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	<title>Presented Without Proof &#187; engineering</title>
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	<description>(an exercise for the reader)</description>
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		<title>Curve-fit with zero offset</title>
		<link>http://301south.net/2010/02/curve-fit-with-zero-offset/</link>
		<comments>http://301south.net/2010/02/curve-fit-with-zero-offset/#comments</comments>
		<pubDate>Tue, 09 Feb 2010 05:38:04 +0000</pubDate>
		<dc:creator>brandon</dc:creator>
				<category><![CDATA[engineering]]></category>
		<category><![CDATA[research]]></category>
		<category><![CDATA[calcium]]></category>
		<category><![CDATA[curve fitting]]></category>
		<category><![CDATA[ionomycin]]></category>
		<category><![CDATA[latex]]></category>
		<category><![CDATA[pstricks]]></category>

		<guid isPermaLink="false">http://301south.net/?p=175</guid>
		<description><![CDATA[Say you are measuring some signal, which you expect to show an exponential decay, but you are unsure of the zero-point on your measuring device.  If you want to measure the time constant of the decay you need to know what it&#8217;s decaying to.  For a signal s,

In order to do conventional least-squares [...]]]></description>
			<content:encoded><![CDATA[<p>Say you are measuring some signal, which you expect to show an exponential decay, but you are unsure of the zero-point on your measuring device.  If you want to measure the time constant of the decay you need to know what it&#8217;s decaying to.  For a signal s,</p>
<p><img src='http://s.wordpress.com/latex.php?latex=s%3Ds_%7B0%7D%2BA_0%5Cexp%28-t%2F%5Ctau%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='s=s_{0}+A_0\exp(-t/\tau)' title='s=s_{0}+A_0\exp(-t/\tau)' class='latex' /></p>
<p>In order to do conventional least-squares fitting to <img src='http://s.wordpress.com/latex.php?latex=A_0%5Cexp%28-t%2F%5Ctau%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='A_0\exp(-t/\tau)' title='A_0\exp(-t/\tau)' class='latex' /> you need to determine <img src='http://s.wordpress.com/latex.php?latex=s_0&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='s_0' title='s_0' class='latex' />.</p>
<p><div id="attachment_180" class="wp-caption aligncenter" style="width: 650px"><a href="http://301south.net/files/2010/02/step-response.png"><img class="size-medium wp-image-180" title="step-response" src="http://301south.net/files/2010/02/step-response-640x400.png" alt="Step response of NIH 3T3 fibroblasts to ionomycin" width="640" height="400" /></a><br />
<p class="wp-caption-text">Step response of NIH 3T3 fibroblasts to ionomycin</p></div><br />
<span id="more-175"></span></p>
<p>For instance, the figure above shows the fluorescent intensity of some 3T3 cells to ionomycin.   The traces are normalized to the initial values.  Notice that the cells don&#8217;t recover to the levels at which they began.  I wanted to know the time constant of the decay on the tail of the trace.</p>
<p>I solved this problem by performing the fit inside an iterative minimum-finding function.  First I identified the point of fastest decline and selected that as the <img src='http://s.wordpress.com/latex.php?latex=t%3D0&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='t=0' title='t=0' class='latex' /> point.  Then I iterated to find the value for <img src='http://s.wordpress.com/latex.php?latex=y_0&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='y_0' title='y_0' class='latex' /> which minimized the norm of the error:</p>
<p><img src='http://s.wordpress.com/latex.php?latex=%7C%7C%20%28y%20-%20y_0%29%20-%20A%5Cexp%28-t%2F%5Ctau%29%20%7C%7C&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='|| (y - y_0) - A\exp(-t/\tau) ||' title='|| (y - y_0) - A\exp(-t/\tau) ||' class='latex' /></p>
<p>for best-fit A and <img src='http://s.wordpress.com/latex.php?latex=%5Ctau&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\tau' title='\tau' class='latex' />.</p>
<pre>
y0_fit = fminbnd( @(y0) ...
                 norm( y-y0 - exp( polyval( polyfit( x, log(y-y0), 1), x))) ,...
