1.4.5: Lanthanide Contraction
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The Lanthanide Contraction describes the atomic radius trend that the Lanthanide series exhibit. Another important feature of the The Lanthanide Contraction refers to the fact that the 5s and 5p orbitals penetrate the 4f sub-shell so the 4f orbital is not shielded from the increasing nuclear change, which causes the atomic radius of the atom to decrease. This decrease in size continues throughout the series.
Introduction
The Lanthanide Contraction applies to all 14 elements included in the Lanthanide series. This series includes Cerium(Ce), Praseodymium(Pr), Neodymium(Nd), Promethium(Pm), Samarium(Sm), Europium(Eu), Gadolinium(Gd), Terbium(Tb), Dysprosium(Dy), Holmium(Ho), Erbium(Er), Thulium(Tm), Ytterbium(Yb), and Lutetium(Lu). The atomic radius, as according to the Lanthanide Contraction, of these elements decreases as the atomic number increases. We can compare the elements Ce and Nd by looking at a periodic table . Ce has an atomic number of 58 and Nd has an atomic number of 60. Which one will have a smaller atomic radius? Nd will because of its larger atomic number.
The graph shows the atomic radius decreasing as the atomic number is increasing, Lanthanide Contraction.
Shielding and its Effects on Atomic Radius
The Lanthanide Contraction is the result of a poor shielding effect of the 4f electrons. The shielding effect is described as the phenomenon by which the inner-shell electrons shield the outer-shell electrons so they are not effected by nuclear charge. So when the shielding is not as good, this would mean that the positively charged nucleus has a greater attraction to the electrons, thus decreasing the atomic radius as the atomic number increases. The s orbital has the greatest shielding while f has the least and p and d in between the two with p being greater than d.
The Lanthanide Contraction can be seen by comparing the elements with f electrons and those without f electrons in the d block orbital. Pd and Pt are such elements. Pd has 4d electrons while Pt has 5d and 4f electrons. These 2 elements have roughly the same atomic radius. This is due to Lanthanide Contraction and shielding. While we would expect Pt to have a significantly larger radius because more electrons and protons are added, it does not because the 4f electrons are poor at shielding. When the shielding is not good there will be a greater nuclear charge, thus pulling the electrons in closer, resulting in a smaller than expected radius.
Row 1 of Periodic Table D block Row 2 Row 3
The graphs depict the atomic radii of the first three rows of transition metals. We can apply the same principle as applied with the elements Pd and Pt to whole rows and columns. As we can see by comparing Row 1 with Row 2, the atomic radii differ greatly between the elements, but if we compare Row 2 with Row 3, the atomic radii do not have much difference. Elements with atomic number 23 and 41 lie in the same column of the periodic table and have a significantly large difference in atomic radii (atomic radii increases from Row 1 to Row 2), but elements 41 and 73, also in the same column, only differ slightly. This is the cause of introducing 4f electrons in Row 3. In Row 3, we would expect the elements to carry on the same trend as was witnessed between Rows 1 and 2 (large increase in atomic radii) but we do not. This is because the 4f orbitals are not doing a great job of shielding.
D Block Contraction (Scandide Contraction)
The d block contraction, also known as the Scandide Contraction, describes the atomic radius trend that the d block elements (Transition metals) experience. Normally the trend for atomic radius, moving across the periodic table is that the atomic radius decreases significantly. In the transition metals with D electrons as we move from left to right across the periodic table, the element’s atomic radius only decreases slightly. This is because they have the same amount of s electrons, but are only differing in d electrons. These d electrons are in an inner shell (penultimate shell) and electrons are getting added to this shell, another shell is not created. The d electrons are not good at shielding the nuclear charge, so the atomic radius does not change much as electrons are added. Almost like disregarding the D electrons being added.
Effects on Ionization Energy and Properties
As the proton number increases and the atomic radius decreases, the ionization energy increases. This is due to a more positively charged nucleus and a greater pull on the electrons by the nucleus. A greater pull is the result of an increased effective nuclear charge. Effective nuclear charge is caused by the nucleus having a more positive charge than the negative charge on the electron (net positive charge). The density, melting point, and hardness increase from left to right throughout the Lanthanide Series. The Lanthanide Contraction makes chemical separation of the Lanthanides easier. The Lanthanide Contraction, while making the chemical separation of Lanthanides easier, it makes the separation of elements following the series a bit more difficult.
References
- Petrucci, Harwood, Herring, and Madura. General Chemistry Principles & Modern Applications . 9th ed. Prentice Hall. Print.
- Cotton, Simon. Lanthanide and Actinide Chemistry . Wiley. Print.
Problems
- What is the cause of Lanthanide Contraction?
- Which element has a greater atomic radius Gd or Tb and why?
- Which element has a smaller atomic radius Dy or Yb and why?
- Why do elements in Rows 2 and 3 that are in the same column have similar atomic radii?
- Place the following elements in order of increasing atomic radius: Eu, Ce, Pr, Ho.
- Which one of the following has the lowest density: Tb, Pm, Er, Tm, Ce
Answers:
- The Lanthanide Contraction is caused by a poor shielding effect of the 4f electrons.
- Gd because as atomic number increases, the atomic radius decreases.
- Yb because it has a larger atomic number.
- Because the elements in Row 3 have 4f electrons. These electrons are do not shield good, causing a greater nuclear charge. This greater nuclear charge has a greater pull on the electrons.
- Ho, Eu, Pr, Ce
- Ce
Contributors and Attributions
- Amrit Paul Bains (UCD)


12.10: Nova, Novae, and Supernova
The word nova comes from the Latin, for ‘new. ’ Novae can brighten as much as 100,000 Suns, 100,000L sun . Material from the Nova fusion event radiates outward from the White Dwarf; we can view this outward radiating material as a planetary nebula. The nova process can repeat itself, but not infinitely . When the White Dwarf accreting mass reaches the low-mass star 1. 33M sun limit, a Type 1a Supernova is possible. First the White Dwarf begins to collapse, with fast, quick heating. Carbon fusion begins (12 in the stellar core) and then the Star “ignites,” exploding as a White Dwarf Supernova. These are usually one-time events.
Type 1a supernovae produce elements up to iron on the atomic table, and also produce elements heavier than iron, like gold, silver, and uranium. These type of supernovae have the same characteristics, such as how bright they will become and length of maximum brightness. Therefore astronomers can use them as a “standard candle,” that is a standard brightness.
Supernovae may also produce cosmic rays , which are composed of electrons, protons and neutrons and move at close to the speed of light. Ultra-high energy cosmic ray particles were discovered in 2005. These are the brightest and fastest radio blasts ever seen on the sky, seen as radio light that appear more than 1000 times brighter than the Sun and almost a million times faster than normal lightning.
The very bright star in the lower left is Supernova SN1994D in Spiral Galaxy NGC4526. Compare the brightness of the supernova to its home galaxy.


