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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.

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

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.

Best,

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

Bruce,

There are a variety of languages one can compare for this type of analysis, it is clear based on the number of unique caractors that reformed Egyptian is not an alphabetic or even syllabic language, but is heavily logographic, so any language comparison hoping for accuracy to calculate the number of plates should take that into account (See pages 4 and 5 of the Caractors Document book at http://www.caractors.org). You chose to compare Arabic, which is an alphabetical language, so maybe is not very accurate in that respect, but I think the point that all are looking at here is that the Book of Mormon could clearly have fit on the calculated number of plates in the plate stack, which you have done with your Arabic comparison as well.

As far as the bulk density (I’m assuming you were referring to my void space comment), I should have been more clear, I was referring to the void space inherent (even in modern plate stacks) in the box of tin plate. It is small but may have improved the accuracy of the calculation.

As far as the calculations involving the plate thickness, there was another parameter that has to be considered and that is that Emma Smith said “They seemed to be pliable like thick paper, and would rustle with a metalic [sic] sound when the edges were moved by the thumb, as one does sometimes thumb the edges of a book.” If you go through the Ziff book I wrote, I actually did physical testing of metal plates with metallic characteristics (hardness, modulus of elasticity, etc.) close to what one would expect in ancient cold-worked metal plates. I found that .01 inches was pretty much the maximum thickness where one could generate the effect that Emma described. Thus the plates could not have been thicker than .01 inches. This parameter is a better control than the common tin description, which as you pointed out, can be a variety of thicknesses. The description of the plate thickness given by some as equivalent to window glass was actually much more reliable to determine thickness than the “common tin” description (the calculations there are also included in the Ziff book). This would bring your minimum number of plates to 112 plates (actually 111.52, unless one has a one-sided plate!).

Anyway, please don’t take my comments as hypercritical, I appreciate it when scientists engage in Book of Mormon research, we need more of it. I think your follow-up work on the tin plate thickness is especially new and valuable, I somewhat ‘punted’ on that in my Ziff book.

Hi Jerry,

I did read the part in your “Ziff” article about the Book of Mormon plates not being greater than 0.01 inches in thickness, and I made that reference in my article. I agree with that conclusion.

I had two purposes in my article: first, to see if these two independent approaches to the question of how much surface area was required to write the Book of Mormon agreed with each other and second, to see if the answers were physically reasonable (i.e., certainly not one plate and probably not one thousand plates). They do to (within 8%) and they are.

I was not trying for a high degree of precision, which may not be possible. I was only aiming to do an analysis that would satisfy this particular engineer and I did that. I am glad that you and others found it useful and thought-provoking. So I hope to have more interactions with you in the future.

I do not think you are hypercritical at all. I welcome well-founded critiques and analysis. Your work is very much well-founded and I am grateful for your attention to detail.

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:

https://journal.interpreterfoundation.org/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.

I had in my notes that there were 491 manuscript pages in the Original Manuscript, which is what I used in the calculation above (1 title page (2 sided) + 489 OM pages plus 116 pages). I need to go recheck that, as your reference from Royal says 492 instead of 491. Small difference, but I am a bit OCD when it comes to accuracy. I ignored the title page which is part of the small difference we have in word count. On the 24 Jaredite plates, I am actually working on the same premise as you as one is talking about sets of plates instead of individual plates and am also doing research along those lines. Let me know when you get something published.

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

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.

Hi Benjamin:

Thank you for your comments.

I agree. We both estimate about 100 plus plates (not 10 plates and not 500 plates) were enough to engrave the text of the Book of Mormon. That is close enough for this engineer. 🙂

Best,

Bruce

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.pdfSince 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

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

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