weapon spec repository - dimensions, material, weight, speed, inertia, energy...

I figure it's good for people to be able to look at other robots for a baseline when designing their own, so here's some (very quick, rough and elementary) calculations. I guessed on the dimensions for Copperhead and Knightrous, but I have accurate data for Roflcopter and Touro. If you have better data, feel free to correct me, or add info for another robot's weapon. I was going to calculate the acceleration time, but I'm not going to bother until I'm more sure of the data I have. My estimated dimensions for Copperhead and Knightrous are probably high, and Knightrous' gearing ratio looked right but seems like it might be off. Rofl's dimensions are accurate, and seem like overkill because they were.



Roflcopter's bar:

Dimensions: 4” x .75” x 36” = 108 in^3
Material: A2 tool steel
Density: .283 lb/in^3
Weight: 30 lbs
Moment of inertia: I = (1/12)m(w^2 + l^2) = (1/12)*30*(16 + 36^2) = 3280 lbs * in^2

2 Magmotor S28-400's, 2:1 gearing
Max blade RPM: 2000
2000 rotations/min = 2000*(2pi) radians/min = 2000*2*pi/60 rad/sec = 209.4 rad/s = w
Blade tip speed: pi*36” * 2000/min = 314 FPS

Weapon KE: ½ Iw^2 = (1/2)(3280 lbs*in^2)(209.4 /s)^2 = 21,044 joules


Acceleration time to 90% max speed:

Torque(one magmotor) = 12.74 - .025w

T = I dw/dt
(# motors)(gear ratio)(12.74 - .025(gear ratio * w)) = I (dw/dt)
2*2(12.74 - .05w) = (.960) (dw/dt)
dt = (.960)/ (2*2(12.74 - .05w)) dw
integrating from 0 to (209 * .9) = 188.5 rad/s:

t = -4.8 (ln(12.74 - .05*188.5) – ln(12.74)) = 6.5 sec





Copperhead's Drum:

Dimensions: 6” OD, 5” ID, 1/6th of surface area drilled out. 16.5” long
Material: 4130 chrome moly steel?
Density: .283 lb/in^3
W = .283 * (5/6)*pi*(3^2 – 2.5^2) * 16.5 = 33.6 lbs
W (side pieces) = .283 * (5/6) * pi * 2.5^2 * .5 * 2 = 4.6 lbs
Moment of inertia = ½ M(Ri^2 + Ro^2) = 33.6/2 * (9 + 6.25) = 256.2 lbs * in^2
I (side pieces) = ½ MR^2 = ½ * 4.6 * 2.5^2 = 14.375
I (total) = 271 lbs * in^2

1 Magmotor S28-400, 1:1 gearing
Max drum RPM: 4000
4000 rotations/min = 418.8 rad/s = w
Blade tip speed = pi*6” * 4000/min = 105 FPS

Weapon KE: ½ Iw^2 = (1/2)(277 lbs*in^2)(418.8 /s)^2 = 7,109 joules





Knightrous' Thresher:

Dimensions: 8” height, 5” gap/dead space, 2.5” thickness
Material: S7 tool steel
Density: .283 lb/in^3
W = 54.7 lbs
I = 460 lbs * in^2

1 Magmotor S28-400, ~4:1 gear ratio
Max drum RPM: 1000
1000 rotations/min = 104.7 rad/s = w
Blade tip speed = pi*8” * 1000/min = 35 FPS

Weapon KE: ½ Iw^2 = (1/2)(2483 lbs*in^2)(104.7 /s)^2 = 3,983 joules





Touro's Drum:

Dimensions: 5.1” OD, 3.15” ID, 7” long
Material: mild steel, S7 tool steel inserts
Density: .283 lb/in^3

W = .283 * pi*(2.55^2 – 1.575^2) * 7 = 25.03 lbs
W (side pieces) = .283 * pi * 1.575^2 * .5 * 2 = 2.2 lbs
Moment of inertia = ½ M(Ri^2 + Ro^2) = 25.03/2 * (6.5 + 2.481) = 112.4 lbs * in^2
I (side pieces) = ½ MR^2 = ½ * 2.2 * 2.48 = 2.728
I (total) = 115 lbs * in^2

1 Magmotor S28-400, 3:2 gearing
Max drum RPM: 6000
6000 rotations/min = 628.3 rad/s = w
Blade tip speed = pi*5.1” * 6000/min = 134 FPS

Weapon KE: ½ Iw^2 = (1/2)(115 lbs*in^2)(628.3 /s)^2 = 6,643 joules

well.... nice to meet you

looks good. It's good to have reference for other bots. We did copperhead's drum calculations a while back but I think this is probably a better place for them. Oh, and I uploaded the video of Copperhead vs Roflcopter to youtube cause I found it like three days ago...

