Arghh!! Another long post lost forever in cyberspace! This thread is becoming too frustrating and time consuming for me, not because of the posts disagreeing with me but because the hamsters seem to hate me and it takes longer for me to get my posts to “take” than it takes to write them. Then it takes forever to refresh the page. Too frustrating. I think the information has been presented and now each one can make up his mind. I do not think there is too much point in repeating the same things over and over but I’d like to hear from anyone who will take up my challenge. That would be a fun experiment and the wager would make it doubly interesting. I wish I had a high speed camera as I would be willing to do some experiments for free. If anyone has one I will be glad to design the experiments.
I think you are misunderstanding the “counterweights”. The point is that the parabounce requires neutral or very close to neutral bouyancy so, rather than adjust the volume of the balloon for each user, it is easier to provide ample bouyancy and then cancel some of it with ballast but you could do the same thing by adjusting the amount of helium for each user. The object is neutral or very close to neutral bouyancy so that the least shove up or down will send you up or down. Note also that with neutral bouyancy you have no traction on the ground so you cannot propel yourself sideways, only give yourself an upward push. It is similar to having neutral bouyancy in the water: the least push up or down will send you off in that direction a long way.
If you use too much helium, you just float, use too little and you you come right back down without much benefit. I guess you’d have to use an amount of helium that slightly exceeds the the minimum amount to cause weightlesness.
I ask this on behalf of a friend (seriously) who wants to make a sturdy vest of some sort that would allow him to “jump 20 feet in the air”. Any figures on how many cc’s of helium required to lift a 175lb man?
My thinking says that if you could defeat the slack in the line caused by leaping, and get 20 feet off the ground, you’d just keep floating.
Your response:The point is that the parabounce requires neutral or very close to neutral bouyancy so, rather than adjust the volume of the balloon for each user, it is easier to provide ample bouyancy and then cancel some of it with ballast but you could do the same thing by adjusting the amount of helium for each user. The object is neutral or very close to neutral bouyancy so that the least shove up or down will send you up or down. Note also that with neutral bouyancy you have no traction on the ground so you cannot propel yourself sideways, only give yourself an upward push.
I guess I’m obviously a retarded brain-dead moron, I’ll just keep spitting in the wind and dealing with the consequences.
Good thing I didn’t submit my “fart in a bag” theory to y’all, prolly would have gotten shot down in flames.
Sheesh another lost OP. this is getting boring. Mangetout, I have posted here a design for a possible experiment. By adjusting the weight and height you drop on the right you can adjust the height to which the mass on the left is propelled. You could even have the mass on the left sliding on a couple of wires so that it would not fly about.
OK, dnooman, how about this: air has a density at standard temperature and pressure of about 1.293 grams/liter; heluim has a density of 0.178 grams/liter. That means that a liter of helium provides a buoyancy force equal to the weight of a 1.115 gram mass (= 0.0109N = 0.00224 pounds). To lift a 175 pound man, you’d need 175/0.00224 = 78,000 liters (which is 78 cubic meters) of helium. Of course the balloon fabric adds some additional weight, so you’d need a little more helium to account for that. This calculation scales with the cite given by Fear Itself: the helium in a 180 cubic meter balloon gives 500 pounds of lift (the baloon weighing in at 100 pounds)
Anyway, I think everyone would agree to the following:
If your 175 pound man strapped on a balloon giving more than 175 pounds of bouyancy, he’d drift upward…and keep going until something else stopped him.
If your 175 pound man strapped on a balloon giving slightly less than 175 pounds of buoyancy (let’s say the balloon gives 157.5 pounds of buoyancy, just to pick a number), and the man’s harness was rigidly attached to the balloon (to avoid the slack string argument), then the man would be able to jump much higher, as his jump would “push” the balloon upwards.
In the case of the balloon having 157.5 pounds of buoyancy, I suspect (using a potential energy argument) that the man would be able to jump nearly ten times the height that he normally would, since he effectively weighs 1/10 of his normal weight (175-157.5 = 17.5). I say “nearly 10 times” because he has to accelerate the mass of the balloon also.
