Derigible Lift Capacity: Calculations and Accuracy Explained

[Jadson]

Alright. I submit. I'm lost. I've been working on a plan for a derigible, or airship, and found that I've run into a problem. I've been using the equation:

[(H x W x L) x Pi] / 3

to calculate the capacity of the airbags within the derigible. I don't think this is correct so please help me there if you can. Anyway, the real problem is I don't think my results are acurate. I have looked up helium's lift capacity and found that it is 1 m3 can lift 91.82566798 grams, or 0.2024409 pounds. This seems logical, but that's not the problem either. After calculating the entire derigibles helium lift capacity (4869.468161 m3), I have found that it can only lift about 4471.4216659116718478 grams, or 985.77973565 pounds. This does not seem acurate seeing as the airship has 20 airbags ranging from 7x7x7 m to 12x8x7 m. I would think it could lift a bit more but, I am rather unexperienced in this field.

Can someone please confirm my calculations?

 

[faust9]

Are you air bags shaped like cones? I would use spherical airbags V=\frac{4\pi r^3}{3} or cylindrical V=\pi r^2 h or even cubic air bags V=L \times W \times H

For example: let's say the average radius of 20 spherical bags is 8.5 m thus:

V_{one sphere}=\frac{4\pi (8.5m)^3}{3}\approx 2572 m^3

V_{all sphere}=V_{one sphere}\times 20 \approx 51448 m^3

which can lift \approx4682 kg, or 10322 lb.

 

[Jadson]

Thanks a lot Faust9, I was really worried that all of my work had gone to crap. Anyway, there are some spherical airbags, but some are not spheres, more oval shape. circular shapes that are such sizes as 12x8x7 as an example. To be more specific, there are 6 7x7x7 airbags, 2 12x8x7 airbags, and 12 9x3x8 airbags. I can't seem to find an equation for oval shaped ones:

does anyone have an equation for that?

 

[enigma]

Volume of a spheroid is π*a*b*c

Where a, b, and c are the semiaxes (like a radius in a single direction)

 

[wj]

How are you planing to steer the airship?