Re: calculating airship volume

"The Older Gentleman" <chateau.murrayTAKETHISOUT@xxxxxxxxxxxxx> wrote
in message
news:1i2hcrg.skxuni2i6ggoN%chateau.murrayTAKETHISOUT@xxxxxxxxxxxxx
I'm no mathematician, but I need to know the answer to this.

The Hindenburg airship was just over 800ft long and 135ft in diameter
and contained seven million cu ft of hydrogen and had a payload of 123
tonnes.

Assuming that it's a cylinder (which it isn't, quite), what dimesnions
would it need to be, to carry double that quantity?

Simplest answer - Twice as long. However I am sure there are more elegant
engineering solutions!

Also, assuming it carried 14 million cu ft of helium rather than
hydrogen, what would the useful payload be?

A very rough (and very likely incorrect) calculation - 214 tonnes, based on
2x123 = 246 tonnes for payload with Hydrogen less the additional weight of
the helium (approx twice the weight of hydrogen) at approx 32 tonnes.

It doesn't help that the volume is given in imperial measure and the payload
in metric. Nor does it help that, after doing the caculation, I found that
the actual payload was 123.5 tons or 125.5 (approx) tonnes. However the
difference is only small.

HTH

--
Tony
'04 XL1200C, '95 LS650
OMF#24


.

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