It’s morning, I’m back, lots of good stuff last night. And by the way **sailor[B/], I’m an Ocean Engineer, I have a little idea of what buoyancy is about. You’re suprisingly quick to dismiss a bunch of people you’ve never even met, don’t assume you’re the smartest one in the room. I’ll concede that the balloon won’t accelerate faster than g (never said it did, I was giving the theoretical maximum), however I still believe that conservation of momentum will allow it to work, yes with inelastic cords - not in every case, but where a balloon provides 98% or so of the lift needed to get you off the ground. I actually went looking for balloons last night, but the stores were closed, I’ll leave earlier today. **Mangetout[B/] and I will do your homework for you, and I will accept the answer, even if wrong.
**Achernar[B/], I’d love to see your derivations (or at least a little more explanation) for your calculations. I’m not at all saying your wrong, I’m just not following exactly how you got there, because it’s entirely possible that I’m not the smartest person in the room.
**Dnooman[B/], I think your numbers are provided above in the link - 180m^3 to provide 180kg of lift. You’re looking for roughly half of that. And yes, it doesn’t matter whether you’ve got a professionally manufacturered balloon, or a bunch of party balloons (be careful!), lift is lift.
Thanks Sailor, oddly enough, I had something similar in mind.
I plan to test it first launching the weight alone a number of times to verify that we are indeed getting a consistent launch force, then a number of times with the balloon attached.
I am sure I’ve seen the scenario I described earlier, where the string going taut causes the weight to rise again, higher than before, but it has just occurred to me that the string could go taut before the weight has started falling again, if the (accelerating)rising balloon manages to take up the slack while the weight is still rising(but decelerating) - would this make any difference at all to the calculations?
You might find that to get to launch to 3 - 4’ you need more force than provided by your weight.
Well, this is what i am saying I do not think will happen but let us know how it goes. Can you take some video? That would be cool.
Not at all. I was only asking because I know Chronos and Achernar from past science related threads and I remember they have a background in science so I was asking about the rest who I do not know.
In that case you might want to explain it to The Ryan who was the one who was asking. Not all my posts are directed at you.
What I meant was, it was pointed out that the balloon couldn’t possibly rise faster than gravity, and I agreed–but what I attempted to explain was that it would be possible for you (with balloon attached however you want) to crouch down low and “slowly” jump up, and (in the case of the numbers I used before) you would have upwards inertia of your 80 lbs and 20 lbs of balloon, but the helium in the balloon would offset all but .05 lbs of that weight as far as the downward pull of gravity was concerned. So you would jump much higher into the air than normal.
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Yeah, sure, I can do that, but I may have made a mistake. Maybe somebody can help me.
When the balloon is going up and the weight is going down, there’s also a third thing in motion - the air. Does this factor in to the momentum conservation equation or not? That is, do we treat it as two bodies in motion or as three? I did it out both ways and got the same qualitative answer, but I’m not clear on which way is correct.
I read the article, and it is unclear about what they mean by “net bouyancy of 2 lbs”. That ill-defined statement, by itself, seems to support your idea, but the videos which show people standing on the ground, jumping and falling back to the ground, plus the statement that “It’s the same identical principle as taking a toy helium balloon, tying a string to it, then adding a small weight so the balloon sits on the table until, with just a gentle tap, it floats up to the ceiling,” which you yourself quoted, support the idea that they actually mean “bouyancy which reduces your effective net weight to 2 lbs”.
Wrong. If the net force is 2lbs down, all of your protestations about this “not being jumping” are wrong. You can’t “stay up there for an hour tethered to the ground” as you keep saying. You fall.
Well, first, you refuse to acknowledge a situation where the balloon+rider system is slightly heavier than air, which allows you to dismiss this as “not jumping.” Second, you’ve ignored my point 1 in my first post above, which is that you can jump slowly if you’re only slightly heavier than air; an idea you ridiculed and dismissed out of hand before.
