Some Puzzles from the John Gee Paper

[Mortal Man]

According to John Gee, the scroll of Horos was originally 41 feet longer than what remains today. http://ispart.byu.edu/publications/review/...d=699#_ednref25

How big were the rolls?

...

One can take a more scientificâ??that is, mathematicalâ??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. 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

[PacMan]

I was with you until here:

The unknown here is the length of the final (Nth) winding, WN. An upper bound on the length of the papyrus (L) is obtained by assuming that the scroll was wound tightly to the core. If the papyrus was rolled up by first folding the edge in half, then a reasonable estimate for the final (innermost) winding is WN = 2*T = 0.166 cm, which yields AN = 0.0022 cm2 and Ar = 7.48 cm2. The maximum length of the papyrus is

L = Ar/T = 90 cm

Please explain again how the maximum bounds of any such length is determinable.

PacMan

[Mortal Man]

I was with you until here:

Please explain again how the maximum bounds of any such length is determinable.

PacMan

Since AN is negligible, I'll edit my OP to remove this distraction...

For simplicity, assume that the original scroll was tightly wound all the way to the center; i.e., that it was a solid cylinder (AN=0).

Viewed from the end, the scroll appears as a circle. Unwinding the scroll it becomes a rectangle. Since no mass is gained or lost in the process of rolling/unrolling, the area of the circle must equal the area of the rectangle. The area of the rectangle is L (length) times (T) thickness. The area of the circle is pi (3.14159) times R1², where R1 is the radius of the rolled-up scroll. Thus L*T = pi*R1² or L = pi*R1²/T

Is this what you were asking?

[PacMan]

Since AN is negligible, I'll edit my OP to remove this distraction...

For simplicity, assume that the original scroll was tightly wound all the way to the center; i.e., that it was a solid cylinder (AN=0).

Viewed from the end, the scroll appears as a circle. Unwinding the scroll it becomes a rectangle. Since no mass is gained or lost in the process of rolling/unrolling, the area of the circle must equal the area of the rectangle. The area of the rectangle is L (length) times (T) thickness. The area of the circle is pi (3.14159) times R1², where R1 is the radius of the rolled-up scroll. Thus L*T = pi*R1² or L = pi*R1²/T

Is this what you were asking?

Mass is one thing, but you're doing area calculations. You assume that constant mass equal constant area. That's incorrect. With a line of definite dimension, you can make multiple shapes with varying areas.

Moreover, the 'tightness' of a rolled up scroll can be significantly different to the 'tightness' of the individual sides of the rectangle. But to be honest, I'm not even sure if you're comparing the circle (the end of the wound-up scroll) to the rectangle of the end of the wound-up scroll, the rectangle created by the side of the wound-up scroll, or the rectangle of the papyri completely unwound. All the same, I think the distinction is moot given your erroneous assumption above (if I understand you correctly).

Best,

PacMan

[PacMan]

Mass is one thing, but you're doing area calculations. You assume that constant mass equal constant area. That's incorrect. With a line of definite dimension, you can make multiple shapes with varying areas.

Moreover, the 'tightness' of a rolled up scroll can be significantly different to the 'tightness' of the individual sides of the rectangle. But to be honest, I'm not even sure if you're comparing the circle (the end of the wound-up scroll) to the rectangle of the end of the wound-up scroll, the rectangle created by the side of the wound-up scroll, or the rectangle of the papyri completely unwound. All the same, I think the distinction is moot given your erroneous assumption above (if I understand you correctly).

Best,

PacMan

P.S. An example:

Take a square with 1 inch sides. That's 4 inches, with an area of 1.

Take that same 4 inch line and make a circle. The area will be (C/2pi)^2*pi. That's (4/6.282)^2*3.141=~1.27. The areas are obviously different. And that's why I hated geometry.

[Mortal Man]

Mass is one thing, but you're doing area calculations. You assume that constant mass equal constant area. That's incorrect. With a line of definite dimension, you can make multiple shapes with varying areas.

I am assuming conservation of volume. The width of the papyrus cancels out, thus if volume is conserved then area is also conserved.

Moreover, the 'tightness' of a rolled up scroll can be significantly different to the 'tightness' of the individual sides of the rectangle.

If significant squeezing occurred as the papyrus was rolled up, then T would decrease and L would increase but L*T and Ar would remain constant. As the papyrus was unrolled, T would increase and L would decrease back to their original values.

But to be honest, I'm not even sure if you're comparing the circle (the end of the wound-up scroll) to the rectangle of the end of the wound-up scroll, the rectangle created by the side of the wound-up scroll, or the rectangle of the papyri completely unwound. All the same, I think the distinction is moot given your erroneous assumption above (if I understand you correctly).

