DREAD weapon system, power requirements

  • Thread starter SK
  • Start date

SK

14
0
Hello everyone,
I joined this forum as I wanted to check if my calculations concerning the above mentioned system are correct and to gain further insight :)

I might just start:
This Thread is all about a new (over)hyped centrifuge weapon system, called the DREAD. You can find advertising info about it here:
http://www.defensereview.com/modules.php?name=News&file=article&sid=526 [Broken]
http://www.defensereview.com/modules.php?name=News&file=article&sid=539 [Broken]
I suggest to read the first link, as it gives sufficient insight into the basic idea. As you can already see though, there are unrealistic assumptions made, like recoillessnes and the like.

The patent of the system can be found here for people interested in details:
http://patft.uspto.gov/netacgi/nph-Parser?patentnumber=6,520,169

Short summary of the patent:
The spinning disk has 20 feed trays extending from the middle, each holding X ball shaped rounds. When the DREAD is spinning and the fire button is pressed, the outmost bullet in the tray gets released at a certain point to fly out of the disk tangentially.

I wrote a short Matlab program to calculate the energy/power requirements. The data I used are:
VELOCITY0 = 800; % m/s
BULLETDIA = 7.62; % mm
ROUNDS_PER_MINUTE = 50000; % rounds per minute
DENSITY = 20; % g/cm^3 ,Tungsten
NUM_SPOKES = 20; % number of feed trays/spokes
BULLETS_PER_FEED = 50; % bullets in each feed tray

All other data like diameter of the disk was calculated from this constants. The weight of the disk is not considered, only the MOI of the balls, with them considered single point masses and added up.
Output of my program is:
Bullet @ 800 m/s
Mass: 4.633 g KE: 1482.667 J
--------------------------
Fired in 1 Second at 50000 rounds per minute:
Mass: 3.861 kg KE: 1.236 MJ
--------------------------
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 is 0.51 MJ.
Expending these 1000 Rounds @ 50000 rounds/minute takes 1.20 s.
The corresponding KE of all those bullets @ 800 m/s is 1.48 MJ.
This leaves the DREAD with the need to "generate" 0.97 MJ of energy in 1.20
seconds, or an average power output of 0.81 Megawatt.
For all I know, these are correct calculations, with 0.81 MW being an insanely high power requirement.
Of course 0.81 MW is just average constant power usage, so if we would want to use a electric motor or other system we certainly would need one with > 0.81 MW peak power output.
I´m arguing with a guy about that for pages on another forum, he keeps telling me I can´t use energy calculations like that and that I´m wrong yadda yadda, so I thought I might ask here to get a third, fourth.. opinion.

The Matlab file can be downloaded here:
http://rapidshare.de/files/817378/DREAD.m.html
(scroll down, at "Bitte den Download-Typ wählen:" select "Free". Then scroll down again and click " Download: DREAD.m")


Source Code:
function DREAD

VELOCITY0 = 800; % m/s
BULLETDIA = 7.62; % mm
ROUNDS_PER_MINUTE = 50000; % rounds per minute
DENSITY = 20; % g/cm^3 ,Tungsten
NUM_SPOKES = 20; % number of feed trays/spokes
BULLETS_PER_FEED = 50; % bullets in each feed tray

RPMtoRAD = 0.105; % for converting RPM into Rad/s

r=BULLETDIA*BULLETS_PER_FEED; % radius is equal to
% length of bullets in feed tray
bulletMass=4/3*pi*(BULLETDIA/20)^3 * DENSITY; % bulletmass[g] from Diameter and Density
bulletMass=bulletMass/1000; % bulletmass[kg]
rps=ROUNDS_PER_MINUTE/60; % rounds per second from RPM, trivial
totalMass=bulletMass*rps; % total mass to be accelerated
bulletEnergy=0.5*bulletMass*VELOCITY0^2; % E0, single bullet
totalEnergy=0.5*totalMass*VELOCITY0^2; % total kintetic Energy of bullets
diskRadius=BULLETDIA*BULLETS_PER_FEED/1000; % Diameter of the DREAD disk
totalBulletNumber=NUM_SPOKES*BULLETS_PER_FEED; % Total Number of Bullets in Trays

diskRevolutions=(VELOCITY0/(2*pi*diskRadius))*60;

