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Animal Behavior

by: Mrs. Sierra Bailey
Mrs. Sierra Bailey
GPA 3.85


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This 126 page Class Notes was uploaded by Mrs. Sierra Bailey on Tuesday September 8, 2015. The Class Notes belongs to NPB 102 at University of California - Davis taught by Staff in Fall. Since its upload, it has received 26 views. For similar materials see /class/191827/npb-102-university-of-california-davis in Neurobiology,Physio & Behavior at University of California - Davis.

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Date Created: 09/08/15
NPE Tum Mudme 3 JANUme GaHs Jumping Galls Part I Introduction eoaTs R ewew hTuTugy u Neurozema saIazowsw The CaTTTuThTaTumpThg gaHWasp WaTeh mewes uHumpmg gaHs Th aeTTeh TTh Th seees w Thp h Th m TThw pTh head TT yuu sTepTe TeehTeT a Tew seeehes yuu H heTTeeThanhese seeds aTe heppThg amuth MaEL These aTeh T seeds aT aH hm gaHs GaHs T e TesuTT eTa eempTew hm squTTsThgTy cummun ThTeTaeTTeh heTweeh ThseeTs aha pTahTs eTheT Dvgamsms eah mdUEE gaHs asweTT GenevaHy a gaH e CaTTTeThTaTumpThg gaHWasp The gaH wasp TamTTy CythTeaeT N saIamws Ts a memheh eehTaThs abum Tana knDWn speeTes The maTeTTTy e1 These speeTes ThTesT uakhees The eggs e1 N saIamnus aTe TaTe Th The W 5W3 0m 3quot Egg hamm m quotTTeweTeuemeamTeeTs hmweTTTawah Tam Sm mng WWW TM saveTTTTTssTTTTTehea phaapywmhweah TheueesThe uak Teane huTTe ah eheTesuTe TeTThe ava w The gaH The pTeeess by theh gaH TheueTTeh uccms Ts peeTTy uneeTsTeee Anhuugh TT Ts knDWn ThanhTs Ts hm a geheTaT Tespehse hyThe TTee Te wasp mvadEYS Eaeh gaHWasp spee s eausesThe hes1 Te mm a VEVy speeTTTe gaH shape and eeTeT ThTs suggesTsThaT gaH TeTmaTTeh Ts cumquEd Te a TaTge dEgVEE by The wasp Rthc ASvmed mvban Tm Ammndougbm ths w Mm wTTsah MudmeUS JumpTheoaTTsjhweTeewmm NPE Tum Mudme 3 Jumpmu GaHs summev Tame yapTeTy muves TTs heeyWTThTh The seaTee 33 Te 52nd The 33 eTsTahees eTupTe Tem mate and Theh muvmg yum budy sueh ThaT mate Thee 2n eTseTeeT ThTe The am The TeueWThg muwes shewTumpThg aeTwTTy The muwe eh The TeTT Ts Th veamme The muwe Em The hghT TSWTTh Thhe Tapse phutugvaphy Ts T Thhe cumpvessmn R eamme Muwe ThheTapse Muwe veeasw Mmhweah VTdeaChp VTdeaChD The puvpuse eTau ThTsTumpThg amuhe Ts heT ehTheTy e T eTspeysaT A evaH uncethewaspsemevge ummew gaHs Th The TeueWThg sphhe They can mspeyse by TTyThg Deemed ayumvmv vauva stM meme Tm geheyaT hypeTheSTsTmTheTumphg behawuv TsThaT hambTW V mmh ENVY eh V2 Tame Tshappad Theme The seaTee gah hew eah TT pessmh knuW U see Wheve TT sheuTeTumpv gm Mm manna 5th My Mame Va th WW V levanv MudmeUS JumpmuGaHsJAnwEvedxmm NPB 102q Module 3 Jumping Galls This module will explore the specific hypothesis that jumping gall behavior is due to heat aversion on the part of the larvae To do this we will make a single key assumption That the jumping frequency of the gall is temperaturedependent That is at hot temperatures the gallsjump more often than at cold temperatures Whetherjumping frequency is temperaturedependent or not is unknown This is an outstanding question that can only be answered with experiments presumably by a cecidologist a biologist who studies galls However without doing experiments we can ask whether temperaturedependent jumping would lead to shadeseeking behavior on the part of the gall wasp larvae This is the hypothesis we will test in this module We will assume that all gall jumping is stochastic random For convenience we39ll assume that the distance jumped is always the same 1cm As a further simplification we will only consider a 1dimensional environment That is galls will jump to the left orto the right along an imaginary line In your own words below explain the hypothesis that we will test in this module What are the alternative outcomes We39ll be testing whetherjumping galls get out of the hot locations if the frequency oftheir random jumping is temperaturedependent The alternatives are 1 galls will leave hot areas 2 galls will leave cold areas or 3 galls won39t show a preference for hot or cold areas What do you thinkthe result of this model will be Will the galls appeartojump towards the shade Or not Any answer is OK here as long as you explain your reasoning I think galls will get out of the heat 80 long as galls are in the heat they39ll jump around They won39t know how to get to the shade But they will know that if they keep jumping eventually they39ll end up in a cooler place Once they arrive at that cooler place then their jumping frequency will decrease This will cause them to largely stay put in the shade Module03 3 JumpingGallsAnsweredxmcd NPB 102d Module 3 Jumping Galls Part II Temperatureindependent Jumping Single Gall Goals Create movement vectorto trackthe movement ofa hypothetical jumping gall through time Create x vector to track the gall39s position Make a graph of the the gall39s position through time In this Part we will begin to construct our model ofjumping gall behavior For simplicity we39ll start with a single gall In Part III we39ll expand this to 30 galls We will arbitrarily decide to divide up time in 1sec increments timesteps At each timestep there will be some probability that our gall will jump jumpprob We will ask Mathcad to generate a random number that determines whether the gall jumps or not during that timestep If the gall does jump then we will move its position 1cm jumpdistance either to the left orto the right each with a 50 chance A second random numberwill determine whetherthejump is to the left orto the right This sort of model can be described as quotdiscretetimequot because we39re dividing time up into discrete chunks In contrast space is described continuously We39ll start the gall at position quotzeroquot but once the jumping starts the galls39 positions could be anything This characteristic ofthe model could be described as quotcontinuous in spacequot This sort of model lends itself very well to being stored in an array In the case of a single gall we39ll use a singlecolumn array a vector Each row of this vector will represent a different timestep Each element of the vector will contain the position ofthe gall a number with units of cm Because this vectortracks position along an imaginary line let39s simply call the vector x Let39s begin by making some assignments forwhat we already know about the model In the space below use the assignment operator to assign jumpdistance to be 1cm Don39t leave out the units Thisjump distance is consistent with that published in Russo 2006 201 jumpidistance 1cm Assign timestep below to be 1sec timestep lsec The rate at which jumping gallsjump is not documented in the literature So we39ll make an educated guess Our guess doesn39t have to be superaccurate because we won39t put any time constraints on our galls39 journey to the shade That is if it takes them all day to get to the shade that39s still OK We should however make a mental note that this guess will affect the output of our model For example we shouldn39t draw any quantitative conclusions about how long it would takes the average gall to get into the shade Our jump rate is just a guess so the galls39 quotaverage speedquot will just be a guess too Module03 4 JumpingGallsAnsweredxmcd NPB 102d Module 3 Jumping Galls 1 105cc once every 10 seconds regardless of temperature In Part IV you39ll change this assumption to include temperature dependence but for now this fixed jump rate is fine Assign jumprate below to be This means we39re assuming that all gallsjump on average 39um rate J p 105cc We said above that at each timestep there would be some probability that our gall will jump Thus we must somehow infer a probability ofjumping based on the fact that the average rate ofjumping is known ie jump rate as you assigned above Approximating this probability is not hard It is helpful to point out that all probabilities are unitless whereas rates typically have units of L Thus just based on the units you might guess that we sec can approximate our probability by mulitplying jumprate and timestep together the units will cancel Assign jumpprob to be julnpiratetimestep below Below your assignment use the regular equals sign to ask Mathcad what jumpprob is 204 jump erb jumpiratetimestep jmnpjrob 01 You should get a unitless answer of 01 Don39t continue until you get this answer Note that this approach to deriving probability from a rate is only valid for small timesteps To see the error ofthis approach imagine that our timestep is not 1sec but instead 10sec Given the expression above we would estimate thejumping probability to be jmnpirateIOSec 1 That is the probability of a gall jumping within a 10second timestep would be 100 This cannot be correct No matter how long of a timestep we choose there must be some nonzero probability that our gall will notjump To obtain the precise probability we would would need to use the quotPoisson distributionquot which is beyond the scope of this module If you take Statistics 100 you39ll learn all about the Poisson distribution That said this approximation is fine for our purposes For a 1sec timestep the approximation is within 1 of the true value Ourjumping rate wasjust a guess to begin with so this amount of error is negligible In your own words explain briefly what the jumpprob value of 01 means and how it will be used A jumpprob value of 01 means that each second each timestep there is a 10 chance that our gall will jump If we find that the gall doesjump in that timestep we39ll move it 1cm to the left or to the right If we find that the gall doesn39tjump we39ll leave it where it is Module03 5 JumpingGallsAnsweredxmcd NPE 1mm Mudme 3 Jumpmu GaHs hems We can a ways adjus1 hTs atevTVWE hke Because we u use msva ue sevevammes Ters sssTgh n heme a Vauab E caHed end AssTgh end e be Zhr heTew end m As mEnHunEd pvevmus y e u keep hack enhe pusmun Mum 33H m a vecmv ThTs Tsvavy mush w hke We eaweme exevmse yuu we h Mudu emr Mmdudmn m Mathcad As M Ma pvevmus mscveterume mueeh we u HEEd m 52 sh INNSCondmon unhe mudg m The vs eTemem unhe M Fuy h m h zem h II h T 2m x n mhme Ters aH hTs VangEvaHab El 1 hh end the Tss1 shumd be We expvessmn umestep 2m Z L i umestep EXp am hheny whywe began 1 3H and hen am zem 2m The vst eTemem m x w m Ts aheady 52 Se whehwe Ta ev use Hu assTgh We mhev ETENEMS WE emu wam m EVENmeme mma euhemeh cumamed m deX zem Lers s1up she Mmkmv a mmu E abum haw We ve gumg e we the xvecmv The xvecmnvacks va ues 1 The x pesmeh h 012 pvevmus Theex Wavmusumes ap 2 The msTshee muved ThTs seem he ehe e1 Wee va ues hhe Tumn attuned m the pvevmus hmes1ep4heh tmsvame Ts zem 0Thehmse4hevsme mwgmbe mp distance 1 chHhE jump was e the WM m 4 mp3s 71 cm THHEJump was eihe e MudmeUS Jummheoausjhwevewmm NPE mm Mudme 3 Jumpmu GaHs Se m Wemswe seem sav nah pesmeh nuwpvevmus eau p smunmuvemem Hwe puHh s Mevms e ah ass gnmem mvehmg the xveemh n Imam uuk hke M s h m muv2mem mam hewhe euahmy mnvemem 5 hm de ned Lerswmh eh tha nuw Theve s a cuup e e1 dWEYEM Wayswe seem appmaeh W s ohe stha we seem ca cu ate mnvemem eh the y as n s HEEdEd m the a ass gnmEm Asecund appmaeh 5 thatwe seem pyeeaheuha e mnvemem as a 2mm ahmheh use mdmdua exemems uHhat mwemenlvecmv m SS gnmEM Whmh anpmaeh We eheese 5 WHEN a ma a emeveyehee Eaeh Wm Wuvk Mv pveVeyehee s he ca cu awm muvemem 5 a We cumphcated Ev vata cu a mu m sepava ew can examme We vesuh a each umestep m S a n That 5 eah ExammE msnhe muvemems wnhum euhsmehne nusneh Thwswm make me mme euhmem m dum nus eeneew Them ave mugmymvee pvucessesmat dE EYmmE the va ue Ma smg e exemem m the mnvemem Vaclav 1 Avandum numbev dE EYmmES n a ump has attuned 2 Vamp has eeeuhewhe gaH s pusmun muves by Jmpimsmee 1 cm Otherwwse We gaH s Pusmun muves v avs pm 3 Ha ump has eeeuhee see AHE ump uccms euhene the WM nhe pusmve eheeheh m the e nhe heeawe eheeheh Lers hemh bv dE EYmmm We ump has eeeuhee pvueessm abqu Tu net Mathcad u nenevate a Yandum humhenm us we Wm useme buHHn md when h hhs mudu e We Wm ask Mathcad e qenevate Yandum numbevs ha ave between D ahm Thws 5 accumphshed bv speemhew as he mpuHuthemdmnmmn Fuvexamme mil 1252 x m 3 dWEYEM numbev bemeen u ahm attuned mhm Fuvexamp m m1lt1mpjxab n The abuve smemem 5 a EDD Ean 251 Wm WM S YUE h e Aha Yandum numbev heve 5 essthan We pmhahmw Ma ump hapnehmmweh he hawue unhe testwm he1 che es SMSE h e a We Yandum numbev 5 qreazenhah the pmhahmw u a ump happEmn Lththe va ue unhe 251 MM be u vmrn n has attuned And N uhe has neeunewheh N m x W m O herwwsE Nme Maths 5 henhe ehme s1uw We sun have m mm m pvucesses abuve Em 01 s 5 a 5 3quot MudmeUS Jummheoausjhweveexmm NPB 102d Module 3 Jumping Galls Assign movementb1 below to be md1 ltjump4probju1npidistance Note that we39re using t1 here ratherthan t This is a somewhat arbitrary convention and implies that movemento will contain the gall39s movement that takes it from x0 to x1 Below your assignment ask Mathcad what movement is Change the displayed units from m the default to cm by first clicking on the displayed vector A blank placeholder will appear at the right Type cm into this placeholder and press Enter 210 movement1 mdtl ltjumpjrobjumpidistance movement KOWNCBLnbWNl O H O H H H N H a OOOOHOOOHOOOOOO o B H J H U1 Inspect your movement vector ts elements should contain values of either 0 or 1cm the units are on the outside of the vector There should be many more 039s than 139s Don39t continue until you39re convinced this is true Module03 8 JumpingGallsAnsweredxmcd NPB102q Module 3 Jumping Galls Explain in your own words what the expression mdl ltju1np4probjumpidistance meansdoes Why are 039s more common than 139s in your x vector above 211 That expression first stochastically determines whether a jump has occurred based on the probability that a jump will occur during a timestep lf ajump has occurred the expression generates the value ofjumpdistance Otherwise the value is zero Zeros are more common becausejumpprob is less than 05 This means that during most timesteps nojump occurs so the x value is zero Excellent We now have the gall jumping But because everyjump results in a positive 1cm value within movement all jumps are implicity to the right We39ll need to add process 3 above to get both positive and negative values To do this we39ll use another separate Boolean test Because we considerjumping to the left or right as equally likely we39ll compare a random number again between 0 and 1 to 05 As you can imagine this will yield quottruequot results half ofthe time However there is an additional complication We would like to use this test to change the sign of jumpdistance if ajump occurs The output of a Boolean test is either 1 or 0 for quottruequot or quotfalsequot Using this result we cannot use mulitplication to change the sign To use multiplication which would be convenient we need a test that outputs 1 for quottruequot and 1 for quotfalsequot To create such a test we39ll use the builtin if function The if function includes a Boolean test but rather than just outputting a 1 for quottruequot and a 0 for quotfalsequot it will output anything we tell it to For example the statement ifabc checks whether a the Boolean test is true If a is true then it sets yb If a is false then it sets yc In English this assignment would say quotIf a is true then y is assigned the value b otherwisey is assigned the value 2quot Here are some examples that use the if statement like we want to if23lt2317171 if23lt241711 Knowing this we can combine the if function with the rnd function to produce the effect we seek Ask Mathcad below what ifmd1 lt 051 71 is You should get either 1 or 1 Click on the expression and press F9 several times You should see the answer value change back and forth between 1 and 1 212 Module03 9 JumpingGallsAnsweredxmcd NPB102q Module 3 Jumping Galls ifmd1 lto5171 71 Grand To accomplish process 3 all we need to do now is use this expression to modify your movement assignment from Exercise 210 Copy your assignment from Exercise 210 above and paste it here below Modify it to include the if statement you used Exercise 212 movement11 md j lt jumpprobjjumpdistanceifmd1 lt U51 1j Below your assignment ask Mathcad what movement is movement1 md1 ltjumpjrobyjumpidistanceifmd1 lt 051 71 movement KOWNCBLnBWNHO HOOOOOOOOOOOOOO o B Scroll through and inspect your movement vector Its elements should contain values of either 1cm 0 or 1cm the units are on the outside of the vector There should be many more values of 0 than either 1cm or 1cm The value of 1cm should be roughly as common as that of 1cm Don39t continue until you39re convinced this is true Modue03 10 JumpingGallsAnsweredxmcd NPB 102g Module 3 Jumping Galls Explain in your own words what the expression md1 lt jumpjrobyjumpidistanceifmd1 lt 05171 meansdoes this is the expression you used above in Exercise 213 Why are there roughly the same number of 1cm39s and 1cm39s in your movement vector above 214 That expression first stochastically determines whether a jump has occurred based on the probability tht a jump will occur during a timestep Ifthe jump has occurred the expression generates either the positive or negative value ofjumpdistance each with 50 probability If no jump has occurred then the expression generates a zero 1cm39s and 1cm39s are equally common because assuming a jump has occurred the chance of ajump to the right vs a jump to the left is 5050 Now that the movement vector is finished we can return to the x vector Recall that the x vector contains the position of our gall Assign xt below to be xti movementi Below your assignment ask Mathcad what x is 1 