math and physics play

ECE1254H Modeling of Multiphysics Systems. Lecture 1: Analogies to circuit systems. Taught by Prof. Piero Triverio

September 22, 2014 ece1254

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Disclaimer

Peeter’s lecture notes from class. May not be entirely coherent.

In slides

A review of systematic nodal analysis for a basic resistive circuit was outlined in slides, with a subsequent attempt to show how many similar linear systems can be modeled as circuits so that the same toolbox can be applied. This included blood flow through a body (and blood flow to the brain), a model of antenna interference in a portable phone, heat conduction in a one dimensional conductor under a heat lamp, and a few other systems.

This discussion reminded me of the joke where the farmer, the butcher and the physicist are all invited to talk at a beef convention. After meaningful and appropriate talks by the farmer and the butcher, the physicist gets his chance, and proceeds with “We begin by modeling the cow as a sphere, …”. The ECE equivalent of that appears to be a Kirchhoff circuit problem.

Mechanical structures example

Continuing the application of circuits like linear systems to other systems, let’s consider a truss system as illustrated in fig. 1, or in the simpler similar system of fig. 2.

fig. 1.  A static loaded truss configuratio

fig. 1. A static loaded truss configuratio

fig 2. Simple static load

fig 2. Simple static load

Our unknowns are

  • positions of the joints after deformation \((x_i, y_i)\).
  • force acting on each strut \(\BF_j = (F_{j,x}, F_{j,y})\).

The constitutive equations, assuming static conditions (steady state, no transients)

  • Load force. \(\BF_L = (F_{L, x}, F_{L, y}) = (0, -m g)\).
  • Strut forces. Under static conditions the total resulting force on the strut is zero, so \(\BF’_j = -\BF_j\). For this problem it is redundant to label forces on both ends, so we mark the labeled end of the object with an asterisk as in fig. 3.
fig 3. Strut model

fig 3. Strut model

Consider a simple case

One strut as in fig. 4.

fig. 4.  Very simple static load

fig. 4. Very simple static load

\begin{equation}\label{eqn:multiphysicsL1:20}
\BF^\conj = – \Ba_x
\underbrace{
\epsilon
}_{\text{constant, describes the beam elasticity, given}}
\biglr{
\underbrace{
L
}_{\text{unloaded length \(L = \Abs{x^\conj – 0}\), given}}
– L_0}
\end{equation}

The constitutive law for a general strut as in fig. 5 is

fig 5.  Strut force diagram

fig 5. Strut force diagram

The force is directed along the unit vector

\begin{equation}\label{eqn:multiphysicsL1:40}
\Be = \frac{\Br^\conj – \Br}{\Abs{\Br^\conj – \Br}},
\end{equation}

and has the form
\begin{equation}\label{eqn:multiphysicsL1:60}
\BF^\conj = – \Be \epsilon \lr{ L – L_0 }.
\end{equation}

The value \(\epsilon\) may be related to Hooks’ constant, and \(L_0\) is given by

\begin{equation}\label{eqn:multiphysicsL1:80}
L = \Abs{\Br^\conj – \Br} = \sqrt{(x^\conj – x)^2 + (y^\conj – y)^2}.
\end{equation}

Observe that the relation between \(\BF^\conj\) and position is nonlinear!

Treatment of this system will be used as the prototype for our handling of other nonlinear systems.

Returning to the simple static system, and introducing force and joint labels as in fig. 6, we can examine the \textAndIndex{conservation law}, the balance of forces.

fig 6.  Strut system

fig 6. Strut system

  • At joint 1:\begin{equation}\label{eqn:multiphysicsL1:100}
    \Bf_A + \Bf_B + \Bf_C = 0
    \end{equation}or
    \begin{equation}\label{eqn:multiphysicsL1:120}
    \begin{aligned}
    \Bf_{A,x} + \Bf_{B,x} + \Bf_{C,x} &= 0 \\
    \Bf_{A,y} + \Bf_{B,y} + \Bf_{C,y} &= 0
    \end{aligned}
    \end{equation}
  • At joint 2:\begin{equation}\label{eqn:multiphysicsL1:140}
    -\Bf_C + \Bf_D + \Bf_L = 0
    \end{equation}or
    \begin{equation}\label{eqn:multiphysicsL1:160}
    \begin{aligned}
    -\Bf_{C,x} + \Bf_{D,x} + \Bf_{L,x} &= 0 \\
    -\Bf_{C,y} + \Bf_{D,y} + \Bf_{L,y} &= 0
    \end{aligned}
    \end{equation}

We have an equivalence

  • Force \(\leftrightarrow\) Current.
  • Force balance equation \(\leftrightarrow\) KCL

Sum of digits of small powers of nine.

July 15, 2014 math and physics play , , , ,

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In a previous post I wondered how to prove that for integer \(d \in [1,N]\)

\begin{equation}\label{eqn:numberGame:20}
((N-1) d) \text{mod} N + ((N-1) d) \text{div} N = N-1.
\end{equation}

Here’s a proof in two steps. First for \(N = 10\), and then by search and replace for arbitrary \(N\).

