How are horsepower and standard displacement for a rigid airship calculated according to this book?

This is a follow-up based on a question posted here. I updated my initial question but came to the conclusion that a new post might be more manageable. For bibliographic reference(s) and/or getting a copy please refer to the linked question. Be that as it may, this is my current issue:

Having had some time at hand, I went back to the book. The solution of the problem set above continues like this:

Assuming the weight of the power plant and its cars to be 8.lbs/Hp. and the weight
of the fuel and the fuel system to be .6 lb. per horsepower hour, the total weight
of power plant, fuel system and fuel is

[8 + ( .6 x 60 ) ] Hp. = 44 (Hp.)

combining this with the military load,

15,000 + 44 (Hp.) = .357

No issue so far, I was also able to compute this formula:
enter image description here

However, I cannot make heads or tails of the next part

Combining equations (1) and (2)

enter image description here

My question is how did the author arrive at equation (3)? Where does the value 48D(2/3) come from and how were these formulae combined.
When using the previously computed value of .0357 for D, I am not getting the result of 42,000.
Apart from the "actual" math involved, I fail to see the meaning of the components involved in equation (3):
Could someone help me out here and maybe translate that formula into English sentences?

Aviation Asked by Sasquatch on December 29, 2020

1 Answers

One Answer

The book author just eliminated Hp from 1 by the formula 2. $$ 15000 + 44(.39D^{2/3}) = .357 D$$ Do arthitmatic and rearrange: $$.357 D -17.16 D^{2/3} = 42000$$ Divide by the .357 to get $$ D- 48.067 D^{2/3} = 42016.8 $$ Then round off because he used a slide rule not a calculator.

Correct answer by W H G on December 29, 2020

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