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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:

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:

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:

For all I know, these are correct calculations, with 0.81 MW being an insanely high power requirement.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.

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));

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