I am keeping the volume of the wound and unwound papyrus constant.

[MAsh]

According to John Gee, the scroll of Horos was originally 41 feet long.

This math is way over my head, but are you sure the photos from which you measured are to scale or proportionally accurate? If so, how do you know this to be the case?

Mike

[Mortal Man]

P.S. An example:

Take a square with 1 inch sides. That's 4 inches, with an area of 1.

Take that same 4 inch line and make a circle. The area will be (C/2pi)^2*pi. That's (4/6.282)^2*3.141=~1.27. The areas are obviously different. And that's why I hated geometry.

I am not enclosing an area with a line. I am comparing two areas that are topologically equivalent.

[PacMan]

I am assuming conservation of volume. The width of the papyrus cancels out, thus if volume is conserved then area is also conserved.

The how do you explain the difference in areas between a circle and a square? I think you're basic premise is flawed.

If significant squeezing occurred as the papyrus was rolled up, then Ar would be less than L*T, thus making L even smaller. In other words, as the papyrus was unrolled, its thickness would increase causing it to shrink lengthwise.

No it wouldn't. At the time of the unraveling, it expands in width, but that does it mean it must shrink lengthwise. The idea is that more air is taken in--not material. Thus no material needs to be "shrunk."

I am keeping the volume of the wound and unwound papyrus constant. Any squeezing that may or may not have occurred during the winding would result in a length shorter than 90 cm.

Again, that doesn't make intuitive sense to me.

Cheers,

PacMan

[Chris Smith]

Hi Mortal,

I agree that Gee's measurements are physically impossible, and presented that finding, along with my own analysis, at Sunstone West last week. I posted a preliminary version of my analysis at the Mormon Discussion Boards a while back, in case you're interested. Unfortunately I can't link it here due to board rules.

My analysis agrees very much with your own, though I measured a larger number of the extant wraps and got a slightly longer estimate for the length of the missing portion. Your analysis neglects JSP X, the innermost extant fragment. It's definitely good to have done this analysis independently, so that our results confirm each other.

FWIW, Michael D. Rhodes estimates that each of the missing columns and the closing vignette would have been 20 cm wide, not 10.

Best,

-Chris

[Mortal Man]

This math is way over my head, but are you sure the photos from which you measured are to scale or proportionally accurate? If so, how do you know this to be the case?

Mike

The best photos I have are in my copy of the Nibley-Gee-Rhode's book, The Message of the Joseph Smith Papyri. The proportions there match the jpeg image. I did not know the scale factor of the jpeg image a priori. To obtain the scale factor, I compared the A-B distance in the image to Gee's stated value of 9.7 cm for the outer winding length. If Gee's number is correct, then the scale factor of 1.2 should be accurate. If you can send me an image that is precisely to scale, I would be most grateful. Better yet, just mail me the originals.

[William Schryver]

So have you forwarded your conclusions to John? He's quite easy to contact, you know. I do it all the time. Unless he's traveling to deliver papers or to seminars, etc., he's always been quite prompt to reply to me.

He can be reached here: john_gee@byu.edu

I don't pretend to understand the science behind his claims or yours. I understand the general principle, but my brain is not wired to do that kind of thinking or number crunching. But I'll bet John would either show you where you're going wrong, or acknowledge his mistake if you could persuade him enough to see it. Give it a try.

[Chris Smith]

If you can send me an image that is precisely to scale, I would be most grateful.

Larson indicates the scale of his photos in his book. The Improvement Era photos include rulers in the actual photographs themselves. And the BYU Studies photos indicate that they are exact size, though unfortunately the version you download from the BYU Studies website does not preserve the scale. If PM me your email address I'll send you my copies of the Larson and IE photos, which I have corrected for scale to the best of my ability.

[MAsh]

The best photos I have are in my copy of the Nibley-Gee-Rhode's book, The Message of the Joseph Smith Papyri. The proportions there match the jpeg image. I did not know the scale factor of the jpeg image a priori. To obtain the scale factor, I compared the A-B distance in the image to Gee's stated value of 9.7 cm for the outer winding length. If Gee's number is correct, then the scale factor of 1.2 should be accurate. If you can send me an image that is precisely to scale, I would be most grateful. Better yet, just mail me the originals.

The original papyri is being sent to you FedEX. I've also included two plates from the golden plates, Joseph's seer stone, and the original hat in which he placed the stones.

Mike

[Chris Smith]

I've also included two plates from the golden plates

Don't you mean tin?