% Calculate MOI
% MOI is calculated by using the formula for a point mass (m*r^2) and
% calculating the MOIs for every single bullet, then just adding them up

count=1:1:BULLETS_PER_FEED;
distMOIs=bulletMass * (count*BULLETDIA/1000).^2; % MOI for every consecutive bullet in
% a single tray
totalMOI=0;
for j=1:BULLETS_PER_FEED
totalMOI=distMOIs(j)+totalMOI; % add up all MOIs in a tray
end
totalMOI=totalMOI*NUM_SPOKES; % multiply single tray MOI with number
% of feed trays


diskRad=diskRevolutions*RPMtoRAD; % Convert RPM to rad/s
diskSpinupEnergy=0.5*totalMOI*diskRad^2; % Calculate KE of spinning disk

emptymagTime=totalBulletNumber/rps;
emptymagEnergy=0.5*bulletMass*totalBulletNumber*VELOCITY0^2;


disp(sprintf('Bullet @ %4.0f m/s',VELOCITY0));
disp(sprintf('Mass: %6.3f g KE: %6.3f J', bulletMass*1000, bulletEnergy));
disp('--------------------------');
disp(sprintf('Fired in 1 Second at %6.0f rounds per minute:',ROUNDS_PER_MINUTE));
disp(sprintf('Mass: %6.3f kg KE: %6.3f MJ', totalMass, totalEnergy/10^6));
disp('--------------------------');
disp(sprintf('Disc with Diameter %4.0f mm has to spin at %5.0f RPM to provide %4.0f m/s'...
,diskRadius*2000,diskRevolutions,VELOCITY0));
disp(sprintf('to the outmost bullet on the disk. Energy required to spin the disk up with'));
disp(sprintf('accumulated %5.0f Bullets in the %3.0f Feed trays is %3.2f MJ.'...
,totalBulletNumber,NUM_SPOKES,diskSpinupEnergy/10^6));
disp(sprintf('Expending these %5.0f Rounds @ %4.0f rounds/minute takes %3.2f s.',...
totalBulletNumber,ROUNDS_PER_MINUTE,emptymagTime));
disp(sprintf('The corresponding KE of all those bullets @ %4.0f m/s is %3.2f MJ.',...
VELOCITY0,emptymagEnergy/10^6));
disp(sprintf('This leaves the DREAD with the need to "generate" %3.2f MJ of energy in %3.2f'...
,(emptymagEnergy-diskSpinupEnergy)/10^6,emptymagTime));
disp(sprintf('seconds, or an average power output of %3.2f Megawatt.',...
((emptymagEnergy-diskSpinupEnergy)/10^6)/emptymagTime));
 
Last edited by a moderator:

Mk

1,937
3
How did you find that number? What is an X ball? This thing looks really cool, why hasn't it been made before? It also looks like a great high school science project.
 

pervect

Staff Emeritus
Science Advisor
Insights Author
9,338
711
Bullet @ 800 m/s
Mass: 4.633 g KE: 1482.667 J
--------------------------
Fired in 1 Second at 50000 rounds per minute:
Mass: 3.861 kg KE: 1.236 MJ
Assuming the bullet mass is correct, these calculations are indeed correct. I haven't checked out the rest of your calculations though.

An ordinary 110 v, 15 amp outlet could supply 1650 watts (joules/second) so it could supply 1.236 MJ (assuming 100% efficiency) in 750 seconds, about 12.5 minutes. Of course 100% efficiency is extremely unlikely.

50 000 rounds/minute is 833 rounds/second. If you can fire off 20 rounds in 1 turn, that's an 42 revolutions/second/ To get a velocity of 800 m/s, you need an arm radius of about 3 meters

2*pi*3m * 43 / s = 791 m/s

The acceleration at the end of the arm will be (791 m/s)^2 / 3 m = 208560 m/s^2, about 21,000 gravities. The acceleration could be lowered a little bit with a longer arm, though 3 meters is already pretty big.

