technical proof TP B.1 Squirt angle, pivot length, and tip shape supporting: The Illustrated Principles of Pool and Billiards http://illiards.colostate.edu y David G. Alciatore, PhD, PE ("Dr. Dave") technical proof originally posted: 11/1/007 last revision: 11/8/007 original aiming line cue pivot angle ( α) all radius () contact point offset () cue pivot length (p) ridge pivot point tip radius (r) β a all radius and mass:.5 in := m := 6oz r tip radii: r dime := 0.705 in p α r nickel := 0.835 in r reak := 0.5 in
From the drawings aove, = cos( β) cos( β) a = a = ( p + ) sin( α) ( 1) + r so the contact point tip offset distance "" is related to the cue pivot angle ( α), ridge length (p), and tip radius (r) according to: ( α, p, r) := ( p + ) sin( α) ( ) + r This equation can e rearranged to solve for the natural pivot length for the cue for a given squirt angle (α), tip offset (), and tip radius (r): ( + r) p( α,, r) := ( 3) sin( α) The assumption here is the cue pivot angle (α) exactly cancels the cue squirt angle (α), resulting in a straight shot. From TP A.31 (Equation 13), the squirt angle is a function of tip offset () and the endmass properties of the cue: := atan α m, r 5 1 1 + m r 5 1 + ( 4) where m r is the endmass ratio for the cue defined y: m r = m m e ( 5) where m is the mass of the all and m e is the effective endmass of the cue. The mass ratio m r can e determined experimentally if a squirt value (α exp ) is measured for a known offset ( exp ). From Equation 4, m r (, α) := 5 1 1 tan( α) 1 ( 6)
Now let's look at example data and plots for some real cues. From my Septemer '07 BD article, here is some data for several example cues: Players ("regular" cue): exp := 0.51 in α exp :=.5 deg := m r exp, α exp = 0.151 p exp := p α exp, exp, r dime p exp = 14.31 in Predator Z ("low-squirt" cue): exp := 0.51 in α exp := 1.8 deg m r ( exp, α exp ) = 9.158 = 0.199 in p α exp, exp, r dime Stinger (reak/jump cue): exp := 0.3 in α exp :=.4 deg m r ( exp, α exp ) = 1.008 = 9.3 in p α exp, exp, r reak So, as expected, the natural pivot length (p) for a "low-squirt" cue is longer than for a "regular" cue, and the pivot length for a reak cue is much shorter. These calculated numers are consistent with experimentally-determined pivot lengths (see my Novemer '07 article for more information). Note that the range of endmass ratios reported here (1 to 1) is much smaller and lower than the expected range reported in on Shepard's 001 "Everything you always wanted to know aout cue all squirt, ut were afraid to ask" paper (0 to 100) Also, the pivot lengths are much longer than the values reported y Platinum Billiards (www.platinumilliards.com/rating_deflect.php). The Players cue numers ( and p exp ) are used in the susequent analysis as an example.
So for a fixed ridge length, how does tip offset vary with the cue pivot angle: αp := 0 deg, 0.1 deg.. 3 deg 0.6 αp, p exp, r dime 0.4 0. 0 0 1 3 αp deg As you can see, the offset varies nearly linearly over typical cue pivot angles. So how does a cue's squirt angle vary with tip offset: x := 0 in, 0.01 in.. 0.5 3 α x, deg 1 0 0 0.1 0. 0.3 0.4 x As you can see, the squirt also varies nearly linearly over the range of possile tip offsets. This is why the BHE and FHE aim compensation methods are effective (see my Novemer '07 article).
For the cue pivot angle to exactly cancel the squirt angle, the pivot length needs to e a certain value. Does this value change with tip offset and the size of the tip: x := 0.01 in, 0.0 in.. 0.5 16 ( x ) p α x,,, r dime in ( x ) p α x,,, r nickel 14 in 1 10 0 0.1 0. 0.3 0.4 x The answer is yes... the required pivot length does vary a little with oth tip offset and tip radius. So when comparing different cues with experimental measurements, it is important to use the same tip shape and the same amount of tip offset (English) for each cue. For 50% English, x := 0.5 (, x, r nickel ) ( (, ) x, ) p α x, p α x, r dime 1 = 4.774 % So a cue with a tip of nickel radius has a natural pivot length aout 5% longer than the same cue with a tip of dime radius.
Now, what is the effect of tip size on the accuracy of a near center-all hit? This is an extension of the analysis in TP A.10. We will compare two different ridge lengths: p short := 6in p long := 18 in Let's assume a near cent-all hit, where the shooter has a cue pivot-angle error of up to 1/ a degree: α := 0.5 deg If the ridge distance happens to match the natural pivot length for the cue, the cue pivot angle will cancel the squirt angle, and the cue all will still head in the aiming-line direction. However, the cue pivot angle will create tip offset and English. Here's how the percent English changes with pivot angle error for various cominations of ridge lengths and tip shapes: α, p short, r dime 0.5 = 8.416 % α, p long, r dime 0.5 =.59 % α, p short, r nickel 0.5 = 8.06 % α, p long, r nickel 0.5 = 1.64 % α, p short, r reak 0.5 = 7.653 % α, p long, r reak 0.5 = 0.541 % So a shorter ridge is dramatically etter if you want to minimize English effects with a non-perfect center-all hit. A flatter tip also reduces the impact of error with a center-all hit. However, a shorter ridge can cause stroke prolems (e.g., not enough length for smooth acceleration) and it fails to take advantage of the squirt-canceling natural-pivot-length effect of the cue (see my Novemer '07 article). And a flatter tip is not suitale for applying English on non-center-all hits. See my January '08 article for more information.
So what affect does all mass have on the natural pivot length for a cue and cue all comination? Here is the data for the Players cue with a regulation-weight cue all: exp := 0.51 in α exp :=.5 deg := m r exp, α exp = 0.151 p exp := p α exp, exp, r dime p exp = 14.31 in For a 10% heavier cue all, the mass ratio (m/me) would e slightly larger, creating a smaller squirt angle and a longer pivot length: mr heavy := 1.1 α heavy := α exp, mr heavy α heavy =.3 deg = 15.565 in p α heavy, exp, r dime p α heavy, exp, r dime 1 = 9.381 % 9% longer pivot p exp length For a 10% lighter cue all, the mass ratio (m/me) would e slightly smaller, creating a larger squirt angle and a shorter pivot length: mr light := 0.9 α light := α exp, mr light α light =.738 deg = 1.896 in p α light, exp, r dime p α light, exp, r dime 1 = 9.379 % 9% shorter pivot p exp length