Can enough helium filled balloons make you leap to great heights?

[sethaw]

*Originally posted by sailor *
**I can’t believe we’re seriously discussing this. (I hope I am not being whooshed here but this is getting surreal) “jumping slowly” is not “jumping”, it’s “getting up from your chair” and does not get you airborn. Jumping implies “fast”, the faster you jump, the higher up you go. Try it yourself. Try jumping very slowly and let us know how far up you reach. **

ok fine, jump fast, let the string go slack, it still wont matter since the momentum/energy is conserved. If we start with the balloon pulling with our weight, then gravity alone will not cause us to come down. We add a upward force (from our feet from the ground) and then the velocity of the system (the balloons and us) is always up until we add an outside (downward) force.

[DougC]

• • • Fear Itself’s Parabounce site link above has images of pretty much exactly these types of balloons working as I and others described, allowing “jumps to great heights”. What other proof do non-believers here require?
      ~

[Omphaloskeptic]

*Originally posted by sailor *
**Hint: I do not need a calculator to know your result of an upwards acceleration of a = 137.4 m/s^2 is dead wrong by a factor of more than 15. the reason I do not need a calculator is I have common sense which tells me the upwards acceleration of the balloon can only be a fraction of g. which happens to be in my town 9.8 m/s2. Is gravity especially strong where you happen to be? **

OK, well, your common sense is wrong. The net buoyant force (the upwards buoyant force minus the object’s weight) is not proportional to the mass of the displacing object, so the factor of “m” in F=ma does not cancel out as it does for gravitational acceleration.

More specifically, the net buoyant force F[sub]B[/sub] = (m[sub]fluid[/sub] - m)g, so the acceleration of the object is (m[sub]fluid[/sub]/m - 1)g. If the object weighs less than half that of the displaced fluid, the upward acceleration can exceed g. (I don’t know if this is the case for the situations discussed here since I haven’t visited the links.)

[sailor]

Ok, according to my calculations for the balloon string to become taut at exactly the moment you reach the highest point in the jump (t=0.5714, S=1.2m) the upwards acceleration of the balloon would have to be 7.35 m/s2. For the balloon to be any help in achieving a greater height the upwards acceleration of the balloon would have to be substantially greater than 7.35 m/s2.

Anyone here want to dispute my numbers so far?
Anyone feel like proving a balloon will accelerate faster than that? (Good luck).

[sailor]

*Originally posted by Omphaloskeptic *
**OK, well, your common sense is wrong. The net buoyant force (the upwards buoyant force minus the object’s weight) is not proportional to the mass of the displacing object, so the factor of “m” in F=ma does not cancel out as it does for gravitational acceleration.

More specifically, the net buoyant force F[sub]B[/sub] = (m[sub]fluid[/sub] - m)g, so the acceleration of the object is (m[sub]fluid[/sub]/m - 1)g. If the object weighs less than half that of the displaced fluid, the upward acceleration can exceed g. (I don’t know if this is the case for the situations discussed here since I haven’t visited the links.) **

This is getting tiring. You do realise that the balloon rises because air takes it’s place? Are you saying air would fall faster than g? You really believe that? You really believe a balloon can rise with an acceleration greater than g? And you are willing to put money on that? Yes?

[Trinopus]

This is truly surreal…

Okay: tie the balloons (on strings) to your wrist. Now crouch down real low, and lower your wrist to the floor. Now raise your wrist as fast as you can, but not so fast as to introduce any slack into the strings.

Now straighten up in a jump; the balloons are already moving upward, and will add to the height of your jump.

(Mind you, I don’t believe this is necessary, but it might serve as a thought-experiment to satisfy our resident skeptic…)

Again, get a balloon and DO IT! That’s the real medicine.

(Why am I reminded of the “Is .9999… = 1?” thread?)

Trinopus

[sailor]

Here’s yet another hint: neglecting friction, a balloon with zero mass would rise with an acceleration of -g and a balloon with a mass equal to that of the same volume of air would rise with an acceleration of zero.

If we consider the specific gravity of air as unity and the volume of the balloon as unity and we call the specific gravity of the balloon “d” which is a fraction of 1 (air) then we have that the upwards force is 1-d and the mass which has to be moved is 1+d (air down plus balloon up) and, so, the upwards acceleration is g*(1-d)/(1+d). The following tables gives us the upwards acceleration of a balloon as a function of its bouyancy:

  d      f      m       a   
0.0    1.0    1.0   100.00%
0.2    0.8    1.2    66.67%
0.4    0.6    1.4    42.86%
0.6    0.4    1.6    25.00%
0.8    0.2    1.8    11.11%
1.0    0.0    2.0     0.00% 

[Lemur866]

Sailor, I agree with you, assuming that we are talking about inelastic cords.

A balloon cannot rise faster than g, it is physically impossible, since a balloon rising means that the air it is displacing is falling. The displaced air cannot fall faster than g, and will almost certainly fall much slower than g.

