umap 2005 vol. 25 no. 4

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Paul J. CampbellCampus Box 194Beloit College700 College St.Beloit, WI 53511–[email protected]

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Don AdolphsonChris ArneyRon BarnesArthur BenjaminJames M. CargalMurray K. ClaytonCourtney S. ColemanLinda L. DeneenJames P. FinkSolomon A. GarfunkelWilliam B. GearhartWilliam C. GiauqueRichard HabermanCharles E. LienertWalter MeyerYves NievergeltJohn S. RobertsonGarry H. RodrigueNed W. SchillowPhilip D. StraffinJ.T. SutcliffeDonna M. SzottGerald D. TaylorMaynard ThompsonKen TraversRobert E.D. “Gene” Woolsey

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Vol. 25, No. 4 2004

Table of Contents

EditorialThe Mathematics of Democracy

Paul J. Campbell . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 355

ArticlesOptimal Growth and Productivity in Organizations

Joe Marasco . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 357The Dynamics of Ski Skating

Kenneth R. Driessel, Philip Fink, and Irvin R. Hentzel . . . . . . . . . .375

UMAP ModuleWind Turbine Power Coefficient Optimization

Paul Isihara with Julie Clements, Ronya Kamerlander,William Landry, Jonathan Soyars, and Nathaniel Stapleton . . . . . .411

Reviews . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 439

Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 441

Acknowledgments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 445

Author Electronic Copies of Contributions . . . . . . . . . . 445

Errata . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 446

Editorial 355

EditorialThe Mathematics of DemocracyPaul J. CampbellMathematics and Computer ScienceBeloit CollegeBeloit, WI [email protected]

I write this as the U.S. Presidential and Congressional election is imminent,and I am in the position of explaining to friends abroad how the U.S. electionsystem works—or is supposed to work.My friends find some of the features of our system incompatible with their

own ideas of howademocracy should function, thoughnodoubt their opinionsare formed by familiarity with their own variants. They find our ElectoralCollege bizarrely antidemocratic (well, we are at heart democratic but afterall it is a republic) and our willingness to accept a minority president (anddisempower third parties) absurd (why not a runoff election?).Apart from describing some of the political history of U.S. elections and

offering rationales from the Founding Fathers, I find myself immersed in con-sidering questions of

• voting systems: Would we be better off with a system in which we couldrank candidates, with its “instant runoff” capability? Would we be evenbetter off with approval voting?

• power: Does the Electoral College give more power to small states or tolarge states? Is a vote inWisconsin worth more than a vote in Illinois? Whatis a good index of power in a weighted voting system, such as the ElectoralCollege or the European Union?

• apportionment: Howare thenumbersofCongressional representatives fromthe states determined? Is the method just fine or are their defects in it? Isthere a better alternative? Is the power of state legislatures to draw votingdistricts too often abused in favor of the governing party?

• strategy: What legitimate strategizingdoesour systempermit—what shouldthe Democrats, the Republicans, and other parties do to maximize theirchances of success?TheUMAP Journal 25 (4) (2004) 355–356. c©Copyright 2004byCOMAP, Inc. All rights reserved.

Permission to make digital or hard copies of part or all of this work for personal or classroom useis granted without fee provided that copies are not made or distributed for profit or commercialadvantage and that copies bear this notice. Abstracting with credit is permitted, but copyrightsfor components of this work owned by others than COMAPmust be honored. To copy otherwise,to republish, to post on servers, or to redistribute to lists requires prior permission from COMAP.

356 The UMAP Journal 25.4 (2004)

• hazards: Towhat extent is our voting system subject tomanipulation? How?By people in different states trading votes (“I’ll vote for your Nader here inMassachusetts if you in Floridawill bite the bullet and vote for Kerry insteadof Nader”)? Can we really trust touchscreen voting machines? How comesuch a high percentage of votes are discarded as invalid?

But these are not questions just for me to explain to foreign friends, theyare questions that concern the entire U.S. populace and the national cause ofexporting democracy abroad. Do we really understand our system? Are westuck with a system that the rest of the world thinks is discredited, or can wedefend it? Just what are “emerging” democracies supposed to take from ourexperience?Yes, such questions are taken up in courses in political science. But under-

standing many of the issues involved requires background in the underlyingmathematics—themathematics of elections is not limited to incrementing coun-ters and added totals!From early UMAPModules on—and those are still available—COMAP has

exerted leadership in bringing that underlying mathematics to students. Infact, the topics above almost mirror the contents of chapters in the section onSocial Choice and Decision Making in the COMAP-sponsored For All PracticalPurposes [2003]. In its various editions, that book has been used by half a millionstudents over the past 15 years.As this election draws to a climax, students are once again primed to con-

sider the questions above. It’s a good time to consider altering syllabus or cur-riculum to offer them the mathematical understanding that can lead to someanswers.

ReferencesGarfunkel, Solomon, et al. 2003. For All Practical Purposes: Mathematical Literacy

in Today’s World. 6th ed. New York: W.H. Freeman.

About the AuthorPaul Campbell graduated summa cum laude

from the University of Dayton and received anM.S. in algebra and a Ph.D. in mathematical logicfrom Cornell University. He has been at BeloitCollege since 1977, where he served as Directorof Academic Computing from 1987 to 1990. He isReviews Editor forMathematics Magazine and hasbeen editor of The UMAP Journal since 1984.

Growth and Productivity in Organizations 357

Optimal Growth and Productivityin OrganizationsJoe Marasco3066 Lopez Rd.Pebble Beach, CA [email protected]

There is a science of dynamics in man’s fortunes and nature, as well as ofmechanics.

—Thomas Carlyle, Signs of the Times (1829)

We examine the interaction of growth and productivity. We put forth a verysimple model to enable managers to understand better what occurs as newteam members are integrated into an existing organization. While the modelis straightforward, it does permit some interesting predictions. We show withsome simple graphs how the various factors interact, with a cautionary noteon nonlinear relationships.Attention, managers! Once we have understood the implications of the

model, we need to do whatever we can to mitigate the negative effects ofgrowth. While there are no clear answers to all the problems that growthpresents,we suggest somemeasures that directly affect theparameters towhichthe model is sensitive. Experience shows that paying careful attention to thesefactors can pay big dividends.This article sits at the intersectionof organizational development andproject

management. The organization needs to cope with the problems of growthin order to prosper, while the projects that it does need to be concerned withproductivity and cost. These twoneeds are often seen to be in conflict, requiringdelicate tradeoffs. We attempt here to show areas of commonality and ways inwhich both objectives can be achieved.

IntroductionHealthy organizations have a tendency to growover time. Plainly speaking,

those that are successful in getting donewhat needs to get done are called uponto do more. Usually this translates into growth, as measured by the additionof new people to the team.

TheUMAP Journal 25 (4) (2004) 357–374. c©Copyright 2004 byCOMAP, Inc. All rights reserved.Permission to make digital or hard copies of part or all of this work for personal or classroom useis granted without fee provided that copies are not made or distributed for profit or commercialadvantage and that copies bear this notice. Abstracting with credit is permitted, but copyrightsfor components of this work owned by others than COMAPmust be honored. To copy otherwise,to republish, to post on servers, or to redistribute to lists requires prior permission from COMAP.


A significant challenge facing any successful organization is how to staysuccessful as it grows. This is harder than it looks: The organization needs tocontinue to dowellwhat it has beendoing in the past, it needs to succeed at newchallenges put before it, and it has to do both these things while assimilatingnew teammembers. Scaling become important; what was relatively easy to dowhen the team was small becomes more difficult as the team grows larger.But even in periods of zero growth, there are new hires and their associated

challenges. Organizations that don’t acquire periodic introductions of “newblood” tend to stagnate and succumb to group-think; and there is always ad-dition of team members just to offset attrition. So, although we specificallyaddress growth, there are applications even in no-growth times.However, in the Silicon Valley and in software development, we have seen

the opposite problem: Organizations have a variety of reasons to attempt exces-sive growth. Oftenmanagement seeswindowsof opportunity that theyperceiveas transitory, the corresponding needs sometimes cannot be fulfilled by justadding on contractors or consultants—more organizational staff are essential.In an attempt to capitalize on a leadership situation or a beachhead establishedin a newmarket, organizations rush pell-mell to solidify their position throughultra-rapid growth; middle managers are asked to grow their organizations asfast as they can. The result is usually less than optimal. Certainlywhen growthrates exceed 50% a year, something very difficult is being attempted; we showwhy.As early as 1972, in less politically correct times, Fred Brooks [1995, 25]

talked about adding people to software projects that were late, enunciatingBrooks’s First Law in his classic The Mythical Man-Month:

Adding manpower to a late software project makes it later.

We examine a variant of the same idea. What is new here is a quantitativeexpression of the idea and a generalization to other provocations for growth inorganizations, not just as a knee-jerk reaction to being late. In other words, welook in more detail at Brooks’s notion of the lack of noninterchangeability of“men” for “months.”There are two other interesting data points in Brooks’s notes. Vyssotsky (see

Brooks [1995, 179]) estimates that large projects can sustain growth rates of nomore than 30% per year without suffering. Yet, in the very same note, Corbatois quoted as pointing out that long projects must anticipate a turnover of 20%per year. Given this narrow band of approximately 10% between what mustoccur and what is difficult to accomplish, we need to understand better—in aquantitative fashion—the dynamics of adding people to projects.Other previous work in this area is usually qualitative and anecdotal, prob-

ably because the hard data needed to validate models of this type is difficultto come by. Shooman [1983, 469–479] is a rare exception; his treatment is morecomplex than ours.


The Naıve ModelThemodel is a very simple “one-period”model, with the following assump-

tions:

• At the beginning of the period, the organization is well-integrated, well-trained, and firing on all cylinders. We take the productivity of its teammembers as the baseline for comparison.

• During the period, we add new team members of the same quality as thosealready in place.

• At the end of the period, we have a “new” organization with more peopleand a greater capacity to get work done, because we have more people withthe same average productivity as when we started.

What we deal with in this article is what happens to productivity during theperiod of transition.Note an important assumption: Overall per capita productivity does not

decrease because the organization gets larger. While this is often not the case,we naıvely assume that once we are done with the transition, the organizationhas gained productivity in direct proportion to the growth.For organizations that are constantly growing, we can still use this model:

We just sequence a series of one-period models one after the other.

The Effect of New Hires

ContributionEven during transition, we should add useful productive hours to the cur-

rent total. Even an employee who is only 10% effective during the ramp-upperiod is still adding, for his or her part, 4 hours of “useful” work per week.We must view the remaining 36 hours as training, ramp-up, or “investment inthe future.” We pay for those hours, but they do not show up in the currentproduct1.So, in this most naıve model, we gain hours but lose overall productivity

during the transition period. That’s a reasonable tradeoff, andwe should quan-tify it so that we know what we are trading off for what. However, there is asecondary effect that we should add to the model to increase its fidelity: drag.

1We use product development as our main focus in this article, but the analysis applies to othertypes of organization as well. For a sales organization, substitute the words “hours spent in directselling” for “hours worked on the product”; everything else follows in parallel fashion. When weadd new salespeople, revenues rise but so does the per capita cost of sales.


DragWhile new employees add hours by being productive at some marginal

rate, they also actually detract from the total by being there as well. That is be-cause others in the organization lose productive time in working with the new,less productive employees. This can be time lost in supervision, explanation,“showing themthe ropes,” andanyothermanner of bringing thenewemployeeup to speed, or correcting their mistakes. These are very costly hours, because in-place employees are, by definition, 100% productive. In other words, new em-ployees place a drag on the organization, because otherwise fully-productivein-place employees must interact with them.There are exceptions. If we recruit a new team member with a skill set that

is needed and otherwise absent, there can be a spontaneous increase in overallproductivity. Similarly, if we recruit a superior manager, we may turn aroundan otherwise dysfunctional team. Both these instances are examples of almostimmediate improvement and represent “negative drag,” as it were. But theseare exceptional cases.

The Model and its AssumptionsAs with all models, we make some simplifying assumptions, which we

collect here. We couldmake successively more complicatedmodels to increasefidelity, but at some point we realize diminishing returns; the best models arethe simplest.Our assumptions:

• Existing in-place team members are “100% productive.” We ignore anyloss of productivity due to their training and overhead. We use the existingpeople as the productivity benchmark, and everything is relative to them.Weassume that the organization has a total productivity equal to the productof the average productivity of existing team members times the number ofthem in steady state. Thus, we do not assume that all existing employeeshave the same productivity but rather that their productivities have somedistribution that we can characterize with an average productivity.

• During the ramp-up period (one year), new employees are characterizedby an average fractional productivity P , with 0 ≤ P ≤ 1. We make thesimplifying assumption that they are hired “en masse” at the beginning ofthe year. For example, P = 0.6 indicates that during the ramp-up year theaverage productivity of new team members is 60% of that of existing teammembers, because 60% of the new members’ hours go to “useful” workon the product and 40% of their hours to “getting up to speed.” You canchoose any ramp-up period; my observation is that it is usually longer thanmanagers think it is.


A simple extension of the model could allow for “learning-curve” behavior.That is, we could use the typical “S–curve” to model new-hire ramp-up. Inthe interest of simplicity, we choose instead to assume a constant averageproductivity over the entire period.

• We measure the annual growth of the team by the fraction G of new em-ployees, usually between 0 and 1. For example, G = 0.1 indicates a 10%growth rate, or the addition of 1 new team member for each existing 10;G = 1.0 would correspond to 100% growth, or attempting to add one newperson for each existing person. As we will see later, growth rates of thismagnitude are highly risky. Once again, we make the simplifying assump-tion that new team members are hired “en masse” at the beginning of theyear. More complex models would stage, or phase, the arrival of new hires;doing so leads once again to more complicated mathematics but nothingconceptually novel.

• The effect of new hires is characterized by a drag ratioD, with 0 ≤ D ≤ 1.For example, a D = 1.0 means that for every hour spent in nonproductivework by a new team member, there is an associated loss of an hour by anexisting team member. The smaller D, the better for the organization.

• New team members have the same average hourly pay rate as existingteam members. This is reasonable because we are assuming that once theramp-up period is done the new teammembers will be just as productive asthe existing ones. In other words, new hires are of the same average qualityas those already in place, hence are paid at the same rate.

Consequences of the ModelWefirst considerwhetherwe aremakingmore hours available for the prod-

uct. There are two competing factors:

• On the one hand, we add hours due to the productivity, however small, ofthe new team members.

• Ontheotherhand, training themcostsproductivityon thepart of the existingteam members.

The equation that relates H , the number of useful hours, to G, P , and D is:

H = 1 + G[P − (1 − P )D].

Note some quick checks:

• With G = 0, we have H = 1: There is no difference in useful hours.

• If P = 0, we have H = 1 − DG: Useful hours are decreased by the trainingdrag on the rest of the organization.


• If P = 1, we have H = 1 + G: Useful hours increase as fast as we grow.

• If D = 0, we have H = 1 + PG: The gain in productive hours equals thegrowth rate times the productivity of the new team members.

• Finally, when D = 1, we see that H = 1 + G(2P − 1): We gain hours if Pis greater than 0.5; otherwise, we lose hours. For every hour that new teammembers contribute to the product, they also consume an hour ramping up;the hour that must be matched by an existing team member exactly cancelsout the hour gained by the new person.

Lowerdragmeans that the organization ismore efficient at assimilatingnewpeople; if, for example,D = 0.5, then the new people need only be productiveat slightly greater than 33% in order to increase the total effort put into theproduct.In fact, we increase the number of useful hours on the product if and only if

P > (1 − P )D.

Note that this condition is independent of the growth rate G. That is, you canalways get more useful hours by growing faster if you observe this conditionon P and D.

Nailing the MultiplierWe can rewrite the equation

H = 1 + G[P − (1 − P )D]

asH = 1 + GM,

whereM = P − (1 − P )D.

In Figure 1, we compute the value for M , which when multiplied by thegrowth rate G gives the percentage increase in useful hours on the product.(Since we are using “1” as the baseline, the term GM represents the increase“over one,” so it is actually a percentage increase.)

• The five white bands to the left correspond to negative multipliers. In thisregion, we see a decrease in the number of useful hours per unit of growth;growing faster in this case actually results in fewer and fewer hours devotedto the product.

• The other shaded bands represent progressively better scenarios; for exam-ple, thefirst bandnext to thewhite region corresponds to anM between0and0.2. In this regionwe get an average of 0.1 times the growth rate in incremen-tal useful hours; with 30% growth, for example, we achieve 0.1× 30% = 3%in incremental useful hours.

10.90.80.70.60.50.40.30.20.100

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Productivity of New Team Members, P

Dra

g, D

-1--0.8 -0.8--0.6 -0.6--0.4 -0.4--0.2

-0.2-0 0-0.2 0.2-0.4 0.4-0.6

0.6-0.8 0.8-1


Figure 1. Contour plot of the multiplierM as a function of dragD and productivity P .

• At the opposite extreme, the dark band at the very right of the chart cor-responds to an M between 0.8 and 1.0, so in this region we can assume anaverage value of around 0.9. For the same 30% growth rate, we now attain0.9 × 30% = 27%more useful hours on the product.

To summarize, this multiplier depends on two factors in a nonlinear way:

• P is characteristic of the population of new hires; it estimates their relativeproductivity during ramp-up.

• D characterizes the existing organization; it is a measure of the ability tointegrate new people efficiently.

What the chart tells us is that if we can keep D small, we can tolerate fairlylow P . On the other hand, the higherD, the more sensitive we are to P . Theseconclusions agree with our intuition.

Useful Hours Added to ProductNow that we understand the multiplier effect, we plot a simple graph of

percentage increase in useful hours as a function of growth (Figure 2). Therelationship is simply linear. OnceyouknowM , you can readoff thepercentageincrease in useful hours by looking up G on the horizontal axis and using theappropriate sloped line for M . The result is read off the vertical axis. Or, youcan do the reverse: For a required percentage increase in useful hours, you canread off the growth rate required for your calculated value ofM .

0%

5%

10%

15%

20%

25%

30%

35%

40%

45%

50%

5% 10% 15% 20% 25% 30% 35% 40% 45% 50%

Growth Rate, G

Use

ful H

ou

rs A

dd

ed, D

elta

H

M = 0.2 M = 0.4 M = 0.6 M = 0.8 M = 1.0


Figure 2. Useful hours added to product, ∆H , as a function of growth rate G, for various valuesof the multiplierM .

What About Cost?Remember that you have to pay for all hours, productive or not. As soon as

you have hours that are nonproductive, due either to the new teammembers orthe drag that they cause, the overall productivity of the organization decreases.Whatmanagementmust consider iswhat constitutes a reasonable trade-off.

You produce your product faster because you are devoting more hours to it,but in so doing you are making the product more costly to produce, because theoverall productivity of the organization is lower due to training the added teammembers. The management balancing act consists of deciding howmuch timeis worth how much cost. For many projects, especially in software, the laborcontent represents the preponderance of the cost, so these arguments are cogent2.The total number of useful hours dedicated to the product is

H = 1 + GM.

On the other hand, the new total number of hours we are paying for is

T = 1 + G.

So the new overall productivity of the organization is just

N =H

T=

1 + GM

1 + G.

2And, of course, one has to be sure that all this happens under conditions of “equal or betterquality.” A more costly product brought to market faster may not be a good thing if the overallquality suffers.

0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 15%

10%

15%

20%

25%

30%

35%

40%

45%

50%

Multiplier, M

Gro

wth

, G

70-75 75-80 80-85 85-90 90-95 95-100


We plot this relationship in Figure 3 to see how the overall productivity variesas a function ofM and G.

Figure 3. Contour plot of overall productivity (as a percentage) as a function of the growth rateG

and the multiplierM .

