tp a.28 throw plots for all types of shots · throw plots for all types of shots supporting: ......
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technical proof
technical proof
TP A.28Throw plots for all types of shots
supporting:“The Illustrated Principles of Pool and Billiards”
http://billiards.colostate.eduby David G. Alciatore, PhD, PE ("Dr. Dave")
originally posted: 4/3/2007 last revision: 9/21/2007
Note: These results are based on the analysis in TP A.14.
R1.125 in⋅
m:= ball radius converted to meters
Model of friction based on Marlow data (Table 10 on p. 245 in "The Physics of Pocket Billiards," 1995)
vd1 .1 sin 45 deg⋅( )⋅:= μd1 0.11:=
vd2 1 sin 45 deg⋅( )⋅:= μd2 0.06:=
vd3 10 sin 45 deg⋅( )⋅:= μd3 0.01:=
We can solve for the coefficients from the set of data above using:
initial guesses:
a μd3:= b μd1 μd3−:= c1−
vd2ln
μd2 a−
b
⋅:=
a 0.01= b 0.1= c 0.98=
Given
μd1 a b ec− vd1⋅
⋅+=
μd2 a b ec− vd2⋅
⋅+=
μd3 a b ec− vd3⋅
⋅+=
a
b
c
Find a b, c, ( ):=
a
b
c
9.951 10 3−×
0.108
1.088
=
μ v( ) a b e c− v⋅⋅+:=
this is the "model" of friction that will be used
MathCAD formulations of Equations 15 through 17, using the above formulation for friction:
vrel v ωx, ωz, ϕ, ( ) v sin ϕ( )⋅ R ωz⋅−( )2 R ωx⋅ cos ϕ( )⋅( )2+:=
θthrow v ωx, ωz, ϕ, ( ) atan
min
μ vrel v ωx, ωz, ϕ, ( )( ) v⋅ cos ϕ( )⋅
vrel v ωx, ωz, ϕ, ( )17
v sin ϕ( )⋅ R ωz⋅−( )⋅
v cos ϕ( )⋅
:=
cut angle range:ϕ 0 deg⋅ 1 deg⋅, 75 deg⋅..:=
various speeds (slow, medium, fast):
v1 0.5ms
⋅:= v2 1.5ms
⋅:= v3 4.5ms
⋅:=
v1 1.118 mph= v2 3.355 mph= v3 10.066 mph=
v1v1m
s
:= v2v2m
s
:= v3v3m
s
:=
spin rate from percent English (from TP A.25):
ω v PE, ( ) SRF ωroll⋅=54
vR
⋅ PE⋅=
ω v PE, ( )54
vR
⋅ PE⋅:=
throw angle equation variable info
θthrow v ωx, ωz, ϕ, ( )
ωx, ωr: vertical plane spin rate (follow or draw) ωz, ωe: English spin rate
PEr: % topspin (roll) PEe: % English (side spin)
topspin: ωr, PEr > 0 outside English (OE): ωe, PEe > 0