Back of the envelope calculations indicate that kevlar could probably support itself under these sorts of accelerations with some room to spare for supporting the actual load. Kevlar is supposed to have about a 3 gigapascal yield strrength, and around a 1.5 gm/cm^3 density. The figure of merit, yield strength / density, is around 2e6 m^2 / s^2, the length*acceleration product for the design is less that 600,000 m^2/s^2 (quite a bit less, because the acceleration isn't constant along the length). Steel will probably not be strong enough to hold up - assuming a 1 gigapascal yield strength and 7.8 gm/cm^3 density, steel has a much lower figure of merit of 128,000 m^2/s^2.

Air resistance is going to be large, though I don't have any idea of how to quickly calculate the drag.

Basically I wouldn't hold my breath waiting for this thing to come out - it's not going to be a simple build or design. It doesn't look totally impossible if one doesn't demand sustained fire. Sustained fire would demand some very large and powerful motors, and would also demand that the rotor be able to handle large compression forces as well as large tension forces to support the large power transfers. Sustaiend fire is pretty iffy. Of course, without sustained fire, the device isn't all that useful. Also, getting the spin up time to something reasonably short will be difficult for similar reasons. A weapon that takes 10 minutes to "power up" before it can be fired won't be all that practical either.

I personally wouldn't want to be anywhere near the thing when it operated - if it failed, it would fail quite spectacularly, being at least as dangerous to anyone operating it as it was to the enemy. Being shot at, or even having rocks thrown at it that hit the rotating armature, could easily induce a catastrophic mechanical failure.
 

Mk

1,937
3
What kind of material could sustain its same shape through a sustained 21,000 gravities? If nothing, that would make it safe to say DREAD doesn't work.
 

SK

14
0
How did you find that number? What is an X ball? This thing looks really cool, why hasn't it been made before? It also looks like a great high school science project.
Sorry, I meant there is a number of X balls in every one of the feed trays. In my program it´s 50, but you can change all constants to your liking.


Assuming the bullet mass is correct, these calculations are indeed correct. I haven't checked out the rest of your calculations though.
Bullets mass is derived from Volume(of 7.62mm ball) X Density (of Tungsten), so it should be correct.


50 000 rounds/minute is 833 rounds/second. If you can fire off 20 rounds in 1 turn, that's an 42 revolutions/second/ To get a velocity of 800 m/s, you need an arm radius of about 3 meters
I didn´t assume you have to fire them in one turn, just calculated the length of 50 consecutive bullets of 7.62 mm, leading to a radius of 381mm. (of course this is a simplification as the innermost balls would intersect).
To accelerate the outermost bullet to 800 m/s I then need ~20000 RPM.
That means a feed tray moves past the "bullet release point" 400,000 times per minute, or 6667 times per second. About every eighth feed tray moving past has to release a bullet then (which of course isn´t trivial as some trays might stay full, and other expend more ammunition) to fire 833 rounds/sec.
I ignored that timing problem for my calculations though.

Your acceleration calculation is pretty interesting though, I take with acceleration you mean centrifugal acceleration? Would only get worse with a lower diameter like in my version.

You can look at the thread that got me started here:
http://soldner.jowood.com/forum/showthread.php?t=30406&page=1&pp=20 [Broken]
My nick there is APFSDS. But be warned, it´s long and annoying to read as a certain person just can´t accept that the DREAD won´t really work. :p
 
Last edited by a moderator:

pervect

Staff Emeritus
Science Advisor
Insights Author
9,338
711
Mk said:
What kind of material could sustain its same shape through a sustained 21,000 gravities? If nothing, that would make it safe to say DREAD doesn't work.
If the material is in tension, 21,000 G's for 3 meters isn't too bad. Compare this to the demands on the proposed http://www.spaceelevator.com/docs/472Edwards.pdf [Broken] , for instance.