13: Pendulum Dynamics
The undriven pendulum has only a two dimensional space, forming a phase plane where it is easy to visualize the phase portraits. The phase portrait of the small amplitude simple pendulum is particularly easy to draw. The dimensionless form of the simple pendulum equation is
\[\ddot{\theta}+\theta=0, \nonumber \]
and the solutions for \(\theta=\theta(t)\) and \(u=\dot{\theta}(t)\) are given by
\[\theta(t)=\theta_{m} \cos (t+\varphi), \quad u(t)=-\theta_{m} \sin (t+\varphi) \nonumber \]
The phase portrait in the \(\theta-u\) phase plane consists of concentric circles, with clockwise motion along these circles, as seen in the small diameter circles of Fig. 13.1. As the amplitude increases, the approximation \(\sin \theta \approx \theta\) loses validity, and the relevant dimensionless equation becomes
\[\ddot{\theta}+\sin \theta=0 . \nonumber \]
Beyond the small diameter circles of Fig. 13.1, one observes that the connected lines become elongated in \(u\) as the amplitude increases, implying the pendulum is slowed down when the amplitude becomes large (i.e., the period of the pendulum is lengthened). Eventually, a final closed curve is drawn, called the separatrix, that separates the pendulum’s motion from periodic to rotary, the latter motion corresponding to a pendulum that swings over the top in a circular motion. Exactly on the separatrix trajectory, the pendulum comes to rest at the angle \(\pi\) , or \(180^{\circ}\) , which is an unstable fixed point of the pendulum.
The simple pendulum is a conservative system, exhibiting a conservation law for energy, and this implies a conservation of phase space area (or volume). If one evolves over time a given initial area of the phase space of Fig. 13.1, the area of this phase space will be conserved. When the pendulum is damped, however, the area of this phase space will shrink to zero.
Basin of attraction of the undriven pendulum
We consider the damped, undriven pendulum with governing equation
\[\ddot{\theta}+\frac{1}{q} \dot{\theta}+\sin \theta=0 \nonumber \]
and the stable fixed point given by \((\theta, \dot{\theta})=(0,0)\) . How can we determine the basin of attraction of this fixed point? Of course, with \(\dot{\theta}=0\) , only the values \(-\pi<\theta<\pi\) will lie in the basin of attraction. But we also need to compute the basin of attraction for nonzero initial values of \(\dot{\theta}\) .
As is often the case, to devise a numerical algorithm it is best to appeal directly to the physics. We want to find the borderline of initial conditions between either the attractive fixed points \((0,0)\) and \((2 \pi, 0)\) , or the attractive fixed points \((0,0)\) and \((-2 \pi, 0)\) . The initial conditions just on the border result in the pendulum reaching either the unstable fixed points \((\pi, 0)\) or \((-\pi, 0)\) . A small
perturbation from \((\pi, 0)\) will result in one of the attractive points \((0,0)\) or \((2 \pi, 0) ;\) from \((-\pi, 0)\) , one of the attractive points \((0,0)\) or \((-2 \pi, 0)\) .
The algorithm then initializes the calculation at either the unstable fixed point \((\pi, 0)\) or \((-\pi, 0)\) , with a small perturbation in either the position or velocity that results in the final attracting point being \((0,0)\) . We can call these values final conditions because the differential equations are then integrated backward in time to determine all possible initial conditions that can result in these final conditions. You can convince yourself that the four final conditions that should be used are the two pairs \((\pi,-\epsilon),(\pi-\epsilon, 0)\) , and \((-\pi, \epsilon),(-\pi+\epsilon, 0)\) , with \(\epsilon\) very small. The first of each pair corresponds to the immediately previous motion being rotary, and the second of each pair corresponds to the immediately previous motion being oscillatory.
A graph of the basins of attraction of \((0,0)\) for \(q=4\) , corresponding to an underdamped pendulum (critical damping is \(q=1 / 2\) ) is shown in Fig. 13.2. The attracting fixed point is marked by an ’ \(x\) ’, and the region inside the two curved lines is the basin of attraction.
Spontaneous symmetry-breaking bifurcation
An interesting supercritical pitchfork bifurcation occurs in the pendulum equations, where at the bifurcation point one stable limit cycle splits into two. The single limit cycle displays a symmetry that is no longer respected individually by the two new limit cycles. This type of pitchfork bifurcation is a manifestation of what is called spontaneous symmetry breaking.
The pendulum equation \((11.14)\) , written again here, is given by
\[\ddot{\theta}+\frac{1}{q} \dot{\theta}+\sin \theta=f \cos \omega t . \nonumber \]
Using \(\sin (-\theta)=-\sin \theta\) and \(\cos (\omega t-\pi)=-\cos \omega t\) , the pendulum equation can be seen to be invariant under the transformation
\[\theta \rightarrow-\theta, \quad t \rightarrow t-\pi / \omega \nonumber \]
with the physical interpretation that the equations of motion make no distinction between the right and left sides of the vertical, a consequence of the symmetry of both the pendulum and the external force.
Consider again the asymptotic small amplitude solution given by \((11.10)\) , which in dimensionless variables is
\[\theta(t)=\frac{f}{\sqrt{\left(1-\omega^{2}\right)^{2}+(\omega / q)^{2}}} \cos (\omega t+\phi), \nonumber \]
with
\[\tan \phi=\frac{\omega / q}{\omega^{2}-1} \nonumber \]
We can observe that this solution is also invariant under \((13.6)\) : the small amplitude solution obeys \(\theta(t)=-\theta(t-\pi / \omega) .\) We say that this solution is symmetric, meaning it obeys the same symmetry as the governing equations; that is, the motion of the small amplitude pendulum is symmetrical about the vertical.
also satisfies (13.5) by the following calculation:
\[\begin{aligned} \ddot{\theta}_{2}(t)+\frac{1}{q} \dot{\theta}_{2}(t)+\sin \left(\theta_{2}(t)\right) &=-\ddot{\theta}_{1}(t-\pi / \omega)-\frac{1}{q} \dot{\theta}_{1}(t-\pi / \omega)-\sin \left(\theta_{1}(t-\pi / \omega)\right) \\ &=-f \cos (\omega t-\pi) \\ &=f \cos \omega t \end{aligned} \nonumber \]
If the two solutions \(\theta_{1}(t)\) and \(\theta_{2}(t)\) are equal, then we say that this solution is symmetric. Here, we are considering asymptotic solutions that are independent of the initial conditions, since the initial conditions themselves can also break the symmetry. However, if \(\theta_{1}(t)\) and \(\theta_{2}(t)\) are not equal, we say that these solutions are asymmetric, and that spontaneous symmetry breaking has occurred. Evidently, spontaneous symmetry breaking is a decidedly nonlinear phenomena. After symmetry breaking occurs, the asymmetric solutions must occur in pairs, and the bifurcation point looks like a pitchfork bifurcation, and can be super- or sub-critical. Any subsequent bifurcation that occurs to one of the asymmetric solutions must also be mirrored by the other.
Spontaneous symmetry breaking occurs in the pendulum dynamics at the dimensionless parameter values \(f=1.5, \omega=2 / 3\) , and approximately \(q=1.246 .\) For \(q\) just less than \(1.246\) , the single stable asymptotic solution for \(\theta=\theta(t)\) is symmetric, and for \(q\) just greater than \(1.246\) , there exists a pair of stable asymptotic solutions that are asymmetric.
By projecting the phase-space trajectory onto the \(\theta-\dot{\theta}\) plane, in Fig. \(13.3\) we display both the symmetric solution when \(q=1.24\) and the two asymmetric solutions when \(q=1.30\) . To explicitly observe the symmetry of the solutions, we mark the value of \((\theta, \theta)\) at the time \(t=\) \(n 2 \pi / \omega\) , with \(n\) a positive integer (equivalent to the time \(t=0)\) , and the value of \((-\theta,-\dot{\theta})\) at the time \(t=n 2 \pi / \omega+\pi / \omega\) (equivalent to the time \(t=\pi / \omega\) ). For a symmetric solution, these two points mark the same point on the trajectory, and for asymmetric solutions they mark points on different trajectories. Notice that after the occurrence of symmetry breaking, one of the asymptotic solutions undergoes an oscillation centered to the right of the vertical, and the other, centered to the left.
We can graph a bifurcation diagram associated with the symmetry breaking of the solutions. We fix \(f=1.5\) and \(\omega=2 / 3\) and vary \(q\) across the bifurcation point \(q=1.246\) . We need to distinguish the symmetric from the asymmetric limit cycles, and one method is to compute the average value of \(\theta(t)\) over one period of oscillation; that is,
\[\langle\theta\rangle=\frac{1}{T} \int_{0}^{T} \theta d t \nonumber \]
where \(T=2 \pi / \omega .\) For the symmetric solution, \(\langle\theta\rangle=0\) , whereas \(\langle\theta\rangle\) takes on both positive and negative values after symmetry breaking occurs. In Fig. 13.4, we plot the value of \(\langle\theta\rangle\) versus \(q\) . At a value of approximately \(q=1.246\) , spontaneous symmetry breaking occurs and the stable symmetric limit cycle splits into two asymmetric limit cycles, in what is evidently a supercritical pitchfork bifurcation.
The symmetric limit cycle still exists after the bifurcation point, and though unstable, can also be computed. To compute the unstable cycle, one could determine the values of \(\theta\) and \(\dot{\theta}\) at \(t=0\) that lie on this cycle, and then integrate over one period (over which the instability doesn’t have sufficient time to develop). The problem of determining the correct initial conditions can be cast as a problem in multidimensional root-finding. The key idea is that a symmetric limit cycle satisfies \(\theta(t)=-\theta(t-\pi / \omega)\) . We therefore determine the solution vector of initial conditions \((\theta(0), \theta(0))\) that satisfies the two equations
\[\begin{aligned} &\theta(0)+\theta(\pi / \omega)=0 \\ &\dot{\theta}(0)+\dot{\theta}(\pi / \omega)=0 \end{aligned} \nonumber \]
where \(\theta(\pi / \omega)\) and \(\dot{\theta}(\pi / \omega)\) are determined by integrating from \(t=0\) the differential equations using ode \(45 . \mathrm{m}\) with the initial conditions \((\theta(0), \dot{\theta}(0))\) . Root-finding can be done using either a two-dimensional version of the secant rule or, more simply, the built-in MATLAB function fsolve.m. Convergence to the roots is robust, and an initial guess for the roots can be taken, for instance, as \((\theta(0), \dot{\theta}(0))=(0,0)\) . A plot of the two stable asymmetric limit cycles, and the unstable symmetric limit cycle when \(f=1.5, \omega=2 / 3\) and \(q=1.3\) is shown in Fig. 13.5.
Period-doubling bifurcations
The period-doubling bifurcations continue with increasing \(q\) . The second doubling from period two to period four occurs at approximately \(q=1.370\) and the third doubling from period four to period eight occurs at approximately \(q=1.375\) . The \(\theta-\dot{\theta}\) phase-space projections for period four and period eight are shown in Fig. 13.8.
A bifurcation diagram can be plotted that illustrates these period-doubling bifurcations. We use a Poincaré section to plot the value of \(\theta\) at the times corresponding to \(2 \pi n / \omega\) , with \(n\) an integer. The control parameter for the bifurcation is \(q\) , and in Fig. 13.9, we plot the bifurcation diagram for \(1.34<q<1.38\) . In general, the angle \(\theta\) should be mapped into the interval \(-\pi<\) \(\theta<\pi\) , but for these parameter values there is no rotary motion. Period-doubling bifurcations are observed, and eventually the pendulum becomes aperiodic. Additional windows of periodicity in the aperiodic regions of \(q\) are also apparent.
The aperiodic behavior of the pendulum observed in Fig. \(13.9\) corresponds to a chaotic pendulum. If only the pendulum exhibited the period-doubling route to chaos, then the results shown in Fig. 13.9, though interesting, would be of lesser importance. But in fact many other nonlinear systems also exhibit this route to chaos, and there are some universal features of Fig. \(13.9\) , first discovered by Feigenbaum in 1975. One of these features is now called the Feigenbaum constant, \(\delta\)
Note that for the pendulum, we can already compute
\[\delta_{1}=\frac{1.370-1.348}{1.375-1.370}=4.400 \nonumber \]
The meaning of \(\delta\) can be better elucidated by writing
\[q_{n+2}-q_{n+1}=\frac{q_{n+1}-q_{n}}{\delta_{n}} \nonumber \]
and by continuing to iterate this equation, we obtain
\[q_{n+2}-q_{n+1}=\frac{q_{2}-q_{1}}{\delta_{1} \delta_{2} \cdots \delta_{n}} . \nonumber \]
With all the \(\delta_{n}\) ’s approximately equal to \(\delta\) , we then have the scaling
\[q_{n+2}-q_{n+1} \propto \delta^{-n} \nonumber \]
A \(\delta\) larger than one would then insure that the bifurcations occur increasingly closer together, so that an infinite period (and chaos) is eventually attained. The value of \(\delta\) can be computed to high accuracy and has been found to be
\[\delta=4.669201609102990 \ldots, \nonumber \]
and our value of \(\delta_{1}=4.4\) is a rough approximation.
Period doubling in the logistic map
Feigenbaum originally discovered the period-doubling route to chaos by studying a simple onedimensional map. A one-dimensional map with a single control parameter \(\mu\) can be written as
\[x_{n+1}=f_{\mu}\left(x_{n}\right) \nonumber \]
where \(f_{\mu}(x)\) is some specified function. A one-dimensional map is iterated, starting with some initial value \(x_{0}\) , to obtain the sequence \(x_{1}, x_{2}, x_{3}, \ldots .\) If the sequence converges to \(x_{*}\) , then \(x_{*}\) is a stable fixed point of the map.
The specific one-dimensional map we will study here is the logistic map, with
\[f_{\mu}(x)=\mu x(1-x) \nonumber \]
The logistic map is perhaps the simplest nonlinear equation that exhibits the period-doubling route to chaos. To constrain the values of \(x_{n}\) to lie between zero and unity, we assume that \(0<\mu<4\) and that \(0<x_{0}<1\) .
A period-1 cycle for the logistic map corresponds to a stable fixed point. Stable fixed points are solutions of the equation \(x=f_{\mu}(x)\) , or
\[x=\mu x(1-x) . \nonumber \]
The two solutions are given by \(x_{*}=0\) and \(x_{*}=1-1 / \mu\) . The first fixed point \(x_{*}=0\) must be stable for \(0<\mu<1\) , being the only fixed point lying between zero and one that exists in this range. To determine the stability of the second fixed point, we make use of the linear stability analysis discussed in \(\S 12.1\) .
For a one-dimensional map, \(x_{*}\) is a stable fixed point of \((13.17)\) if \(\left|f_{\mu}^{\prime}\left(x_{*}\right)\right|<1 .\) For the logistic map given by (13.18), \(f_{\mu}^{\prime}(0)=\mu\) so that \(x_{*}=0\) is stable for \(0<\mu<1\) as we have already surmised, and for the second fixed point
\[f_{\mu}^{\prime}(1-1 / \mu)=2-\mu . \nonumber \]
Therefore, we find that \(x_{*}=1-1 / \mu\) is stable for \(1<\mu<3\) .
What happens when \(\mu\) becomes larger than three? We will now show that the first perioddoubling bifurcation occurs at \(\mu=3\) . Because of the simplicity of the logistic map, we can determine explicitly the period-2 cycle. Consider the following composite map:
\[g_{\mu}(x)=f_{\mu}\left(f_{\mu}(x)\right) \nonumber \]
Fixed points of this map will consist of both period-1 cycles and period-2 cycles of the original map (13.18). If \(x=x_{*}\) is a fixed point of the composite map (13.21), then \(x_{*}\) satisfies the equation
\[x=g_{\mu}(x) \nonumber \]
A period-2 cycle of the map \(f_{\mu}(x)\) necessarily corresponds to two distinct fixed points of the composite map (13.21). We denote these two fixed points by \(x_{0}\) and \(x_{1}\) , which satisfy
\[x_{1}=f_{\mu}\left(x_{0}\right), \quad x_{0}=f_{\mu}\left(x_{1}\right) . \nonumber \]
We will call \(x_{0}, x_{1}\) the orbit of the period-2 cycle, with the later generalization of calling \(x_{0}, x_{1}\) , \(\ldots, x_{2^{n}}-1\) the orbit of the period \(-2^{n}\) cycle.
The period-2 cycle can now be determined analytically by solving
\[\begin{aligned} x &=f_{\mu}\left(f_{\mu}(x)\right) \\ &=f_{\mu}(\mu x(1-x)) \\ &=\mu(\mu x(1-x))(1-\mu x(1-x)) \end{aligned} \nonumber \]
which is a quartic equation for \(x\) . Two solutions corresponding to period-1 cycles are known: \(x_{*}=0\) and \(x_{*}=1-1 / \mu\) . Factoring out these two solutions, the second by using long division, results in the quadratic equation given by
\[\mu^{2} x^{2}-\mu(\mu+1) x+(\mu+1)=0 . \nonumber \]
The period-2 cycle, then, corresponds to the two roots of this quadratic equation; that is,
\[x_{0}=\frac{1}{2 \mu}((\mu+1)+\sqrt{(\mu+1)(\mu-3)}), \quad x_{1}=\frac{1}{2 \mu}((\mu+1)-\sqrt{(\mu+1)(\mu-3)}) . \nonumber \]
These roots are valid solutions for \(\mu \geq 3\) . Exactly at \(\mu=3\) , the period-2 cycle satisfies \(x_{0}=x_{1}=\) \(2 / 3\) , which coincides with the value of the period-1 cycle \(x_{*}=1-1 / \mu=2 / 3 .\) At \(\mu=3\) , then, we expect the fixed point of the composite map corresponding to a period-1 cycle to go unstable via a supercritical pitchfork bifurcation to a pair of stable fixed points, corresponding to a period-2 cycle.
When does this period-2 cycle become unstable? At \(\mu=3\) and \(x_{*}=2 / 3\) , we have
\[\begin{aligned} f_{\mu}^{\prime}\left(x_{*}\right) &=\mu\left(1-2 x_{*}\right) \\ &=-1, \end{aligned} \nonumber \]
so that the period-1 cycle becomes unstable when the derivative of the map function attains the value of \(-1\) , and it is reasonable to expect that the period-2 cycle also becomes unstable when
\[g_{\mu}^{\prime}\left(x_{0}\right)=-1, \quad g_{\mu}^{\prime}\left(x_{1}\right)=-1 . \nonumber \]
Now,
\[\begin{aligned} g_{\mu}^{\prime}\left(x_{0}\right) &=f_{\mu}^{\prime}\left(f_{\mu}\left(x_{0}\right)\right) f_{\mu}^{\prime}\left(x_{0}\right) \\ &=f_{\mu}^{\prime}\left(x_{1}\right) f_{\mu}^{\prime}\left(x_{0}\right), \end{aligned} \nonumber \]
and
\[\begin{aligned} g_{\mu}^{\prime}\left(x_{1}\right) &=f_{\mu}^{\prime}\left(f_{\mu}\left(x_{1}\right)\right) f_{\mu}^{\prime}\left(x_{1}\right) \\ &=f_{\mu}^{\prime}\left(x_{0}\right) f_{\mu}^{\prime}\left(x_{1}\right) . \end{aligned} \nonumber \]
The two equations of \((13.26)\) are therefore identical, and the period-2 cycle will go unstable at the value of \(\mu\) satisfying
\[f_{\mu}^{\prime}\left(x_{0}\right) f_{\mu}^{\prime}\left(x_{1}\right)=-1 \nonumber \]
where \(x_{0}\) and \(x_{1}\) are given by (13.25).
The equation for \(\mu\) , then, is given by
\[\mu^{2}\left(1-2 x_{0}\right)\left(1-2 x_{1}\right)=-1 \nonumber \]
Or
\[\mu^{2}\left(1-2\left(x_{0}+x_{1}\right)+4 x_{0} x_{1}\right)+1=0 \nonumber \]
Now, from (13.25),
\[x_{0}+x_{1}=\frac{\mu+1}{\mu}, \quad x_{0} x_{1}=\frac{\mu+1}{\mu^{2}} . \nonumber \]
Substitution of (13.30) into (13.29) results, after simplification, in the quadratic equation
\[\mu^{2}-2 \mu-5=0, \nonumber \]
with the only positive solution given by
\[\begin{aligned} \mu &=1+\sqrt{6} \\ & \approx 3.449490 . \end{aligned} \nonumber \]
We therefore expect the period-2 cycle to bifurcate to a period-4 cycle at \(\mu \approx 3.449490\) .
CHAPTER 13. PENDULUM DYNAMICS
| \(n\) | \(\mu_{n}\) |
|---|---|
| 1 | 3 |
| 2 | \(3.449490 \ldots\) |
| 3 | \(3.544090 \ldots\) |
| 4 | \(3.564407 \ldots\) |
| 5 | \(3.568759 \ldots\) |
| 6 | \(3.569692 \ldots\) |
| 7 | \(3.569891 \ldots\) |
| 8 | \(3.569934 \ldots\) |
Table 13.1: The first eight values of \(\mu_{n}\) at which period-doubling bifurcations occur.
If we define \(\mu_{n}\) to be the value of \(\mu\) at which the period-2 \(^{n-1}\) cycle bifurcates to a period-2 \(^{n}\) cycle, then we have determined analytically that \(\mu_{1}=3\) and \(\mu_{2}=1+\sqrt{6}\) . We list in Table 13.1, the first eight approximate values of \(\mu_{n}\) , computed numerically.
We can compute a bifurcation diagram for the logistic map. For \(2<\mu<4\) , we plot the iterates from the map, discarding initial transients. The bifurcation diagram is shown in Fig. 13.10. Notice the uncanny similarity between the bifurcation diagram for the logistic map, and the one we have previously computed for the damped, driven pendulum equation, Fig. \(13.9 .\) Also note that the computation of Fig. 13.10, being on the order of seconds, is substantially faster than that of Fig. 13.9, which took about an hour, because a one-dimensional map is much faster to compute than the Poincaré section of a pair of coupled first-order differential equations.
Computation of the Feigenbaum constant
Period doubling in the logistic map enables an accurate computation of the Feigenbaum constant \(\delta\) , defined as
\[\delta=\lim _{n \rightarrow \infty} \delta_{n} \nonumber \]
where
\[\delta_{n}=\frac{\mu_{n+1}-\mu_{n}}{\mu_{n+2}-\mu_{n+1}} \nonumber \]
Table \(13.1\) lists the known first eight values of \(\mu_{n}\) at the bifurcation points. These values, and those at even larger values of \(n\) , are in fact very difficult to compute with high precision because of the slow convergence of the iterates at the bifurcation points. Rather, we will instead compute the values of \(\mu\) at what are called superstable cycles. This now well-known method for computing \(\delta\) was first described by Keith Briggs (1989).
Recall that for the general one-dimensional map
\[x_{n+1}=f_{\mu}\left(x_{n}\right), \nonumber \]
a perturbation \(\epsilon_{n}\) to a fixed point \(x_{*}\) decays as
\[\epsilon_{n+1}=f_{\mu}^{\prime}\left(x_{*}\right) \epsilon_{n} . \nonumber \]
At a so-called superstable fixed point, however, \(f_{\mu}^{\prime}\left(x_{*}\right)=0\) and the perturbation decays very much faster as
\[\epsilon_{n+1}=\frac{1}{2} f_{\mu}^{\prime \prime}\left(x_{*}\right) \epsilon_{n}^{2} . \nonumber \]
What are the superstable fixed points of the logistic map? Now, the values \(x_{i}\) that are in the orbit of a period \(-2^{n}\) cycle are fixed points of the composite map
\[x_{n+1}=g_{\mu}\left(x_{n}\right), \nonumber \]
where \(g_{u}=f_{u} \circ f_{u} \circ \cdots \circ f_{u}\) , where the composition is repeated \(2^{n}\) times. The orbit of a superstable cycle, then, consists of superstable fixed points of \((13.37)\) . If the period- \(2^{n}\) cycle has orbit \(x_{0}, x_{1}, \ldots, x_{2^{n}-1}\) , we have \(f_{\mu}\left(x_{0}\right)=x_{1}, f_{\mu}\left(x_{1}\right)=x_{2}, \ldots, f_{\mu}\left(x_{2^{n}-1}\right)=x_{0}\) , and by the chain rule,
\[g_{\mu}^{\prime}\left(x_{i}\right)=f_{\mu}^{\prime}\left(x_{0}\right) f_{\mu}^{\prime}\left(x_{1}\right) \cdots f_{\mu}^{\prime}\left(x_{2^{n}-1}\right), \nonumber \]
for all \(x_{i}\) in the orbit of the period- \(2^{n}\) cycle. With
\[f_{\mu}^{\prime}(x)=\mu(1-2 x) \nonumber \]
we have \(g_{\mu}^{\prime}(x)=0\) for \(x=1 / 2\) . Therefore, if \(x_{0}=1 / 2\) , say, is in the orbit of a period- \(2^{n}\) cycle, then this cycle is superstable. At the bifurcation point creating a period-2 \(^{n}\) cycle, the cycle has marginal stability and \(g_{\mu}^{\prime}\left(x_{0}\right)=1 .\) As \(\mu\) increases, \(g_{\mu}^{\prime}\left(x_{0}\right)\) decreases, and eventually the period-2 \(^{n}\) cycle loses stability when \(g_{\mu}^{\prime}\left(x_{0}\right)=-1\) . At some intermediate value of \(\mu\) , then, there exists a value of \(x_{0}\) with \(g_{\mu}^{\prime}\left(x_{0}\right)=0\) , and here we may assign \(x_{0}=1 / 2 .\) Therefore, every period- \(2^{n}\) cycle contains a value of \(\mu\) for which \(x_{0}=1 / 2\) is in the orbit of the cycle.
Accordingly, we define \(m_{n}\) to be the value of \(\mu\) at which \(x_{0}=1 / 2\) is in the orbit of the period- \(2^{n}\) cycle. We can modify the definition of the Feigenbaum constant (13.33) to be
\[\delta_{n}=\frac{m_{n+1}-m_{n}}{m_{n+2}-m_{n+1}} . \nonumber \]
Though the values of \(\delta_{n}\) computed from (13.33) and (13.40) will differ slightly, the values as \(n \rightarrow \infty\) should be the same.
We can easily determine the first two values of \(m_{n}\) . For the period-1 cycle, we have
\[\frac{1}{2}=m_{0} \frac{1}{2}\left(1-\frac{1}{2}\right) \nonumber \]
or \(m_{0}=2\) , as confirmed from Fig. 13.10. To determine \(m_{1}\) , we make use of the period-2 cycle given by \((13.21)\) , and Fig. \((13.10)\) , which shows that the smaller root passes through \(x_{0}=1 / 2\) . Therefore,
\[\frac{1}{2}=\frac{1}{2 m_{1}}\left(\left(m_{1}+1\right)-\sqrt{\left(m_{1}+1\right)\left(m_{1}-3\right)}\right) ; \nonumber \]
and solving for \(m_{1}\) , we obtain the quadratic equation
\[m_{1}^{2}-2 m_{1}-4=0 \nonumber \]
with solution \(m_{1}=1+\sqrt{5} \approx 3.2361\) . Further values of \(m_{n}\) will be computed numerically.
To determine \(m_{n}\) , we need to solve the equation \(G(\mu)=0\) , where
\[G(\mu)=g_{\mu}(1 / 2)-\frac{1}{2} \nonumber \]
and where as before, \(g_{\mu}(x)\) is the composition of \(f_{\mu}(x)\) repeated \(2^{n}\) times. The roots of (13.44) are given by \(m_{0}, m_{1}, \ldots m_{n}\) , so that the desired root is the largest one.
We shall use Newton’s method, \(\S 2.2\) , to solve (13.44). To implement Newton’s method, we need to compute both \(G(\mu)\) and \(G^{\prime}(\mu)\) . Define \(N=2^{n}\) . Then using the logistic map
\[x_{n+1}=\mu x_{n}\left(1-x_{n}\right), \nonumber \]
and iterating with \(x_{0}=1 / 2\) , we obtain \(x_{1}, x_{2}, \ldots, x_{N} .\) This orbit is superstable if \(x_{N}=1 / 2\) . Therefore, we have
\[G(\mu)=x_{N}-1 / 2 \nonumber \]
Moreover,
\[G^{\prime}(\mu)=x_{N^{\prime}}^{\prime} \nonumber \]
where the derivative is with respect to \(\mu\) . From (13.45), we have
\[\begin{aligned} x_{n+1}^{\prime} &=x_{n}\left(1-x_{n}\right)+\mu x_{n}^{\prime}\left(1-x_{n}\right)-\mu x_{n} x_{n}^{\prime} \\ &=x_{n}\left(1-x_{n}\right)+\mu x_{n}^{\prime}\left(1-2 x_{n}\right) . \end{aligned} \nonumber \]
Since we always choose \(x_{0}=1 / 2\) , independent of \(\mu\) , we have as a starting value \(x_{0}^{\prime}=0 .\) Therefore, to compute both \(x_{N}\) and \(x_{N^{\prime}}^{\prime}\) we iterate \(2^{n}\) times the coupled map equations
\[\begin{aligned} &x_{n+1}=\mu x_{n}\left(1-x_{n}\right) \\ &x_{n+1}^{\prime}=x_{n}\left(1-x_{n}\right)+\mu x_{n}^{\prime}\left(1-2 x_{n}\right) \end{aligned} \nonumber \]
| \(n\) | \(m_{n}\) | \(\delta_{n}\) |
|---|---|---|
| 0 | 2 | \(4.7089430135405\) |
| 1 | \(1+\sqrt{5}\) | \(4.6807709980107\) |
| 2 | \(3.4985616993277\) | \(4.6629596111141\) |
| 3 | \(3.5546408627688\) | \(4.6684039259164\) |
| 4 | \(3.5666673798563\) | \(4.6689537409802\) |
| 5 | \(3.5692435316371\) | \(4.6691571813703\) |
| 6 | \(3.5697952937499\) | \(4.6691910014915\) |
| 7 | \(3.5699134654223\) | \(4.6691994819801\) |
| 8 | \(3.5699387742333\) | \(4.6692010884670\) |
| 9 | \(3.5699441946081\) | \(4.6692015881423\) |
| 10 | \(3.5699453554865\) | \(4.6692023902759\) |
| 11 | \(3.5699456041111\) | \(4.6691974782669\) |
| 12 | \(3.5699456573588\) | \(4.6693329633696\) |
| 13 | \(3.5699456687629\) | |
| 14 | \(3.5699456712052\) |
Table 13.2: The first fourteen values of \(m_{n}\) , and estimates of the Feigenbaum delta.
with initial values \(x_{0}=1 / 2\) and \(x_{0}^{\prime}=0 .\) Newton’s method then solves for \(m_{n}\) by iterating
\[\mu^{(i+1)}=\mu^{(i)}-\frac{x_{N}-1 / 2}{x_{N}^{\prime}}, \nonumber \]
until convergence of \(\mu^{(i)}\) to \(m_{n}\) . In double precision, we have been able to achieve a precision of about 14 digits, which we find can be obtained in fewer than 5 iterations of Newton’s method.
For Newton’s method to converge to \(m_{n}\) , we need a good guess for \(\mu^{(0)} .\) We can use the previous best estimate for the Feigenbaum delta to predict \(m_{n}\) . From \((13.40)\) , we find
\[\mu^{(0)}=m_{n-1}+\frac{m_{n-1}-m_{n-2}}{\delta_{n-2}} . \nonumber \]
Although we can not compute \(\delta_{n-2}\) without knowing \(m_{n}\) , we can nevertheless use the estimate \(\delta_{n-2} \approx \delta_{n-3} .\) The computation, then, starts with \(n=2\) , and we can begin by taking \(\delta_{-1}=4.4\) , so that, for example,
\[\begin{aligned} \mu^{(0)} &=3.2361+\frac{3.2361-2}{4.4} \\ &=3.5170 . \end{aligned} \nonumber \]
Using this algorithm, we have produced Table \(13.2\) for \(m_{n}\) , with corresponding calculations of \(\delta_{n} .\) As the values of \(m_{n}\) converge, the corresponding values of \(\delta_{n}\) begin to lose precision. It would appear that our best estimate from the table is \(\delta \approx 4.66920\) , compared to the known value of \(\delta=4.669201609102990 \ldots\) , computed by a different algorithm capable of achieving higher precision.
Strange attractor of the chaotic pendulum
After the period-doubling cascade, the pendulum motion becomes chaotic. We choose parameter values
\(q=4, f=1.5\)
, and
\(\omega=2 / 3\)
in the chaotic regime, and after discarding an initial transient of 256 forcing periods, compute a Poincaré section of the phase-space trajectory in the
\(\theta-\dot{\theta}\)
plane, sampling points every forcing period. The values of the periodic variable
\(\theta\)
are mapped onto the interval
\(-\pi<\theta<\pi\)
. The full Poincaré section consisting of 50,000 points is shown in the top drawing of Fig. 13.11, and a blowup of the points within the drawn rectangle (from a sample of 200,000 points over the entire attractor) is shown in the bottom drawing.
From Fig. 13.11, it appears that the Poincaré section has structure on all scales, which is reminiscent of the classical fractals discussed in \(\S 12.7\) . The set of points shown in Fig. 13.11 is called a strange attractor, and will be seen to have a fractional correlation dimension.
Using the algorithm for computing a correlation dimension discussed in \(\S 12.7\) , we draw a
log-log plot of the correlation integral \(C(r)\) versus \(r\) , shown in Fig. 13.12. A least-squares fit of a straight line to the middle region of the plot yields a correlation dimension for the Poincaré section of approximately \(D=1.25\) .


