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Re: [Phys-L] Dirigible Flight Question

Ok, John, just for you I'll give it a little push up when I attach the weight so that the oscillation dies down and all of the new tensions around the balloon have time to equilibrate before the vertical velocity goes to zero. What is the acceleration at zero velocity?


On Nov 5, 2012, at 12:18 PM, "John Clement" <> wrote:

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

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 [] On Behalf Of Paul Nord
Sent: Monday, November 05, 2012 11:49 AM
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?

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