Date   
Re: Where can I buy floater balloon?

J68HZ
 

The balloon leaks less at altitude because the pressure on the inside and the outside come to equilibrium by diffusion.  The contents will leak until equilibrium is reached (if that is, in fact what is happening).  At that point then, the buoyancy balance is reduces like this:  [Pgas*Mwgas/Tgas]/[Pair*Mwair/Tair]… but now Pgas=pair… so the buoyancy ratio is [Mwgas*Tair]/[Mwair*Tgas].  And futher, for the most part, the temperature of the lift gas is approximately the same as that of the air… give or take… of course, this can change in the daytime when the sun warms the balloon and causes the lift gas to get warmer… but at night, the buoyancy equation reduces to Mwgas/Mwair… (2/16) meaning you will always have lift.  Again you need to do the real force balance to determine if there is enough lift gas quantity for the balloon to rise or fall… but what it does shows is the limiting case where the balloon will eventually reach a maximum and minimum buoyancy (day/night) and I think we see this in the altitude data that gets reported.

 

The balloon would actually leak MORE at altitude because the ambient pressure at altitude is lower.

 

 

Dr. William J. Schmidt - K9HZ J68HZ 8P6HK ZF2HZ PJ4/K9HZ VP5/K9HZ PJ2/K9HZ

 

Owner - Operator

Big Signal Ranch – K9ZC

Staunton, Illinois

 

Owner – Operator

Villa Grand Piton – J68HZ

Soufriere, St. Lucia W.I.

Rent it: www.VillaGrandPiton.com

Like us on Facebook! facebook icon

 

Moderator – North American QRO Group at Groups.IO.

 

email:  bill@...

 

 

From: QRPLabs@groups.io [mailto:QRPLabs@groups.io] On Behalf Of Jerry Gaffke via Groups.Io
Sent: Sunday, May 26, 2019 11:19 AM
To: QRPLabs@groups.io
Subject: Re: [QRPLabs] Where can I buy floater balloon?

 

Bill,

In post 34692 Mikael said:

> If I fill a foil balloon (like the one linked to first in this thread) with helium with 6gram of lift, seal it and let it sit inside it have lost its 6 gram of lift within a week and fall to the > floor but when used as a balloon for a radio tracker at +10000m its fine for several month, I had a tracker up for 64 days last year before a storm toke it down and Dave > > have a balloon released in February with this (2) balloon stil flying, whats happening at altitude that prevents the gas from leaking out as it does at ground level?

In post 34705 he states that this is repeatable, so I assume he did not happen to have a leak
in the balloons that remained in the house.

Is there something about mylar that makes it more impermeable to helium at -50C
(as found at 10,000 meters) vs room temperature of 20C?


I don't see PV=nRT     https://en.wikipedia.org/wiki/Ideal_gas_law
causing the reported behavior.
Here is my reasoning, a first attempt at thinking this through:

Let's assume the balloon has "just enough" helium to fill it out at 10,000 meters,
so the pressure inside the balloon is always equal to the ambient pressure,
whether it is at sea level or at 10,000 meters.
At 10,000 meters the air is roughly four times less dense, so the balloon
will be 4 times larger than at sea level.  But at sea level, the air that the balloon displaces
will have 4 times the density, so the lift of the helium is the same as at 10,000 meters.
(The lift is equal to the weight of the air displaced by the balloon minus the weight of the helium.)
Likewise, the temperature change affects the density of helium and air in equal measure.
(Assume the helium eventually reaches the same temperature as the ambient air.)

If the mylar balloon has more helium than needed at 10,000 meters (but does not pop),
then the pressure of the helium at altitude will be greater than the "just enough" case
and thus it is more dense.  The lift will be less at altitude than it is at sea level. 
The balloon will rise to that altitude at which the lift is equal to the weight of the payload.

When Mikael's balloon left behind in the house sinks to the floor after a week, it has
less helium remaining in the balloon than the one that is floating at 10,000 meters.
Even though the balloon at altitude is likely overfilled, with the helium pressing on the mylar walls.

But somehow the balloon leaks less at altitude?  Curious.

Jerry, KE7ER
 

On Sun, May 26, 2019 at 12:21 AM, J68HZ wrote:

Some science here.

 

Balloons lift (are buoyant) on the basis of Archimedes principle (cite: https://en.wikipedia.org/wiki/Archimedes%27_principle) which has the corollary that less dense material will rise above more dense material.  It’s the same principle that causes oil to float on water… and with gases, it causes the lightest one (less dense gas) to rise above a heavier gas.  Even hot air will rise above ambient air just because the density of the air in the balloon is less than that of ambient air.

 

In the buoyancy calculation, the force of lift is proportional to the relative differences between the densities of the gases… and in this case, that would be H2 (or whatever the lift gas is) and air.  Now, the density of a gas can be calculated from the ideal gas law… P*V=n*R*T, or rewriting, P=p*R*T/Mw where “p” is the gas density and “Mw” is the molecular weight of the gas.  The ratio for lift is then [P1*Mw1/T1]/[P2*Mw2/T2].  Plug in the numbers and you can determine where the lift will cease.  Of course you need to do a force balance for the real lift and include the weight of the payload and the balloon.

 

The point here is that the buoyancy can vary based on the ambient temperature and pressure… even of the temperature and pressure of the gas in the balloon stays constant.


Virus-free. www.avg.com

Re: #pa #u3s Extreme power curve across the bands. #pa #u3s

Nik
 

Hi Allison,

I'm using the 40m lpf and the 20m lpf (positions 5 and 3 respectively).  Although, I've double checked the values and the windings on the 40m filter, I've not checked the 20m one!  Think I'd better take them all out and have a check, then test with just the 2 in place. (I've 80m in slot 0, 40m in 5, 30 in 4, 20 in  3, 17 in 2 and 15 in 1).

In fact I think I'll put a signal through them and have a look on the scope.

Thank you.  It's nice to have someone else's thoughts.

Nik.

Re: Where can I buy floater balloon?

Stephen Farthing G0XAR JO92ON97
 

Gerry, 

I’m wondering what you mean by four times larger? Do you mean four times the volume? Or four times the surface area? 

It’s not a trick question. I’m just trying to follow your argument. It’s 50 years since I did any academic physics so I’m a bit rusty. 

Warm regards from a rainy England, 

Steve G0XAR 

Re: Where can I buy floater balloon?

Joe Street
 

K9HZ wrote:
"The balloon leaks less at altitude because the pressure on the inside and the outside come to equilibrium by diffusion."

One has to consider the partial pressure of gasses.  There is still a huge difference in partial pressure of He between inside and outside regardless of altitude so the balloon should continue to leak at altitude all other things being equal.  My guess is that it is what was conjectured previously (sorry I can't remember who wrote it) something happens to mylar at cold temperatures which densifies it at a molecular level thereby lowering the diffusion through the material.


Joe


On Sun, May 26, 2019 at 12:47 PM J68HZ <bill@...> wrote:

The balloon leaks less at altitude because the pressure on the inside and the outside come to equilibrium by diffusion.  The contents will leak until equilibrium is reached (if that is, in fact what is happening).  At that point then, the buoyancy balance is reduces like this:  [Pgas*Mwgas/Tgas]/[Pair*Mwair/Tair]… but now Pgas=pair… so the buoyancy ratio is [Mwgas*Tair]/[Mwair*Tgas].  And futher, for the most part, the temperature of the lift gas is approximately the same as that of the air… give or take… of course, this can change in the daytime when the sun warms the balloon and causes the lift gas to get warmer… but at night, the buoyancy equation reduces to Mwgas/Mwair… (2/16) meaning you will always have lift.  Again you need to do the real force balance to determine if there is enough lift gas quantity for the balloon to rise or fall… but what it does shows is the limiting case where the balloon will eventually reach a maximum and minimum buoyancy (day/night) and I think we see this in the altitude data that gets reported.

 

The balloon would actually leak MORE at altitude because the ambient pressure at altitude is lower.

 

 

Dr. William J. Schmidt - K9HZ J68HZ 8P6HK ZF2HZ PJ4/K9HZ VP5/K9HZ PJ2/K9HZ

 

Owner - Operator

Big Signal Ranch – K9ZC

Staunton, Illinois

 

Owner – Operator

Villa Grand Piton – J68HZ

Soufriere, St. Lucia W.I.

Rent it: www.VillaGrandPiton.com

Like us on Facebook! facebook icon

 

Moderator – North American QRO Group at Groups.IO.

 

email:  bill@...

 

 

From: QRPLabs@groups.io [mailto:QRPLabs@groups.io] On Behalf Of Jerry Gaffke via Groups.Io
Sent: Sunday, May 26, 2019 11:19 AM
To: QRPLabs@groups.io
Subject: Re: [QRPLabs] Where can I buy floater balloon?

 

Bill,

In post 34692 Mikael said:

> If I fill a foil balloon (like the one linked to first in this thread) with helium with 6gram of lift, seal it and let it sit inside it have lost its 6 gram of lift within a week and fall to the > floor but when used as a balloon for a radio tracker at +10000m its fine for several month, I had a tracker up for 64 days last year before a storm toke it down and Dave > > have a balloon released in February with this (2) balloon stil flying, whats happening at altitude that prevents the gas from leaking out as it does at ground level?

