Today's question: how does the number of runs scored depend on where in the batting order the inning starts?
My first guess was to look at the percentages of runs scored, but this is a clear error, as the #1 spot in the lineup gets so many more chances than the other positions. What we really want is the variation in scoring rate - how does the average number of runs scored when this hitter leads off compare to the average number of runs scored overall.
Now, one could look at offenses - but there's some concern that efficient offenses are shaped differently than inefficient ones. So for my baseline, I decided to turn the question about and look at runs allowed - facing an assortment of lineups over the course of a season is a rough approximation of facing an average lineup.
Obviously, the designated hitter looms as a distortion that we need to worry about. In collecting the data, I aggregated AL and NL results independently, but limited my tests to those years when the DH rule was in effect. Interleague play threatens to add another hobgoblin to the mix, but fortunately the dataset I obtained from retrosheet stopped at 1992 anyway.
An overview of the results:
The general curve of results is about the same shape in both leagues. The amplitude differs in that the NL is much higher and lower than the AL. The ranking of the lineup spots is almost identical - in the NL, the 8 spot is slightly below the 5 spot; in the AL, the 8 spot is distinctly higher.
| # | NL | AL |
|---|---|---|
| 1 | 1.189 | 1.109 |
| 2 | 1.169 | 1.089 |
| 3 | 1.133 | 1.069 |
| 4 | 0.962 | 0.982 |
| 5 | 0.884 | 0.912 |
| 6 | 0.816 | 0.892 |
| 7 | 0.798 | 0.881 |
| 8 | 0.873 | 0.945 |
| 9 | 1.004 | 1.035 |
The standard deviation is around 0.1585, which suggests that when looking at offenses, distinguishing between signal (true deviations from the distribution of talent in the lineup) and noise is going to be a bitch.
Whee, not only did the nice people at allmediashop.com send me (via Amazon) the Feynman audio lectures that I wanted, but the folks at The Tuva Trader sent me the Feynman video lectures that I wanted.
Work has bothered me for a while - the idea that the change in energy could be related to the force times a distance seems really weird to me. So I did some number crunching to figure out what was going on.
When I think of energy, I normally end up relating things to kinetic energy. Kinetic energy is (1/2)mv^2. The 1/2 in front is dimensionless, of course, so the units here are
M(L/T)^2
= ML^2/T^2
= ML(L/T^2)
= (ML/T^2)L
= force * distance
So that checks. Easier still would be to consider something like a gravitational potential mgh = (mg)h = force*distance.
Fine. But how does it work?
Assume a mass m has velocity v when it encounters a constant force F. This force acts over a distance x. What is the KE of the mass before and after encountering the force (assuming non relativistic speeds).
Well, the mass travels distance x in some unknown time t, which must satisfy the equation
x = v(0)t + (F/m)t^2/2
This is a simple quadratic, so we can solve for t easily enough
t = [ -v(0) + (v(0)^2+2(F/m)x)^.5 ] / (F/m)
So the velocity at the end of the exercise is
v(t)
= v(0) + (F/m) [ -v(0) + (v(0)^2+2(F/m)x)^.5 ] / (F/m)
= v(0) -v(0) + (v(0)^2+2(F/m)x)^.5
= (v(0)^2+2(F/m)x)^.5
KE(t)
= m v(t)^2/2
= (m/2) v(t)^2
= (m/2) ((v(0)^2+2(F/m)x)^.5)^2
= (m/2) (v(0)^2+2(F/m)x)
= (m/2) v(0)^2 + (m/2)2(F/m)x
= KE(0) + Fx
How about that. having stared at it a bit, I think I failed to fully appreciate that as you go faster, a much smaller change in velocity is required to achieve the change in energy.
dKE = mv dv
Demonstrating the randomness of my corner of the universe, I twigged to working on the scalar and vector potentials recently. I'm not sure how much of my library covers them.
Feynman scatters his material across several chapters of Volume II
II-2 Differential Calculus of Vector Fields
II-15 Vector Potential
II-25 Electrodynamics in Relativistic Notation
Carver Mead also discusses it in Collective Electro Dynamics, and there is some information in Griffiths.
But I wasn't quite finding what I was looking for - a demonstration of the theorems from which the scalar and vector potential follow.
The derivation of the scalar and vector potentials rests on two theorems.
∇ ∙ B = 0 ↔ B = ∇ x A
∇ x F = 0 ↔ F = ∇ ψ
The drill in both cases is the same - right to left is completely straightforward, but left to right requires a process for discovering the potential.