t139 th xti1 movement1 KOWNCBLnBWNHO OOOOOOOOOOOOOOO o B Module03 l l JumpingGallsAnsweredxmcd NPE mm Mudme 3 Jumpmu GaHs he pDSNNE wheyhe aH pusmve El 3 hegawe that s owee bunheve shame am he a vangE e1 va ues heuusm e 5 and 4 s Semmhmugh yuu mspxay e x abuve and make suve W5 SUUE gaH gues CVEa E a new Xry mm be uw Wuh lumen punthexraws 5 aha Em the yraws Senhe mm m vangE mu m n 12m Set 012 v39awstu yahee hem 43H m an 7 Vuuv maph shumd uuk svm anu We uhe anhe mm 5 Heweveh becausE Vuuv maph 5 cvea ed Wuh Yandum h bevs M sunhke yyuuvswmuukexac yme ame mm m 5H n 5u 5n mu tnmzstzp hme m the 2 mm the WM emu7 m The gvaph shuws he pusmun enhe gaH eh We yraws and me Em the mm The 33 muves amuhe qune a hm hm w my gvaph gays mus ymthe Ham enem pusmve X va ues MudmeUS Jummheoausjhwevewmm NPB 102d Module 3 Jumping Galls Press CtrlF9 to get Mathcad to recalculate your worksheet This should change your graph in Exercise 216 Describe briefly below how the graph is now changed You can also get this graph to recalulate by scrolling up to your movement assignment in Exercise 213 clicking on the assignment and pressing F9 Leaving your edit line on that assignment you can scroll backto Exercise 216 to watch your graph as you press F9 repeatedly The gall still moves around quite a bit but now it spends most of its time to the left of zero negative x values The sort of model that you have constructed in this Part is known as a quotRandom Walkquot Indeed a random walk is is an apt description of ourjumping gall39s motion Random walk models are used in a variety of fields including ecology chemistry physics and economics Now that we39ve generated a way to trackthe movement of a single jumping gall overtime in the next Part you will expand this technique to track 30 galls all at once Modu e03 JumpingGallsAnsweredxmcd NPB 102d Module 3 Jumping Galls Part III Temperatureindependent Jumping Multiple Galls Goals Expand the x and movement arrays to trackthe position of multiplejumping galls Graph the x array for all galls using only a subset oftimepoints Our model of a single gall from Part II gives us some information about the path of a single gall However we39ll eventually want to make some generalizations aboutjumping gall movement We want to know if temperaturedependent jumping leads to shadeseeking behavior Knowing that one gall makes it to the shade doesn39t really tell us much That gall mayjust be lucky We want to run a large number of simulations and see what proportion of galls make it to the shade Running multiple simulations like this is not particularly difficult especially on a fast computer Such a set of simulations is often called a quotMonte Carlo Simulationquot after the Monte Carlo Casino in Monaco As in the case of the jumping galls we admit that we do not know how to describe the behavior of the galls generally However we do know or think we know the probabilistic behavior of a single gall That is we39re willing to assume something about thejumping rate and distance of each jump A Monte Carlo Simulation says quotFine Let39sjust run the model a bunch of times and see what tends to happenquot As you can imagine Monte Carlo simulations are helpful in understanding longterm performance in games of chance hence the name Let39s begin by deciding that we want to see the movement of 30 galls Assign totalgals below to be 30 totaligalls 30 To expand yourjumping gall model to include multiple galls we39ll expand the x and movement vectors to be 2D matrices where each column represents a single gall This means that we39ll need a new range variable to handle the column indices ofthose new x and movement arrays Let39s call this range variable gall Assign gall below to be a range variable The first value should be 0 the second value should be 1 and the last value should be totagas1 302 gall 01 tom1 alls 7 1 We39re now going to overwrite the x and movement arrays that you created in Part II When doing such an overwrite it39s always best to explicitly clear out the old array data That way you eliminate the chance that old data will carry over into your new array Module03 14 JumpingGallsAnsweredxmcd NPB102q Module 3 Jumping Galls Clear out the x and movement arrays by assigning them each below to be NaN NaN is a special value in Mathcad that stands for quotNot a Numberquot 303 X NaN movement NaN Because our new x array will have multiple columns we39ll need an initial condition in the whole first row of that array As in Exercise 207 we39ll assume that all galls start out at position 0 Assign x0 gall below to be 0 Below your assignment ask Mathcad what x is You should see a single row and 30 columns of zeros 304 01234 5 6 7 89 0 0 0 0 0 0 0 0 0 0 V Now we39ll create new movement and x assignments based on the assignments we made back in Part II but now expanded to include columns specified by gall Copy your movement assignment from Exercise 213 above not Exercise 210 and paste it here below Change the left side of the assignment from movementi t b 39 t 1 0 e movementt71galllnstead movementkl gall md j ltjwanarubjumpdistanceifmd1j lt EI51 1j movement Lgall md1 ltJmperbJuIHp7dISthCelfmd1 lt 05171 Copy your x assignment from Exercise 215 above and paste it here below Append gall to all the index subscripts xt gauxt1 gaumovementt1 gan Xtaga11 xtil gall movementtilaga11 Assuming you didn39t get any errors let39s graph just the first gall gall 0 If everything went correctly this should give us a graph that looks very similarto the graph you produced in Part II From there we39ll graph all 30 galls Module03 15 JumpingGallsAnsweredxmcd NPE mm Mudme 3 Jumpmu GaHs be uw Vuu Hmma NnE anenuv That sOK chanee be hgeae Thatshuu d ge vvheh Vuu hke We Em he acme Vuuv mph shumd uuk sumemme e at We Ham tmstep 3m 5n xan U 5 nmescep Aaaumme thatwuvked wehwe e haw hke m pm the paths mu an eaua ah the samE mph umanunamw p umnu an hmea mme puma am am mph 5 numq m make the mph a We s uW Sn M H be he pm m hmmhe numbev av pmmswe pm Vuu may vecaH tha umestep 1 s rm s end 2 hrs mu a mn mquot 3 umestep Pmmsmv eaeh eaw Suvew we can make u Wuh mph 1 pumHuv evew SEEunds Anemth meanma uuv gaHgvapthHhave 32quot mmng 72x 13 pmmsnme same humhevuv Pmmswew peeh pxumhu Tu u WS WE need uhh mea e a new vamevanab e m eumym um mph Because W5 yahee vanab e Wm aHuW usm mpxay a subset umumme puma we u can u tsuh 3n and the aswa ue shumd be mquot umestep mp e mu i umestep MudmeUS Jumpmeoahsjhweveexmm NPE mm Mudme 3 Jumpmg GaHs Cupv Wm maph hum Emma 3 ms abuve ah Pasta A have be uw Cha ng m pmh the p and yraws m be sub msmad Nutethanhws duesn sub am aHv chahue Vuuv maph EVEquot huuuhwehe HEW uhw P u m mmh u 012 pmms Nuw change We y awshum 395 m be M the gvaph mm ppm paints and 2012 quotace mp2 Whenvuu ve h shumd mph sumethmn We the uhe ms ead Change hm hhes charm anE Vuuv map at the Ham behavmv uHhe gaH pupu a mn What u yuu 5227 h geneyah huw 312012 gaHs muvmg Wuh shu nee WE uhghhv h genmh n uuks We the gaHs ave muvmg pvugvesswe ymnhev away hmmheu uhgh a XU Huweveh many gaHs End up 51m nee XU a evlwu hams MudmeDS Jumpmgoausjhmveumm NPB102q Module 3 Jumping Galls Part IV Temperaturedependent Jumping Multiple Galls Goals Change the jumping rate of galls to be temperaturedependent Define how temperature depends on space quota sunny spotquot Graph gall behavior In the last Part you showed how a constant rate ofjumping leads to dispersion of the galls This dispersing behavior is analogous to diffusion the tendency of molecules to disperse from a area of high concentration to one of low concentration Note that this behavior doesn39t derive from any inherent quotdesirequot to disperse or get away from one39s neighbors It is simply the result ofjumping at a fixed rate and in random directions This Part will modify the above model to include temperature As mentioned before we will assume that thejumping rate of galls is d n m quott In hot thejumping rate will be high In cold temperatures thejumping rate will below When dealing with the temperaturedependence of rates it is useful to talk about the Q10 the quottemperature coefficientquot of the process Q10 is a unitless measure of how much a rate changes when the temperature is changed by 10 C For example if the Q10 ofa process is 2 then raising the temperature 10 C will double the rate Ifthe Q10 is 3 then raising the temperature 10 C will triple the rate To express ourjumping rate in terms of Q10 we will need three pieces of information 1 The Q10 itself Q10 We will assume this is 20 That is the galls will jump at a rate 20fold greater when the temperature is raised 10 C 2 The reference rate ofjumping to which we39ll compare all other rates refrate We will 1 assume this is the Julnpirate 110 value that we39ve used in preVIous Parts sec 3 The reference temperature that yields our reference rate reftemp We will assume this is 40 C 104 F Note that these values are just guesses We don39t have any experimental evidence to support these values Indeed we don39t have any evidence that the jumping rate is even temperature dependent But it would be surprising ifjumping was truly independent of temperature Assign Q10 below to be 20 Q10 20 Assign refrate below to be jumprate We39re going to change jumprate shortly so this will save our number for later use refirate jumpirate Module03 18 JumpingGallsAnsweredxmcd NPB 102d Module 3 Jumping Galls Assign reftemp below to be 40 C The C is a special type of unit in Mathcad and can be accessed in the quotCustom Charactersquot toolbar To access this toolbar click on quotViewquot in the top menu then quotToolbarsquot and then quotCustom Charactersquot it39s at the bottom of the list refitemp 40 C We39ll now changejump rate to reflect our new assumption that jump rate is now temperaturedependent Back in Part II you assigned jumprate to be a single number 10 sec Here we39ll assign jumprate to be a function jumpratetemp This means thatjumprate will no longer have a particularvalue but rather its value now depends on the input value of temperature temp temperefitemp Assign jumpratetemp below to be refirateQIO IOAOC The A C is a special unit that indicates a change in temperature Don39tjust use C here or your function won39t work right To get the A C quotunitquot click on quotInsertquot in the top menu then quotUnitquot In the Dimension list select temperature and then in the unit list select quotChange C A Cquot temrrefitemp 1 0 AOC iumR rate temp refirateQIO Be careful You are raising Q10 to a powerthat involves temperature temperefitemp rebateQ10 WC CORRECT This is not the same as trying to make Q10 a function of the same temperature term temp 7 refitemp refirateQIO WRONG IOAOC And this is not the same as mulitplying Q10 by the same temperature term ALSO WRONG t 7 ft ref ratteo w 10A C Module03 19 JumpingGallsAnsweredxmcd NPE mm Mudme 3 Jumpmg GaHs Tes yumjump rmuemmmmmn be uw kkMamcadwhatjump rzletd C 5 Van shumd get yum um vate mm Pans H and m Askwhatjumpirzle wcj 5 Van shumd get 21m 2n m msAWc J p l sec m mssn c 2n J p l sec Gwen mm Qm 2n yuu shumd cunvmce yuuvse hanhe abuve Yates ave tuned Name y yuu shuum 522 mum uHqug quanmy SZEI m w yuu dunmhen yum Juran 2 jump rzletlempj mummy s pmbamywmng Dun cummue um saw 255 nu shaderseekmn ehavmv uHhesE uaHs We H ammanw demde m h m 51 um m a mmgmpsyama avea ma 5 sunuunded bv a uw4empevatuve avea Ms mum cunESpund m a surmv spm an 012 mm sunuunded w shade Let s demde ma N s 1 1D F mthe surmv spa and shade 9mme c m M m N w mm m mam We ve nm As mm as m m n u asneeded N n N m tevms u1 F F b291 Eandthe aswa ue shuum b211 F Ana quotmp wwm um MudmeDS Jumpmgoausjnmvmmm NPE mm Mudme 3 Jumpmu GaHs Make a new maph be uw shuwmu jumpirzletlemm and lamp FuHhe vraws unms we mama n cunvemem 1 Im me am 2 cunsw emwnmhevateyuu m use u msee used m Pans H and m l su Shawna the xraws mm s a M emckv because quotF s a Specwa um Vuu can swmp v dmde bv quotF asvuu a uh uthevum s nstead vuu mus usethe nvevse Famequot e mrwmh uuks hke F and can be Dune m We cusmm chavamevsmu bav mu m5 1n temp F W2 l su Vuu shumd be ame m see m yum gvapmha he yam a Lump An c equaxs um Lute 1 m Tm s unewaym eunmm thaHhe gvaph s eunem nu pvubabwmy Because jump Wm jump prnh be a empeya meeepeneem unenen MudmeUS Jumpmeoausjnwevewmm NPE mm Mudme 3 Jumpmg GaHs AsswgnjumpJrnhtlempj wqu be m pvubabmtymat alump Wm uccm m a smgxenmemp Tm 5 OK Ee uw yum asswgnment ask Mathcad Wha jumpiprnhH WH 5 Van 5mm get an away u H mm 4m Wimp mpgmomp umestep Mumbonwn n us Make a gvaph u1jump7prnhtlempvs lamp be uw Be suve m use appmpnate um s 2n JmPJ WbUemp 7 m n 95 mn m5 1n temp F 2n Jumpjmbmmp m Wm a 12W sentences be uw abumwhat yuu 522 m yuuvjumpJrnh gvaph Huw dues tempevatuve mum pmbamm umump happemng m a smg e Hm2512p7 s m cunswsmm W m assumpnun 0m gaHsjump muvehequen y m hut empeyamasv Huw 5m 4m MudmeDS Jumpmgoausjnmvmmm NPB 102g Module 3 Jumping Galls Temperature increases the probability of a gall jumping in a single timestep Yes this is consistent with gallsjumping more frequently in hot temperatures lfthe probability of jumping is higher in hot temps then more jumps will tend to occur in hot temps We mentioned back in Part II that multiplying jumprate and timestep is only an approximation to the true probability of ajump occuring in one timestep Now that temperature affects probability it39s worth asking whether our approximation is still OK over the temperature range we39re interested in 90 F 110 F The answer is yes we39re fine The approximation gets worse as probability gets large But even at 110 F our approximation is within 4 ofthe true value Incidentally this is the reason that we used the tsub range variable in plotting the paths of the 30 galls in Part III As an alternative we could have increased timestep to 30sec But in turn this would have invalidated our approximation ofthejumping probability Let us now return to the design of our quotsunny spotquot on the ground We need a way to associate a certain location an x value with a temperature There39s a few different ways we can accomplish this Let39s decide to recreate our temp variable as a function instead This new function tempposition will use the builtin if function If position is between 25cm and 25cm then the temperature will be 110 F Otherwise the temperature will be 90 F Assign the tempposition function below to be if7250m lt position lt 25cm110 F90 F The square brackets arejust parentheses They39ll automatically change to brackets as you enter the assignment 411 tempposition if7250m lt position lt 25cm110 F90 0F To get an idea of how the tempposition function works let39s make a graph To make that graph let39s create a range variable that represents the positions we39d like to see Let39s NOT use x for this range variable Our gall positions are in x and we don39t want to mess with that Instead let39s create a range variable called position Assign position to be a range variable below The first value should be 100cm the second should be 99cm and the last should be 99cm 412 position 7100cm799cm 990m Module03 23 JumpingGallsAnsweredxmcd NPE mm Mudme 3 Jumpmg GaHs mm a gvaph uHEmp pnsl an vs n be uw Be suve m use appmpnate unns ml 7 temppasmnn PF mu 7 7 an ism U SE 1 n pasmnn m m um wsmn PF mn 7 an ism U SE 1 n pusman Wm a cuup e sentences be uw abumwhanhe abuve gvaph shquvepvesems m Tn have gvaph snuws huw empevatuve vaneswnh ucatmn Spem caHy We have a sunny 5pm between 725cm and 25m Wheve We empevatuve s hwgh w MEI F 0n enney swde uHha sunny spuHheve s shade Wheve We tempevatuve 5 mmquot F We mus abum madth Wuvk an 012 mwemenl and x asswgnmemsmnhe aevuyewen qn q n v m 2 WanHu uuk aHha data agam su M s pveserve n by asswgmng mm 3 Meyerquot nden Tn n newan we u anay quotdaunempeyamannuepe MudmeDS Junnmngoausjnmvmmm NPB102q Module 3 Jumping Galls Assign xind below to be x Xililld X With the temperatureindependent data preserved let39s turn to creating the movement and x arrays forthe temperaturedependent model As in part III the first step is to clear out the old arrays Clear out the x and movement arrays by assigning them each below to be NaN X NaN movement NaN Initialize your x array by assigning x0 gall below to be 0 gall 0 In part III your assignment for the movement array was movementt1 gall mom lt jumpprobjumpdistanceifmdlj lt U51 1 To make the model temperaturedependent we need to focus on modifying where jumpprob appears in the first Boolean test Specifically we need to replace jumpprob with our new functionjump probtemp However this alone will not work We still need to specify the temperature input forthe jumpprobtemp function This is where the tempposition function comes in That is we can use the output of the tempposition function as an input to the jumpprobtemp function 80 the Boolean test would look like this rnd1 lt jump jrobtempposition However we again lack some information Namely we need the position input that the tempposition function requires This we will get from the x array assuming the galls quickly take on the of their 39 quot39 quot e can use that particular gall39s position at time t1 In turn this would make the Boolean test look like this rnd1 lt jUIHPJTObtempXt71 gall That about covers the modification we need Unfortunately there39s one additional technical problem that we39ll have to deal with It39s probably best to demonstrate this problem by attempting to make the movement assignment You39ll get an error and then we can talk about why that error occurs Modue03 25 JumpingGallsAnsweredxmcd NPE mm Mudme 3 Jumpmu GaHs Cupy yum mavemenl i g asswgnmem mm Exevmse 3 us abuve and paste n have be uw Change Mumpm