The zone to the right is “good.” That is, we have a productivity that isbetween 95% and 100% of our base productivity. At the other extreme, at theupper left, overall productivity has dropped to between 70% and 75% of thebaseline.Note that the best way to guard against precipitous drops in productivity

is to grow slowly, say no more than 20%. If we stay under 20% growth, we canguarantee that overall productivity will not drop below 85%, regardless of themultiplier, and it won’t fall below 90% so long asM is at least 0.4. On the otherhand, above 20% growth we start to become very sensitive toM .If you can keep M at 0.85 or above, you can also stay in the “95%” pro-

ductivity zone regardless of the growth rate. Achieving an M of 0.85 is a realchallenge, as you can see from Figure 1.What about the labor cost to produce the product? If we are putting H

useful hours into the product per unit time, then the product will be completedin 1/H the time. But each unit of time now costs (1 + G) dollars, so the newtotal cost is (1 + G)/H . This is just the inverse of the overall productivity. Theidea that cost and productivity are in inverse proportion to each other appealsto our notions of good sense.Figure 4 shows how product cost increases withM and G.

0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 15%

10%

15%

20%

25%

30%

35%

40%

45%

50%

Multiplier, M

Gro

wth

, G

0.0-5.0 5.0-10.0 10.0-15.0

15.0-20.0 20.0-25.0 25.0-30.0

30.0-35.0 35.0-40.0 40.0-45.0


Figure 4. Contour plot of percentage cost increase as a function of growthG and the multiplierM .

An Illustrative ExampleSuppose that we currently are spending $1M in labor per year and we

assume we can get out our new product at the end of the year with the currentteam in place. Management would like to get the product out in 10 monthsinstead of 12. We assume that any new teammembers whomwe recruit will beonly 60% productive during the year, and that our organization has a mediumdrag, D = 0.5. The two numbers we need to come up with are

• how many new people do we need to hire, and• what does this do to the cost of developing the product?To produce in 10 months what would have taken 12 means that over those

10monthswe need 20%more useful hours devoted to the product. SoH = 1.2,which means that GM = 0.2. But M = P − (1 − P )D = 0.6 − (0.4)(0.5) =0.6 − 0.2 = 0.4. That means that Gmust equal 0.2/0.4 = 0.5, that is, we have a50% annual growth rate.The labor cost of the product per unit time goes up by 50% becauseG = 0.5.

We produce it in only ten-twelfths of the time that it would have taken, so thetotal cost goes up by only 25%. That is, we spend at a rate of $1.5M per yearin labor but produce the product in ten-twelfths of a year, so the total cost is$1.25M. This 25% increase is exactly what the increased cost graph of Figure 4predicts.From the overall productivity chart of Figure 3 withM = 0.4 and G = 0.5,

we achieve an overall productivity of 80%. This result agrees with the idea that


we are devoting 1.2 useful hours to the product but paying for 1.5 total hours,since 1.2 ÷ 1.5 = 80%. The inverse of 0.8 is 1.25, which squares with the ideathat the cost to produce the product has gone up by 25%.The bottom line: To gain 2months on the 12-month schedule, wemust grow

at a 50% annual rate and increase our product cost by 25%. This consequence isa result of having only 60% productivity by new teammembers and a trainingdrag of 1 hour of current team-member time for each 2 hours of new team-member training time. Also, we are undertaking substantial increased riskdue to the 50% growth rate.We leave it to the reader to compute the implications of a management

desire to reduce the time to market from 12 months to 9 months.

A Note on NonlinearityWe tend to have good intuition when there is a linear relationship between

the controlling variables. For example, all other things being held equal, weexpect that three times asmuchworkwill take us three times as long. However,our intuition works less well when variables have a nonlinear relationship.In the problem at hand, givenM , there is a linear relationship between the

number of useful additional hours H devoted to the product and the growthrate G. That is, we gain useful hours in direct proportion to growth rate, withM as the constant of proportionality. Unfortunately, this is the only linearrelationship in the whole model.To compute the key ingredientM , the multiplier, we use a relationship that

is nonlinear in the variables P and D:

M = P − (1 − P )D.

This leads to Figure 1, where different bands correspond to different ranges ofM . We have only weak intuition as to how those bands come out as a functionof P and D; we know that low P and high D are bad in combination, but it ishard to judge just how bad. That is why the graph is so helpful.Similarly, both the overall productivity and increased cost to produce the

product are nonlinear in M and G. Recall that the relationship for the newoverall productivity is

N =1 + GM

1 + G,

onceagainanonlinear relationship. Wedon’t haveagood“feel” forhowoverallproductivity goes, other than the notion that largeG and smallM must be bad.The increase in cost is just the inverse of N , so that relationship is nonlinear aswell. The two graphs for these quantities Figures 3–4 show a banded structuresimilar to that when computingM . And, sinceM is nonlinear in P andD, wehave a nonlinearity on top of a nonlinearity! So trying to infer what N mightbe as a function of P , D, and G is a difficult leap indeed!


The relationship that we are trying to intuit is

N =1 + G[P − (1 − P )D]

(1 + G).

Most of us would freely admit that we have little or no intuition for N as afunction of these three variables. Yet that is implicitly the task that we start outwith.The moral of the story:

Even simple natural phenomena are sometimes nonlinear.

When this occurs, our intuition can be weak. In order to understand whatis really happening and make intelligent predictions, having a simple mathe-matical model and some graphical visualization tools can be invaluable.

Call to ActionNow that we understand how the various factors interact to affect produc-

tivity during periods of growth, we are faced with doing something about it.We have three suggestions.

• Whenever possible, use real data to estimate the parameters P and D.These are the ingredients in computing the multiplier M , on which every-thing else depends. Pulling numbers for them out of the air will likely leadyou astray. Not all organizations keep records that are detailed enough tohelp. On the other hand, you can look, for example, at hours explicitly de-voted to training; these go into both P and D, as both new employees andold are involved. All meetings should be looked at to see whether they areprimarily “product-related” meetings or “get-the-new-hires-up-to-speed”meetings. The better you can estimate P andD, the better the numbers thatcome out the other end.

• Think hard about P when evaluating new hire candidates. The model ismost sensitive to P , and nothing can overcome a low value for it.

– Selecting “fast learners” is always a good thing; “slow learners” havelow P and should be avoided.

– Look out for a “missing skill,” such as specific knowledge about a pro-gramming language or technique that allegedly can be picked up easily.Even very bright people take time to learn new things; and if these thingsare to be learned on the job, you are agreeing to reduceP for that person.

– Avoid hiring people who come from a company culture radically differ-ent from yours. There is an implicit and somewhat large decrease in P(and a corresponding increase in D) for hires who have the additionallearning burden of adapting to “how we do it here.”


If the new hire has a tendency toward inflexibility, these factors will bemagnified still further.

• Think of D as amplifying the effects of a low P . Ironically, smaller orga-nizations may have largerD than more-mature organizations, because theyhave most of their organizational memory stored in the brains of a few peo-ple. When new people are brought on board, there is no way to get them upto speed other than in one-on-one dialogs with these key people. There area few things that are obvious:

– Hire and train new people in batches, to get some economy of scale inthe adaptation process.

– To the extent possible, invest in documenting “how things get done,”so that new hires can read materials instead of having to take time fromexisting team members every time a question arises.

– For software development organizations, having well-written andwell-documented source code can dramatically diminish D, as well as helpto increase P .

There is no “silver bullet” here, but understanding the importance of bothPandD can help managers keep damage to a minimum and estimate the returnon investment of activities such as documentation. Modeling reduction ofD bya few tenths can tell us how much we could improve overall productivity andreduce incremental cost during the next growth period. We can then comparethe cost of reducing D in this way with the benefits that would accrue.

ConclusionsWehave explored a very simple three-parametermodel for growth and pro-

ductivity in organizations during a transition period of assimilating new hires.We have modeled “useful hours” and “overall organizational productivity” asfunctions of the annual growth rate, new team-member productivity, and orga-nizational assimilation effectiveness. While the model is simple, it shows thathigh growth rates are risky and that the combination of high growth and lownew-employee productivity can be disastrous, even if the steady-state produc-tivity of new hires equals that of the existing ones. Add to that an organizationthat is inefficient in assimilating new team members, and you have the recipefor a perfect storm.Even when the new hires are as good as the people already in place, there

is yet another danger: The new, larger organization may be less productive,per capita, than before. Overall productivity may suffer due to size and scalingeffects, such as the increased overhead of more communication links, and soon. Our model ignores these effects.What happens in reality is often evenmore insidious: In attempting to grow

faster, we often lower recruiting standards. When this happens, not only is the


combined “growth + training hit” large, but there is a dilution of the talentpool, leading to a permanent lowering of productivity even after the ramp-up is complete. The problem for the organization is a difficult one: During theramp-upperiod,weattribute thedecrease inproductivity to ramp-upactivities;it is only afterwards that we observe that overall productivity has suffered. Bythen, of course, it is too late. A long-term problem has escaped us under theguise of being a transitional or short-term one; there has been “masking.” Thatis why it is crucial when hiring always to bring people on board who, in steadystate, will raise the average productivity of the group.Although our model is very simple, it leads to several nonlinear relation-

ships, for which we have poor instincts. Our graphs to illustrate the interde-pendencies amongst the variables revealed structures thatwe had not foreseen.These “hidden” structures can cause otherwise goodmanagers to be surprised;things diverge on them more rapidly than they expect, because of nonlinear-ities. Cutting “just one more month” off a schedule by adding people canhave disastrous consequences if you are already in the “yellow zone” withoutknowing it. That last seemingly small step pushes you over the cliff and intothe abyss; better to draw some graphs and understand where you really arethan to scream loudly on the way down.We urge paying attention to improving the parametersP andD that govern

everything else; this is our “call to action.”Our model makes several simplifying assumptions, about which we pru-

dently warn the reader. If, based on these assumptions, the model predictsproductivity and cost effects that are unpleasant, it behooves the manager toproceed with caution. The reason is simple: Nature is rarely as kind as themathematical model. Adding to this fact Murphy’s Law (“whatever can gowrong, will”), you can be sure that the results will in general be worse thanwhat the models predict. You need a margin for error. So “taking the modelwith a grain of salt” means this: If the model says that you are OK, proceedwith caution; if the model says that you might have a problem or be in a high-risk situation, back off and reconsider. In most situations, managers are toooptimistic in estimating themodel parameters; that tendency, coupledwith theoptimistic assumptions that go into it, can lead to project Hell.


Appendix A: NomographIt would be nice to be able to get the calculations done quickly, even better

to get them without having to plug numbers into equations. Back in the daysof ships of wood and men of iron, there was a simple solution to capturing re-lations among variables: graphical methods, in particular, the device known asthe nomograph3. The user simply laid a straightedge across specially arrangedand calibrated scales. If serial calculations were required, multiple instances ofthis technique were chained together, all on one sheet of paper. An engineerdid the nontrivial mathematical work behind the nomograph, and a masternomographer produced a chart that was ridiculously simple to use. This tech-nique, documented by d’Ocagne around 1900, was highly developed by theworldwide engineering community between 1925 and 1975, when the adventof pocket calculators and computers caused it to go out of fashion.I created a nomograph for the equations in this article (Figure 5; a higher-

quality rendering is at http://www.barbecuejoe.com/GP.htm ). As with allgood nomographs, a usage example is included right on the chart, the illustra-tive example from earlier.Nomographs have the interesting property of solving equations both ex-

plicitly and implicitly: You can find any missing variable or combination ofvariables by working backwards. Experiment a little with this. For example,start with an “allowable” increase in cost and percentage increase in usefulhours desired and find the growth rate needed and the multiplier that this im-plies. Then see which combinations of P and D give you that M . Often, thistype of exploration is useful in seeing whether your constraints can be met.A quick word on how I constructed the nomograph. Moving left to right,

the first part is done by observing that the relationship between P , D, and Mis a standard form that is amenable to a “Z-type” nomograph; the nonlinearscale for D is constructed manually by finding appropriate common intersec-tion points. Then the M , ∆H , and G nomograph is added, using the notionthat useful hours derive from growth G times the multiplier M . Once again,a slightly different “Z-type” nomograph is appropriate, and the nonlinear Gscale is determined by the method of intersections. Finally, we make the ob-servation that once we know the ∆H that we want and the multiplier M thatwe have already determined, both overall productivity and its inverse, the in-crease in cost, are determined. These, however, turn out to be envelopes, notsimple intersections. These envelopes were constructed graphically by usingmany pairs of values for ∆H and M corresponding to each value of overallproductivity; adding the increase in cost is just a labeling exercise at that point.

3From the Greek nomos (law) and graphein (to write); “nomogram” is used synonymously.


Figure A1. Growth and productivity nomograph.


Appendix B: SpreadsheetThemoremodernapproach to simplifying the calculations is tousea spread-

sheet. Figure B1 is a snapshot of one that I constructed in far less time than ittook to create the nomograph. (The spreadsheet file is available by request [email protected] .)

Figure B1. Growth and productivity spreadsheet.

Starting from this very simple spreadsheet, you can easily develop someof the more sophisticated enhancements mentioned earlier in the article. Forexample, you can use a different wage rate for the new hires, although theexample has them equal so that you can compare the spreadsheet result withyour own calculation using the equations or nomograph. Moreover, once themodel is cast in spreadsheet form, you can use tools such as Goal Seeking toimplicitly solve, or work backwards, as you see fit.


ReferencesBrooks, Frederick P. 1995. The Mythical Man-Month: Essays on Software Engi-

neering. 2nd ed. Boston, MA: Addison-Wesley.

Shooman, Martin L. 1983. Software Engineering: Design, Reliability, and Manage-ment. New York: McGraw-Hill.

About the AuthorJoe Marasco is a retired senior vice-president and busi-

ness-unit manager for Rational Software, now one of thefive brands of the IBM Corporation. He held numerouspositions of responsibility in product development, mar-keting, and the field sales organization, overseeing initia-tives for the Rational Apex product and Visual Modelerfor Microsoft Visual Studio. In 1998, he served as SeniorVP of Operations. He retired from Rational in 2003. Heholds a bachelor’s degree in chemical engineering fromThe Cooper Union, a Ph.D. in physics from the Universityof Geneva, Switzerland, and an M.B.A. from the Univer-sity of California, Irvine. When not writing, he barbecuesand plays golf; his ribs are much tastier than his scores.

The Dynamics of Ski Skating 375

The Dynamics of Ski SkatingKenneth R. Driessel1703 Palmer DriveLaramie, WY [email protected]

Philip FinkCenter for Complex SystemsFlorida Atlantic UniversityBoca Raton, FL [email protected]

Irvin R. HentzelMathematics Dept.400 Carver HallIowa State UniversityAmes, IA [email protected]

AbstractWediscuss the dynamics of ski skating, in particular, the optimization prob-

lem: Maximize average speed for a given power.Webeginwith amathematicalmodel of ski skating. To simplify the analysis,

we limit our attention to ski skating with no poles on a level plane. We alsolimit our attention to the physics of ski skating; we ignore most biomechanicalconsiderations.In the early days of skating, ski instructors often said: “Take a long glide on

a flat ski.” Many skiers interpreted this to mean: “Take a passive glide, thengive a short hard push to the side.” As an application of our theory, we showthat this advice is wrong. We show that the skier should start pushing to theside as soon as possible.

Keywords: cross-country skiing, nordic skiing, freestyle, skating, dynamics, model, ordinarydifferential equation, control, optimization, biomechanics, physics, frictionAMS Subject Class: 34H05 Ordinary differential equations, control problems



IntroductionBy limiting the scope of inquiry, researchers can attain greater certaintyabout questions they do answer. —William C. McGrew [1966, 76]

In the years following the 1980Winter Olympics, cross-country skiing expe-rienced a rapid revolution in skiing technique. The classic technique, whereinthe skis stay in a track parallel to the direction of motion, was replaced by askating technique, where the ski moves at an angle to the direction of motion.This change in technique brought about lower times in races (see Karvonenet al. [1989] and Bilodeau et al. [1992]), possibly lower metabolic costs [Smith1992], and eventually a split of the sport into separate classic and freestyle races.Along with the split in races, a renewed effort to optimize ski performance hasalso occurred.

Traditional ApproachesTraditionally, biomechanists have used two methods to optimize perfor-

mance:

• The first, and most common, is to assume that elite athletes are performingat or near optimal performance. This method has been used extensively toanalyze cross-country skiing. Three variables of interest have been the stridefrequency, stride length, and the angle between the ski and the direction ofmotion. Several authors [Bilodeau et al. 1992; Smith et al. 1989; Smith etal. 1989] have reported values for athletes in classic skiing as well as threedifferent methods of freestyle, or skate skiing.

– Stride frequencies ranged from 0.68 Hz to 0.81 Hz for the three skat-ing techniques on flat surfaces as compared to 0.80 Hz for the classictechnique. (Recall that “Hz” is the standard abbreviation for “hertz” or“cycles per second”.) Higher stride frequencies were found for skatingup an incline (0.81 to 0.84 Hz).

– Stride lengths were greater for the skating techniques (from 7.19 m to8.69 m) as compared to the classic technique (5.95 m), with a decrease inlength when skiing up an incline (2.99 m to 3.84 mwhen on a 7◦ and 11◦

slope, respectively).

– Ski anglewas likewisedependenton slope,with smaller ski angles foundon a lower grade (24◦) than on a steeper grade (28.9◦).

Strong positive correlations were found by Bilodeau et al. [1992] betweenvelocity and the distance traveled per cycle (which they call cycle length) forall strides, while the significant correlations between velocity and frequencywere negative when significant. This analysis seems to suggest that skierswouldgain an advantageby taking longer, slower strides. By contrast, Smith


et al. [1989] found significant positive correlations between frequency andvelocity, although these values were obtained from skiing up a slope. Bothcorrelations were found using elite athletes, which suggests a problem withthis type of analysis: All elite athletes, especially when the techniques arerelatively new, do not show identical movement patterns.

• The second method that can be used to optimize performance is to modelperformance and try to find optimal results that way. An example of thesuccess of this strategy comes from swimming. For a long period of time, itwas assumed that swimmers propelled themselves through the water usingthe drag forces acting on the hands and feet as a propulsive force. The resultof this assumption was that swimmers were taught to move their hands in alinear path through the water. However, when the forces acting on the handwere modeled, a different result was found [Brown and Counsilman 1971]:A more effective means of propulsion is to use a combination of lift anddrag forces acting on the hand, and this discovery had the effect of changingthe instruction to an S-shaped path used by elite swimmers today. Thusfar, theoretical work of this kind has not been done on ski skating. In thispaper we propose to use this method to gain further understanding of howto optimize ski skating performance.

Our Goal and OutlineWe discuss the dynamics of ski skating. In particular, we try to solve the

following optimization problem:

Maximize average speed for a given power.

We first need to develop a mathematical model of ski skating. To simplifythe analysis, we limit our attention to ski skating with no poles. We also limitour attention to the physics of ski skating; we ignore most biomechanical con-siderations.In Preliminaries, we briefly review Newton’s Second Law of Motion and

its application to skiing. We also review some of this law’s elementary conse-quences that we use repeatedly.In Straight-Line Skiing, we consider a simple model of classical (diagonal)

skiing and review the model of friction that we use.In the main section, Ski Skating, we introduce our model of ski skating

with nopoles on a level plane. Our analysis produces a number of relationshipsbetween the variables overwhich the skier has control. These variables includetempo, glide time, step length, push force, push time and the angle between theskis. We then solve the constrained optimization problem: Maximize averagespeed.We include Suggestions for Future Work and two appendices. In Ap-

pendix 1: Measuring theDynamic Coefficient of Friction, we consider sliding


friction. For a skier coasting down an incline, we describe an experiment thatcan be used to measure the dynamic coefficient of friction between the snowand the skis. In Appendix 2: Dimensional Analysis, we show how to formdimensionless parameters from the mass of the skier, the power of the skier,the glide time and the acceleration of gravity.We are deliberately somewhat repetitious in this report, so that the sections

are somewhat independent of one other; we want to try to accommodate the“grasshopper” style of reading.We have tried to minimize the mathematical background needed to un-

derstand this paper, since we hope that skiers and ski coaches will read it.Elementary college physics and calculus are the only prerequisites for under-standingmost of it (see, for example, Finney and Thomas [1990]). We have alsoincluded most of the details in our calculations.