23.1D: Proteins: Sources, Uses in the Body, and Dietary Requirements
Proteins are composed of 20 different amino acids, about half of which are essential, meaning they must be obtained from the diet.
- Evaluate protein sources and uses in the body
Key Points
- Protein -based foods (plant and animal) provide amino acids; however, the best source of essential amino acids is animal.
- If the diet does not provide adequate protein, the body will obtain what it needs from itself, especially from its own muscles.
- While adequate protein is required for building skeletal muscle and other tissues, there is ongoing debate regarding the use and necessity of high-protein diets in anaerobic exercise, in particular weight training and bodybuilding.
- This use of protein as a fuel is particularly important under starvation conditions as it allows the body’s own proteins to be used to support life.
Key Terms
- amino acid : Any of the twenty naturally occurring α-amino acids (having the amino, and carboxylic acid groups on the same carbon atom), and a variety of side chains, that combine, via peptide bonds, to form proteins.
- denaturation : Denaturation is a process in which proteins or nucleic acids lose their tertiary and secondary structure which is present in their native state, by application of some external stress or compound such as a strong acid or base, a concentrated inorganic salt, an organic solvent (e.g., alcohol or chloroform), or heat.
- protein : Any of numerous large, complex naturally-produced molecules composed of one or more long chains of amino acids, in which the amino acid groups are held together by peptide bonds.
Most microorganisms and plants can biosynthesize all 20 standard amino acids, while animals (including humans) must obtain some of the amino acids from the diet. The amino acids that an organism cannot synthesize on its own are referred to as essential amino acids. Key enzymes that synthesize certain amino acids are not present in animals. One example is aspartokinase, which catalyzes the first step in the synthesis of lysine, methionine, and threonine from aspartate.
Amino Acid : Ball-and-stick model of the cystine molecule, an amino acid formed from two cysteine molecules.
In animals, amino acids are obtained through the consumption of foods containing protein. Ingested proteins are then broken down into amino acids through digestion, which typically involves denaturation of the protein through exposure to acid and hydrolysis by enzymes called proteases. Some ingested amino acids are used for protein biosynthesis, while others are converted to glucose through gluconeogenesis, or fed into the citric acid cycle. This use of protein as a fuel is particularly important under starvation conditions as it allows the body’s own proteins to be used to support life, particularly those found in muscle. Amino acids are also an important dietary source of nitrogen.
Protein : Muscle meat, such as steak, is an excellent source of the essential amino acids the body needs to create all necessary proteins.
A high-protein diet is often recommended by bodybuilders and nutritionists to help efforts to build muscle and lose fat. It should not be confused with low-carb diets, such as the Atkins Diet, which are not calorie-controlled and which often contain large amounts of fat. While adequate protein is required for building skeletal muscle and other tissues, there is ongoing debate regarding the use and necessity of high-protein diets in anaerobic exercise in particular weight training and bodybuilding. Extreme protein intake (in excess of 200g per day), coupled with inadequate intake of other calorie sources (fat or carbohydrates), can cause a form of metabolic disturbance and death commonly known as rabbit starvation.
Relatively little evidence has been gathered regarding the effect of long-term high intake of protein on the development of chronic diseases. Increased load on the kidney is a result of an increase in reabsorption of NaCl. This causes a decrease in the sensitivity of tubuloglomerular feedback, which, in turn, results in an increased glomerular filtration rate. This increases pressure in glomerular capillaries. When added to any additional renal disease, this may cause permanent glomerular damage.


