The Dolphins Make Me Cry.com - Forums

Well, I wasn't sure where to put this...I was watching Mythbusters on the Discvery channel.They were doing a myth of a football filled with Helium would fly farther than one filled with air.

They cut a few scenes of a football team playing,and a kicker kicking a couple of field goals.I thought,hmm those look familiar,but they are based out of San Fran California. They showed them again,they are from a football game at the Local College,Missouri State,back when it was Southwest Missouri State.

I thought it pretty cool..

P.S. A helium filed ball doesn't fly as far,less mass. Force=Mass times Acceleration
If the force applied is equal,the one with the heavier mass tends to fly longer.Inertia is conserved.

I'm no engineer,something close to that,it's been over ten years since my physics classes

I watched it. I was more impressed with catching the bullet with the teeth.

Kari is SO hot.

I AM an engineer - so I'll say that "THEORETICALLY" the ball should fly farther.  Mostly because the WEIGHT of the ball is less, thanks to the buoyancy of the helium.  However, due to the buoyancy of the helium in comparison to the weight of the ball itself, the difference is negligible and the distance wouldn't be much more (i'm thinking fractions of an inch).  Its that whole Newton's law of gravity being that all object fall at the same rate.

I could work out the kinematics for you, if you want, but its all theoretical values.  Real world behaves much different than theory.

As for your F=ma - if F is held the same, and mass decreases, accelleration would have to increase.  But, even if accelleration increases, that just means more time in flight, no necessarily more distance.

I can't help but think that your reasoning is flawed. 

F=ma

F= Force given to the ball upon launch.
m= mass of the ball
a= acceleration of the launcher/ball (QB's arm)

The whole point of this would be how to find the difference in F between helium and air-filled.  Why would you assume that F stays constant?

We have 2 variables - m, because the mass of the ball changes with helium, as opposed to air, and possibly a, if the QB would be able to accelerate an object faster that had less mass.

Love that show

No, you're backwards...

F is the force put onto the ball by the propellant (QB's arm, or i'd guess in this case a throwing machine).  The acceleration in the equation is the accelleration of the object as a result of the force.

The force would have to be the same.  If i put a 20 pound force on a block, its going to accellerate faster than if i put a 10 pound force on it.  Thus, if you vary the mass, you keep force constant and see what the accelleration is.  But regardless, they're not looking for accelleration, they're looking for distance.

I'm going to run the numbers out and see what the theories of kinematics and projectile motion have to say about this issue.

Trust me, I have a degree in this stuff.

I agrre with Brian.  F would stay constant as long as the kicker hit the both balls the same.  Therefore acceleration on the ball is the same.  The other part of the formula, though, is that wind resistance on the football would be the same as well since the shape of the ball didn't change.  Therefore using F=ma again, constant wind resistance would equate to a quicker deceleration of the football.  So, the question is, is the increased deceleration greater than the icreased acceleration.  If it is, the ball won't fly as far.  Am I right, Brian?

Looking at this again, I don't even think that F=ma is the correct equation to use in this scenerio.

It seems to me that the helium ball would be in the air longer (increasing t) but may not be moving as quickly on a horizontal path. 
I guess you're playing with two variables working against each other.

Acceleration has the time (t) part of the equation you are talking about.  Remember the units for acceleration is distance/time^2.

There is no question that F=ma is the start of the problem.  The ball would not enter the air without the force of the kicker's foot hitting it and the distance that it goes in the air has a direct relationship with how hard he hits it. 

The other factor that needs to be held constant is the angle at which the kicking force is applied.  It sounds like the mythbusters were trying to test the helium theory so that should be the only variable in the test.

It wouldn't be so confusing if we were talking about a football shaped ping pong ball, right?  I think it is the same idea - the wind resistance kills the flight. 

The fact that it's a throw and not a kick, might make a difference...but I guess not.

The way I remember it, the equation for distance = 1/2 at^2.  Is that right?

Here's what I came up with...

I made certain assumptions:
the force pushing the ball is 100 N = 22.5 lb
the ball is fired at a 45 degree trajectory angle
the time of throwing the ball is 0.1 seconds

The air-filled ball is about 0.12 kg more massive than the He ball.  Thus the accelleration under the 100 N force is higher, and the initial velocity coming off the throwing machine is higher.  Makes sense so far.

Plugging numbers into the equations of Kinematics -

The air ball should travel 454 m in 2.3 seconds (damn, this guy's got a gun)
The He ball should travel 870 m in 2.88 seconds (jesus!)

