Roman Camps and Orientations

Feb 17, 2007 Alun Science, The Past 5 Comments Ancient History, Archaeoastronomy, Archaeology, Mathematics, Roman Archaeology, Statistics

A follow-up to The Ori­ent­a­tion of Roman Camps and Forts. This is an applic­a­tion of the Bino­mial Dis­tri­bu­tion test that I’m using in my own work applied to the data from the ori­ginal paper, which is why you may have the impres­sion you’ve already read this recently. My ana­lysis may not be cor­rect, so I’m put­ting it up on iScience and sub­mit­ting that to Car­ni­val of Math­em­at­ics and Tangled Bank to see if people think the maths is wrong. I’m also put­ting it up on Revise and Dis­sent where it will get sub­mit­ted to the His­tory Car­ni­val and Four Stone Hearth to see if it’s intel­li­gible and sounds reas­on­able to His­tor­i­ans and Archaeologists.

There is cur­rently a debate in the pages of the Oxford Journal of Archae­ology on the ori­ent­a­tions of Roman camps and forts. Richard­son (2005:514–426) argues that the ori­ent­a­tion of these camps is non-random and relied on some form of astro­nom­ical obser­va­tion. He presents data which he argues sup­ports his case. Recently Peterson (2007:103–108) has argued this relies on a flawed use of the Chi-squared test. I accept Peterson’s find­ings that Chi-squared is not a use­ful method. How­ever examin­ing the camps as a bino­mial dis­tri­bu­tion would be feas­ible and would make expli­cit the archae­olo­gical and astro­nom­ical assump­tions made in the argument.

What is a Roman Camp?

The sites being examined are Roman camps and forts in Eng­land. One of the major advant­ages that the Roman army had over the nat­ive oppos­i­tion when occupy­ing new ter­rit­ory was their organ­isa­tion. The Roman army was effect­ively a pro­fes­sional army tak­ing on ama­teurs. Their camps reflect this organ­isa­tion. Typ­ic­ally their early camps a ditch sur­roun­ded by a bank in a playing-card shape. They fol­lowed a set design. The rationale for this was if there were attacked by sur­prise equip­ment and people would be in the same place at each camp, min­im­ising the effects of the surprise.

A Roman fort at Wall­send. Photo from Google Earth

The ancient sources give some detail on how to lay out a Roman camp. The main gate should face the enemy, or the line of advance (Vege­tius 1.23, Hyginus 56). The rear gate should be on the higher ground to aid sur­veil­lance. Sites over­looked by hills were con­sidered a bad idea, as were sites near wood­lands which would allow the enemy to sneak up on the camp. The basic lay­out of the camp could be set up quickly by sur­vey­ors using gro­mae, sur­vey­ing tools for lay­ing out lines at right angles. Hyginus (chapter 12) states that you set up your groma at the junc­tion at the centre of the camp and lay out your roads to the gates from there.

This would appear to be an effi­cient method of lay­ing out a camp. Were obser­va­tions to ori­ent­ate the camp also part of the method? It doesn’t seem neces­sary, but Richard­son (415,422–23) provides quotes from ancient sources which sug­gest this is plaus­ible hypo­thesis in some cir­cum­stances.

Test­ing the camps for astro­nom­ical significance

To test his hypo­thesis Richard­son has examined plans and meas­ured the angle the long axis of a camp makes to the meridian, which yields an azi­muth from true north. The accur­acy of these meas­ure­ments is ques­tion­able, not through Richardson’s work but in that they rely on an accur­ate mark­ing of north. A big­ger dif­fi­culty is that such meas­ure­ments do not include declin­a­tion data. If the local hori­zon is flat then an align­ment will point to a dif­fer­ent part of the celes­tial sphere than if it is moun­tain­ous. Finally Richard­son notes that some Roman camps, being square, lacked a long axis. He states “…these were alloc­ated to the nearest con­veni­ent group.” He then applies the Chi-squared test to his data. This is iden­ti­fied by Peterson as the weak point of Richardson’s ana­lysis, so this should be examined closely.

