Droopy wrote:Chap wrote:I see that Droopy has a thirst for truth about the Book of Abraham.
Does that thirst motivate him to do a little math?
If so, let him go to
this thread, and do the necessary sums to show I am wrong in saying that Gee's attempt to derive an original scroll length of anything like 1200 cm does not hold water at all.
No algebra required, by the way. Just the ability to use a simple spreadsheet.
I am hoping that unlike the apparently absconded Schryver, the master of dialectic that we see in Droopy will not try to justify a mathematical conclusion by an argument from supposed authority.
I'm not an expert on that particular split hair, and I'd have to look at what William has to say about it before taking your word for things.
I do know that we have clear, eyewitness accounts from the nineteenth century bearing witness to large quantities of material not now extent. It clearly existed, and equally clearly is not now in our possession.
You may or may not be correct on about the length of this particular text, but the large sums of missing material that are not a part of either Gee's or your analysis raise serious questions about any attempts to stick the present material in our possession with the entire Book of Abraham.
That argument failed decades ago.
Schryver wrote:Chris Smith has apparently prepared what he believes to be a rebuttal of Gee's measurements and calculations concerning the scroll of Horos. How he presumes to speak to the issue authoritatively without having yet obtained access to the originals is a mystery to me. But he's going to go out on a limb. And, in a way, I admire that. However, the fact remains that the only way Gee's calculations can be wrong is if he either: 1-is lying, or 2-is inept. I've seen no indication, at this point, of either being the case. In fact, if Gee's measurements and calculations don't pan out in terms of this scroll length stuff, he has much more to lose than he ever could have hoped to gain. I think his professional credibility would come into serious question. But, as of now, it's not, the mindless rantings of Dissonance and others notwithstanding.
I tend to feel that in the same way that the Book of Abraham is something of a smoking gun for Joseph Smith, the question of the supposed length of missing papyrus may turn out to be a smoking gun for John Gee. And the nice thing is that anyone who can do simple arithmetic can follow the argument for themselves.
In the
thread I mentioned above, I brought the argument down from the level of algebra to the simpler level of an easy Excel spreadsheet. But from the responses above it looks as though Droopy and Schryver don’t trust an argument in Excel any more than they trust algebra. So let’s go a level further down.
Here is what John Gee had to say in his article for Farms Review:
http://farms.BYU.edu/publications/revie ... #_ednref25One can take a more scientific—that is, mathematical—approach because the circumference of a scroll limits the amount of scroll that can be contained inside it. Thus, we can determine by the size of the circumference and the tightness of the winding how much papyrus can be missing at the interior end of a papyrus roll. Friedhelm Hoffmann has already developed such a formula in calculating the amount of material missing from the end of Papyrus Spiegelberg, from which he was able to determine that there were five columns missing from the text.[24] I will not bore you with the derivation of the formula; it has been in print over a decade. If S = the average difference between the winding measurement, and E = the length of the last winding, then the theoretical length of the missing portion is Z, so that Z=((E^2-6.25)/2S)-E.[25] We can apply this to the Joseph Smith Papyri and obtain some usable results.
For the scroll of Noufianoub, the final winding length is 7.8 cm and the average difference is .33 cm. The formula says that there are 74 cm missing, which is just over 2 feet. Thus this vignette was at the very end of the roll it was on. Unfortunately there is no way of knowing how much was missing from the beginning of the scroll.
For the Semminis scroll, the final winding length is 14 cm and the average difference is .25 cm. Thus there were 365.5 cm left in the scroll. This is the equivalent of 143.9 inches, or nearly 12 feet. The vignette in Joseph Smith Papyrus II is the furthest vignette into the Semminis scroll and normally occurs about halfway through the Book of the Dead,[26] which means that the total scroll would be about 20 to 24 feet long. This is longer than some scrolls[27] but shorter than others.[28]
For the scroll of Horos, the initial winding length is 9.7 cm, the last winding is 9.5 cm, and there are seven windings in total. This leaves us with an average value of .03333 for S. E is, as already stated, 9.5 cm. Plugging this into the equation gives 1250.5 cm of missing papyrus. This is the equivalent of 492.3 inches, or 41 feet of missing papyrus. Would such a thing be unusual? No. P. Turin 1791 is 57 feet 3 inches (= 1745 cm) long,[29] P. Nesmin is 1280 cm long,[30] and the ritual roll of Imouthes in the Metropolitan Museum of Art is 1088.6 cm,[31] 428.6 inches, or 35 feet 8 inches long, while his Book of the Dead is longer still. So such a length is not out of the question. While we know that the scroll of Horos had a text of the Document of Breathings Made by Isis, about one in four of those documents contain additional texts.[32] So the presence of additional texts would not be unusual.