The drum:
49 lbs
6.625 OD
6.213 ID
Teeth are 1" x 1" x 2.5" 3 per side
the max measured RPM was ~3850 but it was legit

Dan Curhan wrote:

Okay. You guys haven't taken AP physics yet, so you're solving for linear Kinetic Energy, which this isn't. We are trying to find the ROTATIONAL kinetic energy of our drum. The drum weighs 49 lbs, and Eddie was right about the cylinder. The OD of the cylinder in 6.625, with 1" teeth and a wall thickness of .412".

The equation for rotational kinetic energy is a parallel to (1/2)MV^2, it's (1/2) I (omega)^2 where I is the moment of inertia of the drum and omega is its angular velocity in radians/second. We can find the I of a hollow cylinder with some calculus and the accepted I formula for a hoop... and we get I = (1/2) M (outer radius ^ 2 + inner radius ^ 2).

If we combine all this into one equation (ignoring the teeth for now), we get:

RKE = (1/2) ( (1/2)M(outer radius ^ 2 + inner radius ^ 2) )(omega ^ 2)

Now. Let's start converting stuff into SI units.
The mass of our cylinder is 49 lbs minus teeth. We can do this pretty easily. The density of steel is .283 lbs per cubic inch. The teeth are 2.5" x 1" x 1" = 2.5 cubic inches each. There are six, which means there are 15 cubic inches of teeth, or 4.245 lbs. 49 - 4.245 = 44.755 lbs / 2.2 = 20.343 Kg.

M(cylinder) = 20.343 Kg

The OD is 6.625" * 2.54 cm/in = 16.8275 cm. We need the IR, so 16.8275 / 2 = 8.4137 - (.412 * 2.54) = 7.36722 cm.

IR(cylinder) = 7.36722 cm = .0736722 m
OR(cylinder = 8.4137 cm = .084137 m

The angular velocity = 3850 RPM * 2pi radians/rev = 24190.3 rad/min / 60 seconds = 403.1711 rad/sec.

OMEGA(cylinder) = 403.1711 rad/sec.

Now we can plug all this in:

RKE(cylinder) = (1/2) ( (1/2)(20.343 Kg)(8.4137 cm ^ 2 + 7.36722 cm ^ 2) )(403.1711 rad/sec ^ 2)
RKE(cylinder) = 10338.8925 J = 10.339 KJ

Now. We can add in the energies of the teeth (we can approximate and assume that they are point masses located at their centers of mass).

CM of each tooth is at (1.25,.5,.5) within the tooth [the center, because it's a regular rectangle of uniform density]. We disregard length because it cancels out in the moment of inertia equations, so the CM of the teeth lies .5" beyond the OD of the cylinder or 3.8125" from the center of the cylinder.

I for a point mass = (1/2) M r^2 and I for multiple point masses is the summation of mr^2's: M1 * r1^2 + M2 * r2^2 + M3 * r3^2... and so on.
M = (4.245 / 2) = 2.1225 lbs / 2.2 = .965 Kg.
R = 3.8125 * 2.54 = 9.68375 cm = .0968375 m

((.965)(.0968375)^2 )*2 = .0181

RKE(teeth) = (1/2) (.0181) (513.1268)^2
RKE(teeth) = 2382.86 J

TOTAL RKE = 10338.89 + 2382.9 = 12721.75 J or 12.722 KJ.

Just for the sake of us newbies , not in the know...could you put the weight class of the bots in the specs as well?

Otherwise, it's hard to get a comparison of who is packing more power to weight

Thanks

fwiw: depending on which motor we use, our 3 pounder 'Sting' is creating between 1 and 2.5k joules at full speed

all the robots listed are in the 120 lb weight class.

colin wrote:

all the robots listed are in the 120 lb weight class.

Any links to roflcopter's fights?

Seems like it must pack one heck of a punch?

rjw wrote:

colin wrote:

all the robots listed are in the 120 lb weight class.

Any links to roflcopter's fights?

Seems like it must pack one heck of a punch?

check youtube.

and not really. the problem was we didn't know enough about E&M, motors or batteries because we were mostly self taught. we did some calculations, but not enough and we didn't have any other robot to compare to. I edited my first post - you can see that the weapon spin up time was 6.5 seconds, and that's assuming 100% gearbox efficiency and assuming the batteries can supply the necessary current, which they couldn't. The magmotors actually drew enough current that the robot couldn't drive properly while the weapon was speeding up. It was pretty effective if we could evade the other robot until it had spun up, though - in our first match against a series of tubes, we cracked one of their gearboxes.

http://video.google.com/videoplay?docid=-2693338874763437924

I see what you mean....it's one thing to have the potential and another to put it down effectively....

