Date: Sat, 16 Sep 2000 01:46:30 -0700
To: balloon@balloonhq.com
From: Mark Balzer <mbalzer@balloonhq.com>
Subject: Re: twists adding weight.
>Yes, in fact, each twist *does* add weight to the balloon design!
Don't take this personally, but it really bothers me it when amateurs
*emphatically* make incorrect statements when playing "self-appointed
science teacher" in public. Science is hard enough to learn when
your teacher gets it right, let alone when different people
masquerading as experts tell you opposite things.
I don't know about you, but in high school I learned that
Weight = mass x acceleration due to gravity (often called "g")
As long as we don't leave the surface of the earth, the acceleration
due to gravity "g" doesn't really change... it is pretty constant.
Barring any E=mc^2 nuclear reactions (which not even the fastest
twisters seem capable of producing :-), neither the mass of the latex
balloon, nor the mass of the helium enclosed changes.
So then, weight = the product of two constants, which not
surprisingly IS CONSTANT!!!
>(For those who are uninterested in Physics or Trivia, please ignore
>this post.)
... and those who are very interested in Physics or Trivia would also
do best to ignore that post!
>Each twist that is added to a balloon increases the pressure of
>the gas inside it. (helium or air, it dosen't matter!)
Only if the balloon is fully inflated "hard" to begin with (and who
would be twisting a fully inflated hard balloon?). Even so, the
pressure increase is not appreciable... you are talking about
fractions of one "psi" or "pound per square inch" (the air we breathe
is about 14-15 psi).
>If latex was an "ideal" elastic (in terms of a Physics professor),
>then the increased pressure would result in an equal change in the
>volume of the balloon.
That is COMPLETELY, 100% FALSE!
Steel is an elastic material... Heck, it's even "linearly elastic"
and that's as "ideal" as elastic materials come. You are telling us
that if we double the pressure in our steel helium tanks, the tanks
should double in volume, ie, that they should get twice as long. How
can you post blatantly false statements like that, and then refer to
"Physics Professor terms" in an attempt to give it phoney credence?
Have you no shame? Impressionable children read this list!
>Latex is *not* an "ideal" elastic.
Is too, is too!
> When you are filling a latex balloon, you start out fighting the
> un-tensioned balloon. It takes a relatively large pressure to get
> the balloon started. Once you get that first bubble started,
> however, the pressure necessary to continue the inflation is much
> less. As the balloon nears the maximum safe inflation point, the
> necessary pressure increases!
>
> Meaning: When the balloon gets tight, you need to force in more air
> to get it to expand!
Arghhhh!!!!! You are really confusing a lot of separate effects.
When you strain (apply pressure, stress or force to) an object, it
deflects (stretches, sags, etc.).
An "ideal" elastic material is one for which the deflection (stretch)
disappears when you remove the strain (pressure).
Springs are elastic. Clay is not.
In fact, latex is an excellent "elastic" material. Now, after you
release the pressure from inside a latex balloon it may take some
time before it shrinks completely back to original size, but it
_will_ get there. (Technically, this time dependence makes latex a
"viscoelastic" material, but that is neither here nor there... it's
still elastic.)
> This compression, this increased pressure, means that the same mass
> of latex and air is displacing *less* of the surrounding
> uncompressed air. It is this displacement that makes a balloon seem
> light. A helium-filled balloon displaces a greater mass of air than
> the mass of the balloon and helium. This excess displacement is what
> makes the balloon float. If you could cram in an extra few grams of
> helium without changing the size of the balloon, (its displacement.)
> then the balloon would be a few grams heavier, and wouldn't float as
> well. (if at all!)
Ahhhh! If you had only written that and nothing more...
>When you twist a 260, you decrease the displacement of the
>balloon without reducing the mass. This is why an added twist
>actually *does* add weight to a twisted balloon.
But if mass and weight are proportional (see above) how can you
change one but not the other?
Now that you have contradicted your own post, which one of your
*emphatic* statements should we believe?
Too bad you never met Archimedes because he solved this problem in
the third century B.C.
Let's say that when you step on a scale, it reads 180 pounds.
Although you might tell people that you "weigh" 180 pounds, you
really don't. You actually weigh more but don't see it on the dial
because you are being lifted up by the air all around you. Given
your density and the density of the air, it's not too hard to compute
that you are displacing 0.23 pounds of air, hence the air is exerting
a buoyant (vertically upward) force on you of 0.23 pounds. If you
could weigh yourself in a vacuum chamber, you would find your true
weight to be 180.23 pounds.
Now in common, everyday life we don't go around correcting for the
buoyancy of air. 180.23 pounds is close enough to 180 pounds that we
neglect the 0.23 pounds of buoyant force the air exerts on you.
However, in the case of balloons and blimps, the buoyant force
exerted by the air is THE controlling factor in the design and CANNOT
be neglected.
Therein is the answer. The mass AND weight of the balloon and the
helium inside it remain unchanged as you twist the balloon. The
buoyant force that the air exerts on the balloon changes a tiny bit.
But I maintain that it's not significant. As I wrote in the Guide
many years ago, as long as you have enough total bubble length, a
helium inflated 260 will float.
>If anyone with access to a jeweler's scale
>(the most accurate kind of scale I know about)
It doesn't matter how accurate your "jeweler's scale" is (and you'd
be better off with a "laboratory _balance_" than a scale), you still
need to correct for the buoyant force that the air exerts on whatever
you are weighing. When you do that, you will find TWISTS DON'T
CHANGE THE WEIGHT OF A BALLOON!
>and an interest in providing empirical evidence wants
>to help me out here, I would *love* to hear some actual
>numbers. Weigh the inflated balloon both before and
>after twisting, and send me the numbers.
Check the "How helium balloons float" chapter of the guide where some
actual numbers that have been there for 5 years show that:
1. each twist shortens the balloon by using up rubber that could
otherwise be bubbles, and
2. as long as you have a total of at least 41" of helium-inflated
260 (in a nice low humidity environment), it should float. (of
course, 41" is only to be used as a guide. This will vary if your
balloons are heavier/lighter, or your helium is not pure, or your air
is humid, etc.)
Now, as your penance for that misleading post, you must read this web
page and surf every single link on it 10 times:
http://www.amasci.com/miscon/miscon.html
I hope this helps,
Mark
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