be Implpmppmpv gmn msmad Vuu ng an em ThestK WE H mm m a m mavement HM e md ltJumplpmbnemmx i vg nmJumpidnstance mam ltn514 4 1a mawmem i vgm e mm mumppmpp g n mpimpnee mm lt n 514 What eeesme enm say7 Why u yuuthmk yuu ve gemng W5 enuy7 M guesswstha hepmb emws Theenuvsays Thwsawaydesmvahdmnhwsana y m mma cundmunpusmuns aH vuw Hay thanheXanayun yhasunemwmdammuscdav e zems Because yeveyepees vuws beyund vuw these ave quotmam smmexa Vuu mav have amed uuHhat mm x anay un y has me vuw uMata m n 5 av Sn accessmg xH gmumywuvksmnhevevymshuwmx Aneymampeye sm muvevuwsmwuvkwnh mm x anay Em m mm We can un y ca cu ate mm x anay epee We mnvemem anay s eempme Tm cwcmav dapendence s a pmmem Fuvtunate vJMS s a pvemem We can su ve Pemaps We swmme semnep mp meuvpeme um mnvemem asswgnmem mm mm x asswgnmem That s theve s nu pameum veasun Wpywe have easwenu mm abum mnvemem and x as pemgme sepavate pmeesses Hewevempeye s nu veasunwhvwe can ca cu ate muvemem empew amendmu e1 Cupy yum x gmasswgnmen mm Exevmse 3 ms abuve and paste n have be uw Vuu H mmaHy get an em That s OK Cupy 012 new hand We e1 yum mavemenl i gmasswgnmemlmm Exevmse Hngbewaw md ltJumplpmb temmx i vg nmJumpidnslance md j and 42 XLgu Hm Lgm mm mumppmpp g n mpimpnee mm lt n 514 MudmeUS Jumpmeoausjnwevewmm NPE mm Mudme 3 uummnu GaHs yum gvaph mm Exevmse 3 ma abuve and pass u nsns be uw Gwe yum gvaph an appmpnateme m mama s was genevated usmg hetempevamverdependem mudg Tamperatln39cdepemleut assumated Wun um sunny spum q n nm Huwsysn unceme gaHs nsssn enhev edge nuns sunny spumnsy seemsm s1up mm m 255 Thws causesme susm accummate m the snsuy nsgnunsyus adjacem m the sunny 5th By We and an hvs vevyVEW gaHs sssm m be nssn yu MudmeUS uummnuoausjnmveuymm NPB 102d Module 3 Jumping Galls Part V Comparing the models Goals Compare the behavior of the temperatureindependent and temperaturedependent galls Count the proportion of galls remaining in the heat for each model Now that you39ve generated models of both temperatureindependent temperaturedependent galls let39s take some time to compare their behavior sidebyside Copy your graph from Exercise 421 above and paste TWO copies of it here below Change the title ofthe second copy to be quotTemperatureindependent and change the yaxis of that graph to This should recall your data from Part III and allow you to see it sidebyside be Xmdtsub gall cm with your data from Part IV Temperaturedependent tsubtimestep min Temperatureindependent tsubtimestep min Module03 28 JumpingGallsAnsweredxmcd NPB 102g Module 3 Jumping Galls Describe what you see in the two graphs above How are they similar How are they different Both graphs show galls dispersing from their start point at x 0 The movement of the tempindependent galls seems uniform That is the gall movement doesn39t seem to change at all as the galls move away from their origin In contrast the movement ofthe tempdependent galls is not uniform As soon as they hit the shade they seem to stay put moving much less Back in Exercise 102 you made a guess about whether the temperaturedependent galls would appear to jump towards the shade How accurate was that guess Does temperaturedependent jumping appear to lead to shadeseeking behavior on the part of the gall wasp larvae Explain your reasoning 503 In Exercise 102 I guessed that the galls would get out ofthe heat This seems to be the case The gallsjump a lot so long as they39re in the heat but once they get to the shade they stay put This behavior appears to be quotshadeseekingquot but in truth the galls have no idea where the shade is Theyjustjump until they find it the shade And then once they39ve found it they stay put In looking at the graphs above you may get the sense that temperaturedependentjumping leads to a higher proportion of galls getting out of the heat However there39s a lot of dots in those graphs so it39s a little hard to make any confident conclusions In particular it would be helpful if we could quantify this result For example through time what percentage of galls remain in the sun as opposed to the shade And is this statistic affected by the type ofjumping ie temperaturedependent vs independent You will answer these questions presently Module03 29 JumpingGallsAnsweredxmcd Mudme 3 Jumpmg GaHs we m v 5th mm vameshum 725cm m Om una s m mne igzllsidep and sunnyigzlls nd my new agamst tumestep man We yuu mu WNHWE swng ergaH xvecmv back m Pan u Thwswm awe Tha s new mpuHhe pmpumun m neacncasennnsnssnmpweuuauumm and hwswsm We mmmg Dune sunny 5pm Assng exemen 221 Dune sunnyigzllsidep and sunnyigzlls nd vecmvs be uwtu each be mm 5m sunnyigu sidepn mm sunnyigu simdn mm Tu ca cu ate We pvupumun uvgaus m We sun we Wm needm u VuuHhmgs 1 Use a Eun ean Es1m demvmmewhemenhe pusmun Math 33 a a pamcu amme s bemeen 725 m and mam m nm Thwsww awe us a 1 m Sunny 33 and a 22ch m shady H 2012 angeVaname Summatmn 0pmme sumuvaHgaHsaHhe1 s and shum me n We Range an ya 2 Us R Eun eantes Thws WM 2 us 012 mm numbev u gaHs m 012 Sun Vuu canh d Vanab e Summauun Opevamv m the Ca m us Tun bavwheve N uuks We 2 WM 71 undemea h N a 4 a n n m 012 Sun aHhanmeste A Du 52p51 thvuugh m Ahhuugh W5 5 a mumstep pmcess we can accumphsh n W a sma e asswgnmem sswgn We vemawmng exemens D39sunnyigzllsidep and sunnyigzllsi d be uw usmg and a Eun ean my Hm d in 3H umesteps A We Range Vanamg Summanun Opevatumhe 1 and gall vangevanab es mne tuned asswgnmen uv sunnyigzllsidep The Synmm asswgmng sunnyigzllsi s e m yuu S we Mm 25cm sunnyiga sidep mam MudmeDS Junnmngoausjnmvmmm NPB 102q sunny Agallsidep t sunny Agallsiindt Ask Mathcad to display your sunnygallsdep and sunnygallsind vectors below sunnyigallsidep 0 0 100 1 100 2 100 3 100 4 100 5 100 6 100 7 100 8 100 9 100 10 100 11 100 12 100 13 100 14 100 15 Z 7250m lt xt gall lt 250m gal total galls totalig alls sunny Agallsiind Z 7250m lt xiindt gall lt 250m gall 0 0 100 1 100 2 100 3 100 4 100 5 100 6 100 7 100 8 100 9 100 10 100 11 100 12 100 13 100 14 100 15 Module 3 Jumping Galls Checkthat both of your vectors are in fact vectors only one column Each should have several thousand rows The values within each element should begin at 100 and range from 0 to 100 throughout the vector you39ll need to append the quotunitquot to each array Don39t continue until you39re convinced this is true Modu e03 JumpingGallsAnsweredxmcd NPE Tum Cveam a maph heTew shewme hmh sunnv qzllsidep aha sunnyigzllsiind vs thmestep Be suve m use appmpnam um s Senhe hahsm yahee mu m 12H chahee the We uaee 5 13 W5 3 suhd hhe Wheh Vuu vE anE we maph shumd Teeh sumathmg We the gvaph emhe Ham Mudme 3 Jumpmu GaHs sunny my 2 6 sun usuam 40 7 2n 5m sun unnst 2n 5n sunny ns md 40 TempeyaTmeeepeheemTumphg Teaesm appavem shaderseekmg hehawmv 5m made The The shew Ted hhe shews gaHs hat have aTempeyamedepeheehuumpmg Tate The hme gaHs hat have a xadjumpmg Tate empeyamyeheepeheehg The empeya uyeeepeheem gaHs get um e the Sunny 5pm mueh mme Tammy And a evZ hvs ehTy abuutZEWa enhe empeyamyeeepeheem gaHs ave 51m m the Sun The cumva s he e1 wnh ahumam the empeya mehheepeheem gaHs 51m Tahgmshmg mhe heat The gvaph suppunsthe M23 ma empevatuverdependemmmpmg eadstu appavem shaderseekmg behawuv MudmeUS Jummheoausjhwevewmm NPB 102q Module 3 Jumping Galls References Russo R 2006 Field guide to plant galls of California and other Western states University of California Press Berkeley p156157 Module 3 Jumping Galls Written by Eric Leaver 2009 Martin VWson Last modified March 11 2009 at 245pm Modu e03 JumpingGallsAnsweredxmcd NPB 102d Module 3 Jumping Galls Help Improve This Module Completing this section is not required but will be helpful for future development ofthis module Your answers here will not affect your grade About how much time did you spend completing this module Were there any particular parts of this module that you found too difficult too confusing or too tedious How might these sections be improved Were there any particular parts that you found especially useful or helpful in understanding the material from NPB 102 Other suggestions Module03 34 JumpingGallsAnsweredxmcd NPE mm Mama 8 Catag vpmsH Foraging in Cataglyphis The Journey Home Part I Introduction 305 R ewew Camgyphrsmvagmg behawuv Am spemes uHhe genus CareyWm ave heaHu evam msymupmhs scavengevs mum m We day CareyWm s spemaHy adamed m Withstand budy empevamves up m smowmsh suuws w M succumbed m We heat CareyWm s u enmund neauveesswas12ands cumammg sw andmavks The CareyWm q Fm mm omy am geneva mags as a gmup aw ng down phevemunesm mavkme shunes path bsmssn We nss1 and a snnsu 01212 5 no n cuwuvkevsm me my ucauun me now vecuveved and Ham Newtheywuu d vecaH um am s umbuu path be usemHunhe mum va home a an vawgh me 12mm m 012 He Wm Careyw AS 5 heaHu evanL n WWW m cannm suvvwemv ung penuds umms abuvsgmum Funhemhenes seeks m mawmze ms vate mum acqmsmun n N w m my mscwmma mnmmum cuc e mwv 7quotme mummy mmmmmssmm rmmmcm WWW 2 Spam M s 39 mm m Wehnevand Wehnev 1990 5mm saw s Careywhwuvaama smmzzm 1 Once and smunm haw dues awuvkev 12mm hum27 2 What shaggy u Wuvkevs use m nd and m We m1 p ac27 Mudme catagwpmstAnmvswmm NPE mm Mama 8 Catag vpmsH 2 3n mm m ends vamev smupny mm m Inas oLre ducumemavy hus12d by Dawd Anenbumugh Czlzglygh39s Viden uuumssmkm ng Wm uvagmg Catag yphws vegu av y s1ups m check us unsmsnun mg dwedmn N s mmssy uw examy amswumd descnbe thewwsmns uHhe pu anzed sky By ana ugym smm hamwave engm such a wsmn uHhe sky rm m uukhke a may pmwhee ysmsuksmks un uvm hm hm We cummun y assuma e Wumks dayhme sky sss ame Whamvev heaven ywsmnsme ems may psyssys ms pvuwdes unsmmguuus mmymsuun abuunksu cunem unsmauun Funhev Catag yphwcan measumks I ov 1m 35 dws ance n hastvaw ed smss n 351 Ms pasnwmma Wm ask w mm mm m swam by WWW mm mm m evend m me my Wham m Wm use mm m p mummy VA1E mamaanunmm t y m1auma avExpeumemawamawzmmzz 2942 Catag yphws Eu huwduestmssys em Wuvk7 Andhuwduwe knuwms nawgauun sys12m Wuvks w W Way7 Tksss ave sums uHhe queshuns yuu H EXp uVE m MS mm s Mudme catagkpmstAnmvsukmm NPB 102g Module 8 Cataglyphis Part II A Complete Solution Goals Review output of guassianbased turn model of Cataglyphis foraging Calculate complete solution for homeward journey Later in this module you will see that navigation system of Catagyphis is not entirely accurate It is only an approximation ofthe complete solution for the homeward journey Bad for Catagyphis but good for us Examining the errors that ants produce allows us to infer the mechanistic details ofthe navigational system However before we examine the errors of Catagyphis we must spot them That is we must know the complete and accurate solution before we can understand the approximation calculated by Catagyphis We don39t have a lot of real data on outbound Cataglyphis foraging so we39ll use a model to generate some quotdataquot The regions below define a Gaussianbased turn model of outbound Cataglyphis foraging You may have seen this model in an earlier module but if you haven39t don39t worry We39ll take the time to examine it now speed 051 end 5mm sec 0 10deg timestep lsec X01 0 y0 0 or1ent0 n1nlf17180deg180deg0 end 1 1 2 7 ttmestep turn Inorm 0deg0 ttmestep A X1 1 X171 speedtnnestepsm or1entlil y y1 speedtimestepcos orientiil 1 1 orient orient turn 1 17 1 17 1 A odule08 Catagyphisl Answeredxmcd NPB 102g Module 8 Cataglyphis II This is a discretetime model of Cataglyphis foraging over a period of 5 minutes The model tracks the path of a single quotantquot using three vectors There is an element in each ofthe vectors for each timestep ofthe model intervals of 1sec The first two vectors x and y contain the spatial coordinates ofthe ant at each timestep The nest is assumed to be at the origin 00 The third vector orient contains the ant39s orientation at each timestep In the orient vector Odeg is assumed to be due north 90deg to be due east 90deg to be due west etc Ask Mathcad what x y and orient are below While both the x and yvectors will have units of m meters the orient vector should appear unitess The orient vector contains angles and by default Mathcad treats these in terms of radians Change the displayed units for orient to be deg This will display orient in terms of degrees which is how we39ll work with angles in this module Scroll through the model39s output Each vector should have 300 elements indices 0 through 299 201 o o o o o o o 0 179543 1 3985103 1 05 1 172 2 0074 2 0995 2 15619 3 0275 3 1453 3 135452 4 0626 4 1809 4 132168 5 0997 5 2145 5 137247 6 1336 6 2512 6 12484 x 7 1747In y 7 2797 m orient 7 124134 39ng 8 216 8 3078 8 110686 9 2628 9 3255 9 117245 10 3073 10 3483 10 11978 11 3507 11 3732 11 132223 12 3877 12 4068 12 126757 13 4278 13 4367 13 127065 14 4677 14 4668 14 128035 15 15 15 The spatial coordinates of the ant are all we need to accurately calculate the ant39s homeward journey However before we do that let39s make a graph ofthese two paths ie outbound and homeward odule08 Cataglyphisl Answeredxmcd Mama 8 Catag vpmsH NPE mm Cveate a new mph be uw p u mg l 333 Duub erchck uh 012 mph and theme the am sMem be crnssed Wevev because W5 5 w uhhheh Wu u ve HDHE vum mph shuum uuk sumethmu 1 When w We the mph aHhE mm Hu m h sb 52d Em yahqu numbev ah pamwwuukexad ythe same 2m OK veamycheck VVhatduesyuuvgvaph yepyesem Whateveyuu p umng39 Whamuthetwu empth c1me sqmgg y hnE yepyesem 2m vepvesehmhe nestwheve We am when The mh a sumevuu 7 Thws s mudg am paw u uu buund magma m p u mg the am u w eryde h W 212012 am smpped a ev mmu Es maybe Muuh a Muumena cataghph smhmvewmm NPB 102g Module 8 Cataglyphis ll Curiously plotting the return home is actually easier than calculating the orientation that our quotantquot must adopt We39ll do both eventually but for now let39sjust do the plot To create a plot of the path home we39ll create two new vectors xhome and yhome Each vector will have two elements and have units of m meter The first element will be the coordinate of where the ant ended its foraging trip The second element will be the coordinate ofthe nest Because each vector contains only a couple elements We39ll make each of these vectors by hand ie we don39t need a range variable Let39s think about this for a minute What is the xcoordinate ofthe quotantquot at the end of its foraging trip What is the ycoordinate at the end of the foraging trip Write each ofthe numbers below AND describe how you got them Of course you could inspect the vectors from Exercise 201 and xlastx 5688m ylastlt 12566m copy the last numbers here y However here are some alternate ways of arriving at the same Xrowsx71 5688 m yrowsy71 1239566m answer There are a variety of ways to get Mathcad to return the last value in the x and y vectors One way is to use the builtin last function This function will return the last index of any vector Once you combine this function with the subscript operator you have a handy way of determining the last value of any vector Ask Mathcad what xlastx is below You should get the same answer as you did in Exercise 204 above 205 Xlastx 5688m X1 t Now create you xhome vector by assigning xhome to be as X You39ll need to create the 0 vector using the insert matrix dialog box This can be found in the Vector and Matrix Toolbar or alternatively by pressing Ctrlm After your assignment is complete ask Mathcad what xhome is Make sure Mathcad returns numbers with units that make sense X 5688 Xihome laStX Xihome m 0 0 206 odule08 Cataglyphisl Answeredxmcd NPE 1mm Meme 8 Catau vpmsH The ass gnmEnHm yihume Ts made swhahy The exact SyMaX Ts e e yuu yhsm W 12555 m n T n cepv Wu maph hum Exevmse 2 n2 and paste m heve heTuw Add an addmuna mm u Umquot m We yeah y and mm m the mm L m w vvheh vuu ve anE VDUY maph shumd uuk sumemmu m We the maph anhe hem Huwevev because he maph Ts based Em vandum numbevs n s uhhheh Wm am paw Wm Teeh exac ythe same Wme a eeupTe semehees abuutwhat yuu see 7 yum gvaph Du the heme vecmvs T e xihnme aha yihnme appeaHu vepvesem the tuned and emehesn YDU E hum27 2m The heme vemuvs made a NEE mm hemhe as cuuvdmate e We anHD n hes1 anhe u gm up Ves hTs Teehs hke 0T2 EDHEEl YDU E heme Muemena cataehphTsTLAhwevewmm NPB 102d Module 8 Cataglyphis ll At this point it may seem like we39ve come up with a simple solution to our ant39s navigation problem That is so long as we know the ant39s last coordinate we can easily mark a straight line back to the nest Unfortunately this impression is wrong fortwo reasons 1 Even though we know our ant39s position explicitly it39s not at all clear that members of Catagyphis have access to this information In fact experimental data suggest that do not While it39s clear that real ants keep track of their orientation and distance between orientation stops we should not assume they use this information to calculate position There are other ways of navigating that do not involve explicit knowledge of position 2 As we shall see shortly even if we know our ant39s position calculating an appropriate homeward orientation ie which direction the ant should start walking requires trigonometry Calculating a homeward orientation Let39s say you39re foraging in the desert And you39ve just stumbled across some food You know your campsite is at coordinate 00 and you know your present coordinate is 5200m1000m and you can39t really see your camp Further let39s assume that you can tell by looking at the sky or a compass or whatever what your absolute orientation is eg whether you39re facing due north Odeg or any other orientation angle Which orientation should you turn to and start walking Kind of hard eh You know you could walk 5200m due east 90deg orientation and then 1000m due north Odeg orientation