Little Stories from the History of Science and Technology

Posted on 27-December-2010 by mathscinotes

Introduction

I have always been interested in the connectedness of people. The theory that each of us is separated by only six degrees from anyone else is a theory I like to test. If you examine things carefully, you may find that you have more personal connections to history than you think. I have already written one blog post that alluded to this connectedness. I thought I would write a few more stories.

I am no Wernher Von Braun

Wernher von Braun was one of the great engineers of the 20th century. He clearly was a god to those that worked for him. My story begins when I was working as a contractor on a US Army missile program in the 1990s and was responsible for presenting some work my company was doing. I only had 10 minutes to present, so I simply presented some overview material. After my presentation an old gentleman came up to me and told me that:

"You gave a good presentation, but we are going need much more detail. It would be one thing if you were Wernher von Braun . But you are not Werhner von Braun, at least not yet, and we need much more detail."

While I was a bit shocked, I told the gentleman that next time I would have more detail. Later, I laughed hard about this incident. I found out that the old gentleman had worked under von Braun and held him in total awe. It reminded me of when Lloyd Bentson told Dan Quayle that "You are no Jack Kennedy." Anyway, I now feel like I am two degrees of separation from Wernher von Braun.

My High School Physics Teacher and Wernher von Braun

When I was in high school, my physics teacher (Tom Truax) related a Wernher von Braun story. During the 1960s, there was a huge push to strengthen science and math education in the United States. As part of this effort, Wernher von Braun would occasionally visit teacher conferences around the country. During one conference, he made quite a spectacular arrival. The teachers were all sitting in an auditorium, wondering where their speaker was. Precisely when the speech was scheduled to begin, there was the sound of a helicopter landing on the school grounds. A man in a suit ran out of the helicopter and right onto the podium. He then gave a rousing speech, which concluded by asking "Any questions?" The crowd was silent for a few seconds, so he immediately put his speech away and ran out to the waiting helicopter and flew away. It was quite an impressive display. My physics teacher used to love imitating von Braun's accent in class by saying "Any qvestions?" Anyway, it was the 1960s. Anything could happen.

Earl Bakken, Founder of Medtronic

I spent a summer in 1978 as an engineering intern at Medtronic. It was a large building and one day I got lost. A kindly old man asked me if he could help and I said I was lost. He laughed and asked where I wanted to go. I told him and he guided me to my work area. I thanked him and went about my work. My coworkers were stunned to see me with this gentleman. I asked why and they said, "That is Earl Bakken and he is the founder of Medtronic." Bakken had created largest medical electronics company in the world. All I know was that he was very helpful to a lost young man. While helping me find my way, he also mentioned that he was holding a dancing class and I was welcome to attend. I later found out that he loved to dance and taught classes for people who were interested.

Wozniak and HP

I joined HP in 1979. At one point, I was part of group that was looking at the Apple 2 and trying to determine if there was anything there that HP should be considering in terms of future products. I look back at that effort now and laugh. There were all these stories about Wozniak (who I have never met) trying to convince HP to build home and small business computers. Apparently, Wozniak was adamant that there was an untapped market there. Of course, HP management felt that he was nuts. One of the guys even had an HP internal memo that flat out said that there was no future for home and small business computing (I wish I had kept a copy of that memo). Even after Apple had shown that people wanted computers, HP continued to rationalize their earlier decisions by saying the Apple product was simply too low of quality for a company like HP to build. Sounded like sour grapes to me then and it still does today. I must say that I have great admiration for the work Wozniak did.

The Cavity Magnetron

One of the electrical engineering professors at the University of Minnesota was a man named Wehner (I do not remember his first name) who was a German emigre. He did not teach any of the classes that I had, but some of my friends had him as a professor. One of the stories they related about him was that he was a scientist on the German side during World War II. At that time, he was tasked with analyzing downed aircraft and examining their technology. He was one of the first German scientists to examine a downed British aircraft that carried a cavity magnetron-based radar unit. He described to his students how stunned he was when he encountered it. Here was a very clever piece of the technology that the vaunted German science establishment had not been able to duplicate. It portended bad things to come for Germany. Many historians believe that the development of microwave radar to be an invention that changed the world. In fact, most homes today have a cavity magnetron – it is in their microwave oven.

As long as I am on the topic of radar, I sometimes hear people talking about the effects of electromagnetic radiation from cell phones with so many researching on how to get protection from EMF radiation. I always tell them the story of a man I encountered at Alliant Techsystems. He had been a radar technician in the military for 20 years, mainly working on small, dish-antenna units. When he needed to check if a radar was transmitting, he would put his left hand in front of the antenna. When his hand became warm, he knew the radar was transmitting. He retired from the military and began working in the civilian world. He ended up having a lot of trouble with that left hand over time. It never felt right and often had infections that never seemed to heal. Eventually, he developed gangrene in that hand and he had to have it amputated. This tale has always made me suspicious of the long-term effects of exposure to electromagnetic radiation. I know it is non-ionizing radiation, but ...

Otto Schmitt and the Schmitt Trigger

All electrical engineers are familiar with the "Schmitt Trigger." It was invented by a man named Otto Schmitt. I actually met him when I was at the University of Minnesota. At that time, he was in the Temporary Engineering Annex (old WWII buildings that were still being used in the late 1970s). I was amazed at the dilapidated condition of these facilities, which are gone now. At the time I met him, I had no idea he was such a famous guy. He just seemed like a friendly old man. He sure had a hard charging bunch of graduate students.