And “jumping slowly” doesn’t make any sense. Jumping slowly means “jumping with only a little force”, which means “jumping not very high”. The higher you want to jump the faster you have to move upwards.

If someone is attached to the balloon with elastic bands, or with rigid attachments, then the above analysis is void.

[Mangetout]

I can offer nothing more than a red face when people start throwing about actual formulae and calculations, as it has been quite a few years since I worked with any of this stuff.

What I don’t see in any of the above calculations (and it might just be my complete naiveté here) is the fact that the balloon and its contents have mass and are travelling upward while the string is slack, objects in motion tend to stay in motion etc, so the balloon imparts more than the force of buoyancy upon the weight when the string tightens again.

All I can say is that I have done these experiments numerous times with small weights and balloons that my kids brought home - add a weight to the string that is about 90% countered by the buoyancy, launch it upwards and what you see is that it rises normall as the string is slack, slows and starts to fall, it falls until the string is tight, then the balloon carries it upward again, higher than the point to which it was first thrown, then it slows and falls slowly.

Maybe the effect is not so significant when it is scaled up to human-sized weights.

[sailor]

*Originally posted by DougC *
**- - - Fear Itself’s Parabounce site link above has images of pretty much exactly these types of balloons working as I and others described, allowing “jumps to great heights”. What other proof do non-believers here require?
~ **

This has been addressed already by Chronos and myself so you might want to read the thread. The OP mentions and we are discussing balloons which are floating freely tethered to the jumper by a line. The parabounce does not satisfy these conditions. What we are discussing is whether a balloon tethered by a line will aid in jumping higher. get it? tethered by a line? yes? get it? tethered by a line? get it? yes? good.

PS: don’t forget the tethering line.

[Trinopus]

One more stake in this dracula:

Stand tall, with your arm high overhead, the balloons tied to that wrist.

Jump, and, at the same time, pull your arm downward, keeping the strings taut. Essentially, pull yourself up by the strings.

Gotta shower.
Trinopus

[sailor]

*Originally posted by Lemur866 *
Sailor, I agree with you, assuming that we are talking about inelastic cords.

A balloon cannot rise faster than g, it is physically impossible, since a balloon rising means that the air it is displacing is falling. The displaced air cannot fall faster than g, and will almost certainly fall much slower than g.

And “jumping slowly” doesn’t make any sense. Jumping slowly means “jumping with only a little force”, which means “jumping not very high”. The higher you want to jump the faster you have to move upwards.

If someone is attached to the balloon with elastic bands, or with rigid attachments, then the above analysis is void.

Lemur, you are very correct in all points and note I mentioned elastic bands at the beginning of the thread and you are correct there too.

[sailor]

*Originally posted by Trinopus *
**One more stake in this dracula:

Stand tall, with your arm high overhead, the balloons tied to that wrist.

Jump, and, at the same time, pull your arm downward, keeping the strings taut. Essentially, pull yourself up by the strings.

Gotta shower.
Trinopus **

Well, if we are going to change the conditions of the OP I would suggest to buy a helicopter or just go out on a date.

[Hail_Ants]

I’d just like to add that this very thing (helium balloons) was used when filming the moonscape scenes in the HBO miniseries From the Earth to the Moon to simulate one sixth gravity.

[DougC]

This has been addressed already by Chronos and myself so you might want to read the thread. The OP mentions and we are discussing balloons which are floating freely tethered to the jumper by a line. The parabounce does not satisfy these conditions. What we are discussing is whether a balloon tethered by a line will aid in jumping higher. get it? tethered by a line? yes? get it? tethered by a line? get it? yes? good. PS: don’t forget the tethering line. - sailor

• • • Well, click the link and take a look: each balloon has a number of lines coming off it, attached to a ring. Then there’s a couple lines hanging off opposite sides of that ring, that seem to run down to the harnesses the REAL PEOPLE are wearing. Pick whichever line you want, and then explain why it doesn’t work, even though they are photographed doing it…
      ~

[Achernar]

*Originally posted by Mangetout *
What I don’t see in any of the above calculations (and it might just be my complete naiveté here) is the fact that the balloon and its contents have mass and are travelling upward while the string is slack, objects in motion tend to stay in motion etc, so the balloon imparts more than the force of buoyancy upon the weight when the string tightens again.

Well, I went ahead and did that, so I might as well say what I got. I’ll give you the briefing, but I can show derivations for any of this if you like.

The string will become taut at a time t[sub]TAUT[/sub] = (1 + d)t[sub]APEX[/sub], where t[sub]APEX[/sub] is the time at which the weight reached its highest height. Since d is in the range 0 to 1, this will be during the weight’s descent.

At this time, the balloon will be going upward with a speed of v[sub]0[/sub](1 - d), and the weight will be going downward with a speed of d×v[sub]0[/sub]. (v[sub]0[/sub] is the liftoff speed.)

