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 Copperhead's Drum Calculations

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Eddie Burgess



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PostSubject: Re: Copperhead's Drum Calculations   Mon Mar 16, 2009 4:29 pm

Copperhead's drum is schedule 80 steel pipe with a 6.625" O.D. if i remember correctly. The teeth should be 1" long so 8.625 should be the total O.D.

This is all from memory, I'll have to check with Dan to see if I'm right.

The drum weighs more than 40 lb. as well but I don't know exactly how much.
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Alan Dreher
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PostSubject: Re: Copperhead's Drum Calculations   Mon Mar 16, 2009 4:32 pm

Eddie Burgess wrote:
Copperhead's drum is schedule 80 steel pipe with a 6.625" O.D. if i remember correctly. The teeth should be 1" long so 8.625 should be the total O.D.

This is all from memory, I'll have to check with Dan to see if I'm right.
So I was close...
It is more like an 11 cm radius(22cm diameter)

Will go ahead an modify my calculations
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Alan Dreher
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PostSubject: Re: Copperhead's Drum Calculations   Mon Mar 16, 2009 4:42 pm

Eddie Burgess wrote:

The drum weighs more than 40 lb. as well but I don't know exactly how much.
Ok... I remember being told that it is slightly under 40 lbs. Maybe I am mistaken.
Dan, clear all this up for us.
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Julien Kuzniarek

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PostSubject: Re: Copperhead's Drum Calculations   Mon Mar 16, 2009 5:09 pm

Alan Dreher wrote:
Julien Kuzniarek wrote:
Copperhead's drum spinning at it's fastest (4900rpm) has 1.88 MJ of kinetic energy so it really isn't that surprising that we were able to break knightrous's thresher

I have been double checking your Calculations...
They don't come out right.

Kinetic energy is calculated by the formula
.5(m)(v)²

mass = 40lb / (2.2lb/kg) = 18.18 kg
radius = 10cm (?) I wasn't certain so this is a guess Feel free to correct me with more accurate figure.
velocity = 2πr * 4900 rpms = 3.078 km/min = 184.725 km/hr = 114.783 mph = 51.3127 m/s

.5(18.18kg)(51.3127 m/s)² = 23,936.277 J = 23.936 kJ
For a quick comparison, that is almost 6 times the energy produce by the explosion of 1 gram of TNT...
Which sounds about right.

While I am at it, here are the calculations for The Dethinator:
mass = 4.55lbs(3.85 lb shell + .7lb weapon assembly) / (2.2lb/kg) = 2.068 kg
radius = 6.7295 inches(5.7295 inches [shell radius] + 1 inch [weapon assembly]) = 17.0929 cm
velocity = 2πr * 3071 rpms = 3.298 km/min = 197.891 km/hr = 122.89 mph = 54.9698 m/s

.5(2.068kg)(54.9698 m/s)² = 3,124.416 J = 3.124 kJ
That is slightly less than the energy produce by the explosion of 1 gram of TNT...
Again, seems about right.

Please tell me the true radius for Copperhead's drum though so I can get that perfect.

when i did the calculation i had the drum's radius as 3.5"=8.89cm and the equation i used was mass * radius(in meters)^2 * rpm^2 *.5 so i messed up there probably because it was like 1 in the morning when i did it. and i think that the dethinator's diameter is 14". now that i actually think about it megajoules is something you would expect from a massive explosion so my initial figures are ridiculous. however kilojoules is still a lot of energy so you cant discount it.
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Alan Dreher
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PostSubject: Re: Copperhead's Drum Calculations   Mon Mar 16, 2009 5:15 pm

Julien Kuzniarek wrote:

when i did the calculation i had the drum's radius as 3.5"=8.89cm and the equation i used was mass * radius(in meters)^2 * rpm^2 *.5 so i messed up there probably because it was like 1 in the morning when i did it. and i think that the dethinator's diameter is 14". now that i actually think about it megajoules is something you would expect from a massive explosion so my initial figures are ridiculous. however kilojoules is still a lot of energy so you cant discount it.
A megajoule is the energy of a 1000kg car going 100mph.

So, yeah, it was ridiculous.
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Julien Kuzniarek

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PostSubject: Re: Copperhead's Drum Calculations   Mon Mar 16, 2009 7:11 pm

Alan Dreher wrote:
Julien Kuzniarek wrote:

when i did the calculation i had the drum's radius as 3.5"=8.89cm and the equation i used was mass * radius(in meters)^2 * rpm^2 *.5 so i messed up there probably because it was like 1 in the morning when i did it. and i think that the dethinator's diameter is 14". now that i actually think about it megajoules is something you would expect from a massive explosion so my initial figures are ridiculous. however kilojoules is still a lot of energy so you cant discount it.
A megajoule is the energy of a 1000kg car going 100mph.