\(N = 10\)

Let

\begin{equation}\label{eqn:numberGame:40}
x = 9 d = 10 a + b,
\end{equation}

where \(1 \le a, b < 9\), and let \begin{equation}\label{eqn:numberGame:180} y = a + b, \end{equation} the sum of the digits in a base \(10\) numeral system. We wish to solve the following integer system of equations \begin{equation}\label{eqn:numberGame:60} \begin{aligned} 9 d &= 10 a + b \\ y &= a + b \\ \end{aligned}. \end{equation} Scaling and subtracting we have \begin{equation}\label{eqn:numberGame:80} 10 y - 9 d = 9 b, \end{equation} or \begin{equation}\label{eqn:numberGame:100} y = \frac{9}{10} \lr{ b + d }. \end{equation} Because \(y\) is an integer, we have to conclude that \(b + d\) is a power of \(10\), and \(b + d \ge 10\). Because we have a constraint on the maximum value of this sum \begin{equation}\label{eqn:numberGame:120} b + d \le 2 ( 9 ), \end{equation} we can only conclude that \begin{equation}\label{eqn:numberGame:140} b + d = 10. \end{equation} or \begin{equation}\label{eqn:numberGame:160} \boxed{ b = 10 - d. } \end{equation} Back substitution into \ref{eqn:numberGame:40} we have \begin{equation}\label{eqn:numberGame:200} \begin{aligned} 10 a &= 9 d - b \\ &= 9 d - 10 + d \\ &= 10 d - 10 \\ &= 10 \lr{ d - 1 }, \end{aligned} \end{equation} or \begin{equation}\label{eqn:numberGame:220} \boxed{ a = d - 1. } \end{equation} Summing \ref{eqn:numberGame:220} and \ref{eqn:numberGame:160}, the sum of digits is \begin{equation}\label{eqn:numberGame:240} a + b = d - 1 + 10 - d = 9. \end{equation}

For arbitrary \(N\)

There was really nothing special about \(9, 10\) in the above proof, so generalizing requires nothing more than some search and replace. I used the following vim commands for this “proof generalization”

:,/For arb/-1 y
:+/For arb/+1
:p
:,$ s/\<9\>/(N-1)/cg
:,$ s/\<10\>/N/cg
:,$ s/numberGame:/&2:/g

Let

\begin{equation}\label{eqn:numberGame:2:40}
x = (N-1) d = N a + b,
\end{equation}

where \(1 \le a, b < N-1\), and let \begin{equation}\label{eqn:numberGame:2:180} y = a + b, \end{equation} the sum of the digits in a base \(N\) numeral system. We wish to solve the following integer system of equations \begin{equation}\label{eqn:numberGame:2:60} \begin{aligned} (N-1) d &= N a + b \\ y &= a + b \\ \end{aligned}. \end{equation} Scaling and subtracting we have \begin{equation}\label{eqn:numberGame:2:80} N y - (N-1) d = (N-1) b, \end{equation} or \begin{equation}\label{eqn:numberGame:2:100} y = \frac{N-1}{N} \lr{ b + d }. \end{equation} Because \(y\) is an integer, we have to conclude that \(b + d\) is a power of \(N\), and \(b + d \ge N\). Because we have a constraint on the maximum value of this sum \begin{equation}\label{eqn:numberGame:2:120} b + d \le 2 ( N-1 ), \end{equation} we can only conclude that \begin{equation}\label{eqn:numberGame:2:140} b + d = N. \end{equation} or \begin{equation}\label{eqn:numberGame:2:160} \boxed{ b = N - d. } \end{equation} Back substitution into \ref{eqn:numberGame:2:40} we have \begin{equation}\label{eqn:numberGame:2:200} \begin{aligned} N a &= (N-1) d - b \\ &= (N-1) d - N + d \\ &= N d - N \\ &= N \lr{ d - 1 }, \end{aligned} \end{equation} or \begin{equation}\label{eqn:numberGame:2:220} \boxed{ a = d - 1. } \end{equation} Summing \ref{eqn:numberGame:2:220} and \ref{eqn:numberGame:2:160}, the sum of digits is \begin{equation}\label{eqn:numberGame:2:260} a + b = d - 1 + N - d = N-1. \end{equation} This completes the proof of \ref{eqn:numberGame:20}.

Pick a number between 1 and 10

July 14, 2014 math and physics play , , , ,

I saw the following on mathfail.com (EDIT: dead link), and thought about it a bit

 

Notice that the +4 here is entirely misdirection.  This is really just a statement that the sum of the digits of any integer power of 9 up to 81, is 9.

It also appears to be true that, for integer a in \([1, N+1]\)

\[(N a) \text{div} (N+1) + (N a) \text{mod} (N + 1) = N.\]

This is demonstrated in the following Mathematica Manipulate

However, I’m unsure how to prove or disprove this?