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Originally posted by AThousandYoung- ballast - "Aeronautics. something heavy, as bags of sand, placed in the car of a balloon for control of altitude and, less often, of attitude, or placed in an aircraft to control the position of the center of gravity."
Ballast doesn't counter descent.
I don't understand how he can descend from that kind of altitude
without slamming into the ground.
Assuming each balloon he pops creates a downward force of 20N,
that's an acceleration of 0.2 m/s^2. I understand that wind resistance
and pressure will be in his favour but if they were'nt he'd be slamming
into the ground at some 600m/s?
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Originally posted by Thequ1ckmaybe he could find a local thundercloud and use the updrafts to slow his decent...pretty frickin bumpy ride though. I saw McGyver do it once on TV. I wish I had friends like McGyver.....come back McGyver!
- ballast - "Aeronautics. something heavy, as bags of sand, placed in the car of a balloon for control of altitude and, less often, of attitude, or placed in an aircraft to control the position of the center of gravity."
I don't understand how he can descend from that kind of altitude
without slamming into the ground.
Assuming each balloon he pops cr ill be in his favour but if they were'nt he'd be slamming
into the ground at some 600m/s?
Originally posted by AThousandYoung..unless we clone ourselves.
But women seem to keep finding ways to make us breed anyway!
But of course we are all aware of the importance of genetic variations/diversities to survive, so better to stick to the traditional ways.. and so men should look after their women carefully and sincerely😉
Originally posted by Thequ1ckMy option would be to have a helium tank and some spare balloons with me. Pop one, start descending, then blow one up slowly to maintain a slow desent as increasing air pressure compresses the balloons. Assuming neutral buoyancy at the start, he could bring himself to a feather touchdown with enough practice... how he gets the practice is another question.
- ballast - "Aeronautics. something heavy, as bags of sand, placed in the car of a balloon for control of altitude and, less often, of attitude, or placed in an aircraft to control the position of the center of gravity."
I don't understand how he can descend from that kind of altitude
without slamming into the ground.
Assuming each balloon he pops cr ...[text shortened]... ill be in his favour but if they were'nt he'd be slamming
into the ground at some 600m/s?
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Originally posted by Thequ1ckTo have a controlled descent, you want ballast equal to or slightly less than the bouyant force of the air on the balloon I would think. If you got it exactly right you could descend at an unchanging speed with a gentle push to start you off.
- ballast - "Aeronautics. something heavy, as bags of sand, placed in the car of a balloon for control of altitude and, less often, of attitude, or placed in an aircraft to control the position of the center of gravity."
I don't understand how he can descend from that kind of altitude
without slamming into the ground.
Assuming each balloon he pops cr ill be in his favour but if they were'nt he'd be slamming
into the ground at some 600m/s?
EXCEPT that the air gets more dense as you get closer to the ground...and gravity is slightly stronger there too...
Originally posted by AThousandYoung"A genetic tendency to take risks"! You can have fun with that kind of language. Do your genetic tendencies affect the brand of toothpaste you purchase?
Are they? I postulate that men have a genetic tendency to take risks compared to women.
Incidentally, having a child is an enormous risk.
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Originally posted by Thequ1ckThe only solution for this problem would be a parametrized solution.
One for the mathematicians :
Assuming the weight of the man and chair were 100kg,
and he was using a total of 45 balloons, how much balast
would he need to counter a descent from 15,000 feet?
For the information given there is no possible unique solution. To formulate this problem we need to make the following observations and assumptions. Firstly, your balloons would have to be inflated with a lighter than air substance to give a buoyant force to the system. The buoyancy of the whole system in equilibrium (ie when it floats at 15,000 ft) is dependent on the volume of air displaced by the balloons. The total weight of the system is the man and chair(Massmc)xg plus the mass of the gas used in the balloon(Mgas)xg. The buoyancy is the mass of air (Mair)xg displaced. If any payload or ballast(Mb) has to be added then it must be added to the total weight of the system.
==> (Massmc + Mgas + Mb)xg = (Mair)xg (1)
We can simplify by dividing through by g and rearranging we have
Mb = Mair - Mgas -Massmc. (2)
By this equation we can see that the ballast we can carry is a function of the mass of air we can displace(buoyancy) minus the mass of the lighter than air gas which displaces this volume of air, minus the weight of the man plus chair.
Given that our only constant is the mass of the man and chair (Massmc=100kg), it would appear that our ballast is a function of the mass of air we displace minus the mass of lighter than air gas we use as buoyant force. If we make an assumption at this point that the lighter than air gas used is helium, we can simplify this puzzle a bit with the following observation, and that is that the mass of a gas is equal to its density times its volume.