2.4: A Look at Standard Industry Practices
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To legally operate a childcare program in California, a center must comply with certain licensing policies and procedures. These state regulations provide “a baseline standard and are primarily focused on protecting children from harm rather than on advancing child development and early learning,” (Workman & Ullrich, 2017, p. 3). In other words, Title 22 Regulations stipulate health and safety standards, space and square footage requirements, supervision standards, and teacher qualifications. Title 22 Regulations do not however, consider curriculum activities, age appropriate materials, or teacher-to-child interactions, nor does it address developmentally appropriate practices such as family culture and perspectives, child development theories, or principles and practices.
To ensure early childhood education programs are compliant with all state policies and procedures, a licensing analyst will conduct an annual inspection, or will evaluate a program as needed. Typically, the analyst will look at the center’s overall cleanliness and they will inspect both the indoor and outdoor environment to certify that the center is safe for the children. The analyst will confirm that the ratios are met, and that there is proper space and square footage available for each child to play and nap. The analyst will also conduct a spot check of the employee files to confirm teacher qualifications, and they will look through the children’s files to validate that the proper paperwork is signed and in complete order. To comply with state regulations and policies, a center director will need to download the California Code of Regulations, Title 22 administered by CDSS (Divisions 12 only) and follow all the mandated requirements. If any criteria are not met, the analyst will cite the program. The program will then be given a certain timeframe to rectify and correct the concern. The analyst will return to verify the problem has been fixed. Families who are considering a program can check the Community Care Licensing website to see if a center has received any “substantiated citations.” [7]