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Test Time and Estimating Bit Error Rate

Posted on 26-December-2010 by mathscinotes

Quote of the Day

Our heads are round so our thoughts can change direction.

— Francis Picabia

Introduction

Figure 1: Production test system. Time on these systems is costly and needs to be kept as short as possible.

Figure 1: Production test system. Time on these systems is costly and needs to be kept as short as possible. (Source)

Test time is expensive. Since our products need to conform to industry standards for Bit Error Rate (BER), we need to test for BER. It is important that we test long enough to ensure that we meet the requirements, yet not so long as to spend more money than we need to.

I was asked to develop a rational approach for determining the amount of test time required. I put together our current procedures years ago, but now they need to be refreshed as we prepare to offer newer, higher speed transports. While I was reviewing these procedures, I saw that the analysis required was interesting and thought I would document it here. Our procedures are based on a couple of papers from Maxim and Lightwave Magazine. In this blog, I generate a Mathcad model of BER based on the results of these papers and examine a couple of transport examples.

For testing purposes, a number of bits must be transferred with the number of errors less than a given amount to provide sufficient confidence of meeting the BER requirement. This analysis will compute the number of bits that must be transferred with less than a given number of errors to provide sufficient confidence that we are meeting the BER requirement. The test time is computed by multiplying the number of bits by the transfer rate.

Analysis

Definitions

As with most technical discussions, it is important to get your terms defined upfront.

Bit Error Rate (BER)
BER is the ratio of the number of bit errors to the total number of bits transferred over an infinite time interval. Mathematically, we can express this definition as BER\triangleq \underset{n\to \infty }{\mathop{\lim }}\,\frac{\varepsilon (n)}{n}, where n is the number of bits transferred and the ε is the number of errors among those n bits.
Confidence Interval (CI)
The confidence interval is a particular kind of interval estimate of a population parameter and is used to indicate the reliability of an estimate. It is an observed interval (i.e it is calculated from the observations), in principle different from sample to sample, that frequently includes the parameter of interest, if the experiment is repeated. The frequency that the observed interval contains the parameter is determined by the confidence level or confidence coefficient.
Confidence Level (CL)
Confidence level refers to the likelihood that the true population parameter lies within the range specified by the confidence interval. In this case, the confidence interval is in the range from 0 to the specified BER limit. For example, a 99% confidence limit tells us that for a given sample size and number of bit errors, 99% of the time the true BER is within the confidence interval. Mathematically, we can express this definition as CL\triangleq P\left( BER<\gamma |\ \varepsilon ,\ n \right), where γ is the confidence limit.

Modeling

Equation 1 gives us the probability of have N or fewer events for test described by a binomial distribution.

Eq. 1 P\left( \varepsilon <N \right)=\sum\limits_{k=0}^{N}{{{P}_{n}}}\left( k \right)=\sum\limits_{k=0}^{N}{{{C}_{n,k}}}\cdot {{p}^{k}}\cdot {{q}^{n-k}}

where Cn,k is the number of combinations of n items taken k at a time and n is the number of samples.

When the probability p is small and the number of observations is large, binomial probabilities are difficult to calculate. Fortunately, the binomial probability distribution in this case is well approximated by the Poisson distribution. For those who want more details on this approximation, please check out one of the web sites that demonstrates the validity of this approximation. Figure 2 shows the results of substituting the Poisson distribution for the binomial distribution.

Eq. 2 P\left( \varepsilon \le N \right)=\sum\limits_{k=0}^{N}{{{C}_{n,k}}}\cdot {{p}^{k}}\cdot {{q}^{n-k}}=\sum\limits_{k=0}^{N}{\frac{{{\left( np \right)}^{k}}}{k!}\cdot {{e}^{-n\cdot p}}}

We can relate the confidence level CL to the Poisson distribution as shown in Equation 3.

Eq. 3 CL=1-P\left( \varepsilon \le N \right)\Rightarrow \sum\limits_{k=0}^{N}{\frac{{{\left( np \right)}^{k}}}{k!}\cdot {{e}^{-n\cdot p}}}=1-CL

We can manipulate Equation 3 to form Equation 4, which is convenient for use with Mathcad's nonlinear numerical solver.

Eq. 4 -n\cdot p=\ln \left( 1-CL \right)-\ln \left( \sum\limits_{k=0}^{N}{\frac{{{\left( np \right)}^{k}}}{k!}} \right)

Note that the number of bits required (n) does not vary with transport speed.

Worked Example

Figure 2 is a screenshot of my Mathcad spreadsheet that I used to work this example. In Figure 2, the variable α represents the number of bits that must be transferred with a given number of errors to meet the required CL.

Figure 1: Illustration of BER Calculation in Mathcad.

Figure 2: Illustration of BER Calculation in Mathcad.

For the example worked here, I assume that

  • CL = 99%
    This is a pretty strict standard. Some folks use 90%, others 60%. The higher the CL you require, the more time you must spend testing.
  • Maximum allowed BER of 1E-10
    This reflects GPON requirements. Other transports have different requirements.
  • Test time is computed by multiplying the required number of bits transferred by the bit time \left( {{T}_{Test}}=\frac{n}{{{f}_{DateRate}}} \right).

I computed results for 1 Gigabit Ethernet (fDataRate = 1.25 Gbps) and GPON (fDataRate = 2.488 Gbps). Table 1 summarizes my results.

