[islandlabs] How Much Helium?

Burns, William burns
Thu Dec 17 18:02:16 EST 2009


John:
 
Here's the math I (um.. excel) used:
MIT:
(4/3) * pi * (5/2)^3 = 65.45 cubic feet.
(am I using the right formula? I just blindly applied what I saw in the e-mail thread)
 
IL:
(4/3) * pi * (6/2)^3 = 113 cubic feet
 
I got the 6-foot diameter figure by assuming that our payload would be the same weight as the MIT payload, and making the assumption that an MIT sized payload would require just-as-much more helium to work w/ a kaymont 800 (percentage-wise) as the example 250gr payload that kaymont uses in their tables.
(scale the diameter up by 20% to move from a kaymont 350 to a kaymont 800)
 
This is just a ballpark calculation.
In the field, we could measure the amount of lift that the balloon provides (w/ a fish scale) and fill it 'till it provides 150% of the payload weight.
If no fish-scale is handy, we can fill a pail 'till it weighs 150% of our payload, and fill w/ helium 'till we get neutral buoyancy.
 
-Bill


________________________________

	From: list-bounces at islandlabs.org [mailto:list-bounces at islandlabs.org] On Behalf Of John Teddy
	Sent: Thursday, December 17, 2009 5:46 PM
	To: Island Labs main mailing list
	Subject: Re: [islandlabs] I have an antenna for the balloon in my office...
	
	
	So sqrt(350) and sqrt(800)....
	
	28.2842712 / 18.7082869 = 1.51185789
	
	So our balloon will be 1.5 times the diamater of MIT. So it will be ~7.5 feet diameter at launch, and ~30 feet diamater at burst (this all depends on how much helium we put in. More helium means it rises faster, but bursts at a lower altitude I believe. Someone tell me if this makes sense logically. We have to use the free lift formula and decide on a proper amount of cubic feet to add to the balloon.
	
	It would be helpful if we knew how much the MIT guys used. I'm guessing the just winged it and used as much as they thought was necessary.
	
	(4 / 3) * pi * (5^3) = 523.59 and (4 / 3) * pi * (7.5^3) = 1767.14
	
	

	1 767.14 / 523.59 = 3.37504536

	So our volume of the balloon will be ~3.3 times what the MIT guys used. So we need ~3.3 more helium.
	
	
	On Thu, Dec 17, 2009 at 5:04 PM, John Teddy <masterjediyoda at gmail.com> wrote:
	

		I see, the person must have been mistaken. Or maybe they were talking about industrial tanks of larger size.
		
		http://www.target.com/gp/detail.html/185-8901296-2975720?ASIN=B000PGQ63O&AFID=Froogle&LNM=B000PGQ63O|Disposable_Helium_Tank&ci_src=14110944&ci_sku=B000PGQ63O&ref=tgt_adv_XSG10001 <http://www.target.com/gp/detail.html/185-8901296-2975720?ASIN=B000PGQ63O&AFID=Froogle&LNM=B000PGQ63O%7CDisposable_Helium_Tank&ci_src=14110944&ci_sku=B000PGQ63O&ref=tgt_adv_XSG10001> 
		
		This is a 12lb tank, and holds 15 cubic feet of helium. It only costs $50.
		
		I don't think 15 cubic feet of helium be nearly enough.
		
		The MIT balloon was 5 feet diameter at launch, and 20 feet at burst. So ours will be under 10 feet diamater at launch, and under 40 feet diamater at burst. My figures are off, need the sphere formula. 4/3*?*radius3
		
		I don't know what the weight in the balloon (we have 800gram balloon, they used 350 gram right?), translates to as far as diameter and cubic feet of helium.
		
		From the guide:          "each cubic foot of helium can lift 28g" and "each pound of free lift would me 300feet per minute of ascension"
		
		
		
		http://www.balloonsandhelium.net/heliumtanksforsale.html <--- they are way to expensive to buy. Rent or launch at a party store are the only options. And the weight should definitely be under 100 pounds as you said Mark.
		
		
		They don't actually say in the guide how much helium they use I believe. Does anyone see a figure on how much helium the MIT guys used? They use an example of "61 cubic feet of helium" to describe an example for a formula, but that doesn't mean that is how much they used. If we need 61 cubic feet of helium this target tank is not nearly big enough.
		
		
		
		
		
		As far as the antenna goes, it was bought with a cellphone modem, it's designed to be used for cell phones. It's omni-directional, long, and quality. And it has the wire connector we need which attaches to the phone already I believe, so no soldering. 




		On Thu, Dec 17, 2009 at 3:30 PM, Mark Drago <markdrago at gmail.com> wrote:
		

			Unless the large helium tank is obscenely large, it's not 500 pounds.
			I worked at a party store that rented tanks in 3 sizes (small, medium,
			large) and had a fourth size for our own uses (jumbo).  The jumbo
			tanks were easy for one person to roll around on their edge, but were
			too heavy to really pick up and carry.  I would say they were easily
			under 100 pounds, probably closer to 60 or 70.  This is a rough guess
			as it has been 10 years since I moved one.  But I can definitely say
			they were not 500 pounds.
			
			Mark.
			

			On Thu, Dec 17, 2009 at 15:13, John Teddy <masterjediyoda at gmail.com> wrote:
			> http://i45.tinypic.com/riaop5.jpg and http://i50.tinypic.com/1htb15.jpg
			>
			> This antenna is already rigged to connect to a usb cellular modem, that is
			> able the size of a large thumb drive. So I believe this connector should
			> work with a lot of cell phones also. When the phone comes we ordered, I will
			> bring this antenna and see if it fits.
			>
			> I believe the remaining parts we need are: parachute and helium (and some
			> string/rope to connect it all, and other little things which are easy to
			> get).
			>
			>
			> We need to calculate how much helium we will need for the baloon. As it was
			> explained to me yesterday, the large helium tank can weigh up to 500 pounds.
			> So this is exceedingly difficult to carry around. If anyone can crunch the
			> numbers and figure out how much helium we would need, and how much each size
			> tank can hold.. that would be helpful.
			>
			> -John
			>
			
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