In post 34705 he states that this is repeatable, so I assume he did not happen to have a leak
in the balloons that remained in the house.

Is there something about mylar that makes it more impermeable to helium at -50C
(as found at 10,000 meters) vs room temperature of 20C?


I don't see PV=nRT     https://en.wikipedia.org/wiki/Ideal_gas_law
causing the reported behavior.
Here is my reasoning, a first attempt at thinking this through:

Let's assume the balloon has "just enough" helium to fill it out at 10,000 meters,
so the pressure inside the balloon is always equal to the ambient pressure,
whether it is at sea level or at 10,000 meters.
At 10,000 meters the air is roughly four times less dense, so the balloon
will be 4 times larger than at sea level.  But at sea level, the air that the balloon displaces
will have 4 times the density, so the lift of the helium is the same as at 10,000 meters.
(The lift is equal to the weight of the air displaced by the balloon minus the weight of the helium.)
Likewise, the temperature change affects the density of helium and air in equal measure.
(Assume the helium eventually reaches the same temperature as the ambient air.)

If the mylar balloon has more helium than needed at 10,000 meters (but does not pop),
then the pressure of the helium at altitude will be greater than the "just enough" case
and thus it is more dense.  The lift will be less at altitude than it is at sea level. 
The balloon will rise to that altitude at which the lift is equal to the weight of the payload.

When Mikael's balloon left behind in the house sinks to the floor after a week, it has
less helium remaining in the balloon than the one that is floating at 10,000 meters.
Even though the balloon at altitude is likely overfilled, with the helium pressing on the mylar walls.

But somehow the balloon leaks less at altitude?  Curious.

Jerry, KE7ER
 

On Sun, May 26, 2019 at 12:21 AM, J68HZ wrote:

Some science here.

 

Balloons lift (are buoyant) on the basis of Archimedes principle (cite: https://en.wikipedia.org/wiki/Archimedes%27_principle) which has the corollary that less dense material will rise above more dense material.  It’s the same principle that causes oil to float on water… and with gases, it causes the lightest one (less dense gas) to rise above a heavier gas.  Even hot air will rise above ambient air just because the density of the air in the balloon is less than that of ambient air.

 

In the buoyancy calculation, the force of lift is proportional to the relative differences between the densities of the gases… and in this case, that would be H2 (or whatever the lift gas is) and air.  Now, the density of a gas can be calculated from the ideal gas law… P*V=n*R*T, or rewriting, P=p*R*T/Mw where “p” is the gas density and “Mw” is the molecular weight of the gas.  The ratio for lift is then [P1*Mw1/T1]/[P2*Mw2/T2].  Plug in the numbers and you can determine where the lift will cease.  Of course you need to do a force balance for the real lift and include the weight of the payload and the balloon.

 

The point here is that the buoyancy can vary based on the ambient temperature and pressure… even of the temperature and pressure of the gas in the balloon stays constant.


Virus-free. www.avg.com

Re: Where can I buy floater balloon?

Joe WB9SBD
 

In actuality, this is Backwards of what is happening with these small floaters.

They are in fact miniature small "Superpressure" Balloons.

When they float it is because there is it lifting gas that has lost it's lifting power because it has become more dense that is was at low altitudes because it is now under pressure. There is a much larger pressure differential inside vs outside of the balloon envelope when at float than there is when at lower altitudes.

so the thought that at float, if it is going to leak it has less chance is 100% totally backwards.

Joe WB9SBD

The Original Rolling Ball Clock
Idle Tyme
Idle-Tyme.com
http://www.idle-tyme.com

On 5/26/2019 2:30 PM, Joe Street wrote:
K9HZ wrote:
"The balloon leaks less at altitude because the pressure on the inside and the outside come to equilibrium by diffusion."

One has to consider the partial pressure of gasses.  There is still a huge difference in partial pressure of He between inside and outside regardless of altitude so the balloon should continue to leak at altitude all other things being equal.  My guess is that it is what was conjectured previously (sorry I can't remember who wrote it) something happens to mylar at cold temperatures which densifies it at a molecular level thereby lowering the diffusion through the material.


Joe

On Sun, May 26, 2019 at 12:47 PM J68HZ <bill@...> wrote:

The balloon leaks less at altitude because the pressure on the inside and the outside come to equilibrium by diffusion.  The contents will leak until equilibrium is reached (if that is, in fact what is happening).  At that point then, the buoyancy balance is reduces like this:  [Pgas*Mwgas/Tgas]/[Pair*Mwair/Tair]… but now Pgas=pair… so the buoyancy ratio is [Mwgas*Tair]/[Mwair*Tgas].  And futher, for the most part, the temperature of the lift gas is approximately the same as that of the air… give or take… of course, this can change in the daytime when the sun warms the balloon and causes the lift gas to get warmer… but at night, the buoyancy equation reduces to Mwgas/Mwair… (2/16) meaning you will always have lift.  Again you need to do the real force balance to determine if there is enough lift gas quantity for the balloon to rise or fall… but what it does shows is the limiting case where the balloon will eventually reach a maximum and minimum buoyancy (day/night) and I think we see this in the altitude data that gets reported.

 

The balloon would actually leak MORE at altitude because the ambient pressure at altitude is lower.

 

 

Dr. William J. Schmidt - K9HZ J68HZ 8P6HK ZF2HZ PJ4/K9HZ VP5/K9HZ PJ2/K9HZ

 

Owner - Operator

Big Signal Ranch – K9ZC

Staunton, Illinois

 

Owner – Operator

Villa Grand Piton – J68HZ

Soufriere, St. Lucia W.I.

Rent it: www.VillaGrandPiton.com

Like us on Facebook! facebook icon

 

Moderator – North American QRO Group at Groups.IO.

 

email:  bill@...

 

 

From: QRPLabs@groups.io [mailto:QRPLabs@groups.io] On Behalf Of Jerry Gaffke via Groups.Io
Sent: Sunday, May 26, 2019 11:19 AM
To: QRPLabs@groups.io
Subject: Re: [QRPLabs] Where can I buy floater balloon?

 

Bill,

In post 34692 Mikael said:

> If I fill a foil balloon (like the one linked to first in this thread) with helium with 6gram of lift, seal it and let it sit inside it have lost its 6 gram of lift within a week and fall to the > floor but when used as a balloon for a radio tracker at +10000m its fine for several month, I had a tracker up for 64 days last year before a storm toke it down and Dave > > have a balloon released in February with this (2) balloon stil flying, whats happening at altitude that prevents the gas from leaking out as it does at ground level?

In post 34705 he states that this is repeatable, so I assume he did not happen to have a leak
in the balloons that remained in the house.

Is there something about mylar that makes it more impermeable to helium at -50C
(as found at 10,000 meters) vs room temperature of 20C?


I don't see PV=nRT     https://en.wikipedia.org/wiki/Ideal_gas_law
causing the reported behavior.
Here is my reasoning, a first attempt at thinking this through:

Let's assume the balloon has "just enough" helium to fill it out at 10,000 meters,
so the pressure inside the balloon is always equal to the ambient pressure,
whether it is at sea level or at 10,000 meters.
At 10,000 meters the air is roughly four times less dense, so the balloon
will be 4 times larger than at sea level.  But at sea level, the air that the balloon displaces
will have 4 times the density, so the lift of the helium is the same as at 10,000 meters.
(The lift is equal to the weight of the air displaced by the balloon minus the weight of the helium.)
Likewise, the temperature change affects the density of helium and air in equal measure.
(Assume the helium eventually reaches the same temperature as the ambient air.)

If the mylar balloon has more helium than needed at 10,000 meters (but does not pop),
then the pressure of the helium at altitude will be greater than the "just enough" case
and thus it is more dense.  The lift will be less at altitude than it is at sea level. 
The balloon will rise to that altitude at which the lift is equal to the weight of the payload.

When Mikael's balloon left behind in the house sinks to the floor after a week, it has
less helium remaining in the balloon than the one that is floating at 10,000 meters.
Even though the balloon at altitude is likely overfilled, with the helium pressing on the mylar walls.

But somehow the balloon leaks less at altitude?  Curious.

Jerry, KE7ER
 

On Sun, May 26, 2019 at 12:21 AM, J68HZ wrote:

Some science here.

 

Balloons lift (are buoyant) on the basis of Archimedes principle (cite: https://en.wikipedia.org/wiki/Archimedes%27_principle) which has the corollary that less dense material will rise above more dense material.  It’s the same principle that causes oil to float on water… and with gases, it causes the lightest one (less dense gas) to rise above a heavier gas.  Even hot air will rise above ambient air just because the density of the air in the balloon is less than that of ambient air.