Table 1: Test Time Summary.
Number of Errors Total Bits Transferred Test Time @ 1.25 Gbps (sec) Test Time @ 2.488 Gbps (sec)
0 4.61E10 36.84 18.51
1 6.64E10 53.11 26.68
2 8.41E10 67.25 33.79
3 1.00E11 80.36 40.37
4 1.16E11 92.84 46.64

Conclusion

I derived an expression for the number of bits that must be transferred to provide a given level of confidence for having achieved a specified BER. One can see from my example that achieving a 99% confidence level requires a lot of test time. Since test time can cost hundreds of dollars per hour, you can see how the costs add up quickly.

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Posted in Electronics, Fiber Optics | Tagged , , | 11 Comments

Calculating the Earth's Age

Posted on 22-December-2010 by mathscinotes

Quote of the Day

Desire to develop mastery; Desire for autonomy; Desire for purpose.

— Daniel Pink, Author of "Drive," talking about the primary employee motivations in today's job environment.

Introduction

Figure 1: Barringer Crater in Arizona. Scientists first used lead-lead dating on fragments from this meteorite. (USGS)

Figure 1: Barringer Crater in Arizona. Scientists first used lead-lead dating on fragments from the meteorite that formed this crater. (USGS)

I have been listening to the audio book The Disappearing Spoon, which is an excellent tale about all of the elements of the periodic table. Of particular interest to me was the discussion of how geologists date the age of the Earth using ratios of uranium and lead. The book also discusses determining the ages of meteorites and the Sun. The discussions were interesting enough that I thought I would look up some additional information. As frequently happens, I was amazed at the amount of information on the web about this subject. This technique has been around since 1956, when it was first used to date meteor fragments from a well-known impact site (Figure 1).

Computing the Age of the Earth

Basic Chemistry

The basic chemistry used in the dating process is nicely described in this About.com article. I summarize the contents of this article as follows.

  • Tiny zirconium crystals (called zircons) are the oldest rocks in the world.
  • The zircons form containing uranium with little or no lead
  • Any lead in the crystals comes from the decay of U235 and U238 into Pb207 and Pb206, respectively.
  • We can use the ratios of Pb207-to-U235 and Pb206-to-U238 to estimate the age of the zircons. Both ratios can be used to estimate the age of the Earth, so we can cross-check our results for consistency. This technique is referred to as lead-lead dating.

Basic Mathematics

We can use Equation 1 as our model for the variation of the isotope ratio over time.

Eq. 1 \text{Ratio}\left( t,{{t}_{\text{Half-Life}}} \right)=\frac{1-{{\left( \frac{1}{2} \right)}^{\frac{t}{{{t}_{\text{Half-Life}}}}}}}{{{\left( \frac{1}{2} \right)}^{\frac{t}{{{t}_{\text{Half-Life}}}}}}}

where t is time and tHalf-Life is the half-life of the isotope in question. Using time as a parameter, we can plot x(t) = Ratio(t,tHalf Life U235) and y(t) = Ratio(t,tHalf Life U238). I threw the equations into Mathcad and plotted the data (Figure 1) assuming a U235 half life of 704 million years and a U238 half life of 4.47 billion years. As I would expect, my plot here agrees with the same plot shown on the About.com web site.

Figure 1: Concordia Plot Showing Uranium and Lead Ratios Versus Time.

Figure 1: Concordia Plot Showing Uranium and Lead Ratios Versus Time.

When geologists measure the ratios and look at the chart, they see that the Earth is about 4.4 billions years old.

Conclusion

I was able to duplicate the results presented on a couple of different web sites on determining the age of the Earth. This was a good exercise and really shows the power of using radioactive decay to determine the age of the Earth.

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Power Dissipated in a Maximum Phone Load During Ringing

Posted on 10-December-2010 by mathscinotes

I am writing a specification this morning and I realized that I have never calculated the maximum power drawn by a phone during ringing. I was surprised by the number – 3 W. This is a lot for unit that must operate from battery backup when AC power fails. Figure 1 summarizes the calculation, which were performed in Mathcad.

Figure 1: Ring Power Calculation.

Figure 1: Ring Power Calculation.

To check the accuracy of this result, I measured the power into the ringer power supply. This power supply has an efficiency of ~85 % (\epsilon \triangleq \frac{{{P}_{Out}}}{{{P}_{In}}}) and I measured 3.4 W into the supply.

Figure 2: Measured Ring Power at the SLIC Power Supply.

Figure 2: Measured Ring Power at the SLIC Power Supply.

My measured result of 2.9 W and my computed result of 3.0 W are close enough considering the potential errors in my load circuit and my efficiency estimate for the power system.

Posted in Electronics | 1 Comment

Thermal Runaway Model of Lead-Acid Battery (Part 2)

Posted on 10-December-2010 by mathscinotes

Derivation of the Output Power Equation

The output power equation (Equation 3) is really a restatement of Newton's law of cooling. Equation 3 states the battery's steady-state power dissipation is a linear function of the battery's temperature and the ambient temperature. This equation assumes that the power generated within the enclosure is dissipated by convection and conduction, which is a good assumption at low battery temperatures.

Eq. 3 {{P}_{OUT}}={{G}_{Thermal}}\cdot \left( {{T}_{Battery}}-{{T}_{Ambient}} \right)

Figure 6 shows a graph of this equation assuming that GThermal = 50 mW/°C and TAmbient = 25 °C (I measured this characteristic on a small battery in my lab).

Figure 6: Battery Power Dissipation Versus Battery Temperature.

Figure 6: Battery Power Dissipation Versus Battery Temperature.