6.8: Carousels
Carousels are typically used to present a series of panels or images that rotate at a particular frequency.
WAI-ARIA roles, states, and properties used in carousels
- role="region"
- aria-live="polite"
- tabindex="0"
- aria-describedby="[id of div with instructions]"
- aria-hidden="(true|false)"
The following JSFiddle presents a typical carousel widget. Review the JavaScript and HTML markup. Test the carousel presented under the Result tab with ChromeVox to understand how it functions without any accessibility features added. You can work in JSFiddle itself by clicking “Edit in JSFiddle.” Copy the accessibility/WAI-ARIA code described below to fix the accessibility of the accordion before completing Activity 11 on the page that follows.
Though instructions are not always required, they can be helpful for screen reader users when there is non-standard keyboard navigation. In our case, we’ll add a few words and assign them to the “instructions” variable in the default settings of the
init()
function for the carousel. The instructions will be rendered in its own

and referenced with
aria-describedby
a little later in the code.
We’ll define a few attributes when the carousel is initialized: give it a
role="region"
to add it to the landmarks, add a
tabindex
to make it keyboard focusable, and reference the ID of the instructions

with
aria-describedby
. Add keyboard operability with
.on('keydown')
and a reference to the
onKeyDown
function, described below.
Screen reader users will not need the Next/Previous controls, so hide them. They will be using the Arrow keys instead, defined in the
onKeyDown
function further below.
Hide images from screen readers. Notice that the
alt
text for the images are defined in the HTML but left empty so it is not read in this case. Screen readers will read the
figcaptions
.
Add screen reader instructions by generating a

that contains the instruction text defined earlier and hide the

by default. The instructions are read when the carousel receives focus, and the
aria-describedby
attribute is dynamically added to reference the instructions.
Add an
aria-live
attribute to the
stopTimer
function. Set its value to
polite
so content updating in the live region announces when a screen reader is not reading elsewhere on the page. The content of the visible carousel panel is read automatically when it is in focus, manually navigating between panels with the Arrow keys.
Remove the live region when focus on the carousel is removed in the
startTimer
function. By doing so, the live region stops reading when the timer is reactivated
onblur
, and it does not interfere with the screen reader reading elsewhere on the page.
Hide the active slide from screen readers with
aria-hidden="true"
. Then, make the next slide visible to screen readers with
aria-hidden="false"
in the
gotoSlide
function.
Adding Keyboard Operability
Add keyboard operations for the carousel, pulling keyboard events from ik_utils.js to use Left and Right arrows for moving between panels in the carousel, and the Esc key to exit the carousel and resume automatic rotation.
Accessible Carousel in Action
Watch the following video to see how ChromeVox interacts with a carousel. The carousel rotates automatically when focus is elsewhere on the page. When it receives focus, rotation stops, and navigation instructions are read. The Left and Right arrow keys are used to move manually between panels in the carousel while it has focus. The contents of each panel are read through a live region, dynamically added to the main container

when the carousel has focus. Using the Tab key while the carousel has focus sends focus to any focusable element within the panel that is in view, a link to the person who shared the photo in this case. Aim to have the carousel you update in the activity on the following page operate and announce like the one in the video.
Video: Accessible Carousel