PreliminariesWe discuss the dynamics of skiing. We base our discussion on Newton’s

Second Law of Motion—force equals mass times acceleration—and apply it tocross-country skiing. (For a discussion of Newton’s law applied to downhillskiing, see Lind and Sanders [1996].)

NotationWe use the following notation:

• r(t) is the position of the center of gravity of the skier; this position is a vectorquantity that is a function of time;

• t denotes time;

• v(t) := dr/dt is the velocity of the center of gravity; a(t) := dv/dt is theacceleration of the center of gravity;

• m is the mass of the skier; and

• F(t) is the vector sum of all the external forces acting on the center of gravity.

We set forth all of our notation for easy reference in Table 1.

Newton’s Law and Its ConsequencesUsing our notation, we can write Newton’s law as F = ma. However, we

prefer to work with the following system of equations:

m dv/dt = F,

dr/dt = v.


Table 1.

Notation, by section.

Symbol Meaning Dimension Unit

Preliminariesr(t) Position of the center of gravity of the skier L m (meter)r(t) Position of the center of gravity of the skier. L m (meter)t Time T s (second)v(t) := dr/dt Velocity of the center of gravity L/T m/sa(t) := dv/dt Acceleration of the center of gravity L/T 2 m/s2

m Mass of the skier M kg (kilogram)F(t) Vector sum of all external forces acting on

the center of gravity of the skier ML/T 2 N (newton)♦ End-of-remark notation

R Force exerted by the groundas a reaction to the skier’s pushing ML/T 2 N

W ba Work done by a force when a particle moves

along a curve from time t = a to time t = b ML2/T 2 J (joule)· The inner product operation between two vectorsv2 := v · v The square of the length of a vectorK(t) := 1

2 mv(t)2 Kinetic energy of a particle at time t ML2/T 2 JP (t) := dW t

a/dt Power, the rate at which work is done,is the derivative of work with respect to time ML2/T 3 W (watt)

T Half the period of a periodic function T s

Straight-Line Skiingv := dy/dt Speed of the skier’s center of gravity L/T m/sy(t) y-coordinate of the skier’s center of gravity L mR(t) Force exerted by the ground as a reaction to pushing ML/T 2 NS(t) Snow friction force ML/T 2 NV Average speed of the skier L/T m/sµ Coefficient of friction —N Force normal to the travel plane ML/T 2 Ng Acceleration of gravity L/T 2 m/s2

p := P/(mg) Power-to-weight ratio of the skier L/T m/sz(t) Vertical height of the center of gravity of the skier L mz′(t) := dz/dt Vertical velocity of the center of gravity of the skier L/T m/sz′′(t) := dz′/dt Vertical acceleration of the center of gravity

of the skier L/T 2 m/s2

Ski SkatingS(t) Snow friction force ML/T 2 Nα Angle between the direction of travel and

the glide direction of the right ski —tr Time of weight transfer to the right foot T stl Time of weight transfer to the left foot T sFr Size of the component of the force

perpendicular to the ski ML/T 2 NFs Size of the component of the force along the ski ML/T 2 Nw Velocity vector that results from the push L/T m/swr Size of the velocity component perpendicular

to the ski L/T m/sws Size of the velocity component along the ski L/T m/sδ The angle π/2minus the angle between the ski tracks —∆t Push time T ses Unit vector in the direction of the right ski glideer Unit vector perpendicular to es and to the

skier’s left sides(t) The “ski speed” or “foot speed” L/T m/sr(t) The “side speed” (effected by the skier’s push

with the right foot) L/T m/sA Average speed in the direction of travel L/T m/sρ Size of the push force relative to the skier’s weight —τ := (T − tp)/T Amount of the push time relative to the half period —f(α, τ) “Normalized speed function” —


Symbol Meaning Dimension Unit

l Size of the step to the left that the skier takes as aresult of the side push by the right leg L m

L Maximum step size L mλ(l) := µ3gl/2p2 Normalized step size —Λ := λ(L) Normalized maximum step size —U Feasible region where we wish to maximize

average speedA

Appendix: Measuring the Dynamic Coefficient of Frictiony = f(x) Function representing the incline down which

the particle is slidinge2 Unit vector (0, 1)T Unit vector tangent to the graphθ Angle between the incline and verticals Distance down the incline from some reference position L ms′ := ds/dt Speed of the particle along the curve y = f(x) L/T m/ss′′ = F (x) Acceleration of the particle due to gravity and friction L/T 2 m/s2

γ Path defined by(x, f(x)

)h(x, y) µgx − gyK(t) := 1

2 ms′(t)2 “Kinetic energy” of the particleb Average slope —

These are the basic ordinary differential equations that determine the trajec-tory of the center of gravity of the skier. We concentrate on the first of theseequations.

Remark Concentrating on the center of gravity of the athlete is a standardviewpoint in biomechanics. See, for example, Alexander [1992] and/orMcMahon [1984]. In particular, this is the viewpoint adopted by Svensson[1994] for skiing. ♦We can decompose the forces affecting themotion of the skier into a number

of main components:

• sliding friction between the snow and the ski,• drag between the air and the skier,• gravity,• muscular action forces, and• ground reaction forces.The force vector F is the vector sum of these forces. In particular, a force Ris exerted by the ground as a reaction to the skier’s pushing. (We ignore thebiomechanical details that produce this force.)In later sections, we use a relation between kinetic energy and work that

follows fromNewton’s law. If a particlemoves along a curve r(t) between timea and time b and is acted on by a force F(t), then the workW b

a is given by

W ba =

∫ b

a

F · v dt,


where the dot denotes the inner (dot) product of the two vectors. (Note, forexample, that if F is constant, then W b

a = F · (r(b) − r(a)), so that in this case

work equals the inner product of force and distance.) Using Newton’s law, wehave

W ba =

∫ b

a

F · v dt =∫ b

a

mdvdt

· v dt

= 12m

∫ a

b

d

dt(v · v)dt = 1

2mv2|bt=a,

where v2 := v · v. In summary, we have

W ba = K(b) − K(a) (*)

where K(t) := 12mv(t)2 is the kinetic energy at time t. Thus the work equals

the difference in the kinetic energies. We refer to the relation (∗) as the relationbetween the work and the kinetic energy.Power is the rate at which work is done, commonly measured in watts or

horsepower. In particular, instantaneous power is the derivative of work withrespect to time. We have

P (t) :=d

dtW t

a.

Using the formula for work, we get

P (t) =d

dt

∫ t

a

F · v dt = F(t) · v(t).

We are particularly interested in the average power of the skier. If R(t) andv(t) are periodic functions with period 2T , then the average power of the skieris defined by

P :=1

2T

∫ 2T

0

R(t) · v(t)dt.

(Defining the average of a function is more complicated if the function is notperiodic.) Note that only the component ofR parallel to v enters here.

Straight-Line SkiingWe consider straight-line skiing on a plane and fashion a mathematical

model of classical (or diagonal-stride) skiing. Later, we apply to ski skatingsome of the results that we derive in this simpler setting. (The mathematicsthat we develop in this section can also be used to model simple double polingor travel by means of a Norwegian kick sled that is propelled like a scooter;these are simpler applications.)We introduce Cartesian coordinates as follows:


• The xy-plane is the travel plane;

• the direction of travel is the positive y-axis;

• the x-axis is in the travel plane and directed to the right of the direction oftravel; and

• the z-axis is perpendicular to the travel plane and pointing up.

We assume that each ski travels in a straight line in the direction of they-axis, and that the skis are on opposite sides of the y-axis. It follows that theskier’s center of gravity remains (approximately) in the yz-plane. In particular,we consider the following scalar differential equation:

mdv

dt= R − S,

where

• m is the mass of the skier,

• v := dy/dt is the speed of the skier’s center of gravity,

• y(t) is the y-coordinate of the skier’s center of gravity,

• R(t) is the force that is exerted by the ground as a reaction to the skier’spushing,

• and S(t) is the snow friction force.

(We are ignoring air drag.) We are interested in solving the following optimiza-tion problem:


We assume that the functions R(t) and v(t) are periodic with period 2T , inparticular, that v(0) = v(2T ). We expect these functions to be approximatelyperiodic if the skier is traveling in a straight line on a plane and is kicking andpoling at a steady rate. We use P to denote the average power of the skier; insymbols,

P :=1

2T

∫ 2T

0

R(t)v(t)dt.

It is clear that for periodic velocity, the average power input by the skier (de-termined by the function R(t)) equals the average power loss (determined bythe function S(t)). The following mathematical argument corresponds to thisintuition. Using the relation between work and kinetic energy, we have

∫ 2T

0

R v dt −∫ 2T

0

S v dt =∫ 2T

0

(R − S)v dt


= 12m v(2T )2 − 1

2m v(0)2 = 0

and hence

P =1

2T

∫ 2T

0

R v dt =1

2T

∫ 2T

0

S v dt.

Model for Constant Snow FrictionWe now consider the case when S is constant. We then obtain the relation

P = SV , where V is the average speed, given by

V :=1

2T

∫ 2T

0

v dt =y(2T ) − y(0)

2T.

From the equation V = P/S, we see that the average speed of the skier iscompletely determined by the skier’s power P and the constant frictional forceS. Thus, average speed does not depend on the shape of the function R(t).Furthermore (on the basis of this model), there is no mathematical reason forthe skier to adopt any particular tempo; the same average speed is achievedby using short fast strides or by using long slow (but more forceful) strides.(We will see that for ski skating, the situation is quite different.) However,there are undoubtedly biomechanical reasons for choosing a certain tempo.For example, bicycling experiments show that power depends on tempo (seeWhitt and Wilson [1974]).We consider several specific examples. The usual model for ski friction is

S = µN whereµ is the coefficient of friction andN is the force normal to the travelplane. (This model appears in physics textbooks; see, for example, Sears [1958,34ff]. It also appears in Svensson [1994]; see especially p. 254 Figure A-2, p. 257,and p. 259 Figure A-4. There are more-sophisticated models of friction; Krim[1996] discusses the work by physicists on friction.) For skiing on a level plane,N equals the weight of the skier, that is, N = mg where g is the accelerationof gravity. Then V = p/µ where p := P/(mg) is the power-to-weight ratio of theskier.

Example We compute the power-to-weight ratio and the average speedfor some typical values. A typical value for µ is 0.05 [Svensson 1994, 257].The coefficient of friction isdimensionless, so there arenoassociatedunits.A typical weight for a male skier is 750 Newtons (= 75 kg × 10 m/s2

≈ 165 lbs since 1 Newton =1 kg × 10 m/s2 ≈ (1/4.46) lbs.). A typicalpower for recreational athletes (bicycle riders) is 75 to 225 watts [Whittand Wilson 1974, 22], which is equivalent to 0.1 to 0.3 hp (since 745 watts≈ 1 hp). Hence, the power-to-weight ratio p is typically between 75/750 =0.1m/s and 225/750 = 0.3m/s. (Recall the following relations betweenunits: Watt = Joule/s, Joule=Newton–meter.) It follows thatV is typicallybetween

0.1/0.05 = 2m/s ≈ 7.2 km/h ≈ 4.5mi/h


and0.3/0.05 = 6m/s ≈ 21.6 km/h ≈ 13.5mi/hr.

Bilodeau et al. [1992] report that the speeds of elite skiers are about4.75 m/s, in this range, though our top theoretical value seems a bit high.

♦Weeasily find the solution of the differential equationm dv/dt = −S, where

S = µmg; we will use this solution later. Substituting for S and canceling mgives dv/dt = −µg. Integrating both sides of this differential equation gives

v(t) − v(t0) =∫ t

t0

(−µg)dt = −µg(t − t0).

From this result, we see that if the skier is freely gliding on a horizontalplane then the change in velocity of the skier is proportional to the glide timewith constant of proportionality equal to µg. We can also easily compute thedistance traveled as function of time. Since v = dy/dt, we have

dy

dt= v(t0) − µg(t − t0).

Integrating, we get

y(t) − y(t0) =∫ t

t0

(v(t0) − µg(t − t0))dt

= v(t0)(t − t0) − 12µg(t − t0)2.

Effect of Vertical MotionWe consider the effect of any vertical motion of the center of gravity of the

skier while the skier is gliding on a horizontal plane. We want to neglect suchvertical motion when we analyze skating. Our argument here shows its effectto be negligible. We just derived formulas for v(t) and y(t) when the normalforce N is constant and equal to the weightmg of the skier. Suppose now thatN = mg + mz′′(t), where z(t) is the vertical height of the center of gravityof the skier, z′ := dz/dt is the vertical velocity and z′′ = dz′/dt is the verticalacceleration. We consider the differential equation

mdv

dt= −µ(mg + mz′′),

ordv

dt= −µ(g + z′′).

Integrating, we get

v(t) − v(t0) =∫ t

t0

−µ(g + z′′)dt


= −µg(t − t0) − µ(z′(t) − z′(t0)).

Note that if z′(t) = z′(t0), then v(t) − v(t0) = −µg(t − t0), which is the sameformula that we got when N = mg. Integrating again, we get

y(t) − y(t0) =∫ t

t0

[v(t0) − µg(t − t0) − µ

(z′(t) − z′(t0)

)]dt

= v(t0)(t − t0) − 12µg(t − t0)2 − µ(z(t) − z(t0)) + µz′(t0)(t − t0).

If z(t) = z(t0) and z′(t0) = 0, then we get

y(t) − y(t0) = v(t0)(t − t0) − 12µg(t − t0)2,

which is the same formula that we got when N = mg. In other words, if theskier starts the glide with no vertical velocity and returns to the same verticalheight, then the skier travels the same distance as if the center of gravity hadstayed at a constant height. This analysis shows that vertical motion generallyhas only a minor effect on forward motion. Consequently, we often ignorethe vertical motion of the center of gravity of the skier when motion is on ahorizontal plane.However, this analysis also shows that there is a way for the skier to use

vertical motion to increase average speed: The skier can lower the center ofgravity during the glide, to reduce friction during this phase, and then quicklyraise it when the ski stops. In other words, the skier can take advantage of thestopping of the ski to increase efficiency. Thismay be the source of the directive“Take long glides” that we hear from a number of ski coaches. Since in order toprovide a propulsive force during straight-line skiing, one of the skismust stop,the skier can take advantage of this stop time to accelerate the body upwards.This procedure reducesN and hence the frictional force during the glide phase.Furthermore, increasingN during the kick phase increases the frictional force,which allows the skier to apply a greater forward propulsive force.

Ski SkatingWe consider ski skating with no poles on a level plane. In particular, we

consider the following vector differential equation (Newton’s law):

mdvdt

= R − S

where

• m is the mass of the skier,

• v(t) is the velocity of the center of gravity of the skier,

• R(t) is the ground reaction force that is the result of the skier’s muscularaction, and


• S(t) is the snow friction force.

(We ignore air drag.) We are interested in the following optimization problem:


We assume that the skier is traveling with a steady rhythm. More precisely,we assume that the velocity vector v(t) and the reaction force vector R(t) areperiodic functionsof time. Weuse2T todenote the cycle time—a timeallotmentT for the right foot and a time allotment T for the left foot. We assume right-left(or bilateral) symmetry of the skier’smotion. We also assume that the right andleft skis travel in straight lineswhile the skis are gliding. We useα to denote theangle between the direction of travel and the glide direction of the right ski. Itfollows from thebilateral symmetry assumption thatα is also the angle betweenthe direction of travel and the glide direction of the left ski (Figure 1). In thisfigure, α is the angle between the line segments Q1Q2 and Q1B2. (This figureis a variation of ones that appear in Svensson [1994], for example, on p. 101.See also the overhead pictures in Caldwell [1987], for example, on pp. 76–81.)

Remark We have only rarely encountered any objection to our assump-tion that the skis travel in a straight line during the glide. However,when one of us (Driessel) mentioned this assumption to Antonina Anikin(a well-known coach of Russian and American cross-country ski racers)during a visit to Duluth, Minnesota in January, 1997, she objected. Sheclaimed that the skis change direction during the push at the end of theglide. We have observed some ski tracks which turn farther away fromthe line of travel in the last part of the glide. We have only rarely observedsuch tracks. ♦

We introduce Cartesian coordinates as follows:

• The xy-plane is the travel plane,

• the centerline of travel as the y-axis directed in the direction of travel,

• the x-axis is in the travel plane and directed to the right of the direction oftravel (see Figure 1), and

• the z-axis is perpendicular to the travel plane and pointing up.

We restrict our attention to the xy-components of the vectors that we con-sider. In other words, we ignore the vertical motion of the skier. We do thismainly to simplify the analysis. However, we also believe that such verticalmotion is not very important. (See also our discussion of vertical motion in thesection Straight-line Skiing.) Throughout the rest of this section all vectorsshould be regarded as vectors in the xy-plane. In particular, we are concernedwith the motion of the projection of the center of gravity of the skier onto thexy-plane.


x

y

A1

R1

B1

Q1

A2

R2

B2

Q2

C1

L1

D1

C2

L2

D2

alpha

alpha

Figure 1. Ski tracks.


Remark The following suggestion by Svensson [1994, 229] seems to berelevant here: “To ski effectively andmaintain speed, the ski skater shouldstrive for minimizing the vertical fluctuation by keeping the level of thecenter of gravity relatively high during the skate cycle.” ♦

The Skate CycleWeanalyze theportion of the skate cycle between the timeofweight transfer

to the right foot and the time of weight transfer to the left foot. We temporarilyuse tr and tl to denote these times of weight transfer to the right and to the left.In Figure 1, we represent the locations of the right and left skis at time tr by theline segments A1B1 and C1D1; we represent the locations of the right and leftskis at time tl by the line segments A2B2 and C2D2.We assume that the skier completes a full periodic cycle by stepping to the

right, to the left, and then again to the right. Using the bilateral symmetryassumption, we then see that the half-cycle time is equal to the time betweenthe weight transfer to the right foot and the weight transfer to the left foot; insymbols, we have T = tl − tr. In other words, T equals the glide time on theright ski. We now assume that tr = 0 and tl = T . We can shift the time scale sothat these two equations are satisfied.During the glide on the right ski, the skier pushes sideways with the right

leg. This push is perpendicular to the right ski. If the ski stops at the end ofthe glide, then the skier can also push backwards along the ski. It appears thatthere are two styles of skate skiing—one for which the ski stops and one forwhich the ski does not stop. We limit our attention to the nonstopping style.In particular, we assume that the skier’s push is perpendicular to the ski.

Aside We believe that a ski-skater’s push is usually perpendicular tothe ski. However when one of us (Driessel) mentioned this assumptionto Scott Hauser during a visit to the Steamboat Ski Touring Center inDecember 1996, Hauser questioned this assumption. We then performedtwo simple informal experiments.

• For the first experiment the skier stood still on level ground, leanedforward and pushed off the right ski. (Of course, poles were notused.) We measured the angle between the left and right ski tracksthat the skier produced. We found this angle to be about 70◦ (whichis significantly different than 90◦).

• For the second experiment the skier skied up an incline (with a slopeof about 3%) with no poles. The skier was instructed to let each skiglide to a momentary stop (or near stop) before stepping off it. Weagain measured the angle between the ski tracks; again we found itto be about 70◦.

