# Helium Ballons?

Hello,

My team has weight issues and I came up with the (albeit stupid) idea to use balloons with Helium to help the weight a little bit. Would it be legal in the first place?

Thank you.

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Think about how much air fits inside your allowable robot volume that has a maximum height of 45 inches. Now calculate how much that air weighs. Your helium filled device would be able to lift slightly less than that.

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If your robot was at maximum volume and consisted entirely of Helium somehow held together with a massless balloon, it would have a buoyancy of just over one pound.

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Did you run the math first? You need a balloon about 3 feet in diameter to lift 1 pound. Is that practical?

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DO IT. You got this, fam.

Remember, once the match begins, you can gain another 12" in every horizontal directions. So thatâ€™s a bunch more unaccounted volume you can use.

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itâ€™s probably not legal if your helium balloons hold any pressure and arenâ€™t pressure rated.

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We find that the Java function Levitate.robot works just as well. The only problem is that you have to weigh with the battery and hope the RI doesnâ€™t realize the robot is on.

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Have you considered using the SLIM CIM? Itâ€™s a powerful alternative to the CIM and weighs a lot less. Anyone using a Slim CIM?

/s

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A few â€śreasonableâ€ť alternates:

1. Negative Mass is a thing, though as far as I can tell only in theories. Find a way to make it more experimental, and Bobâ€™s your uncle.
2. Suspend a large mass of osmium directly overhead of the inspection station. The large amount of mass will alter earthâ€™s gravitational field to reduce the apparent weight measured at the scale. [^1][^2]
3. Build the robot from steel, then suspend a very large magnet over the inspection station. [^3]

[^1] Physics Exercise: Determine the radius of an osmium sphere required to reduce measured robot weight by one pound. Assume sphere is suspended such that the lower extreme of it is 500 ft above the inspection station.

[^2] Psychology Exercise: How do you convince onlookers that such a sphere is â€śOKâ€ť and there is â€śno reasonâ€ť to panic?

[^3] Physics Exercise: Using an electromagnet model of your choice, determine a solution for the geometry of an electromagnet positioned 500ft above the inspection station, that is powered by an existing nuclear generator of your choice, and reduces the apparent robot weight by 1 pound.

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Itâ€™s probably easier to switch to competing at an event with higher elevation or closer to the equator, taking advantage of the fact that the limit is on weight, not mass, and that the Earthâ€™s gravity isnâ€™t the same everywhere.

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Thatâ€™s not an issue if they canâ€™t see it. All you need is a couple of projectors and hope nobody looks too close.

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Heliumâ€¦ Nope for reasons stated.

In all seriousness, you are not the first team to go over weight late in build season. If you explain the general design of your robot, or better yet provide a picture, and tell us how far over you are we can make much better suggestions on where to go looking for weight savings.

This was a constant battle on my last team. I think we ended up over 4 years in a row. Current team has only managed to go over once in the last 5 years.

Unfortunately drilling holes is likely one of the lowest effort vs return options. You donâ€™t get much weight savings, it is hard to do on an assembled robot and you need to do lots of stress calcs to determine how much you can remove, and where. Removing metal after the fact is much harder than designing in â€śskeleton styleâ€ť structural members.

Taking a hard look at your motor choices, and replacing CIMs with MiniCIMs is often an easy way to reduce weight. Drastically reducing lexan thickness for covers, and belly pans is another. Nuclear option is sacrificing mechanisms to the build gods.

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I have heard a legend about an air vent over the scale at an event causing robots to vary by a pound or two depending on whether is was on or off. Hypothetically if your robot has a flat surface to catch the air, then weighing it surrounded by fans blowing up could make it lighter.

When I was in wrestling in high school I found if I raised my arms at the right time I could make the scale register two pounds lighter than if I didnâ€™t. Maybe one of your mechanisms is spring loaded and could be set up to move at the right time during weighing.

Helium balloons, like pneumatic tires or gas springs, are not considered pneumatic devices and could be used legally. Previous posts have stated the obvious volume constraints. As you point out, they are not pressure rated, so they could not be hooked up to the robotâ€™s pneumatic system as a pneumatic device would be.

Good luck with that! weigh-in is not a snapshot, we wait for the scale to reach a steady state reading.

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Slide over and change the scale to KG when they are not looking.

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If you were able to remove all of the air from your robot and have a perfect vacuum(in lieu of helium), I believe you could lose up to 3.3lbs. The good news is you probably have a lot of air in your robot, so there is a good chance you can lose a few lbs there!

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In the context of the rest of this thread, hereâ€™s where my thoughts went:

Editorial: @gerthworm does not condone drugs, physical violence, or any related tomfoolery against robot inspectors.

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It would require an osmium sphere with a radius of approximately 12,600 meters. All things said, reducing your weight with helium is a far more viable solution than the giant osmium sphere.

Work:
Assuming a 126 lb robot and newtonian gravity, and that the density of osmium is 22,590 \frac{kg}{m^3} .
Quick googling reveals that the robot has a mass of 57.1526 kg and the required force to support one pound of mass is 4.45 newtons, the distance from the bottom edge is 152.4m. Gravitational force is given by the equation F_G=\frac{Gm_1m_2}{r^2} where r is the distance between the objectsâ€™ center of mass, m_1 and m_2 are the objectsâ€™ masses, and G is the universal constant of gravitation. Since the osmium is a sphere, the distance between the centers of mass is l+r, and the mass of the sphere is \frac{4\rho\pi r^3}{3} where \rho is the density, r is the radius of the sphere, and l is the distance between the robotâ€™s center of mass and the bottom edge of the sphere. Substituting those into the force of gravity equation I got this:
F_G=\frac{4\rho\pi G m_1 r^3}{3(l+r)^2}.
To solve for r I substituted \frac{4\rho\pi m_1}{3} for a since all are constants to get this:
F_G = \frac{ar^3}{(l+r)^2}
I used synthetic division to break dow the fraction:
F_G=ar-2la+\frac{3l^2 a}{l+r}-\frac{4l^3 a}{(l+r)^2}
I assumed the third and fourth terms had a negligible effect so I could actually solve for r:
F_G = ar-2la
Solved for r:
r=\frac{F_G}{a}+2l
substituted values constants back in for a:
r=\frac{3F_G}{4\rho\pi G m_1}
Throw in numbers:
r = \frac{3*4.45N}{4*22500\frac {kg}{m^3}*\pi*57.1526 kg*6.67*10^{-11}\frac{m^3}{kgs^2}}+2(152.4m)=12641.3392442m=12600m
I did plug my answer back into my original equation and got 4.45195673709N for force, which is accurate to the same number of significant digits I put in.

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