Values for GThermal will vary for each battery design and state (e.g. age, charge history). For example, the large battery used in the example below from the reference paper has GThermal = 690 mW/°C.

Analysis

Thermal Runaway Operating Conditions

Figure 7 shows a combined plot of PIN and POUT. The point of intersection of these two curves (marked by a pink line) is where P_{IN}=P_{OUT}, which means that the battery temperature and float voltage are stable at that power level. When PIN exceeds POUT, the temperature of the battery will begin to increase. Under typical operating conditions, the battery will simply stabilize at a higher temperature that allows the internal power to be dissipated. However, battery temperature can only go so high before the electrolyte boils away, resulting in the destruction of the battery. If the battery operating point is such that PIN always exceeds POUT prior to reaching the destructive temperature, the battery will destroy itself. This is not good.

Figure 7: Combined Plot of Power Generation and Dissipation.

Figure 7: Combined Plot of Power Generation and Dissipation.

Figure 7 is based on my estimate of the parameters for a battery (Y3) from the original paper. Here are the coefficients I used:

  • GTherm=0.690 W/°C
  • float current: \displaystyle {{\text{I}}_{BAT}}\left( {{V}_{BAT}},{{T}_{BAT}} \right)={{e}^{k+\alpha \cdot {{V}_{BAT}}+\beta \cdot {{T}_{BAT}}}}\ \text{mA}
    • k = -25.939187
    • β = 0.096457 1/°C
    • α = 0.909478 1/V
  • battery power: \displaystyle {{P}_{BAT}}\left( {{V}_{BAT}},{{T}_{BAT}} \right)={{V}_{BAT}}\cdot {{I}_{BAT}}\left( {{V}_{BAT}},{{T}_{BAT}} \right)\ \text{mW}

In Figure 7, I actually plotted Watts instead of milliWatts.

Condition for Thermal Runaway

We can derive a criticality condition for thermal runaway by observing that (1) PIN = POUT and (2) the slopes of the two curves must be equal at their points of intersection (see Equations 4 and 5, respectively).

Eq. 4 \frac{\partial {{P}_{OUT}}}{\partial {{T}_{Battery}}}=\frac{\partial {{P}_{IN}}}{\partial {{T}_{Battery}}} \Rightarrow
G_{Thermal}=V_{Float} \cdot \beta \cdot e^{k+\alpha \cdot V_{Float}+\beta \cdot T_{Battery}}
Eq. 5 {{P}_{OUT}}={{P}_{IN}}\Rightarrow
{{G}_{Thermal}}\cdot \left( {{T}_{Battery}}-{{T}_{Ambient}} \right)={{V}_{Float}}\cdot {{e}^{k+\alpha \cdot {{V}_{Float}}+\beta \cdot {{T}_{Battery}}}}

If we make a ratio of these two equations, simplify, and solve for TBattery, we obtain Equation 6.

Eq. 6 {{T}_{Battery}}={{T}_{Ambient}}+\frac{1}{\beta }

We can substitute the constraint of Equation 6 back into Equation 4. After a bit of algebraic manipulation, I get Equation 7. Note that my Equation 7 is different than that of the paper I am reviewing. I have done all sorts of verification and I believe that my version is correct.

Eq. 7 {{V}_{Max}}=\frac{\ln \left( \frac{{{G}_{Thermal}}}{{{V}_{Max}}\cdot \beta} \right)-1-k-\beta \cdot {{T}_{Ambient}}}{\alpha }

Let's examine Equation 7 just a bit.

  • Higher ambient temperature lowers the maximum allowed charging voltage.
    This makes complete sense to me. My experience is that batteries operating under high ambient temperatures are more subject to thermal runaway.
  • Improved battery cooling (i.e. lower β) makes thermal runaway less likely.
    Again, I have rarely seen thermal runaway when the battery has a good thermal design. For example, batteries in a flat package generally have better thermal conductivity and appear to me to be less likely to runaway.
  • Old batteries are more likely to thermally runaway than new batteries.
    Old batteries require higher charging currents (i.e. higher α) and have lower thermal conductivity (i.e. GThermal is smaller), each of which reduce the critical voltage level.

Figure 8 is a graph of Equation 7. You will see that the demarcation line between thermally stable and unstable regions is nearly linear. As the reference paper shows, this linearity can be explained by a slight rearrangement of Equation 7 into Equation 8.

Eq. 8 Eq8

The function K\left( {{V}_{Max}} \right) is roughly constant over a limited voltage region because of the logarithmic dependence of the right-hand side of Equation 8 on VMax. This means that the function appears to be nearly linear when plotted on a graph.

Figure 8: Thermally Stable and Unstable Regions.

Figure 8: Thermally Stable and Unstable Regions.

Figure 8 clearly shows that a combination of high charging current and high ambient temperature makes thermal runaway more likely.

Conclusion

My objective was to duplicate the analysis details discussed the paper I am reviewing here. The paper derived results that I have observed in my work with batteries, which I summarize below.

  • Higher ambient temperature makes thermal runaway more likely.
  • Higher charging voltage makes thermal runaway more likely.
  • Some batteries are more susceptible than others (e.g. old batteries and those with poor thermal design).
Posted in Batteries, Electronics | Tagged , , , | 3 Comments

Thermal Runaway Model of a Lead-Acid Battery (Part 1)

Posted on 9-December-2010 by mathscinotes

Introduction

Very nearly every product we make ships with one or more lead-acid batteries. Since we have built hundreds of thousands of units, that is a lot of batteries. While most people encounter batteries everyday, few really understand the problems that they present.

