Abstract:* We do not have the Book of Mormon metal plates available to us. We cannot heft them, examine the engravings, 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, what more can we learn about the physical nature of the plates without their being present for our inspection? Building on available knowledge, this article estimates the total surface area of the plates using two independent approaches and finds that the likely surface area was probably between 30 and 86 square feet, or roughly 15% of the surface area of the paper on which the English version of the Book of Mormon is now printed.*

There are two questions I seek to address in this article. First, what is the estimated surface area of the plates on which the Book of Mormon was engraved? Second, is this estimate 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 [Page 262]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.

**Estimating the Thickness of the Plates**

I am indebted to Jerry Grover for his interesting and useful paper entitled *Ziff, Magic Goggles and Golden Plates*.^{1} Grover provides a thorough summary of various accounts of the physical properties of the plates.^{2} He also performed an impressive number of experiments and calculations to learn more about the plates*. *I have relied heavily on his work 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 the plates containing the Book of Mormon measured about 6 inches wide by 8 inches long and were “not quite so thick as common tin.”^{3} The engravings were small and filled both sides of the plates.^{4} The plates weighed approximately 40–60 pounds,^{5} and about half of the plates were sealed.^{6} 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, each weighing about a pound.^{7}

[Page 263]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 1 inch thick, meaning 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 the density of the tinplate is roughly equal to the density of iron (491 pounds per cubic foot).

With these assumptions, we can estimate the thickness of a sheet of tinplate. The formulae are:

- Weight = density x volume
- Volume = area (length x width) x thickness

Since each sheet of tinplate weighed about one pound, the thickness of tinplate can be calculated using this formula:

- 1.0 pound = (491 pounds/cubic foot) x (12 inches x 18 inches) x thickness x conversion factor (cubic feet to cubic inches)

Rearranging the equation above to calculate thickness we find:

- Thickness (in inches) = (1.0 pounds/(491 pounds per cubic foot x 12 inches x 18 inches)) x 1728 cubic inches per cubic foot) = 0.0163 inches

There are other ways we can estimate the plate thickness as well. William Smith, Joseph’s brother, stated that the plates were made of gold and copper.^{8} Mesoamericans did use a copper-gold alloy the Spaniards called “tumbaga,” but there was no fixed ratio of copper to gold in the [Page 264]alloy, which could vary from 95% copper to 95% gold.^{9} (Tumbaga also contained some silver that 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 calculations, I assume Grover’s fourth scenario is both realistic and possible.

Ancient American metal workers could form metal to a thickness of about 0.2 millimeters (about 0.008 inches),^{10} 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 plate 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.^{11}

**First Approach: Calculating the Area from the**

Mass and Thickness of the Plates

Mass and Thickness of the Plates

Given the background information considered so far, a reasonable questioner might ask if it is plausible to write a record like the Book of Mormon 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.

The relevant equations are:

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

[Page 265]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 the densities of copper, gold, and silver are additive according to their mass percentages in the mixture (85.2%, 11.4%, and 3.4% respectively). 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 a plate thickness of between 0.008 and 0.016 inches and total weight of plates between 20 and 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 indicates. If so, I estimate the plates contained about 30 square feet for engraving.

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

**Second Approach: Calculating by Word**

Count Compared to the Qu’ran

Count Compared to the Qu’ran

We can also compare the Book of Mormon with the Qu’ran. The Book of Mormon contains about 250,000 words in my English translation, while my English translation of the Qu’ran 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 (Mormon 9:33). 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 framed painting containing the entire text of the Qu’ran. The painting of the text is done in very small but perfectly legible Arabic script. As I looked at the painting, and admired its beauty, the idea for this calculation came into my mind. I asked my hosts to take a picture [Page 266]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 them any offense.)

The hat that I wore to the museum measured 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.^{12}

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 presumed 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 in Arabic as it did to write the entire Qu’ran. To state this assumption in another way: we are assuming for the sake of this calculation that the language in which the Book of Mormon was written is similar to Arabic in its compactness and can express the same ideas in a similar surface area devoted to writing.

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 three 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:

- 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
- [Page 267]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 these conditions 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 primary motivation would be to reduce the weight 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 (Jacob 4:1). 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^{13} and also free space around the edges so 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 length 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.