====> Mb = rho_air*V - rho_he*V - 100kg (2)
we simplify even further to see that
=====> Mb = V ( rho_air - rho_he) -100kg (3)
For a sphere V = 4/3*(pi)*R3 where R is the radius.
So as you can see while the radius of any of the balloons is unknown the solution is undefined so that if we accept Slimjim's 400lb ballast proposal, it simply means that each of the 45 balloons has to be 1.31m (4.29ft) in radius. This solution assumes the balloons to be equal in radius, the density of air, rho_air @ 15,000ft to be 0.771 kg/m3 and the density of helium ,rho_he @15,000 ft to be 0.1065 kg/m3.
But this solution is by no means unique. If I chose a larger radius balloon I could carry more ballast and if I carried more ballast I could control a more variable rate of descent.
This is one aspect of the question that is totally unknown. How fast is this rig going to descend? If you compare the radius of balloon to provide floating at 15,000ft with that of10ft you will notice that if you did not carry any ballast the radii of each of your 45 balloons will shrink from 0.927m(3ft) to 0.795m(2.6ft). For a ballast of 400lbs that shrinkage would be from 1.31m(4.29ft) to 1.12m(3.67ft) for each balloon.
These calcs assume rho_air @ ssl to be 1.225 kg/m3 and rho_he @ssl to be 0.1693kg/m3.
So if you were to assume a steady descent rate of 1000ft/min which equals 16.6ft/s or 5.06m/s the total trip would take 15minutes and given that the volume of air change in the balloons would be (3.34m3-2.1049m3)==> delta_V=1.239m3, for zero ballast, that would mean that you would have to release a volume of 1.239 cubic metres of helium from each of the balloons over the the descent time of 15minutes, which means that you would be bleeding off gas at a rate of 0.00138m3/sec. For the same descent rate with 400lbs ballast you would have (9.423m3 - 5.931m3)==> delta_V = 3.492m3/15min = 0.00388m3/sec
Given that 1m3 = 1000L this would indicate that you would need to bleed off your helium at a rate of 1.4litres/sec off each of your 45 balloons if you had no ballast and you would bleed off your helium at a rate of 3.88litres/sec for each of your balloons had you 400lbs of ballast. Note this this also assumes a linear relationship between altitude and density.
If done without ballast the rate at which you bleed off air is critical. If you pick up to much downward velocity you wont be able to dump ballast to let your buoyant force brake your downward plunge.
Obviously the more ballast you have, the more you can play with your descent profile, such that you can descend really quickly at first and then when you want to slow down your descent you can dump a lot of ballast to reverse thrust and bring yourself back to a nice gentle sink rate with at least 100ft off the deck to play with just so as to give you a nice margin for error should you not dump ballast quick enough to reverse a high sink rate and you find that close to the deck you are still falling fast, at least if you had started with a lot of ballast you would still be able to reign in your plunge with a lot of urgent dumping.
All of this is dependent on you being able to control 45 balloons accurately and simultaneously of course. Failing all this I would advise the Irish solution and that is to free fall the distance to within three feet of the surface, and then step down the rest of the way.😵
Nice maths but the article does specify the width of the balloons as 4ft (0.9m).
Walters and his girlfriend, Carol Van Deusen, purchased 45 four-foot
weather balloons
http://en.wikipedia.org/wiki/Lawnchair_Larry
With this constant added in, here's my calculation :
Va (Volume of Air Displaced) = Vh (Volume of Helium in balloons);
Da = Density of Air, Dh = Density of Helium;
Wm = Weight of man; Wb = Weight of Ballast
g = Gravitational constant
If we assume he is only just able to take off, we can first find
the maximum amount of ballast he can carry.
The volume of each balloon is - (4/3 * 3.14 * 0.9) = 0.9m^3
Total volume of balloons is 40.5m^3
So Va = Vh = 40.5m^3;
Da = 1.293 kg/m^3;
Dh = 0.178 kg/m^3
Wm = 100 kg;
Net upward force = Total upward force - Total downward force
Total upward force - Total downward force = 0 (at equilibrium)
(Va * Da * g) - (Vh * Dh x g + (Wm + Wb) * g = 0
dividing by g, we get :-
(40.3 * 1.293) - (40.3 * 0.178) + (100 + Wb) = 0;
Wb = 145Kg;
So the maximum ballast he can carry is 145kg
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To find out what happens at 15,000 ft (4500m)
We can assume that the balloons are able to expand to a limitation
which we can call Em. Looking at this weather balloon site of modern
balloons, the constant for a 5ft balloon is 2.56 (e.g. it can expand 2.56
times its inflation size before it bursts).
