I’m trying to calculate some torque, but my team is having trouble finding a scale we’d consider accurate for an object the weight of the ball. I tried to do some searching… I saw a post that refered to it as 2lb’s, but I’m not sure if the person actually weighed it or if it was just an estimate. So, has anyone measured the weight of the fully-inflated ball on a device that could be considered accurate on 1-10 lb objects?
We weighed the large 2x yellow ball today and got 3.6 lbs +/- .2 lbs. If you figure on 4 lbs you’d be safe. The 2 lbs seems a bit low. Anyone else confirm this? If all else use some triple beam balances from your schools science department (they usually hold up to 1.1 kg ~ 2.3 lbs), just balance the ball on two balances and add their readings up. Or take the ball to your local Mail Boxes Etc or Post office and have them wiegh it on their scales.
its a lot easier to weigh before you inflate it!
You’d have to weight the air too right?
No, the weight of the air in the ball won’t affect it as long as you are measuring it in air. If you were to measure it in a vaccuum, then yes, it would matter.
There will be a difference in weight, though it’s most likely negligible.
Take for instance a cup. If you weigh it, it’ll have the same weight as if you weigh it with a lid on top to trap the air.
However, let’s say I was to take that cup, and really make it a cylinder, and fill it to 500 PSI… there’s more mass of air in the system.
Because the ball is inflated, there will be a (slight) difference in weight.
10,000 points to the first person that can calculate the weight difference between an inflated and uninflated ball… (30" diameter, 30 PSI, assume 1/8" wall thickness) 
Matt
We came up with 3.4lb agin ±.2lb so that seems accurate, ours was also inflated
well…
volume of a sphere = 4/3 * pi * r^3
30 inches = 0.762 meters
4/3 * pi * (0.762 / 2)^3 = .232 cubic meters = 232 liters
Now, considering 30 pounds per square inch = 2.0413 atm, if you rearrange the ideal gas law pV = nRT to solve for n,
n = pV/RT = (2.0413 atm)(232 l)]/(8.2057 * 10^-2 latm/(molK))((273+25) K)] = 19.4 moles of air
According to
this lab, the molar mass of air is about 29 g/mol, soooo
19.4 moles * 29 g/mol = 560 grams, or 1.23 pounds
That means that the air accounts for about 36% of what seems to be the measured mass of the inflated ball.
NOTE: For those of you who have read this far, I sincerely appologize, but I have a chem midterm next week.
[Edit]
BONUS POINTS to the first person to calculate how relativistic effects change the weight
Yeah, but what was the name of that guy…hmmmm… archimedes or something 
… bouyant force… “A body submerged in a fluid experiences a buoyant force equal to the weight of the displaced fluid.” good thing your are studying for chemistry and not physics. Offhand, I think you need to reduce your calculation by about half since you’ve calculated on about 2 atmosphere and you need to take 1 off.
The ball is inflated at around 30 psi which is a pressure of about 2atm.
Yeah, you’re right, it would weight more. I was thinking that the air inside wasn’t presurized.
Let it be known that on this 15th day of January, 2004, SuperDanman is awarded the
Number Cruncher of the Day Award!
[left]Woo hoo for Dan! He’s getting some rep from me! Join in the fun![/left]
[left]Matt[/left]
HEY! he left out the bouyant force and thus did the calculation wrongly! Don’t I get any credit for pointing that out? :ahh: another award passes me by… darn Nobel, happened with that one too! :yikes:
And he just approximated 
Should’ve used Dalton’s law to solve for moles of the gas components (taking into account the composition of your local air supply) then used Van der Waals modified ideal-gas equation to calculate the partial weights of each gas then added them all together. The bouyant force’s effect on the balls weight is easy to take into account if your assuming 2 ATM’s pressure in the ball- just divide your final air weight by two (keeping in mind the air mass stays the same… for all those acceleration problems).
Don’t forget molecules have volume and experience forces between eachother!
Of course even this is an approximation! You would need an even more complex model to take into account the molecular attaction between the differrent types of molecules in air.
Sorry about the extremely unecassary sarcasm there I got very (very) close to your answer by doing it my way. (good job)
Does anyone find it a bit concerning that we are trying to take the air’s weight into account? A gust of wind or well placed vent could easily contribute that much force. I think we’ve determined the ball is in the 3 - 6 lb range, if I were designing a mechanism to manipulate the big ball I would probably assume 10 lbs.
(good luck on that test dan)
Greg
A factor of 2 is not an approximation!
I am new to the forum and to the First robotics competition so I will give this my best shot. My team, 624, was wondering the same thing about the weight of the big yellow ball. However we decided that the wait wasn’t a problem, with the weight being only a couple of pounds. The real problem we decided was the inertia the ball will have if you are lifting the ball off the four foot goal and then trying to move with it. That could be a real problem, possibly rocking the robot badly or even tipping it over. So we have not really come up with a solution to this, we kinda just went around it and have been trying to get the more basic design of the robot done first, like the components we want as well as our drive train.
I like this thread and yet I wonder why it’s gone on so long! It’s a few pounds and you should design for more because you might have an object in the way such as a robot brushing against it. And as for finding the weight and the difference with air - good work on the math, but I would have just weighed it without the air and with the air, done a little math (a-b) and left it at that:)
Just a quick tip on the big balls… When you try to set them on the ground, you often notice that they roll a little. They aren’t perfectly spherical. I suggest letting the balls come to a rest where they like to stop, and then putting an X on the very top. This way when you reset your field for the 1000th time in week six - you can just slap the ball down on the platform and it won’t move!
THE ANSWER TO THE QUESTION
Here’s the answer that we’ve been waiting for… they didn’t say whether or not this was inflated or uninflated. To play it safe you should assume they weighed it uninflated, so it’s real weight is probably around 3.75 lbs +/- .25 lbs.
Section: 6.1.1.1
**Status: **Answered
Date Answered: 1/16/2004
**Q: **How much does the 34" BIGENS Ball weigh.
**A: **You should have received one in your kit but, in case you do not have a scale, it weighs 3.35 lbs.
Side notes:
The calculation above for the air weight assumed that it was inflated to 30 PSI… and looking back I’m betting that’s a bit too high.
And though I would never take away the number cruncher of the day award… the thickness of the ball wasn’t subtracted from the volume, further reducing the quantity of air.
That’s all folks.
Matt