that would surely get you home But alas it is not the most efficient route What you really want to do is cut the diagonal Let39s think about this for a minute to try to get some intuition about the proper solution In the example above 5200m1000m both yourx and y coordinates are negative You may not know the orientation you should take yet but what are some reasonable bounds for that orientation That is forthis orienthome value what value must it be greaterthan What value must it be less than Hint It may help to grab a piece of paper and make a little drawing You don39t need trigonometry for this answer DoubleHint It may be helpful to think in terms of 180deg to 180deg being the full range of angles you can consider That39s not the answerto this problem But it may be easierto think using that range as opposed to a 0 to 360deg range 210 orienthome must be largerthan 0 and less than 90deg What if your x coordinate is still negative but now your y coordinate is positive What are the bounds of orienthome then orienthome must be larger than 90deg and less than 180deg odule08 Cataglyphisl Answeredxmcd NPE mm Mama 8 Catag vpmsH miem hnme th2n7 212 unemjume mus1 be essmanra deg and gyeamnhamameg And what w x s pusmve but y s negaw What ave m buunds u nrienLhnme henv m unemjume mus1 be essthan deg and gveatenhan ra deg u HEEIdeg Waugh nu necessava m uvdev Dam cummue unm yuuVe gunha Wuvked um Trigonometry Review mgunumeuy Let stake a mmmem vewew Wm yuu need m knuw H s aH abuul quotgm mang es AH Hgmmang e have WEI ang esthat ave essman 9mg Fm swmphmw WE H EDHSMEY unw une unmg ewesquot a weor pv meme a peyspecwawe can assmn namesm me me was mum mang e 1 vamm m dwanuna me 2 OPPUSME bemu Uppusne 8 3 Amacem m vemawmm SME amacennu a and NOT me hvpmenuse Based an m anamementwe ve mtevested We uHuwmu Umnumem dehmtmn 0 am 1mm Whmh can be mampu ated m sum m a usmn m mvevse amemmnmmn amen m oppmu Adjacent HM mm y w Inan Fm nuw M s Dans un dvawmq a mmmanme sucmhat ommsue cunespundstu any x cuuvdmate and Amazem cunespundsm um y cuuvdmate Muumena catagwpmstAnmvewmm NPE mm Meme 8 CatauWPms The ame 3 WM shews heme hhe up yuuvmang e m ca cu ate a M012 ease Wham yuuHEYmma cumdmam 5 Nest hegame m bu h X she y Thmk abuuths eahemuy h hh yuwe ODDOSWe 0 0 cump eted the 015 Catag yphws mudu e eh emheuhe magma H212 We do human anhe quot25 nsmad we Stan Wham ewmagmg EndEd w hupemHyWheve We semev uuhe nee ohemaneh umdeg 5 assumed m be due h Su Wuuv X she y EDDYdma ES ave hmh negawe WEquot he Adjacenf ave a su negawe Vemhe vatm enhese va ues 5 pusmve she 5 We shumd amve Nest at a pusmve va ue em 77 sumewheve bemeen u deg she 0 0 wueeg And m M s case a s he same as um nrienLhnme aheh 5 shhhahwheh X 5 pusmve hm y 5 sun hegame h hhs case 9 Wm he negawe hm that s OK 9 5 sun equwabn u n enLhnme Nme that m 3H eases he hypmehuse Mum Ham mang a eehespehesm the ms1ahee e the quot25 We Wehheeheeyhemseheswuhmsheye am e eemse 01 s 5 sh whensm bu e1 m39 vma n she eah he Db amEd usmg the Pylhaguvean heuvem 4 and 1 We can use M S D tesHhe hmmuh yuu devemped m ExevmsesZ 1n hvuuth 13 ASS gn nrienLhnmem be m Whamhe uHheluuvsnuatmnsduesth scunespund c Du We ahswevmv unem heme make sehse7 m anenLhnme 1 TWS cunespundsm when x and y ave hmh 4 hegame And yes thwsanwevmakessense The anWEY 5 M de Whmh 5 bemeen n she anenLhnme e 45 deg 9mg 9 Muemena cataehph smhwevewmm NPB 102d Module 8 Cataglyphis ll Assign orienthome to be What one ofthe four situations does this correspond to Do the answerfor orient home make sense 215 1 This corresponds to when x is positive but y is W negative And yes this answer makes sense The answer is 45deg which is between 0 orientihome 745deg and 90deg Assign orienthome to be atan What one ofthe four situations does this correspond to Do the answerfor orient home make sense orient home am lj This corresponds to when x and y are both 1 positive But no this answer doesn39t make nse The answer is 45deg but should be orientehome 4539deg somewhere between 90deg and 180deg Assign orienthome to be atan 71 1 What one ofthe four situations does this correspond to Do the answerfor orient home make sense orient home atan1 This corresponds to when x is negative but y 1 positive But no this answer doesn39t make sense The answer is 45deg but should be orientehome T4539deg somewhere between 90deg and 180deg You should have found that the statement Opposite Adjacent only works when the y coordinate is negative Indeed because we39re examining the ratio of Opposite Adjacent we loose information about their absolute signs However that39s not a disaster We can always examinethe sign of y directly and add that information back in Wejust need to figure out how 6 is related to orienthome when y is positive orientihome aunt odule08 Cataglyphisl Answeredxmcd NPB 102d Module 8 Cataglyphis Reexamine your answer to Exercise 216 above What answer did you get What answer do you expect check your answer to Exercise 212 above What might you add to to get the answerthat will take you home You might find it helpful to play around with your calculations here below to try to come up with an answer 217 atm lj 45deg get 45deg but I really want an answer 1 of 135deg I can get this by adding 180deg to atan1 that gives me 225deg but that39s 45deg 180deg 225deg equivalent to 135deg 225deg 7 360deg 7135deg Do the same for Exercise 217 above What answer did you get in Exercise 217 What answer do you expect check your answer to 213 above What might you add to to get the answerthat will take you home Is this consistent with what you added in Exercise 217 above It should be There are multiple solutions so think about how your answer here can be made equivalent to your answer in Exericse 217 orvice versa 213 1 get 45deg but I really want an answer of amn j 45deg 135deg I can get this by adding 180deg to atan1 Yes this is consistent with my addition in 745deg 180deg 135deg Exercise 2 17 odule08 Catagyphisl Answeredxmcd NPB102q Module 8 Cataglyphis ll You should have found that when y is negative orient home can be determined by xihome0 orientihome atan for negative y yihome0 And if y is positive then we must use thefollowing instead Xihome orientihome atan 180deg for positive y LyihomeOJ To write a general assignment for orienthome we must test whether y is greater than zero or not And then based on that Boolean test we must add 180deg if the test result is true and add nothing if the test result is false There are a few different ways we could do this Perhaps the simplest method is to take advantage ofthe fact that for Boolean tests Mathcad returns a 1 if the result is quottruequot and a 0 if the result is false That is the statement yihome0 gt 0 1 should be returning a1 or 0 This is entirely dependent on your model39s output Go backto your graph in Exercise 208 Does the quotantquot end its foraging where y is negative or positive Is this consistent with the Boolean test displayed in the tan section above Yep that39s right The end of the path is negative and the boolean test for ygt0 is returning 0 false Knowing the two different calculations you must make for orient home listed in the tan section above and using a Boolean test create an assignment below for orient home that will calculate correctly regardless of whether yihomeo is negative or positive 220 xihome0 orient home atan yihome gt 0180deg NVWWVVWWVVWlA yihomeo 0 Ask Mathcad what orienthome is below Does this answer match your intuition eg your answers in Exercise 210 through 213 Write out the number you get for orienthome here because we39re going to change it shortly 221 Yep that looks right I get 1362deg My ant is directly below the nest due south from the nest so this heading seems appropriate for the return home orientihome 204354deg odule08 Cataglyphisl Answeredxmcd NPB 102q Module 8 Cataglyphis Make a new copy of the model here below by copying all the faint yellow regions from the beginning of Part II this Part and pasting them here below Also copy your xhome and yhome assignments from Exercises 206 and 207 and your graph from Exercise 208 Finally copy your orienthome assignment and display from Exercises 220 and 221 222 mg 051 m 5mm sec W IOdeg lsec X01 0 y0 0 oriento Iunif17180deg180deg0 1 2 end 1 timestep d rnorm en Ode 0 m g timestep xi 1 X171 speedt1mestepsmor1entii1W yi yii1 speedt1mestepcosor1entiil orient orient turn 1 171 171 Xihome X1a5tx yihome ylaStY 0 0 Module08 Catagyphisl Answeredxmccl NPB 102d Module 8 Cataglyphis ll Xihomeow home gt0 180de LyihomeOJ y 0 g mm m orientihome 57327deg Examine the answer for orienthome you get above Exercise 222 Is this consistent with what you see in the graph above Exercise 222 Scroll up to the turn assignment also in Exercise 222 click on it and press F9 This will generate a new ant path Check again that the orienthome result makes sense Do this several more times to ensure that you have the correct and general solution for orienthome Make sure you run the model enough times that your quotantquot ends up in each of the four quadrants at least once Don39t continue until you39ve worked out all the problems with your model Write a few sentences about what you did in this Part What are the problems that we faced in determining orienthome How did we solve them Graphing the route home was pretty easy wejust grabbed the last point from the x and y vectors output by the model However figuring out what orienthome angle this corresponded to was kind of involved First we had to arrange a right triangle and compute the arctangent to determine our home heading But frustratingly this strategy only worked when y was negative For positive y39s we got the incorrect answer 80 we had to include a Boolean test in the orienthome assignment to determine whether we needed to add 180deg to the answer or not Finally this generated the correct answer odule08 Cataglyphisl Answeredxmcd NPB 102q Module 8 Cataglyphis II Part III Real Ants don39t do trigonometry Goals Review the navigation strategy of Catagyphis Understand how navigation errors can be experimentally measured Implement a model ofthe approximate solution used by Catagyphis Having worked hard in the last Part to completely solve the quothome navigation problemquot it may surprise you to learn that Catagyphis who does this sort of thing for a living literally does nothing of the sort It39s true that Catagyphis navigates home quite well but experimental evidence indicates that Catagyphis only approximates the complete solution There are several ways to view navigation One way which we have so far adopted is a geocentric approach That is once food is found we check our xy coordinates and use trigonometry to determine a homeward heading We say quotwhere in the world are wequot And then compute how that relates to where we want to go ie home Cataglyphis appears to adopt a different strategy one that is egocentric In this strategy positional information is discarded entirely Instead the ant carries what might be called a quotmental threadquot that it attaches to the nest when it sets out to forage As Catagyphis forages this mental thread is kept taut at all times As the straightline distance from the nest increases the ant lets out more thread lfthe ant gets closer to the nest the ant instantly takes up any slack in the thread This mental thread strategy is also known as quotpath Integrationquot Maintaining the taut mental thread gives Catagyphis a simple way of returning home It just follows the thread home no trigonometry required But note while this thread provides easy access to the homeward heading it does not explicitly trackthe ant39s position although in principle one could use trigonometry to infer position from the thread This is probably suits Catagyphisjust fine It39s not clear the ants ever need to know their position in absolute terms Exactly how this mental thread strategy is implemented in the brain of Catagyphis is not known However Cataglyphis makes reproducible navigational errors under experimental conditions These errors lead to predictions that any mathematical model of Catagyphis navigation must meet You will implementjust such a model in this Part The navigational errors of Catagyphis fortis were revealed by experiments shown below Ants walked from their nest N to a food source F but their movement was constrained to a path determined by two connected channels 81 and 82 with lengths of 10m and 5m repectively The ants could see the sky and presumably navigated as usual After reaching F ants were quickly removed to a different location and their freerange movement was taken as an indication oftheir best guess of orienthome Of course this guess was totally wrong because the ants were removed to a location far from the nest However if the ant39s guess was taken in context ie imagining that the ant had not been removed a systematic navigational bias was revealed This error a in the ants39 navigation was dependent on the angle a at which the two channels were joined The black bars indicate histograms of individual trials showing some variability across different ants odule08 Cataglyphisl Answeredxmcd NPE mm Mama 8 Ca ag vphws u S r by r 1 museum veumems eacMswvv heams vesvamesmasmv emmanv e u rmmwghhua shhhmmmz h mememmwspam Helmwur m Je erv mu MuHevLVIehnev gm Vmcedmvs v he thmemm mm USAS55757 52m The gvaph at Ham summanzesme YE a mnsh p bemeen EharmE ang E ahu the nawgatmna enuy ang E mspxayeu by 2 ems ave b m on the amsv Th ES aHMEng k quot25 uca mn when u 5 why 3ng 2 g a a mum uvsmaH2 g dengmpuuva 3 20 myme ang as g Cuhuuswhe 12mmst bemeen u and y g 10 5 skewed such thanhe mus1 gnevuus w enuvs uccm m the 137132 12mm 7 15am on Lam m M s mudu e yuu Wm vepmduce B a a thws guve usmg a mudg unhe 0 5 20 180 Camgypms appmwma mns Huwmh I angle ca m a mg enuvs 5 atwu stEp pvucess quotE d huh hagmxmmm mm m 412 mum nrie ndevthese E ndmmvs and 9mm mmquot quotLhnme u WEquot cumpave Mann the 251mm genevamd hum the Catag yphws mudg s vawgh urwavd Tu use the mahmques yuu devemped h Paihwe musmm de evmme the w mama We Hduthwsbycveatmg Wu HEW mama xichznnel ahuyichznnel CVEa mg 1252 s hm Mam hm h the mevests unmeJhey have been cvea edluv yuu be uw Muumena cataghpmsmhmvewmm NPB 102d Module 8 Cataglyphis ll aistep IOdeg sl 10m 180d j 01 eg 52 5m aistep 01 j OListep Xichanneloaj 0 yichanneloaj 0 Xichannellaj 0 yichannellaj sl Xichannelzaj sZsin01j yichannelz j yichannellaj sZcosocj The above regions have created three new arrays 1 xchanne 2 ychanne 3 11 Ask Mathcad what xchannel is below o 1 2 3 4 5 6 7 Xichamel o o o o o o o o o m o o o o o o o o 2 o 0868 171 25 3214 383 433 You should see that unlike the x vector from Part II xchannel is a 2 D matrix Further unlike the x vectorwhich had 301 rows the xchanne matrix only has 3 rows Your assignment for orienthome from Part II requires only a single terminal location That is given any position xy your orienthome assignment will compute the correct heading to the origin 00 The experimental channel setup for any given value of a can be described by 3 positions 1 The nest N 00 2 The quotelbowquot 0s1 010 m 3 The food source F which requries trigonometry to calculate Those three locations are represented by the three rows in the xchanne matrix Note that we need not make any assumptions about how fast the ants traverse the channels only the final position is important odule08 Cataglyphisl Answeredxmcd NPB102q Module 8 Cataglyphis ll Each column in the xchanne matrix represents a different value of d the amount of bend at the quotelbowquot of the two channels The correspondence between a column in xchanne and d can be determined by examining the a vector Ask Mathcad what u is below You can find the a character in the Greek Symbols toolbar or alternatively by typing actrlg Display the vector in terms of deg 0 1 2 3 4 5 6 7 8 9 You should notice that d has 19 elements ranging from 0 to 180deg Notice too in Exercise 301 above that xchanne has 19 corresponding columns Ask Mathcad to display ychannel below You should see that it is organized similarto xchanne 303 0 1 2 3 4 5 6 yichamel 0 0 0 0 0 0 0 m 10 10 10 10 10 10 10 2 15 14924 14698 1433 1383 13214 19 odule08 Cataglyphisl Answeredxmcd Mama 8 Catag vpmsH we mm mm a new gvaph be uwmat shuwsme yichznnel vs xichznnel when 9qu Tu u thws yuu H needtu use m man upevatuvm se ed a specm cum mm each mum yichznnel and xichznnel mamces y f Whmh cu umr We L uuk m We nvecmy Sam m we m be crnssed Am 012 m mum gvaphmhym makeme x7 and y awsscdesswm av m 2 3 A 5 When yuu ve dune yum gvaph mum uuk sumemmg hke xichanmfw m m yiclunnz w E tuaHy We H make a gvaph shuwmg haw nnenLhnme s mated m aHva ues u n Humveme nuw 2 sjus1wuvkwmh m une mam u Eldeg nPavHL hnmeanuy am Lam yuu emactedme m1 exemem Bleach umese mmva use m yum n39 nlihnm asswgnm m Hm Wewun t mm mumng me humewavd mg 5 We dam need m mam xihnme and yihnmevecmvs ns ead we H cvea etwu newvanamesmaum cumam me 351 x7 and yrcuuvdmams mum expenmema am path WE H caHmese newvanab es um and yilasl H Um wqu be xichannzlz g Thwswm emac he as x cunvdma e m cu umnB Wham n Asswg B deg u1xichznnel m5 Um xichannzlzvg mm a svm av asswgnmen uvyilasl be uw ma yJasL Unarmng Muumena 2n catagwpmstAnmvewmm NPE mm Mama 8 Catag vpmsH Nuw cupy yum nrienl hnme asswgnmem mm Exevmse 2 22 and paste n have be uw Muuwym mm hand We make vuuvnew x 351 and v amanames mm accuum That 5 theme A m be x m jygmgtn Wag m shun X 155LgtU BUS um mg L gt a 39 hnme s ExevmseZ mmvuugh 213 abuve7 m Vesmsmakessense nExemse212xsamma anenLhnme ms 555 deg unemjume 5mm be between Meg and VWEEIdeg mm x and y ave pusmve Hm unen jume 