Estimating/Calculating "Tonnage"

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Estimating/Calculating "Tonnage"

Post by Navclio »

I originally posted this in response to a comment in another thread but thought it would be more useful to readers if it also were under a relevant heading.

You don't need to compare dimensions with the dimensions of British ships if you have measurements that you can convert to British feet. The original British formula for burthen or builder's measure tonnage was (L × B × D)/94 where
L = length on gun deck measured from the outside of the stempost to the outside of the sternpost
B = beam measured from the outside of the planking
D = depth in hold
Later, ½B was substituted for D, so the formula became (L × B × ½B)/94.

"Inside" vs. "outside" was an important detail. The despicable Americans cheated by measuring INSIDE the stem and sternposts and INSIDE the planking (i.e., outside the ribs) and dividing by 95, and then publishing the results. This use of slightly smaller measurements and a slightly larger divisor gave their ships a smaller tonnage than they would have had if the Americans had used the correct formula, and thus defrauded British officers into thinking that the American ships were smaller than they really were.

In the case of Italian or other navies, it would be important, or at least useful, to know how the measurements were made. In the absence of information either way, one can, of course, assume that the measurements had been made correctly, outside the posts and planking. The bigger the ship, the less difference this makes in practice. Moreover, comparing burthen tonnage assumes identical lines, which might not be the case.

I often have to estimate displacement tonnage for ships of the 1740s for which it has not been calculated. At that time, the Swedish and French navies were actually calculating displacements. Jan Glete estimated the displacements of ships in other navies, to the nearest 50 or 100 metric tonnes, by reviewing plans and estimating "block coefficients," and this is what I do for ships of unknown displacement for which I have dimensions. I find a French or Swedish ship of known dimensions and displacement, or several, preferably of the same type (frigate, small two-decker, two-decked SOL, etc.) and calculate the volume of the prism defined by the average of gun-deck length and keel length × beam × depth in hold (data on drafts is available for French ships but I have not seen it for any others), for both the ship of known displacement and the ship of unknown displacement. I don't know whether the French and Swedes used "inside" or "outside" measurements and I don't usually know that for anyone else either, so I have to assume that they were the same. I then multiply the ratio of the volumes by the known ship's displacement. I try to do this for several comparison ships and usually get fairly consistent results. All of this assume that the lines, in terms of such characteristics as flat or rising floors and bluff or sharp bows, are about the same for both the ship being estimated and for the reference ships. So the formula for converting a reference ship's displacement to an estimate of the displacement for another ship is
(RV/EV) × RT = ET
where RV = the volume for the reference ship (whose displacement is known)
EV = the volume for the ship whose displacement is being estimated
RT = the displacement of the reference ship in metric tonnes
ET = the estimated displacement of the ship whose actual displacement is not known

Many of the French displacement figures are listed by Demerliac (I have not yet obtained a copy of Winfield and Roberts) as "environ X," introducing additional uncertainty. For that reason, I usually round everything to the nearest 100 metric tonnes. It's clear enough that 2400-tonne Spanish 70s and 2500-2800-tonne French 74s were significantly larger than 1700-tonne British 70s, for instance.

The French usually calculated or estimated two displacements, one for just the hull and the other for the ship fully fitted out, with rigging, ordnance, ship's stores, water and victuals, and crew (even at just 150 lbs. each for a man and his personal effects, this is significant). I use the "full load" figure.
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Re: Estimating/Calculating "Tonnage"

Post by Cy »

As I understand it, the length of gundeck was never used, it was always the length of the keel, but over time the way that was used changed.