bottom-spin: ωr, PEr < 0 inside English (IE): ωe, PEe < 0
These plots show how the amount of throw varies with the amount of English for various cutangles, speeds, and amounts of vertical plane spin (draw or follow):
PEr 0%:= ϕ 0 deg⋅:= PEe 100− %⋅ 99− %, 100 %⋅..:=
100− 0 100
5−
0
5
θthrow v1 ω v1 PEr, ( ), ω v1 PEe, ( ), ϕ, ( )deg
θthrow v2 ω v2 PEr, ( ), ω v2 PEe, ( ), ϕ, ( )deg
θthrow v3 ω v3 PEr, ( ), ω v3 PEe, ( ), ϕ, ( )deg
PEe
%PEr 0%:= ϕ 30 deg⋅:=
100− 0 100
5−
0
5
θthrow v1 ω v1 PEr, ( ), ω v1 PEe, ( ), ϕ, ( )deg
θthrow v2 ω v2 PEr, ( ), ω v2 PEe, ( ), ϕ, ( )deg
θthrow v3 ω v3 PEr, ( ), ω v3 PEe, ( ), ϕ, ( )deg
PEe
%
PEr 0%:= ϕ 60 deg⋅:=
100− 0 100
5−
0
5
θthrow v1 ω v1 PEr, ( ), ω v1 PEe, ( ), ϕ, ( )deg
θthrow v2 ω v2 PEr, ( ), ω v2 PEe, ( ), ϕ, ( )deg
θthrow v3 ω v3 PEr, ( ), ω v3 PEe, ( ), ϕ, ( )deg
PEe
%PEr 50%:= ϕ 0 deg⋅:=
100− 0 100
5−
0
5
θthrow v1 ω v1 PEr, ( ), ω v1 PEe, ( ), ϕ, ( )deg
θthrow v2 ω v2 PEr, ( ), ω v2 PEe, ( ), ϕ, ( )deg
θthrow v3 ω v3 PEr, ( ), ω v3 PEe, ( ), ϕ, ( )deg
PEe
%
PEr 50%:= ϕ 30 deg⋅:=
100− 0 100
5−
0
5
θthrow v1 ω v1 PEr, ( ), ω v1 PEe, ( ), ϕ, ( )deg
θthrow v2 ω v2 PEr, ( ), ω v2 PEe, ( ), ϕ, ( )deg
θthrow v3 ω v3 PEr, ( ), ω v3 PEe, ( ), ϕ, ( )deg
PEe
%
PEr 50%:= ϕ 60 deg⋅:=
100− 0 100
5−
0
5
θthrow v1 ω v1 PEr, ( ), ω v1 PEe, ( ), ϕ, ( )deg
θthrow v2 ω v2 PEr, ( ), ω v2 PEe, ( ), ϕ, ( )deg
θthrow v3 ω v3 PEr, ( ), ω v3 PEe, ( ), ϕ, ( )deg
PEe
%
PEr 100%:= ϕ 0 deg⋅:=
100− 0 100
5−
0
5
θthrow v1 ω v1 PEr, ( ), ω v1 PEe, ( ), ϕ, ( )deg
θthrow v2 ω v2 PEr, ( ), ω v2 PEe, ( ), ϕ, ( )deg
θthrow v3 ω v3 PEr, ( ), ω v3 PEe, ( ), ϕ, ( )deg
PEe
%
PEr 100%:= ϕ 30 deg⋅:=
100− 0 100
5−
0
5
θthrow v1 ω v1 PEr, ( ), ω v1 PEe, ( ), ϕ, ( )deg
θthrow v2 ω v2 PEr, ( ), ω v2 PEe, ( ), ϕ, ( )deg
θthrow v3 ω v3 PEr, ( ), ω v3 PEe, ( ), ϕ, ( )deg
PEe
%
PEr 100%:= ϕ 60 deg⋅:=
100− 0 100
5−
0
5
θthrow v1 ω v1 PEr, ( ), ω v1 PEe, ( ), ϕ, ( )deg
