also move downwards.

[add]
Note that the new center of mass was already moving downwards to start with, because it is located to the left of the original center of mass. It's original downwards velocity was w*dr, where dr is the amount the center of mass shifts to the left.

The shift of the location of the center of mass of the rotor is the key to solving the problem and getting the correct answer.

Anyway, that's the clearest way I can think of to describe the solution (it's a bit difficult with only limited diagrams and ascii text).

I've also simplified the problem somewhat - a real system would not only have a rotor, but a motor of non-negilgible mass to consider. I've omitted the motor in the simplified analysis.

It may help to draw a digram like the one above, with the velocities of all the parts shown as arrows. (I won't attempt to do this in ascii). You can then see visually how the rotational motion decomposes on the main rotor to the sum of a new rotational motion plus a translational motion downwards - you add up the arrows for the rotational motion around the new axis of rotation, plus the arrows representing the linear motion, and see that they add up to the original velocity vectors of the rotating bar.
 
Last edited:
  • #37
Thank you for these details : I understand better the problem and I agree with you.
May be this device could work, but only at a very low speed, and of course the balls would not have enough power compare to a real weapon. I wonder if they have already build one, and I'd be interested to ear about their conclusions...
 
  • #38
Another interesting point that came out of the last analysis is that the device will be throwing "curve balls". The original projectiles had angular momentum, which they will retain when they are released. Unfortunately the spin axis is oriented vertically, perpendicular to the direction of flight, which isn't a very desirable sort of spin. This is the same sort of spin that pitchers put on a baseball to make it curve, which is why I say that the device will throw curveballs...
 
  • #39
I think that the "inventor" hope to obtain some publicity, and may be some money from army research officies. For the first point, he won, but for the second I am not sure he will get anything !
 
  • #40
Actually, they just recently claimed that a large U.S. "defense entity" signed up with "Leader Propulsion Systems" to fully develop the DREAD ( http://www.defensereview.com/modules.php?name=News&file=article&sid=727 ). I take this message with a huge grain of salt.
Interestingly enough they also say:
This strategic teaming will also enable Leader to begin developing a series of gas-operated lightweight weapons chambered for the .50 BMG, 338 Lapua, 300 Winchester Magnum and .308 Nato. According to Leader, this new development program will result in a family of weapons that will utilize a revolutionary recoil attenuation/mitigation system, which will also be used in the development of a man-portable 25mm lightweight weapon system.
This whole paragraph has no connection to the DREAD whatsoever, but certainly makes the article look more important. Also, you can´t find "Leader Propulsion Systems" anywhere on the net.
 
Last edited by a moderator:
  • #41
If anyone´s interested, I commented on the DREAD on Defensereview.com now. The inventor replied, but didn´t really comment on the issues presented.
You can read the comments here:
http://www.defensereview.com/modules.php?name=News&file=article&sid=727&mode=&order=0&thold=0
 
Last edited by a moderator:
  • #42
SK said:
You can read the comments here:
http://www.defensereview.com/modules.php?name=News&file=article&sid=727&mode=&order=0&thold=0
Oh man! SK, you have got one serious idiot for an opponent there. I can't believe his response to your challenge. (And one hell of a masterfully presented challenge it was.) He sounds like a petulent 6-year-old caught in the act.
What truly reaffirmed my appreciation for PF is that a lot of your arguments were presented here by people who essentially intuited the impossibility of the thing through some basic science knowledge and common sense. I wonder how this clown would feel if he knew that some high-school kids and a drop-out know more than he does. (Maybe you should tell him. :devil: )
 
Last edited by a moderator:
  • #43
Looks like its going pretty far, according to this NewScientist article from 11 May 2005:
http://www.newscientist.com/article.ns?id=dn7365&print=true
 
Last edited by a moderator:
  • #44
Mk said:
Looks like its going pretty far, according to this NewScientist article
It seems that Abrahams and Gander are pretty with it. Crane is out to lunch. How the hell did a guy like that get involved in 'Defense Review'? Isn't that supposed to be some official thing like Jane's?
 