We performed these experiments with only one intermediate skier; weexpect that the angles probably vary between skiers.


We believe that the skier is using the static friction between the ski and thesnow to get a velocity component in the direction of the ski. The followingsample calculation seems to support this belief. We assume that the skierexerts a constant force F during the push. Let Fr denote the size of thecomponent of this force perpendicular to the ski and let Fs denote thesize of the component along the ski. (We use “s” for “snow”.) Let wdenote the velocity vector that results from the push. We assume thatthe skier starts from rest, that is w(0) = 0. From Newton’s law (namely,F = mdv/dt), we getw = (F/m)∆twhere∆t is the push time. In terms ofcomponents we havewr = (Fr/m)∆t andws = (Fs/m)∆t, wherewr andws are respectively the sizes of the velocity components perpendicular tothe ski and along the ski. Let δ be π/2 minus the angle between the skitracks. Then

tan δ =ws

wr=

Fs

Fr.

We assume that Fs = µsmg, where µs is the coefficient of static friction.(In the rest of this report, we use µ to denote the coefficient of slidingfriction.) Then we have

ws = µsg∆t, wr = ws/ tan δ.

For example, take µs := 0.1, g := 10 m/s2, ∆t := 0.1 s, and δ := 0.1 ra-dian. Then

ws = (0.1)(10m/s2)(0.1 s) = 0.1m/s

and (using tan δ � δ)

wr � (0.1m/s)/0.1 = 1m/s.

(Recall that 1 m/s = 3.6 km/h = 2.24 mi/h.) These speeds are reasonable.In particular, we see that the skier can get some speed in the direction ofthe right ski.

The following comment by Svensson [1994, 259] is relevant here: “Duringthe final skate push-off . . . the ski is stationary for a short time.” Whenone of us (Driessel) mentioned this comment to Antonina Anikin duringa visit to Duluth,Minnesota in January 1997, she said: “No, the ski shouldnot stop.” ♦

ConstraintsLet es denote a unit vector in the direction of the right ski glide. The snow

friction force vector is in the opposite direction. Let er denote a unit vectorperpendicular to es and to the skier’s left side. The ground reaction force is inthe direction of the vector er. We resolve the velocity v(t) of the skier’s centerof gravity vector into components in these two directions as follows:

v(t) = r(t)er + s(t)es.


This equation defines the functions r(t) and s(t). We call s(t) the ski speed orfoot speed; it is effected by the snow friction. We call r(t) the side speed or reactionspeed; it is effected by the ground reaction force caused by the skier’s push withthe right foot.We assume that the skier is balanced over the right foot when the weight is

transferred to the right ski. It follows that the velocity of the skier’s center ofgravity equals the velocity of the right foot at this time; in symbols, we havev(0) = s(0)es, or r(0) = 0.Let 2W denote the work done by the skier during a cycle and let P be the

average power of the skier. Then P = 2W/(2T ) = W/T . Now r(T )er is thevelocity vector that results from the skier’s push with the right leg. Since thisvector is the result of the workW done by this leg during the glide of the rightski, we also have (using the relation between work and kinetic energy)

W = 12m r(T )2.

From the two equations involving the work, we get

PT = 12m r(T )2.

We call this relation the power constraint.Since the skier is balanced over the left ski when the skier’s weight is trans-

ferred to this ski, we have

v(T ) = r(T )er + s(T )es.

Using the bilateral symmetry assumption, we have |v(T )| = |v(0)| = s(0).In Figure 2, we represent these various velocities: The line segment AD

represents the vector v(0). The line segment AC represents the vector v(T ).The line segment BC represents the vector r(T )er. Note that α equals angleEAD.In the figure, |AC| = |AD| = s(0), |BC| = r(T ) and sin 2α = |BC|/|AC|.

Consequently, we have

sin 2α =r(T )s(0)

.

We also have 0 < 2α ≤ π/2, or

0 ≤ α ≤ π/4,

since s(T ) ≥ 0. We call these relations the geometric constraints.Recall that s(t) denotes the speed of the right ski during the glide. We

assume that the snow friction is constant and equal to µmg. From the previoussection, we then have

s(t) = s(0) − µgt.

(This formula for the slow-down holds only approximately, since the modelof friction that we use is only an approximation. More important, perhaps, is


x

y

A

B

C DE

alpha

v(T) v(0)

r(T)e r

Figure 2. Velocities.

the fact that the coefficient of friction µ probably changes as the ski is edgedduring the glide. We don’t know of any experimental results that quantify thischange. However, Svensson [1994, 259] does discuss this matter.)From the slow-down formula for s(t), we get |AB| = s(T ) = s(0) − µgT

and hence |BD| = µgT . Note that ∠BCD equals α. Hence,

tanα =µgT

r(T ).

We call this relation the slow-down constraint.In summary, we have the following constraints:

PT = 12m r(T )2, (power)

sin 2α =r(T )s(0)

and 0 ≤ α ≤ π

4, (geometric)

tanα =µgT

r(T ). (slow-down)


Reducing to a Single ParameterWe regard the quantities m, P , µ and g as parameters, the values of which

usually remain unchanged during a ski outing:

• The massm and power P are determined by the skier’s body.

• The coefficient of friction µ is determined by the interface between the skiand snow.

• The acceleration of gravity depends slightly on the location at which theouting takes place.

We regard thequantitiesT , r(T ), s(0), andαasvariables. The skiermayvarythese quantities during the outing. However, they must satisfy the constraints.In fact, we can use the constraints to reduce the number of variables. Since

there are three constraints and four variables, we expect that we might be ableto express all of them in terms of one of them; indeed, we find formulas for allof them in terms of α.We can solve for r(T ) in terms of α as follows. From the power and slow-

down constraints, we get

µgPT = 12µmgr(T )2

andµgPT = Pr(T ) tanα.

Hence, we have12µmgr(T )2 = Pr(T ) tanα

and12µmgr(T ) = P tanα

provided r(T ) �= 0. Thus,

r(T ) =2p

µtanα

where p := P/(mg) is the power-to-weight ratio (introduced in the previoussection).We can now express s(0) in terms of α. From the geometric constraint and

the formula for r(T ) in terms of α, we get

s(0) =r(T )sin 2α

=2p tanα

µ sin 2α,

or

s(0) =p

µ cos2 α

sincetanα

sin 2α=

(sin α/ cos α)2 sin α cos α

=1

2 cos2 α.


We can also express the half-period time T in terms of α. From the formula forr(T ) and the slow-down constraint, we get

T =r(T )µg

tanα =(2p/µ) tan2 α

µg,

or

T =2p

µ2gtan2 α.

Remark Svensson [1994, 232] makes the following statement: “Skierstherefore should strive to skate with optimal smallest angle and longglide which is more efficient.” We don’t understand how it is possible fora skier to have a long glide time T and a small angle 2α between the skis.The formula for T in terms of α shows that if α is small, then T must besmall. ♦

Average SpeedWe now consider the average speed of the skier in the direction of travel.

Recall that v(t) denotes the velocity of the skier. Note that e2 := (0, 1) is a unitvector in the direction of travel. Then the inner product v(t) · e2 is the speed ofthe skier in the direction of travel. The average speed in this direction is

A :=1T

∫ T

0

v(t) · e2dt.

Recall that we represent the velocity of the skier during the glide on the rightski as

v(t) = r(t)er + s(t)es.

Note that er · e2 = sinα and es · e2 = cos α. Hence,

A =

(1T

∫ T

0

r(t)dt

)sin α +

(1T

∫ T

0

s(t)dt

)cos α.

In particular,A is a weighted sum of the average of r(t) and the average of s(t).From above, we have s(t) = s(0) − µgt and hence (using the formulas for

s(0) and T ) we get

1T

∫ T

0

s(t)dt =1T

∫ T

0

(s(0) − µgt)dt

=1T

[s(0)t − µg

t2

2

]T

t=0

=1T

(s(0)T − µg

T 2

2

)


=p

µ cos2 α− 1

2µg

(2p

µ2gtan2 α

)

=p

µ

(1

cos2 α− tan2 α

)=

p

µ.

We also want to compute the average of the function r(t). We assumethat the skier pushes perpendicular against the right ski with a constant forcestarting at time tp. We use the expression ρmger to represent this constantforce. In particular, we have normalized this force using the skier’s weight;the dimensionless number ρ indicates the size of this push force relative to theskier’s weight. In other words, we consider reaction force functions R(t)er

where R(t) := 0 if t is between 0 and tp and R(t) := ρmger if t is between tpand T .

Remark This model of the skier’s side push function is quite simple.More-general functions can be treated using the calculus of variations;see, for example, Alexander[1996] orWeinstock[1974]. We expect that theconclusions using more-general functions will be similar to the ones thatwe derive here. ♦

From Newton’s law (that is, frommdr/dt = ρmg), we get

r(t) = ρg(t − tp)

for tp ≤ t ≤ T . Note thatr(T ) = ρgτT,

where τ := (T − tp)/T , and that 0 ≤ τ ≤ 1. In particular, τ is the percentage ofthe glide time that the skier is pushing.The parameters ρ and τ are related to the power of the skier. From the

power constraint, we getPT = 1

2m(ρgτT )2,

orp =

P

mg= 1

2gρ2τ2T.

There is a trade off between ρ and τ for fixed p and T . If ρ is large, then τ mustbe small. Using the formula for T in terms of α, we get

p = 12gρ2τ2

(2p

µ2gtan2 α

)=

(ρ2τ2

µ2tan2 α

)p.

Hence,

ρ =µ

τ tanα.


We can now compute the average of r(t). We have (using the formulas forT and ρ in terms of α)

1T

∫ T

0

r(t)dt =1T

∫ T

tp

ρg(t − tp)dt

=1T

ρg 12 (t − tp)2|Tt=tp

=1T

ρg 12 (T − tp)2

=1T

ρg 12τ2T 2 =

12ρgτ2T

= 12

( µ

τ tanα

)gτ2

(2p

µ2gtan2 α

)

=τp

µtanα.

Thus,

A =

(1T

∫ T

0

r(t) dt

)sin α +

(1T

∫ T

0

s(t) dt

)cos α

=τp

µtanα sin α +

p

µcos α,

orA =

p

µ(τ tanα sin α + cos α).

OptimizationWe regard p and µ as fixed parameters. We want to maximize the average

speed A = A(α, τ) subject to the constraints 0 ≤ α ≤ π/4 and 0 ≤ τ ≤ 1. Weconsider the function

f(α, τ) := τ tanα sin α + cos α

on the rectangle [0, π/4] × [0, 1]. We call this function the normalized speed func-tion. (Figure 3 shows a plot of this function. Figure 4 shows its level curves;the arrows in this figure point uphill.)Note that if α �= 0, then

∂f

∂τ= tanα sin α > 0.

It follows that themaximumof f occurs on the boundary of the rectangle. Now

f(0, τ) = 1,


0

0.2

0.4

0.6

0.8

1

tau

0

0.2

0.4

0.6

alpha

0.8

1

1.2

1.4

Avg

0.2

0.4

0.6

0.8

1

tau

.8

1

Figure 3. Normalized speed function.

0 0.2 0.4 0.6 0.8alpha

0

0.2

0.4

0.6

0.8

1

tau

Figure 4. Level curves for normalized speed function.


f(π/4, τ) = τ,

f(α, 0) = cos α, and

f(α, 1) = tanα sin α + cos α =1

cos α.

We see that the maximum of f occurs at α = π/4, τ = 1, with f(π/4, 1) =√

2,since cos π/4 =

√2/2. It follows that the maximum of A(α, τ) on the rectangle

is√

2p/µ. Note that τ = 1 implies that tp = 0; in other words, the skier shouldimmediately start pushing to the side.

Examples and Adjustment of ModelExample: Intermediate Athlete We consider an example to see if wehave missed any constraints. Let p := 0.2 m/s, µ := 0.05, a := π/4, andτ := 1. Then

A =p

µ(τ tanα sin α + cos α)

=0.20.05

(tan

π

4sin

π

4+ cos

π

4

)= 4

√2m/s ≈ 20 km/h ≈ 12mi/h.

This value seems rather high. We also compute the half-period time T(with g := 10m/s2); we get

T =2p

µ2gtan2 α =

2(0.2)(0.05)210

= 16 sec .

This value is not reasonable. ♦

We can also compute the size of the step to the left that the skier takes asa result of the side push by the right leg. We use l to denote the length of thisstep. In the figure representing the ski tracks, this step is represented by theline segment R2L2. We have

l =∫ T

tp

r(t)dt =∫ T

tp

ρg(t − tp)dt

= 12ρgτ2T 2

= 12

( µ

τ tanα

)gτ2

(2p tan2 α

µ2g

)2

or

l =2p2τ

µ3gtan3 α.


Example: Intermediate Athlete (continued) With the values in the lastexample, we get

l =2(0.2)2

(0.05)310= 64m.

This step size is clearly impossible. ♦

It is nowobvious thatwe need to add another constraint to our optimizationproblem. We add the constraint l ≤ L, whereLdenotes themaximumstep size.We also introduce a dimensionless version of these quantities; in particular, welet

λ(l) :=µ3gl

2p2and Λ := λ(L).

We can rewrite the equation for l as follows:

λ(l) = τ tan3 α.

We want to maximize A(α, τ) on the region U defined by the constraints0 ≤ α ≤ π/4, 0 ≤ τ ≤ 1, and τ tan3 α ≤ Λ. In Figure 5, we plot such a region.

-0.2 0.2 0.4 0.6 0.8alpha

-0.2

0.2

0.4

0.6

0.8

1

tau

Figure 5. Restricted optimization region.

As before, we see that the maximum must occur on the boundary—in fact,it must occur on the boundary curve

{(α, τ) : Λ = τ tan3 α, 0 ≤ α ≤ π/4, 0 ≤ τ ≤ 1}.

Note, in particular, that τ is a decreasing function of α on this curve. Weconsider the function α → A

(α, τ(α)

), where τ(α) := Λ/ tan3 α. We claim that

this function is decreasing on the interval (0, π/4] := {α : 0 < α ≤ π/4}. It will


follow that the maximum of A(α, τ) on U occurs when τ = 1 and tan3 α = Λ.We have

A(α, τ(α)

)=

p

µ

(τ(α) tanα sin α + cos α

)=

p

µ

(Λ

tan3 αtanα sin α + cos α

)

=p

µh(α),

where h(α) := Λ(1/ sin α − sin α) + cos α, since

sin α

tan2 α=

cos2 α

sin α=

1 − sin2 α

sin α.

We have

h′(a) =dh

dα=

d

dα

[Λ

(1

sin α− sin α

)+ cos α

]

= Λ(− cos α

sin2 α− cos α

)− sin α.

Clearly, h′(α) < 0 on (0, π/4].

Final Model: Intermediate Athlete vs. Elite AthleteExample: Intermediate Athlete (continued) We reconsider the previousexample, with the values p := 0.2 m/s, µ := 0.05, g := 10m/s2, andL := 1m. Then

Λ = λ(1) =µ3g

2p2=

(0.05)3102(0.2)2

=(

14

)3

.

From Λ = τ tan3 α and τ = 1, we get tanα = 14 or α � 1

4 radians ≈ 14◦).We can now easily compute the corresponding average speed A; we get

A =p

µ(τ tanα sin α + cos α)

� 4((0.25)2 + 0.97

)= 4.12m/s ≈ 14.8 km/h ≈ 9.2mi/h.

We can also compute the corresponding half-period T ; we get

T =2p

µ2gtan2 α =

2(0.2)(0.05)210

(14

)2

= 1 s.

♦


Example: Elite Skier The value p := 0.2m/s is appropriate for an inter-mediate athlete. If we repeat the calculation in the previous paragraphwith p := 0.3m/s, which is appropriate for an elite athlete (and using thesame values for µ, g, and L as above), then we get

Λ ≈ 6.94 × 10−3,

tanα ≈ α ≈ 0.191 radians ≈ 11◦,A ≈ 6.10m/s, andT ≈ 0.866 s.

FromBilodeau et al. [1992], we have the following typical observed valuesfor real elite skiers (using poles) traveling on a flat site:

• average velocity = A = 5.8 m/s,

• cycle time = 2T = 1.4 s, and

• cycle length = 2TA = 8.1 m.

Our values for the theoretical elite skier (not using poles) are

• A = 6.10 m/s,

• 2T = 1.7 s, and

• 2TA = 10.6 m.

We see that real elite skiers travel at about the same speed and use asomewhat faster tempo than our theoretical elite skiers. The angle ofabout 1◦ for the theoretical elite appears reasonable for flat plane skiing.Smith et al. [1988] report a mean angle of about 24◦ for skiers climbing a7% slope; angles are generally larger for climbing steeper slopes. ♦

Suggestions for Future WorkWe have discussed a simplemathematical model of ski skating. This model

explains some aspects of ski skating. But we would like to understand more,and there are many possibilities for future work that are related to our model.In this section, we discuss a few of these possibilities.

Testing the TheoryOurmodel agrees qualitatively with data gathered for actual skiers. Unfor-

tunately, we have no data for ski skatingwithout poles on a level plane. Is therereasonable quantitative agreement between our model and the performance ofactual skiers who are not using poles? We need experimental data to answerthis question. The experiments could be done on a snow-covered lake so thatthe level-plane assumption is satisfied.


HillsWe model a skier traveling on a level plane. This situation is rare; skiers

are almost always climbing or descending. We believe that our mathematicalmodel can be generalized to inclined-plane situations.Here is a question related to such a situation: Is there a climbing steepness

at which the skier should glide to a stop with each stride? We conjecture thatthere is such an angle.There are several different hill situations that can be modeled using an

inclined plane. In one situation the line of travel is straight uphill—in otherwords, the line in the travel plane perpendicular to the direction of travel islevel. In this situation, bilateral symmetry is again a reasonable assumption.Here is another situation involving an inclined plane. The line of travel is

level but the line in the travel plane perpendicular to the line of travel is notlevel. This is a simple “side-hill” situation. In this situation, bilateral symmetryis not reasonable.Of course, skiers sometimes climb in side-hill situations. Here the line of

travel is not directly uphill and not level.

More-General Push FunctionsWe consider simple push (that is, reaction force) functions. In particular, we

consider functions that equal zero until a push starting time tp and equal a pos-itive constant value after that time. We find that among these push functions,the one with tp = 0 maximizes the average speed in the direction of travel.The question arises: Can an optimum function be found when more generalfunctions are allowed? (We mentioned this generalization in a remark in thesection Ski Skating.)A related question is: What more-general class of push functions should be

considered?

Three DimensionsWe consider a two-dimensional model of ski skating. In particular, we

regard the center of gravity to be in a plane and the push (that is, groundreaction force) vector to be in this plane. We should consider an analogousthree-dimensional model, with the center of gravity and the push vector inthree-dimensional space.Our two-dimensional analysis implies that the skier should start pushing

to the side at the beginning of the glide. But in three dimensions, if the skier’scenter of gravity is directly above the foot and the push vector is directed alongthe line from the foot to the center of gravity, then a push at the beginning ofthe glide only raises the center of gravity and does not cause a motion to theside.


Multi-linked ChainsZatsiorsky [1998] writes: “A human body can be modeled as a multi-link

system comprising several body segments connected by joints.” A linkage ofrigid bodies is referred to as a multi-linked chain. Zatsiorsky [1998; 2002] dis-cusses extensively the modeling of the human body by means of multi-linkedchains. (See also Alexander [1992], Alexander [1996, Chapter 3: Optimummovements], and McMahon [1984, Chapter 8: Mechanics of locomotion].)We should consider simple multi-linked chain models of ski skating as one

way of incorporating the three-dimensional aspect of ski skating.

PolingWe consider a model of ski skating with no poles. We should find a model

that includes poling.