θthrow v2 ω v2 PEr, ( ), ω v2 PEe, ( ), ϕ, ( )deg
θthrow v3 ω v3 PEr, ( ), ω v3 PEe, ( ), ϕ, ( )deg
PEe
%
The remaining plots show how throw varies with cut angle for various speed shots with various amounts ofEnglish and vertical plane spin (follow or draw):
PEr 0%:= PEe 0%:= ϕ 0 deg⋅ 1 deg⋅, 75 deg⋅..:=
0 15 30 45 60 75
5−
0
5
θthrow v1 ω v1 PEr, ( ), ω v1 PEe, ( ), ϕ, ( )deg
θthrow v2 ω v2 PEr, ( ), ω v2 PEe, ( ), ϕ, ( )deg
θthrow v3 ω v3 PEr, ( ), ω v3 PEe, ( ), ϕ, ( )deg
ϕ
deg
PEr 0%:= PEe 25%:=
0 15 30 45 60 75
5−
0
5
θthrow v1 ω v1 PEr, ( ), ω v1 PEe, ( ), ϕ, ( )deg
θthrow v2 ω v2 PEr, ( ), ω v2 PEe, ( ), ϕ, ( )deg
θthrow v3 ω v3 PEr, ( ), ω v3 PEe, ( ), ϕ, ( )deg
ϕ
deg
PEr 0%:= PEe 50%:=
0 15 30 45 60 75
5−
0
5
θthrow v1 ω v1 PEr, ( ), ω v1 PEe, ( ), ϕ, ( )deg
θthrow v2 ω v2 PEr, ( ), ω v2 PEe, ( ), ϕ, ( )deg
θthrow v3 ω v3 PEr, ( ), ω v3 PEe, ( ), ϕ, ( )deg
ϕ
deg
PEr 0%:= PEe 75%:=
0 15 30 45 60 75
5−
0
5
θthrow v1 ω v1 PEr, ( ), ω v1 PEe, ( ), ϕ, ( )deg
θthrow v2 ω v2 PEr, ( ), ω v2 PEe, ( ), ϕ, ( )deg
θthrow v3 ω v3 PEr, ( ), ω v3 PEe, ( ), ϕ, ( )deg
ϕ
deg
PEr 0%:= PEe 100%:=
0 15 30 45 60 75
5−
0
5
θthrow v1 ω v1 PEr, ( ), ω v1 PEe, ( ), ϕ, ( )deg
θthrow v2 ω v2 PEr, ( ), ω v2 PEe, ( ), ϕ, ( )deg
θthrow v3 ω v3 PEr, ( ), ω v3 PEe, ( ), ϕ, ( )deg
ϕ
deg
PEr 0%:= PEe 25− %:=
0 15 30 45 60 75
5−
0
5
θthrow v1 ω v1 PEr, ( ), ω v1 PEe, ( ), ϕ, ( )deg
θthrow v2 ω v2 PEr, ( ), ω v2 PEe, ( ), ϕ, ( )deg
θthrow v3 ω v3 PEr, ( ), ω v3 PEe, ( ), ϕ, ( )deg
ϕ
deg
PEr 0%:= PEe 50− %:=
0 15 30 45 60 75
5−
0
5
θthrow v1 ω v1 PEr, ( ), ω v1 PEe, ( ), ϕ, ( )deg
θthrow v2 ω v2 PEr, ( ), ω v2 PEe, ( ), ϕ, ( )deg
θthrow v3 ω v3 PEr, ( ), ω v3 PEe, ( ), ϕ, ( )deg
ϕ
deg
PEr 0%:= PEe 75− %:=
0 15 30 45 60 75
5−
0
5
θthrow v1 ω v1 PEr, ( ), ω v1 PEe, ( ), ϕ, ( )deg
θthrow v2 ω v2 PEr, ( ), ω v2 PEe, ( ), ϕ, ( )deg
θthrow v3 ω v3 PEr, ( ), ω v3 PEe, ( ), ϕ, ( )deg
ϕ
deg
PEr 0%:= PEe 100− %:=
