Re: My first ARSG, or a Dangerous Design? Terry Blake speaks

["Tesla list" <tesla-at-pupman-dot-com>]

Original poster: "Terry Blake by way of Terry Fritz <teslalist-at-qwest-dot-net>" <tb3-at-att-dot-net>

Ooops, found that I divided by L for the Newtons.  Should be 1/2 L, so
double all force values.  Bummer, but still pretty safe.

Here is a new ARSG design that was submitted to me by Kent Tinsley.  It is
way cool, and should spark (hehe) another round of discussions.

http://www.tb3-dot-com/tesla/coilers/kt/index.htm

Terry Blake
Coiling in Chicago.

----- Original Message -----
From: "Tesla list" <tesla-at-pupman-dot-com>
To: <tesla-at-pupman-dot-com>
Sent: Thursday, January 23, 2003 7:58 PM
Subject: Re: My first ARSG, or a Dangerous Design? Terry Blake speaks


 > Original poster: "Terry Blake by way of Terry Fritz <teslalist-at-qwest-dot-net>"
<tb3-at-att-dot-net>
 >
 > Who could argue with safety.  An enclosure of some type (plywood) is a
"must
 > have", to block the arcs' UV and to stop flying shrapnel.  Heck, something
 > could fall on the spinning rotor and cause all hell to break lose.  DO NOT
 > RUN WITHOUT ONE.
 >
 > But just how big a deal is the friction fit?  Let us not fear the unknown,
 > but embrace it.
 >
 > Let's do some calculating based on a 5/32" x 7" Tungsten rod.
 >    D = 5/32 * 2.54 = 0.40 cm
 >    L = 7 * 2.54 = 17.80 cm
 >
 > Mass of the 5/32" x 7" Tungsten rod:
 >    Tungsten density: 19.35 g/cc
 >    Volume of a cylinder: Pi * r^2 * h = Pi * (0.2)^2 * 17.80 = 2.24 cc
 >    M = 19.35 g/cc * 2.24 cc = 43.25 g
 >
 > Velocity at the end of the rod at 7500 RPM (1000 BPS on my ARSG)
 >    Circumference = Pi * D = 3.14 * 17.80 cm = 55.89 cm
 >    V = 7500 RPM / 60 * 55.89 cm = 6986 cm / S = 7 m / S
 >
 > Now we have some numbers to work with.
 > D = 0.0004 M (diameter)
 > R = 0.0002 M (radius)
 > L = 0.0180 M (length)
 > M = 0.043 Kg (mass)
 > V = 7 M / S      (velocity)
 >
 > Suppose the rod was slid off center by 0.5 cm (which is huge), there would
 > be an extra 1 cm of mass on one side and not the other.
 >
 > M (1 cm) = 43.25 g / 17.80 = 2.43 g = 0.0024 Kg
 >
 > The centrifugal force pulling on the rod would be
 > F= MA = MV^2/R = 0.0024 Kg * (7 M/S)^2 / 0.018 M = 6.6 N (Newtons)
 >
 > 1 lb = 4.448 N
 >
 > Soooo, 6.6 / 4.448 = 1.48 pounds of force pulling on the rod.
 >
 > Sounds like a lot, but when I go check one of my rotors, I find that I
have
 > to
 > push on it with over 20 pounds of force (using my bathroom scale), to get
 > the
 > rod to move.  So my safety margin in that situation is about 20 / 1.5 = 13
 > factors, or 1300 %.
 >
 > So what about a more likely scenario.  The rod is off-center by 1 mm,
there
 > would be an extra 2mm of mass on one side and not the other.
 >
 > M (2mm) = 0.2 * 43.25 g / 17.80 = 2.43 g = 0.00048 Kg
 >
 > The centrifugal force pulling on the rod would be
 > F= MA = MV^2/R = 0.00048 Kg * (7 M/S)^2 / 0.018 M = 1.3 N (Newtons)
 > Soooo, 1.3 / 4.448 = 0.3 pounds of force pulling on the rod.
 > My safety margin in that situation is about 20 / 0.3 = 67 factors or
6700%.
 >
 > What if we only spin around at 3750 RPM (500 BPS on my ARSG)
 > F= MA = MV^2/R = 0.00048 Kg * (3.5 M/S)^2 / 0.018 M = .33 N (Newtons)
 > Soooo, .33 / 4.448 = 0.073 pounds of force pulling on the rod.
 > My safety margin in that situation is about 20 / 0.073 = 274 factors or
 > 27,400%.
 >
 > It looks pretty safe to me, but could someone check my math, to see what I
 > messed up on.
 >
 > I cannot address the plastic cold-flow issue, but to say that the rotor I
 > just checked (with the 20 pounds of grip), was assembled 9 months ago.
 >
 > As far as heat isues, my rotors have been checked numerous time after TC
 > runs and were found to have cool rods near the center.
 >
 > If you are overly concerned, I would suggest scraping up the rod (with
 > sandpaper) and putting a dab of epoxy on either side of the rotor (next to
 > the poly) to hold the rod in place.  That should be helpful.
 >
 >
 > Terry Blake
 > Coiling in Chicago.
 >
 >
 >