The string becoming taut can be treated as an inelastic collision. A lot of energy is lost to that string, and this is probably what makes it so counter-intuitive. But in the end, the weight and the balloon must be moving at the same velocity. Conserving momentum, and setting upward equal to the positive direction, that velocity is:

V = -d (1 - m[sub]AIR[/sub] / (m[sub]WEIGHT[/sub] + m[sub]BALLOON[/sub])) v[sub]0[/sub]

Now, look at the fraction inside the parentheses. m[sub]AIR[/sub] / (m[sub]WEIGHT[/sub] + m[sub]BALLOON[/sub]). This has to be less than 1, because if the weight and the balloon were together less massive than the displaced air, then it would have taken off without an initial jump, and it will never reach the ground. Therefore, the term in parentheses is positive, and therefore V is negative. That means that after the string becomes taut and all that energy is lost, the weight will continue to drop, taking the balloon with it.

[The_Ryan]

*Originally posted by DougC *
**- - - And you are not reading carefully. I said that you would not be able to jump up faster than the rate of gravity.

• Consider this example:

1. A 80 lb person,
2. A 20 lb empty balloon, and
3. enough helium to lift 99.99 lbs.
   …

• Now if the balloon with helium was attached to the person and they crouched down and jumped up slower than the accelleration of gravity, the person, the balloon and the helium will be moving upwards at X speed with 100 lbs of inertia, but the downward force of gravity on the person, balloon and helium will only be .05 lbs. So you would be able to jump up higher, just not any faster than the accelleration of gravity.
  ~ **

By “slower than the acceleration of gravity”, do you mean that “accelerating faster than g”, or “leave the ground with a velocity more than g”? If the former, I don’t see how initial acceleration would have any direct effect. If the latter, velocity and acceleration are two completely different things. “a velocity greater than g” is meaningless. Similarly, “100 lbs. of inertia” is also meaningless. Inertia is measured in units of massdistance/time. Pounds are units of massdistance/time*time. The units don’t match up.

[The_Ryan]

*Originally posted by Omphaloskeptic *
**OK, well, your common sense is wrong. The net buoyant force (the upwards buoyant force minus the object’s weight) is not proportional to the mass of the displacing object, so the factor of “m” in F=ma does not cancel out as it does for gravitational acceleration.

More specifically, the net buoyant force F[sub]B[/sub] = (m[sub]fluid[/sub] - m)g, so the acceleration of the object is (m[sub]fluid[/sub]/m - 1)g. If the object weighs less than half that of the displaced fluid, the upward acceleration can exceed g. (I don’t know if this is the case for the situations discussed here since I haven’t visited the links.) **

You’re forgetting about m[sub]fluid[/sub]. Remember, this force isn’t being applied to just m, it’s being applied to both masses. So a=f/m=((m[sub]fluid[/sub] - m)g)/(m[sub]fluid[/sub] + m).

[spinky]

First, I think Sailor is right that a balloon’s upward acceleration is g at the most, and in practice, quite a bit less. What I’m not sure about is that this actually matters, for two reasons:

1. given a massive balloon like those used in parabounce, you actually can “jump slowly”. It’s shown in the videos on the web site – people crouching down and the act of just standing up allows the balloon to rise, giving it momentum which picks you up off the ground. It’s hard to tell if the harnesses are completely rigid, but the more massive the balloon, the slower you have to “jump” in order to have it pick you up, so it seems quite conceivable that the upward acceleration your legs provide can be less than the balloon’s natural acceleration from bouyancy, and yet still enough to launch you off the ground.

2. Even if there is a rope which goes slack when you jump and you (by virtue of the balloon accelerating at less than g) reach the top of your jump’s arc before the rope is taut, the massive balloon gains quite a bit of upward momentum while you’re traveling back down along your natural arc. At some point, the rope becomes taut, and the balloon can have enough momentum to yank you back in the upwards direction, and if the balloon is massive enough, it can have enough momentum to lift you past the highest point in your original jump’s arc. This is similar to what would happen if you were attached by a long rope to a very massive balloon, and while standing on the ground, you pulled the balloon down to the ground slowly, then released it. Given the right balloon, this could pick you up off the ground when the rope became taut.

Comments, Sailor?

[Achernar]

*Originally posted by ntucker *
2) Even if there is a rope which goes slack when you jump and you (by virtue of the balloon accelerating at less than g) reach the top of your jump’s arc before the rope is taut, the massive balloon gains quite a bit of upward momentum while you’re traveling back down along your natural arc. At some point, the rope becomes taut, and the balloon can have enough momentum to yank you back in the upwards direction, and if the balloon is massive enough, it can have enough momentum to lift you past the highest point in your original jump’s arc.

No, it can’t. See my previous calculations.

However, a little elasticity can go a long way. The calculations assumed that the rope was perfectly inelastic, and in practice, this is probably a bad assumption.