0 15 30 45 60 75
5−
0
5
θthrow v1 ω v1 PEr, ( ), ω v1 PEe, ( ), ϕ, ( )deg
θthrow v2 ω v2 PEr, ( ), ω v2 PEe, ( ), ϕ, ( )deg
θthrow v3 ω v3 PEr, ( ), ω v3 PEe, ( ), ϕ, ( )deg
ϕ
deg
PEr 50%:= PEe 0%:=
0 15 30 45 60 75
5−
0
5
θthrow v1 ω v1 PEr, ( ), ω v1 PEe, ( ), ϕ, ( )deg
θthrow v2 ω v2 PEr, ( ), ω v2 PEe, ( ), ϕ, ( )deg
θthrow v3 ω v3 PEr, ( ), ω v3 PEe, ( ), ϕ, ( )deg
ϕ
deg
PEr 50%:= PEe 25%:=
0 15 30 45 60 75
5−
0
5
θthrow v1 ω v1 PEr, ( ), ω v1 PEe, ( ), ϕ, ( )deg
θthrow v2 ω v2 PEr, ( ), ω v2 PEe, ( ), ϕ, ( )deg
θthrow v3 ω v3 PEr, ( ), ω v3 PEe, ( ), ϕ, ( )deg
ϕ
deg
PEr 50%:= PEe 50%:=
0 15 30 45 60 75
5−
0
5
θthrow v1 ω v1 PEr, ( ), ω v1 PEe, ( ), ϕ, ( )deg
θthrow v2 ω v2 PEr, ( ), ω v2 PEe, ( ), ϕ, ( )deg
θthrow v3 ω v3 PEr, ( ), ω v3 PEe, ( ), ϕ, ( )deg
ϕ
deg
PEr 50%:= PEe 75%:=
0 15 30 45 60 75
5−
0
5
θthrow v1 ω v1 PEr, ( ), ω v1 PEe, ( ), ϕ, ( )deg
θthrow v2 ω v2 PEr, ( ), ω v2 PEe, ( ), ϕ, ( )deg
θthrow v3 ω v3 PEr, ( ), ω v3 PEe, ( ), ϕ, ( )deg
ϕ
deg
PEr 50%:= PEe 100%:=
0 15 30 45 60 75
5−
0
5
θthrow v1 ω v1 PEr, ( ), ω v1 PEe, ( ), ϕ, ( )deg
θthrow v2 ω v2 PEr, ( ), ω v2 PEe, ( ), ϕ, ( )deg
θthrow v3 ω v3 PEr, ( ), ω v3 PEe, ( ), ϕ, ( )deg
ϕ
deg
PEr 50%:= PEe 25− %:=
0 15 30 45 60 75
5−
0
5
θthrow v1 ω v1 PEr, ( ), ω v1 PEe, ( ), ϕ, ( )deg
θthrow v2 ω v2 PEr, ( ), ω v2 PEe, ( ), ϕ, ( )deg
θthrow v3 ω v3 PEr, ( ), ω v3 PEe, ( ), ϕ, ( )deg
ϕ
deg
PEr 50%:= PEe 50− %:=
0 15 30 45 60 75
5−
0
5
θthrow v1 ω v1 PEr, ( ), ω v1 PEe, ( ), ϕ, ( )deg
θthrow v2 ω v2 PEr, ( ), ω v2 PEe, ( ), ϕ, ( )deg
θthrow v3 ω v3 PEr, ( ), ω v3 PEe, ( ), ϕ, ( )deg
ϕ
deg
PEr 50%:= PEe 75− %:=
0 15 30 45 60 75
5−
0
5
θthrow v1 ω v1 PEr, ( ), ω v1 PEe, ( ), ϕ, ( )deg
θthrow v2 ω v2 PEr, ( ), ω v2 PEe, ( ), ϕ, ( )deg
θthrow v3 ω v3 PEr, ( ), ω v3 PEe, ( ), ϕ, ( )deg
ϕ
deg
PEr 50%:= PEe 100− %:=
0 15 30 45 60 75
5−
0
5
θthrow v1 ω v1 PEr, ( ), ω v1 PEe, ( ), ϕ, ( )deg