Achernar, if you mean the air outside of the balloon, aka atmosphere, I would think that is external to the system we are talking about. If you’re talking about the air inside the balloon, then it’s mass is just accounted along w/ the balloon wouldn’t it?
What’s “d”? I didn’t understand how you came up with this.
If the net force is two pounds down, then you don’t have a net buoyancy. Now, it’s possible that whoever wrote that made a mistake, and meant to say “net weight”, but “net buoyancy of two pounds” means that the force is up.
PeeWee, I have uploaded my notes on this problem - 475kb PDF file. Let me know if you still have any questions. If this file is too big for you, I can do something else.
I see some are mentioning the resistance of the air to the balloon rising but no mention of it resisting the decent. It takes x amount of force to move through the air either left or right or up or down. As far as displacing the air is concerned. Or at least close to the same effort, right?
So, if I make a balloon the weight 100 tons and can be filled to raise 100 tons + 200 pounds and I weight 201 pounds and I tie myself to it ( on a calm day ) and then do as sailor wants and jump hard and normally as one would to hop up onto to a two square foot box the balloon would start rising at least an inch or so I would think even with that massive inertia. Right? So when my 201 pounds started back down, the balloon’s direction would win , right ? For a little bit anyway? Now if the balloon only weighted 5 pounds and was filled to lift 205 pounds, then the 201 pounds of me descending would win, correct?
As long as the shape and size of the balloon did not let it’s resistance to changing direction or even moving through the air enter into it.
Think about the stopping power of an old TT parachute, it did not weigh much but when it was open, it did not slow a person down gently, the air did not move around it easily, ask any jumper that has used one… ‘ouch’ It takes a while to get the airflow moving around it to allow it to accelerate to normal decent speed with things in equilibrium. Right?
The OP =
Now since the OP did not specify the size, weight, shape, number or anything else, just that the jumper would be able to get some UNSPECIFIED amount of slack in the attaching line (s) we can assume my 200 ton balloon is just as viable an idea as 2 pound balloon. Right sailor?
Now if I squat down with enough balloons tied to me that they will lift all but 3oz of my weight in still air and stand up at such a speed as to not quite put slack in the lines and then within the last 2 inches of being straight, I increased my motion so that I put slack in the lines and also would under normal conditions actually leave the floor for at least ½ inch, which satisfies the ‘jump requirement’ IMO, I do think I might exceed ½ inch in altitude before starting back down.
The reversing of the direction of the entire assembly including the body of the jumper and even just a normal round balloon of sufficient size to lift the person ( 201 Pounds) will not be as amiable to changing direction as implied without taking into account the forces need to move the balloon through the air and be accelerating to any great degree at the same time depending quite a bit as to the proportions of mass divide between the balloon and the load…
No, sorry, this point is wrong. What you don’t realize is that a balloon with very little lifting power greater than itself will rise very slowly, and in the time it takes you to reach the apex of your jump and start descending, it will not have been able to build up very much momentum.
d = 200000 / 200200 = 0.999
a = (1 - d) / (1 + d) g = 0.0005g = 0.192 in/sec[sup]2[/sup]
At this acceleration, it will take 3.2 seconds for the balloon to rise one inch. You said you wanted to jump up two feet - this takes about 0.35 seconds. In this much time, the balloon will have risen about one-hundreth of an inch, and will have gained a momentum equal to about 1121 lb · ft / sec. By comparison, the momentum you had when you left the ground was 2274 lb · ft / sec.
Correct, and if you have a net buoyancy, you don’t sit on the ground until you give a slight upward push. So clearly the two statements in the article contradict one another.
Agreed. This is why I wanted clarification on everyone’s definition of “net buoyancy of 2 lbs”, since the parabounce article seems to be using it incorrectly. Based on their statement about sitting on the ground until given a slight upward force, it’s pretty clear that the system is heavier-than-air (unless that statement is incorrect, but given that it’s designed to be a simple and demonstrative, that seems unlikely).