 

In the buoyancy calculation, the force of lift is proportional to the relative differences between the densities of the gases… and in this case, that would be H2 (or whatever the lift gas is) and air.  Now, the density of a gas can be calculated from the ideal gas law… P*V=n*R*T, or rewriting, P=p*R*T/Mw where “p” is the gas density and “Mw” is the molecular weight of the gas.  The ratio for lift is then [P1*Mw1/T1]/[P2*Mw2/T2].  Plug in the numbers and you can determine where the lift will cease.  Of course you need to do a force balance for the real lift and include the weight of the payload and the balloon.

 

The point here is that the buoyancy can vary based on the ambient temperature and pressure… even of the temperature and pressure of the gas in the balloon stays constant.


Virus-free. www.avg.com

Re: LightAPRS Tracker

 

Obviously I meant 144.390  !

Re: Where can I buy floater balloon?

J68HZ
 

Here’s the math.   Again, excuse the bandwidth and delete this straight away if you have no interest.  This will be my last post on the subject because it’s getting old.

 

There is no reason to really talk in terms of partial pressures of gas on the balloon side.  It’s a pure gas for all intent.  Oh the other side, air is a mix and I suppose you could add the force from the partial pressures of N2 + O2, but they are so similar in molecular weight, size, and behavior that they can be treated as the same (or just use the reference below which gives the density of air at various temperatures, pressures, altitudes.)

 

In general a volume V of material of density ρ immersed in a fluid of density ρf experiences a buoyant force of

 

Fb= ρfgV

and a weight of

W=−gVρ.

so the available lifting force is

Fl=gV(ρfρ).

 

Where the object is floating at the surface of a liquid the buoyant force is modified to reflect the volume of liquid displaced Fb=gVdρf  where Vd is enough to cover the weight of the floating object.

 

Similarly, buoyant force on a submerged object (e.g. a balloon submerged/ floating in air) is equal to the weight of the displaced fluid,

Fb=ρfgV

The physical origin of this force is actually the pressure difference between the top and bottom surfaces of the floating object. Pressure in a fluid at a certain height is related to the depth of the fluid above that height by

P(height2)−P(height1)=ρfg(height2−height1),

 

that is, density of fluid times gravitational acceleration times height difference. If you have an object whose top and bottom surfaces are parallel, then it's pretty easy to calculate the buoyant force as the pressure differential times the area of those surfaces,

Fb=(ΔP)(A)=ρfg*Δheight*A=ρfgV

 

For an irregular shape like a baloon, you'll have to do some integration (a sphere for a balloon is a good approximation). You can also take into account variations in density (or gravitational acceleration) over the size of the balloon by doing some additional calculus. But, according to the US standard atmosphere model, the density of the atmosphere takes about 12 miles to drop off to near zero, which corresponds to a fraction of a percent change over the height of a typical hot air balloon (a few tens of meters). That fraction of a percent is negligible, so you're pretty safe just using a single value for the density.  However, you can't neglect differences in density between vastly different altitudes. Remember that the buoyant force on the balloon is equal to the weight of the amount of fluid displaced. As you go higher, the density of the air drops, which means the balloon displaces a lower mass of air. Therefore, as the balloon rises, the buoyant force drops. Eventually it reaches a height at which the buoyant force exactly balances out the weight of the balloon (and payload), and the balloon levitates at that level.  For a controlled balloon, you can adjust the level by either heating the gas inside the balloon (thus making it expand and displace more air) or by letting some gas out (thus making the balloon contract and displace less air).

 

 

 

 

Dr. William J. Schmidt - K9HZ J68HZ 8P6HK ZF2HZ PJ4/K9HZ VP5/K9HZ PJ2/K9HZ

 

Owner - Operator

Big Signal Ranch – K9ZC

Staunton, Illinois

 

Owner – Operator

Villa Grand Piton – J68HZ

Soufriere, St. Lucia W.I.

Rent it: www.VillaGrandPiton.com

Like us on Facebook! facebook icon

 

Moderator – North American QRO Group at Groups.IO.

 

email:  bill@...

 

 

From: QRPLabs@groups.io [mailto:QRPLabs@groups.io] On Behalf Of Joe WB9SBD
Sent: Sunday, May 26, 2019 2:37 PM
To: QRPLabs@groups.io
Subject: Re: [QRPLabs] Where can I buy floater balloon?

 

In actuality, this is Backwards of what is happening with these small floaters.

They are in fact miniature small "Superpressure" Balloons.

When they float it is because there is it lifting gas that has lost it's lifting power because it has become more dense that is was at low altitudes because it is now under pressure. There is a much larger pressure differential inside vs outside of the balloon envelope when at float than there is when at lower altitudes.

so the thought that at float, if it is going to leak it has less chance is 100% totally backwards.

Joe WB9SBD


The Original Rolling Ball Clock
Idle Tyme
Idle-Tyme.com
http://www.idle-tyme.com

On 5/26/2019 2:30 PM, Joe Street wrote:

K9HZ wrote:

"The balloon leaks less at altitude because the pressure on the inside and the outside come to equilibrium by diffusion."

 

One has to consider the partial pressure of gasses.  There is still a huge difference in partial pressure of He between inside and outside regardless of altitude so the balloon should continue to leak at altitude all other things being equal.  My guess is that it is what was conjectured previously (sorry I can't remember who wrote it) something happens to mylar at cold temperatures which densifies it at a molecular level thereby lowering the diffusion through the material.

 

 

Joe

 

On Sun, May 26, 2019 at 12:47 PM J68HZ <bill@...> wrote:

The balloon leaks less at altitude because the pressure on the inside and the outside come to equilibrium by diffusion.  The contents will leak until equilibrium is reached (if that is, in fact what is happening).  At that point then, the buoyancy balance is reduces like this:  [Pgas*Mwgas/Tgas]/[Pair*Mwair/Tair]… but now Pgas=pair… so the buoyancy ratio is [Mwgas*Tair]/[Mwair*Tgas].  And futher, for the most part, the temperature of the lift gas is approximately the same as that of the air… give or take… of course, this can change in the daytime when the sun warms the balloon and causes the lift gas to get warmer… but at night, the buoyancy equation reduces to Mwgas/Mwair… (2/16) meaning you will always have lift.  Again you need to do the real force balance to determine if there is enough lift gas quantity for the balloon to rise or fall… but what it does shows is the limiting case where the balloon will eventually reach a maximum and minimum buoyancy (day/night) and I think we see this in the altitude data that gets reported.

 

The balloon would actually leak MORE at altitude because the ambient pressure at altitude is lower.

 

 

Dr. William J. Schmidt - K9HZ J68HZ 8P6HK ZF2HZ PJ4/K9HZ VP5/K9HZ PJ2/K9HZ

 

Owner - Operator

Big Signal Ranch – K9ZC

Staunton, Illinois

 

Owner – Operator

Villa Grand Piton – J68HZ

Soufriere, St. Lucia W.I.

Rent it: www.VillaGrandPiton.com

Like us on Facebook! facebook icon

 

Moderator – North American QRO Group at Groups.IO.

 

email:  bill@...

 

 

From: QRPLabs@groups.io [mailto:QRPLabs@groups.io] On Behalf Of Jerry Gaffke via Groups.Io
Sent: Sunday, May 26, 2019 11:19 AM
To: QRPLabs@groups.io
Subject: Re: [QRPLabs] Where can I buy floater balloon?

 

Bill,

In post 34692 Mikael said:

> If I fill a foil balloon (like the one linked to first in this thread) with helium with 6gram of lift, seal it and let it sit inside it have lost its 6 gram of lift within a week and fall to the > floor but when used as a balloon for a radio tracker at +10000m its fine for several month, I had a tracker up for 64 days last year before a storm toke it down and Dave > > have a balloon released in February with this (2) balloon stil flying, whats happening at altitude that prevents the gas from leaking out as it does at ground level?

In post 34705 he states that this is repeatable, so I assume he did not happen to have a leak
in the balloons that remained in the house.

Is there something about mylar that makes it more impermeable to helium at -50C
(as found at 10,000 meters) vs room temperature of 20C?


I don't see PV=nRT     https://en.wikipedia.org/wiki/Ideal_gas_law
causing the reported behavior.
Here is my reasoning, a first attempt at thinking this through:

Let's assume the balloon has "just enough" helium to fill it out at 10,000 meters,
so the pressure inside the balloon is always equal to the ambient pressure,
whether it is at sea level or at 10,000 meters.
At 10,000 meters the air is roughly four times less dense, so the balloon
will be 4 times larger than at sea level.  But at sea level, the air that the balloon displaces
will have 4 times the density, so the lift of the helium is the same as at 10,000 meters.
(The lift is equal to the weight of the air displaced by the balloon minus the weight of the helium.)
Likewise, the temperature change affects the density of helium and air in equal measure.
(Assume the helium eventually reaches the same temperature as the ambient air.)

If the mylar balloon has more helium than needed at 10,000 meters (but does not pop),
then the pressure of the helium at altitude will be greater than the "just enough" case
and thus it is more dense.  The lift will be less at altitude than it is at sea level. 
The balloon will rise to that altitude at which the lift is equal to the weight of the payload.