**Conclusion**

The total surface area required to engrave the characters in which the Book of Mormon is written on the plates is unknown. However, we do [Page 268]have a considerable amount of eyewitness testimony as to the dimensions and weight of the plates. We also have a modern language, Arabic, which is likely similar to the language in which the Book of Mormon plates were written. We know approximately how much surface area was required to write the Qu’ran, using very small Arabic characters. Based on this and other information, several questions can be asked: 1) Can we estimate the surface area required to engrave the Book of Mormon? 2) Can we check that estimate using an independent method of calculation? 3) Do these two estimates give physically reasonable results?

Two separate and completely independent calculation approaches were taken to address the question of the surface area of the Book of Mormon plates. The results of the calculations are between about 30 and 86 square feet, a difference of less than three-fold. The average of these two values is about 60 square feet, meaning the Book of Mormon was engraved on about 40 individual plates. This is roughly 15 percent of the surface area of the text of the Book of Mormon in our modern English translation. Thus the two independent calculation approaches give consistent and reasonable values. They also support the idea that the Book of Mormon authors achieved great economy of space in writing the Book of Mormon.

1. Jerry D. Grover, *Ziff, Magic Goggles, and Golden Plates: Etymology of Zyf and a Metallurgical Analysis of the Book of Mormon Plates* (Provo, UT: Grover Publishing, 2015). Available online at https://archive.bookofmormoncentral.org/content/ziff-magic-goggles-and-golden-plates-etymology-zyf-and-metallurgical-analysis-book-mormon

2. Ibid., 67–70.

3. Joseph Smith, “Church History,” in* Times and Seasons *(Nauvoo, IL), 1 Mar. 1842, vol. 3, no. 9 (whole no. 45), pp. 707, online at http://www.josephsmithpapers.org/paper-summary/church-history-1-march-1842/2.

4. “What was the Appearance of the Engravings on the Gold Plates?” *FairMormon Answers*, online at https://www.fairmormon.org/answers/Question:_What_was_the_appearance_of_the_engravings_on_the_gold_plates%3F.

5. Martin Harris interview, *Iowa State Register*, August 1870, as quoted in Richard Lloyd Anderson, *Investigating the Book of Mormon Witnesses* (Salt Lake City: Deseret Book, 1981), 14.

6. Kirk B. Henrichsen, “What Did the Golden Plates Look Like?,” *New Era *(July 2007): 31.

7. Mike Smith, “Tin Plate,” http://mike.da2c.org/igg/rail/12-linind/tinplate.htm.

8. “The Old Soldier’s Testimony. Sermon preached by Bro. William B. Smith, in the Saints’ Chapel, Detroit, Iowa, June 8th, 1884. Reported by C.E. Butterworth,” *Saints’ Herald* 31 (4 October 1884): 643–44; reproduced in Dan Vogel ed., *Early Mormon Documents* (Salt Lake City: Signature Books, 1996), 1:505. https://www.fairmormon.org/answers/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.

9. See “Tumbaga,” *Antique Jewelry University*, online at http://www.langantiques.com/university/index.php/Tumbaga.

10. 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.)

11. “Last Testimony of Sister Emma,” *Saints’ Herald*, 26 (1879): 290; interview conducted between Februay 4 and February 10, 1879.

12. The photograph is in my collection but is not provided here, as it would likely not reproduce well in the printed version of this article.

13. Joseph Smith, “Church History,” in* Times and Seasons *(Nauvoo, IL), 1 Mar. 1842, vol. 3, no. 9 (whole no. 45), 707.

In Jerry Grover’s “Translation of the Caractors Document” (http://bmslr.org/translation-of-the-caractors-document/), in Chapter 1 he compares the number of distinct glyphs (~100) on the ‘Caractors’ document to the number of glyphs in alphabetical scripts (~30). By this he concludes that the BOM used a logographic script. I’m curious to hear ways that might impact the article’s estimate.

It seems to me, with an average word length in the BOM of 4 characters, the length and space could potentially be around 25% of that used by the English translation. Without vowels the average word length is around 2.7 characters. This would make your Quran estimate (86 sqft / 2.7 = 31.8) almost identical to your other (30 sqft). The BOM characters look slightly more complex than Arabic characters, but perhaps use more vertical space, and aren’t written in cursive, so perhaps they would require more space per character than Arabic, though maybe still less per word.