Friction of an Edged SkiIn the section Ski Skating, we allude to the following question: Is the

coefficient of friction for a flat ski different than the coefficient of friction for anedged ski? We don’t know any experimental results concerning this question.We indicate in an appendix how to measure the dynamic coefficient of

friction of a pair of skis. Here is a way to answer the question concerningedged skis: Build a sled to which a pair of skis can be attached at variousedging angles. Then use the theory in the appendix to measure the coefficientof friction for skis edged at various angles.

Appendix 1: Measuring the DynamicCoefficient of FrictionWe consider sliding friction. For a skier coasting down an incline, we de-

scribe an experiment to measure the dynamic coefficient of friction betweenthe snow and the ski.We begin by constructing a mathematical model. We represent the skier

as a particle with mass m. We represent the incline by the graph y = f(x) ofa smooth function f . We choose our coordinate axes so that the gravitationalforce vector is in the direction of the negative y-axis; that is, the gravitationalforce vector is represented by −ge2, where g is the acceleration of gravity ande2 := (0, 1). We assume that the particle moves from left to right down theincline; that is, we assume that the particle moves in the direction of increasingx. If the particle is at a point (x, y) on the incline, then the component of theparticle’s weight in the direction tangent to the incline acts to accelerate (or


decelerate) the particle. Let

T :=

(1, f ′(x)

)[1 + f ′(x)2

]1/2;

note that T is a unit vector tangent to the graph. The size of the weight com-ponent in the direction of T is

mg(−e2 · T) = −mgf ′(x)[

1 + f ′(x)2]1/2

= mg cos θ,

where θ is the angle between the incline and vertical. (We call this force thetangential gravitational force.) In Figure 6, we picture the incline; the slope isrepresented by the curved line. At the point P on the graph, the tangent isrepresented by the line segment AB and the normal is represented by the linesegment CD. The vertical direction e2 at P is represented by the line segmentEF . The angle θ is represented by APF .

x

y

A

B

C

D

E

F

P

Figure 6. Incline.

We assume that the friction force between the particle and the incline isdetermined by the coefficient of friction µ and the component of the particle’sweight in the direction perpendicular to the incline. (We call this component ofthe weight the normal gravitational force.) In other words, the frictional force isgivenby−µmg sin θ. (Thismodel of the frictional force is standard; for example,


it appears in Svensson [1994, 236].) We assume that gravity and friction are theonly two significant forces acting on the particle (in particular, we ignore airdrag).In summary, we use the following differential equation tomodel themotion

of the sliding particle:

ms′′ = mF (x), ors′′ = F (x),

where s is the distance down the incline from some reference position,

F (x) := g cos θ − µg sin θ,

cos θ =−f ′(x)[

1 + f ′(x)2]1/2

, and

sin θ =1[

1 + f ′(x)2]1/2

.

We regard the quantities x and θ as functions of s.Note that the change in speed of the particle is related to the work done. In

particular, from the differential equation, we have

d

dt

(12 (s′)2

)= s′s′′ = F (x)s′

and hence12 (s′)2|t1t=t0 =

∫ t1

t0

Fs′dt.

We can compute the work done and obtain the following simple formula.

Proposition. Let f be a continuously differentiable function. Assume that aparticle moves along the graph of f according to the differential equation givenabove and the particle occupies points (x0, y0) and (x1, y1) at times t0 and t1respectively where t0 < t1 and x0 < x1. Then the work done in moving betweenthese points is given by

−g(y1 − y0) − µg(x1 − x0).

Remark Note that the first term in this formula is positive if the particlemoved downhill (that is, y1 < y0). Note that the second term is negative.

♦

Proof: Recall the following relation between x and s:

ds

dx=

[1 + f ′(x)2

]1/2.


Using the formula for change of variables, we then get∫ t1

t0

Fs′dt =∫ s1

s0

F ds

=∫ x1

x0

F (x)[1 + f ′(x)2

]1/2dx

=∫ x1

x0

(−gf ′(x) − µg)dx

= −g(f(x1) − f(x0)

) − µg(x1 − x0)= −g(y1 − y0) − µg(x1 − x0).

We conclude that the change in speed of the particle inmoving from (x0, y0)to (x1, y1) is determined by the difference in elevation and the horizontal dis-tance traveled. In terms of formulas, we have

12 (s′)2|t1t0 = −g(y1 − y0) − µg(x1 − x0).

In particular, the detailed shape of the graph (that is, of the incline) is notrelevant. Also note that themass of the particle does not appear in this relation.

Aside The work is given by the following line integral:∫ s1

s0

F ds =∫ s1

s0

(−g,−µg) · T ds

=∫

γ

(−µg)dx − g dy

where γ is the path defined by γ(s) :=(x, f(x)

). This line integral also

represents the work done in moving a particle from (x0, y0) to (x1, y1) inthe force field (−µg,−g). Let h(x, y) := −µgx − gy. Then

dh = −µg dx − g dy, or ∇h = (−µg,−g).

It follows that the line integral is path-independent.

Let V (x, y) := −mh(x, y) = µmgx + mgy. Then we have

12m(s′)2|t1t0 = −V (x0, y0) + V (x1, y1)

orK(t0) + V (x0, y0) = K(t1) + V (x1, y1)

where K(t) := 12ms′(t)2 is the “kinetic energy” of the particle. In other

words the quantityK + V is conserved. ♦

Wewant to consider some examples. We rewrite the relation between speedand work as follows:

12 (s′)2|t1t=t0 = g(x1 − x0)(b − µ)


where b := −(y1 − y0)/(x1 − x0) is the “average slope”. Note that if b = µ thenthere is no change in the skier’s speed between (x0, y0) and (x1, y1).

In the following exampleswe assume that g = 10 m/s2 and x1−x0 = 10m.Then g(x1 − x0) = 102 m2/s2. We also assume that t0 = 0 s.

Example Assume that s′(0) = 0m/s, that is, the skier starts at rest. Alsoassume that b = 0.1 (a 10% down slope) and µ = 0.05. Then

12s′(t1)2 = (102 m2/s2)(0.1 − 0.05)

= 5m2/s2

and hence

s′(t1) =√

10m/s = 3.16m/s ≈ 11.4 km/h ≈ 7.08mi/h.

♦

Example Assume that s′(0) = 1.0m/s, b = 0.01, and µ = 0.05. Then

12s′(t1)2 − 1

2 (1m2/s2) = 5m2/s2

and hence

s′(t1) =√

11m/s = 3.32m/s11.9 km/h ≈ 7.44mi/h.

♦

We can solve the relation between speed and work for the coefficient offriction µ:

µ = b − 12

v21 − v2

0

g(x1 − x0),

where v1 := s′(t1) and v0 := s′(t0).We canperform the following experiment to determineµ: Weuse surveying

equipment to measure the difference in elevations y1 − y0 and the horizontaldistance x1 − x0. We have a skier coast down the incline across the points(x0, y0) and (x1, y1). We measure the skier’s speeds v0 and v1 at these points.We can then easily compute µ from the last displayed formula.

Appendix 2: Dimensional AnalysisWe want to form a dimensionless parameter from the quantities m, g, P ,

and T . Recall that m denotes the mass of the skier, g denotes the accelerationof gravity, P denotes the power of the skier and T denotes the half-period ofthe skier’s body motion. We use M , L, and T to denote the mass, length, and


time dimensions. (We hope the reader can easily distinguish by context ourtwo different uses of the symbol T .) We have dimm = M , dim g = L/T 2,dimP = ML2/T 3, and dimT = T . Hence

dim(mwgxP yT z) = Mw(L/T 2)x(ML2/T 3)yT z

= Mw+yLx+2yT−2x−3y+z.

Consequently, we want to solve the following linear equations:

w + y = 0,x + 2y = 0,

2x + 3y −z = 0.

We get

w = −y,

x = −2y,

z = 2x + 3y = −4y + 3y = −y.

If we take y = 1, we get w = −1, x = −2 and z = −1. The correspondingdimensionless quantity (which we call the power parameter) is

m−1g−2PT−1 =p

gT,

where p := P/mg. Recall that in the main part of this report we called p thepower-to-weight ratio. Note that

dim p = dimP/ dim(mg)

= (ML2/T 3)/(ML/T 2)= L/T,

which is a velocity dimension, and

dim(gT ) = (L/T 2)T = L/T,

which is also a velocity dimension. Recall the formula T = 2p tan2 α/µ2gfor the half-period from the section on ski skating. From this formula, weget µ2 = 2(p/gT ) tan2 α, in which we see the dimensionless parameter foundabove.

ReferencesAlexander, R. McNeill. 1992. The Human Machine, New York: Columbia Uni-

versity Press.. 1996. Optima for Animals. Princeton, NJ: Princeton University

Press.


Bilodeau, B., M.R. Boulay, and B. Roy. 1992. Propulsive and gliding phasesin four cross-country skiing techniques. Medicine and Science in Sports andExercise 24: 917–925.

Brown, R.M., and J. Counsilman. 1971. The role of lift in propelling the swim-mer. In Proceedings of the C.I.C. Symposium on Bio-mechanics, edited by J.M.Cooper, 179–188. Chicago, IL: The Athletic Institute.

Caldwell, John. 1987. The New Cross-Country Ski Book. 8th ed. Lexington, MA:Stephen Greene Press.

Finney, Ross L., and George B. Thomas, Jr. 1990. Calculus. Reading, MA:Addison-Wesley.

Karvonen, J., R. Kubica, B. Wilk, J. Wnorowski, S. Krasicki, and S. Kalli. 1989.Effects of skating and diagonal skiing techniques on results and some phys-iological variables. Canadian Journal of Sports Science 14: 117–121.

Krim, J. 1996. Friction at the atomic scale. Scientific American 275 (4): 74–80.

Lind, D., and S.P. Sanders. 1996. The Physics of Skiing: Skiing at the Triple Point.New York: Springer-Verlag.

McGrew, William C. 1966. Moral kin? Scientific American (September 1996).

McMahon, Thomas A. 1984. Muscles, Reflexes, and Locomotion. Princeton, NJ:Princeton University Press.

Sears, Francis Weston. 1958. Mechanics, Wave Motion and Heat. Reading, MA:Addison-Wesley.

Smith, G.A. 1992. Bio-mechanical analysis of cross-country skiing techniques.Medicine and Science in Sports and Exercise 24: 1015–1022.

, J. McNitt-Gray, and R.C. Nelson. 1988. Kinematic analysis of al-ternate stride skating in cross-country skiing. International Journal of SportBio-mechanics 4: 49–58.

Smith, G.A., R.C. Nelson, A. Feldman, A., and J.L. Rankinen. 1989. Analysis ofV1 skating technique of Olympic cross-country skiers. International Journalof Sport Bio-mechanics 5: 185–207.

Svensson, Einar. 1994. Ski SkatingwithChampions, Ski SkatingwithChampions,18405 Aurora Ave. N., Suite H-83, Seattle, WA 98133.

Whitt, F.R., and D.G. Wilson, D.G. 1974. Bicycling Science. Cambridge, MA:MIT Press.

Weinstock, R. 1974. Calculus of Variations. New York: Dover.

Zatsiorsky, V.M. 1998. Kinematics of Human Motion. Champaign, IL: HumanKinetics.

. 2002. Kinetics ofHumanMotion. Champaign, IL:HumanKinetics.


Acknowledgments (Driessel)I did most of my part of the research for this report during 1996 and 1997,

when I was a visiting scholar at Iowa State University. I wish to thank thepeople who made this extended visit possible: Irvin R. Hentzel, Stephen J.Willson (former chairman) and Max D. Gunzberger (chairman).I gave a colloquium talk on this research at the University of Minnesota at

Duluth in January, 1997. I wish to thank Richard F. Green, who arranged thetalk, and the people at UMD who encouraged me to continue this research.I wish to thank others who spoke with me about ski skating during 1996

and 1997 and who offered advice: Ralph Smith (Mathematics Department,Iowa State University), Scott Hauser (Steamboat Touring Center, SteamboatSprings, Colorado), Charles Mann (Spirit Mountain Touring Center, Duluth,Minnesota), Bert Kleerup (Eagle River Nordic, Eagle River, Wisconsin), SteveGaskill (Team Birkie Ski Education Foundation, Minneapolis, Minnesota), andAntonina Anikin (Duluth, Minnesota).We wrote this paper in January, 1998. Joe Keller read it, and I thank him

for his comments. We made a number of minor changes in response to hissuggestions.Wemade further changes in response to a referee’s comments; in particular,

we added the section on suggestions for future work. I thank the referee.The editor also provided significant assistance and encouragement, partic-

ularly suggesting that we add the table of notation. I thank him.I also wish to thank Ruth DeBoer for typing this report.

About the Authors

KennethR.Driessel is an adjunct scholar associatedwiththe Mathematics Dept. at Colorado State University in FortCollins. He was born and raised in southeastern Wiscon-sin. (In his youth, he sometimes walked across snowy farmfields with old skis strapped to his feet.) He completedundergraduate studies in mathematics at the University ofChicago in 1962 and received his Ph.D. from Oregon StateUniversity in 1967. From 1967 to 1971, he taught at the Uni-versity of Colorado (in Boulder and Denver) (and learned

how toput kickwax on cross-country skis). From1971 to 1985, heworked at theAmoco Production Company Research Center in Tulsa, Oklahoma (and con-tinued to cross-country ski for one or twoweeks per year in Colorado—usuallynear Steamboat Springs). From 1987 to 1993, he worked for the MathematicsDept. at Idaho State University (and learned to skate ski). From 1993 to 2000, hewas a visiting scholar at Iowa State University (and started to study the scienceof skiing). In 2000, he moved to Laramie, Wyoming, where he currently lives(because he can reach groomed cross-country trails after a 15-minute drive).


He is currently interested in computational linear algebra and applications ofcontrol theory to biomechanics.

Philip W. Fink is a research assistant professor at the Center for ComplexSystems and Brain Sciences at Florida Atlantic University. He received a B.S.from the University of Connecticut and an M.S. and a Ph.D. from Purdue Uni-versity in biomechanics. After a series of postdoctoral appointments at FloridaAtlantic University, San Francisco State University, and Brown University, hereceived a faculty position at Florida Atlantic University, where he is currentlystudying problems in motor control.

Irvin Roy Hentzel is Professor of Mathematics at IowaState University in Ames, Iowa. He is married with fourgrown children. His pure mathematical interests are incomputation techniques for nonassociative algebras. Healso has interests in strategies for two-person games, codecracking, Judo, Taekwondo, and the history of AkpatokIsland. He is a private pilot and is looking forward to re-tirement, when he can pursue evenmore diverse interests.

UMAPModules inUndergraduateMathematicsand ItsApplications

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Calculus, vector analysis

COMAP, Inc., Suite 210, 57 Bedford Street, Lexington, MA 02420 (781) 862–7878

Module 786Wind Turbine PowerCoefficient Optimization

Paul A. Isihara, Dept. of Mathematicswith Julie Clements, Ronya Kamerlander,William Landry, Jonathan Soyars, andNathaniel Stapleton


INTERMODULAR DESCRIPTION SHEET: UMAP Unit 786TITLE: Wind Turbine Power Coefficient OptimizationAUTHOR: Paul A. Isihara

Dept. of [email protected] Clements, Ronya Kamerlander,William Landry, Jonathan Soyars, andNathaniel StapletonWheaton CollegeWheaton,IL 60187

MATHEMATICAL FIELDS: Calculus, vector analysisAPPLICATION FIELD: Wind turbine technology; alternative energy sourcesTARGET AUDIENCE: Students in calculus, differential equations, or vector

analysis.

ABSTRACT: We consider the important problem of optimizinga wind turbine’s power coefficient—the percentage ofavailablewind power used by the turbine, either by therotor assembly as a whole (Section 4), or more locallyby a cross-sectional area of a rotor blade (Section 6). Weassume only basic physics and explain the specializedconcepts needed for our mathematical results.We describe a simple case of Bernoulli’s equation

from fluid mechanics (Section 2), analyze the momen-tum change of the wind as it passes through the tur-bine’s rotor (Section 3), and obtain an optimal powercoefficient result known as Betz’s Law (Section 4): Arotor-type, horizontal-axis turbine can convert nomorethan 16/27 (≈ 59%) of the kinetic energy in the wind tomechanical energy.Wederive a secondoptimality condition in Section 6:

The power coefficient of an airfoil is maximized whenthe ratio of the airfoil’s headwind speed to the realwindspeed is 2/3 of the ratio of the lift coefficient to the dragcoefficient. This result leads to the final exercise, inwhich the student discovers the startling fact that liftutilization can produce 50 times the power per utiliz-able area than purely drag-driven devices.Simplifying assumptions enable us to obtain these

results using just calculus and elementary vector anal-ysis. References to the broader scope of wind turbinetechnology and advanced research work are includedin the concluding section.

PREREQUISITES: Calculuswith elementary vector analysis, including ac-quaintance with surface integrals.

The UMAP Journal 25 (4) (2004): 411–438.

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Wind Turbine Power Coefficient Optimization 413

Wind Turbine Power CoefficientOptimizationPaul A. IsiharaDepartment of [email protected] ClementsRonya KamerlanderWilliam LandryJonathan SoyarsNathaniel StapletonWheaton CollegeWheaton, IL 60187

Table of Contents1. INTRODUCTION . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

2. BERNOULLI’S EQUATION FOR A HORIZONTAL STREAMLINE . . . . . . 3

3. A STREAMTUBE MOMENTUM ANALYSIS . . . . . . . . . . . . . . . . 5

4. WIND FLOW AND BETZ’S LAW . . . . . . . . . . . . . . . . . . . . . 7

5. AIRFOIL GEOMETRY . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

6. POWER COEFFICIENT AND LIFT UTILIZATION . . . . . . . . . . . . . . 15

7. FURTHER DIRECTIONS . . . . . . . . . . . . . . . . . . . . . . . . . . 18

8. SOLUTIONS TO THE EXERCISES . . . . . . . . . . . . . . . . . . . . . . 20

REFERENCES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22

ACKNOWLEDGMENTS . . . . . . . . . . . . . . . . . . . . . . . . . . 23

DEDICATION . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24


ABOUT THE AUTHORS . . . . . . . . . . . . . . . . . . . . . . . . . . 24

MODULES AND MONOGRAPHS IN UNDERGRADUATEMATHEMATICS AND ITS APPLICATIONS (UMAP) PROJECT

The goal of UMAP is to develop, through a community of users and devel-opers, a system of instructional modules in undergraduate mathematics andits applications, to be used to supplement existing courses and from whichcomplete courses may eventually be built.The Project was guided by a National Advisory Board of mathematicians,

scientists, and educators. UMAP was funded by a grant from the NationalScience Foundation and now is supported by the Consortium for Mathemat-ics and Its Applications (COMAP), Inc., a nonprofit corporation engaged inresearch and development in mathematics education.

Paul J. CampbellSolomon Garfunkel

EditorExecutive Director, COMAP


1. IntroductionThroughout the world, depleted resources and environmental hazards as-

sociated with electricity production have heightened the need for renewableenergy alternatives. Wind power provides a viable alternative to fossil fuel ornuclear means. The practicality of harnessing wind energy is the result of over30 years of research into the design of efficient and cost-effectivewind turbines,combined with an economy of scale associated with increased production. Inthe U.S. alone, wind power currently generates 16 billion kilowatt hours (kWh)of electricity annually, a supply adequate for 1.6 million average-size house-holds. This production reduces CO2 emissions by approximately 10 milliontons [AWEA 2004a].Wind power has its origin in solar energy. Solar radiation heats the earth

unevenly, creating differences in atmospheric temperature, density, and pres-sure; and these variations produce air movement. For example, as warmer aircomes into contact with cooler air, the warmer air rises above the cooler air.Moving air, combined with the Earth’s rotation and differences in topography,results in semi-stable wind patterns around the globe. In particular, the TradeWinds andWesterlies are good sources ofwindpower [Frost andAspliden 1998].Charts showingglobalwindpatterns, extensivewindatlases, anddatabases

provide wind-power developers with important regional and local informa-tion, including the distribution of wind speeds, mean annual and seasonalwind speeds, and the percentage of land suitable for turbine installation. In theU.S., NorthDakota, Texas, andKansas have the greatest wind energy potential.In especially conducive areas (annual average speeds exceeding 13mph), largewind farmswith multiple turbines have been established [AWEA 2004b].