The first of these is outlined as a problem in Griffiths (5.30). We can write out the vector B as ( Bx, By, Bz ).
∇ ∙ B = 0
therefore ∂Bx/∂x + ∂By/∂y + ∂Bz/∂z = 0
therefore ∂Bx/∂x = - ( ∂By/∂y + ∂Bz/∂z )
Now A is vector, and we can discover an A that satisfies the equation by discovering (Ax,Ay,Az). From the definition of the cross product, you can deduce that A ∇ (B+C) = A ∇ B + A ∇ C. Using the result from the second theorem, right to left, it follows that B = ∇ ⨯ A implies B = ∇ ⨯ ( A + ∇ ψ ), because the latter addend has no cross product of its own.
Therefore, there's a degree of freedom available, which is used at once to make life easier. Assume that Ax = 0 ! Now the equations of the cross product are reduce to:
Bx = ∂Ay/∂z - ∂Az/∂y
By = - ∂Az/∂x
Bz = ∂Ay/∂x
The latter two, from the Fundamental Theorem of calculus, produce
Ay = ( ∫ Bz(t,y,z) dt ) + C
Az = D - ( ∫ By(t,y,z) dt )
so now we just need to satisfy one more equation. We don't need both constants, so it's easier at this point to take C=0, and solve the first equation.
Bx(x,y,z)
= ∂Az/∂y - ∂Ay/∂z
= ∂D/∂y - ∂/∂y ∫ By(t,y,z) dt - ∂/∂z [∫ Bz(t,y,z) dt]
= ∂D/∂y - ∫ ∂Bz(t,y,z)/∂z dt - ∫ ∂By(t,y,z)/∂y dt
= ∂D/∂y - ∫ ( ∂Bz(t,y,z)/∂z + ∂By(t,y,z)/∂y ) dt
= ∂D/∂y + ∫ ∂Bx(t,y,z)/∂x dt
= ∂D/∂y + Bx(x,y,z) - Bx(0,y,z)
The last two steps relying on B being divergenceless and the Fundamental Theorem of Calculus.
Bx(0,y,z) = ∂D/∂y
D = ∫ Bx(0,t,z) dt
So our candidate solution vector potential is
A = ( 0, ∫ Bz(t,y,z) dt, ∫ Bx(0,t,z) dt - ∫ By(t,y,z) dt )
Calculating ∇ x A demonstrates the first theorem right to left. QED.
The other problem gave me fits - I'm hopeful that this indicates only that I'm terribly out of practice, rather than incapable of discovering these demonstrations on my own - until I tracked down this entry from MathWorld.
∇ x F = 0 ↔ ∂Fy/∂z = ∂Fz/∂y etc.
So let's discover ψ
Fx(x,y,z) = ∂ψ(x,y,z)/∂x
ψ(x,y,z) = ∫ Fx(t,y,z) dt + C(y,z)
now we plug that result into the next restriction
Fy(x,y,z)
= ∂ψ(x,y,z)/∂y
= ∂/∂y [ ψ(x,y,z) ]
= ∂/∂y [ ∫ Fx(t,y,z) dt + C(y,z) ]
= ∫ ∂/∂y Fx(t,y,z) dt + ∂/∂y C(y,z) ]
= ∫ ∂Fx(t,y,z)/∂y dt + ∂C(y,z)/∂y ]
= ∫ ∂Fy(t,y,z)/∂x dt + ∂C(y,z)/∂y ]
= Fy(x,y,z) - Fy(0,y,z) + ∂C(y,z)/∂y
Fy(0,y,z) = ∂C(y,z)/∂y
C(y,z) = ∫ Fy(0,t,z) dt + D(z)
Running through the same sequence starting from Fz(x,y,z) produces
D(z) = ∫ Fz(0,0,t) dt + K
ψ(x,y,z) = ∫ Fx(t,y,z) dt + ∫ Fy(0,t,z) dt + ∫ Fz(0,0,t) dt + K
K is a completely arbitrary constant, unless you've some additional boundary condition to satisfy
Report from Tim Spector at St. Thomas Hospital that
genetics is a significant factor in a woman's capacity for achieving orgasm.
Introducing the Gauss Rifle. I particularly like the explanation, which strikes me as quite simple and elegant. Which is probably indicative of how much time I've spent pouring over non-relativistic QM problems of late.