I hope I am posting this in the right forum. My current problem is part aviation, part maths. I have been struggling with the following computation ( Burgess, Charles Paine. Airship Design. Honolulu, Hawaii: University Press of the Pacific, 2004. page 14. The pdf is available at http://www.xlta.org/library/burgess.pdf, note that the first edition of this book was written in 1927. The present snippet uses the typographical conventions from the pdf.)

Problem 1: Find the volume and horsepower of a rigid airship to carry a military load of 15,000 lbs.at 60 knots (101.3 ft. / sec.). for 60 hours, 85% of the total volume being filled with helium lifting .064 lb./ft.**3 (94 % pure) in the standard atmosphere. Since the hull is specified in the condition of the problem to be 85% full of gas, the weight of the gas is 85% D multiplied by the difference between the weight of air and the lift of gas per unit volume, and divided by the weight of air per unit volume. The total weight of air and gas is therefore given:

$$ Air\; and\; gas = [ .15 + .85 ( .07635 - .064 ) / .07635 ] \times D = .288 D $$

While this formula is not an issue, the subsequent formula has been giving me headaches.

From existing data, the fixed weights exclusive of power plant, power cars, and fuel system equal -3D, and the crew, stores and ballast amount to .055 D. There remains for power plant, fuel, and military load

$$ ( 1 - .288 - .30 - .055) D = .357 D $$ Question is how did the author arrive at this result? Not matter how I approach solving the equation, I am unable to get the same result. Problem is that the result of this equation is used at a later stage. I also considered that 1 might be an i, but there was no mention of any constant i in the chapter this formula was taken from. I have no clue what I am missing here.

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    $\begingroup$ I agree the "1" can easily be mistaken with a "I", but I also think this is a "1". I'm glad now we can use modern tools to write equation and avoid such ambiguity. $\endgroup$
    – Manu H
    Jun 30, 2020 at 11:42
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    $\begingroup$ A lower-case L "l" was often used as a substitute for the digit one "1" in early typography and typewriters. An off-topic anecdote: My father learned to type on a typewriter without a digit one key and this bit him hard when he was programming in the punch card days. He one spent spent 3 weeks chasing down a bug in a program because he'd used an "l" (ell) when he needed a "1" (one) and the compiler happily created a whole new variable for him. Lesson: always predeclare your variables and shun compilers that will create variables for you on the fly! $\endgroup$
    – FreeMan
    Jun 30, 2020 at 13:57
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    $\begingroup$ @FreeMan Thanks for your note on typography, that actually helps. Apart from dealing with all the math, the print is not great and although I've come across several "old" books with "strange" typography, trying to figure out variables in formulae makes it even harder. Luckily, pdf files can be magnified, another benefit of "the modern era" (not implying that everything is better now). $\endgroup$
    – Sasquatch
    Jul 1, 2020 at 22:45
  • $\begingroup$ I think there is a typo in the second quote which initially threw me: -3D should be .30 D. BTW: I've read some of the old fluid dynamics books of the Saturn 5 era before modern CFD. Excellent read. $\endgroup$
    – tim
    Jul 1, 2020 at 23:56

1 Answer 1

  • $D$ is the total displacement (the weight of the air displaced by the airship)
  • $0.288 \cdot D$ is the weight of helium and air inside the air ship
  • $0.055 \cdot D$ is the weight of crew, storage, and ballast
  • $x \cdot D$ is the unknown weight of power plants, power cars, and the fuel system
  • $0.3 \cdot D$ is anything else (fixed weights)

The total must be one:

$$.288 + .055 + .3 + x = 1$$

or equivalent, the total must be the displacement:

$$D \cdot (.288 + .055 + .3 + x) = 1 \cdot D$$

solve for x and you get:

$$x \cdot D = (1 - .288 - .055 - .3) \cdot D$$

or just

$$x = 1 - .288 - .055 - .3 = .357$$

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    $\begingroup$ Could it be that there is a typo in your formula. I think it should be 0.055 instead of 0.55 (cf. edit(s) above). I tried to edit it, but could not save the changes due to minimum amount of characters required. Anyway, I entered the formula in the calculator and got the result. Still can't tell why I wasn't able to compute this formula in the first place, since it is quite simple, maybe too simple. $\endgroup$
    – Sasquatch
    Jul 1, 2020 at 22:39
  • $\begingroup$ Yes, indeed, thanks! $\endgroup$
    – bogl
    Jul 1, 2020 at 22:47

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