How Big a Book? Estimating the Total Surface Area of the Book of Mormon Plates-Blog Post #20

Note:  After writing the original version of this blog post, I rewrote it to submit as a research article to the journal Interpreter.  Here is a link to the article as it was published by Interpreter.  I got a number of comments on the article, including one showing that I had made a mistake in my calculations.  🙂  Yes, I did. So I fixed my mistake and found out that the two independent calculations gave even better agreement after the error was corrected. The comments are interesting to read, but I have only provided my original article and the corrections in this blog post.

How Big A Book? Estimating the Total Surface Area of the Book of Mormon Plates

How Big a Book? Estimating the Total Surface Area of the Book of Mormon Plates

Abstract

We do not have the Book of Mormon metal plates available to us. We cannot heft them, examine the engravings thereon or handle the leaves of that ancient record, as did the Three Witnesses, the Eight Witnesses and the many other witnesses to both the existence and nature of the plates.  In such a situation, is there anything more we can learn about the physical nature of the plates without them being present for our inspection?

I believe so.

There are two questions addressed here. First, what was the approximate surface area of the plates on which the Book of Mormon was engraved?  Second, is this a reasonable value when compared with the printed surface area of the current English translation of Book of Mormon?  This article provides two separate, independent calculations that estimate the surface area of the plates on which the Book of Mormon was engraved.  These calculations are what engineers and scientists refer to as “order of magnitude” estimates—they are not intended to yield exact results. If the two independent calculations give roughly comparable and physically reasonable results then our confidence in both the calculations and the reality of the plates is strengthened.

The two approaches taken here are: 1) how many square feet of plates were actually used to engrave the Book of Mormon, given what we know about the physical nature of the plates and 2) how many square feet of plates would be required in order to write the Book of Mormon, given what we know or can infer about the language and script used. We will begin with things we already know, and then use that knowledge to learn more.

Background Knowledge

I am indebted to Jerry D. Grover, Jr., PE, PG for his interesting and useful paper entitled Ziff, Magic Goggles and Golden Plates (1).  Grover provides (see pages 68-70) a thorough summary of various accounts of the physical properties of the plates. Grover also performed an impressive number of experiments and calculations to learn more about the plates. I have relied heavily on Grover for portions of my analysis.

Since Joseph Smith, Jr., had more contact with the plates than anyone else, I will use the physical information provided by him whenever possible. Smith said that the plates containing the Book of Mormon measured about 6 inches wide by 8 inches long and were “not quite as thick as common tin” (2). The engravings were small and filled both sides of the plates (3).  The plates weighed approximately 40-60 pounds (4) and about half of the plates were sealed (5). Thus the Book of Mormon as we have it today was written on about 20-30 pounds of thin metal plates.

We have reasonably good estimates of the weight, length and width of the plates, but not the thickness. In the time of Joseph Smith “common tin” was actually tinplate, which was iron covered with a thin layer of tin to prevent corrosion.  A standard wooden box of tinplate sheets was 14 inches by 20 inches and held 112 sheets of tinned iron, each sheet weighing about a pound (6).

Obviously, for the tinplate sheets to fit in the box, they would have to be somewhat smaller than the outside dimensions of the box. The full box of tinplate sheets weighed well over 100 pounds and would need to be quite sturdy to withstand shipping and storage.  Accordingly, I assume that each of the boards from which the box was constructed was about one inch thick, meaning that the tinplate sheets measured about 12 inches by 18 inches, a convenient width and length for construction purposes.  I neglect the contribution of the density of tin to the overall density of a sheet of tinplate and assume that the density of the tinplate is roughly equal to the density of iron (491 pounds per cubic foot).

With these assumptions, I estimate the thickness of a sheet of tinplate.  The formula is weight equals (density x volume).  Volume is area (length x width) times thickness.  Thus the thickness of tinplate is approximately 1.0 pounds divided by 491 pounds per cubic foot times 1728 cubic inches per cubic foot divided by 12 inches wide by 18 inches long, or about 0.0163 inches thick.

Can we estimate the plate thickness using other approaches? Yes, we can.

William Smith, the prophet’s brother, stated that the plates were made of gold and copper (7).  Mesoamericans did use a copper-gold alloy that the Spaniards called “tumbaga”, but there was no fixed ratio of copper to gold in the alloy, which could vary from 95% copper to 95% gold (8). (Some silver was naturally present along with the gold.)

Grover evaluates four different likely scenarios for the composition and construction of the plates. Two of the scenarios exceed the weight limit of 60 pounds and the third applies to gold gilding on a copper base. Plates prepared under the third scenario would have been more susceptible to corrosion and therefore would probably not have been used by Nephi.