https://secure.scientificsales.com/weather/Details.cfm?ProdID=129&category=18
If we take into account that air pressure at sea level is approximately
1.293 kg/m^3 and that this pressure decreases by 3% every 1000 ft
http://www.2-stroke-porting.com/altiden.htm
Then we can say at 15000 ft, it will have decreased by 45%
So Da = 1.293 * 0.55 = 0.7 kg/m^3;
and Dh = 0.178 / 2.56 = 0.07 kg/m^3; (at the point of the balloon
bursting)
Va = Vh = (40.5 * Em) = 103.7 m^3
At equilibrium
Va * Da * g = ( Vh * Dh * g ) + ( 100 + Wb ) * g
so now the equation looks like this :
( 103.7 * 0.7 * 9.8 ) = ( 103.7 * 0.07 * 9.8) + ( 100 + Wb ) * 9.8
dividing both sides by 9.8 and summing
72.6 = 7.3 + 100 + Wb
Wb = 72.6 - 7.3 - 100 = -35Kg
In other words larry and chair would have to have weighed 65Kg
and be carrying no balast for the balloons not to have burst.
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Originally posted by Thequ1ckThanks.
Nice maths but the article does specify the width of the balloons as 4ft (0.9m).
Walters and his girlfriend, Carol Van Deusen, purchased 45 four-foot
weather balloons
http://en.wikipedia.org/wiki/Lawnchair_Larry
The volume of each balloon is - (4/3 * 3.14 * 0.9) = 0.9m^3
Total volume of balloons is 40.5m^3
Wb = 145Kg;
So the maximum ballast he can carry is 145kg
I dont agree with your result only because of your initial volume calc.
I'll use the same variable names as you.
Net upward force = Total upward force - Total downward force
Total upward force - Total downward force = 0 (at equilibrium)
(Va * Da * g) - (Vh * Dh x g + (Wm + Wb) * g = 0
dividing by g, we get :-
====> Va*Da - Vh*Dh -Wm - Wb =0
Because we want to know Wb, make it the subject of the formula and
===> Wb = V*Da - V*Dh - Wm (Va=Vh)
Tidying up we have
===> Wb = V*(Da-Dh) - Wm
This just tidies up the equation and shows you the ballast required to not have any lift. Also written this way you can see that if your Volume remained constant at take off, using a lighter gas like hydrogen, would give you more lift, which if you wanted to float at ground level would mean that you would need to carry more ballast.
So the nett upward force is the difference between the ballast required to keep the chair in equilibrium and the ballast Larry actually had strapped onboard with 'jugs of water'.
Using a conversion factor of 1ft = 0.3048m==> 4ft = 1.2192m
Therefore the radius of each balloon is R = 0.6096 m
Therefore the volume of one balloon is V = 4/3 * pi * (0.6096)^3 = 0.9489 m^3
===> V = 0.95m^3 (2 dp)
Funnily enough even though you start with the wrong diameter ( 4ft is not 0.9m) you end up with close enough to the right volume for each balloon which you give as 0.9 m^3
===> Wb = V*(Da-Dh) - Wm
Da = 1.293 kg/m^3; Using your density values
Dh = 0.178 kg/m^3
So plugging it all into our formula gives
V of 45 balloons = 42.7 m^3 (V of 1 = 0.9489 m^3)
===> Wb = 42.7*(1.293 - 0.178) - 100
===> Wb = -52.388 kg
A negative ballast result indicates that in this configuration the chair would not even lift off as the solution is a negative ballast.
Even using my density values which are for US standard sea level
Da = 1.225 kg/m^3; US SSL @ 15 deg C
Dh = 0.1693 kg/m^3
====> Wb = 42.7* (1.225 - 0.1693) -100
====> Wb = -54.92 kg
Again this would not even get him off the ground as you so rightly point out.
To find out the number of balloons needed you would equate the weight of the man to the net balloon lift
====> Wm = V*(Da-Dh)
====> 100 = V* (1.225 - 0.1693)
====> V = 94.72m^3 or 99.82 balloons using US SSL
or
====> 100 = V* (1.293 - 0.178)
====> V = 89.686m^3 or 94.5 balloons using your densities.
Either way as you worked out the chair would not be able to take off at all.
This pdf off the following webpage ~ http://www.coastal.udel.edu/faculty/jpuleo/CIEG305/homework4.pdf ~ shows that the balloons were probably inflated past their 4ft diameter and at this point almost all bets are off because as I first pointed in my original post that if the radius is not known, then this problem becomes a parameterized one having no definite unique solution, and then the figures you end up with are totally dependent on the diameter at initial inflation that you choose to work with.
In short this was a nice rubber band problem and I have enjoyed chasing my tail trying to disprove it! 😉