When Mikael's balloon left behind in the house sinks to the floor after a week, it has
less helium remaining in the balloon than the one that is floating at 10,000 meters.
Even though the balloon at altitude is likely overfilled, with the helium pressing on the mylar walls.

But somehow the balloon leaks less at altitude?  Curious.

Jerry, KE7ER
 

On Sun, May 26, 2019 at 12:21 AM, J68HZ wrote:

Some science here.

 

Balloons lift (are buoyant) on the basis of Archimedes principle (cite: https://en.wikipedia.org/wiki/Archimedes%27_principle) which has the corollary that less dense material will rise above more dense material.  It’s the same principle that causes oil to float on water… and with gases, it causes the lightest one (less dense gas) to rise above a heavier gas.  Even hot air will rise above ambient air just because the density of the air in the balloon is less than that of ambient air.

 

In the buoyancy calculation, the force of lift is proportional to the relative differences between the densities of the gases… and in this case, that would be H2 (or whatever the lift gas is) and air.  Now, the density of a gas can be calculated from the ideal gas law… P*V=n*R*T, or rewriting, P=p*R*T/Mw where “p” is the gas density and “Mw” is the molecular weight of the gas.  The ratio for lift is then [P1*Mw1/T1]/[P2*Mw2/T2].  Plug in the numbers and you can determine where the lift will cease.  Of course you need to do a force balance for the real lift and include the weight of the payload and the balloon.

 

The point here is that the buoyancy can vary based on the ambient temperature and pressure… even of the temperature and pressure of the gas in the balloon stays constant.

 

Virus-free. www.avg.com

 

QCX party on Monday 27th May #qcx #qrp-dx

Peter GM0EUL
 

Good evening everyone

Remember its the last Monday tomorrow so its the QCX party.  Fun for all and conditions seem reasonable today so fingers crossed it will stay good.  Hope to catch some of you on the air and if you haven't tried it yet please do, its great fun and a good opportunity to get some qcx/qcx contacts.  

Don't worry if you are a beginner at cw or don't consider yourself a contester.  Its not a contest and everyone is very friendly and accommodating.

Rules and info here https://www.qrp-labs.com/party.html

73
Peter GM0EUL

Re: Where can I buy floater balloon?

Joe WB9SBD
 


On 5/26/2019 4:26 PM, J68HZ wrote:

For an irregular shape like a baloon, you'll have to do some integration (a sphere for a balloon is a good approximation). You can also take into account variations in density (or gravitational acceleration) over the size of the balloon by doing some additional calculus. But, according to the US standard atmosphere model, the density of the atmosphere takes about 12 miles to drop off to near zero, which corresponds to a fraction of a percent change over the height of a typical hot air balloon (a few tens of meters). That fraction of a percent is negligible, so you're pretty safe just using a single value for the density.  However, you can't neglect differences in density between vastly different altitudes. Remember that the buoyant force on the balloon is equal to the weight of the amount of fluid displaced. As you go higher, the density of the air drops, which means the balloon displaces a lower mass of air. Therefore, as the balloon rises, the buoyant force drops. Eventually it reaches a height at which the buoyant force exactly balances out the weight of the balloon (and payload), and the balloon levitates at that level.  For a controlled balloon, you can adjust the level by either heating the gas inside the balloon (thus making it expand and displace more air) or by letting some gas out (thus making the balloon contract and displace less air).


You are forgetting, that the gas in the balloon is also getting less dense per unit of volume as it rises due to expansion.

The Balloon expands to equalize the pressure internally vs externally.


Until the balloon stops expanding, the buoyancy amount does not change from 10 feet above sea level to 150,000 feet above sea level NO DIFFERENCE!


Joe WB9SBD

 

 

 

 

Dr. William J. Schmidt - K9HZ J68HZ 8P6HK ZF2HZ PJ4/K9HZ VP5/K9HZ PJ2/K9HZ

 

Owner - Operator

Big Signal Ranch – K9ZC

Staunton, Illinois

 

Owner – Operator

Villa Grand Piton – J68HZ

Soufriere, St. Lucia W.I.

Rent it: www.VillaGrandPiton.com

Like us on Facebook! facebook icon

 

Moderator – North American QRO Group at Groups.IO.

 

email:  bill@...

 

 

From: QRPLabs@groups.io [mailto:QRPLabs@groups.io] On Behalf Of Joe WB9SBD
Sent: Sunday, May 26, 2019 2:37 PM
To: QRPLabs@groups.io
Subject: Re: [QRPLabs] Where can I buy floater balloon?

 

In actuality, this is Backwards of what is happening with these small floaters.

They are in fact miniature small "Superpressure" Balloons.

When they float it is because there is it lifting gas that has lost it's lifting power because it has become more dense that is was at low altitudes because it is now under pressure. There is a much larger pressure differential inside vs outside of the balloon envelope when at float than there is when at lower altitudes.

so the thought that at float, if it is going to leak it has less chance is 100% totally backwards.

Joe WB9SBD


The Original Rolling Ball Clock
Idle Tyme
Idle-Tyme.com
http://www.idle-tyme.com

On 5/26/2019 2:30 PM, Joe Street wrote:

K9HZ wrote:

"The balloon leaks less at altitude because the pressure on the inside and the outside come to equilibrium by diffusion."

 

One has to consider the partial pressure of gasses.  There is still a huge difference in partial pressure of He between inside and outside regardless of altitude so the balloon should continue to leak at altitude all other things being equal.  My guess is that it is what was conjectured previously (sorry I can't remember who wrote it) something happens to mylar at cold temperatures which densifies it at a molecular level thereby lowering the diffusion through the material.

 

 

Joe

 

On Sun, May 26, 2019 at 12:47 PM J68HZ <bill@...> wrote:

The balloon leaks less at altitude because the pressure on the inside and the outside come to equilibrium by diffusion.  The contents will leak until equilibrium is reached (if that is, in fact what is happening).  At that point then, the buoyancy balance is reduces like this:  [Pgas*Mwgas/Tgas]/[Pair*Mwair/Tair]… but now Pgas=pair… so the buoyancy ratio is [Mwgas*Tair]/[Mwair*Tgas].  And futher, for the most part, the temperature of the lift gas is approximately the same as that of the air… give or take… of course, this can change in the daytime when the sun warms the balloon and causes the lift gas to get warmer… but at night, the buoyancy equation reduces to Mwgas/Mwair… (2/16) meaning you will always have lift.  Again you need to do the real force balance to determine if there is enough lift gas quantity for the balloon to rise or fall… but what it does shows is the limiting case where the balloon will eventually reach a maximum and minimum buoyancy (day/night) and I think we see this in the altitude data that gets reported.

 

The balloon would actually leak MORE at altitude because the ambient pressure at altitude is lower.

 

 

Dr. William J. Schmidt - K9HZ J68HZ 8P6HK ZF2HZ PJ4/K9HZ VP5/K9HZ PJ2/K9HZ

 

Owner - Operator

Big Signal Ranch – K9ZC

Staunton, Illinois

 

Owner – Operator

Villa Grand Piton – J68HZ

Soufriere, St. Lucia W.I.

Rent it: www.VillaGrandPiton.com

Like us on Facebook! facebook icon

 

Moderator – North American QRO Group at Groups.IO.

 

email:  bill@...

 

 

From: QRPLabs@groups.io [mailto:QRPLabs@groups.io] On Behalf Of Jerry Gaffke via Groups.Io
Sent: Sunday, May 26, 2019 11:19 AM
To: QRPLabs@groups.io
Subject: Re: [QRPLabs] Where can I buy floater balloon?

 

Bill,

In post 34692 Mikael said:

> If I fill a foil balloon (like the one linked to first in this thread) with helium with 6gram of lift, seal it and let it sit inside it have lost its 6 gram of lift within a week and fall to the > floor but when used as a balloon for a radio tracker at +10000m its fine for several month, I had a tracker up for 64 days last year before a storm toke it down and Dave > > have a balloon released in February with this (2) balloon stil flying, whats happening at altitude that prevents the gas from leaking out as it does at ground level?

In post 34705 he states that this is repeatable, so I assume he did not happen to have a leak
in the balloons that remained in the house.

Is there something about mylar that makes it more impermeable to helium at -50C
(as found at 10,000 meters) vs room temperature of 20C?


I don't see PV=nRT     https://en.wikipedia.org/wiki/Ideal_gas_law
causing the reported behavior.
Here is my reasoning, a first attempt at thinking this through:

Let's assume the balloon has "just enough" helium to fill it out at 10,000 meters,
so the pressure inside the balloon is always equal to the ambient pressure,
whether it is at sea level or at 10,000 meters.
At 10,000 meters the air is roughly four times less dense, so the balloon
will be 4 times larger than at sea level.  But at sea level, the air that the balloon displaces
will have 4 times the density, so the lift of the helium is the same as at 10,000 meters.
(The lift is equal to the weight of the air displaced by the balloon minus the weight of the helium.)
Likewise, the temperature change affects the density of helium and air in equal measure.
(Assume the helium eventually reaches the same temperature as the ambient air.)