5 2m 55mg thch 5 2 way m 453 435w su anenLhnme 7 36Udeg 452 425 deg W 5mg OK a Van mm m an away u 2ue SESdEu whmh s Euwa emm 2cm S Sdegr 36Udeg 152 425 deg Dam cummue unm yuu gum Graphing paths for all values of 1 Wm aHva ues um Fm M s make a gvaph mwha aumuse exp enmems uuk m cm Wm maph mm Emma 3 m abuve and 23512 n have be uw Tu shuw awe new swmpW y cm x cm vemuve m man upmmysnum bum and any yiclunnzl U12 3 A 5 mm Muumena catagwpmstAnmvewmm NPB102q Module 8 Cataglyphis Now that you39ve cauated orienthome for a single value of a calculating all 19 values at once is not much harder You39ll do this in three steps 1 Recreate xast as a vector containing 19 elements values from row 2 of xchanne each value corrsponding to a different value of a 2 Recreate ylast as a vector containing 19 elements values from row 2 of ychanne Recalculate orienthome which will now be you guessed it a vector of 19 elements 0 To do this we39ll need a range variable to step through all 19 indices of the various arrays Conveniently such a range variable already exists j This was created at the beginning of this Part to create the u xchannel and ychanne arrays Confirm below that really ranges from 0 to 18 in integer increments Will it worK 1 Yep this will do fine Use the range variable below to recreate xlast as a vector containing the elements in the last row of xchanne 311 W Xichannelzaj Recreate ylast similarly x yichannelzaj odule08 Catagyphisl Answeredxmcd NPB 102q Module 8 Cataglyphis ll Copy your orienthome assignment from Exercise 307 above and paste it here below You shouldn39t need to modify it to accept the new xlast and ylast vectors 1 t orlent home atan X as yilast gt 0180deg ANVWVWVWWWM yilast Display your Xlast yast and orienthome vectors below Are these what you expected How so 314 0 0 0 0 180 1 1 1 18333 2 2 2 186636 3 3 3 189896 4 4 4 193082 5 5 5 196165 6 6 6 199107 Xilast 7 yilast 7 orientihome 7 201862 deg 8 8 8 204374 9 9 9 206565 10 208334 11 209543 12 210 13 209441 14 207516 15 Yep that looks right They all have 19 elements The first quotrowquot is 015m with a orienthome of 180deg That39s correct when a is Odeg And the last quotrowquot is 05m with a orienthom of 180deg That39s correct as well when a is 180deg odule08 Cataglyphisl Answeredxmccl NPE mm Meme 8 Catag vpmsH CYEa E a gvaph u m 39 mm vs 1 be uw deg deg D sp av hhes and 5mm Wheh Vuu vE anE Vuuv maph shumd uuk sumathmghke hws s m 5n mm 15m 2 dz 3m 2m dz no wu 5n mm 15 n 62 0K veamy eheeh What ave yuu mumhg heye7 What duesthe hm shape EDYYESpDnd m7 EIHEHLHEINEV depends Em the ang E a whmh the EharmE s avejumed u Wheh u 5 senu Eldeg m WE deg ems swmp ywa k du h nh and 012quot due 5 ca e u mhwumueeg nenhev 52 the EDHEE uhemahuh heme 5 Meg 77 due sumh Em a humewam EINEMaHEIn that 5 gveatenhan 18ng a sumhwesuy YEWquot heme Muumena cataghpmsmhwevewmm NPE anzg Magma a caaaaavphysy Part IV A model of Cataglyphis navigation Goa s cyaaaa aghaan aa ayahaaaaa ahgaaaaa a uhygua yahga VWEEldeg aa aaugag cyaaaa HEW magaa aa Caaagayphya hayhawayg appyahyhaayah a paah Magyaayah magaa wa u can ya aha paah yhaagyaayayy magaa Thya magaa ghaayaayayh aha campaaaa aaauayah magaa yuu ayaaaag yyy Pan H yh ahyaa Ways Tha paah Magyaayah magaa ya ah appmwma mn ahg haa ahayyaay aaauyaaa 2 Tha paah Magyaayah magaa yhyyhyahaaha hayygaayahaa ayyaya aayaaa ahaa 3 Tha paah Magyaan magaa yghayaa aha Xry aaaygyhaaaa aa aha aha Thya aaaa yaayh yayaayaaaaaha paM maga yh Fequot m ahaa Caaagayphya appaayaaa navagaae by use my a yhahaaa ahyaag aaaaahag aa aha haaa Navagaamg ya gaha ahayyaay by mahyaayyhg aha aahgah ahg ahgaa aaahaa ahvead ubwatmg ahy haag aa EXp cMy cumpme pusmun Vuuv gaaa yh ahya Pan yaaa uaaaha paah Magyaayah magaaaa yapyaguaa aha enuvs a gahayaaag by yaaa ahaa yh aha ahhaa EXpEHmEM whah aha awn ahahhaaaayaaayhag aa ES EL N V H H 7 Meg SEE ame ha yyyyayaawuyh ya aheadv gayya any Vuu haaaw Thaaa yagyaha ayaaaa Muvecauvsw Ium ahg nr39 a Fm HDW WE H aha ha aahaayhag Wyah aha aaaa whaya gag y a a a a 9n gag ah PavaV yuu H Expand aha magaaaa yhaauga ahaauu yahga aaaha nvecauv m Na mam a Na sa a a Bus a mum Md mm m and may Maguaana caaaahphysaajyywayagxyhcg NPB 102d Module 8 Cataglyphis ll turn 51 9 m 0 speed timestep orient orient turn 1 171 1 Ask Mathcad what orient is below Change the displayed units to deg The orient vector dictates the orientation the quotantquot must make when moving through the twochannel experiment Due north is assumed to be Odeg orientation Each 100ms timestep ofthe model is presented by a separate element in the orient vector As mentioned above we39re starting with element 9 ofthe a vector which is to say that the two channels are joined at an angle of 19 90deg Scroll through the orient vector and write a few sentences about what you see Does this accurately describe how ants will be moving through the twochannel experiment How so 51 speed It will take the ant 200 timesteps to reach the quotelbowquot where the two channels are t1mest joined How many timesteps will it take for ant to exit the channels That looks about right The ant walks due north until element 200 at which point the ant turns to an orientation of 90deg due east Finally the ant exits the channels after element 299 5152 d 39 orient Ispee 3 00 rowsor1ent 300 timestep It takes 300 timesteps for the ant to exit the channels 26 odule08 Cataglyphisl Answeredxmcd NPB 102d Module 8 Cataglyphis ll Before moving on to the path integration model we have one more issue to address One part of the path integration model involves converting angles to an output range of 180deg to 180deg This is not strictly necessary for the twochannel experiment because a is contained within this range However you39ll later use this model on quotfreerange antsquot that aren39t similarly constrained Let39s anticipate that step and introduce a angle converting function Without giving it much thought let39s call this angle converting function g The function itself is a little hairy so it39s simply provided here for your use 9 7 Trunc9360deg 9 97T 9360d itr lax Iunc eg unc lgodeg j360deg The 9 function accepts a domain of all real angles 9 and translates them into a range of 180deg to 180deg For example given an input of 360deg this function will output Odeg Of course these angles are the same the function simply converts redundant angles to a single unique specification Make a graph of the 9 function below Perhaps the easiest way to do this is to make a quickplot using an unassigned variable However if you39d preferto use a range variable that39s fine too Because the g function is discontinuous just plot points Explain what you see in your graph Does this fit the description of 9 above How so 21 73360deg73360deg 10deg3360deg I 100 g22 7 100 UU I I 7 1x103 0 1x103 zl 22 3g 3g Yep this fits The no matter what the input g always returns a value from 180deg to 180deg Further for 6 values of 180deg to 180deg g6 6 eg g0 0 So that fits as well odule08 Cataglyphisl Answeredxmcd NPE mm Mudme a CataEWPmsH mva u vm n m N ma yum humewavd ca cu auun based su e y enme Vma pusnmn e1 yum am The 5 WWW have knuwstwu thmgswnh absu me accuvacyw ds enemsnen and us speed WE H assume mm knuw edge e1 ds speed s mam hvuugh seme pmpnucepwe mechamsm And me enemsnen mmymsuen enmesvyemme am s penudmwews e 012 Sun and sky pu anzatmn N m mm m m N mm x 5 no vandum Fm exammg mv El Va uEs bemeen u and Ended e s a ways pusmve w nave nedsme hm N n m n thew marv a Weed attached e We quot25 w an and e We enemsnen uHhe Weed quotewe am avuund me quot25 A dnds u my he endm uHhe tweed am bemeen the am andme nes1 a an ang e me dwveyenee be weenme am s enemsuen ssvednd m driem and w 2 3 Schema mam mtegvatmn medex Am spath smdmated by yed wnh 12d deds vepvesemmg pusmun a each mes1e ThememaHhvead s h A s orient D vepvesem dde quotmm mded enemsuen n mds1 be emphaswzedma he ang e w and ength A uHhe am s memaHhvead s an an accuva e depmtmn ed the am s pusmunn even Waugh the am mm mm su Tu bedm me path megvatmn medeL M s mmahze me myee yemsmmd VEE DYSW w A and a Mudme catadwpmsmnwevedxmcd NPE mm Meme 8 Catau vpmsH The mma va ue em bemve any muvemem has uncuved sthe same asthe mma enemauen w unemu Make hwsasswgnmen mwube uw 4m lt4 e anen u H mm m n n m anvFVFP w u u u e Assng h be uwm be 1mm 4m gttl e 1mm Fm u unemu m Exevmse 4 U3 Se asswgn an have be uwtu be u zem m g e n We ve nuw veadym me me ca u atmn anay th2 paw megvatmn mudg Aeaxeuxauen my 5 jus a aneyway e1 saymg thatwe ve gumg e ca cu a e w A and a swmuhaneuus y Thws echmque s veqwed because each uHhe va ues depend un each uthev AHhe begmmng mum Pam an asswgnmemwas made unhe vange meme me cumm they H 3 have me same numbev Me emems Fuvexamp e W W x e x a g anentk 7 wk OWE peume pyevene m 2012 smes unthe ngmrhand we and w my me xevehane 5M2 M y m w y n Ana Muemena cataewpmsmnwevewmm NPB 102d Module 8 Cataglyphis ll l l 1 Kp171 1 x171 a J 1 gorientii1 7 kpiilJ Ask Mathcad to display you p A and 6 vectors below Write a few sentences about what you see Is this what you expected from your calculation array above How so 407 o o o 1 o o 1 1 1 1 o 2 2 1 2 o 3 3 1 3 o 4 4 1 4 o 5 5 1 5 o 6 6 1 6 o 7 7 1 5 7 0 deg 8 8 1 8 o 9 9 1 9 0 1o 1 1o 0 11 1 11 o 12 1 12 o 13 1 13 o 14 1 14 o 15 15 This looks reasonable They all have 300 elements just like the orient vector so that39s good The p and vectors are just full of the initial conditions but that39s what the calculation array said to do The 6 vector basically tracks the orient vector because p is always zero but it shifted one element eg the first 90deg value is at element 201 in 6 but element 200 in orient odule08 Cataglyphisl Answeredxmcd NPB 102d Module 8 Cataglyphis II The path integration model has one free parameterthat is used to fit the model to experimental data This parameter is called k and has the rather curious units of deg2 Assign k lowercase below to be 4009 X 10 Sdeg 2 This is the value used by Muller amp Wehner 1988 to match their data 408 k 4009 X 10 Sdeg 2 The full path integration model is provided below However you39ll need to first clear out the p A and 6 vectors first by assigning them each to be NaN NaN stands for quotNot a Numberquot and is a special value in Mathcad Making these assignments will allow us to use your assignments from earlierwithout the possibility of previous data getting mixed in 409 kpl NaN NaN 5 NaN Copy your initializations statements for p A and 6 from Exercises 403 through 405 above and paste them here below 0 oriento A0 1mm 90 0 Here39s the full model If you get any errors don39t continue until you resolve them above kpi 1 k180deg Siil180deg 7 517063 1m P11 7 17 171 x I5 I 1 171 speedtimestep 1 7 a 1 1 90deg 1 gor1entii1 7 kpiil Notice that this model has quite a number of terms and compared to the quotcomplete solutionquot model given in Part II this isn39t exactly simple Why all this regmarole with a complicated model Why notjust use the one from Part II 411 The model from Part II didn39t have any navigational bias it was accurate The ant39s aren39t accurate so we need something else to explain their behavior Hopefully this is it odule08 Cataglyphisl Answeredxmcd NPE mm Mama 8 Catag vpmsH Jus1 as a vsmmusmm sthe Mu charme cun umanun We ve m n n anav shuum swsmsrus sssmw cu swdevmu The abuve ca cu atm have uenevated uenevated a w vemuvwnh am Why Wm mm hume s a ways w18 deg v may quot 1 Kquot 39 J a nu ecaH m Pan H mu thatwe used nnenl hnmem vepvesem 012mm hume unsmauun Hm wswm desmna e nrienLhnmew asthe nrienl hnme appmmanun genevated bvme path mamaan mudg A su quotme thatwe a veadv knuw mm humehnhwsva ue m n J Ge ma hcad m mspxay We tuned nrienLhnmevamgvuym s pamcu avva ue um ExevmseA m may we yuu same hmts Wyuu ve dvawmg a New have unenlihnmeg ms 565 deg Mame av miem m 12mm ghasmj mg 153 4 deg m Ewen an ang e gveatev but c usem 207deg 0y muwa en y an ang e gysam them but c use m 453m Let s see Wyuu ve quotgm Assng unen jumew be uwm be mmw18 deg Then ask Mamcad Wha unem hume s Duesthwsmatchwha vuu armmpated abuve7 m anenLhnmew mg lXUdeg Vep mafswha amed 77 215m 5 may mmLhmwznm deg than bmdusemmmeg Mudme catagwpmstAnmvswmm NPB 102d Module 8 Cataglyphis OK reality check What is orienthomeq How Is this different than orienthome 415 orienthomep is the quotant39s estimationquot of the route home based on the path integration model In contrast orienthome is the quottruequot route home based on trigonome ry Knowing orienthomep and the orienthome calculate the error 8 produced by the path integration model when a is 90deg You39ll have to figure out the syntax on your own 416 E orient homekp 7 orient home MA 7 7 9 E 9845deg You should have determined a to be 9845deg odule08 Catagyphisl Answeredxmcd NPE mm Mama 8 Catag vpmsH Part V The relationship between E and a Goa s Expand path megvauun mude m ca m a e unemjumew uvaHvames u u Gvaph enuvs v2 mama bym path megvauun mude as a mumquot u H mm yuu Wm expand me paw an h a smvauva uesm heu mm avemm names h N ang e awmh mm charme s ave a Jumed SnthwsPanwmewmwckmewuvk 7 Emmy ehnev a LWHU g systemaucaHy changed u and vecuvded v usmg vea ems see ame Q As yuu shuwed m We as Pam ca cu a mg ms easy unce mm hume and unemjumew ave knuwn m Pan W n a a yuu aheady ca m ated mm hume mm 0 30 no 180 enme ummv H212 yuu Wm an m same Turning angle a m unenLhumew am m yuu m aHva ues um Tm s accumphshed mth s eps FusL We H need m expandthe Ium ventuth a 270 mm Secund we H use me new Ium mam m mam m nrienl mam Theve s mg uHhem mdudmg me NaN asswgnmem page 012m 3 be uw uxnNaN 1 nmzsmp um 51 speed nmzsmp Muumena catagwpmstAnmvewmm NPB 102d Module 8 Cataglyphis II The turn array dictates the turns that an ant will make on its journey In the case of an ant coursing through the twochannel experiement it will only make a single turn at the elbow Therefore for the vast majority oftimesteps rows the ant39s quotturnquot will be Odeg ie no turn This condition is created by the initial turn assignments that first assigns turn to be NaN and then assigns tum end to be 0 With turn being empty from the NaN assignment assigning element 299 71 timestep ie 7 1 299 to be 0 also assigns all the other elements ofturn to be zero as well timestep This isjust a property of Mathcad arrays and provides a simple although somewhat sloppy way of creating a vector full of zeros The third turn assignment takes care of the fact that one of the turns is nonzero the turn at the joint of the two channels which has a value of a What element this occurs at depends on the length of the first channel 51 10m the ant39s speed speed 05E and the assumed length S 51 ofa single timestep timestep 01 s Thus a turn of 1 occurs at element 200 speedtimestep 7 in the turn vector In the single case considered in Part IV this element was assigned to be 049 90deg To expand the turn vector to a 2 D matrix where each row is a separate timestep and each column is a separate value of a you will need to use the j range variable from Part II Copy the three turn assignments from Exercise 501 above and paste them here below Use the j range variable which should still range from 0 to 18 as it did in Part III to expand the turn array to include all values a from the a vector 502 tum NaN tum 0 Nwwv end J timestep tum OL 51 J 5 J speed timestep odule08 Cataglyphisl Answeredxmcd NPB 102d Module 8 Cataglyphis ll Ask Mathcad what turn is below Is this what you expect How so KOWNCBLnBWNHO OOOOOOOOOOOOOOO Q In 00 0000000000000000 0000000000000000 0000000000000000 0000000000000000 0000000000000000 0000000000000000 0000000000000000 0000000000000000 0000000000000000 That looks correct The matrix has 299 rows just like the old vector And all elements are zero except for row 200 which contains the a values Again this structure is similar to the old turn vector Now we39ll need to make similar changes to the orient vector assignments Copy the assignments for the orient vector that were given to you at the beginning of Part IV There39s three of them including the NaN assignment paste them both below You39ll initially get an error in the last assignment That39s OK We39ll fix that in a bit 504 orient NaN m 0 orient orient turn 1 171 1 odule08 Cataglyphisl Answeredxmcd NPB 102g Module 8 Cataglyphis II The orient array represents the orientation of the ant along itsjourney Let take a moment to articulate the difference between turn and orient The turn array contains turn angles relative to the orientation at the previous timestep That is a turn of Odeg indicates quotwalk in the same direction you39ve been walkingquot regardless of what direction that happens to be u The orient array contains absolute orienations with Odeg indicating due north This represents the direction that the ant is walking at any given time Each element in the orient vector is the sum of the turns in all previous timesteps plus the initial condition oriento The regions in Exercise 504 accomplish this by use of the i range variable not thej range variable You should take a minute to convince yourself of this ignoring the error for now Here you will expand the orient vector to a 2D matrix where each column represents a different value of d and as before each row represents a separate timestep Doing this will resolve the error above which is essentially a mismatch of 1D orient vector and the now 2D turn matrix Make another copy of the three orient assignments from Exercise 504 above and paste them below As you did in Exercise 502 use the j range variable to expand the turn array to include all values a from the a vector When you finish you should have no errors Hint You39ll need to keep