Of course alot of it has to do with the assumptions.  100 N is a decent force..  The numbers don't really make sense how they came out but I can't find a math error.

I'm gonna try doing it another way, here, in a few....

distance is 1/2 a t^2 + Vo t

Finding Vo, a and t is most of the problem.  All three are different for each ball.

And, since you don't know the initial velocity, but you know a force, you have to use conservation of energy and impulse and momentum to get there.  You use F dt / m to get Vo.

Brian, I see how you're using force...that makes sense to me now.
So that's a constant -- okay.

But what is the time assumption you're using....the time it takes to actually throw the ball?

yeah, that's the big jump.  The time over which the force is applied. 

If anyone else can give me another way to translate that 100N force into an initial velocity, I'm all ears.  I used the equation for impulse and momentum to do it

F*dt=m*Vo

Wouldn't initial velocity just be 0?  The ball is at a standstill when launched, more or less.

no, initial velocity is the velocity of the ball the instant AFTER it is launched.  That's what the initial velocity in this equation needs.

Not only that but the 100N force is in the 45-dgree-angle direction, you have to break it up into horizontal and vertical components to find the time in flight (falling objects theory)

Brian, if you keep assuming that the ball is flying in a vacuum, you will keep getting that the lighter ball flies furthur.  The key is that this is not a vacuum, you have other forces on the ball during flight.  

I know, I was working out the Theoretical (in a vaccuum) stuff.  just for fun.

The wind drag is dependant upon the shape of the ball, inflation pressure, etc.  I don't see how air resistance would differ between the two cases.

But, what is different is the rate at which gravity acts upon the ball, since the helium ball has a buoyancy on it that the air ball doesn't have...

hey brian, got a question for you  I was watching speed the other day and wondering can a bus really jump 50 feet. My question for you is can it really happen and what elements have to be considered in finding the answer scientificly.

you seem like you know these complicated formulas

velocity, acceleration, shape, and mass of the bus.  And the angle of the ramp it is jumping off of.

The amount of air resistance doesn't chage (assuming same shape and inflation pressure).  What does change is the effect on the flight of the ball of this constant reistance force.  If the ball is lighter, it will decelerate quicker. 

You are focusing on the initial force, not the forces throughout the flight of the ball. 

I'm going back to my football shaped ping-pong ball.  If it was in a vacuum, it would fly much furtur than a real football.  But in real life, we know that it won't fly anywhere close to as far. 

And with a ping-pong ball, gravity has much less effect since the mass is so much lighter.

I was thinking about this on my way down to get lunch.

While the ball's in flight, there are 4 things happening...
- The ball has momentum due to the initial force
- drag force acting on the ball
- gravity acting on the ball
- buoyant force acting on the ball (in the case of the air-filled ball, this is zero)

Now, the Helium ball has less mass, thus the force of gravity is less.  In addition, it has a buoant force.  These combined will add to the flight time of the ball.  But, the buoyant force is equal to the weight of the air displaced by the HELIUM (not the ball) which I'm guessing is so small, its negligible in this problem.  Therefore, the same forces acting on both balls, the flight path should be the same.

Beef, I'm not sure I'm getting what you're saying about decelerating quicker.  All objects fall to earth at the same rate, thanks to gravity (and Mr. Newton).

And raptor - that crap from Speed was just that - crap.  There is a way a bus can jump 50 feet, but not on an overpass like that.  They'd need to be going way faster than 55 MPH and need a much bigger ramp.  If you want I can try to solve that problem for you, too (in a vaccuum of course!).

I don't understand a fucking thing in this entire thread.  That's what I know, Joe = ??

That's an easy one, vaccuum or not...

Joe  = dumbass

Please tell me you didn't just call raptor ..Sugarbush? Lol :D

I have no clue what you're talking about...

I got nailed by the quick look at the avatar.  Not sure why I did that...  Damn...

My bad, raptor!

Critical elements to be considered for the bus-jumping equation:

* Drive: If I am at the wheel, that bus can fly. Period.
* Mass: Have I eaten a large meal before take-off?
* Momentum: Do I really need to use the restroom once I land?
* Drag Force: Did I forget to take the smiley-face ball off the bus's antenna prior to launch?
* Gravity: Am I taking my bus-jumping task seriously?
* Angle: If I fly more than 50 feet in the bus, will it help me score with the hotties in the crowd?