  • #45
DefRev seems to be a one-man show run by David Crane more or less. It´s readily apparent from the author list which looks like this:
10 most active authors

· 1: davidc - (551 news published)
· 2: fmadison - (30 news published)
· 3: God - (4 news published)
http://www.defensereview.com/modules.php?name=Top
Don´t know about the other articles, but he definitely seems to have been carried away by the whole "world exclusive on DefRev" stuff when writing about the DREAD.
 
Last edited by a moderator:
  • #46
SK said:
DefRev seems to be a one-man show run by David Crane more or less.
And to think, I almost went there looking for information once. :eek: (Good thing I found it somewhere else first.)
 
  • #47
Power requirements

Without going into detail about your calcs, which we will assume are correct. There still shouldn't be a problem, because the electrical energy can be gradually applied to the disk to get its angular momentum up to horrendous levels. Then as the spheres are ejected, the entire disk will slow down a bit. Of course the motor would then begin slowly accelerating it back to top speed.

Depending on the mass of the disk, the angular velocity would decrease correspondingly. Imagine a heavy tank mounted disk, weighing say 200 kilograms, spinning at 100,000 rpm, with a relatively large electric motor that has been accelerating the disk for several hours. The decrease in angular velocity to eject 1000 tungsten spheres would be negligible, although each batch of spheres would travel a little bit slower in full automatic mode. Whenever the weapon stopped firing, the electric motor would kick in, gradually accelerating the disk back up to top speed.
 
  • #48
Recoil

Yeah, the "no recoil" bit bothered my engineering/physicist friends too. How about the idea that the ball accelerates, starting closer to the center of the disk and achieves max velocity just as it exits? Thus the acceleration or "recoil" is distributed across the entire disk, 360°, around. For example, say it rotates 1000 times during acceleration of a single ball from the center of the disk to the perimeter. Would this in essence be zero, or at least a small distributed recoil?
 
  • #49
more on recoil

Sorry I didn't post this with the previous post, but are you guys CERTAIN there is recoil? Here is why I ask: Newton's first law of motion is the law of inertia... I think it is the first one. "An object at rest remains at rest; an object in motion remains in motion." The ball is already traveling at maximum velocity on the perimeter of the disk when it is released. The velocity of the ball does not change, therefore, there is no net force on the ball. Therefore no recoil. There IS a change in angular momentum of the disk, which is compensated by the disk angular velocity decreasing slightly.

Seems the biggest problem would be the out-of-balance condition before or after the ball is released, since I am assuming the disk is spinning at 50,000 to 100,000 rpm or more. Unless, like an earlier poster said, two balls 180° apart could be released, a split second apart so they go the same direction. So the disk would be seriously out of balance for one-half revolution. Seems this would cause serious vibrations. I suppose these could be damped using electromagnetic fields?
 
  • #50
epepep said:
Sorry I didn't post this with the previous post, but are you guys CERTAIN there is recoil?
Yes. You should be too.

Here is why I ask: Newton's first law of motion is the law of inertia... I think it is the first one. "An object at rest remains at rest; an object in motion remains in motion." The ball is already traveling at maximum velocity on the perimeter of the disk when it is released. The velocity of the ball does not change, therefore, there is no net force on the ball. Therefore no recoil. There IS a change in angular momentum of the disk, which is compensated by the disk angular velocity decreasing slightly.
Here's a short recap of my previous comments.

If the disk was initally non-moving, the whole disk has a total linear momentum of zero.

Now, one piece of it breaks off, a piece with positive momentum. The other piece of the disk must have a negative momentum, because the total momentum of the whole disk was assumed to be zero to start with.

You are right that there are no forces involved - one part of the disk was moving forwards, but the reamining part was already moving backwards. When the piece moving forwards separates from the rest of the disk, the part of the disk that was moving forwards keeps moving forwards - the part that was moving backwards keeps moving backwards (until stopped by external forces).

Seems the biggest problem would be the out-of-balance condition before or after the ball is released, since I am assuming the disk is spinning at 50,000 to 100,000 rpm or more. Unless, like an earlier poster said, two balls 180° apart could be released, a split second apart so they go the same direction. So the disk would be seriously out of balance for one-half revolution. Seems this would cause serious vibrations. I suppose these could be damped using electromagnetic fields?