The space elevator will need carbon nanotubes - nothing else is really strong enough. The proposed system doesn't have nearly the same demands as that.

Thus other materials like those used in other space tether systems such as kevlar can be used.

The situation is complicated by the fact that the system won't be in uniform tension while it's being accelerated. I'm not sure whether or not some composite structure can be built or not.

I definitely wouldn't want to be anywhere near one. Designing the system to hold together is one thing - designing it to hold together if someone throws a rock at the rotor is another thing.
 
Last edited by a moderator:
447
0
How can it make no sound if it is accelerated past the speed of sound? Wouldn't there have to be a sonic boom?
 
It does mention that there is a sonic boom. But there wouldn't be a muzzle crack when a conventional firearm "uncorks" and vents the excess propellent gas. This is usually much louder that the bullet's sonic boom.

What bother's me is the claim of no recoil. If the designer is that uninformed about basic physics, I want to be in a concrete bunker when he fires it. Sure this isn't an early April Fool. The ammo looks an awful lot like a golf ball. :wink:
 
as if the military werent playing around enough allready!!!
 
if that were practicable - how on earth would you ever survive the first shot? I mean, once you shot once, the thing would be out of balance. I sucked a pebble about the size of 2 grains of rice once (judging from the hole in the seal around my air filter.) through the turbo of my truck which runs at a little less that 20000 rotations per minutes at typical driving speeds. It put a nick about the size of a half a pea in the ceramic and the truck constantly vibrated when ever the engine went above 900 rotations per minute. When it drop to 2nd gear on a hard run, the truck would vibrate such that it was almost uncomfortable to hold on to the steering wheel. if maybe two thirds of a gram unbalance in my turbo could cause my 3000 Kg truck to vibrate, what do you think a 2 or 300 gram unbalance in a system not weighing more than a few kilograms, rotating at a speed of 50,000 rotations per minute would do?
 
Most likely explode. Even if it didn't it would go into some pretty wild precessions. It would be impossible to control the aim. You can't even elevate the gun while the rotor is spun up. Ever play with a toy gyroscope?
 

SK

14
0
Well the supposed inventor says:
According to Charles St. George, the DREAD's inventor, the DREAD does NOT create/cause ANY gyroscopic forces or effects, whatsoever. To put it another way, the DREAD does NOT act like a gyroscope. So, the DREAD will NOT have ANY adverse effect on the mounting vehicle's maneuverability, even while the DREAD is being fired on its highest rpm and velocity settings. NO adverse effect. None, whatsoever. I can't make it any clearer than that.
http://www.defensereview.com/modules.php?name=News&file=article&sid=539 [Broken]
Now how that´s supposed to work remains unknown, of course.
I still find it interesting that there´s a real patent on this thing, although it´s pretty obvious that it´s completely impractical.
 
Last edited by a moderator:

Andrew Mason

Science Advisor
Homework Helper
7,472
274
kevinalm said:
What bother's me is the claim of no recoil. If the designer is that uninformed about basic physics.
There is a recoil force but it is countered by the torque that is required to keep the wheel turning at the same speed:

[tex]\tau/R = F = dp/dt = vdm/dt = \omega Rdm/dt[/tex]

One thing that appears to be overlooked here is a mechanism to keep the projectiles stable so they move on a straight course. Normally, rifling or spin provides this. These projectiles are not going to move in a straight path.

AM
 
That just transfers the recoil to the vehicle. Which will rock back on it's chassis. Torquing the rotor axis. Which _is_ a gyro and will either come apart or precess violently, throwing the aimpoint all over creation. Everybody dive into a foxhole. :smile:
 

pervect

Staff Emeritus
Science Advisor
Insights Author
9,338
711
I agree, there has to be a recoil, if something shoots out the front, and nothing shoots out the back, conservation of momentum demands that there be a recoil.

