[islandlabs] How Much Helium?
masterjediyoda at gmail.com
Thu Dec 17 18:18:37 EST 2009
o right, I used the diameter by accident in my formula, not the radius.
So you are right with:
*(4 / 3) * pi * ((5 / 2)^3) = 65.4498469*
yea, it doesn't make sense to add helium proportional to size. So if there
volume of balloon is x and they added b amount of helium, and our volume is
x+y, we should not add x+y*b helium, this is excessive. We actually need
closer to x as well.
I like this fish hook thought very much. We can turn it upside down. good
idea. I have one of these I will bring to the next meeting, it is digital
and accurate. Used for striped bass in the Long Island sound.
2009/12/17 Burns, William <burns at cshl.edu>
> Here's the math I (um.. excel) used:
> (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)
> (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
> (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.
> *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
> 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.
>> 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*π*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
>> 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
>> 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.
>>> On Thu, Dec 17, 2009 at 15:13, John Teddy <masterjediyoda at gmail.com>
>>> > http://i45.tinypic.com/riaop5.jpg and
>>> > This antenna is already rigged to connect to a usb cellular modem, that
>>> > able the size of a large thumb drive. So I believe this connector
>>> > work with a lot of cell phones also. When the phone comes we ordered, I
>>> > bring this antenna and see if it fits.
>>> > I believe the remaining parts we need are: parachute and helium (and
>>> > string/rope to connect it all, and other little things which are easy
>>> > get).
>>> > We need to calculate how much helium we will need for the baloon. As it
>>> > explained to me yesterday, the large helium tank can weigh up to 500
>>> > So this is exceedingly difficult to carry around. If anyone can crunch
>>> > numbers and figure out how much helium we would need, and how much each
>>> > tank can hold.. that would be helpful.
>>> > -John
>>> > _______________________________________________
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