I don't think electromagnetic fields are going to do a very good job of damping. I would still suggest the "washing machine" model as a reasonable way of dealing with the unbalanced loads. The engine (probably a gas turbine) will need to be allowed to vibrate as the center of mass of the device shifts around. It will also have to be designed such that it can withstand the vibrations without (for instance) shattering the turbine blades (if it is a gas turbine).
 
  • #51
epepep said:
Depending on the mass of the disk, the angular velocity would decrease correspondingly. Imagine a heavy tank mounted disk, weighing say 200 kilograms, spinning at 100,000 rpm, with a relatively large electric motor that has been accelerating the disk for several hours. The decrease in angular velocity to eject 1000 tungsten spheres would be negligible, although each batch of spheres would travel a little bit slower in full automatic mode. Whenever the weapon stopped firing, the electric motor would kick in, gradually accelerating the disk back up to top speed.
While true, this somehow contradicts the claims of the supposed makers about the weight, portability and size of the DREAD. I edited my program to take disk mass into account, I´ll further extend it if I find the time. Some example outputs:
Disc with Diameter 762 mm has to spin at 20051 RPM to provide 800 m/s
to the outmost bullet on the disk. Energy required to spin the disk up with
accumulated 1000 Bullets in the 20 Feed trays and mass of the disk 20.0 kg
is 3.73 MJ. KE of bullets is 0.51 MJ then.
Disc with Diameter 762 mm has to spin at 20051 RPM to provide 800 m/s
to the outmost bullet on the disk. Energy required to spin the disk up with
accumulated 1000 Bullets in the 20 Feed trays and mass of the disk 200.0 kg
is 32.68 MJ. KE of bullets is 0.51 MJ then.
Note how grossly inefficient this whole thing gets. To accelerate 1000 round balls (@ 800 m/s , 1,48 MJ) towards your target you have to put 32 MJ of energy into the disk, and this is without taking into account any losses in drivetrain etc. , that will naturally occur. We better don´t start considering gyroscopic effect and the like here.
I´m not into material science, but it´s a good guess that you can´t just increase mass of the disk as you want, as all mass will be subject to centrifugal force, eventually leading to a (catastrophic) failure at the weakest link. Pervect went into this in one of the first posts.
The system mentioned in DefRev´s first article is said to weight in at 28 pounds empty, so I think we should go from there.
 
  • #52
SK said:
While true, this somehow contradicts the claims of the supposed makers about the weight, portability and size of the DREAD. I edited my program to take disk mass into account, I´ll further extend it if I find the time. Some example outputs:


Note how grossly inefficient this whole thing gets. To accelerate 1000 round balls (@ 800 m/s , 1,48 MJ) towards your target you have to put 32 MJ of energy into the disk, and this is without taking into account any losses in drivetrain etc. , that will naturally occur. We better don´t start considering gyroscopic effect and the like here.
I´m not into material science, but it´s a good guess that you can´t just increase mass of the disk as you want, as all mass will be subject to centrifugal force, eventually leading to a (catastrophic) failure at the weakest link. Pervect went into this in one of the first posts.
The system mentioned in DefRev´s first article is said to weight in at 28 pounds empty, so I think we should go from there.

I've got a bunch of sharp buddies interested in this one. Brought another physicist into the mix. Hope he responds. Another dude and I are still in the "no recoil" camp. Others agree with you. I think the recoil of a linear gun is f = ma. I believe the acceleration in the DREAD occurs as the balls are accelerated from the center of the disk to the outside, and is distributed across several revolutions, and thus applies 360° -- in effect no LINEAR recoil. Acceleration equals zero therefore force equals zero when the ball leaves. The recoil occurs by decreasing the Angular momentum. When the ball leaves the perimeter, it exchanges linear momentum for angular momentum. Total system momentum is conserved. (Hey I might be wrong!)

On the weight bit, this guy is obviously a salesman. I think a soldier would be limited to BB's. But a tank-mounted monster disk, weighing 200 kilograms, spinning up over the course of an hour or two, that MIGHT be able to generate some firepower.
 