                 -10*y(1), y(1) );
</pre>
<p>This, along with other analysis:</p>
<p><a href="http://301south.net/files/2010/02/step-analysis.png"><img src="http://301south.net/files/2010/02/step-analysis-640x378.png" alt="step-analysis" title="step-analysis" width="640" height="378" class="aligncenter size-medium wp-image-183" /></a></p>
<p>For a population of cells (Showing mean and standard deviation for each test point.  numbers in parentheses indicate number of cells in the sample.):</p>
<p><a href="http://301south.net/files/2010/02/step-statistics.png"><img src="http://301south.net/files/2010/02/step-statistics-640x300.png" alt="step-statistics" title="step-statistics" width="640" height="300" class="aligncenter size-medium wp-image-187" /></a></p>
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		<title>The Chevy Volt, the EPA, and thermodynamics</title>
		<link>http://301south.net/2009/08/the-chevy-volt/</link>
		<comments>http://301south.net/2009/08/the-chevy-volt/#comments</comments>
		<pubDate>Thu, 13 Aug 2009 09:09:55 +0000</pubDate>
		<dc:creator>brandon</dc:creator>
				<category><![CDATA[engineering]]></category>
		<category><![CDATA[energy]]></category>
		<category><![CDATA[thermodynamics]]></category>

		<guid isPermaLink="false">http://301south.net/?p=81</guid>
		<description><![CDATA[So, the chevy Volt &#8212; that&#8217;s big.
The US environmental protection agency are in the process of developing a fuel economy standard for hybrid and plug-in electric vehicles, and according to a recent draft version of the standard, General Motors&#8217; flagship entry into the &#8220;range-extended electric vehicle&#8221; market would clock in at 230 miles per gallon.
Recently [...]]]></description>
			<content:encoded><![CDATA[<p>So, the chevy Volt &#8212; that&#8217;s big.</p>
<p>The US environmental protection agency are in the process of developing a fuel economy standard for hybrid and plug-in electric vehicles, and according to a recent draft version of the standard, General Motors&#8217; flagship entry into the &#8220;range-extended electric vehicle&#8221; market would clock in at <a href="http://wheels.blogs.nytimes.com/2009/08/11/gm-chevy-volt-gets-230-mpg/">230 miles per gallon</a>.</p>
<p>Recently I entered a debate in defense of the EPA&#8217;s rating.  It was a retaliation against the claim that GM and the EPA have entered into a conspiracy in which the standards are designed to the specific product available and politically palatable, and the net result would be consumers&#8217; further disillusionment at the very notion of &#8220;clean energy&#8221; when they discovered that their gleefully acquired products failed to perform to their expectations.</p>
<p><span id="more-81"></span>It was a Master&#8217;s student, and so naturally I assumed that he hadn&#8217;t thought out his arguments.  Comically, I then undertook the very error I was presuming, and eventually after working through the question, I came to an understanding of the physical conditions of the problem which I now share with you. The essential innovation of GM&#8217;s car is its all-electric drivetrain.  The motors which turn the wheels are all electric, and they&#8217;re powered by an on-board generator which is attached to a traditional internal-combustion engine which burns gasoline.  This is more efficient for two reasons.</p>
<p>First, you dispense with the transmission.  Instead of having to amplify the relatively narrow dynamic operating range of a combustion engine to match the realities of vehicle speed, we make immediate use of the full flexibility of the direct-current motor.  Ultra-high torque at low speeds, minuscule power consumption at high speeds, and a direct connection to the wheels.  Electronic speed control. Second, the gas motor which still supplies all the energy (when off-grid) has gone from the center of the vehicle and the Source of All Things to a specialized component.  It has exactly one purpose&#8211; to run a generator&#8211; which means it can be highly tuned to a specific optimal operating point and switched on when needed.</p>
