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This analysis is probably good for a perfectly stiff dirigible, but remember
a blimp is elastic and will stretch initially so immediately after depends
on how immediately. So immediately afterwards the mass will accelerate at g
and the blimp not at all. Then a little later (how long?) the analysis
would be correct, except for corrections due to the oscillations of the
mass.
I presume you could try this with a Helium filled balloon and using some
video analysis you could confirm the idea. But a balloon would have a short
oscillation period so that might not be noticeable.
I have often seen the Goodyear blimp, but sufficiently far away as to be
unable to see the flexing of the balloon. But I seem to recall there are
some videos showing it. I missed riding in it when it was stationed outside
Houston, but perhaps someone who has been in it can come up with an account.
I suspect the dirigibles were a bit smoother in their "flight".
John M. Clement
Houston, TX
-----Original Message-----
From: Phys-l [mailto:phys-l-bounces@phys-l.org] On Behalf Of Paul Nord
Sent: Monday, November 05, 2012 11:49 AM
To: Phys-L@Phys-L.org
Subject: [Phys-L] Dirigible Flight Question
If I've got a blimp inflated to neutral buoyancy and I hang a
small mass from it, what will the acceleration of the blimp
be immediately after I attach the mass?
Since we're still at zero velocity we can ignore the viscous
effects of the air for just a moment. I believe that I need
to know the un-inflated mass of the balloon and the payload,
the volume of the helium, the mass of the helium, and the
mass of the displaced air. Let's assume a very small pressure
is held by the balloon so that we can think of it as simply a
volume of helium at the ambient pressure. The mathematics of
a simple Attwood's Machine would seem to apply.
The total mass going down:
balloon and payload
helium mass
extra ballast weight (call this 'm')
Total mass going up:
mass of displaced air
Let's call the sum of all of the mass except for the ballast
weight 'M'.
M = balloon + payload + helium + displaced air
The acceleration of the balloon is then:
a = g * m / (m + M)
Question 1: Of course Pascal's Principle says that air
pressure will distribute itself equally on all sides. In the
static case I can ignore the effects of pressure and air
mass. The net force is zero (ignoring the vertical pressure
gradient of the air, yes). However, for a balloon to move
down, an identical volume of air needs to move up the same
distance. The mass of that air cannot be ignored. Is this a
valid assumption?
Question 2: Flow through a vicious fluid is typically modeled
with a term which rises as a function of the square of the
velocity. There is a force resisting the passage of a moving
object. If it use the mass of the displaced fluid in the
calculation above, am I already accounting for some of the
"drag" force which is normally accounted for in this velocity
squared term?
Paul
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