Modern turbines can be operated at just 4–5¢ per kilowatt hour (kWh).Yet without the Wind Energy Production Tax Credit (PTC)—a federal subsidyextended in 2004 only through the end of 2005 [AWEA 2004c] that offers a1.5¢/kWh tax credit and hence reduces costs to the 3–4¢/kWh range—utilitycompanies would not embrace this technology. Researchers are pressing to-wards the goal of reducing operation costs to the 2–3¢/kWh cent range withinthe next 10 years, to offset the need for longer-range subsidies [Toal 2001].In addition to widespread commercial usage in industrialized nations such

as the U.S., Spain, Germany, and Denmark, wind power provides developingnations with an important energy source, especially in more remote areas out-side the existing power grid. As a standalone, or in conjunction with otheralternative energy sources such as photovoltaics (solar panels), wind powerhas played a vital role in the development of community resources such aspumped water supplies, medical clinics and schools.In this Module, we use calculus and vector analysis to optimize a wind

turbine’s power coefficient, the percentage of the available real wind power thatcan be utilized by the turbine. We do this both globally, for the turbine’s rotoras a whole (Section 4), and more locally, for an airfoil-shaped cross-sectionalarea of one of its rotor blades (Section 6).

1


We make numerous simplifying assumptions, such as:

• the turbine blades rotate in a vertical plane, called the rotor plane (which wetake to be the yz plane in xyz coordinate space);

• the real wind, unobstructed, travels horizontally (i.e., in the direction of thepositive x-axis or the vector i);

• the real wind velocity’s horizontal component is constant in planes parallelto the rotor plane;

• the real wind passes through the rotor in such a way that wake effects suchas rotational motion and turbulence may be ignored;

• the mass-density of the air is constant throughout space; and• the wind velocity and pressure are in steady state, that is, they may varywith position but not with time.

We attempt mathematical analysis of a horizontal axis wind turbine (HAWT)only, as opposed to a vertical axis wind turbine (VAWT) (Figure 1).

Figure 1. The rotor blades of a horizontal axis wind turbine, or HAWT (left), rotate about ahorizontal axis. A vertical axis window turbine, or VAWT (right), has a vertical axis of rotation.

Assuming a minimal physics background, in Section 2 we derive a simpleform of Bernoulli’s equation, relating wind pressure and velocity at points alonga horizontal streamline (path along which wind elements travel). Section 3obtains a result based on themomentum change of thewind as it flows througha streamtube (a continuous volume region formed by streamlines). Using boththe simplified Bernoulli’s equation and momentum change result, Section 4analyzes wind flow through the turbine’s rotor modeled simply as a power-extracting filter in the shape of a circular disk (more technically called an actuatordisk). Optimizing the power coefficient yields Betz’s Law: An ideal HAWTwith disc-like rotor can convert at most 16/27 (≈ 59%) of the kinetic energy in thewind to mechanical energy. In Section 5, we look more carefully at how a crosssection of a rotor blade (modeled as an airfoil) translateswith respect to variousvelocities (the real wind, headwind and apparent wind) and forces (lift and

2


drag). We then explain in Section 6 how the power coefficient of a lifting airfoilis maximized when the ratio of the real wind speed to the airfoil’s headwindspeed is approximately two-thirds of the ratio of the lift and drag coefficients(constants related to the lift and drag forces). The significance of lift utilizationis striking, as the reader will discover in the concluding exercise (Exercise 7),which compares the optimum power coefficient of a lift-utilizing airfoil withthe optimum power coefficient for a purely drag-driven device over the sameutilizable area. We conclude in Section 7 by mentioning key references totheoretical and practical aspects of wind-turbine design and performance.

2. Bernoulli’s Equation for a HorizontalStreamline

To construct a simple model of how much of the real wind’s power can beutilized by a turbine, we must analyze the pressure and velocity of the wind.In this section, we obtain a relationship between wind pressure and speed, asimple case of a fundamental result from fluid mechanics known as Bernoulli’sequation.A wind element travels along a path called a streamline. We assume that the

density ρ of the air is constant throughout space. Thus, a small wind elementwith rectangular parallelepiped volume dV = dx dy dz hasmassm = ρ dV. Forsimplicity, we assume that the element moves in the positive direction alongthe x-axis in xyz coordinate space (Figure 2).

Figure 2. An idealized wind element with volume dV = dx dy dz moving along a horizontalstreamline, shown as a dashed line coincident with the x-axis.

We assume that along this streamline the speed of thewind element Vrealwindand static pressure p exerted on the element by the air are both functions of x;

3


that is, Vrealwind=Vrealwind(x) and p = p(x). (Static pressure is the pressure feltby an object that moves along with the wind.) Since the only force acting in thedirection of the element’s motion is static pressure, if the element is sufficientlysmall, then the i component of the force is essentially equal to (Exercise 1)−p′(x)dV.Let x = x(t) denote the position of the wind element at time t. Using

Newton’s Second Law of Motion (force equals mass times acceleration), wehave

−p′(x) dV = dmdVrealwind(x)

dt

= dmdVrealwind(x)

dx

dx

dt= ρ dV V ′

realwind(x) Vrealwind(x).

Cancelling dV in the first and last expressions and then antidifferentiating withrespect to x, we obtain

−p(x) = 12ρ V 2

realwind(x) + C.

We record this result as

Simplified Bernoulli Equation. Along a horizontal streamline, we have

p(x) + 12ρV 2

realwind(x) = constant.

Here p(x) is static pressure and the term 12ρV 2

realwind(x) is the dynamic pressure.This result is a simple instance of Bernoulli’s equation, which in more generalform applies also to nonhorizontal streamlines (see [Young et al. 2001]).The simplifiedBernoulli’s equation reveals that as thewind speed increases,

its static pressure decreases. A simple experiment using two plain sheets ofpaper illustrates this inverse relationship [Stiesdal 1999]. Hold the sheets sothat they extend lengthwise away fromyou, parallel to each other, about four orfive inches apart. Nowblow between the two sheets. Note that the sheets comecloser together rather than separating farther apart. The increased air speedin between the sheets reduces the pressure in the space between them. Thehigher air pressure on the outside of both sheets forces them closer together.In (Section 5), we return to the idea of pressure differences caused by differentwind speeds when we discuss the shape of an airfoil.In Section 4, we will use the simplified Bernoulli’s equation to find a rela-

tionship between the real wind speeds at three different locations on a horizon-tal streamline:

• far upstream (well before the element reaches the wind turbine),

• far downstream (considerably after the element has passed through the tur-bine’s rotor), and

• directly at the plane in which the rotor blades are assumed to rotate.

4


Exercise

1. Show that the i component of the force due to static air pressure on a smalldifferential element with volume dV moving along a horizontal stream-line can be well-approximated by −p′(x) dV. (Note that the approximationimproves as the size of the element decreases.)

3. A Streamtube Momentum AnalysisIn the last section, we considered a single element moving along a horizon-

tal streamline. We now extend our viewpoint to an ensemble of wind elementshaving the same horizontal component of velocity. Interacting with a turbinewith disc-like rotor, these elements form a circular areamovingwith expandingradius inside a streamtube, as shown in Figure 3. By analyzing the totalmomen-tum change of the wind as it passes through the streamtube, we can determinethe total force exerted by the turbine on the elements in the streamtube. Weassume that the rotor plane coincides with the yz plane, so by “horizontal” wemean in the direction of the positive x-axis, or in the direction of the i vector.

rotorplane

Vupstream Vrotorplane Vdownstream

Aupstream

A

Adownstream

Figure 3. Streamtube diagram.

Elements that belong to the same vertical cross section have the same hori-zontal component of velocity, whichwe denote by Vcross section. (These cross sec-tions are denoted “upstream,” “rotorplane,” and “downstream” in Figure 3.)The streamtube diagram indicates that Vcross section decreases as the elementsapproach and pass through the rotor plane. This is understandable, since thewind is slowed down by the turbine. On the other hand, as Vcross section de-

5


creases, the area Across section of the cross section increases. Intuitively, this mustbe the case, since we are assuming that the mass-flow rate of the wind is con-stant; that is, the mass of wind flowing through any vertical cross section of thestreamtube per unit time is assumed to be constant. Technically, the mass-flowrate is obtainable by a flux integral (Exercise 2a):∫ ∫

cross sectionρV · dA = ρVcross sectionAcross section.

In computing the integral, we have assumed that

i) the density ρ is constant throughout the streamtube; and

ii) the horizontal component of the realwind velocity is both constant through-out, and normal to, each vertical cross section.

The first assumption is a good approximation when the radial velocity com-ponent is small compared to the axial component, and the second is as longas pressure variations are small compared to atmospheric pressure (which isusually the case).The further assumption that the mass-flow rate across each vertical cross

section is constant implies that

VupwindAupwind = VrotorplaneArotorplane = VdownwindAdownwind.

(This string of equalities implies that as Vcross section decreases, Across section in-creases.)Linear momentum is the product of mass times velocity. Similar to the

mass-flow rate, the momentum-flow-rate through a vertical cross section of thestreamtube is computed by the vector flux integral∫ ∫

cross sectionV ρ V · dA.

By Exercise 2b, it follows that the rate of outflux of the horizontal componentof linear momentum across a vertical cross section is given by

ρV 2cross sectionAcross section.

If the streamtube is sufficiently long, thewind at the entrance and the exit of thestreamtube is essentially horizontal. (We are assuming that the unobstructedwind is always horizontal.) There is nomomentum outflux through the side ofthe streamtube. Thus, the total rate of change of momentum per unit time inthe streamtube is the difference in the rates of horizontal momentum outfluxthrough the two ends:

ρV 2downstreamAdownstream − ρV 2

upstreamAupstream.

This difference can be expressed as (Exercise 2c)

6


ρVrotorplaneArotorplane(Vdownstream − Vupstream).

Finally, byNewton’s SecondLaw, force is equal to the derivative ofmomentum:

Force =ma = mdvdt

=d

dt(mv) = derivative of momentum.

Themagnitude of the total force exerted by the rotor on the wind in the stream-tube is therefore the absolute value of the total momentum change per unittime of the wind in the streamtube:

Momentum Change Equation. The magnitude of the total force exerted bythe turbine on the wind in the streamtube shown in Figure 3 is given by

ρVrotorplaneArotorplane(Vupstream − Vdownstream).

Exercise

2. a) Show that ∫ ∫cross section

ρV · dA = ρVcross sectionAcross section.

b) Show that the horizontal component of the momentum-flow-rate inte-gral ∫ ∫

cross sectionV ρ V · dA

is given byρV 2

cross sectionAcross section.

c) Show that the difference between the rates of outflux of horizontal mo-mentum across the two ends of the streamtube equals

ρVrotorplaneArotorplane(Vdownstream − Vupstream).

4. Wind Flow and Betz’s LawWe describe an interesting discovery by German physicist Albert Betz in

1919 [Betz 1926]:

Betz’s Law. An ideal HAWT with a disc-like rotor can convert no more than16/27 (or 59%) of the kinetic energy in the wind to mechanical energy.

(Intuitively: The wind must retain some of its kinetic energy to avoid “pilingup” at the rotor.)We continue to use the streamtube model introduced in the last section, in

which we can think of the turbine’s rotor as a “power-extracting air-filter” in

7


the shape of a thin circular disk with area Arotorplane. Though the streamtubemust be long enough so that the real wind velocities at the entrance and exit ofthe tube are essentially horizontal, it is convenient to sketch a shortened, two-dimensional representation of the streamtube, which indicates the real windvelocities at three different places, as shown in Figure 4.

upstreamCL

Rotorsideview

Vrotorplane VdownstreamV

Figure 4. Real wind velocities at the streamtube’s entrance, rotor plane, and exit.

In this diagram, a side view of the disk representing the rotor appears at thecenter of the diagram as a thickened vertical segment. As before, we assumethat the horizontal component of the real wind velocity is constant along eachvertical cross section. These components at the entrance to the streamtube, atthe rotor plane, and at the streamtube’s exit are denoted Vupstream, Vrotorplane, andVdownstream, respectively.We assume further that the center line of the streamtube (marked CL in

Figure 4) is a horizontal streamline along the x-axis, and that the rotor planecoincides with the yz plane (x = 0). This allows us to apply the simplifiedBernoulli’s equation obtained in Section 2 to both the upstream portion (i.e.,the negative x-axis) and to the downstream portion (i.e., the positive x-axis)of the horizontal streamline CL. Let −∞ denote a position “sufficiently farupstream” that the wind can be considered as essentially free from the rotor’sobstructing effect. On the upstream side of the rotor plane, Bernoulli’s equationimplies that

pupstream(−∞) + 12ρV 2

upstream = pupstream(0) + 12ρV 2

rotorplane.

Similarly, on the downstream side, we have

pdownstream(∞) + 12ρV 2

downstream = pdowntream(0) + 12ρV 2

rotorplane.

8


Assuming that pdownstream(∞) = pupstream(−∞) gives

pupstream(0) − pdownstream(0) = 12ρ(V 2

upstream − V 2downstream).

By the symmetry of the streamtube, there is cancellation of the nonhorizontalwind force vectors on each vertical cross section. Assuming that the horizontalcomponent of force due to the air pressure is constant throughout each verticalcross section, since pressure is force per unit area, the magnitude of the totalforce of the wind on the disk representing the rotor is given by

Arotorplane(pupstream(0) − pdownstream(0)

)= 1

2Arotorplaneρ(V 2upstream − V 2

downstream).

Since the disk exerts a force on the air in the streamtube which is equal inmagnitude to the force of the air on the disk, the momentum change equationobtained in Section 3 provides a second expression for the magnitude of thisforce:

ρVrotorplaneArotorplane(Vupstream − Vdownstream).

Equating these expressions for the horizontal force, and simplifying, wefind that

Vrotorplane = 12 (Vupstream + Vdownstream).

In other words, the horizontal wind speed at the rotor plane is the average ofthe horizontal wind speeds at the entrance and exit of the long streamtube.The mass-flow rate (i.e., the mass of air flowing through a cross-sectional

diskperunit time) at the rotorplane (andhence at each cross section) is therefore

mass-flow rate = ρVrotorplaneArotorplane = 12Arotorplaneρ(Vupstream + Vdownstream).

(1)

We now obtain an expression for the power Pcross section of the wind passinghorizontally through any vertical cross section (recall that this is the case atthe entrance and exit to the streamtube). We compute this power as the totalkinetic energy of the wind passing through the cross section per unit time:

Pcross section = 12 (mass-flow rate)V

2cross section. (2)

Substituting from (1), we get

Pcross section = 12

[12Arotorplaneρ(Vupstream + Vdownstream)

]V 2cross section.

Assuming that there is no energy loss due to friction, downstream rotation ofthe wind, etc., the total power Pextracted extracted from the wind by the rotor isthe difference in power of the wind at the two ends of the streamtube:

Pextracted = 12

[12Arotorplaneρ(Vupstream + Vdownstream)

](V 2upstream − V 2

downstream).

Next, we compute thepowerPtotalwind of a horizontalwindflowingwithout anyobstruction (i.e., with the rotor removed) across the same amount of vertical

9


cross-sectional area, namely Arotorplane. In this case, the streamtube becomes ahorizontal right circular cylinder with vertical cross-sectional area Arotorplane.The velocity of the wind throughout this streamtube is both horizontal andconstant. Using the fact that Vupstream, Vdownstream, and Vcross section are all equal,by substitution in (2)we obtain

Ptotalwind = 12ArotorplaneρV 3

upstream.

Using elementary algebra (Exercise 3a), the power coefficient—the ratio betweenthe power extracted by the rotor to the total wind power over the same area—can be expressed as

PextractedPtotalwind

= 12 (1 + r)(1 − r2),

where r = Vdownstream/Vupstream.It is a straightforward calculus problem (Exercise 3b) to show that on the inter-val 0 ≤ r ≤ 1, the function f(r) = 1

2 (1+ r)(1− r2) is maximized when r = 1/3.We then have that f( 1

3 ) = 1627 ≈ 59% of the total power can be extracted, the

result in Betz’s Law.

Exercise

3. a) Show that if

Pextracted = 12

[12Arotorplaneρ(Vupstream+Vdownstream)

](V 2upstream−V 2

downstream)

andPtotalwind = 1

2ArotorplaneρV 3upstream,

thenPextractedPtotalwind

= 12 (1 + r)(1 − r2),

where r = Vdownstream/Vupstream.b) Show that on the closed interval 0 ≤ r ≤ 1, the function

f(r) = 12 (1 + r)(1 − r2)

attains its absolute maximum value at r = 1/3.

5. Airfoil GeometryIn the last section, we obtained Betz’s law for the optimal power coefficient

of a turbine as a whole, under the simplifying assumption that a HAWT’s rotorcan be represented by a power-extracting disc. In this section, we introducea more refined model of how the wind interacts with a rotor blade when itscross sections are shaped in the form of airfoils. Our goal here is only to

10


provide a rough idea why air flowing around an airfoil shape creates a typeof force known as lift. (In Section 6, we give actual equations for lift and foranother important force called drag, which are used to compute a certain powerproduction advantage associated with lift utilization.)In describing the interaction of thewindwith a rotor blade’s cross section, in

addition to real wind velocity, there are three other velocities to be considered,which we call translation, headwind, and apparent wind. (In what follows, weassume all vectors to be free vectors in a reference frame that is fixed withrespect to the earth.)To give an everyday example of these velocities, imagine that you are a

novice skier getting ready to ski straight down a hill with no bumps. You arewearing a thin, light scarf on a blustery afternoon. As you stand at the top of thehill, a strong gust of wind (the real wind) hits you directly from your left side,so that your scarf extends at a right angle to the straight path that you are aboutto take down the hill. You wait until the wind dies down before proceeding.As you begin to ski down your straight path, you acquire a velocity (translation)and feel a wind (headwind) blowing directly against your face. The translationand headwind are by definition equal in magnitude but opposite in direction.Halfway down the hill, with your speed steady, a prevailing gust of real windhits you—as before, from your left side (always at right angles to your path).Now, however, the loose end of your scarf aligns behind you and to your right,at an acute, rather than perfect right angle to your path. Your scarf stretchesin the direction of the apparent wind, which is the vector sum of the real windand the headwind (see Figure 5). If you decrease your speed and/or the realwind picks up, your scarf gets blown farther toward a right angle to your path;in other words, the direction of the apparent wind moves more towards thedirection of the realwind. On the other hand, if you increase your speed and/orthe real wind diminishes, your scarf is blownmore nearly straight behind you;in other words, the apparent windwill movemore towards the direction of theheadwind.We now analyze how these same four velocities occur as the real wind

interacts with a portion of a rotor blade. Figure 6 shows the translationalmotion of a point P belonging to a vertical cross section of the blade. Themagnitude of this velocity is denoted Vtranslation. The magnitude Vheadwind ofthe headwind velocity is equal to Vtranslation, and the direction of the headwindvector is opposite to the translation vector. The apparent or relative wind is thevector sum of the real wind and the headwind, with magnitude Vapparentwind.