As far as gyroscopic action goes, I suppose in theory one could attempt to add in a counter weight rotating in the opposite direction. However, the materials would have to be strong enough to be able to handle the full gyroscopic torque involved if you attempt to tilt the device, which would be considerable. (You'd generate two opposing torques - while the sum of the two torques is zero, each torque still exists in the sense that it will put strain on the rotors and the connections between the two rotors).

I haven't attempted any detailed calculations, but this is obvioulsy going to generate more very large non-tension forces to the rotating components of the device. It's already going to take fairly exotic materials to hold it together (steel won't be strong enough) - this additional requirement is going to be the 2,000 pound straw that broke the camel's back.
 
It's really beyond direct analysis. To many variables.. In a situation like this it would most likely need complex computer modeling. But you can predict generally that it must recoil, that unless you bolt it to a concrete bridge abutment you wind up torquing the plane of rotation of what is an unbalenced gyro. All you know what is going to break lose if you are at the rim velocity they are bragging about. It strikes me as a joke, a con or a real incompetent. And I won't go into what happens if the bullet release mechanizm goes flakey.
 

pervect

Staff Emeritus
Science Advisor
Insights Author
9,338
711
kevinalm said:
It's really beyond direct analysis. To many variables.. In a situation like this it would most likely need complex computer modeling. But you can predict generally that it must recoil, that unless you bolt it to a concrete bridge abutment you wind up torquing the plane of rotation of what is an unbalenced gyro. All you know what is going to break lose if you are at the rim velocity they are bragging about. It strikes me as a joke, a con or a real incompetent. And I won't go into what happens if the bullet release mechanizm goes flakey.
I almost hate to say this, but there might be a solution for this part of the problem.

One could envision suspending the whole contraption from cables (including the motor, which would probably have to be a 1000hp+ gas turbine) from cables,just like a washing machine (washing machines a rather neat mechancial gadgets, there's a diagram at http://home.howstuffworks.com/washer2.htm for instance showing the motor and tub and a little bit of the suspension system)

and just let the whole contraption shake. rattle, roll, and roar. That might complicate the aiming a bit though :-)

Maybe there's some simpler solution that could be borrowed from jet engines (I recall seeing a program about some of the new large engine designs for the new large planes - they were supposed to be able to not fail catastrophically even if they lost a fan blade, though they didn't go into the tehcnical details of how they managed to accomplish this).
 
I think I see what your getting at. A washing machine acheives a dynamic balence by allowing the "rotor" axis to move in a small circle, as veiwed from on top. (Sighting straight down along the axis) This has also been used to stabilize the wheels on cars and on internal combustion engines and probably other devices. But there is a big difference. Hopefully, your washer isn't throwing sneakers out during the spin cycle. The dread _is_ throwing pieces off the rim. The circle will not close, there is a net movement. Recoil.
 

pervect

Staff Emeritus
Science Advisor
Insights Author
9,338
711
I agree that there has to be recoil (though I didn't work out the detailed forces, your description gives me an idea of how the unbalanced torque gets converted into recoil).

I was thinking about the dynamic balance problem. I think the dynamic balance issue may be solvable in terms of keeping the machine mechanically intact in spite of the changing balance, but it's going to cause other problems (such as aim, as the machine shakes around on its mounting). There might be some other solution, but the only one I see is to let the motor vibrate along with the rotor (like the washing machine) - otherwise the shaft joining the motor to the rotor will break.
 

Danger

Gold Member
9,458
244
Hi folks;
Your theoretical counter-weight would also have to reduce itself to match the projectile loss of mass. That might be done with incredibly rapid fluid transfer of something like mercury in a tube, if you can pump it back against centrifugal force. Otherwise, you might have to make it a double-barrelled unit firing in opposite directions at the same time. (As if it weren't unpredictable enough without that.) I haven't looked at the link yet; I'm just going by what I read above. I design weapons as a hobby, and one of the principles that I go by is to cause more damage to the target than the user. This thing doesn't sound like it qualifies.
 

Want to reply to this thread?

"DREAD weapon system, power requirements" You must log in or register to reply here.

Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving
Top