[Mortal Man]

The how do you explain the difference in areas between a circle and a square? I think you're basic premise is flawed.

No it wouldn't. At the time of the unraveling, it expands in width, but that does it mean it must shrink lengthwise. The idea is that more air is taken in--not material. Thus no material needs to be "shrunk."

Again, that doesn't make intuitive sense to me.

Cheers,

PacMan

PacMan,

Perhaps an example would help clarify the 'tightness' issue.

Consider a papyrus of thickness T and length 2*pi*R, such that it winds exactly once around a cylinder of radius R. Now suppose that you stretch the papyrus, doubling its length to 4*pi*R. The thickness of the papyrus will be reduced by half but it will now wind twice around the cylinder. Thus, the side-on area of the wound papyrus does not change; i.e., one winding of thickness T is the same as two windings of thickness T/2.

Cheers,

MM

[William Schryver]

Don't you mean tin?

Speaking of Dan "The Tin Man" Vogel, he must be driving his latest ambitious project to completion, since we've seen so little of him lately.

If you're listening Dan, you should know I was in a Kroger's in Gahanna about a month ago, and I went around a corner and almost thought that it was you walking down the aisle towards me. I was just about to say, "What a small world it is," when I realized it wasn't really you. This guy wasn't nearly as young and handsome. And it's just as well, you probably would have wanted to share my beer.

[Mortal Man]

Hi Mortal,

I agree that Gee's measurements are physically impossible, and presented that finding, along with my own analysis, at Sunstone West last week. I posted a preliminary version of my analysis at the Mormon Discussion Boards a while back, in case you're interested. Unfortunately I can't link it here due to board rules.

My analysis agrees very much with your own, though I measured a larger number of the extant wraps and got a slightly longer estimate for the length of the missing portion. Your analysis neglects JSP X, the innermost extant fragment. It's definitely good to have done this analysis independently, so that our results confirm each other.

FWIW, Michael D. Rhodes estimates that each of the missing columns and the closing vignette would have been 20 cm wide, not 10.

Best,

-Chris

Hi Chris,

Thanks for the Rhodes reference. Can you PM me the link to your Sunstone paper?

I considered using JSP X in my analysis but I couldn't find an electronic image that gave a proper distance for the gap between X and XI. If I can find the time, I may go back and scan the composite image in Nibley's book and extend the analysis to include X.

Best,

MM

[John Williams]

Quick question to anyone who knows: Has anyone else besides Gee come up with an estimated length anywhere near Gee's 41 feet? I'd like to see a citation, if there is one.

[Mortal Man]

So have you forwarded your conclusions to John? He's quite easy to contact, you know. I do it all the time. Unless he's traveling to deliver papers or to seminars, etc., he's always been quite prompt to reply to me.

He can be reached here: john_gee@byu.edu

I don't pretend to understand the science behind his claims or yours. I understand the general principle, but my brain is not wired to do that kind of thinking or number crunching. But I'll bet John would either show you where you're going wrong, or acknowledge his mistake if you could persuade him enough to see it. Give it a try.

Will,

Thanks for the contact info.

MM

[Mortal Man]

The original papyri is being sent to you FedEX. I've also included two plates from the golden plates, Joseph's seer stone, and the original hat in which he placed the stones.

Mike

Thanks, but why can't I have all the plates? And I suppose you're keeping the sword of Laban for yourself?

[William Schryver]

Thanks, but why can't I have all the plates? And I suppose you're keeping the sword of Laban for yourself?

Because Mike exchanged the rest of them at the local branch of Zion's Bank, at face value, in order to pay off an old gambling debt.

Ain't that so, Mr. Shaking Dice Sin Drone?

[Chris Smith]

Quick question to anyone who knows: Has anyone else besides Gee come up with an estimated length anywhere near Gee's 41 feet? I'd like to see a citation, if there is one.

Not that I'm aware.

[MAsh]

Because Mike exchanged the rest of them at the local branch of Zion's Bank, at face value, in order to pay off an old gambling debt.

Ain't that so, Mr. Shaking Dice Sin Drone?

After Guido took out my right knee, I had to do sump'thin.

Tip: Never bet against Bill Hamblin in a ditty-creation contest.

[dblagent007]

Using MM's technique to calculate thickness, the other scrolls are:

Noufianoub = 0.0525 cm, which is about 110-120 lb paper (105 lb paper is 0.0445 cm thick)

Semminis = 0.0398 cm, which is about 85 lb paper

Horus (using Gee's data) = 0.00477 cm, which is about 8 lb paper!