  • #53
Air Drag on a Sphere

The formula (English units -- sorry -- my main modern source is at work) for decrease in velocity of a supersonic sphere is:

Vs = V0 * exp-[12(A/w)*(rho_air)*(Cd)*R]

where A = projected area of sphere in sq in, w = weight of sphere in ounces, rho_air = density of air = 0.00071 oz/in3, Cd = drag coefficient = 0.43, R= distance traveled in feet, V0 = initial velocity ft/sec, Vs = striking velocity ft/sec

so Vs = Vo * exp(-0.0016*R) for a 7.62 mm tungsten sphere (mass = .16 oz, @ density = 19 g/cc)

so at 500 feet distance, if V0 = 8000 ft/sec

Vs = 8000*exp(-.0016*500) = 3600 ft/sec

at 1000 feet, Vs = 1615 ft sec

Still respectable, but 8000 ft/sec is way higher than standard muzzle velocity (~2200 ft/sec).

So I think we need to think way bigger here -- let's go for tank mounted, 120,000 rpm, 200 kilogram disk, 8000 ft/sec initial velocity.

Forget the infantry man and his BB's.
 
  • #54
"Braniac" speaks

This is from my retired physicist friend. He's fantastically intelligent -- nearly half as smart as he thinks he is -- which is a bunch. But he's still wrong sometimes, just like everybody else...

1. Definitely not "no wear". I suspect the bearings will take quite a beating, but let's assume that the technology is available (jet engines, turbines, etc.) Without seeing the video, I would think the ejection port or whatever it is would definitely be subject to wear from mechanical "abrasion" and probably from heat. Do we grant them the technology for that? And whenever a bullet is ejected from a tray, the remaining bullets are going to move out radially and slam into whatever restrains them.

2. On first thinking, "no recoil" seems reasonable. Until release, the centripetal force required to hold the bullets in "orbit" is provided by the mechanical strength of the "gun" itself. When a bullet is released, it seems that there would just be a lessening of the tensile stresses. A "normal" gun experiences recoil because it suddenly imparts momentum to the projectile and must absorb an equal and oppsite impulse to balance that. In this system, the bullet already has attained its final momentum before it is released. The gun does not suddenly "act" on the bullet. Up until the moment of release, the gun has been acting on the bullet... it has been continuously changing the direction of the bullet's momentum vector. Upon release, it simply quits acting on the bullet... but the change which occurs is a lessening of the tensile stresses and it's not obvious that this would produce a recoil impulse.

3. Imbalance would definitely be a factor. Initially, the arrangement is symmetrical. But when the first bullet is released from the first tray, the symmetry is destroyed and the balance continues to change as each bullet is released.

The effect may well become greater as the ammunition is expended, because the mass of one bullet represents a steadily increasing percentage of the total remaining mass. You would actually have to consider how the moment of inertia is changing to verify/refute this.

It would seem that some sort of harmonic would be generated because the system would momentarily return to balance every time all the trays have released their Nth bullet. Can the materials and bearings handle this? I have no idea.

4. I agree with the guy, that if all the bullets are going to attain the same velocity, then the system has to provide a huge amount of power while firing. Why? Because initial spinup gets only the outermost bullet in each tray up to full speed. Thus the total rotational kinetic energy available to be transformed into linear kinetic energy is (far) less than the total linear kinetic energy of all the bullets would be if they were all ejected with the same velocity as the first bullet. Either the drive mechanism must provide a huge amount of power, or else the velocity of the succeeding bullets released from a particular tray will steadily decrease.

5. Certainly will be a gyroscopic effect. Would probably require a very powerful control system if you wanted to be able to change the plane of rotation during the firing sequence. If no such change were desired, then the gyro effect would be a good stabilizing factor.

In summary, an interesting concept, but it appears that the guys are pretty good "spin" doctors.
 
  • #55
pervect said:
The engine (probably a gas turbine)
It's already been established that a turbine is pretty much the only (practical?) power source for this thing. Speaking now as a weapons designer; if I had to use a freakin' jet engine to power a weapon, I'd just channel it into a honkin' big blowpipe and hopper-feed the ammo into the blast stream. (More engineering involved than that, of course, and I'd never really design something that silly, but that's the basic principle.) I bet it'd be a lot more efficient and accurate than this monstrosity.
 
  • #56
epepep said:
This is from my retired physicist friend. He's fantastically intelligent -- nearly half as smart as he thinks he is -- which is a bunch. But he's still wrong sometimes, just like everybody else...


