How Do You Build a Stable Homemade Mini-Blimp?

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SUMMARY

This discussion focuses on the construction of a homemade miniature blimp, emphasizing the importance of using helium instead of hydrogen due to safety concerns. Participants highlight the challenges of stabilizing a blimp with a large surface area and low mass ratio, particularly in windy conditions. The Drag Equation is introduced as a method for estimating drag forces, with a drag coefficient of approximately 1 suggested for initial calculations. Additionally, the volume calculation for an ellipsoid is clarified, noting the distinction between semi-axes and diameters.

PREREQUISITES
  • Understanding of basic aerodynamics and lift principles
  • Familiarity with the Drag Equation and its application
  • Knowledge of gas properties, specifically helium and hydrogen
  • Basic geometry, particularly regarding ellipsoids and volume calculations
NEXT STEPS
  • Research the Drag Equation and its application in airship design
  • Learn about the properties and safety considerations of helium versus hydrogen
  • Study the mathematics of ellipsoids, focusing on volume and surface area calculations
  • Explore stabilization techniques for lightweight airborne crafts in windy conditions
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Aerospace hobbyists, engineering students, and anyone interested in building and stabilizing lightweight airships or blimps.

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I've been having thoughts of building a home-made miniature blimp roughly the size of a coffee table or a small room.

I haven't quite thought out the specifications of lift, or weather to use helium or hydrogen (or weather hydrogen is safe/legal for this kinda thing.) or the dimensions of the balloon (or anything else for that matter)


But when i proposed this idea to a friend, he pointed out that stabilizing something that small with such large surface area/mass ratio would be difficult (especially in the windy conditions at the altitudes i plan to fly this thing.)

So, I've come here to ask if anyone knows any way to theoretically calculate the effects of forces on airbourne bodies of arbitrary shape but given mass. (The mass will NOT be centered at the center of the blimp obviously, (it will be suspended directly below it).
 
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Don't even think about using hydrogen! :eek:
I'm afraid that I can't help with the rest of your question.
 
The upper bound of the drag force will be roughly given by Drag Equation with drag coefficient ~1. Realistically, it will probably be a bit lower, but this is a good place to start your estimates.

Using hydrogen is not safe. If you have any amount of air in there, and with home made blimp you are almost guaranteed, you end up with an extremely unstable mixture. The size you are talking about would create a very serious fireball. Hydrogen also happens to burn very hot, so you definitely don't want to be anywhere near it when it goes off.

Helium is not that hard to come by, in quantities you need, and generates almost the same lift. Use that. It's far safer. Bear in mind that a suffocation risk is still there if you test it indoors.
 
I see...

I dunno, i hear some of the older war-era blimps were hydrogen based, but i guess that's why the hindenburg accident happened.

Thanks for the tip on hydrogen and the lift.
 
That's a big part of why it happened, and Hindenburg is a big part of why they stopped using hydrogen. Modern blimps are all helium-filled.
 
This probably belongs in the math section but,

With that drag formula, it's said that for airships they use the a function of the volume of the thing in order to find the reference area.

Now the shape of the part i want to calculate the reference area for is an ellipsoid. So i need to find the volume of an ellipsoid. This is given on wikipedia though is kind of strange.

They say the volume of any spheroid is 4/3 pi *(a^2) * (b),

However they go on to say that
IF A=2a is the equatorial diameter, and B=2b is the polar diameter the volume is 1/6 pi *(A^2) *(B)

Is the equatorial diameter A or a? Same for the polar diameter.
 
A and B respectively. Easy to check. For a sphere, volume is 4/3 pi R^3. Take a=b=R, and you get formula for sphere. So a is a semi-axis, and you get diameter by taking 2*a.
 

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