# 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:

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

Combining equations (1) and (2)

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

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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