the references to the i range variable orient NaN m 0 5 orient orient 1 171 turn J 1a Ask Mathcad what orient is below Is this what you expect How so Think about howwhy this is different from the turn array 506 odule08 Cataglyphisl Answeredxmcd NPB 102g Module 8 Cataglyphis ll orient deg KOWNCBLnBWNHO H O H H H N H w OOOOOOOOOOOOOOO H J 0000000000000000 0000000000000000 0000000000000000 0000000000000000 0000000000000000 0000000000000000 0000000000000000 0000000000000000 0000000000000000 H U1 That looks right All ants have an initial orientation of Odeg due north but then they turn at timestep 200 Their orientation changes to d at that point and then keep walking straight in that direction 80 for the latter timesteps turn is 0 but orient is 1 We39re now ready to feed this orient array into the path integration model To do this you39ll need to clear out the p A and 6 arrays from Part IV and then reinitialize them To do this copy the six regions total from Exercise 409 and 410 and paste them here below You39ll initially get an error forthe p0 assignment That39s OK We39ll fix that in a bit 507 kpl NaN NaN 5 NaN kpo oriento A0 1mm g0 0 odule08 Cataglyphisl Answeredxmcd NPB 102g Module 8 Cataglyphis ll As you did with the turn and orient arrays the p A and 6 vectors need to be expanded to to 2D matrices It is the mismatch between the 1D p vector and the now 2D orient matrix that is producing the error above You39ll fix that below However remember that we39re only dealing with the initial conditions for p A and 6 at this point Once initialized they39ll have a column for each value of d but only a single row row 0 the initial timestep Make another copy of the three six assignments from Exercise 507 above and paste them below Use the range variable to expand these assignments to in each case create a 1row many column array When you finish you should have no errors 503 kpl NaN NaN 6 NaN wold orientoj bola 1mm gs140 Ask Mathcad what p A and 6 are below Is this what you expect How so 509 WI 0123450789 M 0 0 0 0 0 0 0 0 0 x 0 1 2 3 4 5 6 7 lm 0 1103 1103 1103 1103 1103 1103 1103 0 1 2 3 4 5 6 7 8 9 Yep that looks right Each ofthe arrays has 19 columns but only one row odule08 Cataglyphisl Answeredxmcd NPE mm Mudme a Ca aqums A Hand paste dpeye be uw Ved u mmaHngan enm That sOK Tp sasadpmepnedpeeds m u 525 cu umns m each ease Makemese medmeaneps be uw When yed ye dune yum assdpmem Shuu d uuk hke W5 a managed managed W V H H H m L LJ x e 391 x speedumestep 17 p H u Mam Mammy amended dude p a mm a mm e B a a Mg m ganenLHdr Hy Make suve yed ehmma e any enms bemve cummum Cupy yum asadpmemm dr39 quotgunned mm Exevmse A 14 and paste d have be uw Ved u ge an em mmaHy That s OK WE H what shuv y w unenlihnmew mam madeg VDUY drienLhdmeq asswgnmem has a cuup E e pyemems 1 5ch am needs m be expanded Spem caHv m vemuv swmw a u We unem hume Venn VDU made m Pan m 2 The Iasudmmp un ywuvkswnh vectuvs and p s nuw a 270 mam sd ved u havem use a dmeyepneepmddem spec yt e 351 mwm p Huweveh mm a 1 e 299 Shuu dwmkjus me That ca m atmn SHDUH be yeadmd 299 WVEIU VE dupe evempmd cunec v up m W5 pew Mudme m Catadwpmsmpweyedmed NPE mm Meme 8 Catau vpmsH nrienLhnmeq asswgnmen vum Exevmse 5 M abuve and paste n have be uw Use We ab e and me vuwstechmque deta ed abuve m cveam unemjumew as a Vaclav cunespundmg We a va ues m We nvecmv Cepyme vange Va 512 menLhmew mww4dxzneeg Cepv Wm maph mm Exevmse 3 m and paste n have be uw Add an addmunam Dunn 45 mm Shuwsymbu sunmehace WNW 22 a m 39 g dz 2m 1 m e m u HvuuVe dune evemhmq cunec v m W5 Pam enemrhemevm u Vuuv maph shumd uuk sememme We the ewe 7sz aHhE new Emu 19D 7 5n mm 15 2 n 3s m an m 22n s 21 ammhnmzwmn mead2g wu Muemena cataewpmsmnwevewmm NPE mm Mama 8 Catag vpmsH Wm a 12W sentences abum Wm yuu 522 m yum gvaph Wha 5 U7 Wha s nrienLhnmgv What 5 nnenLhnmwv m 2 ang e at Whmh We WEI charme s mthe expenmem amumeu unenLhume s luv a gwen va ue um We hue unemauun an em shm d mm get backtu me quot25 unemjumew m mum sthe esumanun umnemjume pvuduced bythe path megvatmn mudg Nut suvpnsmg y unenLhume and unemjumew May at a mus aH va ues um AI uva ues u u and Meg n uuks hke unemjume and unemjumew 312012 same unemjumew s a ways gveatenhan m Lhume The paw megvatmn mudg 5 must naccuvate when u s amund 135m We uvemuv 3 aanhnmew 7 unenLhnme Make a gvaph u di agamst u Shaw symbu s an 012 2 quotace E 2n HvuuVe dune mqu cunecw m m Pam Vuuv 3g gvaph 5mm uuk sumemmg me me une aHhe mm mm Muumena u catagwpmstAnmvewmm NPE mm Mama 8 Catag vphwsx mm 5 mn 15m n W Huw dues yum gvaph cumpavewnh ms quamnawe m yum cumpansun Error angle 0 60quot 120 180 Turning angle a m m y va ue an uccmswhen u weuusg Eutthws s vue swam my gvaph and ms 123 data Anhat peak t uuksm be abum ZUdEg m bum cases The curves bum nee 22ch at u va ues mum and WEDdEg mm curve 5 mm skewed m We quotgm lt1 megvatmn mudg kas Knuwmg We M s expmve ms sass Wham me am muve may and Uth assess N s nawgauuna accuvacy dunng a namva 1uvagmgmp Mudme catagwpmstAnmvswmm NPB 102q Module 8 Cataglyphis II Part VI A FreeRange quotAntquot Goal Use the Gaussianbased turn foraging model and the path integration model to assess the accuracy of quotantquot navigation At this point you have confirmed that the path integration model introduced in Part IV mimicks real ant behavior under experimental conditions ie the twochannel experiment But what does this mean for ants that can freely move about David Attenborough claims that navigation skills of freerange Catagyphis are excellent Yet experimental data from Muller amp Wehner reveals that Catagyphis only approximates the navigational quotsolutionquot These observations appear to contradict one another How can we resolve this conflict It39s worth noting that the twochannel experiment forced Catagyphis along paths that were to a large extent quotunnaturalquot Certainly none ofthe outbound foraging trips produced in Part II resemble the path imposed by the twochannel experiment However it39s not clear how the twochannel experiment might bias our interpretation of Catagyphis navigation For example the twochannel experiment includes only one turn d Freerange Catagyphis makes numerous turns on its outbound trips One might suppose that if one turn is inaccurate then many turns would be even worse accumulating evergreater error as the individual errors add up Alternatively if the errors from individual turns tend to be small eg from small turns and in opposite directions the individual errors might tend to quotcancel outquot From this we might suppose that many turns would result in less error than a single turn Clearly we have no intuition for how the path integration model will perform with a freerange ant And this is exactly why modeling techniques are so powerful We have have all the tools necessary to simulate an outbound foraging trip and then assess the nest location both in truth and according to the path integration model How well does our quotantquot perform To begin clear out the turn orient x and y arrays below 601 tum NaN orient NaN X NaN y NaN Copy the Gaussianbased turn model regions including the graph but you can leave the orienthome calculations behind from Exercise 222 and paste them here below 602 IOdeg timesteB lsec 0 NW M 0 X0 0 Iunif17180deg180deg0 odule08 Cataglyphisl Answeredxmcd NPB 102d Module 8 Cataglyphis end 1 E 1 2 timestep d rnorm L de 0 g timestep Ki 1 X171 speedt1mestepsmor1entii Ill y yii1 speedtimestepcosorientiOJ orienti orienti Xihome X1a5tx yihome ylaStY 0 0 1 mum You should see a new outbound path in your graph here Don39t continue until you resolve any errors The above regions include a new orient vector which you can use as an input to the path integration model from Part IV Modue08 Catagyphisl Answeredxmcd NPB 102d Module 8 Cataglyphis ll Clear out the p A and 6 arrays below Then initialize the arrays as vectors You should be able to do this by copying the six regions from Exercise 507 not 508 Any error in the regions should go away once you paste them here below 603 kpl NaN NaN 5 NaN 0 oriento A0 1mm g0 0 Now copy the path integration model39s calculation array that was given just above Exercise 411 and paste it here below 604 kp 1 k180deg 314180deg 7 51715171m 1 P11 171 x 7 I5 I 1 171 speedtimestep 1 7 a 1 1 90deg gorient171 7 kpiil To calculate the believed location ofthe nest as opposed to the nest39s true location we need to generate two new vectors akin to the xhome and yhome vectors that you used in your graph above We39ll call these new vectors xhomeq and yhomqu to indicate they are based on the approximation of 01 Unfortunately calculating xhomeq and yhomqu requires a bit more trigonometry so in the interest oftime these vectors are provided for you here X1ast X Xihomekp Lxlastx l x1astk39smwlast l lgodegu ylast y yihomekp ylastx 1astx39cos quot1asttp lgodeg Note that the use oftrigonometry here does not imply that Catagyphis uses trigonometry On the contrary we assume that Catagyphis does not These vectors are only for our own use to convert the mental thread of Catagyphis which is inherently nonspatial into xy coordinates that we can graph odule08 46 Cataglyphisl Answeredxmcd NPE mm Meme 8 Catau vpmsH Cupv Wm maph hum Exevmse E U2 abuve aha pasten heve be uw Add m the y aws and x ham m We ham chahee hmh unhe heme uaees m shew awhth TMSWM make ME nevcewed ueahuh unhe hes1 mme c eav vvheh vuu ve anE VDUY maph shumd uuk sumethmu hke he uhe anhe mm Athuuuh W5 uhhhew WU Wm have We same umbuund new unhe same hes1 accuvacy x xhnmz xhnmzw am pvedu the ucatmn unhe quot25 m yum abuve swmu a mn Huwvay 5 n W hhaghe manhe am pusamy mach a ung Ms paw The am 57 bad hm N s hm gvea H heads m a veasunab e dHEdmn hm Ms guess 5 abumZSm em Wa kmg Vunhekuu d uhxy mm a We maybe shave aha mam m WEI umws ance Muemena cataehpmsmhweveexmm NPB 102d Module 8 Cataglyphis Press CtrlF9 to recalculate your worksheet Depending on the speed of your computer this may take a minute Just be patient This will generate a new outbound path for your ant How does the ant do this time This run ofthe model was similar The ant would have ended up about 25m away from the nest Walking further wouldn39t have helped Copy your graph from Exercise 605 above and paste it here below Conduct 4 more quotrunsquot of your model using CtrIF9 For each run write out the quotresultsquot of your experiment below 608 X Xihome Xihomeap a a m m m 1 This result seemed pretty bad The ant never got any closer than about 30m Walking further wouldn39t have helped 2 This result was pretty good The ant got to within 5m of the nest Walking furtherwouldn39t have helped though 3 This result was also pretty good The ant got within 10m of the nest Walking further wouldn39t have helped 4 This result was excellent The ant practically walked right over the nest within 1m of its t Ioca ion odule08 Catagyphisl Answeredxmcd NPB 102d Module 8 Cataglyphis ll Based on your six observations above what sort of accuracy would you attribute to our quotantquot What was the mean closest difference between the actual nest site and the mental thread pathway home 25m 25m 30m 5m10m1m 16 m 629921 f 16m 1m 70111161 0696mi 16m 52493ft 1 11 The mean closest distance for my six runs was about 16m 53 feet Frankly that sounds pretty poor If Cataglyphis is 1in long then that39s 630 body lengths 630 body lengths for me would be over half a mile That stinks In all likelyhood you should have found that the path integration model predicts fairly large errors If you didn39t you might want to calculate how many Catagyphis body lengths your error represents you can assume Catagyphis is 1in long and then compare that to your own body length In other words the conflict presented at the beginning ofthis part remains That is real ants find their way homejust fine but experimental data again with real ants imply that they should often miss their home by a large distance What gives What do you think is going on here Assuming Cataglyphis doesn39t frequently get lost and die on their way home why does this conflict exist How many explanations can you think of The path integration model could be wrong It may fit the unnatural behavior in the twochannel experiment but it does not fully capture the calculations made by free range ants The outbound model could be wrong It39s based on random numbers not real recordings of ant foraging Real ants might avoid maneuvers that lead to large andor cummulative navigational errors The navigation system of cataglyphis may be complemented by other sensory cues smells visual memories of terrain communication with outbound coworkers etc If all else fails ants must have a systematic search behaviorstrategy to find the nest once they realize they39re totally lost odule08 Cataglyphisl Answeredxmcd NPB 102g Module 8 Cataglyphis ll Let39s assume for a moment that 1 The path integration model is an accurate representation of Cataglyphis navigation 2 The outbound Gaussianbased turn model is an accurate representation of Cataglyphis foraging This implies that Cataglyphis must have mechanisms other than its path integration calculation to insure it eventually finds its way to the nest What sort of mechanisms do you thinkthose might be 611 Cataglyphis navigation may be complemented by sensory cues smells pheremones laid down by outbound coworkers visual memories of terrain direct communication with outbound coworkers Cataglyphis might even survey the density of surrounding coworkers This should be highest near the nest If all else fails Cataglyphis must have a systematic search behaviorstrategy to find the nest once they realize they39re totally lost References Alcock J 2005 Animal Behavior 8th Edition SinauerAssociates Sunderland MA p130135 Muller M and Wehner R 1988 Path integration in desert ants Catagyphis fortis Procedings of the National Academy of Science USA 8562875290 Wehner R 2003 Desert ant navigation how miniature brains solve complex tasks Journal of Comparative Physiology A 189579588 Wehner and Wehner 1990 Insect navigation use of maps orAdriadne39s thread Ethology Ecology amp Evolution 22748 Module 8 Cataglyphis ll Written by Eric Leaver 2009 Martin Wilson Last update on March 17 2009 at 330pm odule08 Cataglyphisl Answeredxmcd NPE mm Muduie 7 CataENpmsi Foraging in Cataglyphis A Spatial Model of Finding Food Part I Introduction Goai Review oazagyphaieyagihg behaviuv Ant spacias enhe genus Catagyphhs ave heaHuievant thavmuphihc scavengevs Theytend tn iwe day Catagyphhs is speciaiiy adapted to withstand budytempevatuves up to smowhieh aiiuws imt N succumbed tn the heat Catagyphhs is mien ieuhe in ieameiesswasieiahes E ntammg iew iandmavks The Catagyphhs t Fm mni Othev ant geneva image as a gmup now vecuveved and item Neitheywuuid vecaii eiah ahi s outbound path he usemnenhe YEtumtiip heme a because Em C cannut swweim iung pEViudS EIHimE WWW w v r y Aviemhmhg a and Ham ahy deiay in a apeheheiehammhem anddivedhammavdvatMMMVE my aisewehmm itam aivetiiied eheie hm mama EuncEmmgthE spatiaiaspects u seeahmmmmp maiaumampm5921m rimmed Calagypms magma 2m5 mu Wehhey ha Wehhey 1990 5mm Ecotwy a Emtmnzzna 1 Once and iSVDund hew dues awuvkev 12mm heme7 2 What shaiegy en Wuvkevs use to he and h the ihsi pian27 Muduiem caiaehphisijemnhweveemm NPB 102q Module 7 Cataglyphisl This module will deal exclusively with the latter question of food acquisition While the above background is a sufficient review of Cataglyphis you may enjoy the following video 230 min it ends rather abruptly from the Trials ofLife documentary hosted by David Attenborough Cataglyghis Video 1Doubleclick to begin At first Cataglyphis39 strategy forfinding food may seem trivial David Attenborough describes the outbound movements as quotrandomquot And so too the outbound path depicted above may seem random to you as well But how random We may intuitively understand a random walk as quotwalk a little this way and then turn a bit and then walk some in that new direction etcquot But is this a sufficient explanation of all sorts of random walks Surely not How far do we walk How much do we turn Even quotrandomnessquot is a quantitative issue In this module you will generate a quantitative description of Cataglyphis foraging as it relates to their outbound paths Once we generate something that quotlooks rightquot we will attempt to quantify the effectiveness of that strategy One point made in the video that will not be addressed here is the ants39 use ofthe sun39s position to determine their own direction of travel This information helps Cataglyphis later calculate its homeward path once a food item is found However we39ll assume this information isn39t important forthe outbound foraging path Module07 2 CataglyphislSemiAnsweredxmcd NPB 102d Module 7 Cataglyphisl Part II Basic Random Model Goals Create a model of Catagyphis foraging assuming uniformly random turns Create a vector of relative turns and use that to generate a vector of absolute orientations Use the orientation vector to compute x and y vectors representing the ant39s position Use the x and y vectors to graph the ant39s path Examine how the ant39s path is affected by the model39s timestep size We39ll begin where David Attenborough left us with a random walk Here you will generate a model that tracks an ant moving at constant speed 051 speed through a featureless 2D sec terrain Arbitrarily we39ll decide to trackthe ant for 5 minutes end In the simulation time will be divided into discrete timesteps which for now we39ll assume to last 1 second timestep At each timestep the position of the ant will be recorded in terms of a x and