Who needs engineers...

Only in a vacuum which we are not using for this example (I assume). Outside of a vacuum a lighter object is affected by air drag more than a heavier object. Think of the example this way project a ping pong ball and an exact sized steel ball bearing. The ping pong ball will be affected by the air drag sooner and not travel as far.

yeah, that's what I meant when I said the helium ball would be in the air longer.

I'm talking about deceleration in the x, not the y.  Gravity is a downward force, but the ball is traveling in both the x and y.  The air resistance (and ground) are the only things appling an opposite force in the x.  And ultimately we are talking about which ball will fly furtur in the x, not overall flight path.

I try to give a little love to the the Local U,and it turns into a 3 page debate...

A stupid question,I don't think it really applies here,but wouldn't both the Gravity,and the wind resistance be constant,or the same for both balls?

Both balls are inflated to 13 PSI
The Wieghts of the balls are different,but they don't change during flight....

So Newtons law that inertia,applies?

Or have you guy's already explained this,and I just missed it... :-[

Yes both gravity and wind resistance are the same forces.  But the effect of these forces changes with mass - which is changing with the addition of helium in to the ball.  So, if you consider F=ma if F is constant with wind resistance and m is smaller, then a (deceleration) has to be larger.  In the case of gravity and a downward force, F=mg.  Since g is constant and m is smaller, the downward force is smaller.  The problem with the downward force in a non-vacuum system is that there is a maximum (terminal) velocity that the ball can reach.  Once this is reached, it doesn't matter what the gravitational force is on the ball, it won't fall any faster - making air resistance in the x the only differing opposing force.

Thanks,beef

I kind of assumed that they used the sledge hammer,and showed that both balls leave with the same velocity,and the wieghts don't change in flight,it would revert to inertia (being the rest is the same.)

I didn't think about the terminal velocity.

Damn,over Ten years away from doing physics,It all rushes back in one day...My head hurts....

The flight in the x depends upon the flight int he y.  The flight in the x can continue forever if there's no force in the y (mg).  So the time in flight can only be found by simplifying the problem to firing a football straight up in the air and see when it comes down. 

Imagine the example of if you fired a bowling ball and a beach ball out of a cannon at the same time, same shape, same field.  Which would go farther?  I'd think the beach ball would, since its mass is less and it requires less impulse to change its inertia.

The flight in the x only depends on the y since there is a fixed limit due to the ground.  So you want to simplify the problem by calculating the hang time with a straight up kick and then using average velocity to determine distance?  The problem with that is that the average velocity changes due to wind resistance when you kick with a trajectory.  You logic only works in a vacuum.  But I've already said that in a vacuum I believe the lighter object will go furthur.  To make the lighter object go less of a distance you can't be assuming a vacuum. 

Funny you use the bowling ball and beach ball example.  Because I think that in the real world, the beach ball doesn't go furthur.  I've kicked a soccer ball and a beach ball and I can't kick the beach ball nearly as far.  Maybe you are more special than me when it comes to this. 

Brian this is why the air resistance makes such a difference with a ligher object. 

Incorrect. As previously stated, air drag affects a lighter object more than a heavy object.

I'm not disputing the existance of air resistance.

Never have

I also never stated anything about average velocity, nor simplifying the problem to a striaght up kick.  Its hard to explain what I'm trying to communicate over a message board.

At the same time, imagine a beach ball at a football game, being batted around in the crowd.  Imagine someone serves it up, volleyball style, hits the ball and it goes flying.

Now change it to a bowling ball.  Does it go flying?  Or does it move 2 inches and fall to the ground?

Maybe a bowling ball is too extreme, but what's the function of distance vs mass?

Not doubting the theory, but in practice, I can't see this as accurate. If I have a cannonball weighing 10 lbs in one catapult and a beach ball of the same size and shape, but only weighing in at 4 ounces next to it in an identical catapult. Both catapults fire with the same force and at the same angle. The mass of the bowling ball will affect its inertia significantly more than the lesser mass of the beach ball will affect its inertia.

That's it...I'm running away from this topic before I feel even geekier.




modified to add: While I was typing this, Brian posted that example above...the same example exactly that I used with Fausto at about the exact time Brian was typing. Interesting...

You said time of flight could be simplified to a straight up kick.  I guess I didn't see how that was relative except to assume that time of flight is the same for both types of kicks - at which point you can jut multiply by average v to get total distance traveled (in both x and y).  Anyway, that doesn't matter if thats not what you were saying.