Also, it seems likely to me that there are some words/phrases in the English BOM that were either impossible or unnecessary to express in the original language/script and exist only in the inspired translation, further reducing the space required on the plates.

Thank you, Diego. My apologies to you and everyone who has waited for me to engage them in their comments about my article. I just returned from two weeks in Scotland and England on a family history visit and I leave again tomorrow for a week in Alaska fishing with my son and my brother. So I hope you will excuse the somewhat hurried and superficial nature of this reply. When I return from Alaska, and then a trip to Seattle during the week of the Fourth of July, I will try to do a better job of responding.

Several of my readers found a mistake in my calculations. Embarrassing for this engineer, but as I redid the calculations, I was happy to find that the agreement between the two approaches to estimating the surface area of the plates was actually much better than before. The two approaches now agree to within 8%. My revisions/corrections to the article are given below.

However, these new calculations do not fit very well with a reduced page count for the Book of Mormon, as you suggest below. My current estimate is that the Book of Mormon as written on the plates took up about half the number of pages that our current English translation does.

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

The surface area of the Golden Plates is an interesting thing to think about.

You conclude that the current text took up about 60 square feet and that can be written on 40 double sided 6″ x 8″ plates. The math doesn’t add up.

(6/12)*(8/12)*40*2 = 26.67 sqft

It would take 90 double sided plates to reach 60 sqft.

Dear Benjamin:

My apologies to you and everyone who has waited for me to engage them in their comments about my article. I just returned from two weeks in Scotland and England on a family history visit and I leave again tomorrow for a week in Alaska fishing with my son and my brother. So I hope you will excuse the somewhat hurried and superficial nature of this reply. When I return from Alaska, and then a trip to Seattle during the week of the Fourth of July, I will try to do a better job of responding.

You are right that the math doesn’t add up. Embarrassing for this engineer, but as I redid the calculations, I was happy to find that the agreement between the two approaches to estimating the surface area of the plates was actually much better than before. The two approaches now agree to within 8%.

My revisions/corrections to the article are given below. I

hope I got it right this time. 🙂 I am grateful to anyone who takes the time to read and check my work.

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

135 plates. This is close to the calculations that I have previously done. Based on an assumption of the thickness of each plate and the percentage of voids, I came to a conclusion that the 6″ thick Golden Plates contained about 450 individual plates.

The composition of the plates from top to bottom is the following:

Lost 116 pages – 32 plates

Remainder of Mormon & Moroni’s record – 86 plates

Small plates of Nephi – 32 plates

Words of Mormon and Title Page – 1 plate

Sealed Plates – 300 plates

Thus the text of the Book of Mormon that we have now was written on 119 plates.

Does this account for the sealed portion of the plates? We know they constituted 1 to 2 thirds of the entire mass (begging the question how Moroni managed to get enough material to write or copy the entire sealed portion of the plates; but that’s fascinating speculation for another time).

If the entire set of plates was 60 pounds but 2/3rds of it was sealed, that leaves only 20 pounds of plates for the written text we have now.

Ah, rereading the article I see it does take that into account. What about the lost 116 pages? I don’t know how much that would translate to today’s text; likely more than 116 pages.

Hebrew is astonishingly high in information density; at the cost of the vowels. Mormon lamented that if he could have written in Hebrew his record would have been “perfect.” This implies that Reformed Egyptian was very compact indeed. That leads credence to the idea that our current text could in fact have been represented on something like 40-80 (small) pages; as a page of the plates was about half the size of a standard US Letter page.

I imagine it was murder to engrave without slipping and ruining an entire plate, though. I wonder how many times Mormon threw his engraving tool across the room as he slipped on the 3rd line from the bottom of the second page of a plate…..

Trying to write the BOM in Hebrew has been previously looked at with respect to character density:

https://publications.mi.byu.edu/publications/jbms/10/1/S00005-50be48520d86f5Sjodahl.pdf

https://publications.mi.byu.edu/publications/jbms/10/1/S00006-50be4867e89a06Gee.pdf

Thank you, Jerry. I will study this more closely when I return from my travels. Right now, I am feeling the effects of being up for 30 hours straight. 🙂

Hi Vance:

My apologies to you and everyone who has waited for me to engage them in their comments about my article. I just returned from two weeks in Scotland and England on a family history visit and I leave again tomorrow for a week in Alaska fishing with my son and my brother. So I hope you will excuse the somewhat hurried and superficial nature of this reply. When I return from Alaska, and then a trip to Seattle during the week of the Fourth of July, I will try to do a better job of responding.