8.8: Review and Additional Resources
Review
Review and assess your learning. Start with the "Important Terms and Concepts" to ensure you know the terminology related to the topic of the chapter and concepts discussed. Move on to the "Review Questions" to answer critical thinking questions about concepts and processes discussed in the chapter. Finally, test your overall understanding by taking the "Self-assessment quiz".
-
- Air mass
- a vast pool of air having similar temperature and moisture characteristics over its horizontal extent.
-
- Air mass source region
- where air masses are born; where they take on their characteristic temperature and moisture content
-
- cP air mass
- Continental polar (cP) air is considered a cold and dry air mass that is warmer than the arctic air mass located to the north. Continental polar air is typically a stable or conditionally stable mass of air.
-
- mP air mass
- Maritime polar (mP) air is cool and moist air that brings mild weather to coastal locations. Maritime polar air is warmer than continental polar air in the winter as the surface temperature of the ocean is higher.
-
- cA air mass
- Continental arctic air (cA) is typically described as extremely cold and dry.
-
- cAA air mass
- an exceedingly cold air mass and is drier than its arctic counterpart as the source region is the continent of Antarctica.
-
- cT air mass
- Found over subtropical and tropical continents the source region for the hot and dry
-
- Air mass modification
- As air masses move out of their source region the temperature and moisture content of air masses are modified.
-
- Front
- boundaries between contrasting masses of air.
-
- Cold front
- where cold air replaces warm air.
-
- Warm front
- where warm air replaces cold air.
-
- Occluded front
- forms when a cold front catches up with a warm front.
-
- Stationary front
- where no change in air masses or movement of the front occurs.
-
- Quasi-stationary front
- found migrating within a particular latitudinal zone throughout most of the year.
-
- Polar front
- the boundary between polar-type air and tropical-type air.
-
- Arctic front
- the boundary between cold arctic air mass and the warmer air of the polar cell.
-
- Warm front weather
-
- Cold front weather
-
- Warm sector weather
- increase in the humidity of the air; "fair weather" clouds are often created by convection and instability in the warm and humid afternoon air. After a while the winds start to shift to the southwest and the humidity continues to rise.
-
- Cyclogenesis
- the development and intensification of a cyclone
-
- Initial Stage - Cyclogenesis
- opposing streams of air meet and create a cyclone shear
-
- Wave Stage- Cyclogenesis
- Once the air collides and cyclonic circulation commences, warm air from the south invades where cold air was once located north of the polar front. The developing system takes on the characteristic wave form
-
- Occluded Stage- Cyclogenesis
- Being more dense, the air behind the cold front can "bulldoze" the warmer and less dense air out of the way. The warm air rises off the surface and glides up and over the colder more dense air ahead of the warm front. there is less horizontal displacement and the warm front moves slower across the earth than does a cold front. Over time the cold front catches up with the warm front and the cyclone starts to occlude.
-
- Dissolving Stage- Cyclogenesis
- The system enters the dissolving stage after it occludes and the lifting mechanism is cutoff. Without the convergence and uplift, the cyclone dissipates in the atmosphere.
-
- Thunderstorm - Cumulus stage
- The initial stage of development. During this stage warm, moist, and unstable air is lifted from the surface. As the air ascends, it cools and upon reaching its dew point temperature begins to condense into a cumulus cloud. Near the end of this stage precipitation forms.
-
- Thunderstorm - Mature Stage
- The second stage - During the mature stage warm, moist updrafts continue to feed the thunderstorm while cold downdrafts begin to form. The downdrafts are a product of the entrainment of cool, dry air into the cloud by the falling rain. As rain falls through the air it drags the cool, dry air that surrounds the cloud into it. As dry air comes in contact with cloud and rain droplets they evaporate cooling the cloud. The falling rain drags this cool air to the surface as a cold downdraft.
-
- Thunderstorm - Dissipating Stage
- The final stage - the thunderstorm dissolves away. By this point, the entrainment of cool air into the cloud helps stabilize the air. In the case of the air mass thunderstorm, the surface no longer provides enough convective uplift to continue fueling the storm. As a result, the warm updrafts have ceased and only the cool downdrafts are present. The downdrafts end as the rain ceases and soon the thunderstorm dissipates.
-
- Lightning
- a massive discharge of electricity in response to a charge differential.
-
- Thunder
- a shockwave through the atmosphere caused by lightning
-
- Tornado
- an intense system of low pressure with violent updrafts and converging winds.
-
- Hurricane
- a large rotating cyclonic system born in the tropics.
-
- Storm Surge
- the high water level that accompanies a hurricane as it comes ashore.
Describe the characteristics of an air mass source region.
- Answer
-
An air mass source region is one that has flat terrain and dominated by high pressure
Describe two ways in which air masses can be modified.
- Answer
-
Air masses are modified by the surface over which they travel. For example, an mT air mass that passes over a cold ocean current is chilled at the surface enhancing stability. Cold air masses traveling over warm surface are heated enhancing their instability.
Describe the characteristics of a front.
- Answer
-
A front is boundary between contrasting masses of air. They are a zone of converging air and lower pressure.
Compare and contrast mP, cP, mT, cT air masses.
- Answer
-
mP air masses are cool and moist forming over oceans at about 60 o N and S.
cP air masses are cold and dry forming over continents at about 60 o N.
mT air masses are warm and moist air masses forming over oceans at about 30 o N and S.
cT air are warm and dry air masses forming over continents at about 30 o N and S.
Describe the process of cyclogenesis.
- Answer
-
Cyclones form along frontal boundaries like the polar front where air is converging from opposite directions. A twisting motion is created as cyclonic shear takes place between the converging air streams. The surface convergence is supported by upper level divergence in the jet stream. The spiraling storm creates a cold front as cold air pushes into regions once occupied by warm air. A warm front is spawned where warm air intrudes into regions once dominated by cold air. Through time the cold front and warm front merge to form an occlusion. Finally the storm dissipates.
Describe the weather changes (e.g. temperature change, pressure tendency, clouds, wind direction) associated with the passage of a warm front.
- Answer
-
A change in cloud cover from cirrus, to cirrostratus, altostratus, and stratus is typical of the approach of a warm front. Air pressure decreases as the front approach, then increases after passage. Nimbostratus clouds are present along the warm front if it is precipitating. Low intensity/long duration precipitation is typical of nimbostratus clouds. Wind are likely to shift from an easterly direction to a southerly direction as the warm from passes. Air temperatures rise upon the passage of the front.
Describe the weather changes (e.g. temperature change, pressure tendency, clouds, wind direction) associated with the passage of a cold front.
- Answer
-
The air ahead of a cold front is generally from the south in the warm sector. Pressure drops as the front approaches. Cumulus clouds grow into cumulonimbus clouds and intense, but short-lasting precipitation occurs. As the front passes, wind shifts from a southerly/southwesterly direction toward the west and possibly northwest. Temperatures decrease as the front passes. After passage, cloud cover dissipates.
Compare and Contrast the kind of precipitation associated with cold and warm fronts.
- Answer
-
Precipitation along warm fronts tends to be low intensity and long in duration. Precipitation along cold fronts tends to be high intensity and short in duration.
Describe the three stages of thunderstorm development.
- Answer
-
Cumulus stage: Creation of a cumulus cloud by updrafts of warm, moist air
Mature stage: Updrafts of warm moist air fuels the developing storm. Downdrafts of cold dry air is created by precipitation entrainment.
Decay stage: Downdrafts predominate as uplift of warm moist air ceases.
Describe the conditions under which tornadoes form.
- Answer
-
There is much we do not know about tornadoes. Tornadoes are most common during the Spring when greatly contrasting air masses collide to produce severe storm systems. Wind shear within the severe thunderstorm causes rotation of air about a horizontal axis. The rotating circulation is tilted into the vertical by the updrafts in the thunderstorm. As the rotating air increases in height and shrinks in size a mesocyclone is formed. For whatever reason, a tornado funnel is spawned within the mesocyclone.
Under what conditions are hurricanes most likely to form?
- Answer
-
A hurricane develops from a tropical disturbance once it reaches sustained winds in excess of 75 mph (65 knots). Most hurricanes form poleward of 10 o latitude as the Coriolis effect is too weak closer to the equator. Hurricanes form in a uniform mass of warm air over tropical oceans with temperatures of 80 o F (26.5 o C) through a depth of 200 feet (60 meters).
What creates the most damage from a hurricane?
- Answer
-
The high water level that precedes the hurricane called the "storm surge" creates the most damage due to flooding.
-
At a warm front
- warm air replaces cold air
- a trough of lower pressure is found
- stratus - type clouds are found
- all the above
-
Which of the following air masses are colder than the surface over which it is traveling?
- mTw
- mTu
- cPk
- none of the above
-
The cold sector of a midlatitude cyclone
- experiences southerly winds
- generates nimbostratus clouds
- experiences light precipitation of long duration
- none of the above
-
The mature stage of thunderstorm formation
- experiences mostly updrafts of air
- experiences updrafts and downdrafts
- experiences mostly downdrafts
- none of the above
-
Which of the following is not a characteristic of a source region
- a region of little to no windiness
- areas dominated by high pressure
- areas dominated by low pressure
- regions of low relief
-
A front
- is a boundary between contrasting masses of air
- is a zone of lower pressure
- is a boundary between opposing centers of high pressure
- can have all the above characteristics
-
______ is fundamentally responsible for the creation of midlatitude cyclones.
- large pressure gradients
- cyclonic shear
- large temperature gradients
- none of the above
-
Which of the following is not associated with a cold front
- cumulonimbus clouds
- warm air replacing cold air
- possibility of intense rain of short duration
- all the above are associated with a cold front.
-
The wind direction ahead of a warm front is generally
- from the east
- from the south
- from the north
- from the west
-
Which of the following is not true about cP air?
- cP air is generally unstable air
- cP air is generally found behind a cold front
- cP air that affects North America has its source in central Canada
- cP air is found in the cold sector of a midlatitude cyclone
- Answer
-
- D
- C
- D
- B
- C
- D
- B
- B
- A
- A
Additional Resources
Focus on The Physical Environment : "Weather Forecasting through the Ages" ( NASA Earth Observatory )
Connections : " Trees lost to Katrina May Present Climate Challenge" (NPR)
Physical Geography Today : US National Weather Service - NOAA | US National Hurricane Center - NOAA
World of Change : Severe Storms ( NASA Earth Observatory )
Multimedia
"Super Twisters" (National Geographic)
"Tracking Hurricanes" (PBS) The News Hour with Jim Lehrer Oct. 1, 2003 "For years, scientists have worked to develop powerful new tools to predict the paths of hurricanes and monitor their likely impact. Betty Ann Bowser reports on the latest technologies meteorologists are using to keep tabs on these devastating storms and what new tools are being developed."
"Hunting Killer Storms: Flying into the Eye of Isabel WGBH Forum Network Mish Michaels, meteorologist, WBZ-TV4 talks about her adventures chasing tornadoes, flying into hurricanes and trekking to Mt. Washington during the winter. (54:50)
" Tornado Sound " (NPR) Weekend Edition May 18, 2003 "Host Liane Hansen speaks with physicist Al Bedard, who studies the sounds of tornadoes at the Environmental Technology Laboratory of the National Oceanic and Atmospheric Administration.
"Looking at Hurricanes" (NASA/GSFC) Explore the "latest" technology for observing hurricanes.
[" Storm that Drowned a City " from NOVA for an in depth look at Hurricane Katrina's impact on New Orleans, LA.]
Interactivities
- Realtime station model plotting and interpretation (WeatherWise)
Readings
Tornadoes....Nature's Most Violent Storms
Visualization
Weather Map Viewer (M. Ritter) - Current weather map slide show.
"How hurricanes form" (CNN)