The least understood problem with a lead-acid battery may be that they are susceptible to thermal runaway. The Wikipedia has a useful definition of thermal runaway.

Thermal runaway refers to a situation where an increase in temperature changes the conditions in a way that causes a further increase in temperature, leading (in the normal case of an exothermic reaction) to a destructive result. It is a kind of positive feedback.

Thermal runaway can be quite destructive. Figure 1 shows a photograph of a car battery in a trunk that experienced runaway. Not a pretty sight.

Figure 1: Car Battery After Thermal Runaway.

Figure 1: Car Battery After Thermal Runaway (Source: Voltphreaks).

I have struggled with coming up with a good explanation of thermal runaway that is generally understandable. Today, I saw a paper that did a nice job of explaining thermal runaway using a simple mathematical model. I review that paper in this blog and add explanatory notes that will make the material more accessible to a non-specialist audience.

Background

Thermal Runaway Basics

Battery thermal runaway is a positive feedback process.

  1. The charging chemical equations are exothermic (i.e. generate heat). As we charge the battery heat is generated.
  2. Heat accelerates the exothermic chemical reaction within the battery.
  3. The accelerated exothermic reactions generate more heat.
  4. go back to step 1.

This process is illustrated in Figure 2.

Figure 2: Illustration of Thermal Runaway Process (Source: Wikipedia)

Figure 2: Illustration of Thermal Runaway Process (Source: Wikipedia)

Fundamental Equations

Thermal runaway only occurs when a battery is being overcharged. If you want to know how to drive a lead-acid battery into thermal runaway, take a look at the document "Induced Destructive Overcharge Test IEC Standard 952-1:1988." Its procedure works like a champ.

The thermal overload model presented herein uses two equations:

  • Input Power Equation
    This equation calculates the electrical power dissipation within the battery as a function of temperature and voltage.
  • Output Power Equation
    This equation calculates the ability of the battery mechanical design to dissipate the electrical heat generated within as a function of battery temperature and ambient temperature.

The simultaneous solution of the input and output power equations show that there exists an operational region where more power can be generated within the battery than the battery can dissipate to the environment while maintaining a stable temperature. This is the region of thermal runaway.

Derivation of the Input Power Equation

Understanding overcharging requires knowing the functional relationship between charge current, float voltage, and temperature. Figure 3 is an excellent illustration of this relationship for an aged battery, which are more susceptible to thermal runaway than new batteries.

Figure 3: Battery Charging Current Versus Float Voltage and Battery Temperature (Source: Battery Technology Handbook, Kiehne, ISBN 9780824742492).

Figure 3: Battery Charging Current Versus Float Voltage and Battery Temperature (Source: Battery Technology Handbook, Kiehne, ISBN 9780824742492).

Because Figure 3 is a semilog plot, we can convert Figure 3 into an equation of the form shown in Equation 1.

Eq. 1 \ln \left( {{I}_{Charge}} \right)=k+\alpha \cdot {{V}_{Float}}+\beta \cdot {{T}_{Battery}}

Given that we know the charging current, we can compute the power generated within the battery by noting that P_{IN} = V_{Float} \cdot I_{Charging}. Equation 2 computes the power generated in the battery.

Eq. 2 {{P}_{IN}}={{V}_{Float}}\cdot {{I}_{Charge}}={{V}_{Float}}\cdot {{e}^{k+\alpha \cdot {{V}_{Float}}+\beta \cdot {{T}_{Battery}}}}

where k, α, β are constants determined through curve fitting.

I will not go through the details of the curve fitting operation, except to say that I captured the graphic data using a tool called Dagra and I processed the data using Mathcad. Figure 4 contains a screen shot of the Mathcad work.

Figure 4: Curve Fit Work in Mathcad.

Figure 4: Curve Fit Work in Mathcad.

A 3D surface graph (Figure 5) does a nice job of showing what the function looks like.

Figure 5: 3D Graphic of Battery Charging Current Versus Float Voltage and Battery Temperature.

Figure 5: 3D Graphic of Battery Charging Current Versus Float Voltage and Battery Temperature.

This completes Part 1. Part 2 will cover how the power generated within the battery is dissipated. Using equations derived for the power generated and power dissipated within the battery, conditions will be derived for when thermal runaway occurs.

Posted in Batteries, Electronics | 4 Comments

Laser Failure Rate Estimate Example

Posted on 5-December-2010 by mathscinotes

Quote of the Day

If you complain, you make yourself a victim. Leave the situation, change the situation or accept it.

— Unknown

Introduction

Figure 1: Brake undergoing dynomometer tests.

Figure 1: Brake undergoing dynamometer tests. Brake pad failures are often modeled using a lognormal probability distribution. (Source)

I have spent some time lately talking to people about laser failure characteristics. Most electronic component reliability modeling is done using the exponential probability distribution, which assumes the components have a constant failure rate and there is no wear-out mechanism. It turns out that lasers have a wear-out mechanism, which means the exponential probability distribution is not appropriate. Laser failure rates are usually modeled by a lognormal probability distribution, as are the failure rates of brakes (Figure 1) and incandescent light bulbs. These components have reliabilities that are dominated by wear-out mechanisms that accelerate when damage to a small region grows exponentially. A good example would be a hard spot on a brake pad that becomes hot during braking relative to the rest of the pad. This hard spot tends grow quickly because the heat generated during braking concentrates there.

Lognormal Probability Distribution Basics

Equations 1 and 2 give the definitions of the lognormal probability density and cumulative probability distribution, respectively.