Richardson’s hypo­thesis is that there is a delib­er­ate ori­ent­a­tion towards the car­dinal points. There are 360 degrees in a circle and four car­dinal points. A camp can point dir­ectly at a car­dinal point. It can be within one degree of the tar­get, within two degrees, within three degrees and so on. How­ever, by the time you get to forty-five degrees you’re work­ing closer to the next car­dinal point. Richard­son there­fore says there are forty-five pos­sible prob­ab­il­ity bins into which a camp can be placed. If there are sixty-seven camps in the sample, you would expect (67/45 = 1.49) 1 and a bit camps in each bin. Between one and two then. What Richard­son finds that some bins have con­sid­er­ably more. For the Chi-Squared test what you do is go in each bin and sub­rac­ted the expec­ted value from observed value. You take the answer and square it, and then divide it by the expec­ted value. You do this for all the bins and add up the res­ults and call this final value chi-squared. Con­fus­ingly the Greek let­ter χ — chi looks like an X so you’ll often see it writ­ten X².

Now what does this value tell us? What we need to do is look up the value in a χ² table. Richard­son has argued there are 45 dif­fer­ent bins, so we look up the χ² value with 44 degrees of free­dom. This will give us a range of val­ues. As χ² gets higher, so it becomes more and more improb­able. In Richardson’s case he argues that this value of χ² in would only hap­pen less than one time in a thou­sand by chance. The prob­lem is how robust are these results?

This is the key cri­ti­cism of Peterson, who argues that Chi-squared tests can­not reli­ably work on small samples. Wor­ry­ingly he argues that the min­imum expec­ted value in any bin should be five, which means the min­imum sample size for such a test should be 5x45= 225 camps. As a gen­eral rule this sample size is simply lar­ger than could reas­on­ably be hoped for.

There may be an altern­at­ive method. There is a prob­ab­il­ity that the camps will or will not sat­isfy cer­tain cri­teria. This would appear to be suit­able for exam­in­a­tion as a bino­mial dis­tri­bu­tion. A bino­mial dis­tri­bu­tion is the set of cir­cum­stances that an event either will or will not occur in n attempts, where n is the sample size.

Bino­mial distributions

The easi­est example is toss­ing a coin. It either will or will not land on heads. If you tossed sixty coins at the same time, then on aver­age they would have heads thirty times. But not every­one would have thirty heads. There would be a nor­mal dis­tri­bu­tion which means that there would be a spread of res­ults some people would have thirty-one heads, oth­ers only twenty-nine. The for­mu­lae gov­ern­ing a bino­mial dis­tri­bu­tion are simple.

n is the sample size
p is the prob­ab­il­ity the event will hap­pen between 0 — abso­lutely will not hap­pen — and 1 abso­lutely will hap­pen.
q is the prob­ab­il­ity the event will not hap­pen, which will be 1-p

The aver­age res­ult will be np. In out case of sixty coin tosses, the aver­age res­ult will be 60x0.5=30 heads.

The stand­ard devi­ation will be √(npq)

The reason you would want the stand­ard devi­ation is that it tells you how sig­ni­fic­ant a res­ult might be. If you toss sixty coins in this dis­tri­bu­tion 2/3 of the time you will have 30 +/- 1 stand­ard devi­ation heads. Ninety-five per cent of the time you will have 30 +/- two stand­ard devi­ations. Ninety-nine per cent of the time you will have 30 +/- three stand­ard devi­ations. Usu­ally in the social sci­ences people get inter­ested after 2 stand­ard deviations.

Graph of prob­able res­ults of toss­ing 60 coins. Red area is one stand­ard devi­ation. Red and Orange two stand­ard devi­ations. Click for full size.

This method doesn’t require the prob­ab­il­ity to be 50:50. Take a stand­ard six sided die. The prob­ab­il­ity of throw­ing a six is one in six. If you throw sixty dice you should expect about 10 sixes. Some­times you’ll have more other times less. Would it be unusual to throw twelve. In this case the for­mu­lae can show the answer.
Aver­age = np = 60*1/6 = 10
Stand­ard Devi­ation = √(npq) = √ (60x1/6x5/6) = 2.886751
So two thirds of the time you should expect to throw between eight and twelve sixes.

Chance of throw­ing sixes with sixty dice. Red area is one stand­ard devi­ation. Red and Orange two stand­ard devi­ations. Click for full size.

The Bino­mial dis­tri­bu­tion of Roman camps

What has this got to do with Roman camps? The Roman camps have a prob­ab­il­ity asso­ci­ated with them. They could by chance line up they do. So to assign a value to p we need to openly state what the prob­ab­il­ity they align as they do by chance is. This is an expli­citly archae­olo­gical ques­tion. To return to the camps, how to you align a camp to North?