Several of my readers found a mistake in my calculations. Embarrassing for this engineer, but as I redid the calculations, I was happy to find that the agreement between the two approaches to estimating the surface area of the plates was actually much better than before. The two approaches now agree to within 8%.

My revisions/corrections to the article are given below.I am grateful to anyone who takes the time to read and check my work. 🙂 The Book of Mormon deserves the best we can give it.

As to the lost 116 pages,it is a good point. I did consider that issue, but given the uncertainty we already have about the fraction of the plates that were sealed and not translated (one third or one half), I decided to ignore that factor.

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

Hi Vance:

My apologies to you and everyone who has waited for me to engage them in their comments about my article. I just returned from two weeks in Scotland and England on a family history visit and I leave again tomorrow for a week in Alaska fishing with my son and my brother. So I hope you will excuse the somewhat hurried and superficial nature of this reply. When I return from Alaska, and then a trip to Seattle during the week of the Fourth of July, I will try to do a better job of responding.

Several of my readers found a mistake in my calculations. Embarrassing for this engineer, but as I redid the calculations, I was happy to find that the agreement between the two approaches to estimating the surface area of the plates was actually much better than before. The two approaches now agree to within 8%.

My revisions/corrections to the article are given above in my responses to other commentators. No more redundancy needed, even for this engineer. 🙂

I am grateful to anyone who takes the time to read and check my work. 🙂

As to the lost 116 pages, I did consider it, but given the uncertainty we already have about the fraction of the plates that were sealed and not translated, I decided to ignore it.

Best wishes,

Bruce

Hello Vance,

Obviously, I need to practice replying to comments before I actually do it. I managed somehow to reply to you as a separate comment, not as a reply directly to you. Sorry about that.

You can see my reply below. Thanks for your interest.

Bruce

A few comments:

1) It is indicated that scenario 3 in my Ziff book would be subject to corrosion. That is not accurate, scenario 3 is a copper base metal with a gold gilded surface finished by depletion gilding, so has the same corrosion resistance as any other pure gold surface.

2) It is likely that the ‘small’ plates in the plate stack were not the original plates written by Nephi, but were also written in reformed Egyptian based on an interpretation of the original small plates, and manufactured later (see Sumerian Roots of Jaredite-Derived Names and Terminology in the Book of Mormon, pages 296-300, http://bmslr.org/books/Sumerian%20Roots%20of%20Jaredite-Derived%20Names%20and%20Terminology%20in%20the%20Book%20of%20Mormon.pdf). If Mormon knew hieratic Egyptian he would likely have written the plates in hieratic Egyptian.

3) There are some errors in the analysis as it doesn’t take into account the 116 lost pages in the word count, or void space between the plates.

3) I actually did address the issue of character density on a recent Facebook Post involving a comparison with the Darius plates with the following response utilizing the translation of the Caractors Document density which would be the best comparison as it contains the actual characters and translation of them, no need to default to some other language. Using this approach back calculating one arrives at 371 plates for the entire plate stack, which falls within the 300 to 600 plate calculated range:

Caractors Documents and Plate Stack Character Density Discussion

A) Language density analysis

The Caractors Document is 8 inches by 3.25 inches, there are 222 characters, so the character density is 222/(8 x 3.25) = 8.54 characters per square inch. The translation of the Caractors Document rendered 2.1 words per character. The English equivalent would be 17.9 words per square inch. Based on the high number of characters in the Caractors Document that are numbers and calendar markers, it would be expected that the number of words per square inch of most of the text of the Book of Mormon is probably somewhat higher.

Based on a metallurgical analysis there were calculated to be 300 to 600 plates in the plate stack (Grover 2015, Ziff, Magic Goggles, and Golden Plates, Chapter 11, available at http://www.caractors.org). Assuming ½ of the stack is sealed, using the high number of 600 plates, that leaves 300 plates double sided at 6 in. x 8 in. equals 48 in2 per side or 28,800 in2 total available surface area. At 17.9 words per square inch, that makes a total of 516,499 English words space available equivalent for the unsealed plate stack.