y coordinates Although time is divided into discrete steps x and y can take on any real value Thus we could describe this model as being discrete in time but continuous in space There are a number of ways we could capture the quotrandomquot nature of the ant39s movements Let us arbitrarily decide to describe the movement by a series of turns that can range from 7180deg to 180deg relative to the ant39s present orientation From this we can describe the sequence of events during a single timestep from the ant39s perspective 1 Begin walking in a straight line at constant speed 051 in the direction you were facing at sec the end ofthe last timestep Walk in that direction forthe duration ofthe timestep lsec Briefly stop we39ll assume the duration ofthis stop is brief enough to ignore and turn in a random direction 7180deg to 180deg relative to the current orientation 0 Note that this scheme entirely ignores the circular turns described by David Attenborough These turns are important for ants39 navigation scheme but as we39ll see later they have little effect on the general features of outbound foraging Think about this model for a bit In your opinion does it accurately describe a quotrandomquot walk Why or why not Do you thinkthis model will produce the sort of outbound path presented in the figure in Part I Why orwhy not 201 Module07 3 CataglyphislSemiAnsweredxmcd NPB 102g Module 7 Cataglyphisl Let39s begin by making some assignments forwhat we39ve assumed Use the assignment operator to assign speed to be 051 Don39t leave out the units sec 202 speed 051 sec Assign timestep below to be lsec and end to be 5min 203 timestep 1 sec end 5min For each timestep in this model we39ll need to record four pieces of information The ant39s x coordinate x The ant39s y coordinate y The amount to turn at the end of the timestep turn The new orientation of the ant relative to the coordinate axes we39ll assume Odeg is due north straight up orient AQNA We39ll store this information in four separate vectors x y turn and orient with each timestep corresponding to an index within those vectors So for example the x and y coordinates forthe ant at timestep 23 would be stored in x23 and y23 respectively To begin the model we39ll need some initial conditions for x y and orient we39ll deal with turn later For convenience let39s assume the nest is at coordinate 00 and that the ant begins its outbound trip with a Odeg orientation facing due north Make assignments for these assumptions below That is assign x0 to be 0 yOto be 0 and oriento to be Odeg 204 X 0 y0 0 oriento Odeg Note This module deals with angles exclusively in terms of degrees deg not radians In a strict sense degrees and radians are not units however Mathcad essentially treats them as such Mathcad assumes that any angle is in radians unless you associate the deg quotunitquot with the number Thus 27v 6283 which is the same as 27v 360deg As alluded to above the vectors you just initialized will eventually have a large number of elements one for each 1second timestep To control references to these elements we39ll need a range variable Let39s call this range variable i Module07 4 CataglyphislSemiAnsweredxmcd NPE 1mm Mudme 7 cataehphTsT ndEXU Ts aheaey spem edmv uuvvemuv su hegm i as a vangE vanab e heTew aH Make the end seeuhe vame 2 and End 0T2 senes at umestep 2n5 912 ems umestep 48Ddegn lXEIdeg Let usmnhev assume manhe pmhahhuy Many pamemanmh ahgTe Ts he same as 3 uthevs Th mhekuvds we Wm pmk mm aheTeshem a umormpmbabm v msmhuheh r h m hh uheheh Yandumr uhhmm The mnimhmmh accems mpu s 1 The numbev evyaheem numbevs yuu Wam runilcvea es a vecmv cumammg hTs mahy HUN 2 The uwev buund enhe uhwehh pmbabmty ms1hhuhehmxnaegm um ease 3 The uppev buund enhe uhwehh pmbabmty msmhuheh lXUdegm um e ase Cveatethetumvecmvbe uwbyasswgmngIumtube mmf mquot umestep 48ndngnd gj m e my eeuaegmaegj umestep Check i and yum x y nrienm and mm vecmvs heTew Huw mahy eTemems dumey hav27 stms Whatyuu expec WhaTahemThevamesmhevemmsv stmswhatyuuexpem7 Hm Mudmem caTaehphTsLSemAhweveemm NPE mm Mudme 7 Ca auNphws and umestep mu mm y H y Let s sun by bummgm ca cu auun aways Acahtu auun my 5 m1 3 Vancyway uvsaymg ma We ve gumg m ca cu ate x y and might swmuhaneuus y m m muuemms sn smc y necessavy We mum ca m a e y y and mm my sepavam vegmns Huwevev mws s a quad 5mm menu and me nutauun s Bumped Hm m y n nNnrinm y y Wm Fm 2mm X M y anent amm m y m a Mudmem CalauwmsLSemLAnwEvedxmcd NPE mm Mudme 7 Ca agvphwsx m min y X1 14 y z yH anent amm mmH 2m X 14 V 1 anean mem umH 2m Mudmem caagwpmsxjemmmveumm NPE mm Mudme 7 Ca auNPms h su whaw am at each whea1ep Hewevemhe am 57 acmaHy gumg anywheve ye We need c use the Th n Tu cc thaLWe H need c use a bu EIHHgEmEImEUy And mgunumehey 5 3H abum Hgmmang as AH Hgmmang as have Mu ang esthat ave essthan anaeg Opposwte Fm hhhh h mehe a nevspecwe we cah assmh hamesm the 01122 sees cnhe mang a 1 vpmehuse nhe ma chax ace 2 OPPUSME bemu Uppusne 8 3 Amacem nhe hemamme swde amacehnc a and NOT ME hvpmehuse huhe case n um an at each umestap ypmenuse cchespchesm We dws ance the am amuaHyUave ed speed umestep Oppcsue cunespundsm the change m x he weehhmea1eps Amacem cchespchesm the change m y be weenumes ps n Envespundsme va ue c1 nriem anhe begmmng uHhat panmu ammes ep Based Em W s anamementwe ve meves ed m we umuhumemc eehhmehs own 1 a e 390 Hypmenm Adjacent 2 an e Hypat GVEH 01252 we eeuahuhs and a We a uebvai we cah ache mam Iwo unknuwns Fm examme WW2 knuw a ahmhe enmh u vamenuse we cah ache uuhe ehmhs uvhmh Oppusne aha Amacem SpecmcaW oppaeu e Hypatznme ma aha Adjacent e Hypatznme ca Mudmem caaehpmsxjemmweveemm NPE 1mm Mudme 7 Ca angms Whhe these EXpYESSmnS ave mus1 235W uheeysmee m the cumex e1 vmmmam es m mammE a s hm hmnedm pusmve va ues essthan s a u eAuse m hehsh es hppe Th sueh case n mav he cunvEmEnHu assume a Ts sTse nE aHvL hm because she she eeshe ave peheme UHEHDHSJMS Ts hm VEquHEd Nuw backm um am The mamsm heTew shuws he msHhveE pesmehs Mum am Em Ms uumEv The pmme em vepvesems eunhmsT Eundmuns snhe quot25 XZEI yZEI she ehem Idag Accuvdmg m mm m h m h V Thwshh r Them ehee the am amves a may Mums she eeTs veadv enhe hemrhes1ep hTh T y wuwa en y mm Wm The mcky pan Tshguhhg uutwhat AX she Ay shumd he FuHhaL E m H s49 s e s Ax 5 s49 Ay 5 ages 0Ax SAy O e tz 0 tumn tum 0T5 onenh 0 tumn s speed Umestep Mudmem caaehphTsLSemAhweveemm Mudme 7 Ca agvphwsx NPE mm mang e uassmsansspamav E adjacem uppusne uvhypmenus Andwhat abumAy7 And am what abum 57 vepvesemed as x XH spud umestep smmnentk yH spud umestep caswrxentk amm mmH n y w h w 212 x XH spaaaammp mmh y yHaspnaammmspmnh nentk umH Assummu Mathcad mum 12mm an enuv was a maph be uw uHhe mama paw mama bv he mudg Them aysvsw dMevemwavsm u W5 Pevhansthe swmme s u y x mm lagems P umng agams1 asuwuvks bm ma mam 57 as sampam Vuu H a su 522 m Exevmse 4 ma mst muve swmme mam auaws usm mm mumms Pa swnhuut mudmcatmn m addmun theme 012 am am ame maph m be crnssed dunb erdck m m gvaph a hung up We mmamng menu Mswm shuwthe nss1 ucatmn as hem anhe m and yraxes mmsecuun mm x7 n thk x H y yum gvaph Wm uuk examy hke ms gvaph anhe quotgm Mudmem caagwpmsLsammmveumm NPE mm Mudme 7 cataghphm tubewa kmg7 Huwvayawayhumhe hesmuesugw Because Vuuv mph 5 amurscawd m We venue unhe ahrs muvement m mav be mmcmnu puHhe ahrs muvemem mm came u ham us evama e he swmu atmn cupv Wu mph hum Exevmse 213 abuve aha 13512 M1212 be uw Numhe anhesnhe am mum have Wa ked m anv uhe mvechuh s a and lSEIm Tn 522 huw hws mmth cumpavesm We mmth manhe am amuaHv drdwa k chem We x7 and waxesm venue hum eedend u speedend When VUU VE anE Vuuv mph shumd uuk sumethmn We the mph aHhE Ham Mudmem CalaENpmsLsemxAnmvedxmcd m m X m 7 mm H mm 7 m m NPE mm Mudme 7 Ca agvphwsx law 7 mm D Cumpave me gvaph yuu made m ExevmsesZ 13 and 214 abuve m We vea am path mama an 012 um FeeHvee m make yum gvaphs abuve a We may m make u gamma see What s gumg un Be suvem amcma e any svm avmes m dw evencesma yuu see Fm exammg w m yea am smuvemennsm eyenuyum M m mudg anL New how mm muvemem 5F 5 mm V x k V27 m thr m Mudmem caagwpmsxjemmmveumm NPB 102d Module 7 Cataglyphisl You should have been able to spot some differences between your graphs and the real ant path How does this square with David Attenborough39s assertion that outbound foraging is quotrandomquot Just a couple sentences here is fine 217 How do you think might we change the model to produce paths that look more like the real Catagyphis foraging path Be specific about how you would change the model Being correct here is not so important I39m interested in your thoughts 218 One thing you may have noticed about the real ant path is that it seems to involve a straighter outbound path than that predicted by our model This could be a consequence of our assumption that the ants make turns every timestep 1sec The skyward observations that David Attenborough describes occur less frequently on the order of once every 4 seconds Wehner and Wehner 1990 Because we assume that the ant walks in a straight line between turns might increasing the size of timestep produce a less circuitous route that is more similar to the real ant foraging That39s an excellent question for you to answer on your own Module07 13 CataglyphislSemiAnsweredxmcd NPB 102d Module 7 Cataglyphisl Begin by clearing out the data in your x y turn and orient vectors To do this assign each below to be NaN NaN stands for quotNot a Numberquot and is a special value in Mathcad Making these assignments will allow you to reuse your assignments from earlier without the possibility of previous data getting mixed in 219 X NaN y NaN turn NaN orient NaN MAMIV Now copy any necessary assignments from above to recreate your model here below including graphs Once you39ve done that and everything works change timestep to be 4sec instead of 1sec Only make that change here leave the assignment in Exercise 203 the way it is Once you39ve done that there39ll be some questions for you below timesteB 305ec 300 340 m0 end 1 1 2 timestep 39f and 180d 180d nlnl 7 e e W timestep g g Xi xii1 speedtunestepsmor1entii1 yi yii1 speedtimestepcosor1entiil LorientJ L orient turn 1 171 171 Modue07 14 CataglyphislSemiAnsweredxmcd NPB 102q Module 7 Cataglyphisl X m Module07 CatagyphislSemiAnsweredxmcd agtlt NPE mm Huw u yum gvaphs uuk nuw w changmg heva ue ummesmpm va ues uthevthanAsecund What 211m dues changmgme Iimvslep va ue seem m have an yum mudew stheve any va ue uHimeslep mm make m 2 yum nu uukmuveswm anuthevea am paw evate a new 52 u vandumtu Mudmem caagwpmsxjemmmveumm Mudme 7 Ca agvphwsx Mummy uH NPE mzd Medme 7 Ca aENphs Part III Gaussianbased Turn Model 05 s se hw ugvamsm cumpave enwenn and Gaussswan pveeaemy dwwwbuuuns MudWyme medew e mcuvpuvate mms eased en a Gausswan beHrshaped msmbmmn By nuw d shumd be c eavthat uuvanHuvagmg medew nas senees pvemenns n cvea es a yandenn path buHhat path deesn veaHy uuk We a vea am path am awaynemne quot25 mene m 255 buths 57 as he pM as d Imam seem a 3ng Iimvslep vaxeedne am muvemem 51m duesn uuk ngm N stuu enwmns enee evevy naw Fm unE even wdn Jevky Andvenanednenwe have ne bm ugwca veasun e assume the am mmme Vuu shm d a ways Let stake a Week at the vea am new anew Vuu mum nu wce ha wh e the anrs path seems vandum m1act Hsn OvevsmaH i 2 ee The 5 en a secundrbvrsecund basws Shan mms ave rare and tums e1 EldegU e ne mn ave mmmon Tnene ave a numbev uNavswe mmm expvessthws eenawenn a medex Let es ecwdemwmmeme hwsbvchanmnmhe n v e a the pveeaemy dwwwbuuun s Gadasan WWW 4 usswan dw vwbuuuns ave belrshaped and u en eaued normsmsubuuuns Tne eneeaemy e be Lshaped eewe cemeved avuund uded Thewwdm m spvead e1 W5 msmbuuun wm depend en a new pavemetevrr A1h2 standavd dewanen E was OK demunshate Wha We ve ta kmg abum Hyuu have nu cump eted the Hislngrzrrs M Mndule yuu shumd de se new Stup Save yum me cump e e We mwmrmudu e and came backm W5 Fan 51 Mudmem CalaENpmsLsemxAnmvedxmcd NPE 1mm Mudme 7 cataehphTsT ummvm yrmsmbmed nuns and um hew sssumpheh e1 nuvmaHyrdembmed mms Se hum My a 3ng numbev e1 mmsw mun Maddmum sepava etumvemuvs WE HcaH TemIurniunilandluminnrm Cveam the umVEIYmW39dTSWbMEd mm Vaclav heTew by sssTghThg Ium uniHu he 2 me am mum mr1nnn42naeg12naeg Tu cveam 0T2 nuvmaHyrdmnbmed mm veemhwe u use the huum mnrm uheueh LTheThe runil Vunc mn mnrm cveates a Venn e1 yahqu numbevs she acceptsthvee mpms 1 The numb mushqu humhevs Vuu am 2 The name enhe hehshspeu pmbabmty msmhuheh Eldegm hTs ease 3 The s andavd dewauunA1uHh2pvubabMy msmhuheh Eemve cveatmg luminnrm Ters vst make sh sssTghmem my a Vuu H EXp DYE hew ehshgmg hTs pavemetev a emsthe shrs paw a ev Fm HDW Ters assume a Tsauueg TeeThsy Ahematwe y yuu can type SClrlog m e e Z deg N 3m tuninmm mmonnnmew Mme space be uw cveam sepavate hTsmgysms e1 Iurniunil she luminnrm Use 11 bms bms m eseh gvaph Hyuu heeu a vemmdev Em xmg hm mama s gm haehmhe Hislngrzm M Mndule FeeT hee m use the space Em the ght w yuu heeu u Watch nu Du hm yesssTgh anyvaHab E hem yum abuve meueTwhue makmg yum hTsmgysms Fm m Dem Tense end have m mEan yum ehu hTsmgysm hThT Mudmem caaehphTsLSemAhweveumm NPB 102q Module 7 Cataglyphisl bins 11 j 01 bins beginihist 7180deg endihist l80deg d hi t7 b 39 hi t intervals beginihist jw J bins turniunifih histogramintervals turniunif turninormih histogrami11tervals turninorm g lt gt g tumiunifih 1 5 rowstum unif U 2 20 11 IL 0 7 200 7100 0 100 200 tumiunifihlt0gt deg Turn Angle gt o g luminon nihlt1gt 5 rowstum norm 5 2 20 11 11 0 7 200 7100 0 100 200 ruminormihlt0gt deg Turn Angle Module07 CatagyphislSemiAnsweredxmcd NPE mm Vuuv gvaphs mum uuk sumemg hke thesewhen yuu ve dune Wme a cuup e semenees be uw abumwhat yuu see m We gvaphs What ave yuu gyepmngv Whyavethe gvaphsm evenw Huwdues mm W Wha was expxameu abuve7 Assummg152cumesepawhawumd mesa Meyenees mean uyme amwa k mudew Mudmem caagwpmsLSemmmveumm Frequency Frequency Mudme 7 Ca agvphwsx 3 1 Emma u mu m unmannjngt a2 TumAngle NPB 102g Module 7 Cataglyphisl Experimental evidence suggests that Cataglyphls foraging really does rely on gaussian distributed turns Further observations of Cataglyphls fort395 on a hard sandy plain suggest a a value of 10deg Wehner and Wehner 1990 Ethology Ecology amp Evolution 22748 Let39s use this value of o to modify your model from Part II Begin below by assigning a to be 10deg W 10deg Show below how this affects the histogram of turn angle frequencies as you did in Exercise 305 above To do this you need only copy three regions from above The turnnorm assignment which will automatically take the new value of a your assignment creating your Gaussian histogram and the Gaussian histogram graph itself you may need to adjust the yaxis range turn norm Inorm10000deg0 MAMAMAMAMA turn norm h histogramintervalstutninonn turninormihlt1gt 6 0 rows luminorm 40 Frequency in 0 7 200 7100 0 100 200 ruminormihlt0gt deg Turn Angle Write a couple sentences describing how this new histogram looks Module07 21 CataglyphislSemiAnsweredxmcd NPB 102d Module 7 Cataglyphisl To generate the new model let39s begin by clearing out the x y turn and orient vectors as you did in Exercise 219 310 X NaN y NaN turn NaN orient NaN m Now copy your regions including graphs from Exercise 220 and paste them here below After that make these modifications Set the timestep assignment here not in Exercise 220 to be 1sec Make sure end is assigned to be 5min If not make a new assignment here Change the turn assignment to be Inormei0deg0 timestep timesteB lsec end 5min I 0 KO 0 onentS 0 end 1 12 timestep end tuIn Inorm 0deg0 mww timestep Ki 1 X171 speedtimestepsinorientii Iii yi yii1 speedtimestepcosor1entiOJ orient orient turn 1 17 1 17 1 Modue07 CatagyphislSemiAnsweredxmcd Mudme 7 Ca agvphwsx m 3 l 111 7mm H mm 2 1 4n n 2 40 6D 2 mn NPE mm lalw Once cump eted yum gvaphs mum muve c use y vesemb eme sun uhea umbuund magma path shuwn have J Tn 522 mm vuns uwuu mum chck an M yuuHum asswgnmem abuve and pvessF9 The Myquot asswgnmem mm have m be wsmxe WHHE yuu u mquot yuu can watch me gvaphs ms ead Vuu 5mm a su take a uuk aHhe umpm yuuvhvs1 mudg back m Emma 21m cumpave mm m umpm have u nd mm umm m Mudmem caagwpmsxjemmmveumm NPB 102d Module 7 Cataglyphisl Write a few sentences below about how your model seems to mimickthe real outbound ant path Is this model superiorto the one you built in Part II How so 313 Based on your histogram in Exercise 308 the turns in this model are almost nonexistent That is the ant walks straight ahead almost all the time However you may have noticed in various runs ofthe model that the model ant does in fact make some significant turns sometimes even making complete circles How do you explain this apparently conflicting behavior of the model 314 What do you think now about David Attenborough39s statement that Cataglyphis quotforages randomlyquot Assuming your Gaussianbased turn model mimicks ant foraging reasonably well how is he correct How is he wrong 315 Module07 24 CataglyphislSemiAnsweredxmcd NPB 102q Module 7 Cataglyphisl Part IV One Nest Many Ants Goals Review the basics of foraging strategy Extend the model to incorporate many ants Make a movie of the many ants foraging Relax our assumed initial condition allow ants to leave the nest in all directions So far we39ve been focused on reproducing the outbound path of a single ant Let us now step back and think about the biology some more That is what are the goals of a worker ant What is an effective strategy for foraging And how can we measure that effectiveness It is important to note that ant society is communal The ant workers themselves cannot reproduce at least not sexually This makes survival