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Algebra/Table gives vague values for estimating space vessel tonnage. Unable to determine number relationship and find governing equation.


Hello, and thank you for offering to answer questions. Though I am pretty sure the answer will be obvious to you, and therefore my question may be annoying, the plain fact is that I have a serious learning disability and it severely impacts my ability to calculate even the simplest things. My difficulties from this problem even extend to make it hard for me to solve arithmetic of very young children, for example.

I have a certain sourcebook, and it includes what is supposed to be a table for getting "rough and ready" estimates for the tonnage of space vessels. The chart is for use in writing stories, and the like. I realize that values shown in this chart are only ballpark estimates. But judging from the values shown here, how did the author find a way to guess tonnage based on length? Put another way, what method was used? I do not understand the pattern of progression, I have no clue.

Ideally, I would then plug in the length of one of "my" vessels (in one particular case, 384 meters) and arrive at a crude estimate of the vessel's laden tonnage. This would be so much easier if I knew how the chart had been made! Thank you very much for your time and attention.

I took the table and given the length, computed a value that is a rough approximation of the volume as the length * length * length.  This is obviously bigger than the volume, but it should be roughly the same times as large in each case.  When the tons are divided by the cubic yards, it says that the approximate weight is roughly 100 tons per cubic foot (when the average is computed in Excel, I get 104.0).

I basically constructed a table in Excel.  Column A was meters, column B was tons,
column C was yards, column D was yards^3, column E was the weight in pounds (2000*tons),
and column F was the pounds per yard cubed.

Column A: The only entry in Column A is down at the bottom where I put in 384.

Column C: To get yards, multiply this amount by 39.37/36, since there are 39.37 inches in a meter and 36 inches in a yard.  This gives the 419.947.

Column D: Cubing it gives 74,059,780.

Column F: Computing the average number of pounds per cubic yard gives 19.72594752.

Column E: Multiplying column D by column F gives the total pounds, and that is 1,460,899,326.

Column B: Dividing column E by 2000 gives the tons, and that is roughly 740,000.


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Scott A Wilson


Any algebraic question you've got. That includes question that are linear, quadratic, exponential, etc.


I have solved story problems, linear equations, parabolic equations. I have also solved some 3rd order equations and equations with multiple variables.

Documents at Boeing in assistance on the manufacturiing floor.

MS at math OSU in mathematics at OSU, 1986. BS at OSU in mathematical sciences (math, statistics, computer science), 1984.

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