If the mylar balloon has more helium than needed at 10,000 meters (but does not pop),
then the pressure of the helium at altitude will be greater than the "just enough" case
and thus it is more dense.  The lift will be less at altitude than it is at sea level. 
The balloon will rise to that altitude at which the lift is equal to the weight of the payload.

When Mikael's balloon left behind in the house sinks to the floor after a week, it has
less helium remaining in the balloon than the one that is floating at 10,000 meters.
Even though the balloon at altitude is likely overfilled, with the helium pressing on the mylar walls.

But somehow the balloon leaks less at altitude?  Curious.

Jerry, KE7ER
 

On Sun, May 26, 2019 at 12:21 AM, J68HZ wrote:

Some science here.

 

Balloons lift (are buoyant) on the basis of Archimedes principle (cite: https://en.wikipedia.org/wiki/Archimedes%27_principle) which has the corollary that less dense material will rise above more dense material.  It’s the same principle that causes oil to float on water… and with gases, it causes the lightest one (less dense gas) to rise above a heavier gas.  Even hot air will rise above ambient air just because the density of the air in the balloon is less than that of ambient air.

 

In the buoyancy calculation, the force of lift is proportional to the relative differences between the densities of the gases… and in this case, that would be H2 (or whatever the lift gas is) and air.  Now, the density of a gas can be calculated from the ideal gas law… P*V=n*R*T, or rewriting, P=p*R*T/Mw where “p” is the gas density and “Mw” is the molecular weight of the gas.  The ratio for lift is then [P1*Mw1/T1]/[P2*Mw2/T2].  Plug in the numbers and you can determine where the lift will cease.  Of course you need to do a force balance for the real lift and include the weight of the payload and the balloon.

 

The point here is that the buoyancy can vary based on the ambient temperature and pressure… even of the temperature and pressure of the gas in the balloon stays constant.

 

Virus-free. www.avg.com

 


Re: Where can I buy floater balloon?

J68HZ
 

“The Balloon expands to equalize the pressure internally vs externally.”  This is never true.  The only way it could possibly be true if the balloon itself offered no containment force against the gas at any volume (e.g. was infinitely elastic and forceless as if it weren’t there).   The overall force balance includes the force of the gas on the inside… against the restraint force of the balloon plus that of the atmosphere.

 

Dr. William J. Schmidt - K9HZ J68HZ 8P6HK ZF2HZ PJ4/K9HZ VP5/K9HZ PJ2/K9HZ

 

Owner - Operator

Big Signal Ranch – K9ZC

Staunton, Illinois

 

Owner – Operator

Villa Grand Piton – J68HZ

Soufriere, St. Lucia W.I.

Rent it: www.VillaGrandPiton.com

Like us on Facebook! facebook icon

 

Moderator – North American QRO Group at Groups.IO.

 

email:  bill@...

 

 

From: QRPLabs@groups.io [mailto:QRPLabs@groups.io] On Behalf Of Joe WB9SBD
Sent: Sunday, May 26, 2019 4:51 PM
To: QRPLabs@groups.io
Subject: Re: [QRPLabs] Where can I buy floater balloon?

 

 

On 5/26/2019 4:26 PM, J68HZ wrote:

For an irregular shape like a baloon, you'll have to do some integration (a sphere for a balloon is a good approximation). You can also take into account variations in density (or gravitational acceleration) over the size of the balloon by doing some additional calculus. But, according to the US standard atmosphere model, the density of the atmosphere takes about 12 miles to drop off to near zero, which corresponds to a fraction of a percent change over the height of a typical hot air balloon (a few tens of meters). That fraction of a percent is negligible, so you're pretty safe just using a single value for the density.  However, you can't neglect differences in density between vastly different altitudes. Remember that the buoyant force on the balloon is equal to the weight of the amount of fluid displaced. As you go higher, the density of the air drops, which means the balloon displaces a lower mass of air. Therefore, as the balloon rises, the buoyant force drops. Eventually it reaches a height at which the buoyant force exactly balances out the weight of the balloon (and payload), and the balloon levitates at that level.  For a controlled balloon, you can adjust the level by either heating the gas inside the balloon (thus making it expand and displace more air) or by letting some gas out (thus making the balloon contract and displace less air).

 

You are forgetting, that the gas in the balloon is also getting less dense per unit of volume as it rises due to expansion.

The Balloon expands to equalize the pressure internally vs externally.

 

Until the balloon stops expanding, the buoyancy amount does not change from 10 feet above sea level to 150,000 feet above sea level NO DIFFERENCE!

 

Joe WB9SBD

 

 

 

 

Dr. William J. Schmidt - K9HZ J68HZ 8P6HK ZF2HZ PJ4/K9HZ VP5/K9HZ PJ2/K9HZ

 

Owner - Operator

Big Signal Ranch – K9ZC

Staunton, Illinois

 

Owner – Operator

Villa Grand Piton – J68HZ

Soufriere, St. Lucia W.I.

Rent it: www.VillaGrandPiton.com

Like us on Facebook! facebook icon

 

Moderator – North American QRO Group at Groups.IO.

 

email:  bill@...

 

 

From: QRPLabs@groups.io [mailto:QRPLabs@groups.io] On Behalf Of Joe WB9SBD
Sent: Sunday, May 26, 2019 2:37 PM
To: QRPLabs@groups.io
Subject: Re: [QRPLabs] Where can I buy floater balloon?

 

In actuality, this is Backwards of what is happening with these small floaters.

They are in fact miniature small "Superpressure" Balloons.

When they float it is because there is it lifting gas that has lost it's lifting power because it has become more dense that is was at low altitudes because it is now under pressure. There is a much larger pressure differential inside vs outside of the balloon envelope when at float than there is when at lower altitudes.

so the thought that at float, if it is going to leak it has less chance is 100% totally backwards.

Joe WB9SBD


The Original Rolling Ball Clock
Idle Tyme
Idle-Tyme.com
http://www.idle-tyme.com

On 5/26/2019 2:30 PM, Joe Street wrote:

K9HZ wrote:

"The balloon leaks less at altitude because the pressure on the inside and the outside come to equilibrium by diffusion."

 

One has to consider the partial pressure of gasses.  There is still a huge difference in partial pressure of He between inside and outside regardless of altitude so the balloon should continue to leak at altitude all other things being equal.  My guess is that it is what was conjectured previously (sorry I can't remember who wrote it) something happens to mylar at cold temperatures which densifies it at a molecular level thereby lowering the diffusion through the material.

 

 

Joe

 

On Sun, May 26, 2019 at 12:47 PM J68HZ <bill@...> wrote:

The balloon leaks less at altitude because the pressure on the inside and the outside come to equilibrium by diffusion.  The contents will leak until equilibrium is reached (if that is, in fact what is happening).  At that point then, the buoyancy balance is reduces like this:  [Pgas*Mwgas/Tgas]/[Pair*Mwair/Tair]… but now Pgas=pair… so the buoyancy ratio is [Mwgas*Tair]/[Mwair*Tgas].  And futher, for the most part, the temperature of the lift gas is approximately the same as that of the air… give or take… of course, this can change in the daytime when the sun warms the balloon and causes the lift gas to get warmer… but at night, the buoyancy equation reduces to Mwgas/Mwair… (2/16) meaning you will always have lift.  Again you need to do the real force balance to determine if there is enough lift gas quantity for the balloon to rise or fall… but what it does shows is the limiting case where the balloon will eventually reach a maximum and minimum buoyancy (day/night) and I think we see this in the altitude data that gets reported.

 

The balloon would actually leak MORE at altitude because the ambient pressure at altitude is lower.

 

 

Dr. William J. Schmidt - K9HZ J68HZ 8P6HK ZF2HZ PJ4/K9HZ VP5/K9HZ PJ2/K9HZ

 

Owner - Operator

Big Signal Ranch – K9ZC

Staunton, Illinois

 

Owner – Operator

Villa Grand Piton – J68HZ

Soufriere, St. Lucia W.I.

Rent it: www.VillaGrandPiton.com

Like us on Facebook! facebook icon

 

Moderator – North American QRO Group at Groups.IO.

 

email:  bill@...

 

 

From: QRPLabs@groups.io [mailto:QRPLabs@groups.io] On Behalf Of Jerry Gaffke via Groups.Io
Sent: Sunday, May 26, 2019 11:19 AM
To: QRPLabs@groups.io
Subject: Re: [QRPLabs] Where can I buy floater balloon?

 

Bill,

In post 34692 Mikael said:

> If I fill a foil balloon (like the one linked to first in this thread) with helium with 6gram of lift, seal it and let it sit inside it have lost its 6 gram of lift within a week and fall to the > floor but when used as a balloon for a radio tracker at +10000m its fine for several month, I had a tracker up for 64 days last year before a storm toke it down and Dave > > have a balloon released in February with this (2) balloon stil flying, whats happening at altitude that prevents the gas from leaking out as it does at ground level?

In post 34705 he states that this is repeatable, so I assume he did not happen to have a leak
in the balloons that remained in the house.