The size of Horos's scroll at first seems excessive, even though it is not unheard of. When I first plugged the numbers in a few years ago and got the result, I checked the measurements and then checked them again. Then I checked the formula again. Then I rechecked the formula's derivation. Then I rechecked the assumptions behind the formula. Then I simply dismissed them and went back to the standard roll length. I had always assumed that the Semminis roll would be the longer one since the Book of the Dead is a much longer composition than the Document of Breathings Made by Isis and my initial estimates of the length of the Semminis roll, based on the length of the text preserved and the percentage of the Book of the Dead preserved, had been almost twenty feet. It was only after plugging the numbers from the other Joseph Smith Papyri into the formula that I realized that the formula does give reasonable results. I have since realized that having a long roll of Horos brings all the nineteenth-century eyewitnesses into agreement.
One might thoughtlessly suppose that one could make measurements from just any photograph. Most of the photographs, however, are not to scale (and making measurements of the Statue of Liberty from photographs, for example, might lead to the conclusion that it is only two inches high). Even if the photographs were to scale, photography can also introduce other distortions. Measurements from photographs are suspect and extremely susceptible to distortion. Calculations from such sources are therefore problematic.
Here’s how we understand this issue without needing algebra, or even spreadsheets, and certainly without needed to go into the question of whether the formula Gee uses is correct or not:
1. Suppose we take a length of papyrus and roll it up into a scroll. That means that if we cut a cross section we would see a spiral made by the cut edges of the papyrus going round a quite big circle on the outside, with the next layer a bit smaller, the next layer after that a bit smaller, and so on inwards.
2. If the papyrus is wound round a wooden rod, then the innermost and smallest layer will just be long enough to go once round the outside of the rod. If there is no rod, we can get more papyrus into the centre of the scroll. Each layer will go on getting smaller, until the thickness and stiffness of the papyrus means we simply cannot get another layer inside the last one. It doesn’t make much difference to the total length of papyrus whether or not we can cram an extra last turn of papyrus into the centre of the roll, since the innermost few layers will be very short in any case.
3. John Gee and California Kid proceed on the assumption that the scroll is wound with uniform tightness throughout, and that as a result there is a constant difference between the length of each layer of papyrus and the one that fits inside it. Since this point is not in dispute, I shall not try to give a non-mathematical proof of it, but for anyone who realises that the circumference of a circle is 2π times the radius, it is clear that if we approximate the tight spiral by a series of concentric circles, then since the radius of each layer decreases by the thickness of the papyrus, t, its outer circumference must be 2πt less than the outer circumference of the layer just outside it. Anybody who finds this paragraph non-reader friendly can forget it. Just remember that it is common ground that everyone that as the papyrus spirals inwards there is a more less constant difference between the length of successive wraps.
Now let us take an example.
For the scroll of Noufianoub, the final winding length is 7.8 cm and the average difference is .33 cm. The formula says that there are 74 cm missing, which is just over 2 feet.
So the first wrap is 7.8 cm long. If we use Gee’s figure for the constant average difference of 0.33 cm, that means we would expect the next wrap to be
7.8 cm-0.33 cm = 7.47 cm
The next wrap would be
7.47 cm-0.33 cm = 7.14 cm
(Of course the second figures after the decimal point are pretty meaningless in practice – they are a result of taking an average difference between successive wraps that we can probably measure to no better accuracy than 1/10 of a centimetre, i.e. a millimetre)
If we go on like that we will get a series of smaller and smaller figures for the length of each wrap, until we hit a point where the next wrap would come out at less than zero, at which point we have clearly reached the centre of the tightest possible roll.
So here we go (all the measurements are in centimetres, and I have given the wraps numbers, with the outermost one labelled 1):
1: 7.47
2: 7.14
3: 6.81
4: 6.48
5: 6.15
6: 5.82
7: 5.49
8: 5.16
9: 4.83
10: 4.5
11: 4.17
12: 3.84
13: 3.51
14: 3.18
15: 2.85
16: 2.52
17: 2.19
18: 1.86
19: 1.53
20: 1.2
21: 0.87
22: 0.54
23: 0.21
Now if you add up all those numbers, you will get the total length of papyrus in the scroll. Try it. The result comes to 88.32 cm. Please note that this is not some kind of rough approximation, inferior to a result predicted by a formula. It is the real result of the initial assumptions made, that is, we start with an outer roll of length 7.8 cm, and decrease the length of each successive layer at by 0.33 cm until we can’t get any more layers in.
Of course, since this result depends on measurements of limited precision, it will have fuzzy edges. If our estimate of the average difference between the length of wraps was to decrease by about 10% to 0.30 cm, we would find that we could squeeze in 23 layers, and the crude total of the lengths would come to 96.75 cm. A 10% increase in difference to 0.36 cm would reduce the number of possible layers to 21, and decrease the crude total of the lengths to 81.27 cm. So if you want to use normal scientific caution, it would probably be safe for us to say that the total length of the papyrus is likely to be somewhere in between 81 cm and 97 cm.
(Since what Gee calls the “missing length” appears to be the total length of papyrus minus the length of the outer wrap, his figures seem to imply a total length of
74 cm + 7.8 = 81.8 cm, very close to the “brute force” result we have obtained so simply.