Mr. Self Destruct
30lb

28" x 3" x 3/8" S7 bar, ~HRC 45

Assuming 2000rpm from s28-150 magmotor through 2:1 gearbox
3,794j

Colin, I've been doing some work on my own on energy storage comparison tables for a variety of weapon shapes and I've noticed a calculation error you've made. All of the units for those equations should be in metric meters & kilograms - in my recent calculations, a bar of the same size, mass & RPM as the one you've specified for roflcopter has a maximum energy storage of 0.54235 kJ (all bars are miserable at storing energy due to the nature of their moment of inertia).

EDIT: See post below.

Jeff L wrote:

Colin, I've been doing some work on my own on energy storage comparison tables for a variety of weapon shapes and I've noticed a calculation error you've made. All of the units for those equations should be in metric meters & kilograms - in my recent calculations, a bar of the same size, mass & RPM as the one you've specified for roflcopter has a maximum energy storage of 0.54235 kJ (all bars are miserable at storing energy due to the nature of their moment of inertia).

I probably should have used metric, as most of the errors I tend to make come from using standard units, but the calculation I did above was correct. the units you use don't matter, as long as you use the correct conversion coefficient when changing from standard to metric, which google calculator will handle for you if you're lazy. here's the calculations for roflcopter's blade in metric:

Dimensions: .1016 x .01905 x .9144 meters = .00177 m^3
Material: A2 tool steel
Density: 7.75 g/cm^3
Mass: .00177 m^3 * 7.75 g/cm^3 = 13.716 kg
moment of inertia: I = (1/12)m(w^2 + l^2) = (1/12)*13.716*(.1016^2 + .9144^2) = .9675 kg * m^2

2 Magmotor S28-400's, 2:1 gearing
Max blade RPM: 2000
2000 rotations/min = 2000*(2pi) radians/min = 2000*2*pi/60 rad/sec = 209.4 rad/s = w
Blade tip speed: pi*.9144m * 2000/min * (min/60s) = 95.76 meters/sec

Weapon KE: ½ Iw^2 = (1/2)(.9675 kg*m^2)(209.4 /s)^2 = 21,212 joules


you can do a *very rough* sanity check based on the calculations dan posted for copperhead (also done in metric). copperhead uses 1 s28-400 with a total weapon KE of 12.722 KJ. roflcopter uses 2 s28-400's. therefore, assuming equal spin-up time (it's not), roflcopter dumps about twice as much energy into the blade as copperhead, and therefore it's KE will be about 2*12 kJ = 24 kJ.

it's true that bars are one of the worst shapes in terms of storing energy, but that's in terms of moment of inertia/weight given a certain radius. you also have to consider the fact that a bar has a much greater radius than a drum. a good sanity check is, the energy the motor puts out is what's stored in the weapon, so two weapons with the same motor and spin up time will have the same KE.

why don't you post your calculations?

Aha! Figured out a miscalculation of my own, my bad . Somehow I left out the 2*pi

Bar Dimensions:
Length: 36 inches = 0.9144 meters
Width: 4 inches = 0.102 meters
Thickness: .75 inches = 0.0191 meters
Weight & Mass: 30.5 lbs = 13.8 kg (using the volume & average density of steel [7.82 g/cm^3]

Moment of Inertia [rectangular prism shape rotating around the axis perpendicular to the length & width dimensions]:
Formula: I = (1/12)*(mass)*(length^2 + width^2)
Moment of Inertia of Specified Bar: (1/12)*(13.8kg)*(0.9144^2 + .102^2) = 0.97623 kg*m^2

Rotational Kinetic Energy
Formula = (1/2)*(I)*((radians/second)^2)
RPM Specified for Bar: 2000 RPM = 4000*pi Radians Per Minute (2*pi radians per rotation)
Radians/Second for Bar: (4000*pi)/60 = 209.43951 radians/second
Rotational Kinetic Energy for Specified Bar: (1/2)*(.97623)*(209.43951^2) = 21411.1197 Joules

Well, I do have some excellent news for roflcopter then. Its maximum legal output under RFL rules is 53.57353971 kilojoules - however, it's currently running over the BotsIQ legal tip-speed limit.

As an aside, copperhead's drum alone could store in excess of 100 kilojoules if it was running at the maximum permissible RPM.

(Also, one weapon candidate for Tech's new 120 could have somewhere in the neighborhood of a megajoule of stored energy )