So, yeah, it was ridiculous.

no more midnight math for me Sleep
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Eddie Burgess



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PostSubject: Re: Copperhead's Drum Calculations   Mon Mar 16, 2009 8:47 pm

Dan sent me a text a little while ago. The drum weighs 49 lb. and the O.D. with teeth is 8.625".
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Alan Dreher
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PostSubject: Re: Copperhead's Drum Calculations   Mon Mar 16, 2009 8:48 pm

Eddie Burgess wrote:
Dan sent me a text a little while ago. The drum weighs 49 lb. and the O.D. with teeth is 8.625".
Wow... 49 lbs. That is one heavy drum.
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Eddie Burgess



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PostSubject: Re: Copperhead's Drum Calculations   Mon Mar 16, 2009 8:50 pm

It used to weigh around 58 lb., then we put holes in it.
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Alan Dreher
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PostSubject: Re: Copperhead's Drum Calculations   Mon Mar 16, 2009 8:51 pm

Eddie Burgess wrote:
It used to weigh around 58 lb., then we put holes in it.
utter insanity... No wonder it is basically an unstoppable force.
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PostSubject: Re: Copperhead's Drum Calculations   Mon Mar 16, 2009 9:00 pm

my keyboard has browser forward and back buttons right above the left/right arrows, and after I worked most of this out in my reply, i hit the back button instead of the arrow and lost it ALL. SO. I wrote it all out in notepad then copy/pasted lol.

________________

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 = 4900 RPM * 2pi radians/rev = 30787.6 rad/min / 60 seconds = 513.1268 rad/sec.

OMEGA(cylinder) = 513.1268 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) )(513.1268 rad/sec ^ 2)
RKE(cylinder) = 16747.29 J

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 = 16747.3 + 2382.9 = 19130.2 J or 1.913 KJ.


A little less than what you guys were calculating, but still pretty impressive.

[if anyone sees any errors in my math, please correct me]
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Will Bales
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PostSubject: Re: Copperhead's Drum Calculations   Mon Mar 16, 2009 9:04 pm

19 kJ you mean..
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Dan Curhan
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PostSubject: Re: Copperhead's Drum Calculations   Mon Mar 16, 2009 9:08 pm

haha yeah. wow. my bad. WAY higher than alan thought - by tenfold.

ooops.


Rotational Kinetic Energy of the drum = 19 KJ !!!
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Eddie Burgess



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PostSubject: Re: Copperhead's Drum Calculations   Mon Mar 16, 2009 9:10 pm

aw, i just grabbed my physics book and worked it out but Dan beat me to posting. The spread sheet I sent you earlier that calculates this stuff did have some errors, I'll check it again before I send you the new copy. It does drive system calculations now as well.

From what I'm told holes also mess up the accuracy of this calculation quite a bit.
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Jeff L
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PostSubject: Re: Copperhead's Drum Calculations   Mon Mar 16, 2009 9:12 pm

Well, if they're uniformly placed it should cancel out a lot of the error.
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PostSubject: Re: Copperhead's Drum Calculations   Mon Mar 16, 2009 9:13 pm

lol... Dan has holes in his calculations
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PostSubject: Re: Copperhead's Drum Calculations   Mon Mar 16, 2009 9:14 pm

lol!
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Alan Dreher
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PostSubject: Re: Copperhead's Drum Calculations   Mon Mar 16, 2009 9:16 pm

Dan Curhan wrote:
haha yeah. wow. my bad. WAY higher than alan thought - by tenfold.

ooops.


Rotational Kinetic Energy of the drum = 19 KJ !!!
Copperhead 3's drum came out to roughly 29 kJ according to my calculations with the wrong formulas
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Alan Dreher
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PostSubject: Re: Copperhead's Drum Calculations   Mon Mar 16, 2009 9:38 pm

Dan, you have no idea how excited I am for AP Physics C next year after seeing those calculations. And I bet that isn't even all that complicated for Physics C...

I will be very happy next year. I love a challenge, so that is definitely going to work in my favor next year.
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PostSubject: Re: Copperhead's Drum Calculations   Mon Mar 16, 2009 9:41 pm

haha yeah... you do kind of have to think a little bit. And I didn't show you the derivation of the I for a hollow cylinder... you take the integral from OD to ID of the moment of inertia of a hoop (a universally accepted example equation)... it's fun. believe me. Razz
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