The original English measurement, Known after it's originator as "Baker's Tons" and estalished in the 16C during Elizabeth's reign was:
"The length of the keel, leaving out the false post, if there be any. Multiply by the greatest breadth within the plank, and that product by the depth taken from the breadth to the upper edge of the keel produceth a solid number which divided by 100 gives the contents in tons, into which add one third part for tonnage, so have you the tons and tonnage." (FROM Oppenheim: A History of the Administration of the Royal Navy and of Merchant Shipping in Relation to the Navy. Volume 1)

Later still, in the mid 17C, the calculation was changed to use a "length of keel of tonnage" which is a calculated value and not the same as athe actual "touch" length of the physical keel. This was because ship's stems and sterns had become less vertical so the keel was shorter in relation to the overall length of the ship. Here is a link to a page with links to various period descriptions of the methods of calculating tonnage and length of keel. (fromwhich the above was extracted).
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Re: Estimating/Calculating "Tonnage"

Post by Navclio »

Cy is correct about the British (and fraudulent American) formula for "burthen" or "builder's measure" tonnage--the length was the keel as measured between perpendiculars, with the British doing it correctly from the outsides of the perpendiculars and the Americans cheating by measuring between the insides. When I have the data, I prefer to use the mean of the gun deck length and the keel length as the measure of length, since the length at the waterline was longer than the length of the keel. In either case, one must then apply a "block coefficient" to the prism or rectangular solid defined by the length (however one calculates it), beam, and the depth in the hold. Neither the hold itself nor the hull form that causes the displacement of a weight of water equal to the weight of the ship and everything in it is a squared-off rectangular solid
and so the ship occupies only a fraction of that space. Block coefficients varied, with bluff bows and flat floors producing a larger coefficient (closer to 1.0) than sharp bows and rising floors. As I wrote in my original post, for a ship of unknown displacement but known dimensions, I derive a block coefficient from the dimensions of a ship of known displacement and in a similar class at the same time, e.g., a frigate from the same for a frigate, a two-decked ship of the line from the same war for a two-decked ship of the line, a corvette for a corvette, etc. Usually, I am doing this to calculate British or Spanish displacement from French ships or Russian displacement from Swedish ships, and I hope that at any one time period the coefficients were fairly similar. It would be a bad assumption for the Dutch vs. the French, because the Dutch deliberately designed their ships with very flat floors and full hull lines to reduce their draft. Fortunately I have not had to estimate any Dutch displacements; if I did, I would probably use Swedish displacements as the basis.

French and Swedish (and, I suspect, Danish) naval architects calculated the displacement from the lines, the French at least by taking repeated cross-sections at short intervals to determine the actual volume of the hull below the waterline. I think British naval architects must have done so, also, because they had a form for captains to report the weight of everything that had gone into a ship before she began a cruise (e.g., 1680 32-pound shot = 24 tons; 650 men @ 140 lbs./man = 40 tons). There would not have been any point to that if there had not been calculations of the displacement for a set of lines if settled to the waterline, which would be the weight of the hull plus everything that had been added. In other words, British warships, like everyone else's, were designed with a certain total weight and resulting waterline in mind. Adding up what went in on a certain occasion would tell the administrators whether the ship was overloaded, at least in terms of freeboard for the main battery if an immediate engagement ensued. (In 1748, when the Spanish were desperate to get mercury to Mexico to continue silver production, they so overloaded three warships that one of them had only 1½ Spanish feet of freeboard. The Cadiz mercantile regulation agency, the Casa de Contratacion, reported this to the navy minister with some alarm, and some of the cargo, which included more than mercury, was offloaded to another vessel that was added to the convoy of which the mercury ships were a part.)

There is an easier way to calculate displacement for a ship with curving lines, and it was what I assumed that 18th-century naval architects did until I read that the French, at least, used a series of closely-spaced cross-sections. That would be:
(1) carve a scale model of the hull form, with the waterline marked;
(2) immerse the model up to the waterline in a full tank of water with arrangements to collect the overflow;
(3) measure the volume of the overflow;
(4) scale up the volume (e.g., multiply by 192 if the model scale was 1 inch = 16 feet);
(5) multiply the volume by the density (e.g., pounds per gallon) of seawater.
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