nipponbilliards said:
I would like to ask the posters here if you ever felt that:
1. it is easier to go straighter on certain shots with certain cues?
The inertia of a cue is dependent on the mass and center of gravity of the cue. In layman's term, the heavier the cue, the higer is its tendency to stay, hence a higher inertia.
Thank you.
Richard
Here is a post I made to RSB --gasp-- eight years ago....
-mp
***********
The distribution of weight in a cue was discussed some in another thread
titled forward weighted cues, but it's a different issue from location of
the balance point, so I'm starting a new thread.
So let's say you're building a 20 oz 60" cue and you've decided you want
the balance point 20" from the rear. *The issue now is how to distribute
the mass. *Forget joint-type, densities of materials, diameter, etc.
Let's suppose for the sake of argument you built the cue out of materials
with zero density, and that all 20 oz comes from lead you insert at
various locations. *Where, consistent with the balance point, do you put
the lead? *One possibility (a) is to put it all at the balance point.
Another (b) is to try to distribute it evenly. *Another (c) *is to put
some at the balance point, some at the back , and some at the front. And
still another possibility (d) is to put none at the balance point, and
split the mass between the back and the front.
What are the considerations? *Why would one distribution be better than
another? *The *only* two considerations *I* can think of are *(1) squirt
and (2) feedback to the rear hand. * Actually by "feedback," I mean moment
of inertia about the bridge hand, which translates to resistance to moving
off course, and more is better. *I think this is what Ron Shepard calls
feedback to the rear hand, so that's what I'm calling it. *Now Harry
Houdini's wife articulated clearly the importance to squirt of mass near
the tip, so we don't want mass at the tip. *Let's suppose we can safely
put as much mass as we like 10 inches from the tip, and let's further
suppose that 10 inches is our typical bridge length.
So the squirt consideration just says don't put mass more forward than the
10 inch mark. *The feedback consideration suggests we should maximize the
moment of inertia about the bridge hand.
(a) With all the mass at the balance point, the moment of inertia about
the bridge hand is 18,000 oz*in**2 *[I = m* r**2 = 20 oz * (30 in)**2 ].
(b) With 10 oz at the balance point and 5 oz each twenty inches forward
and back, the moment of inertia is 22,000 oz*in**2, somewhat larger than
the first case.
(c) distributing the mass evenly 0.5 oz per inch for 20 in either side of
the balance point gives a moment of inertia of 20,666 oz*in**2.
(d) But case (d) is the one that maximizes the moment of inertia. *Put 12
oz at the butt cap and 8 oz right at the bridge hand. *Then the moment of
inertia is 30,000 oz*in**2!
Now a sideways stroke imperfection (sideways force) that would produce a
1.0 degree deviation in the alignment of the stick at contact for case (c)
would produce the following deviations for the various weight
distributions.
case * * * *deviation
____ * * * *____________
(a) * * * * 1.15 degrees
(b) * * * * 0.94 degrees
(c) * * * * 1.00 degrees
(d) * * * * 0.69 degrees
so (d) is the most stable
These come from T =I*a
T= torque about the bridge hand, which for a given grip position is
proportional to the force
I = moment of inertia
a = angular acceleration (the alignment screwup is proportional to this)