Is there something about mylar that makes it more impermeable to helium at -50C
(as found at 10,000 meters) vs room temperature of 20C?


I don't see PV=nRT     https://en.wikipedia.org/wiki/Ideal_gas_law
causing the reported behavior.
Here is my reasoning, a first attempt at thinking this through:

Let's assume the balloon has "just enough" helium to fill it out at 10,000 meters,
so the pressure inside the balloon is always equal to the ambient pressure,
whether it is at sea level or at 10,000 meters.
At 10,000 meters the air is roughly four times less dense, so the balloon
will be 4 times larger than at sea level.  But at sea level, the air that the balloon displaces
will have 4 times the density, so the lift of the helium is the same as at 10,000 meters.
(The lift is equal to the weight of the air displaced by the balloon minus the weight of the helium.)
Likewise, the temperature change affects the density of helium and air in equal measure.
(Assume the helium eventually reaches the same temperature as the ambient air.)

If the mylar balloon has more helium than needed at 10,000 meters (but does not pop),
then the pressure of the helium at altitude will be greater than the "just enough" case
and thus it is more dense.  The lift will be less at altitude than it is at sea level. 
The balloon will rise to that altitude at which the lift is equal to the weight of the payload.

When Mikael's balloon left behind in the house sinks to the floor after a week, it has
less helium remaining in the balloon than the one that is floating at 10,000 meters.
Even though the balloon at altitude is likely overfilled, with the helium pressing on the mylar walls.

But somehow the balloon leaks less at altitude?  Curious.

Jerry, KE7ER
 

On Sun, May 26, 2019 at 12:21 AM, J68HZ wrote:

Some science here.

 

Balloons lift (are buoyant) on the basis of Archimedes principle (cite: https://en.wikipedia.org/wiki/Archimedes%27_principle) which has the corollary that less dense material will rise above more dense material.  It’s the same principle that causes oil to float on water… and with gases, it causes the lightest one (less dense gas) to rise above a heavier gas.  Even hot air will rise above ambient air just because the density of the air in the balloon is less than that of ambient air.

 

In the buoyancy calculation, the force of lift is proportional to the relative differences between the densities of the gases… and in this case, that would be H2 (or whatever the lift gas is) and air.  Now, the density of a gas can be calculated from the ideal gas law… P*V=n*R*T, or rewriting, P=p*R*T/Mw where “p” is the gas density and “Mw” is the molecular weight of the gas.  The ratio for lift is then [P1*Mw1/T1]/[P2*Mw2/T2].  Plug in the numbers and you can determine where the lift will cease.  Of course you need to do a force balance for the real lift and include the weight of the payload and the balloon.

 

The point here is that the buoyancy can vary based on the ambient temperature and pressure… even of the temperature and pressure of the gas in the balloon stays constant.

 

Virus-free. www.avg.com

 

 

Re: Where can I buy floater balloon?

Joe WB9SBD
 

In most balloons the pressure differential between inside and outside the envelope is soo extremely small that it took many many flights to try to measure it.

After six flights where they did get a measurable difference it still was soo small that noise in the sensor is in the accuracy pressure window.

the differential is less than 0.001 Psi.

And until it reaches the point where it can not stretch any more is there a spike in pressure differential. And then it was still very small, like 0.004 or so psi.
I have been doing these flights for 30+ years. 70+ of them. from small balloons like the foil pico's
to the latex medium size ones.
https://youtu.be/HJ0IT4ZwtSo

How about 118 Thousand feet?
https://youtu.be/EdAuHr-bZ1M

and done some test flights for NASA guys, some really BIG Balloons.


Joe WB9SBD

The Original Rolling Ball Clock
Idle Tyme
Idle-Tyme.com
http://www.idle-tyme.com

On 5/26/2019 5:00 PM, J68HZ wrote:

“The Balloon expands to equalize the pressure internally vs externally.”  This is never true.  The only way it could possibly be true if the balloon itself offered no containment force against the gas at any volume (e.g. was infinitely elastic and forceless as if it weren’t there).   The overall force balance includes the force of the gas on the inside… against the restraint force of the balloon plus that of the atmosphere.

 

Dr. William J. Schmidt - K9HZ J68HZ 8P6HK ZF2HZ PJ4/K9HZ VP5/K9HZ PJ2/K9HZ

 

Owner - Operator

Big Signal Ranch – K9ZC

Staunton, Illinois

 

Owner – Operator

Villa Grand Piton – J68HZ

Soufriere, St. Lucia W.I.

Rent it: www.VillaGrandPiton.com

Like us on Facebook! facebook icon

 

Moderator – North American QRO Group at Groups.IO.

 

email:  bill@...

 

 

From: QRPLabs@groups.io [mailto:QRPLabs@groups.io] On Behalf Of Joe WB9SBD
Sent: Sunday, May 26, 2019 4:51 PM
To: QRPLabs@groups.io
Subject: Re: [QRPLabs] Where can I buy floater balloon?

 

 

On 5/26/2019 4:26 PM, J68HZ wrote:

For an irregular shape like a baloon, you'll have to do some integration (a sphere for a balloon is a good approximation). You can also take into account variations in density (or gravitational acceleration) over the size of the balloon by doing some additional calculus. But, according to the US standard atmosphere model, the density of the atmosphere takes about 12 miles to drop off to near zero, which corresponds to a fraction of a percent change over the height of a typical hot air balloon (a few tens of meters). That fraction of a percent is negligible, so you're pretty safe just using a single value for the density.  However, you can't neglect differences in density between vastly different altitudes. Remember that the buoyant force on the balloon is equal to the weight of the amount of fluid displaced. As you go higher, the density of the air drops, which means the balloon displaces a lower mass of air. Therefore, as the balloon rises, the buoyant force drops. Eventually it reaches a height at which the buoyant force exactly balances out the weight of the balloon (and payload), and the balloon levitates at that level.  For a controlled balloon, you can adjust the level by either heating the gas inside the balloon (thus making it expand and displace more air) or by letting some gas out (thus making the balloon contract and displace less air).

 

You are forgetting, that the gas in the balloon is also getting less dense per unit of volume as it rises due to expansion.

The Balloon expands to equalize the pressure internally vs externally.

 

Until the balloon stops expanding, the buoyancy amount does not change from 10 feet above sea level to 150,000 feet above sea level NO DIFFERENCE!

 

Joe WB9SBD

 

 

 

 

Dr. William J. Schmidt - K9HZ J68HZ 8P6HK ZF2HZ PJ4/K9HZ VP5/K9HZ PJ2/K9HZ

 

Owner - Operator

Big Signal Ranch – K9ZC

Staunton, Illinois

 

Owner – Operator

Villa Grand Piton – J68HZ

Soufriere, St. Lucia W.I.

Rent it: www.VillaGrandPiton.com

Like us on Facebook! facebook icon

 

Moderator – North American QRO Group at Groups.IO.

 

email:  bill@...

 

 

From: QRPLabs@groups.io [mailto:QRPLabs@groups.io] On Behalf Of Joe WB9SBD
Sent: Sunday, May 26, 2019 2:37 PM
To: QRPLabs@groups.io
Subject: Re: [QRPLabs] Where can I buy floater balloon?

 

In actuality, this is Backwards of what is happening with these small floaters.

They are in fact miniature small "Superpressure" Balloons.

When they float it is because there is it lifting gas that has lost it's lifting power because it has become more dense that is was at low altitudes because it is now under pressure. There is a much larger pressure differential inside vs outside of the balloon envelope when at float than there is when at lower altitudes.

so the thought that at float, if it is going to leak it has less chance is 100% totally backwards.

Joe WB9SBD


The Original Rolling Ball Clock
Idle Tyme
Idle-Tyme.com
http://www.idle-tyme.com

On 5/26/2019 2:30 PM, Joe Street wrote:

K9HZ wrote:

"The balloon leaks less at altitude because the pressure on the inside and the outside come to equilibrium by diffusion."

 

One has to consider the partial pressure of gasses.  There is still a huge difference in partial pressure of He between inside and outside regardless of altitude so the balloon should continue to leak at altitude all other things being equal.  My guess is that it is what was conjectured previously (sorry I can't remember who wrote it) something happens to mylar at cold temperatures which densifies it at a molecular level thereby lowering the diffusion through the material.

 

 

Joe

 

On Sun, May 26, 2019 at 12:47 PM J68HZ <bill@...> wrote:

The balloon leaks less at altitude because the pressure on the inside and the outside come to equilibrium by diffusion.  The contents will leak until equilibrium is reached (if that is, in fact what is happening).  At that point then, the buoyancy balance is reduces like this:  [Pgas*Mwgas/Tgas]/[Pair*Mwair/Tair]… but now Pgas=pair… so the buoyancy ratio is [Mwgas*Tair]/[Mwair*Tgas].  And futher, for the most part, the temperature of the lift gas is approximately the same as that of the air… give or take… of course, this can change in the daytime when the sun warms the balloon and causes the lift gas to get warmer… but at night, the buoyancy equation reduces to Mwgas/Mwair… (2/16) meaning you will always have lift.  Again you need to do the real force balance to determine if there is enough lift gas quantity for the balloon to rise or fall… but what it does shows is the limiting case where the balloon will eventually reach a maximum and minimum buoyancy (day/night) and I think we see this in the altitude data that gets reported.