Grover’s fourth scenario uses an upper limit of plate thickness of 0.01 inches and estimates a total weight of the plates of 53.6 pounds with a composition of 85.2% copper, 11.4% gold and 3.4% silver.  For purposes of my calculation, I assume that Grover’s fourth scenario is both realistic and possible.

Mesoamerican craftsmen could form metal to a thickness of about 0.2 millimeters (about 0.008 inches) (9), agreeing well with Joseph’s statement that the plates on which the Book of Mormon were written were “not quite as thick as common tin” and also with Grover’s estimate that the plates may have been up to 0.01 inches thick. (Grover’s experiments actually indicate a thickness less than 0.01 inches for ease of manipulation.)  The fact that the plates could be manipulated with the thumb and would make a noise like paper does when ruffled also argues strongly for a thin, somewhat pliable sheet of metal (10).

Thus we can ask: is it plausible to write a record like the Book of Mormon containing some 250,000 words on 20-30 pounds of plates, each plate being about between about 0.008 to 0.016 inches thick by 6 inches wide and 8 inches long?

First approach to the question of the area of the Book of Mormon plates

The relevant equations are:

  1. Mass of plates = density x volume of plates = density x (plate thickness x plate width x plate length x number of plates)
  2. Total surface area for writing = 2 x area per plate (accounts for the front and back sides of a plate) x number of plates

We want to calculate the total surface area available for writing on 20 to 30 pounds of this metal. The math is straightforward if the thickness of the plates and the density of the metal in the plates are known.  The thickness is estimated at between 0.008 to 0.016 inches, and the density can be estimated from Grover’s calculations, assuming that the densities of copper, gold and silver are additive according to their mass percentages (85.2, 11.4 and 3.4%), respectively, in the mixture.  Applying this assumption, the density of the metal in the plates is about 646 pounds per cubic foot.

We solve Equation 1 for the number of plates using plate thickness of between 0.008-0.016 inches and total weight of plates between 20-30 pounds and then multiply the number of plates by 2 x the area per plate (48 square inches) and divide by 144 square inches per square foot to get the total surface area for writing.

The result is that 10-31 square feet would be available for writing on these plates.  The estimate of 31 square feet is probably closer to being correct than the lower estimate, because a thinner plate is needed to provide the necessary pliability, as Grover concludes.  If so, I estimate the plates contained about 30 square feet for engraving.

This is one estimate.  Is there an independent way of checking this calculation? Yes, there is. We can also try to estimate how many square feet of plates would be needed to write the Book of Mormon.

Second approach to the question

We can also compare the Book of Mormon (approximately 250,000 words) with the Qu’ran, which contains about 77,500 words.  Why the Qu’ran? Because Hebrew and Arabic are both Semitic languages and thus have no vowels and no punctuation.  As a result they are very compact.  The Book of Mormon was apparently written in some system that allowed for a more compact script than even Hebrew (11). The combination of a compact language written in a compact script would help Mormon write a long book on relatively few plates.

Several years ago I visited Kuala Lumpur, Malaysia and was taken by my hosts to tour the Museum of Islam.  In this museum there is a beautiful painting of the Qu’ran in very small but perfectly legible Arabic script.  As I looked at the painting the idea for this calculation came into my mind.  I asked my hosts to take a picture of me standing by the painting. (I did not want to ask for a tape measure and measure the painting. My hosts were very friendly and kind people but I did not want to risk causing any offense to their sacred book.)

The hat that I wore to the museum measures is 12 inches front to back and about 10.8 inches side to side.   By proportion with my hat in the photograph, and by my own visual estimates while looking closely at the painting, this painting is about 4 feet high by 8 feet wide, or 32 square feet.  There are four decorative circles in the painting that I estimate are about 6 inches in diameter (0.8 square feet in total for the four of them) and a decorative strip running lengthwise that is about 8 inches tall and 7 feet long (4.7 square feet). So the entire text of the Qu’ran can be written on about 32-4.7-0.8 = 26.5 square feet.  (The photograph is available from the author but is not provided in this article because it would likely not reproduce well.)

How about the Book of Mormon?  If we are willing to make some assumptions and approximations, how many square feet of plates would it take to write the Book of Mormon?

Given the similarities of the languages and the size and compactness of both scripts, one approach is to assume it would take proportionally the same square footage of plates to write the Book of Mormon as it did to write the entire Qu’ran on this painting.

Since the painting required about 26.5 square feet to write 77,500 words of Arabic it would take approximately (250,000/77,500) x 26.5 square feet or about 86 square feet of plates to write the Book of Mormon in Arabic assuming that as many words can be written per square foot of plates in Reformed Egyptian as in Arabic.

Thus the two independent estimates of the writing area required to engrave the Book of Mormon differ by a factor of two or less. One estimate is about 30 square feet and the other estimate is about 86 square feet.