ofthe nest and its queens paramount To pass on their genetic information indirect as the process may be workers must focus on the health ofthe nest even over their own individual health This suggests that any evaluation of foraging strategy must adopt the nest39s perspective That is what worker foraging strategy is best for the nest Two factors immediately come to mind First the nest must maximize the the rate at which food is acquired But second because each worker ant is another mouth to feed the nest must minimizethe number of worker ants required to achieve this rate of acquisition That is worker efficiency is important How might you design an efficient foraging strategy Certainly there are some basic worker characteristics that would be desireable fast movement good eyesight keen sense of smell etc But in terms of deciding where and how to look forfood what should a worker do quotGet away from the nestquot This is one important strategy All workers leave from the nest so the area near the nest is necessarily wellsearched While any item discovered near the nest can be quickly returned to the nest the likelihood of finding something there is low Many coworkers have already passed over that same ground This consideration could be framed as maximizing the probability of finding food quotDon39t search where other coworkers have searchedquot This follows naturally from the first strategy If you simply follow a coworker around you39re not contributing to the search You might help that coworker haul in a big food item but then again so could anyone else that happens to be in the area when it39s found Do your own searching quotFind the food nearthe nestquot This strategy seems to run counterto the first strategy but it39s important to consider The nest seeks to maximize the rate of food acquisition Once found it takes some time to transport the food backto the nest The length ofthis return trip depends on where the food item was found Thus food items near the nest while potentially more rare are more valuable This consideration could be framed as minimizingthe return path Module07 25 CataglyphislSemiAnsweredxmcd NPE mm Mudme 7 Ca angMsx Gwen 01252 cunswdevauuns m suve yuu can mm Ma Vewwab eshategwes Have sune Wa kawaylmmme quot25 m a wuchback smusmda ashmn makmg avgev swnchbacks as yuu ge unhev umme quot25 Hevevy an e e n2 emu mate a wee panem e seavch uvevage Huwevev them ave pvamma pvub emswnh W5 sus egy Fm une N s nm c eav ems can cummumca e summerva as m a u mpgwhatmma unematmn shm d an em adupHu make suve she s m duphcaung anmhevcuwuvkev s vecem seavch That s e m amphmde and msmee vumthe nest be7 Suva yme mme39mm quot 5 epnmsx yexsnensmp depends my me numbev e1 Wuvkevs seavchmg Du ems have seeessm W5 mvmmsnenv Funhemhe smusmda s va egy vehes un enquee nawgatmnmensuvenunrvedundamcuvevage Em amma muvemem s eneppmne s he neuva cwcumy veqwed m mmvmze nawgauun enms an emmem use e1 hmned bvam quEy7 O s n benene anew a 255 ccuvacyrveham Sha egy A enhanced mess Msmsy mmes emmmemhe Mme m pm we mmew s va egy s we ueume man N s WWW quot mm m n ms Panwe u extend We anHuvagmg mudg mm Fequot m m mcuvpuvate 5n ems Each amwm Ms eehmeue e1 evamatmg a meeex s behawuy by exammmg numevuus vandumWna s s u en knuwn as a MeMeCsne swmu atmn Exevmse 3 m 451 x e um y e mu m um mem e um We saw abuve hanms mudg shumd mdude ms 5 asswgn ants m be 5n be uw 4m Ants m 25 Mudmem caaewpmsxjemmweveemm NPB 102g Module 7 Cataglyphisl In principle expanding the model from Part III to include multiple ants is straightforward At present the model39s quotdataquot is described by four vectors x y turn and orient To expand our view to multiple ants we need only convert those 1D vectors to 2 D matrices instead In the new scheme each row will still represent a particular timestep but in addition each column will now represent a particular ant So for example in Part III the x coordinate of our single ant at timestep 17 was contained in element x17 In the new scheme The x coordinate of ant 23 at time 17 would be contained in element x1123 In practice this is fairly simple to implement Wejust need another range variable to control our columns Let39s call this range variable ant Then wherever our model previously called for for example xi we will replace that with xi am Begin by assigning ant no 39s39 to be a range variable It should begin at 0 have a second value of 1 and end at ants1 403 am 01anm 71 Now initialize your x y and orient matrices This is similarto what you did way back in Exercise 204 but now you need to use both the i and ant range variables For example xu am should be assigned to be 0 The other assignments are made similarly 404 M nt 0 IXtant 0 Want Odeg To create the turn matrix recall that the output of the rnorm function is a vector A vector by itself is no different from a column within a 2 D matrix 80 we can use the ant range variable within a column operator to fill in the turn matrix one column at a time To do this assign tumltan below to be rnorm 0o timestep Wang morm 00 timestep Module07 27 CataglyphislSemiAnsweredxmcd NPB 102d Module 7 Cataglyphisl Copy your calculation array from one of your previous instances of the model eg Exercise 212 and paste it here below You should initially get an error That39s OK We39ll fix that in a bit If you don39t get an error that39s not OK and probably means that you didn39t properly initialize your matrices in Exercise 404 406 XLam W lrxiil mt speedtimestepsmor1engil ant yiil ant speedtimestepcosor1entii1a ant orlentiil ant tumiil ant yiant orient 1 ant In the above calculation array change all the vector references to include a column index of ant This is the same technique you used to expand the intializations you made in Exercise 404 Make sure you do this to all the vector references on both sides of the assignment operator including the ones referencing i1 use ant forthe column on these too not ant1 Completing this should eliminate the previous errors 407 Copy your quotbigquot graph from a previous instance of the model the graph that shows the furthest points that the ant could have traveled eg Exercise 215 and paste it here below Notice that we don39t have to do anything special to get all the x and y columns to plot properly 408 I X m Module07 CataglyphislSemiAnsweredxmcd NPE mm Mudme 7 ca agmnms Fem m m make yum gvaph a mug bwggev su N s gamma see What s gumg un Ham s a smaH vevsmn uNha yum gvaph shm d uuk m muve m 255 Becauseme mudg 5 dwen by 1 vandum numbevs n s unhke y yuuvda awm uuk m examy Me Me data gvaphed at Me Ham 7 Wm a 12W sentences be uw abumwhat yuu 522 M Ma gvaph Based mm Mscussmn anhe begmmng mum Pam dam ems 522mm be dumg a guudjub 1 Vuvagmg7 Whyuvwhynuw m w lt1 n m w w be mce WW2 mum see huwthese ems muve abum m veamme Thwswe can du wnh a Mamcad muwe m va Wth h Me ems muve m Mme asswgn a vanab e m be FRAME Let s an tha have Because FRAME Wm be cummng Mme My Buy muwe Ms aHWs newvanab el Assng 1 uwevcase wqu be FRAME aH Caps 1 FRAME Mudmem caagwpmsxjemmmveumm NPE mm Mudme 7 Ca angMsx New m make a muwe u Tu Watchme mdmdua ems asthav muve vuu H havem theme We menu msmav unw puma nm hnes Dem use a svmbm Thuse avemu hm Thwswun twswb v ehanee Vuuv mm because 3 We path pmms ave vew c usemuethev uwevev vuu shumd nunee Ma Ma v39aws abe themes mm a hne m twee dms mdmawm v u ve nuW meme puma Y an Tuduthws quot and change We 3st m be quot m Thwswm make We gvaph Hank That s OK By dEVauH FRAME The gvaph s swmp y shuwmgme mma cundnmn DVaH ems m We quot25 w m 7 mm D Mudmem caaewpmsxjemmweveemm NPE mm Mudme 7 Ca angMsx chck un Tnnk Animmmn Thwswm hung up a meme buxwheve yuu can de ne FRAME Fm FRAME 1 Set me e be u the dEVauH 2 Sena 353quotquot 012mm numbevummesmps mquot umestep e mu 3 SetM e be FumesSec vwmm dusmu the meme bux Wu can muve Mu sumewheve 2 52 en Vuuv 5122quot Wuu need my use me muuse m Nuw chck un Animate m We mama bux Vuu shumd 522012 gvaph ammate mme mama bux 2 mesh eeumshee Mame Wm upen a new WmdDW eemamme VDUY muwe Chck my me mav buI un m 015 new Wmeewmwamh Vuuv muwe Once yuu ve saus ed Wm yum meme chck un SaveAs m We FRAME mama bux Gwe yum w m mudu e Vuu H need m submw thwsme a unu w m Vuuv cummeted mudu e Mhe Mme space be uw drzgrzndrdmp yum muwe aw me Tu u we 1 Usmg the VWvdqu Exp uvev upen me meey cumammg the awme yuujust made 2 Le hck my me we me n e W5 Mame Wmduw and ve ease n m We vegmn be uw m 7 mm H mu 7 m n m Madu eEILCataq YDMsL nswevedMawe aw Vuu canwew yum muwe Wm Mame nuw by enumeehekmg my me abuve mspxay Muwe eunumswm amumaucaHy appeav Mudmem caaewpmsxjemmweveemm NPB 102g Module 7 Cataglyphisl DON39T DELETE your avi file The above appearance of your movie is only a linkto yourfile not a quotrealquot copy ofthe movie You still need your avi file to view the movie and you39ll still need to turn in your avi file along with this module to receive full credit for your work Describe what you see in your movie How do the ants move as time progresses Does the pattern of their movement appear to change as the movie progresses How so You may have noticed that the present model is biased in the terrain that the ants search That is because all ants initially move due north Odeg it is this northern region that is most well searched Certainly this is inconsistent with quotFind the food near the nestquot lf food exists in the southern region close to the nest it needs to be found To search the terrain more evenly let39s relax our assumption about the initial condition of the orient array That is let39s allow ants to have an initial orientation otherthan Odeg It would be nice if all the ants could agree to leave the nest in different directions This would make their outgoing coverage near the nest most efficient However it39s not cearthat ants can coordinate this Our model shows ants all leaving at the same time but in reality their exits happen at different times How would they communicate their exit orientations And how would they identify and correct redundant orientations Let39s assume they don39t bother Let39s assume that the initial orientation of each ant is determined randomly with each initial orientation value from 180deg to 180deg being equally likely You should recognize from Part II that this language implies the random initial orientations are from a uniform distribution Begin below by clearing out the x y turn and orient vectors as you did before eg Exercise 401 416 X NaN y NaN tum NaN orient NaN Now copy the intialization assignments from Exercise 404 forthe x and y matrices and paste them here below Don39t copy the orient assignment you39ll create a new one shortly 417 Module07 32 CataglyphislSemiAnsweredxmcd NPB 102d Module 7 Cataglyphisl There a few different ways we could initialize the orient matrix If we stick with our above notation of assigning oriento am to be something we39ll need a way to generate random numbers one at a time This is somewhat different from how you39ve used the runif function in the past We could ask for runif17180deg180deg 743825deg and even though this gives us a single random number it is contained within a vector note the parentheses Indeed the output of runif is always a vector even if that vector contains only a single number This is a problem because if you try to assign oriento ant get what you expect Try it below Make the assignment and then below that ask what orient is 418 to be Iunif17180deg180deg you won39t m ant Iunif17180deg180deg 012345678 11 11 11 11 11 11 11 11 orient I I 0 You should see a single row with each element containing 11 This means that each element of the orient vector contains a subvectorthat in turn has one element This will not do We want numbers not subvectors Fortunately the solution to this problem is simple You need only use the subscript operatorto specify element 0 in the vector generated by runif Copv your initialization assignment from above and paste it here below Change the right side to be Iut f17180deg180deg0 using the subscript operator Below the assignment ask Mathcad what orient is just to make sure it39s correct onth an Iut11f1 7180deg180deg0 0 t orientI I 0 I 1 I 2 I 3 I 4 I 5 I deg 0 154673 151204 114308 85422 170109 Module07 33 CataglyphislSemiAnsweredxmcd NPB 102d Module 7 Cataglyphisl You may be scratching your head wondering why this technique works or why the technique in Exercise 418 doesn t As mentioned the output of runif is a vector as indictated by the parentheses in its output runif17180deg180deg 7121527 deg In this case we want a number not a vector so we use the subscript operator to extract element 0 from the vector runif17180deg180deg0 749128deg Notice that the parentheses are now gone With the orient matrix properly initialized you need only copy the turn assignment Exercise 405 calculation array Exercise 406 and graph Exercise 408 and paste them below 420 d tumltantgt morm en 00 m timestep Xi am 1 lrxii1 ant speedtimestep 5111or1entiil ant speed timestep cosorient 7 yiant yiilant i 1ant ant orlentiil ant 171 ant orient 1 s Modue07 34 CataglyphislSemiAnsweredxmcd NPB 102q Module 7 Cataglyphisl By clicking on the turn assignment above and pressing F9 you can View different quotrunsquot ofthe model Write a sentence or two about how this ouput looks different from your previous model Exercise 408 odue07 35 CatagyphislSemiAnsweredxmcd NPB102q Module 7 Cataglyphisl Part V The Effect of a Goal Explore how the value of o affects foraging strategy We introduced the a variable back in Part III as a descriptor of the Gaussian distribution giving rise to the turn array At the time I mentioned there is some evidence that Catagyphis foraging can be modeled by assuming a a value of about 10deg And indeed the output you created in Part IV is a reasonable model of Catagyphis foraging However we should recognize that o is something of a free parameter in our model A o of 10deg makes the model quotlook rightquot but that doesn39t mean we should leave othervalues unexplored For one the model may be rather insensitive to changes in a That is we might get the same output regardless of what we set a to That would be an important observation and would have to be reconciled with the existing experimental data Alternatively the model may be very sensitive to changes in o In this case the ants39 apparent choice of a 10deg value might indicate that this is an optimal value for the conflicting goals they face when foraging We should then ask Why Why is that value of a good but othervalues bad You will attempt to answer these questions in this Part Copy your last model from Part IV and paste it here below You39ll need regions from Exercises 416 417 419 420 501 X NaN y NaN tum NaN orient NaN Jim 0 0 ant 0 onthi ant Iumf17180deg180deg0 Man morm i 0 a timestep Lam X171 ant speedtimestepsmor1entii13am speedtimestepcosorient yiant yiilant 171ant orlenti ant orlentk1 ant tumiilaant Module07 36 CataglyphislSemiAnsweredxmcd NPE mm Mudme 7 Ca auvphwsx Pan w That s guud New because me me e a m genevatmg mms may m be dean M s begm by exammmg a hwsmgvam uHhe tums yuu used m yum as1 mudg Cun vmmat a s 52 e meeg NW D m nuL yuu needm we um Why bemve cummumg Vuu v asswgned um m mn mu 2 n be meeg back m Exevmse 3 U7 ltngt mm he dz Cveate a megvam gvaph u yum Ium ma Tum Angle 3 us Tm s swm anu me hwsmgvam uHhe m innrm Venn yu made n Exevmse ans The un y eweyenee hwsume ysmame cunvevHu hequency yuu must ewe by avtun caluxn because We Ium anay mm mmensmns nm ene ms ead uHus avtun Mudmem CalagwmsLSemxAnwEvedxmcd NPB 102q Module 7 Cataglyphisl o 10deg tumihist histogramintervals tum tumihistlt1gt rows tum colstum 40 Frequency in l 0 7 200 7100 0 100 200 111117111350gt deg Turn Angle Now make a new copy of everything in Exercise 501 and 502 and paste it here below This is where you39ll change a 503 X NaN y NaN tum NaN orient NaN 303m 0 amt0 mam Iunif17180deg180deg0 W 60deg tumltantgt morm 00 m timestep Xi ant X171 ant speednmestep 5111or1entii1 ant yi ant yii1 ant speedt1mestepcosor1entiila ant Lonenti antJ L onentiil ant tuml ant Module07 CataglyphislSemiAnsweredxmcd NPB102q Module 7 Cataglyphisl 100 l m 7 100 100 7 100 x m 0 60deg histogramintervalstmn 100 80 gt o g tumihistlt1gt 393 rows tum c olstum 2 11 ll 20 3 0 7 200 7100 0 100 200 turnihistlt0gt deg Turn Angle 39 Module07 CatagyphislSemiAnsweredxmcd NPB 102g Module 7 Cataglyphisl In the spacejust above your turn assignment in Exercise 503 above create an assignment for o Assign a to be something larger like 60deg Your graph and histogram above should change you may need to set the histogram yaxis to range from 0 to 100 Write a couple sentences below about what you see Is the model sensitive to changes in o How similar is this result to the result displayed in Exercise 501 when a was 10deg Does the output ofthe model remind you of anything you39ve seen earlier in this module It should Why do you think this is 505 How effective do thinkthis foraging strategy would be ie with a set to 60deg More effective than a set to 10deg Less effective Why 506 Copy all your regions from Exercise 503 and paste them here below In your new a assignment below change a to be something really small like 2deg 507 Module07 40 CataglyphislSemiAnsweredxmcd Module 7 Cataglyphisl NPB102q X NaN y NaN tum NaN orient NaN M nt 0 Jim 7 0 orlentiaant run1f17180deg180deg0 W 2deg ltamgt end tum morm 00 m nmestep Xi ant x171 ant speedtunestepsmor1entii13am yi ant yii1 ant speedtunestepcosor1entii1 ant tumiilant Lorlenti antJ orlentk1 ant 2 r 4quot X m laquot Module07 CatagyphislSemiAnsweredxmcd


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