Eq. 1 f(t,\mu ,\sigma )\triangleq \frac{1}{\sqrt{2\cdot \pi }\cdot \sigma \cdot t}\cdot {{e}^{-\frac{{{\left( \ln (t)-\mu \right)}^{2}}}{2\cdot {{\sigma }^{2}}}}}
Eq. 2 F(t,\mu ,\sigma )=\frac{1}{2}\cdot \text{erfc}\left( -\frac{\ln \left( t \right)-\mu }{\sigma \cdot \sqrt{2}} \right)

where μ is the mean and and σ is the standard deviation of the random iterate t.

The idea behind the distribution is simple – a random variable T is said to be lognormal distributed if log(T) is normally distributed. Figure 1 shows the shape of the lognormal density and distribution for μ = 1 and σ=1.

Figure 1: Lognormal Example (Density and Cumulative Distribution).

Figure 1: Lognormal Example (Density and Cumulative Distribution) – μ = 1 and σ=1.

Empirical reliability studies have shown that the lognormal distribution provides a good model for components whose reliability is dominated by a wear-out mechanism.

Laser Reliability Physics

So why does nature demand the use of the lognormal distribution for wear-out problems? To motivate the use of the lognormal distribution, we need to understand how lasers fail. The laser puts out a specified amount of optical power (i.e. light) when driven with electrical current. As a laser ages, it takes more current to generate the required optical power. Eventually, the current required to obtain the required optical power exceeds the current limits of the electrical circuitry.

One way to model this behavior is to assume that there are one or more light-absorbing defects that are growing within the optical channel of the laser. This is a credible theory because these optical defects will be absorbing light, which provides the power needed for the defects to grow. To model this behavior, let a0, a1, …, an be a sequence of independent random variables representing the successive growth of the defect area. Let a0 be the initial defect area and an be the defect area that results in component failure. As the defect area increases, more optical power is absorbed and the rate of defect growth will increase. Equation 3 can be used to model this characteristic.

Eq. 3 {{a}_{i+1}}-{{a}_{i}}={{\delta }_{i}}\cdot {{a}_{i}}\ \Rightarrow {{a}_{i+1}}={{a}_{i}}\cdot \left( 1+{{\delta }_{i}} \right)

This means that we can express an as a product in terms of a0 and the δi's as shown in Equation 4.

Eq. 4 {{a}_{n}}={{a}_{0}}\cdot \left( 1+{{\delta }_{0}} \right)\ \cdots \ \left( 1+{{\delta }_{n}} \right)={{a}_{0}}\cdot \prod\limits_{i=0}^{n}{\left( 1+{{\delta }_{i}} \right)}

If we take the logarithm of both sides of Equation 3 and assume that the δi's are much less than 1 , we obtain Equation 5.

Eq. 5 \ln \left( {{a}_{n}} \right)={{a}_{0}}+\sum\limits_{i=0}^{n}{\ln \left( 1+{{\delta }_{i}} \right)}\doteq {{a}_{0}}+\sum\limits_{i=0}^{n}{{{\delta }_{i}}}

We can use the central limit theorem and the assumption that that the δi's are independently distributed random variables to state that Equation 5 will converge to a normal distribution. Since the logarithm of the random variable an has a normal distribution, we see that an has a lognormal distribution.

A Real-Life Laser Reliability Example

Typical Laser Reliability Parameter Example

I will walk through an actual laser reliability specification and show where all the terms come from. I have used the laser with the reliability characteristics shown in Table 1. Note that wear-out specifications from laser vendors assume the laser is on all the time. For many systems, like PONs, the laser is only on a fraction of the time.

Table 1: Example of Typical Laser Reliability Parameters.

Table 1: Example of Typical Laser Reliability Parameters.

The key data in this table are the following items:

  • σ
  • Ea
  • median life estimate at one temperature

Given this data, the rest of the information in the table can be computed, as I will show below. These key pieces of data are determined in an accelerated life study. This study takes a large number of lasers and stresses them under temperature to simulate operation over a long time interval. The key assumption here is that component aging is modeled by Equation 6, the Arrhenius equation.

Arrhenius Equation

Because it takes many years to get lasers at normal temperature to fail, we use temperature to accelerate the aging process for lasers. The aging of lasers is governed by the Arrhenius equation (Equation 6).

Eq. 6 K\left( {{T}_{1}},{{T}_{0}},{{E}_{a}} \right)\triangleq {{e}^{\frac{{{E}_{a}}}{{{k}_{b}}}\cdot \left( \frac{1}{{{T}_{1}}}-\frac{1}{{{T}_{0}}} \right)}}

where T1 is the temperature at which we desire the failure rate, T0 is the reference temperature, and Ea is the activation energy.

The Arrhenius equation is used to estimate the failure rate at relatively low temperatures based on the test results from a large numbers of parts operating at high temperature. In Table 2, I have included some commonly seen activation energies for different failure modes.

Table 2: Examples of Activation Energies
Mechanism Activation Energy (eV)
Gross Coarsening in Solder 0.42
Intermetallic Defect 0.3
Metallization Defect 0.5
Line Electromigration 0.45 – 1.2
Dark Line Defects in AlGaAs/GaAs Lasers 0.2
Gradual Degradation in AlGaAs/GaAs LEDs 0.5

For the laser used in my example here, the activation energy is 0.4 eV. This number is always listed by the laser vendor and it varies a bit depending on the characteristics of their processes.

Definition of Failure Rate

There are a number of terms that are often erroneously used interchangeably to describe failure rates.