The two estimates would tend to converge if:

  1. the plates weighed closer to 40 pounds rather than 60 pounds as Martin Harris indicated (4), requiring more plate surface area for a given weight and thus also increasing the number of plates available from the first calculation
  2. the reformed Egyptian characters used by Mormon were more compact than the Arabic characters used in the painting so that more words would fit on one square foot of plates, reducing the number of plates in the second calculation
  3. the characters used by Mormon were placed together on the plates even more closely than the Arabic script was on the painting, again allowing more words per square foot of plates and also decreasing the number of plates in the second calculation

I believe conditions 2 and 3 could be achieved and likely were achieved in the construction of the plates and their engraving with the Book of Mormon.  In each case, the two primary motivations would be to reduce the weight and increase the durability of the plates that Mormon and Moroni (and later Joseph Smith) would be required to carry around.

Engraving on a hard metal is well-suited to producing small characters and is very difficult work, as Jacob attests (12).  While the Arabic characters of the painting in the museum were compact, I believe they could have been placed even more closely than they were without loss of readability.

Therefore, to a first approximation, the Book of Mormon was engraved on about 60 square feet of plates. This figure splits the difference between the two independent estimates and allows some room for the three rings by which the plates were bound and also free space around the edges so that the engravings did not fill the entire plate.

Using the 60 square feet estimate, if each plate measured 6 inches by 8 inches (roughly the page size of the modern Book of Mormon) and was engraved on both sides, then the entire Book of Mormon was engraved on approximately 40 individual plates.  In other words, it was about 80 pages long (two pages per plate), roughly fifteen percent of the size of our modern English copies of the Book of Mormon (531 pages).

These calculations and estimates all pass the test of reasonableness. They are two completely independent estimates of a single variable: the total surface area on which the Book of Mormon was engraved. And the different estimates vary by a factor of about three or less.

This may be only a small coincidence.  But perhaps it is a useful addition to the many other correspondences, large and small, with which the Book of Mormon is filled. Cumulatively these correspondences gain great force as their number increases.

 

References

  1. https://archive.bookofmormoncentral.org/content/ziff-magic-goggles-and-golden-plates-etymology-zyf-and-metallurgical-analysis-book-mormon
  2. http://www.josephsmithpapers.org/paper-summary/church-history-1-march-1842/2
  3. http://en.fairmormon.org/Question:_What_was_the_appearance_of_the_engravings_on_the_gold_plates%3F
  4. Martin Harris interview, Iowa State Register, August 1870, as quoted in Milton V. Backman Jr., Eyewitness Accounts of the Restoration (Salt Lake City: Deseret Book, 1986), 226
  5. 1878 Interview between Orson Pratt and David Whitmer. In Book of Mormon Compendium. Sydney B. Sperry, 55-56.  Salt Lake City: Bookcraft/Desert Book, 1970.
  6. http://mike.da2c.org/igg/rail/12-linind/tinplate.htm
  7. http://en.fairmormon.org/Source:William_Smith:The_Old_Soldier%27s_Testimony:1884:When_the_plates_were_brought_in_they_were_wrapped_up_in_a_tow_frock._My_father_then_put_them_into_a_pillow_case
  8. http://www.langantiques.com/university/index.php/Tumbaga
  9. Warwick Bray, “Gold-Working in Ancient America” Gold Bulletin 11/4 (1978): 137-38. (My thanks to Dr. John Sorenson and his book “Mormon’s Codex” for this valuable reference)
  10. Saints’ Herald, 26 (1879), p. 230
  11. Book of Mormon. Mormon 9:33
  12. Book of Mormon. Jacob 4:1

Correction and Additional Calculations for “How Big a Book?” by Bruce E. Dale

Summary

The thickness of tin plate in my article was estimated based on a link that no longer exists.  I didn’t copy that document or make a scanned PDF it for later distribution.  Lacking that source, I did some additional research to find out more about the thickness of tinplate, now and in the early 1800s.

In doing so, I found an error in my original calculations that is corrected below. Several readers of my article also noted this error.  I sincerely thank them.

In brief, the total surface area of the plates based on the corrected calculations and the new data on plate thickness is about 93 square feet.  This compares well with the 86 square feet estimate from the word count of the Koran compared to the Book of Mormon.  The Book of Mormon was therefore engraved on approximately 90 square feet of plates.

New Calculations

Recall that the article assumes that each plate of tinplate weighed one pound and that the width and length of the tinplate sheets were 12 and 18 inches, respectively.

With these assumptions taken from the now-defunct link, the estimated tinplate thickness was:  1.0 pounds divided by 491 pounds per cubic foot divided by width (12 inches) divided by length (18 inches) times 1728 cubic inches per cubic foot equals 0.0163 inches thickness.

What other data are available?