You could align hold up a plumb line to the North star and let it drop. If you do this then there’s a prob­lem. Polaris isn’t exactly on the North Celes­tial Pole. In AD 250 it was at 89° 17′. So there’s a one and a half degree spread on either side of the North Celes­tial Pole. If you factor in at least some human error then the min­imum accur­acy you can argue for is 2/360. If you bring in evid­ence from the his­tor­ical data then you could point out these camps were built dur­ing the day when Polaris wasn’t vis­ible. This could make the align­ments less accur­ate and any­thing +/- five degrees might have been thought to have been aligned north, in which case p is 10/360. Oth­ers might argue for other accuracies, but to get a value for p, you have to be expli­cit about which archae­olo­gical or his­tor­ical evid­ence you’re using. Once you have a value you can plug it in, and if you like use a few other val­ues too to see how robust your res­ults are.

In the case of Richardson’s data he gives the data in 10° wide blocks between 0° and 180° from north. This gives a camp a 1/18 chance of being in any bin and allows us to plot this prob­ab­il­ity graph. The aver­age is four. I’ve shaded to six green and to eight orange. In real­ity the stand­ard devi­ation is 1.9, so the shad­ing should stop just short, but his res­ult of six camps facing within ten degrees of north does not look sig­ni­fic­ant enough to be troubled with.

Camps aligned to north. Click for full size.

His other hypo­thesis is that the camps were aligned within 10 to car­dinal points. In this instance this makes p equal to 4/18. This is an inter­est­ing res­ult. On aver­age we should expect fif­teen camps to match. The stand­ard devi­ation of 3.4 means that 15+/-6.8 between nine and twenty-two camps should face car­dinal points. There is an ele­ment of doubt. If you throw enough tests at the data then you’d expect about one in twenty to have sig­ni­fic­ance at two stand­ard devi­ations. Non­ethe­less as a fil­ter to ask if an idea is worth invest­ig­at­ing fur­ther it is very useful.

Camps aligned to car­dinal points. Click for full size.

Part of the prob­lem is that it’s clear that local con­sid­er­a­tions, like the pres­ence of a hos­tile force or dir­ec­tion of travel took pre­ced­ence in plan­ning. With closer sur­vey it would be pos­sible to elim­in­ate some of these known non-astronomical align­ments. The easi­est cat­egory to remove would be those whose align­ments matches the align­ments of local roads as it seems highly likely that the army would have stopped here while mov­ing along this route. The next step would be to sur­vey the sites and see if other lim­it­a­tions like local topo­graphy define the ori­ent­a­tions of camps. If pre-defined ori­ent­a­tions can be elim­in­ated then it may be pos­sible to show astro­nom­ical align­ment was used as a guide when there wasn’t a local enemy to face, or a route to follow.

There are a couple of advant­ages in using a bino­mial test for stat­ist­ical sig­ni­fic­ance. One is that it is com­par­at­ively sim­pler to use than Chi-squared. Chi-squared may not be arcane math­em­at­ics, but the sug­ges­tion by Peterson that help needs to be sought can be given to most archae­olo­gists. Even in cases where Chi-squared is appro­pri­ate, if archae­olo­gists can­not fol­low the argu­ment, then per­haps another method is needed. Hope­fully the above method is slightly more transparent.

The other advant­age is that it can be applied to smal­ler samples. If the sample is too small to be stat­ist­ic­ally sig­ni­fic­ant then this will become appar­ent. Take for instance toss­ing two coins. If they both come up heads is that sig­ni­fic­ant? np states the expec­ted aver­age is one, two stand­ard devi­ations would be 1.4, so we would expect 95% of samples to fall between 1 +/- 1.4. Given the max­imum res­ult will be 2, we will always be in this range. The test tells us the sample is too small to be helpful.

Also because the test is about eval­u­at­ing the prob­ab­il­ity some­thing will hap­pen it can be used for non-spatial data. For example recently there have been claims that the Iron Age peoples of Fisker­ton have been able to pre­dict lunar eclipses using Saros cycles. This is data dis­trib­uted in time, but non­ethe­less is should be pos­sible to use the archae­olo­gical record and the data to argue for a value of p for the prob­ab­il­ity of pre­dict­ing an eclipse in a Saros cycle within a given time­frame. That would prob­ably be the other example used if I expand this into a full paper.

Bib­li­o­graphy

Peterson, J.W.M. 2007. ‘Ran­dom Ori­ent­a­tion of Roman Camps’. Oxford Journal of Archae­ology 26(1). 103–108.
doi:10.1111/j.1468–0092.2007.00275.x

Richard­son, A. 2005. ‘The Ori­ent­a­tion of Roman Camps and Forts’. Oxford Journal of Archae­ology 24(4). 415–426.
doi:10.1111/j.1468–0092.2005.00244.x