 

The balloon would actually leak MORE at altitude because the ambient pressure at altitude is lower.

 

 

Dr. William J. Schmidt - K9HZ J68HZ 8P6HK ZF2HZ PJ4/K9HZ VP5/K9HZ PJ2/K9HZ

 

Owner - Operator

Big Signal Ranch – K9ZC

Staunton, Illinois

 

Owner – Operator

Villa Grand Piton – J68HZ

Soufriere, St. Lucia W.I.

Rent it: www.VillaGrandPiton.com

Like us on Facebook! facebook icon

 

Moderator – North American QRO Group at Groups.IO.

 

email:  bill@...

 

 

From: QRPLabs@groups.io [mailto:QRPLabs@groups.io] On Behalf Of Jerry Gaffke via Groups.Io
Sent: Sunday, May 26, 2019 11:19 AM
To: QRPLabs@groups.io
Subject: Re: [QRPLabs] Where can I buy floater balloon?

 

Bill,

In post 34692 Mikael said:

> If I fill a foil balloon (like the one linked to first in this thread) with helium with 6gram of lift, seal it and let it sit inside it have lost its 6 gram of lift within a week and fall to the > floor but when used as a balloon for a radio tracker at +10000m its fine for several month, I had a tracker up for 64 days last year before a storm toke it down and Dave > > have a balloon released in February with this (2) balloon stil flying, whats happening at altitude that prevents the gas from leaking out as it does at ground level?

In post 34705 he states that this is repeatable, so I assume he did not happen to have a leak
in the balloons that remained in the house.

Is there something about mylar that makes it more impermeable to helium at -50C
(as found at 10,000 meters) vs room temperature of 20C?


I don't see PV=nRT     https://en.wikipedia.org/wiki/Ideal_gas_law
causing the reported behavior.
Here is my reasoning, a first attempt at thinking this through:

Let's assume the balloon has "just enough" helium to fill it out at 10,000 meters,
so the pressure inside the balloon is always equal to the ambient pressure,
whether it is at sea level or at 10,000 meters.
At 10,000 meters the air is roughly four times less dense, so the balloon
will be 4 times larger than at sea level.  But at sea level, the air that the balloon displaces
will have 4 times the density, so the lift of the helium is the same as at 10,000 meters.
(The lift is equal to the weight of the air displaced by the balloon minus the weight of the helium.)
Likewise, the temperature change affects the density of helium and air in equal measure.
(Assume the helium eventually reaches the same temperature as the ambient air.)

If the mylar balloon has more helium than needed at 10,000 meters (but does not pop),
then the pressure of the helium at altitude will be greater than the "just enough" case
and thus it is more dense.  The lift will be less at altitude than it is at sea level. 
The balloon will rise to that altitude at which the lift is equal to the weight of the payload.

When Mikael's balloon left behind in the house sinks to the floor after a week, it has
less helium remaining in the balloon than the one that is floating at 10,000 meters.
Even though the balloon at altitude is likely overfilled, with the helium pressing on the mylar walls.

But somehow the balloon leaks less at altitude?  Curious.

Jerry, KE7ER
 

On Sun, May 26, 2019 at 12:21 AM, J68HZ wrote:

Some science here.

 

Balloons lift (are buoyant) on the basis of Archimedes principle (cite: https://en.wikipedia.org/wiki/Archimedes%27_principle) which has the corollary that less dense material will rise above more dense material.  It’s the same principle that causes oil to float on water… and with gases, it causes the lightest one (less dense gas) to rise above a heavier gas.  Even hot air will rise above ambient air just because the density of the air in the balloon is less than that of ambient air.

 

In the buoyancy calculation, the force of lift is proportional to the relative differences between the densities of the gases… and in this case, that would be H2 (or whatever the lift gas is) and air.  Now, the density of a gas can be calculated from the ideal gas law… P*V=n*R*T, or rewriting, P=p*R*T/Mw where “p” is the gas density and “Mw” is the molecular weight of the gas.  The ratio for lift is then [P1*Mw1/T1]/[P2*Mw2/T2].  Plug in the numbers and you can determine where the lift will cease.  Of course you need to do a force balance for the real lift and include the weight of the payload and the balloon.

 

The point here is that the buoyancy can vary based on the ambient temperature and pressure… even of the temperature and pressure of the gas in the balloon stays constant.

 

Virus-free. www.avg.com

 

 


Re: #pa #u3s Extreme power curve across the bands. #pa #u3s

SkipF, NT1G
 

I'll bet my 'ticket', your backwards coil is FINE. Leave it for last.
Transformers for phasing are a different issue.
Extra winding (and too few) can be repaired without unwinding.
Don't make extra work for yourself.
73 es gud luck!
SkipF, NT1G

On Sun, May 26, 2019, 7:45 PM <nik@...> wrote:
Just spent the last few hours scrutinising the low pass filters.  Found a couple of mistakes.  One choke with a winding too many, and another wound backwards(!).... 

I rebuilt the 20m unit just to be sure.  The results are the same.

If I select the 17m filter and transmit on 20m, the result is as expected, an increase of around 300mW, and approaching 1W through the 15m filter.

I then made a bypass filter and tried that on a dummy load.  That was around 2W.

I guess the filter is doing it's job well.  Although the output on the fundamental is still very low at 200mW through the 5W amplifier.  I'm now thinking of experimenting with something compatible with the BC170, but with a wider frequency range.  Something like a BC184.

Re: QCX Message Trigger

KE0GYC
 

I had this same confusion when I first started mine up!  It's spelled out in the manual, but I guess it's easy to overlook.

Re: #pa #u3s Extreme power curve across the bands. #pa #u3s

Nik
 

Just spent the last few hours scrutinising the low pass filters.  Found a couple of mistakes.  One choke with a winding too many, and another wound backwards(!).... 

I rebuilt the 20m unit just to be sure.  The results are the same.

If I select the 17m filter and transmit on 20m, the result is as expected, an increase of around 300mW, and approaching 1W through the 15m filter.

I then made a bypass filter and tried that on a dummy load.  That was around 2W.

I guess the filter is doing it's job well.

Re: QCX party on Monday 27th May #qcx #qrp-dx

Steve in Okinawa
 

European friends, please aim your 20 meter antenna toward Japan for part of the 1300-1400z session, and I'll be listening around .060. I'll also be checking 40-meters for NA same session. JS6TMW 

Re: Where can I buy floater balloon?

Jerry Gaffke
 

Stephen,

You are asking what this statement means:

>  At 10,000 meters the air is roughly four times less dense, so the balloon
>  will be 4 times larger than at sea level.

By "four times larger", I mean larger in volume.
The ideal gas law says that if the amount of gas and the temperature
are both held constant, then the volume will be inversely proportional
to any change in pressure.  So if the pressure increases by a factor
of four, the volume decreases by a factor of four.

Likewise, if the temperature changes but the pressure is held constant, 
then the volume changes proportional to the change in temperature 
expressed in degrees Kelvin.  Zero degrees Kelvin is -273 degrees Centigrade,
so an increase in temperature from -50 C (-223 K) to 20 C (293 K) with pressure
held constant will cause the volume of our gas to increase by a factor of 293/223 = 1.31.
If we had 1 liter of gas at -50 C, it expands to 1.31 liters at 20 C.

Generally, all three can change simultaneously, for example compressing a gas causes it to heat up. 
Both relationships can be expressed simultaneously by the formula    PV = nRT,
where P is pressure, V is volume, T is temperature, and you can think of
the nR factor as a constant for a given amount of gas.
Digging a bit deeper, the n is a count of the number of molecules of gas,
and the R is a physical constant called "the ideal gas constant".
    https://en.wikipedia.org/wiki/Ideal_gas_law


The first line of this passage might be a bit confusing:
<  If the mylar balloon has more helium than needed at 10,000 meters (but does not pop),
<  then the pressure of the helium at altitude will be greater than the "just enough" case
<  and thus it is more dense.  The lift will be less at altitude than it is at sea level.  
<  The balloon will rise to that altitude at which the lift is equal to the weight of the payload.

best to edit it as follows:
>  If at 10,000 meters the balloon has more helium than what fits without stretching the mylar 
>  then the pressure of the helium at altitude will be greater than the "just enough" case
>  and thus it is more dense.  The lift will be less at altitude than it is at sea level.  
>  The balloon will rise to that altitude at which the lift is equal to the weight of the payload.

As Joe, WB9SBD, has also stated here, the lifting power of the balloon remains the same as
the balloon rises so long as the balloon can expand without exerting extra pressure on our lifting gas.
Think of a big floppy weather balloon about to be set free, the hydrogen has plenty of room to expand
before the balloon envelope reaches its maximum size and starts pressing back.
Once it gets high enough that the balloon envelope is tight and causes the hydrogen to be at a
higher pressure than the ambient air, the lifting power starts to drop. 