  • failure rate
  • instantaneous failure rate
  • hazard rate

Each of these terms actually has a distinct meaning. This paper will use the term hazard rate, which is the standard method for lasers.

The hazard rate, h(t), is actually a conditional failure rate defined in Equation 7.

Eq. 7 h\left( t \right)\triangleq P\left( \text{failure prob}\text{. in dt }|\text{ laser survived to time t } \right)=\frac{f(t,\mu ,\sigma )}{1-F(t,\mu ,\sigma )}

The term 1-F(t, µ, σ) is sometimes referred to as the survival function.

Mathcad Calculations

I grabbed all the equations from the Wikipedia. Figure 2 shows the equations as implemented in Mathcad. Note that Mathcad has built-in functions for the lognormal density and cumulative distributions. I chose not to use them just to illustrate how these functions could be implemented in other tools.

Figure 2: Mathcad Implementation of Equations.

Figure 2: Mathcad Implementation of Equations.

Figure 3 shows how to calculate all the terms in Table 1.

Figure 3: Reproduction of the Laser Reliabilty Results in Table 1.

Figure 3: Reproduction of the Laser Reliabilty Results in Table 1.

Conclusion

I was able to demonstrate how laser reliability results are calculated. Hopefully, this post will remove some of the mystery behind how laser reliability numbers are computed.

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Posted in Electronics, Fiber Optics | Tagged , , | 3 Comments

Calculating Drawer Heights

Posted on 3-December-2010 by mathscinotes

Introduction

As you can tell from all the recent posts on construction, I am in the middle of planning some remodeling on my home. One of my projects includes designing a chest of drawers. It turns out that there are some common ways of determining pleasing drawer heights. I thought I would review them here. After a quick survey around the web, I found the following mathematical approaches to drawer sizing. They are all based on some form of numerical series.

  • uniform progression
    The simplest approach is to just make all the drawers the same size.
  • arithmetic progression
    This approach uses an arithmetic series to determine the drawer sizes. I did find an online calculator for this progression.
  • geometric progression
    This approach uses a geometric series to determine the drawer sizes. I also found an online calculator this one.
  • Fibonacci series
    This approach uses a Fibonacci series to determine the drawer sizes.
  • Hambridge progression
    A commonly used approach that can be drawn using a geometric construction that allows calculations to be avoided. I found an online calculator for this progression as well.

Comparison of Drawer Appearances

To determine the approach I wanted to take, I decided to use a 6-drawer, 50" tall chest of drawers as an example. Figure 1 illustrates this example.

Figure 1: Illustration of a 6-Drawer, 50 inch set of drawers.

Figure 1: Illustration of a 6-Drawer, 50 inch set of drawers.

Calculation Details

I am a big Mathcad fan and I coded these sequences into Mathcad, which I show in the following screen shots.

Uniform Progression

Figure 2: Computation of the Uniform Progression Drawer Heights.

Figure 2: Computation of the uniform progression drawer heights.

Arithmetic Progression

Figure 3: Computation of the arithmetic progression drawer heights.

Figure 3: Computation of the arithmetic progression drawer heights.

Geometric Progression

Figure 4: Computation of the geometric progression drawer heights.

Figure 4: Computation of the geometric progression drawer heights.

Fibonacci Series

Figure 5: Computation of the Fibonacci series drawer heights.

Figure 5: Computation of the Fibonacci series drawer heights.

Hambridge Progression

Figure 6: Computation of the Hambridge progression drawer heights.

Figure 6: Computation of the Hambridge progression drawer heights.

The Hambridge progression can be drawn using a compass in the field. This process is illustrated in Figure 7.

Figure 7: Illustration of the Hambridge progression construction.

Figure 7: Illustration of the Hambridge progression construction.

Other Examples

Nice Example of a Well-Designed Set of Drawers
Good Forum Discussion
Yet Another Site with Graduated Drawer Information

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The Mathematics of Rafter and Collar Ties

Posted on 29-November-2010 by mathscinotes

Introduction

As a boy, I remember seeing a dilapidated farm building that had a set of rafter ties in its attic. I noticed that the rafter ties had pulled away from the rafters they were nailed to. The rafter ties were clearly pulling away from their rafters because of the tension forces they were experiencing when under snow load. I noticed that steel cables has been hidden behind the rafter ties to try to reinforce them, but the screws attaching the cables to the rafters were also pulling out. This also was a sign of tremendous tension.

I had forgotten about this old farm building until I saw an article in the Journal of Light Construction (JLC) on the problems associated with using rafter ties on a roof. I became curious where the equation used in this article came from and I decided to derive it myself. I also noticed that the same equation is also built into an online calculator. Note that this post does not presume to train anyone on how to do structural engineering. Instead, I am reviewing the basic mathematics and physics used in doing this sort of work. See a structural engineer if you have a real problem that needs to be solved.

For those who are not familiar with rafter ties, I have included Figure 1 as an illustration of what they are and how they are connected to the rafters. The number of ties required per rafter will vary by situation. The rafter ties can generate tremendous force on the rafter to which it is connected. Purlins are often used in conjunction with rafter ties (see Appendix below for a figure).

Collar Tie Illustration and Installation

Rafter Tie Illustration and Installation
Figure 1: Illustration and Implementation Examples of Rafter and Collar Ties

Analysis

Some Definitions

Mathematically, rafter and collar ties are handled the same way. However, each type of tie has a different purpose. Unfortunately, the terms collar and rafter tie are often conflated. I found a good set of definitions from a construction forum that I will use here.

