Received: 18 May 1995 12:11:52                    Sent: 18 May 1995 12:11:02
From:"Christopher Smith " <chris@flowserver.stem.com>
To: Ultimate_Spastic@flowserver.stem.com,Ultimate_OtherSpazes@flowserver.stem.c
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Subject: You will be tested later...
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This is part of a planned series written by Jim Parinella, who  
played with the 1994 UPA Champion, Death or Glory.
Thu May 11 14:54:16 MST 1995

Tip of the Week #5

Physics 101

Hey all, this is Cork. I am filling in for Jim Parinella this week  
because he has a broken hand and can't use his keyboard any more. Jim  
will be back next week to discuss individual defensive coverage.  
This week's topic is catching the disc using the parallel axis  
theorem when you are forced to use one hand.

When catching the disc one handed, many players talk about  
"matching" the spin of the disc or catching the correct side of the  
disc. Many players do this naturally and don't realize why they are  
doing this or even that they are doing it at all.

Many inside-outs (inverts) bounce off the receivers hands and into  
their chests because they don't adjust their catch to the unexpected  
reverse spin of the disc.If you have ever played on a "mac" line, you  
will have realized that there is a correct side to mac the disc on,  
as well as an incorrect side (which stops the disc dead). The perfect  
catch is the opposite of the perfect mac.

There is a firm physical basis for doing this and it relates to the  
parallel axis theorem.

This theorem (actually a related theorem) states that the total  
angular momentum of the disc (about the receiver's hand) is the  
vector sum of two components;

1. The angular momentum of the disc about its own center of mass  
(rotational),

2. The component due to the disc's translational movement with  
respect to your hand.

If you catch the disc in the appropriate position along its rim,  
these two components will cancel each other (yielding zero net  
angular momentum) and the center of mass of the disc will push itself  
directly into the palm of your hand.

OBVIOUSLY, If can grab the disc firmly, just grab it and the disc  
won't go anywhere (two hands, pancake, etc).

The method described here is only for very specific situations (when  
you can only get one hand on a fast moving, hard to catch disc) and  
is simply another tool to use.

The following derivation is used to indicate where this position is  
on the rim of the disc (by the way, you don't have to calculate these  
equations every time you catch it).

This will be hard without a visual, but here goes,

If the disc is traveling directly towards you (the receiver), label  
the point on the rim of the disc closest to yourself as theta = 0  
degs. Angles to the right (viewed by yourself) are labeled positive  
angles, left indicates negative.

If the disc is traveling with velocity V (with respect to you) and  
spin w (lefty forehand, righty backhand), the disc will have angular  
momentum (about your hand) resulting from two components; rotational  
and translational;

L = Lrotational + Ltranslational

These components are vectorially summed to yield the total angular  
momentum;

L = -I*w*z + R x (mV)*z
L = -I*w*z + m*v*r*sin (theta)*z
where;

I = moment of inertia of the disc
w = angular velocity of the disc
z = unit vector in the z hat direction pointing
vertically upwards (couldn't find a hat)
R = Radius vector from your initial point of contact to
the center of the disc
m = mass of the disc
V = Velocity vector of the disc
v = velocity of the disc (magnitude)
r = radius of the disc
theta = angle between the centerline of the disc's path and
the point of contact by the receiver
x indicates the vector cross product

To determine the moment of inertia I of the disc; realize that for a  
perfectly flat disc with a uniform mass distribution, I is equal to;
I = (0.5)*m*r*r
For a perfect hoop with all of the mass concentrated at the rim, I  
is equal to;
I = (1.0)*m*r*r
So, we can approximate the momentum of inertia of a Discraft  
ultra-star as being somewhere in-between these two;
I = (0.7)*m*r*r
If we take a "typical" forehand as being that which would roll along  
the floor without slipping (if thrown sideways) we obtain;
w = v/r
Plugging these into above; we obtain;
L = -(0.7*m*r*r)*(v/r)*z + m*v*r*sin (theta)*z
L = -0.7*m*r*v*z + m*v*r*sin (theta)*z
Now, if we choose an angle where the total angular momentum (about  
your hand) is zero, then the disc will not try to rip itself out of  
your hand as you grab it.

Setting L equal to zero;
L = 0
we get;
0.7*m*r*v*z = m*v*r*sin (theta)*z
or;
sin (theta) = 0.7
theta = 45 degs.
for the throw described. Or;
theta = - 45 degs
for a lefty backhand or a righty forehand.
These approximations are very loose (obviously) and need to be  
adjusted to the particular throw. You can experiment by catching the  
disc at various positions along the rim and deciding where you find  
it easiest to grab. It doesn't matter which hand you catch with.

If you can't get to the optimal side of the disc, you can sweep your  
hand (with twice the disc's velocity) past the disc as it passes you  
and grab it at 180 degs away from the above calculated position.  
Players often refer to this grab as catching the "trailing edge" of  
the disc. If you and the disc are moving at the same speed, you can  
adjust the velocity of your hand relative to the disc to accomplish  
the same effect.

Just a different way to look at things on the field.

Cork