Jerry


On Sun, May 26, 2019 at 10:11 AM, Stephen Farthing G0XAR JO92ON97 wrote:
I’m wondering what you mean by four times larger? Do you mean four times the volume? Or four times the surface area? 
It’s not a trick question. I’m just trying to follow your argument. It’s 50 years since I did any academic physics so I’m a bit rusty. 
 

Intro..

Graham W
 

Hi..I'm Graham VE3WGW residing on the shores of Lake Eire in Canada.
I have just built 2 of the prog rocks.
Did the change of the 39k to 27k in the filter circuit to get closer to .33v.
I installed TCOX's in place of the 27mhz crystal.
After setting the calibration correctly. Mine set to 27,000,025 to get it correct.
They have been sitting on 10 mhz for the past number of hours.
Not moved more than .3 hz......
I have a gps calibrated oscillator... so it is dead on.....
One of them is going to be used in a Kendecom 220 mhz repeater as the RX crystal died... At first tests everything is excellent..
The other is going to be the crystal bank in a GE MSTR II ... for 8 chans on 6 meters.
Heck... The prog rock plus the TCXO is less than the price of 1 crystal !!
Love them..
Graham VE3WGW

Re: Minimally modified QCX-17 has been heard on 80, 60, 40, 30, 20, 17, 15, 12, 10m this UTC day !

Guido PE1NNZ
 

Michael,
With C1, C5, C8 removed and with T1 configured for 40m RX is working well for all bands, but the lower bands (80m, 160m) are a bit deaf. So initially wind T1 for 80m would do probably better.
With a LPF installed, higher band harmonics on RX are not really an issue in practice in my experience; but a LPF is recommended to prevent FM broadcast mixing products.
For LPF switching, I like KD1JVs minimalistic approach by using a few series DPDT slider switches to select the appropriate LPF. Another interesting way to switch LPFs is to use 1N4007 for swiching [1], were you can use the rectified RF + 12V at C30 to bias the diodes either forward or in reverse.
Guido

[1] https://www.qsl.net/in3otd/electronics/PIN_diodes/1N4007.html

Re: Where can I buy floater balloon?

Joe WB9SBD
 

Thank You Jerry!

Buoyancy can be a fun thing to play with!  Check out this design of flight train we have flown twice, and seemed to have worked, sadly the payloads did not survive the ultra cold long night.  But the theroy is there, just to get a payload to withstand that night time temps!

Check out 'Earth Breeze"

https://www.qsl.net/nss/earthbreeze.html

Joe WB9SBD

Near Space Sciences
KB9KHO

The Original Rolling Ball Clock
Idle Tyme
Idle-Tyme.com
http://www.idle-tyme.com

On 5/26/2019 10:56 PM, Jerry Gaffke via Groups.Io wrote:
Stephen,

You are asking what this statement means:

>  At 10,000 meters the air is roughly four times less dense, so the balloon
>  will be 4 times larger than at sea level.

By "four times larger", I mean larger in volume.
The ideal gas law says that if the amount of gas and the temperature
are both held constant, then the volume will be inversely proportional
to any change in pressure.  So if the pressure increases by a factor
of four, the volume decreases by a factor of four.

Likewise, if the temperature changes but the pressure is held constant, 
then the volume changes proportional to the change in temperature 
expressed in degrees Kelvin.  Zero degrees Kelvin is -273 degrees Centigrade,
so an increase in temperature from -50 C (-223 K) to 20 C (293 K) with pressure
held constant will cause the volume of our gas to increase by a factor of 293/223 = 1.31.
If we had 1 liter of gas at -50 C, it expands to 1.31 liters at 20 C.

Generally, all three can change simultaneously, for example compressing a gas causes it to heat up. 
Both relationships can be expressed simultaneously by the formula    PV = nRT,
where P is pressure, V is volume, T is temperature, and you can think of
the nR factor as a constant for a given amount of gas.
Digging a bit deeper, the n is a count of the number of molecules of gas,
and the R is a physical constant called "the ideal gas constant".
    https://en.wikipedia.org/wiki/Ideal_gas_law


The first line of this passage might be a bit confusing:
<  If the mylar balloon has more helium than needed at 10,000 meters (but does not pop),
<  then the pressure of the helium at altitude will be greater than the "just enough" case
<  and thus it is more dense.  The lift will be less at altitude than it is at sea level.  
<  The balloon will rise to that altitude at which the lift is equal to the weight of the payload.

best to edit it as follows:
>  If at 10,000 meters the balloon has more helium than what fits without stretching the mylar 
>  then the pressure of the helium at altitude will be greater than the "just enough" case
>  and thus it is more dense.  The lift will be less at altitude than it is at sea level.  
>  The balloon will rise to that altitude at which the lift is equal to the weight of the payload.

As Joe, WB9SBD, has also stated here, the lifting power of the balloon remains the same as
the balloon rises so long as the balloon can expand without exerting extra pressure on our lifting gas.
Think of a big floppy weather balloon about to be set free, the hydrogen has plenty of room to expand
before the balloon envelope reaches its maximum size and starts pressing back.
Once it gets high enough that the balloon envelope is tight and causes the hydrogen to be at a
higher pressure than the ambient air, the lifting power starts to drop. 

Jerry


On Sun, May 26, 2019 at 10:11 AM, Stephen Farthing G0XAR JO92ON97 wrote:
I’m wondering what you mean by four times larger? Do you mean four times the volume? Or four times the surface area? 
It’s not a trick question. I’m just trying to follow your argument. It’s 50 years since I did any academic physics so I’m a bit rusty. 
 

Re: Is the CLK0 pin of si5361a supposedly up to driving an HEF4013 direct?

Chris Wilson
 

Hello Allison,

Thanks for the reply. What we are doing is driving this modified to
4013 flip flop, pair of IXD_614 fast FET driver IC's, 1kW Class D amp
direct from the Si5351a output via a 220nF poly cap. So what we are
seeing is the Si5351a is generating just enough to fire the 4013 but
it's far from ideal or reliable? What about using a logic level flip
flop like a 74HCT74? That should reliably and correctly trigger from
the Si, right? The LF version of the amp runs exactly as the build
notes later show, no issues. The MF version ran the FET driver mad
hot, with poor gate and drain waveforms, so we modded things...

The other thing we are seeing is unlike the ultra reliable, perfect FET
gate and drain waveforms on the similar LF amp, the MF one shows one
fairly decent gate and drain waveform on one pair of FET's, but the
other pair show incredibly spiky drain waveforms and poor gate
waveforms when power to the FET's is applied, into a good Bird water
colled dummy load or the antenna system. We have tried changing the FET
driver IC's over, tried replacing the driver IC's and the FET's even
to expensive ultra low gate capacitance ones, and tried swapping the
gate leads from the drivers to the FET's over but the same "side" of
the output shows bad waveforms. We also swapped out the 1500pF and 10
Ohm R1, R2, C3 and C4 components, and still the same side shows bad
waveforms. The transformer wiring is symmetrical, we even changed the
whole transformer T1 out to a newly wound one on on a different core
from the FT-240-79 one recommended, still one pair of FET's exhibit
the poor gate and drain waveforms.

https://www.buerklin.com/en/Products/Passive-Components/Inductors/Ferrites/Ring-core%2C-outer-%C3%98-x-inner-%C3%98-x-height-%3D-63-x-38-x-25-mm%2C-N30/p/84D258

but still the same side has bad waveforms... We can't see why this
should be... The circuit is as below in regard to after the FET
driver. We were using an HEF4013 flip flop and a pair of IXD_614 FET
driver IC's. The interconnections are pretty short. The FET driver to
FET gate wires are short and symmetrical.

Amp schematic link and info:

http://www.w1vd.com/137-500-KWTX.html

we use the W1VD LPF's for LF and MF.



www.ixysic.com/home/pdfs.nsf/www/IXD_614.pdf

A bit off topic and definitely not QRP but I do know a few here are
using U3S's to operate on LF and MF into Class D amps, so hopefully
not TOO off topic :)




Best regards,
Chris 2E0ILY mailto:chris@...


aK> BS170 and 2n7000 are not identical.

aK> While the 2n7000 can handle a bit more power it has higher capacitances
aK> and one has to watch the gate threshold.

aK> For class D and E one has to hit the gate hard as in 5-6V to insure the device
aK> is turned on hard (acts like a switch) and turned off hard.

aK> As to the HEF4013 you need the correct drive voltage to get the logic to switch.
aK> Generally using 3 or 3.3V logic (SI5351) to drive 5V CMOS is bad practice
aK> (also unreliable) and you need a level converter to go from 3V to 5V levels.

aK> Going down in frequency its generally easier to drive power MOSFETS
aK> so doing class D or E at 630m or 2200m is switch mode power technology.
aK> There are low cost parts that can easily hit KW power levels at 630M.
aK> You still need the right drive levels.

aK> Allison



--
Best regards, Chris Wilson (2E0ILY)