3.2.6: Summary
3.2.2 Entrepreneurial Opportunity
Entrepreneurial opportunity exists when consumer demand meets the feasibility of bringing the product or service to market. Joseph Schumpeter, an early scholar of entrepreneurship, identified creative destruction—when innovation is disruptive and creates potential entrepreneurial opportunities. Schumpeter’s focus on economics led him to categorize potential opportunities based on supply, demand, or changes in how technology is used.
Burgeoning entrepreneurs should be engaged with industries of interest to remain informed and aware of opportunities to research. Additionally, they should be attentive to drivers of opportunity, such as emerging funding options, technological advancements, and economic factors.
3.2.3 Researching Potential Business Opportunities
Researching the viability of your entrepreneurial idea will help you determine whether the venture is currently worth pursuing. Use primary and secondary research to determine whether there is adequate demand, market structure and size, and the needed margins and resources to launch and sustain the enterprise. Entrepreneurs researching options should also be aware of industry and consumer sources of opportunities.
3.2.4 Competitive Analysis
A competitive analysis helps determine the viability of your idea by considering how feasible it is within a specific competitive landscape. Tools such as a competitive analysis grid, a SWOT analysis, a PEST analysis, and the three circles approach can help you further examine issues and refine your plans.


Chapter 6: Proofs
Consider two arguments in SL:
Argument A
\(P\) ∨ \(Q\)
¬\(P\)
.˙. \(Q\)
Argument B
\(P\) → \(Q\)
\(P\)
.˙. \(Q\)
Clearly, these are valid arguments. You can confirm that they are valid by constructing four-line truth tables. Argument A makes use of an inference form that is always valid: Given a disjunction and the negation of one of the disjuncts, the other disjunct follows as a valid consequence. This rule is called disjunctive syllogism .
Argument B makes use of a different valid form: Given a conditional and its antecedent, the consequent follows as a valid consequence. This is called modus ponens .
When we construct truth tables, we do not need to give names to different inference forms. There is no reason to distinguish modus ponens from a disjunctive syllogism. For this same reason, however, the method of truth tables does not clearly show why an argument is valid. If you were to do a 1024-line truth table for an argument that contains ten sentence letters, then you could check to see if there were any lines on which the premises were all true and the conclusion were false. If you did not see such a line and provided you made no mistakes in constructing the table, then you would know that the argument was valid. Yet you would not be able to say anything further about why this particular argument was a valid argument form.
The aim of a proof system is to show that particular arguments are valid in a way that allows us to understand the reasoning involved in the argument. We begin with basic argument forms, like disjunctive syllogism and modus ponens. These forms can then be combined to make more complicated arguments, like this one:
(1) ¬\(L\) → (\(J\) ∨ \(L\))
(2) ¬\(L\)
.˙. \(J\)
By modus ponens, (1) and (2) entail \(J\)∨\(L\). This is an intermediate conclusion . It follows logically from the premises, but it is not the conclusion we want. Now \(J\) ∨ \(L\) and (2) entail \(J\), by disjunctive syllogism. We do not need a new rule for this argument. The proof of the argument shows that it is really just a combination of rules we have already introduced.
Formally, a proof is a sequence of sentences. The first sentences of the sequence are assumptions; these are the premises of the argument. Every sentence later in the sequence follows from earlier sentences by one of the rules of proof. The final sentence of the sequence is the conclusion of the argument.
This chapter begins with a proof system for SL, which is then extended to cover QL and QL plus identity.