We now suggest how the airfoil shape of the rotor blade’s cross sectioncreates a special force known as lift. Figure 7 shows a portion of a rotor bladewithout any headwind. When real wind hits the stationary turbine blade,to maintain a constant horizontal speed, the wind must actually move morequickly over the top of the blade than underneath. The reason for this is thatthe top of the blade is curved more than the bottom so that the distance thewind must travel over the top of the blade is longer than the distance alongthe bottom. Recalling the inverse relationship between speed and pressure

11


Headwind

Headwind Apparent Wind

Real Wind

Translation Velocity

Real Wind

Apparent Wind

scarf

Skier’s Direction

Figure 5. Real Wind + Headwind = Apparent Wind.

P

translationV

apparentwindV

Vrealwind

Vheadwind

Figure 6. The same velocities in relationship to a translating section of a rotor blade.

described in Section 2, the pressure above the rotor blade will be lower thanthe pressure underneath the blade. This pressure difference creates a lift force,which sets the blade in motion.To orient ourselves to certain geometric features of an airfoil, suppose that

we are looking down at an airplane wing from the tip towards the cabin. Ifwe were to imagine taking a thin vertical cross-sectional slice of the wing, theresulting shape is that of an airfoil as shown in Figure 8.The front edge of the wing is called the leading edge, while the back edge

is the trailing edge. The wing is more rounded at the leading edge and morepointed at the trailing edge. On each airfoil-shaped cross section, there isa single point Pleadingedge where the streamlines split, going to one side of theairfoil or the other, and a single pointPtrailingedge at which the streamlines rejoin.The straight line joining these two points is called the airfoil’s chordline, and thedistance between these points is known as the chord length.The angle between the chordline and the apparent wind velocity vector is

12


Figure 7. A lift force is created by pressure differences above and below the airfoil cross sectionsof a stationary rotor blade.

airfoil shape

leadingedgeP

Ptrailingedge

chordline

leading edge

trailing edge

Figure 8. Similar to an airplane wing, a rotor blade has airfoil shaped cross sections.

Headwind

trailing edge

leading edge

Realwind

ApparentwindAngle of attack chordline

Figure 9. The angle of attack.

13


called the angle of attack (Figure 9). Both the angle of attack and the sharp pointat the trailing edge of the airfoil combine to create a a lift force.If the angle of attack is too large (greater than 15◦), the streamlines above

the airfoil separate from the upper part of the wing (Figure 10).

Angle of attack

Headw

ind

Realwind

separation

Apparent wind

Figure 10. Stall occurs when the angle of attack is sufficiently large that the streamlines separatefrom the airfoils.

The resulting disappearance of the lift force is called stall. For airplanes inflight, the headwind speed is usually much greater than the real wind, so thatthe direction of the apparentwind is close to thedirection of the headwind. As aresult, airplanes seldomoperate at anglesof attackabove30◦ [Stiesdal 1999]. Onthe other hand, wind turbines generally operate with angles of attack between0◦ and 90◦. Thus, for the purposes of wind turbine design, understanding ofairfoil behavior in the “fully-stalled regime” is important (see Lissaman [1998]).

Exercises

4. Recall that if a point is moving with constant angular speed ω at a fixeddistance a from the center of rotation, the magnitude of its translationalvelocity is given by Vtranslation= aω. Referring to Figure 11, suppose that thelength of a rotor blade is L = 10meters, fix ω at 2 rad/sec, and let a be thedistance from a point on the blade to the center of rotation.a) Find Vheadwind at the root of the blade (a = 0.1L), at the midsection(a = 1

2L) and at the tip (a = L).b) Find Vapparentwind at these three places on the blade if Vrealwind = 10m/s.(Assume that the real wind and headwind are perpendicular.)

c) Find the angle of attack at each of these places. For simplicity, assumethat the chordline is in the same direction as the headwind velocity.

14


a = .1 L

headwindV

realwindV

Vheadwind

apparentV

angle of attack

chordline

a = .5L

a = L

Figure 11. Diagram for Exercise 4.

6. Power Coefficient and Lift UtilizationHistorically, the earliest documented use of windmills was in Persia c. 950

A.D. [Shepherd 1998], vertical-axis devices designed to extract power fromthe wind utilizing only drag. Drag is a resultant force in the same directionas the velocity of the translating body; for example, the drag force createdby a constant wind horizontal to the ground can propel an iceboat forwardalong a straight line. The idea of a purely drag-driven wind device with flat“sails” is shown in Figure 12. (A rotating blade element can be thought ofas being instantaneously translating.) It was not until the 12th century thathorizontal-axis wind mills were developed in Europe, evidently as a result ofexperimentation rather than from a theoretical understanding of lift.We give a mathematical explanation why the ability to extract power from

thewind is greatly enhanced if the rotor blades in a horizontal axiswind turbineare designed so that their cross sections utilize lift. First, we compute theoptimal power coefficient for an airfoil-shaped cross section of a rotor blade, inwhich both lift and drag forces are utilized. We then ask the reader (Exercise 7)to perform a similar computation of the optimal power coefficient for the sameutilizable area in which only the drag force is employed (i.e., the lift force isremoved). As we shall see, the significance of lift in terms of optimum powercoefficient per utilizable area is remarkable.Let Vtranslation by the magnitude of the translation velocity of a point belong-

ing to an airfoil-shaped cross section of a rotor blade, and let Vrealwind be themagnitude of the real wind at that point. For simplicity, we assume that thereal wind and translation velocities are both constant throughout the airfoiland perpendicular to each other at each point of the airfoil. The power coeffi-cient of the airfoil is defined to be the ratio of the power extracted by the airfoil

15


Figure 12. A drag-driven wind device.

to the available real wind power over a comparable area. We show that thepower coefficient is maximized when the ratio Vrealwind/Vtranslation is approxi-mately two-thirds of the lift-to-drag-ratio CL/CD. The constants CL and CD arecalled the lift coefficient and the drag coefficient. These coefficients appear in theexpressions for the magnitudes L of the total lift force and D of the total dragforce acting on the airfoil:

L = 12ρV 2

apparentCLA,

D = 12ρV 2

apparentCDA.

In these expressions (based on Bernoulli’s Law, discussed in Section 2), thedynamic pressuredue to thewind is 1

2ρV 2apparent, and the area that can beutilized

per unit time to extract power from thewind is taken to beA = cVrealwind, wherec is the length of the chordline. The coefficients CL and CD are determinedexperimentally and depend on factors such as the shape of the airfoil andangle of attack, as well as the composition of the material with which theblade is manufactured. We resolve the force on the wind into perpendicularcomponents, with the one in the direction of the apparent wind termed thedrag force and the other the lift force(see Figure 13).In general, power is computed as the dot product of a force vector with a

velocity vector. (Recall that the dot product of twovectors is the product of theirmagnitudes and the cosine of the angle between them.) To compute the totalpower extracted by the airfoil, we treat the airfoil as a free body and computethe dot product of the resultant force due to the lift and drag, together with theairfoil’s translation velocity. In other words, if lift is utilized, the power Plift

16


lift

V

translationV

chordline

headwindV

realwind

apparent

V

Fdrag

F

Figure 13. Velocity and force vectors.

extracted by the airfoil can be computed from the formula

Plift = (lift force + drag force) · (translation velocity)= LVtranslation cos(γ) + DVtranslation cos

(γ +

π

2

)= LVtranslation cos(γ) − DVtranslation sin(γ),

where γ is the angle shown in Figure 13. Replacing L and D with the expres-sions involving the lift and drag coefficients, we obtain

Plift = 12ρAV 2

apparentVtranslation[CL cos γ − CD sin γ]

= 12ρV 3

realwindA√

1 + r2 r[CL − CDr].

where r = Vtranslation/Vrealwind. (You are asked to supply the details of thiscomputation in Exercise 5.)The power coefficientCP is defined to be the ratio of thepower extractedby the

translating airfoil to the available wind power over the same utilizable area A,the latter power being equal to 1

2ρV 3realwindA (we obtained a similar expression

for Ptotalwind at the end of Section 4). Hence,

CP = CP(r) =√

1 + r2 r[CL − CDr].

Given specific values for CL and CD (obtainable experimentally using windtunnels), you can use calculus and a numerical utility (Exercise 6) to show thatthe power coefficientCP(r) is maximizedwhen r is approximately (2/3)CL/CDas claimed at the outset of this section. Exercise 7 then outlines the steps thatsubstantiate the following remarkable fact [Wilson 1998]:

Lift Utilization. A lift device can readily produce 50 as much power per unitof utilizable area as can be produced by a purely drag-driven device.

To keep the advantages of a lifting device in perspective, however, we shouldrecall Betz’s Law (Section 4), which limits the overall efficiency of any HAWTwith disc-like rotor.

17


Exercises

5. Provide the missing details that give the expression

Plift = 12ρV 3

realwindA√

1 + r2 r[CL − CDr],

where r = Vtranslation/Vrealwind. (Hint: The magnitudes of Vtranslation andVheadwind are equal.)

6. Given that CL = 1 and CD = 0.1 are realistic values, (the drag value islowered due to a reverse drag force, as well as shape and edge effects) , usea computational utility suchasMaple to show thatCP(r) ismaximizedwhenr is approximately (2/3)CL/CD. Then find the maximum power coefficientCP,max.

7. a) Let A be the area over which power is being extracted by a purely drag-driven device (i.e., L = 0). Show that the power coefficient CPpuredrag isgiven by

CPpuredrag = r(1 − r)2CDpuredragwhere r = Vtranslation/Vrealwind.

b) Show that the maximum value of CPpuredrag occurs at r = 1/3 and hencethemaximumpower coefficient is 4/27 of themaximumdrag coefficientCD,max for a purely drag-driven device.

c) Given thatCD,max is 2.0 for adrag-drivendevice, show that themaximumpower coefficient for the lifting device in Exercise 6 is 50 times that ofthe purely drag-driven device.

7. Further DirectionsWe have relied on several simplifying assumptions to optimize the power

coefficient of a horizontal axis wind turbine; many additional factors are in-volved in designing an efficient wind turbine. Some are easy to think of, suchas the effects of turbulent wind conditions [Muljadi et al. 1996] or wake ef-fects caused by complex terrain [Lissaman 1998]. Others, such as the effectsof nonuniformities in velocity flow near the end of a rotor blade (i.e., tip loss)[Wilson 1998], are identifiable only by specialized research. An excellent com-prehensive technical survey of the many facets of wind turbine technology ispresented in Spera [1998].Further development of the mathematical theory, though fascinating in it-

self, must ultimately be guided by practical questions, such as:

• What are the best construction materials for each part of the turbine? (i.e.,strength, manufacturing costs, etc.)

• What makes a particular site (local or regional, level or hilly, on the grid oroff the grid, etc.) suitable or unsuitable for a particular model wind turbine?

18


• Should awind turbine system be used as a standalone or in conjunctionwithother power sources?

• What is the most cost-efficient way to operate the turbine?• How is performance affected by unstable wind and severe weather condi-tions?

• What are the key maintenance considerations?• What environmental hazards are created by utilization of wind turbines?

In the U.S., a great deal of wind-power research takes place at the NationalWind Center of the National Renewable Energy Laboratory (NREL) in Golden,Colorado. A wealth of technical research papers is available to the public attheir Website

http://www.nrel.gov/wind/library.html .Many governmental and private research centers are making considerable

investments to develop this technology and to educate the public. A goodexample of the latter is theDanishWindTurbine IndustryAssociation’sGuidedTour on Wind Energy at the Website

http://www.windpower.dk/tour/index.htm .The importance of wind turbine research is highlighted by the fact that

wind power has had the largest percentage growth in total capacity (about30% annually) of any power source worldwide during the last decade [AWEA2004b]. The significance of such a technology to the global village cannot beoverlooked, and an understanding of its basic features can enrich an under-graduate mathematics curriculum.

19


8. Solutions to the Exercises

1. Let x be the position of the center of a small differential element of the airmodelled as a rectangular solidwith dimensions dx, dy and dz. The force onthe element in thedirectionof the streamline is due to thedifference betweenthe pressure on the element’s “leading” face (approximated by p(x + dx

2 ))and thepressure on the “trailing” face (approximatedby p(x− dx

2 )). Pressureis force per area and acts on the faces of the element in an inward direction.Hence, the i component of the force is approximated by

[p

(x − dx

2

)− p

(x +

dx

2

)]dy dz =

−[p(x + dx2 ) − p(x − dx

2 )] dx dy dx

dx.

The result follows by observing that dV = dx dy dz and that dx is assumedto be small.

2. a) ∫ ∫cross section

ρV · dA = ρ

∫ ∫cross section

V · −→i dA

= ρ


Vcross sectiondA

= ρVcross section


dA

= ρVcross sectionAcross section.

b) The horizontal component of the momentum flux across a cross sectionis obtained by

[∫ ∫cross section

VρV · dA]· i =


(V · −→i )ρ V · dA

= Vcross section


ρV · dA

= ρV 2cross sectionAcross section.

(Use part a) to evaluate the last integral.)c) ρV 2

downstreamAdownstream − ρV 2upstreamAupstream

= ρVdownstreamVrotorplaneArotorplane − ρVupstreamVrotorplaneArotorplane

= ρVrotorplaneArotorplane(Vdownstream − Vupstream).

20


3. a) PextractedPtotalwind

=12

[ 12Arotorplaneρ(Vupstream + Vdownstream)](V 2upstream − V 2

downstream)12ArotorplaneρV 3

upstream

=12

(Vupstream + Vdownstream

Vupstream

) (V 2upstream − V 2

downstream

V 2upstream

)

=12

(1 +

Vupstream

Vdownstream

) (1 − V 2

upstream

V 2downstream

)

= 12 (1 + r)(1 − r2).

b) Since f ′(r) = 12 (1 − 2r − 3r2), in the interior of the closed interval [0,1]

f has a single critical point at r = 1/3. Moreover, since f( 13 ) = 16/27,

and the values at the endpoints of the closed interval are f(0) = 1/2 andf(1) = 0, the function attains its absolute maximum value at r = 1/3.

4. a) 2 m/s, 10 m/s, 20 m/s.b) 10.2 m/s, 14.14 m/s, 22.36 m/s.c) 78.69◦, 45◦, 26.56◦.

5. Plift = 12ρAV 2

apparentVtranslation[CL cos(γ) − CD sin(γ)]

= 12ρAV 2

apparentVtranslation

[CL

VrealwindVapparent

− CDVtranslationVapparent

]= 1

2ρAVapparentVtranslation[CLVrealwind − CDVtranslation]

= 12ρA

√V 2realwind + V 2

translation Vtranslation[CLVrealwind − CDVtranslation]

= 12ρV 3

realwindA√

1 + r2 r[CL − CDr].

6. Solving the equation C ′P(r) = 0 numerically, we obtain the critical point

value r ≈ 6.642. This is close to the value (2/3)(1/0.1) ≈ 6.67. It follows thatthe maximum value of the power coefficient is approximated by CP,max ≈CP(6.642) ≈ 15.

7. a) Let A be utilizable area over which power is extracted by a purely drag-driven device. The magnitudeD of the drag force exerted on the deviceis

D = 12ρV 2

apparentwindCDpuredragA,

where CDpuredrag is the drag coefficient of the purely drag-driven device.Note that for a purely drag-driven device, the real wind, apparent windand translation velocity are parallel, so that Vrealwind = Vapparentwind +Vtranslation.

21


Thus, we can write

D = 12ρ(Vrealwind − Vtranslation)2CDpuredragA.

The power extracted by the drag device is computed bydotting this dragforce with the translation velocity, giving

12ρ(Vrealwind − Vtranslation)2VtranslationCDpuredragA.

As before, the available wind power is 12ρAV 3

realwind, so that the powercoefficient CPpuredrag for the purely drag driven device is

CPpuredrag = (1 − r)2rCDpuredrag

where r = Vtranslation/Vrealwind.b) It is an easy computation to showthat g(r) = (1−r)2r attains amaximumvalue on 0 ≤ r ≤ 1 of 4/27 at r = 1/3. It follows that the maximumpower coefficient for the purely-drag driven device is (4/27)CDpuredrag .

c) The maximum power coefficient per utilizable area A for the purelydrag-driven device is only (4/27) × 2 ≈ 0.3, whereas Exercise 6 givesa maximum power coefficient of roughly 15 for the lifting airfoil. Thepower coefficient for the lifting airfoil is therefore roughly 50 times thatof the purely drag-driven device for the same utilizable area!

ReferencesAmericanWindEnergyAssociation (AWEA). 2004a. WindWeb tutorial—Wind

energy and the environment.http://www.awea.org/faq/tutorial/wwt_environment.html .

. 2004b. Wind Web tutorial—Wind industry statistics.http://www.awea.org/faq/tutorial/wwt_statistics.html .

. 2004c. News (24 September 2004). Wind energy tax incentive winsextension to end of 2005. www.awea.org/news/index.html.

Betz, Albert. 1926. Wind-Energie. Göttingen, Germany: Vandenhoek andRuprecht.

DanishWindTurbine IndustryAssociation. 2002. Guided tour onwind energy.http://www.windpower.dk/tour/index.htm .

Frost, Walter, and C. Aspliden. 1998. Characteristics of the wind. In Spera[1998], 371–445.

Lissaman, Peter B.S. 1998. Wind turbine airfoils and rotor wakes. In Spera[1998], 283–321.

22


Muljadi, Eduard, C.P. Butterfield, andM.L. Buhl, Jr. 1996. Effects of turbulenceon power generation for variable-speed wind turbines: Technical Report.Golden, Colorado: National Renewable Energy Laboratory.

Nakayama, Yasuki, and R.F. Boucher. 1999. Introduction to Fluid Mechanics.London: Arnold.

Piggott, Hugh. 1999. Small wind turbine design notes.http://users.aber.ac.uk/iri/WIND/TECH/WPcourse/index.html .

Shepherd, Dennis G. 1998. Historical development of the windmill. In Spera[1998], 1–46.

Spera, David A. (ed.). 1998. Wind Turbine Technology: Fundamental Concepts ofWind Turbine Engineering. New York: ASME Press.

Stiesdal, Henrik. 1999. The aerodynamics of the wind turbine. Bonus-InfoNewsletter (Autumn 1999): 5–10. http://www.windmission.dk/workshop/BonusTurbine.pdf .

Toal, Brian A. 2001. Renewables: Future shock.http://www.nrel.gov/docs/gen/fy02/31353.pdf .

Wilson, Robert E. 1998. Aerodynamic behavior of wind turbines. In Spera[1998], 215–282.

Young, Donald, Bruce Munson, and Theodore Okiishi. 2001. A Brief Introduc-tion to Fluid Mechanics. 2nd ed. New York: John Wiley and Sons.

AcknowledgmentsThe authors would like to thank the referee for making numerous helpful

corrections.

23


Dedication

The authors, all of whom benefited di-rectly from the teaching and wisdom of ap-pliedmathematicianDr. James E.Mann, Jr.,would like to dedicate this Module in hishonor. Dr. Mann retired in May 2002 hav-ing served the Lord faithfully for the last 20years of his career as Professor of Mathe-matics at Wheaton College (IL).

About the AuthorsPaul Isihara, Professor ofMathematics atWheatonCollege (IL), has a special

interest indoingexpositoryandcreative researchprojectswithundergraduates.This Module resulted from undergraduate research in his “Advanced Topicsin Mathematics” course in Spring 2002.Nat Stapleton is pursuing his Ph.D. in pure mathematics at University of

Illinois (Urbana–Champaign). The other student authors are all working inthe area of financial mathematics: Ronya Kamerlander and Jonathan Soyarsat Bank of America in Chicago, Julie Clements at Standard and Poor’s in NewYork City, and Will Landry at InvestorTools Inc. in Yorkville, IL.