One interesting point, that John Gee does not mention, is that this calculation implies a maximum thickness of the papyrus, on the assumption that the scroll is tightly wound. If the outer wrap has a circumference of 7.8 cm, that means that its radius must be
7.8 cm /2π = 1.24 cm
Now that is a space within which we have to cram 23 layers of papyrus. That means that the maximum thickness of one layer of papyrus must be:
1.24 cm / 23 = 0.05 cm, i.e 0.5 mm, half a millimetre.
That is consistent with what one source says about ancient papyrus thickness (0.35 - 0.45 mm thick):
http://lists.ibiblio.org/pipermail/corp ... 04858.htmlNow let us turn to the Horos scroll discussed above by John Gee. Adopting his figures, we start with an first wrap of length 9.5 cm, with a difference between wraps of 0.03333 cm. (I continue to use John Gee’s choice of precision here).
So let’s get subtracting. Off we go:
1: 9.5
2: 9.46667
3: 9.43334
4: 9.40001
5: 9.36668
6: 9.33335
7: 9.30002
8: 9.26669
9: 9.23336
10: 9.20003
11: 9.1667
12: 9.13337
13: 9.10004
14: 9.06671
15: 9.03338
16: 9.00005
17: 8.96672
18: 8.93339
That is just the first 18 layers. Since the difference between layers is only 0.03333 cm, the length of each wrap is decreasing very slowly. Some readers may like to work out how long it is all going to go on, but for those who can’t manage all that subtracting, here are the last few rows:
278: 0.26759
279: 0.23426
280: 0.20093
281: 0.1676
282: 0.13427
283: 0.10094
284: 0.06761
285: 0.03428
286: 0.00095
Now I don’t think Doctor Shades would like me to put in all 286 rows of figures, and perhaps some readers of this board might not feel like trusting an atheist when he says that the total length of all the wraps comes to 1358.63585 cm to full precision.
But you really don’t have to trust me. If there are 286 layers, and the first one is 9.5 cm long while the last one is pretty well zero, then the average length is clearly close to:
9.5 cm /2 = 4.75 cm
and since there are 286 layers, the total length must be about
286 x 4.75 cm = 1358.5.
Good enough? Well it is certainly pretty consistent in magnitude with Gee’s figure for “missing papyrus” of 1250.5 cm, given the precision problems we have already discussed.
But in fact there is a GIANT problem of physical implausibility here. Applying the reasoning set out above, if the circumference of the outer layer of papyrus is 9.5 cm, which gives a radius for the roll of
9.5 cm /2π = 1.5 cm to the same precision.
Within that space we somehow have to get 286 layers of papyrus! If that is the case, then the maximum thickness of each layer of papyrus can only be
1.5 cm /286 = 0.0053 cm, i.e. 0.053 mm, about 1/20 mm.
This is thinner than any modern-day paper you are ever likely to meet, and is certainly much thicker than ancient papyrus (or modern papyrus for that matter) could ever be. On the basis that ancient papyrus was around half a millimetre thick, it is 10 times too thin.
If John Gee had considered this simple physical point, I do not believe he could possibly have made the claim he has for the total length of the scroll, assuming it to have been tightly wound within an outer circumference of 9.5 cm. Please note that if the scroll was not tightly wound, the situation would be much worse, since the papyrus would have to be thinner still.
If we take a more reasonable value for thickness of papyrus, such as the 0.5 mm value that we get by assuming the scroll of Noufianoub to have been tightly wound, then within the 1.5 cm radius of the roll we are discussing, we could get a maximum of 30 layers. Since the outer layer is 9.5 cm in circumference, and the innermost one is effectively zero, that gives us 30 layers each with an average length of 4.75 cm, a total of 142.75 cm.
Readers who look back at
this thread will see that California Kid and I, based on his careful re-measurement of photographs, agree in producing figures of this order of magnitude. In his Farms Review piece, John Gee makes great play of the difficulty of being sure about measurements taken from photographs. He is right only to the extent that it is difficult to be sure to what scale the photographs are reproduced; but even a 20% uncertainty will only suffice to make 100 cm length estimates vary between 80 cm and 120 cm. There is no way that the scale of the published photographs can be so uncertain as to produce anything like the length John Gee proposes.
To sum it all up:
The question of the rightness or the wrongness of the formula used by John Gee is a complete red herring. We do not need any formula to produce a total length of a papyrus scroll from the kind of measurements that are being discussed here, since we can produce the result by repeated subtraction and adding up the resultant sequence of lengths, as done here. John Gee’s claim that the length of successive wraps decreases on average by 0.03333 cm implies that the scroll was made of ludicrously thin papyrus, and is contradicted by measurements that anybody with a ruler can make on the basis of the published photographs.
I don’t know why Gee is as wrong as he obviously is. I don’t know why he hasn’t noticed that he is so obviously wrong, particularly in view of the implied thickness of papyrus following from his results. But wrong he is.