3.8: Day 2- Medical Ward
Day: 2
Time: 07h00
Place: Medical Ward
Tracie walks into the nursing station and heads to her usual chair. She moves the assignment book towards her and see that she has the same assignment as yesterday. That’s ok , she thinks. At least I know the patients . She looks up at the patient board and sees that there have been only two admissions since yesterday and one discharge. “Looks like it could be a busy discharge and admissions day.”
Jim comes down and pulls his chair close to Tracie’s “Hi Tracie, how’s it going?”
“So far so good, how was your night?”
“Night was quiet. Mrs. Johns required some adjustments in her Optiflow which the RTs took care of. Mr Alex had a bit of sundowner syndrome and was quite restless, but otherwise the night was pretty quiet.”
“Great. Ok, let’s do the report so you can get home.”
“Yeah, I have to get out of here quick. I’m taking the kids on a field trip later in the afternoon. Twenty-six Grade 2 kids. I can’t believe I volunteered to chaperone after a night shift.”
Tracie laughs, “Lucky you.”
Jim then walks Tracie through the night and updates her on changes to her patients. After reporting, Jim heads out, while Tracie quickly looks through the charts to plan her morning.
Tracie notes she has a lot of 08h00 medications so decides to prepare these meds and then do her patient checks at the same time.
Tracie makes her rounds and drops off all her medications. With that done, she notes to herself, Everyone seems all right now, having breakfast. I didn’t see any issues off the top. I’ll give everyone a few minutes and then do vitals, assessments and discharges.
Tracie sits down to check the RT’s note on Erin Johns when Erin’s call bell rings. She gets up from her chair and walks down the hallway to her room.
“Hi, Mrs. Johns. How can I help?”
“I can’t catch my. Breath. I took. My oxygen off. Just to eat. Put it. Back on. I am. Short of breath.”
“I want you to take some deep breaths.  Remember to close your lips slightly as you breathe out. I’m going to get the vitals machine and be right back.”
Tracie grabs a vital signs machine from the hallway charger and brings it into Erin’s room to take her vital signs.
“Ok, let’s start checking you out.” Tracie wraps the blood pressure cuff around Erin’s left arm, pulse ox on the right forefinger, and temperature probe under the tongue. About 30 seconds later she notes that the blood pressure is up a bit and oxygen saturation is down. Temperature is unchanged with the same low grade fever. Respiratory rate is up as well and heart rate is above 110. This is not going in the right direction, she thinks. I may need the RT to come see her.
| Day: 2 | Pulse Rate | Blood Pressure | Respiratory Rate | Temperature | O 2 Saturation |
| Time: 07h30 | 112 | 165/90 | 22 | 36.5°C | 83% on RA |
“Ok, Mrs. Johns, let’s listen to your heart and lungs.”
Tracie systematically auscultates, thinking, Not any change from what I remember from yesterday.
“How are you feeling now?”
“A bit better. Not great.”
“I’m going to ask the RT to come and see you right away.”
“Oh my. Am I dying?”
“Gosh, no, Mrs. Johns. The respiratory therapist is here to help both you and me. They manage your oxygen and help you with breathing.”
“Oh. Good.”
“I will be right back.” Tracie moves quickly out to the main hallway to the nursing station. She then asks for the RT to be paged and to come to Erin John’s room.
The unit coordinator looks up. “What should I tell them?”
“Mrs. Johns is quite short of breath and has low sats on Optiflow. I’m not sure what to do next.”
“Done. If they have any further questions, I’ll transfer their call to the phone outside the room.”
“Thank you.”
Tracie heads back to Erin’s room.
Tracie has the vitals machine do another cycle of vital signs but not the temperature. She reads the machine, thinking, Heart rate is down to slightly less than 100. Respiratory rate is still up and sats have only improved slightly. Tracie records all this information and her assessment into Erin’s chart. Just then Alexa, the RT, enters the room.
| Day: 2 | Pulse Rate | Blood Pressure | Respiratory Rate | Temperature | O2 Saturation |
| Time: 07h45 | 98 | 165/90 | 22 | – | 85% on Optiflow |
“Hi… Tracie?”
“Yes, you remembered. So you must also remember Mrs. Johns.”
“Yes I do, so her sats have not improved? You asked her to deep breath like yesterday?”
“Yes. She has improved slightly but not very much with the deep breathing.”
“Ok, let’s take a look. Hi, Mrs. Johns do you remember me?”
“Yes, you were. In Emergency?”
“That is right. I understand you are a bit short of breath?”
“Yes. A bit more. Than just a.
Erin just nods. Alexa methodically assesses Erin’s respiratory system.
“Ok, Mrs. Johns, your breath sounds don’t seem much different they did in Emergency, but obviously you are not feeling your best. I think I am going to have to change your oxygen treatment to a face mask.
 It allows me to give you a bit more oxygen than the system you are currently on. I am going to get the equipment to do so and we will see how you do with the new oxygen treatment and whether you need another needle poke in your wrist.”
Alexa steps out to the storeroom and selects a high flow mask with humidifier. Back in Erin’s room she sets the equipment up and turns the flow meter up as high as it goes. Looking at the humidifier bottle, Alexa notes a good amount of bubbling.
“Mrs. Johns, I’m going to take away those nasal prongs and place a mask on your face.”
Alexa then confidently changes the therapy and places the high flow mask on Erin.
 How does that feel now?”
“A bit better. Thank you.”
Turning to Tracie, Alexa states: “I have placed her on .65Fi02 high flow face mask. Can you do another set of vitals for me?” Tracie hits start on the vitals machine.
Both health professionals and Erin look at the numbers as they appear on the screen.
| Day: 2 | Pulse Rate | Blood Pressure | Respiratory Rate | Temperature | O2 Saturation |
| Time: 08h05 | 98 | 165/90 | 22 | – | 092% on High Flow FM |
“Ok, Mrs. Johns, your oxygen level is better and your other vitals look the same as they did when I first came on shift. Are you feeling better?”
“Yes thank you.”
Alexa looks at both Erin and Tracie. “I have another patient to see next door but before I leave I will come back and see how Mrs. Johns is doing.”
Tracie nods and follows Alexa out of the room. “What do you think is going on? Why did her saturations drop, Alexa?”
“Her pneumonia might be progressing a bit. Sometimes that happens even with antibiotics. Antibiotics take about three days to really work well. In the meantime if she could stand it, we should see if Physio can see her and maybe get her moving, which will help with chest expansion. One of the worse things to do is to just lie about. Is she due for a CXR?”
“Yes, I saw that she will be getting one around 2 PM this afternoon. Can she go down with that mask on?”
“No problem. I will make sure there is a full tank available. Tony is the RT in Emergency and he also covers X-ray if need be. I will let him know in case there is an issue while she is down there.”
“Ok, I will talk with the doctor about a physio order and let him know you have adjusted her O 2 .
“Thanks. I will look in on her after I see my other patients on this floor. If her sats stay up where they are, I don’t believe she needs a blood gas.”
“Awesome. Ok, see you later.”
Time: 09h00
“Hi Dr. Hunicutt, I’m Tracie. One of my patients, Erin Johns, is under your care.”
“Aw, yes the older lady with COPD and pneumonia. How was her night?”
“Night was pretty good, needed some adjustment on her Optiflow but otherwise slept well. My concern is what happened this morning. She became quite short of breath and anxious. I called the RT and we made some adjustments and moved her to high flow face mask at .65. Sats responded and she improved to 93% and has stayed there for the past hour. Appears quite relaxed now. No other changes in breath sounds or temperature. Very little sputum production.”
“Very good. Let’s go and see her now and see if I need to make any further adjustments.”
Both Tracie and Dr. Hunicutt walk down the hall to Erin’s room. As they enter, they see Erin sitting in bed with a green oxygen mask on, flicking through the TV channels.
“You have a lot more channels than I do at home, but nothing is on. This darn mask blocks some of my view as well. How much longer do I keep the mask on?”
“Hi, Mrs. Johns, my name is Dr. Hunicutt, and I am the doctor helping to care for you. I see you are less short of breath right now. Would you mind if I listen to your chest?”
“Sure, but I am feeling a bit tired of all the people looking at my chest.”
“I understand Mrs. Johns, but this is one way we can assess how well you are doing and if we need to change treatments.”
Dr. Hunicutt takes his stethoscope out and systematically listens to Erin’s heart and lungs. Then he proceeds to examine the rest of Erin.
“Very good. Thank you, Mrs. Johns.”
“If it was very good, I wouldn’t be here.”
“My apologies. It is ok. Your lungs are not the healthiest due to your COPD, and by listening to your chest I can see that you do have lots of consolidation in both lower lobes.” Dr. Hunicutt points to where he heard the bronchial breath sounds.
Erin looks down at her chest to see where Dr. Hunicutt is pointing. “Oh my, that is about half of each lung.”
“That is about right. Not all is lost. I believe you are on the right antibiotics. We are still waiting for the sputum test to tell me that, and you will also get a chest X-ray today to confirm that the pneumonia is not spreading more. Tracie, is there something more you would like to do for Mrs. Johns?”
“I would like to have Physio come and see her and get her moving a bit more.”
“I think that is a very good idea. I will write the order. They may not come today, but that doesn’t mean Mrs. Johns cannot sit in a chair or stand at the bedside. I don’t want her doing too much, but she does need to move a bit more.”
“We can do the chair. What about bathroom?”
“Commode at bedside until she is below 50% oxygen. Anything else?”
“Alexa the RT said she would wait to do another blood gas as her sats are ok right now.”
“I agree. But I will write an order for another blood gas tomorrow morning to see how we stand with CO 2 and PaO 2 . I want to ensure that her COPD is not getting worse as well. At this point, our other option is BiPAP which I’m pretty sure Mrs. Johns will not be happy with, so let’s see if we can manage her without going to intensive care.”
“Thank you. I think that will do. Mrs. Johns, do you have any questions?”
Erin, looks back at them from the TV. “Can my dog visit?”
Dr. Hunicutt looks at Tracie who shrugs her shoulders “Yes I think that will be ok as long as it’s well behaved and ok with your roommate to have her in this room.”
“Trixie is very well behaved and a small dog. She will be so pleased to see me.”
Both Dr. Hunicutt and Tracie smile, nod at Erin and leave the room.
Outside the room, Dr. Hunicutt asks if anything else needs to be done. Tracie points at a couple of other rooms where she has patients and explains that two patients need discharge orders, and two others need lab work looked at, as pharmacy is stating that antibiotic coverage may not be appropriate. Dr. Hunicutt nods his head and gives her the thumbs up sign as he heads to the chart rack to begin the process of discharge and looking at labs.
Tracie begins the process of recording the interaction with Dr. Hunicutt and Erin.
Over the course of the day, Tracie was able to get Erin up twice to the bedside commode and twice to a chair for 30 minutes each time. Erin’s saturation stayed stable along with her other vitals signs.
When the shift changed rolled around, Tracie was very happy to see Jim come into the unit. “Wow, I did not think I would see you tonight.”
“Me neither. They called me for overtime during my chaperoning of a school trip.”
“Man, you must be exhausted.”
“Yeah but no more than usual. Got some shut-eye before coming in. Anyway, how was your day?”
Tracie gives an overall report of two discharges and two new admissions. She updates Jim on changes to Erin’s oxygen levels and her increased orders for activity.
“Awesome. Thanks. Will I see you in the AM?”
“I believe you will. I’m doing four days in a row. I traded my night shifts so I could attend a play with my husband. We have had so little time together with him finishing his PhD, and my shift work. He promised not to talk research, and take me out on a real date if I would change my night shifts.”
“Wow, I am so jealous. Well, have a good sleep and we will see you in the morning.”
“Good night.”


1.2.7: Small Business, Large Impact
What Are the Advantages and Disadvantages Facing Owners of Small Businesses?
An uncertain economy has not stopped people from starting new companies. The National Federation of Independent Businesses reports that 85 percent of Americans view small businesses as a positive influence on American life. This is not surprising when you consider the many reasons why small businesses continue to thrive in the United States:
- Independence and a better lifestyle: Large corporations no longer represent job security or offer the fast-track career opportunities they once did. Mid-career employees leave the corporate world—either voluntarily or as a result of downsizing—in search of the new opportunities that self-employment provides. Many new college and business school graduates shun the corporate world altogether to start their own companies or look for work in smaller firms.
- Personal satisfaction from work: Many small-business owners cite this as one of the primary reasons for starting their companies. They love what they do.
- Best route to success: Business ownership provides greater advancement opportunities for women and minorities, as we will discuss later in this chapter. It also offers small-business owners the potential for profit.
- Rapidly changing technology: Technology advances and decreased costs provide individuals and small companies with the power to compete in industries that were formerly closed to them.
- Major corporate restructuring and downsizing: These force many employees to look for other jobs or careers. They may also provide the opportunity to buy a business unit that a company no longer wants.
- Outsourcing: As a result of downsizing, corporations may contract with outside firms for services they used to provide in-house. Outsourcing creates opportunities for smaller companies that offer these specialized goods and services.
- Small businesses are resilient: They are able to respond fairly quickly to changing economic conditions by refocusing their operations.
There are several cities and regions that are regarded as the best locations for start-up businesses and entrepreneurs. Among them are Tulsa, Oklahoma; Tampa, Florida; Atlanta, Georgia; Raleigh, North Carolina; Oklahoma City, Oklahoma; Seattle, Washington; Minneapolis, Minnesota; and Austin, Texas.
Why Stay Small?
Owners of small businesses recognize that being small offers special advantages. Greater flexibility and an uncomplicated company structure allow small businesses to react more quickly to changing market forces. Innovative product ideas can be developed and brought to market more quickly, using fewer financial resources and personnel than would be needed in a larger company. And operating more efficiently keeps costs down as well. Small companies can also serve specialized markets that may not be cost-effective for large companies. Another feature is the opportunity to provide a higher level of personal service. Such attention brings many customers back to small businesses such as gourmet restaurants, health clubs, spas, fashion boutiques, and travel agencies.
Steve Niewulis played in baseball’s minor leagues before an injury to his rotator cuff cut short his career. Niewulis decided to combine his love of the game with a clever idea that has elevated him to the big leagues. The fact that players had trouble keeping their hands dry while batting inspired his big idea: a sweat-busting rosin bag attached to a wristband so that a player can dry the bat handle between pitches. In less than two years, Niewulis’s Fort Lauderdale, Florida, company, Tap It! Inc., sold thousands of Just Tap It! wristbands. The product, which retails for $12.95, is used by baseball players, basketball players, tennis players, golfers, and even rock climbers. His secret to success? Find a small distribution network that allows small companies, with just one product line, to succeed.
On the other hand, being small is not always an asset. The founders may have limited managerial skills or encounter difficulties obtaining adequate financing, potential obstacles to growing a company. Complying with federal regulations is also more expensive for small firms. Those with fewer than 20 employees spend about twice as much per employee on compliance than do larger firms. In addition, starting and managing a small business requires a major commitment by the owner. Long hours, the need for owners to do much of the work themselves, and the stress of being personally responsible for the success of the business can take a toll.
But managing your company’s growing pains doesn’t need to be a one-person job. Four years after he started DrinkWorks (now Whirley DrinkWorks), a company that makes custom drinking cups, Richard Humphrey was logging 100-hour weeks. “I was concerned that if I wasn’t there every minute, the company would fall apart.” Humphrey got sick, lost weight, and had his engagement fall apart. When forced by a family emergency to leave the company in the hands of his five employees, Humphrey was amazed at how well they managed in his absence. “They stepped up to the plate and it worked out,” he says. “After that, the whole company balanced out.”
CONCEPT CHECK
- Why are small businesses becoming so popular?
- Discuss the major advantages and disadvantages of small businesses.