Weight of fully inflated ball

[DanL]

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?

[Doug G]

Originally Posted by SuperDanman

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.

[KenWittlief]

its a lot easier to weigh before you inflate it!

[kevin.li.rit]

You'd have to weight the air too right?

[Paul Marshall]

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.

[Matt Adams]

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

[DanL]

Originally Posted by Matt Adams

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

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 l*atm/(mol*K))((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

[briholton]

Originally Posted by SuperDanman

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 l*atm/(mol*K))((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.

[DanL]

Originally Posted by briholton

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.

[Paul Marshall]

Yeah, you're right, it would weight more. I was thinking that the air inside wasn't presurized.

[Matt Adams]

Originally Posted by SuperDanman

1.23 pounds

Let it be known that on this 15th day of January, 2004, SuperDanman is awarded the

Number Cruncher of the Day Award!

Woo hoo for Dan! He's getting some rep from me! Join in the fun!

Matt

[briholton]

Originally Posted by Matt Adams

Let it be known that on this 15th day of January, 2004, SuperDanman is awarded the

Number Cruncher of the Day Award!

Woo hoo for Dan! He's getting some rep from me! Join in the fun!

Matt

HEY! he left out the bouyant force and thus did the calculation wrongly! Don't I get any credit for pointing that out? another award passes me by.... darn Nobel, happened with that one too!

[GregT]

Originally Posted by briholton

HEY! he left out the bouyant force and thus did the calculation wrongly! Don't I get any credit for pointing that out? another award passes me by.... darn Nobel, happened with that one too!

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

[briholton]

A factor of 2 is not an approximation!

Originally Posted by GregT

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 force is easy to take into account if your assuming 2 ATM's pressure in the ball- just devide 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 of 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 I would assume 10 lbs.

(good luck on that test dan)
Greg

[Justin Stiltner]

Originally Posted by Doug G

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.

We came up with 3.4lb agin +-.2lb so that seems accurate, ours was also inflated