Tension Tie
A structural member that is subject to net tension.
Collar Tie
A tension tie in the upper third of opposing gable rafters that is intended to resist rafter separation from the ridge because of wind or unbalanced roof loads. Collar tie is a colloquial term for collar beam.
Rafter Tie
A tension tie in the lower third of opposing gable rafters that is intended to resist the outward thrust of the rafter under load.

Figure 2 is an illustration of a roof constructed using rafter ties.

Figure 2: Illustration of Roof with Rafter Tie.

Figure 2: Illustration of Roof with Rafter Tie.

Calculating Roof Loading

Figure 3 shows how the loading on an individual rafter is defined. Observe that all the loads are specified as if they were projected onto the horizontal plane.

Figure 3: Illustration of a Rafter's Tributory Area.

Figure 3: Illustration of a Rafter's Tributory Area.

The total load on a rafter is given by W, which is calculated using Equation 1.

Eq. 1 W=L\cdot \sigma \cdot d

where L is the rafter length projected onto the horizontal plane, d is the rafter tributory width , and σ is the roof loading projected onto the horizontal plane. The use of projections onto the horizontal plane seems to be the most commonly used approach.

Rafter and Collar Tie Tension Equation

To establish the relationship discussed in the article, we need to observe that the sum of the angular momentums must be 0. Consider the following free body diagram (Figure 4).

Figure 4: Free Body Diagram of the Rafter.Moments are summed about a point on the ridge of the roof. For details on computing the moment due to the distributed load, see the Appendix at the bottom of this post.

Figure 4: Free Body Diagram of the Rafter.Moments are summed about a point on the ridge of the roof. For details on computing the moment due to the distributed load, see the Appendix at the bottom of this post.

For a static (i.e. non-rotating) situation, the moment about ridge beam connection exerted by the tension tie must equal the moment exerted by the support reaction and the distributed load (Equation 2).

Eq. 2 W\cdot L-W\cdot \frac{L}{2}-T\cdot h=0\Rightarrow T=\frac{W}{2}\cdot \frac{L}{h}

This form is correct, but a more useful form for this discussion substitutes the roof slope and height for L.

Eq. 3 {{\left. T=\frac{W}{2}\cdot \frac{L}{h} \right|}_{L=H\cdot \frac{run}{rise}}}=\frac{W}{2}\cdot \frac{H}{h}\cdot \frac{run}{rise}

Equation 3 is presented in the JLC article. This equation can be modified to produce Equation 4, another commonly seen result.

Eq. 4 T={{\left. \frac{W}{2}\cdot \frac{H}{h}\cdot \frac{run}{rise} \right|}_{W=w\cdot L}}=\frac{w\cdot L}{2}\cdot \frac{H}{h}\cdot \frac{L}{H}=\frac{w\cdot {{L}^{2}}}{2\cdot h}

where w is the unit loading (lbf/ft) along the rafter, expressed as force projected horizontally per unit length.

Worked Example

I plugged the JLC example into Mathcad to verify that I get the result that JLC obtained (Figure 5). Note that this example assumed that the dead load had already been projected onto the horizontal plane.

Figure 5: JLC Example in Mathcad.

Figure 5: JLC Example in Mathcad.

Conclusion

Using basic statics, I have derived the equations used in in determining the tension in rafter and collar ties. I have duplicated a worked example from JLC magazine. This exercise has shown me that rafter and collar ties are subject to enormous tension forces. These forces make securing the ties to the rafters an engineering challenge. The tension is so high that a large number of nails/screws would be needed for each joint. In fact, so many fasteners are needed that the joint may weaken because of having so many fasteners in a relatively small area.

I also learned that rafter and collar ties have different purposes: collar ties help secure the rafters to the ridge beam during periods of unbalanced loads (i.e. from wind or asymmetrical snow load); rafter ties are used to counteract the outward thrust of the roof.

Note that this post reviews the mathematics performed by a licensed structural engineer in a construction industry publication. It is not intended to be a tutorial on designing rafter or collar ties. That is a job for a professional.

References

  1. 18th Annual Eastern Conference of the Timber Framers Guild (Article: Pity the Poor Rafter Pair)
  2. Interesting forum discussion
  3. The mechanics of architecture -- free Google Ebook that actually works this problem on pg 121.

Appendix

Calculation Detail on Moment from a Distributed Load

There is a bit of calculus work associated with modeling the moment due to the distributed load along roof. Figure 6 illustrates this computation.

Figure 6: Illustration of Moment Calculation for a Distributed Load.

Figure 6: Illustration of Moment Calculation for a Distributed Load.

Illustration Showing Purlins and Collar Ties

The following figure is from this source.

Figure 7: Illustration of Collar Ties with Purlins.

Figure 7: Illustration of Collar Ties with Purlins.

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Carpentry Math – Drawing a Circular Arch

Posted on 22-November-2010 by mathscinotes

I must admit that I find a certain satisfaction in the geometry that pops up in general carpentry. Fine Homebuilding has a nice video on drawing a circular arch that uses a basic geometric construction. This is something that I expect to be doing soon so I thought I would document it here. The interesting part to me is the basic geometry. Basically, the construction uses a tape measure as a compass. I will just add the following figure, which shows the basic approach.

Basic Arch Construction

Basic Arch Construction

This discussion focused on using geometric methods to determine the radius. You can use algebra to determine the radius. The following figure illustrates how to go about the derivation and provides an algebraic result.

Algebraic Derivation of the Radius

Algebraic Derivation of the Radius

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