# 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?

## 1 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.

• Dividing your second line by 0.357 should increase the 42,000 value on the righthand side by roughly a factor of 3, rather than they very slight change shown in the 3rd line. Is 42,016.8 really what you intended there? Or is 42k not really what you wanted in the 2nd line (15k, perhaps)? – Ralph J Jul 23 '20 at 3:07
• @RalphJ I'll surely take your comment into account. At the moment, my goal is merely to understand the equations and how the author arrived at some of the results. The present question is a actually a good example for some "missing" steps in solving the equation, at least in my opinion, since I haven't done any "sophisticated" maths for quite a while. For the time being, I'll stick with the values taken from the book Appreciate the additional details. – Sasquatch Jul 23 '20 at 21:47
• My mistake in copying notes. The 42000 in second equation should be 15000 (from eq 1). Then divide by .357 gives 42016.8. Just for fun, I figured out that D=213600 (approximately). – W H G Jul 24 '20 at 14:33