θthrow v2 ω v2 PEr, ( ), ω v2 PEe, ( ), ϕ, ( )deg
θthrow v3 ω v3 PEr, ( ), ω v3 PEe, ( ), ϕ, ( )deg
ϕ
deg
PEr 100%:= PEe 0%:=
0 15 30 45 60 75
5−
0
5
θthrow v1 ω v1 PEr, ( ), ω v1 PEe, ( ), ϕ, ( )deg
θthrow v2 ω v2 PEr, ( ), ω v2 PEe, ( ), ϕ, ( )deg
θthrow v3 ω v3 PEr, ( ), ω v3 PEe, ( ), ϕ, ( )deg
ϕ
deg
PEr 100%:= PEe 25%:=
0 15 30 45 60 75
5−
0
5
θthrow v1 ω v1 PEr, ( ), ω v1 PEe, ( ), ϕ, ( )deg
θthrow v2 ω v2 PEr, ( ), ω v2 PEe, ( ), ϕ, ( )deg
θthrow v3 ω v3 PEr, ( ), ω v3 PEe, ( ), ϕ, ( )deg
ϕ
deg
PEr 100%:= PEe 50%:=
0 15 30 45 60 75
5−
0
5
θthrow v1 ω v1 PEr, ( ), ω v1 PEe, ( ), ϕ, ( )deg
θthrow v2 ω v2 PEr, ( ), ω v2 PEe, ( ), ϕ, ( )deg
θthrow v3 ω v3 PEr, ( ), ω v3 PEe, ( ), ϕ, ( )deg
ϕ
deg
PEr 100%:= PEe 75%:=
0 15 30 45 60 75
5−
0
5
θthrow v1 ω v1 PEr, ( ), ω v1 PEe, ( ), ϕ, ( )deg
θthrow v2 ω v2 PEr, ( ), ω v2 PEe, ( ), ϕ, ( )deg
θthrow v3 ω v3 PEr, ( ), ω v3 PEe, ( ), ϕ, ( )deg
ϕ
deg
PEr 100%:= PEe 100%:=
0 15 30 45 60 75
5−
0
5
θthrow v1 ω v1 PEr, ( ), ω v1 PEe, ( ), ϕ, ( )deg
θthrow v2 ω v2 PEr, ( ), ω v2 PEe, ( ), ϕ, ( )deg
θthrow v3 ω v3 PEr, ( ), ω v3 PEe, ( ), ϕ, ( )deg
ϕ
deg
PEr 100%:= PEe 25− %:=
0 15 30 45 60 75
5−
0
5
θthrow v1 ω v1 PEr, ( ), ω v1 PEe, ( ), ϕ, ( )deg
θthrow v2 ω v2 PEr, ( ), ω v2 PEe, ( ), ϕ, ( )deg
θthrow v3 ω v3 PEr, ( ), ω v3 PEe, ( ), ϕ, ( )deg
ϕ
deg
PEr 100%:= PEe 50− %:=
0 15 30 45 60 75
5−
0
5
θthrow v1 ω v1 PEr, ( ), ω v1 PEe, ( ), ϕ, ( )deg
θthrow v2 ω v2 PEr, ( ), ω v2 PEe, ( ), ϕ, ( )deg
θthrow v3 ω v3 PEr, ( ), ω v3 PEe, ( ), ϕ, ( )deg
ϕ
deg
PEr 100%:= PEe 75− %:=
0 15 30 45 60 75
5−
0
5
θthrow v1 ω v1 PEr, ( ), ω v1 PEe, ( ), ϕ, ( )deg
θthrow v2 ω v2 PEr, ( ), ω v2 PEe, ( ), ϕ, ( )deg
θthrow v3 ω v3 PEr, ( ), ω v3 PEe, ( ), ϕ, ( )deg
ϕ
deg
PEr 100%:= PEe 100− %:=
0 15 30 45 60 75
5−
0
5
θthrow v1 ω v1 PEr, ( ), ω v1 PEe, ( ), ϕ, ( )deg
θthrow v2 ω v2 PEr, ( ), ω v2 PEe, ( ), ϕ, ( )deg
θthrow v3 ω v3 PEr, ( ), ω v3 PEe, ( ), ϕ, ( )deg
ϕ
deg