2. On first thinking, "no recoil" seems reasonable.

Actually, it isn't at all reasonable. I'm surprised that someone who is a retired physicist could make this mistake. I was actually surprised that there is so much confusion about this very simple issue even by non-physicisits. Especially after the correct answer was pointed out a number of times.

If you shoot bullets out of a gun, they carry momentum. This causes recoil, because momentum is conserved. That's really all there is too it!

Working out the details in terms of the motion of the body is slightly tricker, but not a lot so, and has already been done by myself in this thread a couple of times.

I think I've already suggested thinking about a rotating rod on a frictionless surface a couple of times already. Imagine the rod initially has a zero average velocity, i.e. it's not moving except the fact that it's spinning. If a small piece of this rod breaks off, the small piece goes forwards, and the large piece goes backwards!

There are a number of ways of seeing this, of which the fact that the center of mass of any closed system stays at the origin is only the simplest.

A more complex way of seeing this may be simpler for some people (?). If one actually draws out the velocity of each piece of the rod as a function of position, one finds that it's a linear function of position, i.e.

v = -5*w at x=-5, v=0 at x=0, v=5*w at x=5

which can be summarized by the simple equation

v = w*x

So what happens if the piece from x=4 to x=5 breaks off? It obviously moves forwards. What happens to the remeaning parts of the rod?

Well, each small piece of the rod maintains its velocity. We can decompose the velocity of the large rod segment which strteches from (x=-5 to x=+4) as the sum of a rotation around it's new center of mass (at x=-.5) and a linear velocity as follows:

v(x) = w*x = (x+.5)*w - .5*w

The first term, (x+.5)*w, represents the velocity due to the rotation around the new center of mass at x=-.5, note that this term is zero at x=-.5 as desired, because x+.5 represents the distance between 'x' and the center of mass which is at -.5. x - (-.5) = x+.5.

You can think of this as

v(x') = w*x' - .5*w, where x' is the distance of a point from the new center of mass, so that x' = x+.5.

The second term, -.5*w, represents a constant linear velocity that the rod segement as a whole has (a negative velocity).

This result, which could have been and was predicted from the conservation of momentum, should not be a surprise, especially after it's been explained so many times.

The small segment of the rod shoots forward, and the large segement recoils backwards. The center of mass of the enitre system stays at the origin. The total linear momentum of the small piece in the forwards direction is equal but opposite to the total linear momentum of the larger piece, because the total momentum of the system is zero.
 
Last edited:
  • #57
there are probably millions of ways that we could try to counter act gyroscopic force and tensions but to look at this simply with the little information we have (hey they may invented some top secret material which is indestructible and all that) the costs for this one gun is going to be around the same as sending america to space so is the benefits that good to produce one? judging by the calculations u have made no the gun itself has no aiming capabilities whatsoever so really it is just fire and hope which isn't what u would want when having a barrel of a machine gun down your neck. the materials to be used are either so expensive or short supply so u would probably not get many of these weapons made and also the production costs are going to be huge and probably come out of tax payers pockets so i conclude that the money that will be spent on producing this is not worth the benefits (if any) that this weapon will provide.
 
  • #58
There will be no 'recoil' per say (a sudden force backwards when the weapon is fired), instead there will be a constant (or almost constant, probably oscillate slightly) force pushing back on you.

Because the bullets are 'slowly' accelerated then released you simple have the force spread out over a longer time.

It would definitely require some kind of mount. Because if you're firing continuously all of that inertia is being transferred to whatever's holding the weapon.
 
  • #59
The timing of when the forces appear depends somewhat on how one measures the forces - the vibration of the device may mask the appearance of a force by some reasonable defintions of "force" for up to half a turn of the rotor. Any such masking effect is eventually lost as the rotor turns, however. When you average over the entire turn of the rotor to "damp out" the vibrational effects and concentrate on the average force, the momentum to propel the projectile(s) has to come from somewhere - and the only place it can come from is from the weapons mount. The average net force will be v*dm/dt, where v is the velocity of the projectiles, and dm/dt is the mass flow rate.
 
  • #60
As far as weight goes, the biggest omission seems to be the weight of the power supply. This is a huge part of any railgun, and would be a huge part of this as well.