Well, another estimate of the thickness of tin-plated steel is given on page 12 of this document, “Tinplate and Tin Free Steel”.  According to this article, the normal thickness of tinplate is 0.15 to 0.25 mm or 0.00591 inches to 0.00984 inches.  Here is the link.  http://www.jfe-steel.co.jp/en/products/sheets/catalog/b1e-006.pdf

Since about 1965, tinplate thickness for “tin cans” has ranged (been reduced) from 0.25 to 0.10 millimeters or from 0.00984 to 0.00394 inches.  Thus presumably before 1965, tinplate thickness was about 0.01 inches. https://www.britannica.com/technology/tin-processing#ref623354

From an article entitled “Basic Tinsmithing” I found this important information: “The material to use is called tin-plated steel, or tinplate (.012″ thick). This is basically the same stuff that was used 200 years ago.”  http://www.northwestjournal.ca/XIV122.htm

In a book on Architectural Metals, page 316, I found the same datum supplied in my article, namely that in the 19th century tinplate was sold in boxes of 112 sheets. Furthermore, this book states that the tin plate was sold in two gauges, 28 gauge and 30 gauge, for different roofing applications.  https://books.google.co.uk/books?id=57jzHvkZrCQC&pg=PA308&lpg=PA308&dq=thickness+of+tin+plate+in+the+1800s&source=bl&ots=iumYMcBQ3P&sig=KFDg7-rnhj6InFPlisrfua39hbU&hl=en&sa=X&ved=0ahUKEwjs_-DR2MvUAhUJKsAKHdmJCW0Q6AEIVDAI#v=onepage&q=thickness%20of%20tin%20plate%20in%20the%201800s&f=false

With this information on the gauge numbers of tinplate, I searched for information on the thickness of different gauges.  I found that 28 gauge sheet metal is 0.015625 inches thick and 30 gauge sheet metal is 0.0125 inches thick.  I assume that these gauge thicknesses used in the late 1800s are the same as those used in the early 1800s.   https://www.tedpella.com/company_html/gauge.htm

For purposes of this calculation, therefore, I assume that the plate thickness was between 0.008 inches (which we know that Mesoamerican craftsmen could produce) and 0.016 inches, the estimated thickness of the tinplate calculated from the original reference (link no longer available) as well as the 28 gauge metal thickness of 19th century tinplate (see above).  The average of these two values, 0.012 inches, is the thickness of the “same stuff that was used 200 years ago” (again, see above), so it seems reasonable to use these two extremes to approximate the thickness of tinplate in the early 1800s.

Thus the extremes of the plate thickness are assumed to be 0.008 inches and 0.016 inches. The extremes of the weight of the plates are 20 pounds and 30 pounds.  Therefore the maximum number of possible plates will be when the weight of the plates is greatest (30 pounds) and the plate thickness is least (0.008 inches).  The minimum number of possible plates will be when the weight of the plates is lowest (20 pounds) and the thickness of the plates is greatest (0.016 inches).

Thus the total weight of plates (either 20 pounds or 30 pounds) is equal to the density of the metal (646 pounds per cubic foot) times the total actual volume of plates.  The total volume of the plates is equal to the total number of plates times plate thickness (either 0.008 inches or 0.016 inches) times the width (6 inches) times the length (8 inches) divided by the conversion factor of 1728 cubic inches per cubic foot.

Solving for the number of plates at 30 pounds total weight and 0.008 inches thickness, the maximum total number of plates is about 209 plates.  Solving for the number of plates at 20 pounds total weight and 0.016 inches thickness, the minimum number of plates is about 69.7 plates.

Multiplying the maximum and minimum number of plates by the total surface area (front and back) per plate (96 square inches) and dividing by the conversion factor of 144 square inches per square foot, the minimum surface area of the plates is 46.4 square feet and the maximum is 139.3 square feet. The average of these two values is 92.9 square feet, less than 10% different from the independent estimate of 86 square feet based on the word count of the Koran versus the Book of Mormon.

Summary

In round numbers, therefore, the Book of Mormon was written on approximately 90 square feet ((86 + 92.9)/2) of total plate surface area.  Each plate contained about (6 inches x 8 inches x 2)/144 square inches per square foot or 0.667 square feet per plate for engraving. The total number of plates was therefore about 90 square feet total divided by 0.667 square feet per plate, or 135 plates. Each plate contained two sides for a total of 270 pages of text, or roughly half of the page count (531 pages) of the current English version of the Book of Mormon.

 

Bruce E. Dale.  June 22, 2017

 

 



2 thoughts on “How Big a Book? Estimating the Total Surface Area of the Book of Mormon Plates-Blog Post #20”

    1. Thanks so much. I really appreciate the feedback. I hope you liked the final result of my attempt to estimate the surface area of the Book of Mormon plates. My next effort will be an attempt to apply Bayesian statistics to specific cultural, physical and other features of the Book of Mormon and compare it these with Dr. Michael Coe’s book “The Maya”.

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