<p>My misstep in the debate was in overestimating the energy in a gallon of gas.  Gasoline has a higher heating value of about <a href="http://en.wikipedia.org/wiki/Heat_of_combustion">47 megajoules per kilogram</a> [MJ/kg], which comes out to roughly 130 MJ or 36 kilowatt-hours per gallon.  The conventional four-stroke gas engine implements the <a href="http://en.wikipedia.org/wiki/Otto_cycle#The_Otto_cycle">Otto cycle</a>, a well-characterized and carefully orchestrated alternation of isochoric and adiabatic processes.  Isochoric means &#8220;constant volume&#8221; (i.e. top dead center and bottom dead center of the piston stroke) and adiabatic means &#8220;no heat transfer&#8221; (i.e. compression and power strokes).  Maybe I should write another post about that sometime.</p>
<p>The thermodynamic efficiency of the Otto cycle is known; it&#8217;s a function of the working fluid (in this case a mixture of gasoline and air) and the compression ratio of the engine, and its upper limit is defined by the detonation temperature of the fuel (think anti-<a href="http://en.wikipedia.org/wiki/Engine_knocking">knock</a>, <a href="http://en.wikipedia.org/wiki/Octane_rating">octane rating</a>).  In the case of modern gasoline engines, it&#8217;s about <a href="http://igss.wikidot.com/otto-cycle:efficiecy-vs-compression-ratio">47%</a>.  This means that if you&#8217;re burning gasoline, you give up 18 kWh/gal &#8220;when you drive it off the showroom floor.&#8221;  Add to that mechanical losses and imperfect assumptions and you take off another (say) 22-23%, which puts the gas-tank-to-flywheel efficiency of the well-tuned engine at a practical maximum of about 36%. In the case of the Volt-style transmission-free hybrids in which the combustion engine drives a special purpose alternator, something close to that efficiency might be achievable.  Perhaps an optimistic assumption would be that 13 kWh of electric power can be harnessed from the combustion of a gallon of gasoline, where it can be stored into a battery and then run through copper windings and onto the pavement and into the winds on the open highway.</p>
<p>According to Chevy&#8217;s engineers, their car takes <a href="http://www.chevrolet.com/pages/open/default/future/volt.do">25 kWh of electric power to go 100 miles</a>, which comes out to 52 miles per gallon of gas at 36% efficiency.  In order to get the 230-mpg figure they must have assumed it drives 40 miles ( on a / free of ) charge and then another 12 miles at 52 mpg.  This bears resemblance to a hypothetical test in which the fuel consumption of a vehicle is measured over a 52-mile course. The interesting thing to note here is the bias in measuring gasoline consumption but ignoring electric grid power as &#8220;free.&#8221;  For me, the marginal cost of 25 kWh is about $3.50, compared to $6.50 for two gallons of gas, but if I were charging my car every day that would go up to $6-7, because Southern California Edison uses tiered pricing (which I heartily endorse).</p>
<p>As far as <a href="http://www.epa.gov/RDEE/energy-resources/egrid/index.html">greenhouse gas emissions</a> are concerned, it&#8217;s hard to do worse than coal.  If you get your grid power from a coal plant you&#8217;re putting out about 1 kg of CO2 per kWh.  US average is more like 0.62 kg/kWh; in California we&#8217;re down around 0.45 kg/kWh, due largely to the fact that we have no coal plants, plus we&#8217;ve got all that nuclear and hydro power.  The internal combustion engine, running at its &#8220;practical maximum&#8221; efficiency of 36%, comes in at around 0.86 kg/kWh.</p>
<p>So any move to plug-in vehicles is an improvement, as long as it&#8217;s coupled with a move away from coal.  That&#8217;s fortunate, because the people who love coal power are the same people who are unlikely to buy plug-in hybrids, so advocacy for these two things together won&#8217;t anger any additional people.</p>
<p>The more interesting question surrounds the 25 kWh/100 miles figure.  What speed of travel does that represent? I wonder what theoretical lower bounds can be produced on the amount of energy required to move a car 100 miles?  Because the 13 kWh/gallon of gas applies to <em>any</em> power plant burning gas in a 4-stroke engine, the only way to improve energy efficiency is to reduce the energy needed to go the distance.  Or to run off the grid.</p>
<p>Pity how badly I botched this topic in my public discussion.  Frankly, I&#8217;m embarrassed.  The whole thing can&#8217;t go away fast enough.</p>
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