John Gee Actually Talks About The Papyri - Oh My!

[noel00]

Phelps and Parrish were his scribes. Why would they use symbols from the BOB rather than the still extant BOA?

[SolarPowered]

Phelps and Parrish were his scribes. Why would they use symbols from the BOB rather than the still extant BOA?

Any statement I or anyone else would make about why they did what they did would be a guess. No one has presented any documents wherein they purport to explain what they were doing. Therefore, we can only guess at it.

My guess for today is that they were making wall decorations for the Gold and Green Ball.

What is your guess for today?

We both get to make new, completely different guesses tomorrow, because they simply didn't tell us what they were doing.

[Chris Smith]

OK, I aways did have a bit of a problem with "show your work"...

Number of wraps = (R2-R1)*P

Average radius of a wrap = (R2+R1)/2

Note that the length of a wrap is linear WRT radius, so we can get the total length by multiplying the number of wraps by the averge length of a wrap. The average length of a wrap is:

Average length of a wrap = 2*pi*Average radius of a wrap

= 2*pi*(R2+R1)/2

= pi*(R2+R1)

Thus:

Total length = Number of wraps * Average length of a wrap

= [(R2-R1)*P] * [pi*(R2+R1)]

= pi*P*(R2^2 - R1^2)

Hey SP,

Thanks for explaining that. I thought you were multiplying P by total area minus inside area. That didn't make any sense to me. What you explained above makes more sense, though your Number of Wraps variable should be R2*P.

I spent some time thinking about this this afternoon and derived the following formula:

length of roll = sum(i = 1 to P*R2) 2*pi*(1/P)*i

where P = the number of wraps between R1 and R2

This seems to be saying roughly the same thing you did, except that rather than using average length I used the sum of actual lengths.

I'm still not convinced, however, of a few things:

1) That the tightness of the wrapping would be consistent. I don't know to what extent papyrus is like or unlike paper, but certainly when you wrap paper it is tighter near the outside.

2) That there would have been no "gap" in the middle of the roll, as for example when you roll paper.

Browsing google offers some suggestive results:

http://www1.istockphoto.com/file_thumbview...yrus_scroll.jpg

http://www1.istockphoto.com/file_thumbview...yrus_scroll.jpg

http://www.ancientegypt.co.uk/writing/expl...ages/scroll.jpg

http://www.humnet.ucla.edu/humnet/classics...us/papyrus.html

http://www.rapturechrist.com/scroll21.jpg

http://dragonchasers.com/wp-content/upload...6/12/scroll.jpg

I conclude that either could in fact be the case. Note that according to the second link the carbonized rolls from Herculaneum have been difficult to unroll precisely because they are rolled so tightly. Most of the other scrolls from the same dig were unrolled in the 18th or 19th centuries.

In any case, Gee's equation as he described it to me over the telephone had something to do with the height of the papyrus, as well, if I recall correctly. Gee told me that the papyrus gets smaller (shorter, as in less tall) near the end of the roll. I'm not entirely sure how this factors in, so I think we'll have to wait for Gee's publication.

-CK

[SolarPowered]

Hey SP,

Thanks for explaining that. I thought you were multiplying P by total area minus inside area. That didn't make any sense to me. What you explained above makes more sense, though your Number of Wraps variable should be R2*P.

I spent some time thinking about this this afternoon and derived the following formula:

length of roll = sum(i = 1 to P*R2) 2*pi*(1/P)*i

where P = the number of wraps between R1 and R2

This seems to be saying roughly the same thing you did, except that rather than using average length I used the sum of actual lengths.

I'm still not convinced, however, of a few things:

1) That the tightness of the wrapping would be consistent. I don't know to what extent papyrus is like or unlike paper, but certainly when you wrap paper it is tighter near the outside.

2) That there would have been no "gap" in the middle of the roll, as for example when you roll paper.

Browsing google offers some suggestive results:

http://www1.istockphoto.com/file_thumbview...yrus_scroll.jpg

http://www1.istockphoto.com/file_thumbview...yrus_scroll.jpg

http://www.ancientegypt.co.uk/writing/expl...ages/scroll.jpg

http://www.humnet.ucla.edu/humnet/classics...us/papyrus.html

http://www.rapturechrist.com/scroll21.jpg

http://dragonchasers.com/wp-content/upload...6/12/scroll.jpg

I conclude that either could in fact be the case. Note that according to the second link the carbonized rolls from Herculaneum have been difficult to unroll precisely because they are rolled so tightly. Most of the other scrolls from the same dig were unrolled in the 18th or 19th centuries.

In any case, Gee's equation as he described it to me over the telephone had something to do with the height of the papyrus, as well, if I recall correctly. Gee told me that the papyrus gets smaller (shorter, as in less tall) near the end of the roll. I'm not entirely sure how this factors in, so I think we'll have to wait for Gee's publication.

-CK

R1 in my formula is the inside diameter if the roll; that is, the space that is either empty, or contains some sort of a core.

The total number of wraps is in fact P*(R2-R1). (Again, assuming that the pitch of the wrapping is constant.)

It appears that your formula will produce the same result as mine produces when one sets R1 to 1/P.

1) Again, I agree that the assumption of a constant pitch may not be valid. But I suspect that it's close enough to constant that the constant assumption results in a reasonably close estimate of the true length using my formula.

2) As I explained just above, I did not assume that there was no gap in the center.

It appears that we're pretty much on the same page.

[Chris Smith]

R1 in my formula is the inside diameter if the roll; that is, the space that is either empty, or contains some sort of a core.

Oh. I was defining R1 differently, then. I was taking R1 to be the inside radius of the roll as it is at present. You apparently are taking it as the original inside radius of the roll before the inner portion was lost. That makes sense. The only problem for implementation purposes, then, would be guessing at the value of R1.

Yes, I'm guessing that our fomulas will produce the same result.

-CK