24

Reviews 439

ReviewsNahin, Paul J. 2004. When Least is Best: How Mathematicians Discovered Many

Clever Ways to Make Things as Small (or as Large) as Possible. Princeton, NJ:Princeton University Press; xviii + 370 pp, $29.95. ISBN 0–691–07078–4.

Nahin has written several books on the lighter side of mathematics, andthis book is another along that line. The theme of this book is the history ofoptimization, and the author has put together a delightful story. He traceshow various ideas led to the formulation of the concept of the derivative, thecalculus of variation, and modern-day problems such as linear programming.I will be forthright and confess that I am not an expert on the history of

mathematics. I cannot judge the correctness of the author’s history. All I cansay is that his tale is fun to read. I learned a few tidbits, I enjoyed a few goodlaughs, and I had several moments of philosophical reflections. That is quite alot to get from one book.However, the author could have done better in several aspects, including

the organization, the history, and the mathematics.In terms of organization, some of the chapters read like random collections

of puzzles. While they are charming, they fail to illustrate the history of themathematics or the lives of the mathematicians, which are supposed to be thetheme of the book. The significance of the problems described is unclear.On the history side, the author has written much on the discovery of the

derivative, outlining the works of Descartes, Fermat, Leibniz, Newton, et al.But he has said nothing on the further refinement of the notion of the derivativein the nineteenth century by Cauchy, Weierstrass, and their contemporaries.Such efforts of formulating calculus rigorously represented one of the majordevelopments in mathematics in the 1800s. In this sense, the story presentedin the book is incomplete.The author’s discussion on themodern agegives the impression that this era

is characterized by a completely new set of problems, mainly of combinatorialnature, such as the traveling salesperson problem and linear programming.But in fact, many old problems take on a new nature when we attempt to solvethem numerically. Some of the classical methods have to be modified whenwe implement them on a computer. One of the rules we learn in numericalminimization is not to rely too much on the derivative information, because ofits computational cost and volatility; such issuesmake a good story and alwaysfascinate students. I am therefore disappointed that the author did not touch



on the rich topic of numerical solution of optimization problems, which is anactive area of current research.The book’s treatment on the simplex method for linear programming is un-

satisfactory. It does not mention Smale’s work while discussing the method’sefficiency. There is also a lack of details on the algorithm, which the authorclaims to be too advanced. However, the basic plan behind the simplexmethodis rather accessible and can be presented without assuming any prior knowl-edge on linear algebra and calculus. The bookkeeping details certainly requirea lot of careful work; but the underlying idea, namely, exchanging the con-straints to generate a series of successively better extreme points, is not hardto grasp. To illustrate the method, the author could have worked out an ex-ample of small size, similar to what he did for other topics such as dynamicprogramming.The book’s mathematics also leaves room for improvement; there are a few

mistakes. Here are two examples.

• The author claims (p.4) that calculus is unable to solve the problem of min-imizing S(x) =

∑ni=1 |x − xi| for given real numbers x1 < x2 < · · · < xn.

That is not true. Using the fact that ddx |x| = sgn(x), except at 0, we can differ-

entiate S(x) and look for critical points. We need to examine points wherethe derivative fails to exist, namely, at x1, . . . , xn; and also points where thederivative is zero, namely, in the open interval (xn/2, x(n/2)+1).We can thenconclude that S(x) is minimized either at the median of the xi (if n is odd)or at all points between xn/2 and x(n/2)+1, inclusive (if n is even).

• The author’s proof (p. 173) of l’Hospital’s rule for computing the indetermi-nate form 0/0 relies on the statement that limx→0 R′(x) = 0 if limx→0 R(x)exists. That statement is false. A counterexample is provided by g(x) =x2 sin(1/x), h(x) = x, so thatR(x) = g(x)/h(x) = x sin(1/x). Then as x → 0,R(x) → 0while R′(x) oscillates without bound.

In summary, this book is good entertainment. I was amused by the interest-ing anecdotes. I would heartily recommend this book to those of us teachingcalculus, optimization, or related subjects. It contains many stories that wecould relay to our students to spice up the sometimes dull lectures. The book isalso a good source of word problems. Students are more motivated if they arepresented problems that come with a story, such as hikers lost in the woods,rather than just equations to solve. However, this book has some omissionsin the history and is imprecise on technical matters. I would not recommendthis book to students learning calculus for the first time, at least not withoutsupervision. The studentswill be confused. As the author stated in the preface,it takes certain mathematical maturity to read the book. Such maturity wouldalso be necessary to steer clear of the occasional erroneous arguments.

M.K. Stephen Yeung, Dept. of Mathematics and Statistics, 3900 Old Main Hill, UtahState University, Logan, UT 84322; [email protected] .

Annual Index 441

Annual Index for Vol. 25, 2004Author IndexAcknowledgments. 24(4): 445.

Amery, Steven G., Eric Thomas Hartley, and Eric J. Malm. The myth of “the myth offingerprints.” 25(3): 215–230.

Annual Index. 24(4): 441–444.

Anspach, Peter, and Kathleen M. Shannon. Judges’ Commentary: The Quick PassFusaro Award paper. 25(3): 354.

Aravkin, Aleksandr Yakovlevitch. See Lovejoy, Tracy Clark.

Arney, Chris. Results of the 2004 Interdisciplinary Contest in Modeling. 25(2): 97–114.

Author electronic copies of contributions. 25(4): 445.

Beeton,Mary. Practitioner’s Commentary: TheOutstanding Fingerprints papers. 25(3):267–272.

Berkove, Ethan, ThomasHill, and ScottMoor. Getting the salt out. UMAP/ILAPModules2003–04: Tools for Teaching: 95–110.

Bogart, Eli, Cal Pierog, and Lori Thomas. It’s all about the bottom line. 25(2): 115–128.

Brega, Moorea L., Alejandro L. Cantarero, and Corry L. Lee. Developing improvedalgorithms for QuickPass systems. 25(3): 319–336.

Bressoud, David, and Joy Laine. Parallel developments in philosophy andmathematicsin India. UMAP/ILAP Modules 2003–04: Tools for Teaching: 133–166.

Burgess, Stephen. See Frolkin, Alexander V.

Camley, Brian, Pascal Getreuer, and Bradley Klingenberg. Not such a small whorl afterall. 25(3): 245–258.

Campbell, Paul J. Acknowledgments. 25(4): 445.

. Author electronic copies of contributions. 25(4): 445.

. Editor’s Commentary: Fingerprint identification. 25(3): 273–280.

. Editor’s note regarding submissions. 25(2): 180.

. The mathematics of democracy. 25(4): 355–356.

. MathServe lives on. 25(1): 3–4.

. “Party like it’s 1999!” 25(1): 1–2.

. Guide for Authors. 25(1): 91–92; UMAP/ILAP Modules 2003–04: Tools forTeaching: 195–196.

. See Garfunkel, Sol.

Cantarero, Alejandro L. See Brega, Moorea L.

Clements, Julie. See Isihara, Paul.

Clancey, Dennis, Daniel Kang, and Jeffrey Glick. Firewalls and beyond: Engineering ITsecurity. 25(2): 143–156.

Corwin, Ivan, Sheel Ganatra, and Nikita Rozenblyum. A myopic aggregate-decisionmodel. 25(3): 281–300.

Crangle, Aidan. See O’Ceallaigh, Seamus.


Crannell, Annalisa. See Habecker, Brian.

Dodge, Ronald C., Jr., and Daniel J. Ragsdale. Authors’ Commentary: The OutstandingInformation Technology papers. 25(2): 171–174.

Driessel, Kenneth R., Philip Fink, and Irvin R. Hentzel. The dynamics of ski skating.25(4): 375–410.

Driscoll, Patrick J. Building a modeling community: MATHmodels.org. 25(2): 93–96.

Editor’s note regarding submissions. 25(2): 180.

Errata. 25(4): 446.

Fink, Philip. See Driessel, Kenneth R.

Fitzkee, Thomas L. Tuition prepayment plan. UMAP/ILAP Modules 2003–04: Tools forTeaching: 69–94.

Frolkin, Alexander V., Frederick D.W. van derWyck, and Stephen Burgess. Theme-parkqueueing systems. 25(3): 301–318.

Ganatra, Sheel Ganatra. See Corwin, Ivan.

Garfunkel, Solomon A. The good fight. 25(3): 185–188.

, and Paul J. Campbell. Introduction. UMAP/ILAP Modules 2003–04: Tools forTeaching: v–vi.

Giordano, Frank. Results of the 2004Contest inMathematicalModeling. 25(3): 199–214.

Glick, Jeffrey. See Clancey, Dennis.

Getreuer, Pascal. See Camley, Brian.

Guide for Authors. 25(1): 91–92; UMAP/ILAP Modules 2003–04: Tools for Teaching: 195–196.

Habecker, Brian, andAnnalisa Crannell. Using fractals tomotivate linear algebra. 25(1):47—82.

Hartley, Eric Thomas. See Amery, Steven G.

Hentzel, Irvin R. See Driessel, Kenneth R.

Hill, Thomas. See Berkove, Ethan.

Isihara, Paul, with Julie Clements, Ronya Kamerlander, William Landry, Jonathan So-yars, and Nathaniel Stapleton. Wind turbine power coefficient optimization. 25(4):411–438.

Kamerlander, Ronya. See Isihara, Paul.

Kang, Daniel. See Clancey, Dennis.

Katzenstein, Warren, Tara Martin, and Michael Vrable. Making the CIA work for you.25(2): 129–142.

Kim, Yoonsoo. See White, Caleb Z.

Klingenberg, Bradley. See Camley, Brian.

Laine, Joy. See Bressoud, David.

Landry, William. See Isihara, Paul.

Lee, Corry L. See Brega, Moorea L.

Lee, Youngbae. See White, Caleb Z.

Lovejoy, Tracy Clark, Aleksandr Yakovlevitch Aravkin, and Casey Schneider-Mizell.KalmanQueue: An adaptive approach to virtual queueing. 25(3): 337–352.

Annual Index 443

Malm, Eric J. See Amery, Steven G.

Marasco, Joe. Optimal growth and productivity in organizations. 25(4): 357–374.

Martin, Tara. See Katzenstein, Warren.

Miller, Seth, Dustin Mixon, and Jonathan Pickett. Rule of thumb: Prints beat DNA.25(3): 259.

Mixon, Dustin. See Miller, Seth.

Moe, Karine, andKaren Saxe. Fixed-point theorems in economics. UMAP/ILAPModules2003–04: Tools for Teaching: 167–194.

Moor, Scott. See Berkove, Ethan.

Neller, Todd W., and Clifton G.M. Presser. Optimal play of the dice game Pig. 25(1):25–46.

O’Ceallaigh, Seamus, Alva Sheeley, and Aidan Crangle. Can’t quite put our finger onit. 25(3): 231–244.

Perkins, Patrick. See White, Caleb Z.

Pickett, Jonathan. See Miller, Seth.

Pierog, Cal. See Bogart, Eli.

Presser, Clifton G.M. See Neller, Todd W.

Puckette, Emily E. The spread of forest fires. UMAP/ILAP Modules 2003–04: Tools forTeaching: 51–68.

Ragsdale, Daniel J. See Dodge, Ronald C., Jr.

Reviews. 25(1): 83–90; 25(2): 181–184; 25(4): 439–440.

Reviews Index. 25(4): 444.

Rozenblyum, Nikita. See Corwin, Ivan.

Saxe, Karen. See Moe, Karine.

Schneider-Mizell, Casey. See Lovejoy, Tracy Clark.

Shannon, Kathleen M. See Anspach, Peter.

Sheeley, Alva. See O’Ceallaigh, Seamus.

Sigmon, Neil P. Determination of satellite orbits with vector calculus. UMAP/ILAPModules 2003–04: Tools for Teaching: 25–50.

Small, Don. See Yu, Lei.

Song, Yunji. See Zhao, Qian.

Soyars, Jonathan. See Isihara, Paul.

Spann, Andrew, Daniel Gulotta, and Daniel Kane. Theme park simulationwith a Nash-equilibrium-based visitor behavior model. 25(3): 353.

Stapleton, Nathaniel. See Isihara, Paul.

Statement of Ownership, Management, and Circulation. 24(3): [391]–[392]. ???

Su, Xueyuan. See Zhao, Qian.

Teets. Donald A. Predicting opportunities for viewing the International Space Station.UMAP/ILAP Modules 2003–04: Tools for Teaching: 1–24.

Thomas, Lori. See Bogart, El.

Thomas, Rekha. See White, Caleb Z.


Tortorella, Michael. Judge’s Commentary: The Outstanding Fingerprints papers. 25(3):261–266.

van der Wyck, Frederick D.W. See Frolkin, Alexander V.

Vrable, Michael. See Katzenstein, Warren.

Wattenberg, Frank. Judge’s Commentary: The Outstanding Information Technologypapers. 25(2): 175–179.

White, Caleb Z., Youngbae Lee, Yoonsoo Kim, Rekha Thomas, and Patrick Perkins.Creating weekly timetables to maximize employee preferences. 25(1): 5–24.

Yu, Lei, and Don Small. Vehicle emissions. UMAP/ILAP Modules 2003–04: Tools forTeaching: 111–132.

Zhao, Qian, Su Xueyuan, and Song Yunji. Catch thieves online: IT security. 25(2):157–170.

Reviews Index(Names of authors are in plain type; names of reviewers are in bold.)

Adams, Colin, Abigail Thompson, and Joel Hass. How to Ace the Rest of Calculus: TheStreetwise Guide. James M. Cargal. 25(1): 90.

Albert, Jim. Teaching Statistics Using Baseball. James M. Cargal. 25(2): 181–184.

Albert, Jim, and Jay Bennett. Curve Ball: Baseball, Statistics, and the Role of Chance in theGame. Rev. ed. James M. Cargal. 25(2): 181–184.

Bennett, Jay. See Albert, Jim.

Brannan, David A., Matthew F. Esplen, and Jeremy J. Gray. Geometry. JamesM. Cargal.25(1): 85–88.

Esplen, Matthew F. See Brannan, David A.

Farin, Gerald, and Dianne Hansford. The Geometry Toolbox for Graphic and Modeling.James M. Cargal. 25(1): 85–88.

Fusaro, B.A., and P.C. Kenschaft (eds.). Environmental Mathematics in the Classroom.Daniel Kaplan. 25(1): 83–85.

Gray, Jeremy J. See Brannan, David A.

Hansford, Dianne. See Farin, Gerald.

Hartshorne, Robin. Geometry: Euclid and Beyond. James M. Cargal. 25(1): 89.

Hass, Joel. See Adams, Colin.

Kenschaft, P.C. See Fusaro, B.A.

Lewis, Michael. Moneyball: The Art of Winning an Unfair Game. James M. Cargal. 25(2):181–184.

Mlodinow, Leonard. Euclid’s Window: The Story of Geometry from Parallel Lines to Hyper-space. James M. Cargal. 25(1): 88–89.

Nahin, Paul J. When Least is Best: How Mathematicians Discovered Many Clever Ways toMake Things as Small (or as Large) as Possible. M.K. Stephen Yeung. 25(4): 439–440.

Thompson, Abigail. See Adams, Colin.

Annual Index 445

AcknowledgmentsI here express my great appreciation for the help of the associate editors, whose

names appear on the masthead. Not only do they do the bulk of work of evaluatingmanuscripts, but they also solicit newworks andencourage andguidepotential authors.

I am also indebted to the additional individuals listed below who have reviewedmanuscripts during thepast year. Their careful evaluation and judgmenthave enhancedthe quality of the articles and Modules that have appeared in the Journal and in Toolsfor Teaching. (Reviewers of some papers considered for Vol. 25 of the Journal wereacknowledged already in Vol. 24, No. 4. Reviewers and editors of the ILAPModules inUMAP/ILAP Modules 2003–04: Tools for Teaching are acknowledged on the frontispieceof the corresponding ILAP Module.)

The Journal offers many opportunities for participation in COMAP’s work. If youwould like to

• referee manuscripts—please contact me;• review books, software, or films—please contact the Reviews Editor;• write a self-contained expository essay about an area ofmathematics, however grandor small—an era, a concept, a theorem, an idea, a term—please contact the On JargonEditor;

• contribute an Interdisciplinary Lively Applications Project (ILAP) Module—pleasecontact the ILAP Editor;

• contribute an article, UMAP Module, or Minimodule—please contact me;• encourage and stimulate colleagues to prepare and submit suitable material—pleasecontact me about being nominated to join the Editorial Board.

Contact information for the associate editors in charge of Reviews, On Jargon, andILAPs, as well as my own information, are on the masthead of every issue.

Finally, the associate editors and I would like to thank the Journal’s authors, withoutwhom none of this would be possible, and its readers, whose benefit and enjoyment arethe culmination of our enterprise.

Paul J. Campbell, Editor

Bruce Atwood, Beloit CollegeEric Bach, University of WisconsinScott Berry, Berry Consultants, LLCGlenn Appleby, Beloit CollegeRichard Kleber, St. Olaf CollegeFrederick Solomon, Warren Wilson CollegePaul Stanley, Beloit CollegeCory Vandenberg, Oakwood, ILJack Winn, SUNY at Farmingdale

Author Electronic Copies of ContributionsAn author of a contribution to The UMAP Journal from Vol. 16, No. 4 (1995) forward

may obtain from the editor an electronic copy of the contribution in PDF format. If thefile is small enough, it will be sent uncompressed; however, just in case, please advisedesired compression format: .sit (StuffIt), .zip (ZipIt), .tgz (Unix tar), or specify other.


ErrataVol. 23, No. 2 (2002)p. 125, l. 2: K −→ kp. 125, l. 2: M −→ mp. 133, l. 10: see below for Vol. 24, No. 3 (2003), p. 484

Vol. 23, No. 3 (2002)p. 187, l. 22: sixteen −→ 17

Vol. 24, No. 2 (2003)p. 102, l. 11: Cassidy −→ Cassady

Vol. 24, No. 3 (2003)p. 199, l. 13: 18th −→ 19thp. 199, l. 23: sixteen −→ 18

Vol. 24, No. 4 (2003)p. 479, l. 16: insert: 24(2): 195–196;p. 484, l. 17 (apropos of Vol. 23, No. 2, p. 133, l. 10):

Basket[a_} := Decreaser[[Decreaser[a]]−→Basket[a_] := Decreaser[Decreaser[a]]

Vol. 25, No. 1 (2004)p. 5, l. 7, and p. 23, ll. 11–12: Youngbae Lee is no longer at LG Chemical Ltd. but in the

Dept. of Mathematics at the University of Washington.

p. 41, ll. 5–8:

While the single-roll, odds-based analyses of Falk and Tadmor-Troyanski [1999]and Knizia [1999] yield policies optimizing expected score for a single turn, theydo not yield policies optimizing scores over an arbitrary number of turns.

−→While the single-roll, odds-based analysis of Falk and Tadmor-Troyanski [1999]yields a policy optimizing expected score for a single turn, it does not yield apolicy optimizing scores over an arbitrary number of turns.

p. 42, ll. 10–12: delete: Similarly examining Knizia’s Pig analysis for maximizing ex-pected score, we find the same deviation of optimal versus odds-based policiesfor Pig games longer than eight turns.

[Author’s remark: Knizia’s similar odds-based analysis for maximizing expected scoreis indeed optimal, because decisions within a turn are conditionally independent ofother turns. The initial false conclusion was based on significant error in numerical ap-proximation. The Tadmor-Troyanski-related computations have been double-checkedand still stand.]

p. 65, l. 28: delete spurious ]p. 68, l. 13: ((a0)) −→ {(a0)}p. 73, l. 15: It −→ (Itp. 77, l. 2: Exercise 7.. −→ Exercise 7.

umap 2005 vol. 25 no. 4

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