As for serving a bowling ball.... The bowling ball won't fly as far because the initial force from the hand is not sufficient to project it.  The beach ball obviously is.  So air resistance doesn't even become a factor.  You need to compare two object that can be projected with the kicking force to allow air resistance to enter the equation.  If you compare a volleyball to a beach ball, which can you serve furthur?

OK, I looked it up - you guys are peaking my curiosity.

drag force is independant of mass.  Drag force is defined as:

F= ( Cd A p v^2 ) / 2

where
Cd = drag coefficient
A = frontal cross sectional area
p (rho) = density of the fluid its traveling thru (in this case, air)
v = velocity of the object

So, please tell me how the drag force differs for different mass objects.

If the drag force is the same on both, how does it affect one differently than the other?

The only thing I can see is the resultant force in the y (mg - Fd) of the falling object is different for each cause each has different weight, thus different net accelleration towards the ground.  But for the helium one, its mg - Fd - Fb, so there should be a small difference.

Am I not thinking of this right?

There are 2 functions here.  First the impact force and the resulting acceleration on the object due to that force.  The second is the air resistance force and the resulting deceleration once the object is in motion.  Obviously if the first force is not sufficient to incude significant motion, the second factor never even enters the problem.

Force doesn't change.  deceleration changes - apply it to F=ma.  If m is bigger, a is smaller.

I've always stated that air restistance is the same for the same size balls.

I don't buy it.  Let's go back to the cannon then.  Fire a bowling ball and a soccer ball out of a cannon, straight up.  

Here's the difference - the velocity is lower on the bowling ball, therefore the drag force is less.  However, lower initial velocity, I gotta think would have more impact on the flight of the projectile, than lower drag.

Yes, agreed.  But, now I'm thinking the drag force is not the same for both, based on what I just posted about different velocities.

So you are saying the lighter object will have even MORE drag force.  So that supports my arguement.  Don't forget that the result of the test was that the He ball flew less distance.  You might want to try to figure out why this is more than try to prove otherwise.

I'm not arguing, nor trying to disprove.  This is not a debate to me, I'm just trying to understand the problem - more for my own curiosity than anything else.  I've flip-flopped three times in this thread, if you notice...

I'm not trying to say you're wrong, or anyone else is, I'm just trying to fully understand the phenomena.

By bad.  It just seems like you are so close and grasping at ways to be unconvinced.

This was a very interesting thread.  Physics rocks!

yeah try.and What would the formula be.

Going with the bowling ball/beach ball scenario again:

Supply the required force to launch a bowling ball 5 feet through the air.
Now apply the same force to a beach ball.
Will the beach ball fly more or less than 5 feet?  I'd guess much, much more.

There's some relationship between the density of the projectile and the wind resistance.

In a vacuum the beach ball goes farther. 

In reality, the initial force is also a factor.  A moderate force would mean a lower initial velocity which means less wind resistance so the beach ball would travel farther.  A larger force would make the bowling ball go farther, and straighter, as  the density of the bowling ball is less affected by wind resistance.

You can kick a beach ball farther than you can kick a bowling ball, but a cannon could shoot a bowling ball farther than a beach ball.

There's some relationship between the density of the object, velocity of the object and wind resistance.

I think this is essentially the point Brian is getting at.  There is definitely a point where a given amount of force will push the less dense item a greater distance than the denser item, and there is another point where a different, much larger amount of force will push the denser item a greater distance than the less dense item.

The question is whether a kick from a player fits in the first category, or the second... and it would appear that it falls in the second.  However, to say that the helium-filled ball doesn't go as far "because it's lighter" is somewhat misleading, in that it implies that the denser football simply travels farther, period.  The results might not be the same if the ball were thrown instead of kicked.

You kick a bowling ball,You break your foot,You kick a Beach ball as hard as you can..you pull your groin...My advise...Don't try to kick anything.

Back to the serious discussion at hand....

I've kicked my bowling ball many a times in anger after not striking or missing a spare, and i've never broken my foot.  And I used 16lb bowling balls.

But it's fun to watch those 16lb balls get slammed on to the ground, and see how high those suckers can bounce back up.

I finally saw the episode of Mythbusters in question.  I realize exactly where this thread went wrong:

It's not "if the force applied is equal"... it's "if the speed is equal".  The exact word they used was "speed," which instantly clarifies this entire debate.

If two objects have equal speed, the one with more mass will go farther.