Assuming the smaller number of plates (300) would render half of that, or available space for 258,249 words.

B) Calculate the number of expected English words based on the current Book of Mormon

Factors: Total number of words in current Book of Mormon (that includes the small plates): Approximately 268,000 words

Calculation of additional words from 116 lost pages (exclude the title page): (1 + 116/605 pages (Skousen, Original Manuscript)) x 268,000 = 319,385 total words in the unsealed portion.

So the expected number of plates based on back calculation is 371 plates in the plate stack.

C) Darius plates comparison

Darius plates: Old Persian section (not including the word divider marks) has 190 “characters” in a space 13 inches by 1.9 inches which equates to 7.69 characters per square inch.

Babylonian section has 150 characters (counting singles which may be determinatives) in a space 13 inches by 1.2 inches which equates to 9.62 characters per square inch. Elamite section has 129 characters in a space 13 inches by 1.5 inches which equates to 6.62 characters per square inch.

There are 59 English words in the translation of the paragraph of Darius plates, so one can also calculate the “density” of equivalent English for each language (59 words/surface area used):

Old Persian: 2.39 English words per square inch

Babylonian: 3.78 English words per square inch

Elamite: 3.03 English words per square inch

Hi Jerry:

First of all, I an really grateful for the work you have done. It is fascinating to read, and I obviously need to read it again, and more closely this time. Thanks for your corrections, including the corrosion resistance of gilding. 🙂

If you have time, I would welcome your comments on the my new calculation set (see above) using the physical properties of the plates and the plate materials.

When I return from my next round of travel, I hope to engage more with your detailed comments.

In the meantime, I would just like to point out that I am not trying working with the bulk density of the plates, but with their estimated weight and the intrinsic density of the copper/gold/silver mixture which the plates were composed of. Without your scenarios, I would not have had a basis for estimating that intrinsic density based on the composition of the plates. Thank you again.

Best wishes,

Bruce

Hi Brother Grover,

My calculations for the 116 pages differ a little from yours. As I understand it, there are 492 pages in the Original Manuscript. I base this off of an answer Royal Skousen gave to a comment I made on one of his articles here:

http://www.mormoninterpreter.com/the-original-text-of-the-book-of-mormon-and-its-publication-by-yale-university-press/

If there were approximately 268,000 words written on 492 manuscript pages, that yields about 545 English words per page. 545 words/page x 116 lost pages = approximately 63,000 English words on the lost 116 pages. So everything Joseph translated from the plates = 268,000 + 63,000 = 331,000 words.

I’m interested in this minutia as part of an article I am writing about the 24 Jaredite plates. I will be proposing that they were actually 24 sets of plates instead of 24 individual plates. I’ve been working on it for a while, and I’m happy to learn about yours and Brother Dale’s work on the number of gold plates.

Our calculations on the 116 pages were close, and since I don’t have Volume 1 of the Critical Text, I may be misunderstanding Royal’s comment about the 492 manuscript pages in O. I’d love to hear how you arrived at your conclusion.

One other resource note, Reference 7 that is provided is actually a dead link. An internet search did not provide an alternative link. It would be nice to see how he arrived at the thickness of common tin in the early 1800’s. In my evaluation of common tin from my Ziff book I determined that in the early 1800’s there was were variable thicknesses for common tin plate, some as thin as .001667 inches.(Encyclopedia!Britannica 1797,12:118). Do you have an alternate link for this reference?

Jerry,

My new calculation set (see my response to the other comments above) has several references that I think provide good input for the thickness of common tin in the early 1800s. I presume that Joseph would have been most familiar with architectural tin, either 28 gauge or 30 gauge.

Perhaps to save you some time, here is the relevant portion of my new calculation set. I am looking for ranges, not exact values.

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

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

Best,

Bruce

And Labans’ “brass plates”–

The Book of Mormon has about 275,000 words. The King James Bible, from Genesis to Jeremiah, has about 500,000 words. So whatever we calculate for the Gold Plates, double it, at least, for the Brass Plates.